3.1.59 \(\int \frac {1}{a+b \tan (c+d \sqrt [3]{x})} \, dx\) [59]

Optimal. Leaf size=176 \[ \frac {x}{a+i b}+\frac {3 b x^{2/3} \log \left (1+\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{\left (a^2+b^2\right ) d}-\frac {3 i b \sqrt [3]{x} \text {PolyLog}\left (2,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{\left (a^2+b^2\right ) d^2}+\frac {3 b \text {PolyLog}\left (3,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{2 \left (a^2+b^2\right ) d^3} \]

[Out]

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

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Rubi [A]
time = 0.20, antiderivative size = 176, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {3824, 3813, 2221, 2611, 2320, 6724} \begin {gather*} \frac {3 b \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 )}{2 d^3 \left (a^2+b^2\right )}-\frac {3 i b \sqrt [3]{x} \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 )}{d^2 \left (a^2+b^2\right )}+\frac {3 b x^{2/3} \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}{a+i b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2221

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/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], 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 2320

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 2611

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 3813

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^Si
mp[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 3824

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 6724

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]

Rubi steps

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

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Mathematica [A]
time = 1.53, size = 163, normalized size = 0.93 \begin {gather*} \frac {2 a d^3 x-2 i b d^3 x+6 b d^2 x^{2/3} \log \left (1+\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )-6 i b d \sqrt [3]{x} \text {PolyLog}\left (2,-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )+3 b \text {PolyLog}\left (3,-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{2 \left (a^2+b^2\right ) d^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.80, size = 0, normalized size = 0.00 \[\int \frac {1}{a +b \tan \left (c +d \,x^{\frac {1}{3}}\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 446 vs. \(2 (147) = 294\).
time = 0.61, size = 446, normalized size = 2.53 \begin {gather*} \frac {3 \, {\left (\frac {2 \, {\left (d x^{\frac {1}{3}} + c\right )} a}{a^{2} + b^{2}} + \frac {2 \, b \log \left (b \tan \left (d x^{\frac {1}{3}} + c\right ) + a\right )}{a^{2} + b^{2}} - \frac {b \log \left (\tan \left (d x^{\frac {1}{3}} + c\right )^{2} + 1\right )}{a^{2} + b^{2}}\right )} c^{2} + \frac {2 \, {\left (d x^{\frac {1}{3}} + c\right )}^{3} {\left (a - i \, b\right )} - 6 \, {\left (d x^{\frac {1}{3}} + c\right )}^{2} {\left (a - i \, b\right )} c - 6 \, {\left (i \, {\left (d x^{\frac {1}{3}} + c\right )}^{2} b - 2 i \, {\left (d x^{\frac {1}{3}} + c\right )} b c\right )} \arctan \left (\frac {2 \, a b \cos \left (2 \, d x^{\frac {1}{3}} + 2 \, c\right ) - {\left (a^{2} - b^{2}\right )} \sin \left (2 \, d x^{\frac {1}{3}} + 2 \, c\right )}{a^{2} + b^{2}}, \frac {2 \, a b \sin \left (2 \, d x^{\frac {1}{3}} + 2 \, c\right ) + a^{2} + b^{2} + {\left (a^{2} - b^{2}\right )} \cos \left (2 \, d x^{\frac {1}{3}} + 2 \, c\right )}{a^{2} + b^{2}}\right ) - 6 \, {\left (i \, {\left (d x^{\frac {1}{3}} + c\right )} b - i \, b c\right )} {\rm Li}_2\left (\frac {{\left (i \, a + b\right )} e^{\left (2 i \, d x^{\frac {1}{3}} + 2 i \, c\right )}}{-i \, a + b}\right ) + 3 \, {\left ({\left (d x^{\frac {1}{3}} + c\right )}^{2} b - 2 \, {\left (d x^{\frac {1}{3}} + c\right )} b c\right )} \log \left (\frac {{\left (a^{2} + b^{2}\right )} \cos \left (2 \, d x^{\frac {1}{3}} + 2 \, c\right )^{2} + 4 \, a b \sin \left (2 \, d x^{\frac {1}{3}} + 2 \, c\right ) + {\left (a^{2} + b^{2}\right )} \sin \left (2 \, d x^{\frac {1}{3}} + 2 \, c\right )^{2} + a^{2} + b^{2} + 2 \, {\left (a^{2} - b^{2}\right )} \cos \left (2 \, d x^{\frac {1}{3}} + 2 \, c\right )}{a^{2} + b^{2}}\right ) + 3 \, b {\rm Li}_{3}(\frac {{\left (i \, a + b\right )} e^{\left (2 i \, d x^{\frac {1}{3}} + 2 i \, c\right )}}{-i \, a + b})}{a^{2} + b^{2}}}{2 \, d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 746 vs. \(2 (147) = 294\).
time = 0.45, size = 746, normalized size = 4.24 \begin {gather*} \frac {4 \, a d^{3} x + 6 \, b c^{2} \log \left (\frac {{\left (i \, a b + b^{2}\right )} \tan \left (d x^{\frac {1}{3}} + c\right )^{2} - a^{2} + i \, a b + {\left (i \, a^{2} + i \, b^{2}\right )} \tan \left (d x^{\frac {1}{3}} + c\right )}{\tan \left (d x^{\frac {1}{3}} + c\right )^{2} + 1}\right ) + 6 \, b c^{2} \log \left (\frac {{\left (i \, a b - b^{2}\right )} \tan \left (d x^{\frac {1}{3}} + c\right )^{2} + a^{2} + i \, a b + {\left (i \, a^{2} + i \, b^{2}\right )} \tan \left (d x^{\frac {1}{3}} + c\right )}{\tan \left (d x^{\frac {1}{3}} + c\right )^{2} + 1}\right ) + 6 i \, b d x^{\frac {1}{3}} {\rm Li}_2\left (\frac {2 \, {\left ({\left (i \, a b - b^{2}\right )} \tan \left (d x^{\frac {1}{3}} + c\right )^{2} - a^{2} - i \, a b + {\left (i \, a^{2} - 2 \, a b - i \, b^{2}\right )} \tan \left (d x^{\frac {1}{3}} + c\right )\right )}}{{\left (a^{2} + b^{2}\right )} \tan \left (d x^{\frac {1}{3}} + c\right )^{2} + a^{2} + b^{2}} + 1\right ) - 6 i \, b d x^{\frac {1}{3}} {\rm Li}_2\left (\frac {2 \, {\left ({\left (-i \, a b - b^{2}\right )} \tan \left (d x^{\frac {1}{3}} + c\right )^{2} - a^{2} + i \, a b + {\left (-i \, a^{2} - 2 \, a b + i \, b^{2}\right )} \tan \left (d x^{\frac {1}{3}} + c\right )\right )}}{{\left (a^{2} + b^{2}\right )} \tan \left (d x^{\frac {1}{3}} + c\right )^{2} + a^{2} + b^{2}} + 1\right ) + 6 \, {\left (b d^{2} x^{\frac {2}{3}} - b c^{2}\right )} \log \left (-\frac {2 \, {\left ({\left (i \, a b - b^{2}\right )} \tan \left (d x^{\frac {1}{3}} + c\right )^{2} - a^{2} - i \, a b + {\left (i \, a^{2} - 2 \, a b - i \, b^{2}\right )} \tan \left (d x^{\frac {1}{3}} + c\right )\right )}}{{\left (a^{2} + b^{2}\right )} \tan \left (d x^{\frac {1}{3}} + c\right )^{2} + a^{2} + b^{2}}\right ) + 6 \, {\left (b d^{2} x^{\frac {2}{3}} - b c^{2}\right )} \log \left (-\frac {2 \, {\left ({\left (-i \, a b - b^{2}\right )} \tan \left (d x^{\frac {1}{3}} + c\right )^{2} - a^{2} + i \, a b + {\left (-i \, a^{2} - 2 \, a b + i \, b^{2}\right )} \tan \left (d x^{\frac {1}{3}} + c\right )\right )}}{{\left (a^{2} + b^{2}\right )} \tan \left (d x^{\frac {1}{3}} + c\right )^{2} + a^{2} + b^{2}}\right ) + 3 \, b {\rm polylog}\left (3, \frac {{\left (a^{2} + 2 i \, a b - b^{2}\right )} \tan \left (d x^{\frac {1}{3}} + c\right )^{2} - a^{2} - 2 i \, a b + b^{2} - 2 \, {\left (-i \, a^{2} + 2 \, a b + i \, b^{2}\right )} \tan \left (d x^{\frac {1}{3}} + c\right )}{{\left (a^{2} + b^{2}\right )} \tan \left (d x^{\frac {1}{3}} + c\right )^{2} + a^{2} + b^{2}}\right ) + 3 \, b {\rm polylog}\left (3, \frac {{\left (a^{2} - 2 i \, a b - b^{2}\right )} \tan \left (d x^{\frac {1}{3}} + c\right )^{2} - a^{2} + 2 i \, a b + b^{2} - 2 \, {\left (i \, a^{2} + 2 \, a b - i \, b^{2}\right )} \tan \left (d x^{\frac {1}{3}} + c\right )}{{\left (a^{2} + b^{2}\right )} \tan \left (d x^{\frac {1}{3}} + c\right )^{2} + a^{2} + b^{2}}\right )}{4 \, {\left (a^{2} + b^{2}\right )} d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(4*a*d^3*x + 6*b*c^2*log(((I*a*b + b^2)*tan(d*x^(1/3) + c)^2 - a^2 + I*a*b + (I*a^2 + I*b^2)*tan(d*x^(1/3)
+ c))/(tan(d*x^(1/3) + c)^2 + 1)) + 6*b*c^2*log(((I*a*b - b^2)*tan(d*x^(1/3) + c)^2 + a^2 + I*a*b + (I*a^2 +
I*b^2)*tan(d*x^(1/3) + c))/(tan(d*x^(1/3) + c)^2 + 1)) + 6*I*b*d*x^(1/3)*dilog(2*((I*a*b - b^2)*tan(d*x^(1/3)
+ c)^2 - a^2 - I*a*b + (I*a^2 - 2*a*b - I*b^2)*tan(d*x^(1/3) + c))/((a^2 + b^2)*tan(d*x^(1/3) + c)^2 + a^2 + b
^2) + 1) - 6*I*b*d*x^(1/3)*dilog(2*((-I*a*b - b^2)*tan(d*x^(1/3) + c)^2 - a^2 + I*a*b + (-I*a^2 - 2*a*b + I*b^
2)*tan(d*x^(1/3) + c))/((a^2 + b^2)*tan(d*x^(1/3) + c)^2 + a^2 + b^2) + 1) + 6*(b*d^2*x^(2/3) - b*c^2)*log(-2*
((I*a*b - b^2)*tan(d*x^(1/3) + c)^2 - a^2 - I*a*b + (I*a^2 - 2*a*b - I*b^2)*tan(d*x^(1/3) + c))/((a^2 + b^2)*t
an(d*x^(1/3) + c)^2 + a^2 + b^2)) + 6*(b*d^2*x^(2/3) - b*c^2)*log(-2*((-I*a*b - b^2)*tan(d*x^(1/3) + c)^2 - a^
2 + I*a*b + (-I*a^2 - 2*a*b + I*b^2)*tan(d*x^(1/3) + c))/((a^2 + b^2)*tan(d*x^(1/3) + c)^2 + a^2 + b^2)) + 3*b
*polylog(3, ((a^2 + 2*I*a*b - b^2)*tan(d*x^(1/3) + c)^2 - a^2 - 2*I*a*b + b^2 - 2*(-I*a^2 + 2*a*b + I*b^2)*tan
(d*x^(1/3) + c))/((a^2 + b^2)*tan(d*x^(1/3) + c)^2 + a^2 + b^2)) + 3*b*polylog(3, ((a^2 - 2*I*a*b - b^2)*tan(d
*x^(1/3) + c)^2 - a^2 + 2*I*a*b + b^2 - 2*(I*a^2 + 2*a*b - I*b^2)*tan(d*x^(1/3) + c))/((a^2 + b^2)*tan(d*x^(1/
3) + c)^2 + a^2 + b^2)))/((a^2 + b^2)*d^3)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{a + b \tan {\left (c + d \sqrt [3]{x} \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{a+b\,\mathrm {tan}\left (c+d\,x^{1/3}\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*tan(c + d*x^(1/3))),x)

[Out]

int(1/(a + b*tan(c + d*x^(1/3))), x)

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