3.7.74 \(\int \frac {1}{\tan ^{\frac {5}{3}}(c+d x) (a+b \tan (c+d x))} \, dx\) [674]

Optimal. Leaf size=525 \[ \frac {b \text {ArcTan}\left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )}{2 \left (a^2+b^2\right ) d}-\frac {b \text {ArcTan}\left (\sqrt {3}+2 \sqrt [3]{\tan (c+d x)}\right )}{2 \left (a^2+b^2\right ) d}+\frac {\sqrt {3} b^{8/3} \text {ArcTan}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} \sqrt [3]{\tan (c+d x)}}{\sqrt {3} \sqrt [3]{a}}\right )}{a^{5/3} \left (a^2+b^2\right ) d}+\frac {\sqrt {3} a \text {ArcTan}\left (\frac {1-2 \tan ^{\frac {2}{3}}(c+d x)}{\sqrt {3}}\right )}{2 \left (a^2+b^2\right ) d}-\frac {b \text {ArcTan}\left (\sqrt [3]{\tan (c+d x)}\right )}{\left (a^2+b^2\right ) d}-\frac {3 b^{8/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{\tan (c+d x)}\right )}{2 a^{5/3} \left (a^2+b^2\right ) d}+\frac {a \log \left (1+\tan ^{\frac {2}{3}}(c+d x)\right )}{2 \left (a^2+b^2\right ) d}+\frac {\sqrt {3} b \log \left (1-\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{4 \left (a^2+b^2\right ) d}-\frac {\sqrt {3} b \log \left (1+\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{4 \left (a^2+b^2\right ) d}+\frac {b^{8/3} \log (a+b \tan (c+d x))}{2 a^{5/3} \left (a^2+b^2\right ) d}-\frac {a \log \left (1-\tan ^{\frac {2}{3}}(c+d x)+\tan ^{\frac {4}{3}}(c+d x)\right )}{4 \left (a^2+b^2\right ) d}-\frac {3 a}{2 \left (a^2+b^2\right ) d \tan ^{\frac {2}{3}}(c+d x)}-\frac {3 b^2}{2 a \left (a^2+b^2\right ) d \tan ^{\frac {2}{3}}(c+d x)} \]

[Out]

-1/2*b*arctan(-3^(1/2)+2*tan(d*x+c)^(1/3))/(a^2+b^2)/d-1/2*b*arctan(3^(1/2)+2*tan(d*x+c)^(1/3))/(a^2+b^2)/d-b*
arctan(tan(d*x+c)^(1/3))/(a^2+b^2)/d-3/2*b^(8/3)*ln(a^(1/3)+b^(1/3)*tan(d*x+c)^(1/3))/a^(5/3)/(a^2+b^2)/d+1/2*
a*ln(1+tan(d*x+c)^(2/3))/(a^2+b^2)/d+1/2*b^(8/3)*ln(a+b*tan(d*x+c))/a^(5/3)/(a^2+b^2)/d-1/4*a*ln(1-tan(d*x+c)^
(2/3)+tan(d*x+c)^(4/3))/(a^2+b^2)/d+b^(8/3)*arctan(1/3*(a^(1/3)-2*b^(1/3)*tan(d*x+c)^(1/3))/a^(1/3)*3^(1/2))*3
^(1/2)/a^(5/3)/(a^2+b^2)/d+1/2*a*arctan(1/3*(1-2*tan(d*x+c)^(2/3))*3^(1/2))*3^(1/2)/(a^2+b^2)/d+1/4*b*ln(1-3^(
1/2)*tan(d*x+c)^(1/3)+tan(d*x+c)^(2/3))*3^(1/2)/(a^2+b^2)/d-1/4*b*ln(1+3^(1/2)*tan(d*x+c)^(1/3)+tan(d*x+c)^(2/
3))*3^(1/2)/(a^2+b^2)/d-3/2*a/(a^2+b^2)/d/tan(d*x+c)^(2/3)-3/2*b^2/a/(a^2+b^2)/d/tan(d*x+c)^(2/3)

________________________________________________________________________________________

Rubi [A]
time = 0.39, antiderivative size = 525, normalized size of antiderivative = 1.00, number of steps used = 30, number of rules used = 18, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.783, Rules used = {3655, 3610, 3619, 3557, 335, 215, 648, 632, 210, 642, 209, 281, 298, 31, 3715, 53, 60, 631} \begin {gather*} \frac {\sqrt {3} a \text {ArcTan}\left (\frac {1-2 \tan ^{\frac {2}{3}}(c+d x)}{\sqrt {3}}\right )}{2 d \left (a^2+b^2\right )}+\frac {b \text {ArcTan}\left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )}{2 d \left (a^2+b^2\right )}-\frac {b \text {ArcTan}\left (2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}\right )}{2 d \left (a^2+b^2\right )}-\frac {b \text {ArcTan}\left (\sqrt [3]{\tan (c+d x)}\right )}{d \left (a^2+b^2\right )}-\frac {3 b^2}{2 a d \left (a^2+b^2\right ) \tan ^{\frac {2}{3}}(c+d x)}-\frac {3 a}{2 d \left (a^2+b^2\right ) \tan ^{\frac {2}{3}}(c+d x)}+\frac {\sqrt {3} b \log \left (\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )}{4 d \left (a^2+b^2\right )}-\frac {\sqrt {3} b \log \left (\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )}{4 d \left (a^2+b^2\right )}+\frac {a \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )}{2 d \left (a^2+b^2\right )}-\frac {a \log \left (\tan ^{\frac {4}{3}}(c+d x)-\tan ^{\frac {2}{3}}(c+d x)+1\right )}{4 d \left (a^2+b^2\right )}+\frac {\sqrt {3} b^{8/3} \text {ArcTan}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} \sqrt [3]{\tan (c+d x)}}{\sqrt {3} \sqrt [3]{a}}\right )}{a^{5/3} d \left (a^2+b^2\right )}-\frac {3 b^{8/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{\tan (c+d x)}\right )}{2 a^{5/3} d \left (a^2+b^2\right )}+\frac {b^{8/3} \log (a+b \tan (c+d x))}{2 a^{5/3} d \left (a^2+b^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(Tan[c + d*x]^(5/3)*(a + b*Tan[c + d*x])),x]

[Out]

(b*ArcTan[Sqrt[3] - 2*Tan[c + d*x]^(1/3)])/(2*(a^2 + b^2)*d) - (b*ArcTan[Sqrt[3] + 2*Tan[c + d*x]^(1/3)])/(2*(
a^2 + b^2)*d) + (Sqrt[3]*b^(8/3)*ArcTan[(a^(1/3) - 2*b^(1/3)*Tan[c + d*x]^(1/3))/(Sqrt[3]*a^(1/3))])/(a^(5/3)*
(a^2 + b^2)*d) + (Sqrt[3]*a*ArcTan[(1 - 2*Tan[c + d*x]^(2/3))/Sqrt[3]])/(2*(a^2 + b^2)*d) - (b*ArcTan[Tan[c +
d*x]^(1/3)])/((a^2 + b^2)*d) - (3*b^(8/3)*Log[a^(1/3) + b^(1/3)*Tan[c + d*x]^(1/3)])/(2*a^(5/3)*(a^2 + b^2)*d)
+ (a*Log[1 + Tan[c + d*x]^(2/3)])/(2*(a^2 + b^2)*d) + (Sqrt[3]*b*Log[1 - Sqrt[3]*Tan[c + d*x]^(1/3) + Tan[c +
d*x]^(2/3)])/(4*(a^2 + b^2)*d) - (Sqrt[3]*b*Log[1 + Sqrt[3]*Tan[c + d*x]^(1/3) + Tan[c + d*x]^(2/3)])/(4*(a^2
+ b^2)*d) + (b^(8/3)*Log[a + b*Tan[c + d*x]])/(2*a^(5/3)*(a^2 + b^2)*d) - (a*Log[1 - Tan[c + d*x]^(2/3) + Tan
[c + d*x]^(4/3)])/(4*(a^2 + b^2)*d) - (3*a)/(2*(a^2 + b^2)*d*Tan[c + d*x]^(2/3)) - (3*b^2)/(2*a*(a^2 + b^2)*d*
Tan[c + d*x]^(2/3))

