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

Optimal. Leaf size=203 \[ \frac {a x^2}{2}+\frac {45 i b \text {Li}_6\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{4 d^6}+\frac {45 b \sqrt [3]{x} \text {Li}_5\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^5}-\frac {45 i b x^{2/3} \text {Li}_4\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^4}-\frac {15 b x \text {Li}_3\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}+\frac {15 i b x^{4/3} \text {Li}_2\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^2}-\frac {3 b x^{5/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+\frac {1}{2} i b x^2 \]

[Out]

1/2*a*x^2+1/2*I*b*x^2-3*b*x^(5/3)*ln(1+exp(2*I*(c+d*x^(1/3))))/d+15/2*I*b*x^(4/3)*polylog(2,-exp(2*I*(c+d*x^(1
/3))))/d^2-15*b*x*polylog(3,-exp(2*I*(c+d*x^(1/3))))/d^3-45/2*I*b*x^(2/3)*polylog(4,-exp(2*I*(c+d*x^(1/3))))/d
^4+45/2*b*x^(1/3)*polylog(5,-exp(2*I*(c+d*x^(1/3))))/d^5+45/4*I*b*polylog(6,-exp(2*I*(c+d*x^(1/3))))/d^6

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Rubi [A]  time = 0.27, antiderivative size = 203, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {14, 3747, 3719, 2190, 2531, 6609, 2282, 6589} \[ \frac {a x^2}{2}+\frac {15 i b x^{4/3} \text {Li}_2\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^2}-\frac {45 i b x^{2/3} \text {Li}_4\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^4}-\frac {15 b x \text {Li}_3\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}+\frac {45 b \sqrt [3]{x} \text {Li}_5\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^5}+\frac {45 i b \text {Li}_6\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{4 d^6}-\frac {3 b x^{5/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+\frac {1}{2} i b x^2 \]

Antiderivative was successfully verified.

[In]

Int[x*(a + b*Tan[c + d*x^(1/3)]),x]

[Out]

(a*x^2)/2 + (I/2)*b*x^2 - (3*b*x^(5/3)*Log[1 + E^((2*I)*(c + d*x^(1/3)))])/d + (((15*I)/2)*b*x^(4/3)*PolyLog[2
, -E^((2*I)*(c + d*x^(1/3)))])/d^2 - (15*b*x*PolyLog[3, -E^((2*I)*(c + d*x^(1/3)))])/d^3 - (((45*I)/2)*b*x^(2/
3)*PolyLog[4, -E^((2*I)*(c + d*x^(1/3)))])/d^4 + (45*b*x^(1/3)*PolyLog[5, -E^((2*I)*(c + d*x^(1/3)))])/(2*d^5)
+ (((45*I)/4)*b*PolyLog[6, -E^((2*I)*(c + d*x^(1/3)))])/d^6

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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 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 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 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 3747

