3.64 \(\int \frac{1}{(a+b \tan (c+d \sqrt [3]{x}))^2} \, dx\)

Optimal. Leaf size=610 \[ -\frac{6 b^2 \sqrt [3]{x} \text{PolyLog}\left (2,-\frac{(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{d^2 \left (a^2+b^2\right )^2}-\frac{3 i b^2 \text{PolyLog}\left (2,-\frac{(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{d^3 \left (a^2+b^2\right )^2}-\frac{3 i b^2 \text{PolyLog}\left (3,-\frac{(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{d^3 \left (a^2+b^2\right )^2}+\frac{6 b \sqrt [3]{x} \text{PolyLog}\left (2,-\frac{(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{d^2 (-b+i a) (a-i b)^2}+\frac{3 b \text{PolyLog}\left (3,-\frac{(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{d^3 (a-i b)^2 (a+i b)}+\frac{6 b^2 \sqrt [3]{x} \log \left (1+\frac{(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{d^2 \left (a^2+b^2\right )^2}-\frac{6 i b^2 x^{2/3} \log \left (1+\frac{(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{d \left (a^2+b^2\right )^2}-\frac{6 i b^2 x^{2/3}}{d \left (a^2+b^2\right )^2}-\frac{4 b^2 x}{\left (a^2+b^2\right )^2}+\frac{6 b^2 x^{2/3}}{d (a+i b) (b+i a)^2 \left ((b+i a) e^{2 i \left (c+d \sqrt [3]{x}\right )}+i a-b\right )}+\frac{6 b x^{2/3} \log \left (1+\frac{(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{d (a-i b)^2 (a+i b)}+\frac{4 b x}{(-b+i a) (a-i b)^2}+\frac{x}{(a-i b)^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 1.40305, antiderivative size = 610, normalized size of antiderivative = 1., number of steps used = 19, number of rules used = 11, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.688, Rules used = {3739, 3734, 2185, 2184, 2190, 2531, 2282, 6589, 2191, 2279, 2391} \[ -\frac{6 b^2 \sqrt [3]{x} \text{Li}_2\left (-\frac{(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{d^2 \left (a^2+b^2\right )^2}-\frac{3 i b^2 \text{Li}_2\left (-\frac{(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{d^3 \left (a^2+b^2\right )^2}-\frac{3 i b^2 \text{Li}_3\left (-\frac{(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{d^3 \left (a^2+b^2\right )^2}+\frac{6 b^2 \sqrt [3]{x} \log \left (1+\frac{(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{d^2 \left (a^2+b^2\right )^2}-\frac{6 i b^2 x^{2/3} \log \left (1+\frac{(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{d \left (a^2+b^2\right )^2}-\frac{6 i b^2 x^{2/3}}{d \left (a^2+b^2\right )^2}-\frac{4 b^2 x}{\left (a^2+b^2\right )^2}+\frac{6 b^2 x^{2/3}}{d (a+i b) (b+i a)^2 \left ((b+i a) e^{2 i \left (c+d \sqrt [3]{x}\right )}+i a-b\right )}+\frac{6 b \sqrt [3]{x} \text{Li}_2\left (-\frac{(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{d^2 (-b+i a) (a-i b)^2}+\frac{3 b \text{Li}_3\left (-\frac{(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{d^3 (a-i b)^2 (a+i b)}+\frac{6 b x^{2/3} \log \left (1+\frac{(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{d (a-i b)^2 (a+i b)}+\frac{4 b x}{(-b+i a) (a-i b)^2}+\frac{x}{(a-i b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-6*I)*b^2*x^(2/3))/((a^2 + b^2)^2*d) + (6*b^2*x^(2/3))/((a + I*b)*(I*a + b)^2*d*(I*a - b + (I*a + b)*E^((2*I
)*(c + d*x^(1/3))))) + x/(a - I*b)^2 + (4*b*x)/((I*a - b)*(a - I*b)^2) - (4*b^2*x)/(a^2 + b^2)^2 + (6*b^2*x^(1
/3)*Log[1 + ((a - I*b)*E^((2*I)*(c + d*x^(1/3))))/(a + I*b)])/((a^2 + b^2)^2*d^2) + (6*b*x^(2/3)*Log[1 + ((a -
I*b)*E^((2*I)*(c + d*x^(1/3))))/(a + I*b)])/((a - I*b)^2*(a + I*b)*d) - ((6*I)*b^2*x^(2/3)*Log[1 + ((a - I*b)
*E^((2*I)*(c + d*x^(1/3))))/(a + I*b)])/((a^2 + b^2)^2*d) - ((3*I)*b^2*PolyLog[2, -(((a - I*b)*E^((2*I)*(c + d
*x^(1/3))))/(a + I*b))])/((a^2 + b^2)^2*d^3) + (6*b*x^(1/3)*PolyLog[2, -(((a - I*b)*E^((2*I)*(c + d*x^(1/3))))
/(a + I*b))])/((I*a - b)*(a - I*b)^2*d^2) - (6*b^2*x^(1/3)*PolyLog[2, -(((a - I*b)*E^((2*I)*(c + d*x^(1/3))))/
(a + I*b))])/((a^2 + b^2)^2*d^2) + (3*b*PolyLog[3, -(((a - I*b)*E^((2*I)*(c + d*x^(1/3))))/(a + I*b))])/((a -
I*b)^2*(a + I*b)*d^3) - ((3*I)*b^2*PolyLog[3, -(((a - I*b)*E^((2*I)*(c + d*x^(1/3))))/(a + I*b))])/((a^2 + b^2
)^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 3734