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 53

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*((m + n + 2)/((b*c - a*d)*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 60

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[{q = Rt[-(b*c - a*d)/b, 3]}, Simp[-
Log[RemoveContent[a + b*x, x]]/(2*b*q^2), x] + (Dist[3/(2*b*q), Subst[Int[1/(q^2 - q*x + x^2), x], x, (c + d*x
)^(1/3)], x] + Dist[3/(2*b*q^2), Subst[Int[1/(q + x), x], x, (c + d*x)^(1/3)], x])] /; FreeQ[{a, b, c, d}, x]
&& NegQ[(b*c - a*d)/b]

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 215

Int[((a_) + (b_.)*(x_)^(n_))^(-1), x_Symbol] :> Module[{r = Numerator[Rt[a/b, n]], s = Denominator[Rt[a/b, n]]
, k, u, v}, Simp[u = Int[(r - s*Cos[(2*k - 1)*(Pi/n)]*x)/(r^2 - 2*r*s*Cos[(2*k - 1)*(Pi/n)]*x + s^2*x^2), x] +
Int[(r + s*Cos[(2*k - 1)*(Pi/n)]*x)/(r^2 + 2*r*s*Cos[(2*k - 1)*(Pi/n)]*x + s^2*x^2), x]; 2*(r^2/(a*n))*Int[1/
(r^2 + s^2*x^2), x] + Dist[2*(r/(a*n)), Sum[u, {k, 1, (n - 2)/4}], x], x]] /; FreeQ[{a, b}, x] && IGtQ[(n - 2)
/4, 0] && PosQ[a/b]

Rule 281

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rule 298

Int[(x_)/((a_) + (b_.)*(x_)^3), x_Symbol] :> Dist[-(3*Rt[a, 3]*Rt[b, 3])^(-1), Int[1/(Rt[a, 3] + Rt[b, 3]*x),
x], x] + Dist[1/(3*Rt[a, 3]*Rt[b, 3]), Int[(Rt[a, 3] + Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3
]^2*x^2), x], x] /; FreeQ[{a, b}, x]

Rule 335

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 632

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 648

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 3557

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[b/d, Subst[Int[x^n/(b^2 + x^2), x], x, b*Tan[c + d
*x]], x] /; FreeQ[{b, c, d, n}, x] &&  !IntegerQ[n]

Rule 3610

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b
*c - a*d)*((a + b*Tan[e + f*x])^(m + 1)/(f*(m + 1)*(a^2 + b^2))), x] + Dist[1/(a^2 + b^2), Int[(a + b*Tan[e +
f*x])^(m + 1)*Simp[a*c + b*d - (b*c - a*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c
- a*d, 0] && NeQ[a^2 + b^2, 0] && LtQ[m, -1]

Rule 3619

Int[((b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*T
an[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Tan[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x] && NeQ
[c^2 + d^2, 0] &&  !IntegerQ[2*m]

Rule 3655

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)/((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[1/
(c^2 + d^2), Int[(a + b*Tan[e + f*x])^m*(c - d*Tan[e + f*x]), x], x] + Dist[d^2/(c^2 + d^2), Int[(a + b*Tan[e
+ f*x])^m*((1 + Tan[e + f*x]^2)/(c + d*Tan[e + f*x])), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c -
a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] &&  !IntegerQ[m]

Rule 3715

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.)*((A_) + (C_.)*
tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[A/f, Subst[Int[(a + b*x)^m*(c + d*x)^n, x], x, Tan[e + f*x]], x]
/; FreeQ[{a, b, c, d, e, f, A, C, m, n}, x] && EqQ[A, C]

