3.7.72 \(\int \frac {\sqrt [3]{\tan (c+d x)}}{a+b \tan (c+d x)} \, dx\) [672]

Optimal. Leaf size=465 \[ -\frac {b \text {ArcTan}\left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )}{2 \left (a^2+b^2\right ) d}+\frac {b \text {ArcTan}\left (\sqrt {3}+2 \sqrt [3]{\tan (c+d x)}\right )}{2 \left (a^2+b^2\right ) d}+\frac {\sqrt {3} \sqrt [3]{a} b^{2/3} \text {ArcTan}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} \sqrt [3]{\tan (c+d x)}}{\sqrt {3} \sqrt [3]{a}}\right )}{\left (a^2+b^2\right ) d}-\frac {\sqrt {3} a \text {ArcTan}\left (\frac {1-2 \tan ^{\frac {2}{3}}(c+d x)}{\sqrt {3}}\right )}{2 \left (a^2+b^2\right ) d}+\frac {b \text {ArcTan}\left (\sqrt [3]{\tan (c+d x)}\right )}{\left (a^2+b^2\right ) d}-\frac {3 \sqrt [3]{a} b^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{\tan (c+d x)}\right )}{2 \left (a^2+b^2\right ) d}-\frac {a \log \left (1+\tan ^{\frac {2}{3}}(c+d x)\right )}{2 \left (a^2+b^2\right ) d}-\frac {\sqrt {3} b \log \left (1-\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{4 \left (a^2+b^2\right ) d}+\frac {\sqrt {3} b \log \left (1+\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{4 \left (a^2+b^2\right ) d}+\frac {\sqrt [3]{a} b^{2/3} \log (a+b \tan (c+d x))}{2 \left (a^2+b^2\right ) d}+\frac {a \log \left (1-\tan ^{\frac {2}{3}}(c+d x)+\tan ^{\frac {4}{3}}(c+d x)\right )}{4 \left (a^2+b^2\right ) d} \]

[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*a
rctan(tan(d*x+c)^(1/3))/(a^2+b^2)/d-3/2*a^(1/3)*b^(2/3)*ln(a^(1/3)+b^(1/3)*tan(d*x+c)^(1/3))/(a^2+b^2)/d-1/2*a
*ln(1+tan(d*x+c)^(2/3))/(a^2+b^2)/d+1/2*a^(1/3)*b^(2/3)*ln(a+b*tan(d*x+c))/(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+a^(1/3)*b^(2/3)*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^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

________________________________________________________________________________________

Rubi [A]
time = 0.39, antiderivative size = 465, 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, 3609, 3619, 3557, 335, 215, 648, 632, 210, 642, 209, 281, 298, 31, 3715, 52, 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 {\sqrt {3} \sqrt [3]{a} b^{2/3} \text {ArcTan}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} \sqrt [3]{\tan (c+d x)}}{\sqrt {3} \sqrt [3]{a}}\right )}{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 {\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 {4}{3}}(c+d x)-\tan ^{\frac {2}{3}}(c+d x)+1\right )}{4 d \left (a^2+b^2\right )}-\frac {3 \sqrt [3]{a} b^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{\tan (c+d x)}\right )}{2 d \left (a^2+b^2\right )}+\frac {\sqrt [3]{a} b^{2/3} \log (a+b \tan (c+d x))}{2 d \left (a^2+b^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 31

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

Rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && 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 3609

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

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

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 0.43, size = 204, normalized size = 0.44 \begin {gather*} \frac {2 \sqrt [3]{a} b^{2/3} \left (2 \sqrt {3} \text {ArcTan}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} \sqrt [3]{\tan (c+d x)}}{\sqrt {3} \sqrt [3]{a}}\right )-2 \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{\tan (c+d x)}\right )+\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} \sqrt [3]{\tan (c+d x)}+b^{2/3} \tan ^{\frac {2}{3}}(c+d x)\right )\right )+12 b \, _2F_1\left (\frac {1}{6},1;\frac {7}{6};-\tan ^2(c+d x)\right ) \sqrt [3]{\tan (c+d x)}+3 a \, _2F_1\left (\frac {2}{3},1;\frac {5}{3};-\tan ^2(c+d x)\right ) \tan ^{\frac {4}{3}}(c+d x)}{4 \left (a^2+b^2\right ) d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^(1/3)*b^(2/3)*(2*Sqrt[3]*ArcTan[(a^(1/3) - 2*b^(1/3)*Tan[c + d*x]^(1/3))/(Sqrt[3]*a^(1/3))] - 2*Log[a^(1/
3) + b^(1/3)*Tan[c + d*x]^(1/3)] + Log[a^(2/3) - a^(1/3)*b^(1/3)*Tan[c + d*x]^(1/3) + b^(2/3)*Tan[c + d*x]^(2/
3)]) + 12*b*Hypergeometric2F1[1/6, 1, 7/6, -Tan[c + d*x]^2]*Tan[c + d*x]^(1/3) + 3*a*Hypergeometric2F1[2/3, 1,
5/3, -Tan[c + d*x]^2]*Tan[c + d*x]^(4/3))/(4*(a^2 + b^2)*d)

