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

Optimal. Leaf size=352 \[ -\frac {45 i b \text {Li}_6\left (-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{4 d^6 \left (a^2+b^2\right )}-\frac {45 b \sqrt [3]{x} \text {Li}_5\left (-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{2 d^5 \left (a^2+b^2\right )}+\frac {45 i b x^{2/3} \text {Li}_4\left (-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{2 d^4 \left (a^2+b^2\right )}+\frac {15 b x \text {Li}_3\left (-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{d^3 \left (a^2+b^2\right )}-\frac {15 i b x^{4/3} \text {Li}_2\left (-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{2 d^2 \left (a^2+b^2\right )}+\frac {3 b x^{5/3} \log \left (1+\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{d \left (a^2+b^2\right )}+\frac {x^2}{2 (a+i b)} \]

[Out]

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

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Rubi [A]  time = 0.41, antiderivative size = 352, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {3747, 3732, 2190, 2531, 6609, 2282, 6589} \[ -\frac {15 i b x^{4/3} \text {Li}_2\left (-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{2 d^2 \left (a^2+b^2\right )}+\frac {45 i b x^{2/3} \text {Li}_4\left (-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{2 d^4 \left (a^2+b^2\right )}+\frac {15 b x \text {Li}_3\left (-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{d^3 \left (a^2+b^2\right )}-\frac {45 b \sqrt [3]{x} \text {Li}_5\left (-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{2 d^5 \left (a^2+b^2\right )}-\frac {45 i b \text {Li}_6\left (-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{4 d^6 \left (a^2+b^2\right )}+\frac {3 b x^{5/3} \log \left (1+\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{d \left (a^2+b^2\right )}+\frac {x^2}{2 (a+i b)} \]

Antiderivative was successfully verified.

[In]

Int[x/(a + b*Tan[c + d*x^(1/3)]),x]

[Out]