Rubi steps

\begin {align*} \int \frac {1}{\tan ^{\frac {5}{3}}(c+d x) (a+b \tan (c+d x))} \, dx &=\frac {\int \frac {a-b \tan (c+d x)}{\tan ^{\frac {5}{3}}(c+d x)} \, dx}{a^2+b^2}+\frac {b^2 \int \frac {1+\tan ^2(c+d x)}{\tan ^{\frac {5}{3}}(c+d x) (a+b \tan (c+d x))} \, dx}{a^2+b^2}\\ &=-\frac {3 a}{2 \left (a^2+b^2\right ) d \tan ^{\frac {2}{3}}(c+d x)}+\frac {\int \frac {-b-a \tan (c+d x)}{\tan ^{\frac {2}{3}}(c+d x)} \, dx}{a^2+b^2}+\frac {b^2 \text {Subst}\left (\int \frac {1}{x^{5/3} (a+b x)} \, dx,x,\tan (c+d x)\right )}{\left (a^2+b^2\right ) d}\\ &=-\frac {3 a}{2 \left (a^2+b^2\right ) d \tan ^{\frac {2}{3}}(c+d x)}-\frac {3 b^2}{2 a \left (a^2+b^2\right ) d \tan ^{\frac {2}{3}}(c+d x)}-\frac {a \int \sqrt [3]{\tan (c+d x)} \, dx}{a^2+b^2}-\frac {b \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)} \, dx}{a^2+b^2}-\frac {b^3 \text {Subst}\left (\int \frac {1}{x^{2/3} (a+b x)} \, dx,x,\tan (c+d x)\right )}{a \left (a^2+b^2\right ) d}\\ &=\frac {b^{8/3} \log (a+b \tan (c+d x))}{2 a^{5/3} \left (a^2+b^2\right ) d}-\frac {3 a}{2 \left (a^2+b^2\right ) d \tan ^{\frac {2}{3}}(c+d x)}-\frac {3 b^2}{2 a \left (a^2+b^2\right ) d \tan ^{\frac {2}{3}}(c+d x)}-\frac {a \text {Subst}\left (\int \frac {\sqrt [3]{x}}{1+x^2} \, dx,x,\tan (c+d x)\right )}{\left (a^2+b^2\right ) d}-\frac {b \text {Subst}\left (\int \frac {1}{x^{2/3} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{\left (a^2+b^2\right ) d}-\frac {\left (3 b^{7/3}\right ) \text {Subst}\left (\int \frac {1}{\frac {a^{2/3}}{b^{2/3}}-\frac {\sqrt [3]{a} x}{\sqrt [3]{b}}+x^2} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{2 a^{4/3} \left (a^2+b^2\right ) d}-\frac {\left (3 b^{8/3}\right ) \text {Subst}\left (\int \frac {1}{\frac {\sqrt [3]{a}}{\sqrt [3]{b}}+x} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{2 a^{5/3} \left (a^2+b^2\right ) d}\\ &=-\frac {3 b^{8/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{\tan (c+d x)}\right )}{2 a^{5/3} \left (a^2+b^2\right ) d}+\frac {b^{8/3} \log (a+b \tan (c+d x))}{2 a^{5/3} \left (a^2+b^2\right ) d}-\frac {3 a}{2 \left (a^2+b^2\right ) d \tan ^{\frac {2}{3}}(c+d x)}-\frac {3 b^2}{2 a \left (a^2+b^2\right ) d \tan ^{\frac {2}{3}}(c+d x)}-\frac {(3 a) \text {Subst}\left (\int \frac {x^3}{1+x^6} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{\left (a^2+b^2\right ) d}-\frac {(3 b) \text {Subst}\left (\int \frac {1}{1+x^6} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{\left (a^2+b^2\right ) d}-\frac {\left (3 b^{8/3}\right ) \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1-\frac {2 \sqrt [3]{b} \sqrt [3]{\tan (c+d x)}}{\sqrt [3]{a}}\right )}{a^{5/3} \left (a^2+b^2\right ) d}\\ &=\frac {\sqrt {3} b^{8/3} \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{b} \sqrt [3]{\tan (c+d x)}}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{a^{5/3} \left (a^2+b^2\right ) d}-\frac {3 b^{8/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{\tan (c+d x)}\right )}{2 a^{5/3} \left (a^2+b^2\right ) d}+\frac {b^{8/3} \log (a+b \tan (c+d x))}{2 a^{5/3} \left (a^2+b^2\right ) d}-\frac {3 a}{2 \left (a^2+b^2\right ) d \tan ^{\frac {2}{3}}(c+d x)}-\frac {3 b^2}{2 a \left (a^2+b^2\right ) d \tan ^{\frac {2}{3}}(c+d x)}-\frac {(3 a) \text {Subst}\left (\int \frac {x}{1+x^3} \, dx,x,\tan ^{\frac {2}{3}}(c+d x)\right )}{2 \left (a^2+b^2\right ) d}-\frac {b \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{\left (a^2+b^2\right ) d}-\frac {b \text {Subst}\left (\int \frac {1-\frac {\sqrt {3} x}{2}}{1-\sqrt {3} x+x^2} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{\left (a^2+b^2\right ) d}-\frac {b \text {Subst}\left (\int \frac {1+\frac {\sqrt {3} x}{2}}{1+\sqrt {3} x+x^2} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{\left (a^2+b^2\right ) d}\\ &=\frac {\sqrt {3} b^{8/3} \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{b} \sqrt [3]{\tan (c+d x)}}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{a^{5/3} \left (a^2+b^2\right ) d}-\frac {b \tan ^{-1}\left (\sqrt [3]{\tan (c+d x)}\right )}{\left (a^2+b^2\right ) d}-\frac {3 b^{8/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{\tan (c+d x)}\right )}{2 a^{5/3} \left (a^2+b^2\right ) d}+\frac {b^{8/3} \log (a+b \tan (c+d x))}{2 a^{5/3} \left (a^2+b^2\right ) d}-\frac {3 a}{2 \left (a^2+b^2\right ) d \tan ^{\frac {2}{3}}(c+d x)}-\frac {3 b^2}{2 a \left (a^2+b^2\right ) d \tan ^{\frac {2}{3}}(c+d x)}+\frac {a \text {Subst}\left (\int \frac {1}{1+x} \, dx,x,\tan ^{\frac {2}{3}}(c+d x)\right )}{2 \left (a^2+b^2\right ) d}-\frac {a \text {Subst}\left (\int \frac {1+x}{1-x+x^2} \, dx,x,\tan ^{\frac {2}{3}}(c+d x)\right )}{2 \left (a^2+b^2\right ) d}-\frac {b \text {Subst}\left (\int \frac {1}{1-\sqrt {3} x+x^2} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{4 \left (a^2+b^2\right ) d}-\frac {b \text {Subst}\left (\int \frac {1}{1+\sqrt {3} x+x^2} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{4 \left (a^2+b^2\right ) d}+\frac {\left (\sqrt {3} b\right ) \text {Subst}\left (\int \frac {-\sqrt {3}+2 x}{1-\sqrt {3} x+x^2} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{4 \left (a^2+b^2\right ) d}-\frac {\left (\sqrt {3} b\right ) \text {Subst}\left (\int \frac {\sqrt {3}+2 x}{1+\sqrt {3} x+x^2} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{4 \left (a^2+b^2\right ) d}\\ &=\frac {\sqrt {3} b^{8/3} \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{b} \sqrt [3]{\tan (c+d x)}}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{a^{5/3} \left (a^2+b^2\right ) d}-\frac {b \tan ^{-1}\left (\sqrt [3]{\tan (c+d x)}\right )}{\left (a^2+b^2\right ) d}-\frac {3 b^{8/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{\tan (c+d x)}\right )}{2 a^{5/3} \left (a^2+b^2\right ) d}+\frac {a \log \left (1+\tan ^{\frac {2}{3}}(c+d x)\right )}{2 \left (a^2+b^2\right ) d}+\frac {\sqrt {3} b \log \left (1-\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{4 \left (a^2+b^2\right ) d}-\frac {\sqrt {3} b \log \left (1+\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{4 \left (a^2+b^2\right ) d}+\frac {b^{8/3} \log (a+b \tan (c+d x))}{2 a^{5/3} \left (a^2+b^2\right ) d}-\frac {3 a}{2 \left (a^2+b^2\right ) d \tan ^{\frac {2}{3}}(c+d x)}-\frac {3 b^2}{2 a \left (a^2+b^2\right ) d \tan ^{\frac {2}{3}}(c+d x)}-\frac {a \text {Subst}\left (\int \frac {-1+2 x}{1-x+x^2} \, dx,x,\tan ^{\frac {2}{3}}(c+d x)\right )}{4 \left (a^2+b^2\right ) d}-\frac {(3 a) \text {Subst}\left (\int \frac {1}{1-x+x^2} \, dx,x,\tan ^{\frac {2}{3}}(c+d x)\right )}{4 \left (a^2+b^2\right ) d}+\frac {b \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,-\sqrt {3}+2 \sqrt [3]{\tan (c+d x)}\right )}{2 \left (a^2+b^2\right ) d}+\frac {b \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,\sqrt {3}+2 \sqrt [3]{\tan (c+d x)}\right )}{2 \left (a^2+b^2\right ) d}\\ &=\frac {b \tan ^{-1}\left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )}{2 \left (a^2+b^2\right ) d}-\frac {b \tan ^{-1}\left (\sqrt {3}+2 \sqrt [3]{\tan (c+d x)}\right )}{2 \left (a^2+b^2\right ) d}+\frac {\sqrt {3} b^{8/3} \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{b} \sqrt [3]{\tan (c+d x)}}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{a^{5/3} \left (a^2+b^2\right ) d}-\frac {b \tan ^{-1}\left (\sqrt [3]{\tan (c+d x)}\right )}{\left (a^2+b^2\right ) d}-\frac {3 b^{8/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{\tan (c+d x)}\right )}{2 a^{5/3} \left (a^2+b^2\right ) d}+\frac {a \log \left (1+\tan ^{\frac {2}{3}}(c+d x)\right )}{2 \left (a^2+b^2\right ) d}+\frac {\sqrt {3} b \log \left (1-\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{4 \left (a^2+b^2\right ) d}-\frac {\sqrt {3} b \log \left (1+\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{4 \left (a^2+b^2\right ) d}+\frac {b^{8/3} \log (a+b \tan (c+d x))}{2 a^{5/3} \left (a^2+b^2\right ) d}-\frac {a \log \left (1-\tan ^{\frac {2}{3}}(c+d x)+\tan ^{\frac {4}{3}}(c+d x)\right )}{4 \left (a^2+b^2\right ) d}-\frac {3 a}{2 \left (a^2+b^2\right ) d \tan ^{\frac {2}{3}}(c+d x)}-\frac {3 b^2}{2 a \left (a^2+b^2\right ) d \tan ^{\frac {2}{3}}(c+d x)}+\frac {(3 a) \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,-1+2 \tan ^{\frac {2}{3}}(c+d x)\right )}{2 \left (a^2+b^2\right ) d}\\ &=\frac {b \tan ^{-1}\left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )}{2 \left (a^2+b^2\right ) d}-\frac {b \tan ^{-1}\left (\sqrt {3}+2 \sqrt [3]{\tan (c+d x)}\right )}{2 \left (a^2+b^2\right ) d}+\frac {\sqrt {3} b^{8/3} \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{b} \sqrt [3]{\tan (c+d x)}}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{a^{5/3} \left (a^2+b^2\right ) d}+\frac {\sqrt {3} a \tan ^{-1}\left (\frac {1-2 \tan ^{\frac {2}{3}}(c+d x)}{\sqrt {3}}\right )}{2 \left (a^2+b^2\right ) d}-\frac {b \tan ^{-1}\left (\sqrt [3]{\tan (c+d x)}\right )}{\left (a^2+b^2\right ) d}-\frac {3 b^{8/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{\tan (c+d x)}\right )}{2 a^{5/3} \left (a^2+b^2\right ) d}+\frac {a \log \left (1+\tan ^{\frac {2}{3}}(c+d x)\right )}{2 \left (a^2+b^2\right ) d}+\frac {\sqrt {3} b \log \left (1-\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{4 \left (a^2+b^2\right ) d}-\frac {\sqrt {3} b \log \left (1+\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{4 \left (a^2+b^2\right ) d}+\frac {b^{8/3} \log (a+b \tan (c+d x))}{2 a^{5/3} \left (a^2+b^2\right ) d}-\frac {a \log \left (1-\tan ^{\frac {2}{3}}(c+d x)+\tan ^{\frac {4}{3}}(c+d x)\right )}{4 \left (a^2+b^2\right ) d}-\frac {3 a}{2 \left (a^2+b^2\right ) d \tan ^{\frac {2}{3}}(c+d x)}-\frac {3 b^2}{2 a \left (a^2+b^2\right ) d \tan ^{\frac {2}{3}}(c+d x)}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 0.25, size = 104, normalized size = 0.20 \begin {gather*} -\frac {3 \left (b^2 \, _2F_1\left (-\frac {2}{3},1;\frac {1}{3};-\frac {b \tan (c+d x)}{a}\right )+a \left (a \, _2F_1\left (-\frac {1}{3},1;\frac {2}{3};-\tan ^2(c+d x)\right )+2 b \, _2F_1\left (\frac {1}{6},1;\frac {7}{6};-\tan ^2(c+d x)\right ) \tan (c+d x)\right )\right )}{2 a \left (a^2+b^2\right ) d \tan ^{\frac {2}{3}}(c+d x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(Tan[c + d*x]^(5/3)*(a + b*Tan[c + d*x])),x]