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Int[ExpandIntegrand[(
c + d*x)^m, (1/(a - I*b) - (2*I*b)/(a^2 + b^2 + (a - I*b)^2*E^(2*I*(e + f*x))))^(-n), x], x] /; FreeQ[{a, b, c
, d, e, f}, x] && NeQ[a^2 + b^2, 0] && ILtQ[n, 0] && IGtQ[m, 0]

Rule 2185

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

Rule 2184

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[(c
+ d*x)^(m + 1)/(a*d*(m + 1)), x] - Dist[b/a, Int[((c + d*x)^m*(F^(g*(e + f*x)))^n)/(a + b*(F^(g*(e + f*x)))^n)
, x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, 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 2191

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

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]

Rubi steps

\begin{align*} \int \frac{1}{\left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right )^2} \, dx &=3 \operatorname{Subst}\left (\int \frac{x^2}{(a+b \tan (c+d x))^2} \, dx,x,\sqrt [3]{x}\right )\\ &=3 \operatorname{Subst}\left (\int \left (\frac{x^2}{(a-i b)^2}-\frac{4 b^2 x^2}{(i a+b)^2 \left (i a \left (1+\frac{i b}{a}\right )+i a \left (1-\frac{i b}{a}\right ) e^{2 i c+2 i d x}\right )^2}+\frac{4 b x^2}{(a-i b)^2 \left (i a \left (1+\frac{i b}{a}\right )+i a \left (1-\frac{i b}{a}\right ) e^{2 i c+2 i d x}\right )}\right ) \, dx,x,\sqrt [3]{x}\right )\\ &=\frac{x}{(a-i b)^2}+\frac{(12 b) \operatorname{Subst}\left (\int \frac{x^2}{i a \left (1+\frac{i b}{a}\right )+i a \left (1-\frac{i b}{a}\right ) e^{2 i c+2 i d x}} \, dx,x,\sqrt [3]{x}\right )}{(a-i b)^2}-\frac{\left (12 b^2\right ) \operatorname{Subst}\left (\int \frac{x^2}{\left (i a \left (1+\frac{i b}{a}\right )+i a \left (1-\frac{i b}{a}\right ) e^{2 i c+2 i d x}\right )^2} \, dx,x,\sqrt [3]{x}\right )}{(i a+b)^2}\\ &=\frac{x}{(a-i b)^2}+\frac{4 b x}{(i a-b) (a-i b)^2}+\frac{\left (12 b^2\right ) \operatorname{Subst}\left (\int \frac{x^2}{i a \left (1+\frac{i b}{a}\right )+i a \left (1-\frac{i b}{a}\right ) e^{2 i c+2 i d x}} \, dx,x,\sqrt [3]{x}\right )}{(i a-b) (a-i b)^2}-\frac{(12 b) \operatorname{Subst}\left (\int \frac{e^{2 i c+2 i d x} x^2}{i a \left (1+\frac{i b}{a}\right )+i a \left (1-\frac{i b}{a}\right ) e^{2 i c+2 i d x}} \, dx,x,\sqrt [3]{x}\right )}{a^2+b^2}-\frac{\left (12 b^2\right ) \operatorname{Subst}\left (\int \frac{e^{2 i c+2 i d x} x^2}{\left (i a \left (1+\frac{i b}{a}\right )+i a \left (1-\frac{i b}{a}\right ) e^{2 i c+2 i d x}\right )^2} \, dx,x,\sqrt [3]{x}\right )}{a^2+b^2}\\ &=-\frac{6 b^2 x^{2/3}}{(a-i b)^2 (a+i b) d \left (i a-b+(i a+b) e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}+\frac{x}{(a-i b)^2}+\frac{4 b x}{(i a-b) (a-i b)^2}-\frac{4 b^2 x}{\left (a^2+b^2\right )^2}+\frac{6 b x^{2/3} \log \left (1+\frac{(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{(a-i b)^2 (a+i b) d}-\frac{\left (12 b^2\right ) \operatorname{Subst}\left (\int \frac{e^{2 i c+2 i d x} x^2}{i a \left (1+\frac{i b}{a}\right )+i a \left (1-\frac{i b}{a}\right ) e^{2 i c+2 i d