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

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

[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)

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Rubi [A]  time = 0.276943, antiderivative size = 176, normalized size of antiderivative = 1., 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 = {3739, 3732, 2190, 2531, 2282, 6589} \[ -\frac{3 i b \sqrt [3]{x} \text{Li}_2\left (-\frac{\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{d^2 \left (a^2+b^2\right )}+\frac{3 b \text{Li}_3\left (-\frac{\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{2 d^3 \left (a^2+b^2\right )}+\frac{3 b x^{2/3} \log \left (1+\frac{\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{d \left (a^2+b^2\right )}+\frac{x}{a+i b} \]

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 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 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 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]

Rubi steps

\begin{align*} \int \frac{1}{a+b \tan \left (c+d \sqrt [3]{x}\right )} \, dx &=3 \operatorname{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) \operatorname{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) \operatorname{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) \operatorname{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) \operatorname{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*}

Mathematica [A]  time = 0.754657, size = 165, normalized size = 0.94 \[ \frac{6 i b d \sqrt [3]{x} \text{PolyLog}\left (2,\frac{(-a-i b) e^{-2 i \left (c+d \sqrt [3]{x}\right )}}{a-i b}\right )+3 b \text{PolyLog}\left (3,\frac{(-a-i b) e^{-2 i \left (c+d \sqrt [3]{x}\right )}}{a-i b}\right )+6 b d^2 x^{2/3} \log \left (1+\frac{(a+i b) e^{-2 i \left (c+d \sqrt [3]{x}\right )}}{a-i b}\right )+2 a d^3 x+2 i b d^3 x}{2 d^3 \left (a^2+b^2\right )} \]

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

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Maple [F]  time = 0.206, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b\tan \left ( c+d\sqrt [3]{x} \right ) \right ) ^{-1}\, dx \end{align*}

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]  time = 2.50773, size = 599, normalized size = 3.4 \begin{align*} \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 +{\left (-6 i \,{\left (d x^{\frac{1}{3}} + c\right )}^{2} b + 12 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 ) +{\left (-6 i \,{\left (d x^{\frac{1}{3}} + c\right )} b + 6 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{align*}

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 + 12*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 + 6*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*po
lylog(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 [C]  time = 1.80085, size = 1901, normalized size = 10.8 \begin{align*} \text{result too large to display} \end{align*}

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 + 2*b^2)*tan(d*x^(1/
3) + c)^2 + 2*a^2 - 2*I*a*b + (2*I*a^2 + 4*a*b - 2*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 + 2*b^2)*tan(d*x^(1/3) + c)^2 + 2*a^2 + 2*I*a*b + (-2*
I*a^2 + 4*a*b + 2*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 + 2*b^2)*tan(d*x^(1/3) + c)^2 + 2*a^2 - 2*I*a*b + (2*I*a^2 + 4*a*b - 2*I*b^2)*tan
(d*x^(1/3) + c))/((a^2 + b^2)*tan(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 +
2*b^2)*tan(d*x^(1/3) + c)^2 + 2*a^2 + 2*I*a*b + (-2*I*a^2 + 4*a*b + 2*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 - 4*a*b - 2*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 - 4*a*b + 2*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., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{a + b \tan{\left (c + d \sqrt [3]{x} \right )}}\, dx \end{align*}

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

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)