[Out]

(-3*(b^2*Hypergeometric2F1[-2/3, 1, 1/3, -((b*Tan[c + d*x])/a)] + a*(a*Hypergeometric2F1[-1/3, 1, 2/3, -Tan[c
+ d*x]^2] + 2*b*Hypergeometric2F1[1/6, 1, 7/6, -Tan[c + d*x]^2]*Tan[c + d*x])))/(2*a*(a^2 + b^2)*d*Tan[c + d*x
]^(2/3))

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Maple [A]
time = 0.20, size = 350, normalized size = 0.67

method	result	size
derivativedivides	\(\frac {-\frac {3}{2 a \tan \left (d x +c \right )^{\frac {2}{3}}}+\frac {-\frac {3 \left (-\sqrt {3}\, b +a \right ) \ln \left (1-\sqrt {3}\, \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )\right )}{4}-3 \left (2 b +\frac {\left (-\sqrt {3}\, b +a \right ) \sqrt {3}}{2}\right ) \arctan \left (-\sqrt {3}+2 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )\right )+\frac {3 \left (-\sqrt {3}\, b -a \right ) \ln \left (1+\sqrt {3}\, \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )\right )}{4}+3 \left (-2 b -\frac {\left (-\sqrt {3}\, b -a \right ) \sqrt {3}}{2}\right ) \arctan \left (\sqrt {3}+2 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )\right )}{3 a^{2}+3 b^{2}}+\frac {\frac {3 a \ln \left (1+\tan ^{\frac {2}{3}}\left (d x +c \right )\right )}{2}-3 b \arctan \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )}{3 a^{2}+3 b^{2}}-\frac {3 \left (\frac {\ln \left (\tan ^{\frac {1}{3}}\left (d x +c \right )+\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}-\frac {\ln \left (\tan ^{\frac {2}{3}}\left (d x +c \right )-\left (\frac {a}{b}\right )^{\frac {1}{3}} \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}\right ) b^{3}}{a \left (a^{2}+b^{2}\right )}}{d}\)	\(350\)
default	\(\frac {-\frac {3}{2 a \tan \left (d x +c \right )^{\frac {2}{3}}}+\frac {-\frac {3 \left (-\sqrt {3}\, b +a \right ) \ln \left (1-\sqrt {3}\, \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )\right )}{4}-3 \left (2 b +\frac {\left (-\sqrt {3}\, b +a \right ) \sqrt {3}}{2}\right ) \arctan \left (-\sqrt {3}+2 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )\right )+\frac {3 \left (-\sqrt {3}\, b -a \right ) \ln \left (1+\sqrt {3}\, \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )\right )}{4}+3 \left (-2 b -\frac {\left (-\sqrt {3}\, b -a \right ) \sqrt {3}}{2}\right ) \arctan \left (\sqrt {3}+2 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )\right )}{3 a^{2}+3 b^{2}}+\frac {\frac {3 a \ln \left (1+\tan ^{\frac {2}{3}}\left (d x +c \right )\right )}{2}-3 b \arctan \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )}{3 a^{2}+3 b^{2}}-\frac {3 \left (\frac {\ln \left (\tan ^{\frac {1}{3}}\left (d x +c \right )+\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}-\frac {\ln \left (\tan ^{\frac {2}{3}}\left (d x +c \right )-\left (\frac {a}{b}\right )^{\frac {1}{3}} \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}\right ) b^{3}}{a \left (a^{2}+b^{2}\right )}}{d}\)	\(350\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/tan(d*x+c)^(5/3)/(a+b*tan(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