x}} \, dx,x,\sqrt [3]{x}\right )}{(a+i b)^2 (i a+b)}-\frac{(12 b) \operatorname{Subst}\left (\int x \log \left (1+\frac{\left (1-\frac{i b}{a}\right ) e^{2 i c+2 i d x}}{1+\frac{i b}{a}}\right ) \, dx,x,\sqrt [3]{x}\right )}{(a-i b)^2 (a+i b) d}+\frac{\left (12 b^2\right ) \operatorname{Subst}\left (\int \frac{x}{i a \left (1+\frac{i b}{a}\right )+i a \left (1-\frac{i b}{a}\right ) e^{2 i c+2 i d x}} \, dx,x,\sqrt [3]{x}\right )}{(a-i b)^2 (a+i b) d}\\ &=-\frac{6 i b^2 x^{2/3}}{\left (a^2+b^2\right )^2 d}-\frac{6 b^2 x^{2/3}}{(a-i b)^2 (a+i b) d \left (i a-b+(i a+b) e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}+\frac{x}{(a-i b)^2}+\frac{4 b x}{(i a-b) (a-i b)^2}-\frac{4 b^2 x}{\left (a^2+b^2\right )^2}+\frac{6 b x^{2/3} \log \left (1+\frac{(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{(a-i b)^2 (a+i b) d}-\frac{6 i b^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 )}{\left (a^2+b^2\right )^2 d}+\frac{6 b \sqrt [3]{x} \text{Li}_2\left (-\frac{(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{(i a-b) (a-i b)^2 d^2}-\frac{(6 b) \operatorname{Subst}\left (\int \text{Li}_2\left (-\frac{\left (1-\frac{i b}{a}\right ) e^{2 i c+2 i d x}}{1+\frac{i b}{a}}\right ) \, dx,x,\sqrt [3]{x}\right )}{(i a-b) (a-i b)^2 d^2}-\frac{\left (12 b^2\right ) \operatorname{Subst}\left (\int \frac{e^{2 i c+2 i d x} x}{i a \left (1+\frac{i b}{a}\right )+i a \left (1-\frac{i b}{a}\right ) e^{2 i c+2 i d x}} \, dx,x,\sqrt [3]{x}\right )}{(a-i b) (a+i b)^2 d}+\frac{\left (12 i b^2\right ) \operatorname{Subst}\left (\int x \log \left (1+\frac{\left (1-\frac{i b}{a}\right ) e^{2 i c+2 i d x}}{1+\frac{i b}{a}}\right ) \, dx,x,\sqrt [3]{x}\right )}{\left (a^2+b^2\right )^2 d}\\ &=-\frac{6 i b^2 x^{2/3}}{\left (a^2+b^2\right )^2 d}-\frac{6 b^2 x^{2/3}}{(a-i b)^2 (a+i b) d \left (i a-b+(i a+b) e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}+\frac{x}{(a-i b)^2}+\frac{4 b x}{(i a-b) (a-i b)^2}-\frac{4 b^2 x}{\left (a^2+b^2\right )^2}+\frac{6 b^2 \sqrt [3]{x} \log \left (1+\frac{(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{\left (a^2+b^2\right )^2 d^2}+\frac{6 b x^{2/3} \log \left (1+\frac{(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{(a-i b)^2 (a+i b) d}-\frac{6 i b^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 )}{\left (a^2+b^2\right )^2 d}+\frac{6 b \sqrt [3]{x} \text{Li}_2\left (-\frac{(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{(i a-b) (a-i b)^2 d^2}-\frac{6 b^2 \sqrt [3]{x} \text{Li}_2\left (-\frac{(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{\left (a^2+b^2\right )^2 d^2}+\frac{(3 b) \operatorname{Subst}\left (\int \frac{\text{Li}_2\left (-\frac{(a-i b) x}{a+i b}\right )}{x} \, dx,x,e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{(a-i b)^2 (a+i b) d^3}-\frac{\left (6 b^2\right ) \operatorname{Subst}\left (\int \log \left (1+\frac{\left (1-\frac{i b}{a}\right ) e^{2 i c+2 i d x}}{1+\frac{i b}{a}}\right ) \, dx,x,\sqrt [3]{x}\right )}{\left (a^2+b^2\right )^2 d^2}+\frac{\left (6 b^2\right ) \operatorname{Subst}\left (\int \text{Li}_2\left (-\frac{\left (1-\frac{i b}{a}\right ) e^{2 i c+2 i d x}}{1+\frac{i b}{a}}\right ) \, dx,x,\sqrt [3]{x}\right )}{\left (a^2+b^2\right )^2 d^2}\\ &=-\frac{6 i b^2 x^{2/3}}{\left (a^2+b^2\right )^2 d}-\frac{6 b^2 x^{2/3}}{(a-i b)^2 (a+i b) d \left (i a-b+(i a+b) e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}+\frac{x}{(a-i b)^2}+\frac{4 b x}{(i a-b) (a-i b)^2}-\frac{4 b^2 x}{\left (a^2+b^2\right )^2}+\frac{6 b^2 \sqrt [3]{x} \log \left (1+\frac{(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{\left (a^2+b^2\right )^2 d^2}+\frac{6 b x^{2/3} \log \left (1+\frac{(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{(a-i b)^2 (a+i b) d}-\frac{6 i b^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 )}{\left (a^2+b^2\right )^2 d}+\frac{6 b \sqrt [3]{x} \text{Li}_2\left (-\frac{(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{(i a-b) (a-i b)^2 d^2}-\frac{6 b^2 \sqrt [3]{x} \text{Li}_2\left (-\frac{(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{\left (a^2+b^2\right )^2 d^2}+\frac{3 b \text{Li}_3\left (-\frac{(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{(a-i b)^2 (a+i b) d^3}+\frac{\left (3 i b^2\right ) \operatorname{Subst}\left (\int \frac{\log \left (1+\frac{\left (1-\frac{i b}{a}\right ) x}{1+\frac{i b}{a}}\right )}{x} \, dx,x,e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{\left (a^2+b^2\right )^2 d^3}-\frac{\left (3 i b^2\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2\left (-\frac{(a-i b) x}{a+i b}\right )}{x} \, dx,x,e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{\left (a^2+b^2\right )^2 d^3}\\ &=-\frac{6 i b^2 x^{2/3}}{\left (a^2+b^2\right )^2 d}-\frac{6 b^2 x^{2/3}}{(a-i b)^2 (a+i b) d \left (i a-b+(i a+b) e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}+\frac{x}{(a-i b)^2}+\frac{4 b x}{(i a-b) (a-i b)^2}-\frac{4 b^2 x}{\left (a^2+b^2\right )^2}+\frac{6 b^2 \sqrt [3]{x} \log \left (1+\frac{(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{\left (a^2+b^2\right )^2 d^2}+\frac{6 b x^{2/3} \log \left (1+\frac{(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{(a-i b)^2 (a+i b) d}-\frac{6 i b^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 )}{\left (a^2+b^2\right )^2 d}-\frac{3 i b^2 \text{Li}_2\left (-\frac{(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{\left (a^2+b^2\right )^2 d^3}+\frac{6 b \sqrt [3]{x} \text{Li}_2\left (-\frac{(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{(i a-b) (a-i b)^2 d^2}-\frac{6 b^2 \sqrt [3]{x} \text{Li}_2\left (-\frac{(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{\left (a^2+b^2\right )^2 d^2}+\frac{3 b \text{Li}_3\left (-\frac{(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{(a-i b)^2 (a+i b) d^3}-\frac{3 i b^2 \text{Li}_3\left (-\frac{(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{\left (a^2+b^2\right )^2 d^3}\\ \end{align*}