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Maple [A]
time = 0.20, size = 328, normalized size = 0.71

method	result	size
derivativedivides	\(\frac {\frac {-\frac {3 a \ln \left (1+\tan ^{\frac {2}{3}}\left (d x +c \right )\right )}{2}+3 b \arctan \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )}{3 a^{2}+3 b^{2}}-\frac {3 \left (\frac {\ln \left (\tan ^{\frac {1}{3}}\left (d x +c \right )+\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}-\frac {\ln \left (\tan ^{\frac {2}{3}}\left (d x +c \right )-\left (\frac {a}{b}\right )^{\frac {1}{3}} \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}\right ) a b}{a^{2}+b^{2}}+\frac {\frac {3 \left (\sqrt {3}\, b +a \right ) \ln \left (1+\sqrt {3}\, \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )\right )}{4}+3 \left (2 b -\frac {\left (\sqrt {3}\, b +a \right ) \sqrt {3}}{2}\right ) \arctan \left (\sqrt {3}+2 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )\right )-\frac {3 \left (\sqrt {3}\, b -a \right ) \ln \left (1-\sqrt {3}\, \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )\right )}{4}-3 \left (-2 b +\frac {\left (\sqrt {3}\, b -a \right ) \sqrt {3}}{2}\right ) \arctan \left (-\sqrt {3}+2 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )\right )}{3 a^{2}+3 b^{2}}}{d}\)	\(328\)
default	\(\frac {\frac {-\frac {3 a \ln \left (1+\tan ^{\frac {2}{3}}\left (d x +c \right )\right )}{2}+3 b \arctan \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )}{3 a^{2}+3 b^{2}}-\frac {3 \left (\frac {\ln \left (\tan ^{\frac {1}{3}}\left (d x +c \right )+\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}-\frac {\ln \left (\tan ^{\frac {2}{3}}\left (d x +c \right )-\left (\frac {a}{b}\right )^{\frac {1}{3}} \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}\right ) a b}{a^{2}+b^{2}}+\frac {\frac {3 \left (\sqrt {3}\, b +a \right ) \ln \left (1+\sqrt {3}\, \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )\right )}{4}+3 \left (2 b -\frac {\left (\sqrt {3}\, b +a \right ) \sqrt {3}}{2}\right ) \arctan \left (\sqrt {3}+2 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )\right )-\frac {3 \left (\sqrt {3}\, b -a \right ) \ln \left (1-\sqrt {3}\, \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )\right )}{4}-3 \left (-2 b +\frac {\left (\sqrt {3}\, b -a \right ) \sqrt {3}}{2}\right ) \arctan \left (-\sqrt {3}+2 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )\right )}{3 a^{2}+3 b^{2}}}{d}\)	\(328\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(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/(a^2+b^2)*b+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))))

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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(tan(d*x+c)^(1/3)/(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 = 3.24, size = 74435, normalized size = 160.08 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(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*sqr
t(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)) + sq
rt(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)) + 9
*sqrt(-72*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) + 36*a^6 + 216*I*a^5*b - 636*a^4*b^2 - 1104*I*a^3*b^3 + 118
0*a^2*b^4 + 728*I*a*b^5 - 196*b^6 - 32*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) - 8*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*(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) + 18*(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*(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*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)) + 9*sqrt(-72*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) + 36*a^6 + 216*I*a^5*b - 636*a^4*
b^2 - 1104*I*a^3*b^3 + 1180*a^2*b^4 + 728*I*a*b^5 - 196*b^6 - 32*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) - 8*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*(71*a^8 - 76*I*a^7*
b - 44*a^6*b^2 - 108*I*a^5*b^3 - 182*a^4*b^4 + ...

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.71, size = 424, normalized size = 0.91 \begin {gather*} \frac {a b \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^{3} d + a b^{2} 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 \, \left (-a b^{2}\right )^{\frac {1}{3}} \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^{2} + \sqrt {3} b^{2}\right )} d} - \frac {\left (-a b^{2}\right )^{\frac {1}{3}} \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^{2} + b^{2}\right )} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 11.41, size = 2111, normalized size = 4.54 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