1/d*(-3/2/a/tan(d*x+c)^(2/3)+3/(3*a^2+3*b^2)*(-1/4*(-3^(1/2)*b+a)*ln(1-3^(1/2)*tan(d*x+c)^(1/3)+tan(d*x+c)^(2/
3))-(2*b+1/2*(-3^(1/2)*b+a)*3^(1/2))*arctan(-3^(1/2)+2*tan(d*x+c)^(1/3))+1/4*(-3^(1/2)*b-a)*ln(1+3^(1/2)*tan(d
*x+c)^(1/3)+tan(d*x+c)^(2/3))+(-2*b-1/2*(-3^(1/2)*b-a)*3^(1/2))*arctan(3^(1/2)+2*tan(d*x+c)^(1/3)))+3/(3*a^2+3
*b^2)*(1/2*a*ln(1+tan(d*x+c)^(2/3))-b*arctan(tan(d*x+c)^(1/3)))-3*(1/3/b/(a/b)^(2/3)*ln(tan(d*x+c)^(1/3)+(a/b)
^(1/3))-1/6/b/(a/b)^(2/3)*ln(tan(d*x+c)^(2/3)-(a/b)^(1/3)*tan(d*x+c)^(1/3)+(a/b)^(2/3))+1/3/b/(a/b)^(2/3)*3^(1
/2)*arctan(1/3*3^(1/2)*(2/(a/b)^(1/3)*tan(d*x+c)^(1/3)-1)))/a*b^3/(a^2+b^2))

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Maxima [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/tan(d*x+c)^(5/3)/(a+b*tan(d*x+c)),x, algorithm="maxima")

[Out]

Timed out

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Fricas [C] Result contains complex when optimal does not.
time = 3.66, size = 74494, normalized size = 141.89 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/tan(d*x+c)^(5/3)/(a+b*tan(d*x+c)),x, algorithm="fricas")

[Out]