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

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 3732

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(c + d*x)^(m + 1)/(d*
(m + 1)*(a + I*b)), x] + Dist[2*I*b, Int[((c + d*x)^m*E^Simp[2*I*(e + f*x), x])/((a + I*b)^2 + (a^2 + b^2)*E^S
imp[2*I*(e + f*x), x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[a^2 + b^2, 0] && 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 \frac {x}{a+b \tan \left (c+d \sqrt [3]{x}\right )} \, dx &=3 \operatorname {Subst}\left (\int \frac {x^5}{a+b \tan (c+d x)} \, dx,x,\sqrt [3]{x}\right )\\ &=\frac {x^2}{2 (a+i b)}+(6 i b) \operatorname {Subst}\left (\int \frac {e^{2 i (c+d x)} x^5}{(a+i b)^2+\left (a^2+b^2\right ) e^{2 i (c+d x)}} \, dx,x,\sqrt [3]{x}\right )\\ &=\frac {x^2}{2 (a+i b)}+\frac {3 b x^{5/3} \log \left (1+\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{\left (a^2+b^2\right ) d}-\frac {(15 b) \operatorname {Subst}\left (\int x^4 \log \left (1+\frac {\left (a^2+b^2\right ) e^{2 i (c+d x)}}{(a+i b)^2}\right ) \, dx,x,\sqrt [3]{x}\right )}{\left (a^2+b^2\right ) d}\\ &=\frac {x^2}{2 (a+i b)}+\frac {3 b x^{5/3} \log \left (1+\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{\left (a^2+b^2\right ) d}-\frac {15 i b x^{4/3} \text {Li}_2\left (-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{2 \left (a^2+b^2\right ) d^2}+\frac {(30 i b) \operatorname {Subst}\left (\int x^3 \text {Li}_2\left (-\frac {\left (a^2+b^2\right ) e^{2 i (c+d x)}}{(a+i b)^2}\right ) \, dx,x,\sqrt [3]{x}\right )}{\left (a^2+b^2\right ) d^2}\\ &=\frac {x^2}{2 (a+i b)}+\frac {3 b x^{5/3} \log \left (1+\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{\left (a^2+b^2\right ) d}-\frac {15 i b x^{4/3} \text {Li}_2\left (-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{2 \left (a^2+b^2\right ) d^2}+\frac {15 b x \text {Li}_3\left (-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{\left (a^2+b^2\right ) d^3}-\frac {(45 b) \operatorname {Subst}\left (\int x^2 \text {Li}_3\left (-\frac {\left (a^2+b^2\right ) e^{2 i (c+d x)}}{(a+i b)^2}\right ) \, dx,x,\sqrt [3]{x}\right )}{\left (a^2+b^2\right ) d^3}\\ &=\frac {x^2}{2 (a+i b)}+\frac {3 b x^{5/3} \log \left (1+\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{\left (a^2+b^2\right ) d}-\frac {15 i b x^{4/3} \text {Li}_2\left (-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{2 \left (a^2+b^2\right ) d^2}+\frac {15 b x \text {Li}_3\left (-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{\left (a^2+b^2\right ) d^3}+\frac {45 i b x^{2/3} \text {Li}_4\left (-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{2 \left (a^2+b^2\right ) d^4}-\frac {(45 i b) \operatorname {Subst}\left (\int x \text {Li}_4\left (-\frac {\left (a^2+b^2\right ) e^{2 i (c+d x)}}{(a+i b)^2}\right ) \, dx,x,\sqrt [3]{x}\right )}{\left (a^2+b^2\right ) d^4}\\ &=\frac {x^2}{2 (a+i b)}+\frac {3 b x^{5/3} \log \left (1+\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{\left (a^2+b^2\right ) d}-\frac {15 i b x^{4/3} \text {Li}_2\left (-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{2 \left (a^2+b^2\right ) d^2}+\frac {15 b x \text {Li}_3\left (-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{\left (a^2+b^2\right ) d^3}+\frac {45 i b x^{2/3} \text {Li}_4\left (-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{2 \left (a^2+b^2\right ) d^4}-\frac {45 b \sqrt [3]{x} \text {Li}_5\left (-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{2 \left (a^2+b^2\right ) d^5}+\frac {(45 b) \operatorname {Subst}\left (\int \text {Li}_5\left (-\frac {\left (a^2+b^2\right ) e^{2 i (c+d x)}}{(a+i b)^2}\right ) \, dx,x,\sqrt [3]{x}\right )}{2 \left (a^2+b^2\right ) d^5}\\ &=\frac {x^2}{2 (a+i b)}+\frac {3 b x^{5/3} \log \left (1+\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{\left (a^2+b^2\right ) d}-\frac {15 i b x^{4/3} \text {Li}_2\left (-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{2 \left (a^2+b^2\right ) d^2}+\frac {15 b x \text {Li}_3\left (-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{\left (a^2+b^2\right ) d^3}+\frac {45 i b x^{2/3} \text {Li}_4\left (-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{2 \left (a^2+b^2\right ) d^4}-\frac {45 b \sqrt [3]{x} \text {Li}_5\left (-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{2 \left (a^2+b^2\right ) d^5}-\frac {(45 i b) \operatorname {Subst}\left (\int \frac {\text {Li}_5\left (-\frac {\left (a^2+b^2\right ) x}{(a+i b)^2}\right )}{x} \, dx,x,e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{4 \left (a^2+b^2\right ) d^6}\\ &=\frac {x^2}{2 (a+i b)}+\frac {3 b x^{5/3} \log \left (1+\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{\left (a^2+b^2\right ) d}-\frac {15 i b x^{4/3} \text {Li}_2\left (-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{2 \left (a^2+b^2\right ) d^2}+\frac {15 b x \text {Li}_3\left (-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{\left (a^2+b^2\right ) d^3}+\frac {45 i b x^{2/3} \text {Li}_4\left (-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{2 \left (a^2+b^2\right ) d^4}-\frac {45 b \sqrt [3]{x} \text {Li}_5\left (-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{2 \left (a^2+b^2\right ) d^5}-\frac {45 i b \text {Li}_6\left (-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{4 \left (a^2+b^2\right ) d^6}\\ \end {align*}