symsum(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*(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)*((6561*(44*a^2*b^10*d^3 + 84*a^4*b^8*d^3 + 36*a^6*b
^6*d^3 - 4*a^8*b^4*d^3))/d^6 + 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*((6561*tan(c + d*x)^(1/3)*(64*a*b^13*d^
6 + 240*a^3*b^11*d^6 + 320*a^5*b^9*d^6 + 160*a^7*b^7*d^6 - 16*a^11*b^3*d^6))/d^7 - (6561*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)*(64*a*b^14*d^6 + 192*a^3*b^12*d^6 + 128*a^5*b^10*d^6 - 128*a^7*b^8*d^6 - 192*a^9*b^6*d^6 - 64
*a^11*b^4*d^6))/d^6)) - (6561*tan(c + d*x)^(1/3)*(50*a^2*b^9*d^3 - 58*a^4*b^7*d^3 + 22*a^6*b^5*d^3 + 2*a^8*b^3
*d^3))/d^7) - (6561*(a*b^8 + 5*a^3*b^6))/d^6) + (6561*tan(c + d*x)^(1/3)*(a*b^7 - 2*a^3*b^5))/d^7))*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(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)) + log(((((419904*a*b^4*(a^2 - b^2)*(a^2 + b^2)^4*(-(a*b^2)/(d^3*(a^2 + b^2)
^3))^(1/3) - (104976*a*b^3*tan(c + d*x)^(1/3)*(a^2 - 4*b^2)*(a^2 + b^2)^4)/d)*(-(a*b^2)/(d^3*(a^2 + b^2)^3))^(
2/3) - (26244*a^2*b^4*(a^2 - 11*b^2)*(a^2 + b^2)^2)/d^3)*(-(a*b^2)/(d^3*(a^2 + b^2)^3))^(1/3) - (13122*a^2*b^3
*tan(c + d*x)^(1/3)*(a^6 + 25*b^6 - 29*a^2*b^4 + 11*a^4*b^2))/d^4)*(-(a*b^2)/(d^3*(a^2 + b^2)^3))^(2/3) - (656
1*a*b^6*(5*a^2 + b^2))/d^6)*(-(a*b^2)/(d^3*(a^2 + b^2)^3))^(1/3) - (6561*a*b^5*tan(c + d*x)^(1/3)*(2*a^2 - b^2
))/d^7)*(-(a*b^2)/(a^6*d^3 + b^6*d^3 + 3*a^2*b^4*d^3 + 3*a^4*b^2*d^3))^(1/3) + log(((3^(1/2)*1i)/2 - 1/2)*(((3
^(1/2)*1i)/2 + 1/2)*(((3^(1/2)*1i)/2 - 1/2)*(((3^(1/2)*1i)/2 + 1/2)*(419904*a*b^4*((3^(1/2)*1i)/2 - 1/2)*(a^2
- b^2)*(a^2 + b^2)^4*(-(a*b^2)/(d^3*(a^2 + b^2)^3))^(1/3) - (104976*a*b^3*tan(c + d*x)^(1/3)*(a^2 - 4*b^2)*(a^
2 + b^2)^4)/d)*(-(a*b^2)/(d^3*(a^2 + b^2)^3))^(2/3) + (26244*a^2*b^4*(a^2 - 11*b^2)*(a^2 + b^2)^2)/d^3)*(-(a*b
^2)/(d^3*(a^2 + b^2)^3))^(1/3) + (13122*a^2*b^3*tan(c + d*x)^(1/3)*(a^6 + 25*b^6 - 29*a^2*b^4 + 11*a^4*b^2))/d
^4)*(-(a*b^2)/(d^3*(a^2 + b^2)^3))^(2/3) - (6561*a*b^6*(5*a^2 + b^2))/d^6)*(-(a*b^2)/(d^3*(a^2 + b^2)^3))^(1/3
) - (6561*a*b^5*tan(c + d*x)^(1/3)*(2*a^2 - b^2))/d^7)*((3^(1/2)*1i)/2 - 1/2)*(-(a*b^2)/(a^6*d^3 + b^6*d^3 + 3
*a^2*b^4*d^3 + 3*a^4*b^2*d^3))^(1/3) - log(- ((3^(1/2)*1i)/2 + 1/2)*(((3^(1/2)*1i)/2 - 1/2)*(((3^(1/2)*1i)/2 +
1/2)*(((3^(1/2)*1i)/2 - 1/2)*(419904*a*b^4*((3^(1/2)*1i)/2 + 1/2)*(a^2 - b^2)*(a^2 + b^2)^4*(-(a*b^2)/(d^3*(a
^2 + b^2)^3))^(1/3) + (104976*a*b^3*tan(c + d*x)^(1/3)*(a^2 - 4*b^2)*(a^2 + b^2)^4)/d)*(-(a*b^2)/(d^3*(a^2 + b
^2)^3))^(2/3) + (26244*a^2*b^4*(a^2 - 11*b^2)*(a^2 + b^2)^2)/d^3)*(-(a*b^2)/(d^3*(a^2 + b^2)^3))^(1/3) - (1312
2*a^2*b^3*tan(c + d*x)^(1/3)*(a^6 + 25*b^6 - 29*a^2*b^4 + 11*a^4*b^2))/d^4)*(-(a*b^2)/(d^3*(a^2 + b^2)^3))^(2/
3) - (6561*a*b^6*(5*a^2 + b^2))/d^6)*(-(a*b^2)/(d^3*(a^2 + b^2)^3))^(1/3) - (6561*a*b^5*tan(c + d*x)^(1/3)*(2*
a^2 - b^2))/d^7)*((3^(1/2)*1i)/2 + 1/2)*(-(a*b^2)/(a^6*d^3 + b^6*d^3 + 3*a^2*b^4*d^3 + 3*a^4*b^2*d^3))^(1/3)

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