-1/48*(12*(a^3 + a*b^2)*d*(2*sqrt(-1/2*(I*a^2 + I*b^2)*d*(2*a/(a^4*d^3 + 2*a^2*b^2*d^3 + b^4*d^3) - (3*a^2 - b
^2)*a/((a^4*d^2 + 2*a^2*b^2*d^2 + b^4*d^2)*(a^2*d + b^2*d)) + a^3/(a^2*d + b^2*d)^3)/b - 1/4*(3*a^2 - b^2)/(a^
4*d^2 + 2*a^2*b^2*d^2 + b^4*d^2) - 1/2*(a^2 - b^2)/(a^4*d^2 + 2*a^2*b^2*d^2 + b^4*d^2) + 1/2*a^2/(a^2*d + b^2*
d)^2) + a/(a^2*d + b^2*d) + I*b/((a^2 + b^2)*d))*log(1/4*(a^6 + 2*a^4*b^2 + a^2*b^4)*d^3*(2*sqrt(-1/2*(I*a^2 +
I*b^2)*d*(2*a/(a^4*d^3 + 2*a^2*b^2*d^3 + b^4*d^3) - (3*a^2 - b^2)*a/((a^4*d^2 + 2*a^2*b^2*d^2 + b^4*d^2)*(a^2
*d + b^2*d)) + a^3/(a^2*d + b^2*d)^3)/b - 1/4*(3*a^2 - b^2)/(a^4*d^2 + 2*a^2*b^2*d^2 + b^4*d^2) - 1/2*(a^2 - b
^2)/(a^4*d^2 + 2*a^2*b^2*d^2 + b^4*d^2) + 1/2*a^2/(a^2*d + b^2*d)^2) + a/(a^2*d + b^2*d) + I*b/((a^2 + b^2)*d)
)^3 - 1/4*(3*a^5 + 2*a^3*b^2 - a*b^4)*d^2*(2*sqrt(-1/2*(I*a^2 + I*b^2)*d*(2*a/(a^4*d^3 + 2*a^2*b^2*d^3 + b^4*d
^3) - (3*a^2 - b^2)*a/((a^4*d^2 + 2*a^2*b^2*d^2 + b^4*d^2)*(a^2*d + b^2*d)) + a^3/(a^2*d + b^2*d)^3)/b - 1/4*(
3*a^2 - b^2)/(a^4*d^2 + 2*a^2*b^2*d^2 + b^4*d^2) - 1/2*(a^2 - b^2)/(a^4*d^2 + 2*a^2*b^2*d^2 + b^4*d^2) + 1/2*a
^2/(a^2*d + b^2*d)^2) + a/(a^2*d + b^2*d) + I*b/((a^2 + b^2)*d))^2 - a^3 - a*b^2 + 1/2*(3*a^4 - 4*a^2*b^2 + b^
4)*d*(2*sqrt(-1/2*(I*a^2 + I*b^2)*d*(2*a/(a^4*d^3 + 2*a^2*b^2*d^3 + b^4*d^3) - (3*a^2 - b^2)*a/((a^4*d^2 + 2*a
^2*b^2*d^2 + b^4*d^2)*(a^2*d + b^2*d)) + a^3/(a^2*d + b^2*d)^3)/b - 1/4*(3*a^2 - b^2)/(a^4*d^2 + 2*a^2*b^2*d^2
+ b^4*d^2) - 1/2*(a^2 - b^2)/(a^4*d^2 + 2*a^2*b^2*d^2 + b^4*d^2) + 1/2*a^2/(a^2*d + b^2*d)^2) + a/(a^2*d + b^
2*d) + I*b/((a^2 + b^2)*d)) - (3*a^2*b - b^3)*tan(d*x + c)^(1/3))*tan(d*x + c) + 2*(a^3 + a*b^2)*(2*(1/2)^(2/3
)*(-I*sqrt(3) + 1)*(6*(sqrt(3)*a^3*d*sqrt(-(a^2 + 2*I*a*b - b^2)/((a^4 + 2*a^2*b^2 + b^4)*d^2)) - I*sqrt(3)*a^
2*b*d*sqrt(-(a^2 + 2*I*a*b - b^2)/((a^4 + 2*a^2*b^2 + b^4)*d^2)) + sqrt(3)*a*b^2*d*sqrt(-(a^2 + 2*I*a*b - b^2)
/((a^4 + 2*a^2*b^2 + b^4)*d^2)) - I*sqrt(3)*b^3*d*sqrt(-(a^2 + 2*I*a*b - b^2)/((a^4 + 2*a^2*b^2 + b^4)*d^2)) -
3*a^2 + 4*I*a*b + b^2)/(a^4*d^2 + 2*a^2*b^2*d^2 + b^4*d^2) + (sqrt(3)*a^2*d*sqrt(-(a^2 + 2*I*a*b - b^2)/((a^4
+ 2*a^2*b^2 + b^4)*d^2)) + sqrt(3)*b^2*d*sqrt(-(a^2 + 2*I*a*b - b^2)/((a^4 + 2*a^2*b^2 + b^4)*d^2)) - 3*a + I
*b)^2/(a^2*d + b^2*d)^2)/(27*(3*sqrt(3)*a^4*d^3*(-(a^2 + 2*I*a*b - b^2)/((a^4 + 2*a^2*b^2 + b^4)*d^2))^(3/2) +
3*sqrt(3)*b^4*d^3*(-(a^2 + 2*I*a*b - b^2)/((a^4 + 2*a^2*b^2 + b^4)*d^2))^(3/2) - 2*I*sqrt(3)*a*b*d*sqrt(-(a^2
+ 2*I*a*b - b^2)/((a^4 + 2*a^2*b^2 + b^4)*d^2)) - 7*sqrt(3)*b^2*d*sqrt(-(a^2 + 2*I*a*b - b^2)/((a^4 + 2*a^2*b
^2 + b^4)*d^2)) + (6*sqrt(3)*b^2*d^3*(-(a^2 + 2*I*a*b - b^2)/((a^4 + 2*a^2*b^2 + b^4)*d^2))^(3/2) + 7*sqrt(3)*
d*sqrt(-(a^2 + 2*I*a*b - b^2)/((a^4 + 2*a^2*b^2 + b^4)*d^2)))*a^2 - 4*a + 4*I*b)/(a^4*d^3 + 2*a^2*b^2*d^3 + b^
4*d^3) + 2*(sqrt(3)*a^2*d*sqrt(-(a^2 + 2*I*a*b - b^2)/((a^4 + 2*a^2*b^2 + b^4)*d^2)) + sqrt(3)*b^2*d*sqrt(-(a^
2 + 2*I*a*b - b^2)/((a^4 + 2*a^2*b^2 + b^4)*d^2)) - 3*a + I*b)^3/(a^2*d + b^2*d)^3 + 18*(sqrt(3)*a^3*d*sqrt(-(
a^2 + 2*I*a*b - b^2)/((a^4 + 2*a^2*b^2 + b^4)*d^2)) - I*sqrt(3)*a^2*b*d*sqrt(-(a^2 + 2*I*a*b - b^2)/((a^4 + 2*
a^2*b^2 + b^4)*d^2)) + sqrt(3)*a*b^2*d*sqrt(-(a^2 + 2*I*a*b - b^2)/((a^4 + 2*a^2*b^2 + b^4)*d^2)) - I*sqrt(3)*
b^3*d*sqrt(-(a^2 + 2*I*a*b - b^2)/((a^4 + 2*a^2*b^2 + b^4)*d^2)) - 3*a^2 + 4*I*a*b + b^2)*(sqrt(3)*a^2*d*sqrt(
-(a^2 + 2*I*a*b - b^2)/((a^4 + 2*a^2*b^2 + b^4)*d^2)) + sqrt(3)*b^2*d*sqrt(-(a^2 + 2*I*a*b - b^2)/((a^4 + 2*a^
2*b^2 + b^4)*d^2)) - 3*a + I*b)/((a^4*d^2 + 2*a^2*b^2*d^2 + b^4*d^2)*(a^2*d + b^2*d)) + 18*sqrt(-18*sqrt(3)*(I
*a^10*b + 5*I*a^8*b^3 + 10*I*a^6*b^5 + 10*I*a^4*b^7 + 5*I*a^2*b^9 + I*b^11)*d^5*(-(a^2 + 2*I*a*b - b^2)/((a^4
+ 2*a^2*b^2 + b^4)*d^2))^(5/2) + 9*a^6 - 54*I*a^5*b - 159*a^4*b^2 + 276*I*a^3*b^3 + 295*a^2*b^4 - 182*I*a*b^5
- 49*b^6 - 8*sqrt(3)*(12*I*a^8*b + 15*a^7*b^2 + 23*I*a^6*b^3 + 45*a^5*b^4 - 3*I*a^4*b^5 + 45*a^3*b^6 - 27*I*a^
2*b^7 + 15*a*b^8 - 13*I*b^9)*d^3*(-(a^2 + 2*I*a*b - b^2)/((a^4 + 2*a^2*b^2 + b^4)*d^2))^(3/2) - 2*sqrt(3)*(39*
I*a^6*b + 48*a^5*b^2 + I*a^4*b^3 - 24*a^3*b^4 + 21*I*a^2*b^5 - 72*a*b^6 + 59*I*b^7)*d*sqrt(-(a^2 + 2*I*a*b - b
^2)/((a^4 + 2*a^2*b^2 + b^4)*d^2)) - 9/4*(71*a^8 + 76*I*a^7*b - 44*a^6*b^2 + 108*I*a^5*b^3 - 182*a^4*b^4 - 12*
I*a^3*b^5 + 52*a^2*b^6 - 44*I*a*b^7 + 119*b^8)*(a^2 + 2*I*a*b - b^2)/(a^4 + 2*a^2*b^2 + b^4) + 9/2*(49*a^10 +
18*I*a^9*b + 139*a^8*b^2 + 72*I*a^7*b^3 + 66*a^6*b^4 + 108*I*a^5*b^5 - 146*a^4*b^6 + 72*I*a^3*b^7 - 179*a^2*b^
8 + 18*I*a*b^9 - 57*b^10)*(a^2 + 2*I*a*b - b^2)^2/(a^4 + 2*a^2*b^2 + b^4)^2 - 279/4*(a^12 + 6*a^10*b^2 + 15*a^
8*b^4 + 20*a^6*b^6 + 15*a^4*b^8 + 6*a^2*b^10 + b^12)*(a^2 + 2*I*a*b - b^2)^3/(a^4 + 2*a^2*b^2 + b^4)^3)/((a^2
+ b^2)^3*d^3))^(1/3) + (1/2)^(1/3)*(I*sqrt(3) + 1)*(27*(3*sqrt(3)*a^4*d^3*(-(a^2 + 2*I*a*b - b^2)/((a^4 + 2*a^
2*b^2 + b^4)*d^2))^(3/2) + 3*sqrt(3)*b^4*d^3*(-(a^2 + 2*I*a*b - b^2)/((a^4 + 2*a^2*b^2 + b^4)*d^2))^(3/2) - 2*
I*sqrt(3)*a*b*d*sqrt(-(a^2 + 2*I*a*b - b^2)/((a^4 + 2*a^2*b^2 + b^4)*d^2)) - 7*sqrt(3)*b^2*d*sqrt(-(a^2 + 2*I*
a*b - b^2)/((a^4 + 2*a^2*b^2 + b^4)*d^2)) + (6*...

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (a + b \tan {\left (c + d x \right )}\right ) \tan ^{\frac {5}{3}}{\left (c + d x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/tan(d*x+c)**(5/3)/(a+b*tan(d*x+c)),x)

[Out]

Integral(1/((a + b*tan(c + d*x))*tan(c + d*x)**(5/3)), x)

