3.54 \(\int (a+b \tan (c+d \sqrt [3]{x}))^2 \, dx\)

Optimal. Leaf size=206 \[ \frac{6 i a b \sqrt [3]{x} \text{PolyLog}\left (2,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}-\frac{3 a b \text{PolyLog}\left (3,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}-\frac{3 i b^2 \text{PolyLog}\left (2,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}+a^2 x-\frac{6 a b x^{2/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+2 i a b x+\frac{6 b^2 \sqrt [3]{x} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}+\frac{3 b^2 x^{2/3} \tan \left (c+d \sqrt [3]{x}\right )}{d}-\frac{3 i b^2 x^{2/3}}{d}-b^2 x \]

[Out]

((-3*I)*b^2*x^(2/3))/d + a^2*x + (2*I)*a*b*x - b^2*x + (6*b^2*x^(1/3)*Log[1 + E^((2*I)*(c + d*x^(1/3)))])/d^2
- (6*a*b*x^(2/3)*Log[1 + E^((2*I)*(c + d*x^(1/3)))])/d - ((3*I)*b^2*PolyLog[2, -E^((2*I)*(c + d*x^(1/3)))])/d^
3 + ((6*I)*a*b*x^(1/3)*PolyLog[2, -E^((2*I)*(c + d*x^(1/3)))])/d^2 - (3*a*b*PolyLog[3, -E^((2*I)*(c + d*x^(1/3
)))])/d^3 + (3*b^2*x^(2/3)*Tan[c + d*x^(1/3)])/d

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Rubi [A]  time = 0.354806, antiderivative size = 206, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 11, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.688, Rules used = {3739, 3722, 3719, 2190, 2531, 2282, 6589, 3720, 2279, 2391, 30} \[ a^2 x+\frac{6 i a b \sqrt [3]{x} \text{Li}_2\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}-\frac{3 a b \text{Li}_3\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}-\frac{6 a b x^{2/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+2 i a b x-\frac{3 i b^2 \text{Li}_2\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}+\frac{6 b^2 \sqrt [3]{x} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}+\frac{3 b^2 x^{2/3} \tan \left (c+d \sqrt [3]{x}\right )}{d}-\frac{3 i b^2 x^{2/3}}{d}-b^2 x \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Tan[c + d*x^(1/3)])^2,x]

[Out]

((-3*I)*b^2*x^(2/3))/d + a^2*x + (2*I)*a*b*x - b^2*x + (6*b^2*x^(1/3)*Log[1 + E^((2*I)*(c + d*x^(1/3)))])/d^2
- (6*a*b*x^(2/3)*Log[1 + E^((2*I)*(c + d*x^(1/3)))])/d - ((3*I)*b^2*PolyLog[2, -E^((2*I)*(c + d*x^(1/3)))])/d^
3 + ((6*I)*a*b*x^(1/3)*PolyLog[2, -E^((2*I)*(c + d*x^(1/3)))])/d^2 - (3*a*b*PolyLog[3, -E^((2*I)*(c + d*x^(1/3
)))])/d^3 + (3*b^2*x^(2/3)*Tan[c + d*x^(1/3)])/d

Rule 3739

Int[((a_.) + (b_.)*Tan[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(1/n - 1)*(a + b*Ta
n[c + d*x])^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, p}, x] && IGtQ[1/n, 0] && IntegerQ[p]

Rule 3722

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Int[ExpandIntegrand[
(c + d*x)^m, (a + b*Tan[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[m, 0] && IGtQ[n, 0]

Rule 3719

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[(I*(c + d*x)^(m + 1))/(d*(m + 1)), x
] - Dist[2*I, Int[((c + d*x)^m*E^(2*I*(e + f*x)))/(1 + E^(2*I*(e + f*x))), x], x] /; FreeQ[{c, d, e, f}, x] &&
 IGtQ[m, 0]

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
 (f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
 - Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2531

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> -Simp[((
f + g*x)^m*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)])/(b*c*n*Log[F]), x] + Dist[(g*m)/(b*c*n*Log[F]), Int[(f + g*x)
^(m - 1)*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 6589