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

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 6609

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

Rubi steps

\begin {align*} \int x \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right ) \, dx &=\int \left (a x+b x \tan \left (c+d \sqrt [3]{x}\right )\right ) \, dx\\ &=\frac {a x^2}{2}+b \int x \tan \left (c+d \sqrt [3]{x}\right ) \, dx\\ &=\frac {a x^2}{2}+(3 b) \operatorname {Subst}\left (\int x^5 \tan (c+d x) \, dx,x,\sqrt [3]{x}\right )\\ &=\frac {a x^2}{2}+\frac {1}{2} i b x^2-(6 i b) \operatorname {Subst}\left (\int \frac {e^{2 i (c+d x)} x^5}{1+e^{2 i (c+d x)}} \, dx,x,\sqrt [3]{x}\right )\\ &=\frac {a x^2}{2}+\frac {1}{2} i b x^2-\frac {3 b x^{5/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+\frac {(15 b) \operatorname {Subst}\left (\int x^4 \log \left (1+e^{2 i (c+d x)}\right ) \, dx,x,\sqrt [3]{x}\right )}{d}\\ &=\frac {a x^2}{2}+\frac {1}{2} i b x^2-\frac {3 b x^{5/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+\frac {15 i b x^{4/3} \text {Li}_2\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^2}-\frac {(30 i b) \operatorname {Subst}\left (\int x^3 \text {Li}_2\left (-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt [3]{x}\right )}{d^2}\\ &=\frac {a x^2}{2}+\frac {1}{2} i b x^2-\frac {3 b x^{5/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+\frac {15 i b x^{4/3} \text {Li}_2\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^2}-\frac {15 b x \text {Li}_3\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}+\frac {(45 b) \operatorname {Subst}\left (\int x^2 \text {Li}_3\left (-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt [3]{x}\right )}{d^3}\\ &=\frac {a x^2}{2}+\frac {1}{2} i b x^2-\frac {3 b x^{5/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+\frac {15 i b x^{4/3} \text {Li}_2\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^2}-\frac {15 b x \text {Li}_3\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}-\frac {45 i b x^{2/3} \text {Li}_4\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^4}+\frac {(45 i b) \operatorname {Subst}\left (\int x \text {Li}_4\left (-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt [3]{x}\right )}{d^4}\\ &=\frac {a x^2}{2}+\frac {1}{2} i b x^2-\frac {3 b x^{5/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+\frac {15 i b x^{4/3} \text {Li}_2\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^2}-\frac {15 b x \text {Li}_3\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}-\frac {45 i b x^{2/3} \text {Li}_4\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^4}+\frac {45 b \sqrt [3]{x} \text {Li}_5\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^5}-\frac {(45 b) \operatorname {Subst}\left (\int \text {Li}_5\left (-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt [3]{x}\right )}{2 d^5}\\ &=\frac {a x^2}{2}+\frac {1}{2} i b x^2-\frac {3 b x^{5/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+\frac {15 i b x^{4/3} \text {Li}_2\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^2}-\frac {15 b x \text {Li}_3\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}-\frac {45 i b x^{2/3} \text {Li}_4\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^4}+\frac {45 b \sqrt [3]{x} \text {Li}_5\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^5}+\frac {(45 i b) \operatorname {Subst}\left (\int \frac {\text {Li}_5(-x)}{x} \, dx,x,e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{4 d^6}\\ &=\frac {a x^2}{2}+\frac {1}{2} i b x^2-\frac {3 b x^{5/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+\frac {15 i b x^{4/3} \text {Li}_2\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^2}-\frac {15 b x \text {Li}_3\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}-\frac {45 i b x^{2/3} \text {Li}_4\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^4}+\frac {45 b \sqrt [3]{x} \text {Li}_5\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^5}+\frac {45 i b \text {Li}_6\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{4 d^6}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 203, normalized size = 1.00 \[ \frac {a x^2}{2}+\frac {45 i b \text {Li}_6\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{4 d^6}+\frac {45 b \sqrt [3]{x} \text {Li}_5\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^5}-\frac {45 i b x^{2/3} \text {Li}_4\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^4}-\frac {15 b x \text {Li}_3\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}+\frac {15 i b x^{4/3} \text {Li}_2\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^2}-\frac {3 b x^{5/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+\frac {1}{2} i b x^2 \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*x^2)/2 + (I/2)*b*x^2 - (3*b*x^(5/3)*Log[1 + E^((2*I)*(c + d*x^(1/3)))])/d + (((15*I)/2)*b*x^(4/3)*PolyLog[2
, -E^((2*I)*(c + d*x^(1/3)))])/d^2 - (15*b*x*PolyLog[3, -E^((2*I)*(c + d*x^(1/3)))])/d^3 - (((45*I)/2)*b*x^(2/
3)*PolyLog[4, -E^((2*I)*(c + d*x^(1/3)))])/d^4 + (45*b*x^(1/3)*PolyLog[5, -E^((2*I)*(c + d*x^(1/3)))])/(2*d^5)
+ (((45*I)/4)*b*PolyLog[6, -E^((2*I)*(c + d*x^(1/3)))])/d^6