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Giac [A]
time = 1.40, size = 457, normalized size = 0.87 \begin {gather*} \frac {b^{3} \left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left ({\left | -\left (-\frac {a}{b}\right )^{\frac {1}{3}} + \tan \left (d x + c\right )^{\frac {1}{3}} \right |}\right )}{a^{4} d + a^{2} b^{2} d} - \frac {3 \, \left (-a b^{2}\right )^{\frac {1}{3}} b^{2} \arctan \left (\frac {\sqrt {3} {\left (\left (-\frac {a}{b}\right )^{\frac {1}{3}} + 2 \, \tan \left (d x + c\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{{\left (\sqrt {3} a^{4} + \sqrt {3} a^{2} b^{2}\right )} d} - \frac {\left (-a b^{2}\right )^{\frac {1}{3}} b^{2} \log \left (\left (-\frac {a}{b}\right )^{\frac {2}{3}} + \left (-\frac {a}{b}\right )^{\frac {1}{3}} \tan \left (d x + c\right )^{\frac {1}{3}} + \tan \left (d x + c\right )^{\frac {2}{3}}\right )}{2 \, {\left (a^{4} + a^{2} b^{2}\right )} d} + \frac {{\left (\sqrt {3} a - b\right )} \arctan \left (\sqrt {3} + 2 \, \tan \left (d x + c\right )^{\frac {1}{3}}\right )}{2 \, {\left (a^{2} d + b^{2} d\right )}} - \frac {{\left (\sqrt {3} a + b\right )} \arctan \left (-\sqrt {3} + 2 \, \tan \left (d x + c\right )^{\frac {1}{3}}\right )}{2 \, {\left (a^{2} d + b^{2} d\right )}} - \frac {b \arctan \left (\tan \left (d x + c\right )^{\frac {1}{3}}\right )}{a^{2} d + b^{2} d} - \frac {a \log \left (\tan \left (d x + c\right )^{\frac {4}{3}} - \tan \left (d x + c\right )^{\frac {2}{3}} + 1\right )}{4 \, {\left (a^{2} d + b^{2} d\right )}} - \frac {3 \, b \log \left (\sqrt {3} \tan \left (d x + c\right )^{\frac {1}{3}} + \tan \left (d x + c\right )^{\frac {2}{3}} + 1\right )}{4 \, {\left (\sqrt {3} a^{2} d + \sqrt {3} b^{2} d\right )}} + \frac {3 \, b \log \left (-\sqrt {3} \tan \left (d x + c\right )^{\frac {1}{3}} + \tan \left (d x + c\right )^{\frac {2}{3}} + 1\right )}{4 \, {\left (\sqrt {3} a^{2} d + \sqrt {3} b^{2} d\right )}} + \frac {a \log \left (\tan \left (d x + c\right )^{\frac {2}{3}} + 1\right )}{2 \, {\left (a^{2} d + b^{2} d\right )}} - \frac {3}{2 \, a d \tan \left (d x + c\right )^{\frac {2}{3}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/tan(d*x+c)^(5/3)/(a+b*tan(d*x+c)),x, algorithm="giac")

[Out]

b^3*(-a/b)^(1/3)*log(abs(-(-a/b)^(1/3) + tan(d*x + c)^(1/3)))/(a^4*d + a^2*b^2*d) - 3*(-a*b^2)^(1/3)*b^2*arcta
n(1/3*sqrt(3)*((-a/b)^(1/3) + 2*tan(d*x + c)^(1/3))/(-a/b)^(1/3))/((sqrt(3)*a^4 + sqrt(3)*a^2*b^2)*d) - 1/2*(-
a*b^2)^(1/3)*b^2*log((-a/b)^(2/3) + (-a/b)^(1/3)*tan(d*x + c)^(1/3) + tan(d*x + c)^(2/3))/((a^4 + a^2*b^2)*d)
+ 1/2*(sqrt(3)*a - b)*arctan(sqrt(3) + 2*tan(d*x + c)^(1/3))/(a^2*d + b^2*d) - 1/2*(sqrt(3)*a + b)*arctan(-sqr
t(3) + 2*tan(d*x + c)^(1/3))/(a^2*d + b^2*d) - b*arctan(tan(d*x + c)^(1/3))/(a^2*d + b^2*d) - 1/4*a*log(tan(d*
x + c)^(4/3) - tan(d*x + c)^(2/3) + 1)/(a^2*d + b^2*d) - 3/4*b*log(sqrt(3)*tan(d*x + c)^(1/3) + tan(d*x + c)^(
2/3) + 1)/(sqrt(3)*a^2*d + sqrt(3)*b^2*d) + 3/4*b*log(-sqrt(3)*tan(d*x + c)^(1/3) + tan(d*x + c)^(2/3) + 1)/(s
qrt(3)*a^2*d + sqrt(3)*b^2*d) + 1/2*a*log(tan(d*x + c)^(2/3) + 1)/(a^2*d + b^2*d) - 3/2/(a*d*tan(d*x + c)^(2/3
))

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Mupad [B]
time = 5.62, size = 2500, normalized size = 4.76 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(tan(c + d*x)^(5/3)*(a + b*tan(c + d*x))),x)

[Out]