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rule 3720

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

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
 e, n}, x] && EqQ[c*d, 1]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

\begin{align*} \int \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right )^2 \, dx &=3 \operatorname{Subst}\left (\int x^2 (a+b \tan (c+d x))^2 \, dx,x,\sqrt [3]{x}\right )\\ &=3 \operatorname{Subst}\left (\int \left (a^2 x^2+2 a b x^2 \tan (c+d x)+b^2 x^2 \tan ^2(c+d x)\right ) \, dx,x,\sqrt [3]{x}\right )\\ &=a^2 x+(6 a b) \operatorname{Subst}\left (\int x^2 \tan (c+d x) \, dx,x,\sqrt [3]{x}\right )+\left (3 b^2\right ) \operatorname{Subst}\left (\int x^2 \tan ^2(c+d x) \, dx,x,\sqrt [3]{x}\right )\\ &=a^2 x+2 i a b x+\frac{3 b^2 x^{2/3} \tan \left (c+d \sqrt [3]{x}\right )}{d}-(12 i a b) \operatorname{Subst}\left (\int \frac{e^{2 i (c+d x)} x^2}{1+e^{2 i (c+d x)}} \, dx,x,\sqrt [3]{x}\right )-\left (3 b^2\right ) \operatorname{Subst}\left (\int x^2 \, dx,x,\sqrt [3]{x}\right )-\frac{\left (6 b^2\right ) \operatorname{Subst}\left (\int x \tan (c+d x) \, dx,x,\sqrt [3]{x}\right )}{d}\\ &=-\frac{3 i b^2 x^{2/3}}{d}+a^2 x+2 i a b x-b^2 x-\frac{6 a b x^{2/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+\frac{3 b^2 x^{2/3} \tan \left (c+d \sqrt [3]{x}\right )}{d}+\frac{(12 a b) \operatorname{Subst}\left (\int x \log \left (1+e^{2 i (c+d x)}\right ) \, dx,x,\sqrt [3]{x}\right )}{d}+\frac{\left (12 i b^2\right ) \operatorname{Subst}\left (\int \frac{e^{2 i (c+d x)} x}{1+e^{2 i (c+d x)}} \, dx,x,\sqrt [3]{x}\right )}{d}\\ &=-\frac{3 i b^2 x^{2/3}}{d}+a^2 x+2 i a b x-b^2 x+\frac{6 b^2 \sqrt [3]{x} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}-\frac{6 a b x^{2/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+\frac{6 i a b \sqrt [3]{x} \text{Li}_2\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}+\frac{3 b^2 x^{2/3} \tan \left (c+d \sqrt [3]{x}\right )}{d}-\frac{(6 i a b) \operatorname{Subst}\left (\int \text{Li}_2\left (-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt [3]{x}\right )}{d^2}-\frac{\left (6 b^2\right ) \operatorname{Subst}\left (\int \log \left (1+e^{2 i (c+d x)}\right ) \, dx,x,\sqrt [3]{x}\right )}{d^2}\\ &=-\frac{3 i b^2 x^{2/3}}{d}+a^2 x+2 i a b x-b^2 x+\frac{6 b^2 \sqrt [3]{x} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}-\frac{6 a b x^{2/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+\frac{6 i a b \sqrt [3]{x} \text{Li}_2\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}+\frac{3 b^2 x^{2/3} \tan \left (c+d \sqrt [3]{x}\right )}{d}-\frac{(3 a b) \operatorname{Subst}\left (\int \frac{\text{Li}_2(-x)}{x} \, dx,x,e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}+\frac{\left (3 i b^2\right ) \operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}\\ &=-\frac{3 i b^2 x^{2/3}}{d}+a^2 x+2 i a b x-b^2 x+\frac{6 b^2 \sqrt [3]{x} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}-\frac{6 a b x^{2/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}-\frac{3 i b^2 \text{Li}_2\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}+\frac{6 i a b \sqrt [3]{x} \text{Li}_2\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}-\frac{3 a b \text{Li}_3\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}+\frac{3 b^2 x^{2/3} \tan \left (c+d \sqrt [3]{x}\right )}{d}\\ \end{align*}

Mathematica [A]  time = 1.91886, size = 185, normalized size = 0.9 \[ \frac{b \left (3 i \left (b-2 a d \sqrt [3]{x}\right ) \text{PolyLog}\left (2,-e^{-2 i \left (c+d \sqrt [3]{x}\right )}\right )-3 a \text{PolyLog}\left (3,-e^{-2 i \left (c+d \sqrt [3]{x}\right )}\right )+\frac{6 i b d^2 x^{2/3}-4 i a d^3 x}{1+e^{2 i c}}+6 d \sqrt [3]{x} \left (b-a d \sqrt [3]{x}\right ) \log \left (1+e^{-2 i \left (c+d \sqrt [3]{x}\right )}\right )\right )}{d^3}+x \left (a^2+2 a b \tan (c)-b^2\right )+\frac{3 b^2 x^{2/3} \sec (c) \sin \left (d \sqrt [3]{x}\right ) \sec \left (c+d \sqrt [3]{x}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Tan[c + d*x^(1/3)])^2,x]

[Out]