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fricas [F]  time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left (b x \tan \left (d x^{\frac {1}{3}} + c\right ) + a x, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.84, size = 0, normalized size = 0.00 \[ \int x \left (a +b \tan \left (c +d \,x^{\frac {1}{3}}\right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 1.23, size = 618, normalized size = 3.04 \[ \frac {5 \, {\left (d x^{\frac {1}{3}} + c\right )}^{6} a + 5 i \, {\left (d x^{\frac {1}{3}} + c\right )}^{6} b - 30 \, {\left (d x^{\frac {1}{3}} + c\right )}^{5} a c - 30 i \, {\left (d x^{\frac {1}{3}} + c\right )}^{5} b c + 75 \, {\left (d x^{\frac {1}{3}} + c\right )}^{4} a c^{2} + 75 i \, {\left (d x^{\frac {1}{3}} + c\right )}^{4} b c^{2} - 100 \, {\left (d x^{\frac {1}{3}} + c\right )}^{3} a c^{3} - 100 i \, {\left (d x^{\frac {1}{3}} + c\right )}^{3} b c^{3} + 75 \, {\left (d x^{\frac {1}{3}} + c\right )}^{2} a c^{4} + 75 i \, {\left (d x^{\frac {1}{3}} + c\right )}^{2} b c^{4} - 30 \, {\left (d x^{\frac {1}{3}} + c\right )} a c^{5} - 30 \, b c^{5} \log \left (\sec \left (d x^{\frac {1}{3}} + c\right )\right ) - {\left (96 i \, {\left (d x^{\frac {1}{3}} + c\right )}^{5} b - 300 i \, {\left (d x^{\frac {1}{3}} + c\right )}^{4} b c + 400 i \, {\left (d x^{\frac {1}{3}} + c\right )}^{3} b c^{2} - 300 i \, {\left (d x^{\frac {1}{3}} + c\right )}^{2} b c^{3} + 150 i \, {\left (d x^{\frac {1}{3}} + c\right )} b c^{4}\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 (-240 i \, {\left (d x^{\frac {1}{3}} + c\right )}^{4} b + 600 i \, {\left (d x^{\frac {1}{3}} + c\right )}^{3} b c - 600 i \, {\left (d x^{\frac {1}{3}} + c\right )}^{2} b c^{2} + 300 i \, {\left (d x^{\frac {1}{3}} + c\right )} b c^{3} - 75 i \, b c^{4}\right )} {\rm Li}_2\left (-e^{\left (2 i \, d x^{\frac {1}{3}} + 2 i \, c\right )}\right ) - {\left (48 \, {\left (d x^{\frac {1}{3}} + c\right )}^{5} b - 150 \, {\left (d x^{\frac {1}{3}} + c\right )}^{4} b c + 200 \, {\left (d x^{\frac {1}{3}} + c\right )}^{3} b c^{2} - 150 \, {\left (d x^{\frac {1}{3}} + c\right )}^{2} b c^{3} + 75 \, {\left (d x^{\frac {1}{3}} + c\right )} b c^{4}\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 ) + 360 i \, b {\rm Li}_{6}(-e^{\left (2 i \, d x^{\frac {1}{3}} + 2 i \, c\right )}) + 90 \, {\left (8 \, {\left (d x^{\frac {1}{3}} + c\right )} b - 5 \, b c\right )} {\rm Li}_{5}(-e^{\left (2 i \, d x^{\frac {1}{3}} + 2 i \, c\right )}) - {\left (720 i \, {\left (d x^{\frac {1}{3}} + c\right )}^{2} b - 900 i \, {\left (d x^{\frac {1}{3}} + c\right )} b c + 300 i \, b c^{2}\right )} {\rm Li}_{4}(-e^{\left (2 i \, d x^{\frac {1}{3}} + 2 i \, c\right )}) - 30 \, {\left (16 \, {\left (d x^{\frac {1}{3}} + c\right )}^{3} b - 30 \, {\left (d x^{\frac {1}{3}} + c\right )}^{2} b c + 20 \, {\left (d x^{\frac {1}{3}} + c\right )} b c^{2} - 5 \, b c^{3}\right )} {\rm Li}_{3}(-e^{\left (2 i \, d x^{\frac {1}{3}} + 2 i \, c\right )})}{10 \, d^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/10*(5*(d*x^(1/3) + c)^6*a + 5*I*(d*x^(1/3) + c)^6*b - 30*(d*x^(1/3) + c)^5*a*c - 30*I*(d*x^(1/3) + c)^5*b*c
+ 75*(d*x^(1/3) + c)^4*a*c^2 + 75*I*(d*x^(1/3) + c)^4*b*c^2 - 100*(d*x^(1/3) + c)^3*a*c^3 - 100*I*(d*x^(1/3) +
c)^3*b*c^3 + 75*(d*x^(1/3) + c)^2*a*c^4 + 75*I*(d*x^(1/3) + c)^2*b*c^4 - 30*(d*x^(1/3) + c)*a*c^5 - 30*b*c^5*
log(sec(d*x^(1/3) + c)) - (96*I*(d*x^(1/3) + c)^5*b - 300*I*(d*x^(1/3) + c)^4*b*c + 400*I*(d*x^(1/3) + c)^3*b*
c^2 - 300*I*(d*x^(1/3) + c)^2*b*c^3 + 150*I*(d*x^(1/3) + c)*b*c^4)*arctan2(sin(2*d*x^(1/3) + 2*c), cos(2*d*x^(
1/3) + 2*c) + 1) - (-240*I*(d*x^(1/3) + c)^4*b + 600*I*(d*x^(1/3) + c)^3*b*c - 600*I*(d*x^(1/3) + c)^2*b*c^2 +
300*I*(d*x^(1/3) + c)*b*c^3 - 75*I*b*c^4)*dilog(-e^(2*I*d*x^(1/3) + 2*I*c)) - (48*(d*x^(1/3) + c)^5*b - 150*(
d*x^(1/3) + c)^4*b*c + 200*(d*x^(1/3) + c)^3*b*c^2 - 150*(d*x^(1/3) + c)^2*b*c^3 + 75*(d*x^(1/3) + c)*b*c^4)*l
og(cos(2*d*x^(1/3) + 2*c)^2 + sin(2*d*x^(1/3) + 2*c)^2 + 2*cos(2*d*x^(1/3) + 2*c) + 1) + 360*I*b*polylog(6, -e
^(2*I*d*x^(1/3) + 2*I*c)) + 90*(8*(d*x^(1/3) + c)*b - 5*b*c)*polylog(5, -e^(2*I*d*x^(1/3) + 2*I*c)) - (720*I*(
d*x^(1/3) + c)^2*b - 900*I*(d*x^(1/3) + c)*b*c + 300*I*b*c^2)*polylog(4, -e^(2*I*d*x^(1/3) + 2*I*c)) - 30*(16*
(d*x^(1/3) + c)^3*b - 30*(d*x^(1/3) + c)^2*b*c + 20*(d*x^(1/3) + c)*b*c^2 - 5*b*c^3)*polylog(3, -e^(2*I*d*x^(1
/3) + 2*I*c)))/d^6

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int x\,\left (a+b\,\mathrm {tan}\left (c+d\,x^{1/3}\right )\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a + b*tan(c + d*x^(1/3))),x)

[Out]

int(x*(a + b*tan(c + d*x^(1/3))), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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