Mathematica [A]  time = 3.2678, size = 538, normalized size = 0.88 \[ \frac{\frac{b \left (\frac{3 b \left (a \left (1+e^{2 i c}\right )-i b \left (-1+e^{2 i c}\right )\right ) \text{PolyLog}\left (2,\frac{(-a-i b) e^{-2 i \left (c+d \sqrt [3]{x}\right )}}{a-i b}\right )}{d^2 \left (a^2+b^2\right )}+\frac{3 a \left (a \left (1+e^{2 i c}\right )-i b \left (-1+e^{2 i c}\right )\right ) \left (2 d \sqrt [3]{x} \text{PolyLog}\left (2,\frac{(-a-i b) e^{-2 i \left (c+d \sqrt [3]{x}\right )}}{a-i b}\right )-i \text{PolyLog}\left (3,\frac{(-a-i b) e^{-2 i \left (c+d \sqrt [3]{x}\right )}}{a-i b}\right )\right )}{d^2 \left (a^2+b^2\right )}+\frac{6 a x^{2/3} \left (a \left (1+e^{2 i c}\right )-i b \left (-1+e^{2 i c}\right )\right ) \log \left (1+\frac{(a+i b) e^{-2 i \left (c+d \sqrt [3]{x}\right )}}{a-i b}\right )}{(a+i b) (b+i a)}+\frac{6 b \sqrt [3]{x} \left (a \left (1+e^{2 i c}\right )-i b \left (-1+e^{2 i c}\right )\right ) \log \left (1+\frac{(a+i b) e^{-2 i \left (c+d \sqrt [3]{x}\right )}}{a-i b}\right )}{d (a+i b) (b+i a)}+\frac{4 a d x}{a-i b}+\frac{6 b x^{2/3}}{a-i b}\right )}{d \left (-i a \left (1+e^{2 i c}\right )+b \left (-e^{2 i c}\right )+b\right )}+\frac{3 b^2 x^{2/3} \sin \left (d \sqrt [3]{x}\right )}{d (a \cos (c)+b \sin (c)) \left (a \cos \left (c+d \sqrt [3]{x}\right )+b \sin \left (c+d \sqrt [3]{x}\right )\right )}+\frac{x (a \cos (c)-b \sin (c))}{a \cos (c)+b \sin (c)}}{a^2+b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.278, 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(1/(a+b*tan(c+d*x^(1/3)))^2,x)