(b*(((6*I)*b*d^2*x^(2/3) - (4*I)*a*d^3*x)/(1 + E^((2*I)*c)) + 6*d*(b - a*d*x^(1/3))*x^(1/3)*Log[1 + E^((-2*I)*
(c + d*x^(1/3)))] + (3*I)*(b - 2*a*d*x^(1/3))*PolyLog[2, -E^((-2*I)*(c + d*x^(1/3)))] - 3*a*PolyLog[3, -E^((-2
*I)*(c + d*x^(1/3)))]))/d^3 + (3*b^2*x^(2/3)*Sec[c]*Sec[c + d*x^(1/3)]*Sin[d*x^(1/3)])/d + x*(a^2 - b^2 + 2*a*
b*Tan[c])

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Maple [F]  time = 0.245, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b\tan \left ( c+d\sqrt [3]{x} \right ) \right ) ^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*tan(c+d*x^(1/3)))^2,x)

[Out]

int((a+b*tan(c+d*x^(1/3)))^2,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a^{2} x + \frac{6 \, b^{2} x^{\frac{2}{3}} \sin \left (2 \, d x^{\frac{1}{3}} + 2 \, c\right ) -{\left (b^{2} d \cos \left (2 \, d x^{\frac{1}{3}} + 2 \, c\right )^{2} + b^{2} d \sin \left (2 \, d x^{\frac{1}{3}} + 2 \, c\right )^{2} + 2 \, b^{2} d \cos \left (2 \, d x^{\frac{1}{3}} + 2 \, c\right ) + b^{2} d\right )} x - \frac{{\left (-8 i \,{\left (d x^{\frac{1}{3}} + c\right )}^{3} a b + 4 \,{\left (6 i \, a b c + 3 i \, b^{2}\right )}{\left (d x^{\frac{1}{3}} + c\right )}^{2} + 12 \, a b{\rm Li}_{3}(-e^{\left (2 i \, d x^{\frac{1}{3}} + 2 i \, c\right )}) + 2 \,{\left (-12 i \, a b c^{2} - 12 i \, b^{2} c\right )}{\left (d x^{\frac{1}{3}} + c\right )} +{\left (24 i \,{\left (d x^{\frac{1}{3}} + c\right )}^{2} a b + 24 i \, a b c^{2} + 24 i \, b^{2} c + 2 \,{\left (-24 i \, a b c - 12 i \, b^{2}\right )}{\left (d x^{\frac{1}{3}} + c\right )}\right )} \arctan \left (\sin \left (2 \, d x^{\frac{1}{3}} + 2 \, c\right ), \cos \left (2 \, d x^{\frac{1}{3}} + 2 \, c\right ) + 1\right ) +{\left (-24 i \,{\left (d x^{\frac{1}{3}} + c\right )} a b + 24 i \, a b c + 12 i \, b^{2}\right )}{\rm Li}_2\left (-e^{\left (2 i \, d x^{\frac{1}{3}} + 2 i \, c\right )}\right ) + 12 \,{\left ({\left (d x^{\frac{1}{3}} + c\right )}^{2} a b + a b c^{2} + b^{2} c -{\left (2 \, a b c + b^{2}\right )}{\left (d x^{\frac{1}{3}} + c\right )}\right )} \log \left (\cos \left (2 \, d x^{\frac{1}{3}} + 2 \, c\right )^{2} + \sin \left (2 \, d x^{\frac{1}{3}} + 2 \, c\right )^{2} + 2 \, \cos \left (2 \, d x^{\frac{1}{3}} + 2 \, c\right ) + 1\right )\right )}{\left (d \cos \left (2 \, d x^{\frac{1}{3}} + 2 \, c\right )^{2} + d \sin \left (2 \, d x^{\frac{1}{3}} + 2 \, c\right )^{2} + 2 \, d \cos \left (2 \, d x^{\frac{1}{3}} + 2 \, c\right ) + d\right )}}{4 \, d^{3}}}{d \cos \left (2 \, d x^{\frac{1}{3}} + 2 \, c\right )^{2} + d \sin \left (2 \, d x^{\frac{1}{3}} + 2 \, c\right )^{2} + 2 \, d \cos \left (2 \, d x^{\frac{1}{3}} + 2 \, c\right ) + d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(c+d*x^(1/3)))^2,x, algorithm="maxima")

[Out]

a^2*x + (6*b^2*x^(2/3)*sin(2*d*x^(1/3) + 2*c) - (b^2*d*cos(2*d*x^(1/3) + 2*c)^2 + b^2*d*sin(2*d*x^(1/3) + 2*c)
^2 + 2*b^2*d*cos(2*d*x^(1/3) + 2*c) + b^2*d)*x - (d*cos(2*d*x^(1/3) + 2*c)^2 + d*sin(2*d*x^(1/3) + 2*c)^2 + 2*
d*cos(2*d*x^(1/3) + 2*c) + d)*integrate(-4*(a*b*d*x*sin(2*d*x^(1/3) + 2*c) - b^2*x^(2/3)*sin(2*d*x^(1/3) + 2*c
))/((d*cos(2*d*x^(1/3) + 2*c)^2 + d*sin(2*d*x^(1/3) + 2*c)^2 + 2*d*cos(2*d*x^(1/3) + 2*c) + d)*x), x))/(d*cos(
2*d*x^(1/3) + 2*c)^2 + d*sin(2*d*x^(1/3) + 2*c)^2 + 2*d*cos(2*d*x^(1/3) + 2*c) + d)