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Mathematica [A]  time = 1.38, size = 310, normalized size = 0.88 \[ \frac {12 b d^5 x^{5/3} \log \left (1+\frac {(a+i b) e^{-2 i \left (c+d \sqrt [3]{x}\right )}}{a-i b}\right )+30 i b d^4 x^{4/3} \text {Li}_2\left (\frac {(-a-i b) e^{-2 i \left (c+d \sqrt [3]{x}\right )}}{a-i b}\right )+60 b d^3 x \text {Li}_3\left (\frac {(-a-i b) e^{-2 i \left (c+d \sqrt [3]{x}\right )}}{a-i b}\right )-90 i b d^2 x^{2/3} \text {Li}_4\left (\frac {(-a-i b) e^{-2 i \left (c+d \sqrt [3]{x}\right )}}{a-i b}\right )-90 b d \sqrt [3]{x} \text {Li}_5\left (\frac {(-a-i b) e^{-2 i \left (c+d \sqrt [3]{x}\right )}}{a-i b}\right )+45 i b \text {Li}_6\left (\frac {(-a-i b) e^{-2 i \left (c+d \sqrt [3]{x}\right )}}{a-i b}\right )+2 a d^6 x^2+2 i b d^6 x^2}{4 d^6 \left (a^2+b^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*d^6*x^2 + (2*I)*b*d^6*x^2 + 12*b*d^5*x^(5/3)*Log[1 + (a + I*b)/((a - I*b)*E^((2*I)*(c + d*x^(1/3))))] + (
30*I)*b*d^4*x^(4/3)*PolyLog[2, (-a - I*b)/((a - I*b)*E^((2*I)*(c + d*x^(1/3))))] + 60*b*d^3*x*PolyLog[3, (-a -
I*b)/((a - I*b)*E^((2*I)*(c + d*x^(1/3))))] - (90*I)*b*d^2*x^(2/3)*PolyLog[4, (-a - I*b)/((a - I*b)*E^((2*I)*
(c + d*x^(1/3))))] - 90*b*d*x^(1/3)*PolyLog[5, (-a - I*b)/((a - I*b)*E^((2*I)*(c + d*x^(1/3))))] + (45*I)*b*Po
lyLog[6, (-a - I*b)/((a - I*b)*E^((2*I)*(c + d*x^(1/3))))])/(4*(a^2 + b^2)*d^6)

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fricas [F]  time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {x}{b \tan \left (d x^{\frac {1}{3}} + c\right ) + a}, x\right ) \]

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(x/(b*tan(d*x^(1/3) + c) + a), x)

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

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(x/(b*tan(d*x^(1/3) + c) + a), x)

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maple [F]  time = 1.03, size = 0, normalized size = 0.00 \[ \int \frac {x}{a +b \tan \left (c +d \,x^{\frac {1}{3}}\right )}\, dx \]

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 = 1.32, size = 810, normalized size = 2.30 \[ \text {result too large to display} \]

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*(15*(2*(d*x^(1/3) + c)*a/(a^2 + b^2) + 2*b*log(b*tan(d*x^(1/3) + c) + a)/(a^2 + b^2) - b*log(tan(d*x^(1/
3) + c)^2 + 1)/(a^2 + b^2))*c^5 - (5*(d*x^(1/3) + c)^6*(a - I*b) - 30*(d*x^(1/3) + c)^5*(a - I*b)*c + 75*(d*x^
(1/3) + c)^4*(a - I*b)*c^2 - 100*(d*x^(1/3) + c)^3*(a - I*b)*c^3 + 75*(d*x^(1/3) + c)^2*(a - I*b)*c^4 + (-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((2*a*b*cos(2*d*x^(1/3) + 2*c) - (a^2 - b^2)*sin(2*d*x^(1/3) + 2*c)
)/(a^2 + b^2), (2*a*b*sin(2*d*x^(1/3) + 2*c) + a^2 + b^2 + (a^2 - b^2)*cos(2*d*x^(1/3) + 2*c))/(a^2 + b^2)) +
(-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((I*a + b)*e^(2*I*d*x^(1/3) + 2*I*c)/(-I*a + b)) + (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)*
log(((a^2 + b^2)*cos(2*d*x^(1/3) + 2*c)^2 + 4*a*b*sin(2*d*x^(1/3) + 2*c) + (a^2 + b^2)*sin(2*d*x^(1/3) + 2*c)^
2 + a^2 + b^2 + 2*(a^2 - b^2)*cos(2*d*x^(1/3) + 2*c))/(a^2 + b^2)) - 360*I*b*polylog(6, (I*a + b)*e^(2*I*d*x^(
1/3) + 2*I*c)/(-I*a + b)) - 90*(8*(d*x^(1/3) + c)*b - 5*b*c)*polylog(5, (I*a + b)*e^(2*I*d*x^(1/3) + 2*I*c)/(-
I*a + b)) + (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, (I*a + b)*e^(2*I*
d*x^(1/3) + 2*I*c)/(-I*a + b)) + 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, (I*a + b)*e^(2*I*d*x^(1/3) + 2*I*c)/(-I*a + b)))/(a^2 + b^2))/d^6

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

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

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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