(log(tan(c + d*x)^(1/3) + 1i)*1i)/(2*(a*d*1i - b*d)) + log(tan(c + d*x)^(1/3)*1i + 1)/(2*(a*d - b*d*1i)) + sym
sum(log(root(32*a^2*b^2*d^4*z^4 + 16*b^4*d^4*z^4 + 16*a^4*d^4*z^4 + 16*a*b^2*d^3*z^3 + 16*a^3*d^3*z^3 - 4*b^2*
d^2*z^2 + 12*a^2*d^2*z^2 + 4*a*d*z + 1, z, k)*(root(32*a^2*b^2*d^4*z^4 + 16*b^4*d^4*z^4 + 16*a^4*d^4*z^4 + 16*
a*b^2*d^3*z^3 + 16*a^3*d^3*z^3 - 4*b^2*d^2*z^2 + 12*a^2*d^2*z^2 + 4*a*d*z + 1, z, k)*(root(32*a^2*b^2*d^4*z^4
+ 16*b^4*d^4*z^4 + 16*a^4*d^4*z^4 + 16*a*b^2*d^3*z^3 + 16*a^3*d^3*z^3 - 4*b^2*d^2*z^2 + 12*a^2*d^2*z^2 + 4*a*d
*z + 1, z, k)^2*(tan(c + d*x)^(1/3)*(314928*a^13*b^13*d^11 - 419904*a^15*b^11*d^11 + 118098*a^17*b^9*d^11 + 39
366*a^19*b^7*d^11 + 39366*a^21*b^5*d^11 + 13122*a^23*b^3*d^11) - root(32*a^2*b^2*d^4*z^4 + 16*b^4*d^4*z^4 + 16
*a^4*d^4*z^4 + 16*a*b^2*d^3*z^3 + 16*a^3*d^3*z^3 - 4*b^2*d^2*z^2 + 12*a^2*d^2*z^2 + 4*a*d*z + 1, z, k)*(root(3
2*a^2*b^2*d^4*z^4 + 16*b^4*d^4*z^4 + 16*a^4*d^4*z^4 + 16*a*b^2*d^3*z^3 + 16*a^3*d^3*z^3 - 4*b^2*d^2*z^2 + 12*a
^2*d^2*z^2 + 4*a*d*z + 1, z, k)^2*(tan(c + d*x)^(1/3)*(419904*a^14*b^15*d^14 + 1259712*a^16*b^13*d^14 + 944784
*a^18*b^11*d^14 - 419904*a^20*b^9*d^14 - 629856*a^22*b^7*d^14 + 104976*a^26*b^3*d^14) - root(32*a^2*b^2*d^4*z^
4 + 16*b^4*d^4*z^4 + 16*a^4*d^4*z^4 + 16*a*b^2*d^3*z^3 + 16*a^3*d^3*z^3 - 4*b^2*d^2*z^2 + 12*a^2*d^2*z^2 + 4*a
*d*z + 1, z, k)*(419904*a^16*b^14*d^15 + 1259712*a^18*b^12*d^15 + 839808*a^20*b^10*d^15 - 839808*a^22*b^8*d^15
- 1259712*a^24*b^6*d^15 - 419904*a^26*b^4*d^15)) + 419904*a^13*b^14*d^12 + 419904*a^15*b^12*d^12 - 288684*a^1
7*b^10*d^12 - 131220*a^19*b^8*d^12 + 183708*a^21*b^6*d^12 + 26244*a^23*b^4*d^12)) - 26244*a^12*b^12*d^9 + 6561
*a^16*b^8*d^9 + 6561*a^18*b^6*d^9) + tan(c + d*x)^(1/3)*(13122*a^12*b^11*d^8 - 6561*a^14*b^9*d^8)))*root(32*a^
2*b^2*d^4*z^4 + 16*b^4*d^4*z^4 + 16*a^4*d^4*z^4 + 16*a*b^2*d^3*z^3 + 16*a^3*d^3*z^3 - 4*b^2*d^2*z^2 + 12*a^2*d
^2*z^2 + 4*a*d*z + 1, z, k), k, 1, 4) + log((-b^8/(a^11*d^3 + a^5*b^6*d^3 + 3*a^7*b^4*d^3 + 3*a^9*b^2*d^3))^(1
/3)*((tan(c + d*x)^(1/3)*(314928*a^13*b^13*d^11 - 419904*a^15*b^11*d^11 + 118098*a^17*b^9*d^11 + 39366*a^19*b^
7*d^11 + 39366*a^21*b^5*d^11 + 13122*a^23*b^3*d^11) - (-b^8/(a^11*d^3 + a^5*b^6*d^3 + 3*a^7*b^4*d^3 + 3*a^9*b^
2*d^3))^(1/3)*((tan(c + d*x)^(1/3)*(419904*a^14*b^15*d^14 + 1259712*a^16*b^13*d^14 + 944784*a^18*b^11*d^14 - 4
19904*a^20*b^9*d^14 - 629856*a^22*b^7*d^14 + 104976*a^26*b^3*d^14) - (-b^8/(a^11*d^3 + a^5*b^6*d^3 + 3*a^7*b^4
*d^3 + 3*a^9*b^2*d^3))^(1/3)*(419904*a^16*b^14*d^15 + 1259712*a^18*b^12*d^15 + 839808*a^20*b^10*d^15 - 839808*
a^22*b^8*d^15 - 1259712*a^24*b^6*d^15 - 419904*a^26*b^4*d^15))*(-b^8/(a^11*d^3 + a^5*b^6*d^3 + 3*a^7*b^4*d^3 +
3*a^9*b^2*d^3))^(2/3) + 419904*a^13*b^14*d^12 + 419904*a^15*b^12*d^12 - 288684*a^17*b^10*d^12 - 131220*a^19*b
^8*d^12 + 183708*a^21*b^6*d^12 + 26244*a^23*b^4*d^12))*(-b^8/(a^11*d^3 + a^5*b^6*d^3 + 3*a^7*b^4*d^3 + 3*a^9*b
^2*d^3))^(2/3) - 26244*a^12*b^12*d^9 + 6561*a^16*b^8*d^9 + 6561*a^18*b^6*d^9) + tan(c + d*x)^(1/3)*(13122*a^12
*b^11*d^8 - 6561*a^14*b^9*d^8))*(-b^8/(a^11*d^3 + a^5*b^6*d^3 + 3*a^7*b^4*d^3 + 3*a^9*b^2*d^3))^(1/3) - log(ta
n(c + d*x)^(1/3)*(13122*a^12*b^11*d^8 - 6561*a^14*b^9*d^8) - ((3^(1/2)*1i)/2 + 1/2)*(-b^8/(a^11*d^3 + a^5*b^6*
d^3 + 3*a^7*b^4*d^3 + 3*a^9*b^2*d^3))^(1/3)*((tan(c + d*x)^(1/3)*(314928*a^13*b^13*d^11 - 419904*a^15*b^11*d^1
1 + 118098*a^17*b^9*d^11 + 39366*a^19*b^7*d^11 + 39366*a^21*b^5*d^11 + 13122*a^23*b^3*d^11) + ((3^(1/2)*1i)/2
+ 1/2)*(-b^8/(a^11*d^3 + a^5*b^6*d^3 + 3*a^7*b^4*d^3 + 3*a^9*b^2*d^3))^(1/3)*(((3^(1/2)*1i)/2 + 1/2)^2*(tan(c
+ d*x)^(1/3)*(419904*a^14*b^15*d^14 + 1259712*a^16*b^13*d^14 + 944784*a^18*b^11*d^14 - 419904*a^20*b^9*d^14 -
629856*a^22*b^7*d^14 + 104976*a^26*b^3*d^14) + ((3^(1/2)*1i)/2 + 1/2)*(-b^8/(a^11*d^3 + a^5*b^6*d^3 + 3*a^7*b^
4*d^3 + 3*a^9*b^2*d^3))^(1/3)*(419904*a^16*b^14*d^15 + 1259712*a^18*b^12*d^15 + 839808*a^20*b^10*d^15 - 839808
*a^22*b^8*d^15 - 1259712*a^24*b^6*d^15 - 419904*a^26*b^4*d^15))*(-b^8/(a^11*d^3 + a^5*b^6*d^3 + 3*a^7*b^4*d^3
+ 3*a^9*b^2*d^3))^(2/3) + 419904*a^13*b^14*d^12 + 419904*a^15*b^12*d^12 - 288684*a^17*b^10*d^12 - 131220*a^19*
b^8*d^12 + 183708*a^21*b^6*d^12 + 26244*a^23*b^4*d^12))*((3^(1/2)*1i)/2 + 1/2)^2*(-b^8/(a^11*d^3 + a^5*b^6*d^3
+ 3*a^7*b^4*d^3 + 3*a^9*b^2*d^3))^(2/3) - 26244*a^12*b^12*d^9 + 6561*a^16*b^8*d^9 + 6561*a^18*b^6*d^9))*((3^(
1/2)*1i)/2 + 1/2)*(-b^8/(a^11*d^3 + a^5*b^6*d^3 + 3*a^7*b^4*d^3 + 3*a^9*b^2*d^3))^(1/3) + log(tan(c + d*x)^(1/
3)*(13122*a^12*b^11*d^8 - 6561*a^14*b^9*d^8) + ((3^(1/2)*1i)/2 - 1/2)*(-b^8/(a^11*d^3 + a^5*b^6*d^3 + 3*a^7*b^
4*d^3 + 3*a^9*b^2*d^3))^(1/3)*((tan(c + d*x)^(1/3)*(314928*a^13*b^13*d^11 - 419904*a^15*b^11*d^11 + 118098*a^1
7*b^9*d^11 + 39366*a^19*b^7*d^11 + 39366*a^21*b^5*d^11 + 13122*a^23*b^3*d^11) - ((3^(1/2)*1i)/2 - 1/2)*(-b^8/(
a^11*d^3 + a^5*b^6*d^3 + 3*a^7*b^4*d^3 + 3*a^9*b^2*d^3))^(1/3)*(((3^(1/2)*1i)/2 - 1/2)^2*(tan(c + d*x)^(1/3)*(
419904*a^14*b^15*d^14 + 1259712*a^16*b^13*d^14 + 944784*a^18*b^11*d^14 - 419904*a^20*b^9*d^14 - 629856*a^22*b^
7*d^14 + 104976*a^26*b^3*d^14) - ((3^(1/2)*1i)/...

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