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Fricas [C]  time = 1.54544, size = 903, normalized size = 4.38 \begin{align*} \frac{6 \, b^{2} d^{2} x^{\frac{2}{3}} \tan \left (d x^{\frac{1}{3}} + c\right ) + 2 \,{\left (a^{2} - b^{2}\right )} d^{3} x - 3 \, a b{\rm polylog}\left (3, \frac{\tan \left (d x^{\frac{1}{3}} + c\right )^{2} + 2 i \, \tan \left (d x^{\frac{1}{3}} + c\right ) - 1}{\tan \left (d x^{\frac{1}{3}} + c\right )^{2} + 1}\right ) - 3 \, a b{\rm polylog}\left (3, \frac{\tan \left (d x^{\frac{1}{3}} + c\right )^{2} - 2 i \, \tan \left (d x^{\frac{1}{3}} + c\right ) - 1}{\tan \left (d x^{\frac{1}{3}} + c\right )^{2} + 1}\right ) +{\left (-6 i \, a b d x^{\frac{1}{3}} + 3 i \, b^{2}\right )}{\rm Li}_2\left (\frac{2 \,{\left (i \, \tan \left (d x^{\frac{1}{3}} + c\right ) - 1\right )}}{\tan \left (d x^{\frac{1}{3}} + c\right )^{2} + 1} + 1\right ) +{\left (6 i \, a b d x^{\frac{1}{3}} - 3 i \, b^{2}\right )}{\rm Li}_2\left (\frac{2 \,{\left (-i \, \tan \left (d x^{\frac{1}{3}} + c\right ) - 1\right )}}{\tan \left (d x^{\frac{1}{3}} + c\right )^{2} + 1} + 1\right ) - 6 \,{\left (a b d^{2} x^{\frac{2}{3}} - b^{2} d x^{\frac{1}{3}}\right )} \log \left (-\frac{2 \,{\left (i \, \tan \left (d x^{\frac{1}{3}} + c\right ) - 1\right )}}{\tan \left (d x^{\frac{1}{3}} + c\right )^{2} + 1}\right ) - 6 \,{\left (a b d^{2} x^{\frac{2}{3}} - b^{2} d x^{\frac{1}{3}}\right )} \log \left (-\frac{2 \,{\left (-i \, \tan \left (d x^{\frac{1}{3}} + c\right ) - 1\right )}}{\tan \left (d x^{\frac{1}{3}} + c\right )^{2} + 1}\right )}{2 \, d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(c+d*x^(1/3)))^2,x, algorithm="fricas")

[Out]

1/2*(6*b^2*d^2*x^(2/3)*tan(d*x^(1/3) + c) + 2*(a^2 - b^2)*d^3*x - 3*a*b*polylog(3, (tan(d*x^(1/3) + c)^2 + 2*I
*tan(d*x^(1/3) + c) - 1)/(tan(d*x^(1/3) + c)^2 + 1)) - 3*a*b*polylog(3, (tan(d*x^(1/3) + c)^2 - 2*I*tan(d*x^(1
/3) + c) - 1)/(tan(d*x^(1/3) + c)^2 + 1)) + (-6*I*a*b*d*x^(1/3) + 3*I*b^2)*dilog(2*(I*tan(d*x^(1/3) + c) - 1)/
(tan(d*x^(1/3) + c)^2 + 1) + 1) + (6*I*a*b*d*x^(1/3) - 3*I*b^2)*dilog(2*(-I*tan(d*x^(1/3) + c) - 1)/(tan(d*x^(
1/3) + c)^2 + 1) + 1) - 6*(a*b*d^2*x^(2/3) - b^2*d*x^(1/3))*log(-2*(I*tan(d*x^(1/3) + c) - 1)/(tan(d*x^(1/3) +
 c)^2 + 1)) - 6*(a*b*d^2*x^(2/3) - b^2*d*x^(1/3))*log(-2*(-I*tan(d*x^(1/3) + c) - 1)/(tan(d*x^(1/3) + c)^2 + 1
)))/d^3

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \tan{\left (c + d \sqrt [3]{x} \right )}\right )^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(c+d*x**(1/3)))**2,x)

[Out]

Integral((a + b*tan(c + d*x**(1/3)))**2, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \tan \left (d x^{\frac{1}{3}} + c\right ) + a\right )}^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(c+d*x^(1/3)))^2,x, algorithm="giac")

[Out]

integrate((b*tan(d*x^(1/3) + c) + a)^2, x)