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

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

[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 = 0.54, 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 = {3574, 3528, 3538, 3476, 329, 209, 634, 618, 204, 628, 203, 275, 292, 31, 3634, 50, 58, 617} \[ -\frac {b \tan ^{-1}\left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )}{2 d \left (a^2+b^2\right )}+\frac {b \tan ^{-1}\left (2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}\right )}{2 d \left (a^2+b^2\right )}+\frac {\sqrt {3} \sqrt [3]{a} b^{2/3} \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} \sqrt [3]{\tan (c+d x)}}{\sqrt {3} \sqrt [3]{a}}\right )}{d \left (a^2+b^2\right )}-\frac {\sqrt {3} a \tan ^{-1}\left (\frac {1-2 \tan ^{\frac {2}{3}}(c+d x)}{\sqrt {3}}\right )}{2 d \left (a^2+b^2\right )}+\frac {b \tan ^{-1}\left (\sqrt [3]{\tan (c+d x)}\right )}{d \left (a^2+b^2\right )}-\frac {a \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )}{2 d \left (a^2+b^2\right )}-\frac {\sqrt {3} b \log \left (\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )}{4 d \left (a^2+b^2\right )}+\frac {\sqrt {3} b \log \left (\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )}{4 d \left (a^2+b^2\right )}+\frac {a \log \left (\tan ^{\frac {4}{3}}(c+d x)-\tan ^{\frac {2}{3}}(c+d x)+1\right )}{4 d \left (a^2+b^2\right )}-\frac {3 \sqrt [3]{a} b^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{\tan (c+d x)}\right )}{2 d \left (a^2+b^2\right )}+\frac {\sqrt [3]{a} b^{2/3} \log (a+b \tan (c+d x))}{2 d \left (a^2+b^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^(1/3)/(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]*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) + Ta
n[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 50

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 58

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[{q = Rt[-((b*c - a*d)/b), 3]}, -Sim
p[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 203

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

Rule 204

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

Rule 209

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*Int[1/(r^2 +
s^2*x^2), x])/(a*n) + 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 275

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 292

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 329

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 617

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 618

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 628

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 634

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 3476

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 3528

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 3538

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 3574

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 3634

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 \operatorname {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) \operatorname {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 \operatorname {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 ) \operatorname {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 ) \operatorname {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 \operatorname {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) \operatorname {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 ) \operatorname {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) \operatorname {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) \operatorname {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 \operatorname {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 \operatorname {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 \operatorname {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 \operatorname {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 \operatorname {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 \operatorname {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 \operatorname {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 ) \operatorname {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 ) \operatorname {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 \operatorname {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) \operatorname {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 \operatorname {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 \operatorname {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) \operatorname {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]  time = 0.41, size = 204, normalized size = 0.44 \[ \frac {2 \sqrt [3]{a} b^{2/3} \left (\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} \sqrt [3]{\tan (c+d x)}+b^{2/3} \tan ^{\frac {2}{3}}(c+d x)\right )+2 \sqrt {3} \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} \sqrt [3]{\tan (c+d x)}}{\sqrt {3} \sqrt [3]{a}}\right )-2 \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{\tan (c+d x)}\right )\right )+3 a \tan ^{\frac {4}{3}}(c+d x) \, _2F_1\left (\frac {2}{3},1;\frac {5}{3};-\tan ^2(c+d x)\right )+12 b \sqrt [3]{\tan (c+d x)} \, _2F_1\left (\frac {1}{6},1;\frac {7}{6};-\tan ^2(c+d x)\right )}{4 d \left (a^2+b^2\right )} \]

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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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tan \left (d x + c\right )^{\frac {1}{3}}}{b \tan \left (d x + c\right ) + a}\,{d x} \]

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]

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

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maple [A]  time = 0.28, size = 527, normalized size = 1.13 \[ -\frac {a \ln \left (\tan ^{\frac {1}{3}}\left (d x +c \right )+\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{d \left (a^{2}+b^{2}\right ) \left (\frac {a}{b}\right )^{\frac {2}{3}}}+\frac {a \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 )}{2 d \left (a^{2}+b^{2}\right ) \left (\frac {a}{b}\right )^{\frac {2}{3}}}-\frac {a \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 )}{d \left (a^{2}+b^{2}\right ) \left (\frac {a}{b}\right )^{\frac {2}{3}}}+\frac {3 \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 ) \sqrt {3}\, b}{4 d \left (3 a^{2}+3 b^{2}\right )}+\frac {3 \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 ) a}{4 d \left (3 a^{2}+3 b^{2}\right )}-\frac {3 \arctan \left (\sqrt {3}+2 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )\right ) \sqrt {3}\, a}{2 d \left (3 a^{2}+3 b^{2}\right )}+\frac {3 \arctan \left (\sqrt {3}+2 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )\right ) b}{2 d \left (3 a^{2}+3 b^{2}\right )}-\frac {3 \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 ) \sqrt {3}\, b}{4 d \left (3 a^{2}+3 b^{2}\right )}+\frac {3 \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 ) a}{4 d \left (3 a^{2}+3 b^{2}\right )}+\frac {3 \arctan \left (-\sqrt {3}+2 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )\right ) b}{2 d \left (3 a^{2}+3 b^{2}\right )}+\frac {3 \arctan \left (-\sqrt {3}+2 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )\right ) \sqrt {3}\, a}{2 d \left (3 a^{2}+3 b^{2}\right )}-\frac {3 a \ln \left (1+\tan ^{\frac {2}{3}}\left (d x +c \right )\right )}{2 d \left (3 a^{2}+3 b^{2}\right )}+\frac {3 b \arctan \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )}{d \left (3 a^{2}+3 b^{2}\right )} \]

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)

[Out]

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

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maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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mupad [B]  time = 11.41, size = 2111, normalized size = 4.54 \[ \text {result too large to display} \]

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

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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