[Out]

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

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Maxima [B]  time = 3.82389, size = 2363, normalized size = 3.87 \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)))^2,x, algorithm="maxima")

[Out]

3*((2*a*b*log(b*tan(d*x^(1/3) + c) + a)/(a^4 + 2*a^2*b^2 + b^4) - a*b*log(tan(d*x^(1/3) + c)^2 + 1)/(a^4 + 2*a
^2*b^2 + b^4) + (a^2 - b^2)*(d*x^(1/3) + c)/(a^4 + 2*a^2*b^2 + b^4) - b/(a^3 + a*b^2 + (a^2*b + b^3)*tan(d*x^(
1/3) + c)))*c^2 + ((a^3 - I*a^2*b + a*b^2 - I*b^3)*(d*x^(1/3) + c)^3 - (3*a^3 - 3*I*a^2*b + 3*a*b^2 - 3*I*b^3)
*(d*x^(1/3) + c)^2*c - (6*(I*a*b^2 + b^3)*c*cos(2*d*x^(1/3) + 2*c) - (6*a*b^2 - 6*I*b^3)*c*sin(2*d*x^(1/3) + 2
*c) + 6*(I*a*b^2 - b^3)*c)*arctan2(-b*cos(2*d*x^(1/3) + 2*c) + a*sin(2*d*x^(1/3) + 2*c) + b, a*cos(2*d*x^(1/3)
+ 2*c) + b*sin(2*d*x^(1/3) + 2*c) + a) + ((-6*I*a^2*b + 6*a*b^2)*(d*x^(1/3) + c)^2 + (-6*I*a*b^2 + 6*b^3 + (1
2*I*a^2*b - 12*a*b^2)*c)*(d*x^(1/3) + c) + ((-6*I*a^2*b - 6*a*b^2)*(d*x^(1/3) + c)^2 + (-6*I*a*b^2 - 6*b^3 + (
12*I*a^2*b + 12*a*b^2)*c)*(d*x^(1/3) + c))*cos(2*d*x^(1/3) + 2*c) + (6*(a^2*b - I*a*b^2)*(d*x^(1/3) + c)^2 + (
6*a*b^2 - 6*I*b^3 - 12*(a^2*b - I*a*b^2)*c)*(d*x^(1/3) + c))*sin(2*d*x^(1/3) + 2*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)) + ((a^3 - 3*I*a^2*b - 3*a*b^2 + I*b^3)*(d*x^(1/3) + c)^3 + (-6*
I*a*b^2 - 6*b^3 - (3*a^3 - 9*I*a^2*b - 9*a*b^2 + 3*I*b^3)*c)*(d*x^(1/3) + c)^2 - 12*(-I*a*b^2 - b^3)*(d*x^(1/3
) + c)*c)*cos(2*d*x^(1/3) + 2*c) + (-3*I*a*b^2 + 3*b^3 + (-6*I*a^2*b + 6*a*b^2)*(d*x^(1/3) + c) + (6*I*a^2*b -
6*a*b^2)*c + (-3*I*a*b^2 - 3*b^3 + (-6*I*a^2*b - 6*a*b^2)*(d*x^(1/3) + c) + (6*I*a^2*b + 6*a*b^2)*c)*cos(2*d*
x^(1/3) + 2*c) + (3*a*b^2 - 3*I*b^3 + 6*(a^2*b - I*a*b^2)*(d*x^(1/3) + c) - 6*(a^2*b - I*a*b^2)*c)*sin(2*d*x^(
1/3) + 2*c))*dilog((I*a + b)*e^(2*I*d*x^(1/3) + 2*I*c)/(-I*a + b)) - ((3*a*b^2 - 3*I*b^3)*c*cos(2*d*x^(1/3) +
2*c) + 3*(I*a*b^2 + b^3)*c*sin(2*d*x^(1/3) + 2*c) + (3*a*b^2 + 3*I*b^3)*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)) + (3*(a^2*b + I*a*b^2)*(d*x^(1/3) + c)^2 + (3*a*b^2 + 3*I*b^3 - 6*(a^2*b + I*a*b^2)*c)*(d*x^
(1/3) + c) + (3*(a^2*b - I*a*b^2)*(d*x^(1/3) + c)^2 + (3*a*b^2 - 3*I*b^3 - 6*(a^2*b - I*a*b^2)*c)*(d*x^(1/3) +
c))*cos(2*d*x^(1/3) + 2*c) + ((3*I*a^2*b + 3*a*b^2)*(d*x^(1/3) + c)^2 + (3*I*a*b^2 + 3*b^3 + (-6*I*a^2*b - 6*
a*b^2)*c)*(d*x^(1/3) + c))*sin(2*d*x^(1/3) + 2*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*a^2*b + 3*I*a*b^2 + 3*(a^2*b - I*a*b^2)*cos(2*d*x^(1/3) + 2*c) + (3*I*a^2*b + 3*a*b^2)*sin(2*d*x^
(1/3) + 2*c))*polylog(3, (I*a + b)*e^(2*I*d*x^(1/3) + 2*I*c)/(-I*a + b)) + ((I*a^3 + 3*a^2*b - 3*I*a*b^2 - b^3
)*(d*x^(1/3) + c)^3 + (6*a*b^2 - 6*I*b^3 + (-3*I*a^3 - 9*a^2*b + 9*I*a*b^2 + 3*b^3)*c)*(d*x^(1/3) + c)^2 - (12
*a*b^2 - 12*I*b^3)*(d*x^(1/3) + c)*c)*sin(2*d*x^(1/3) + 2*c))/(3*a^5 + 3*I*a^4*b + 6*a^3*b^2 + 6*I*a^2*b^3 + 3
*a*b^4 + 3*I*b^5 + (3*a^5 - 3*I*a^4*b + 6*a^3*b^2 - 6*I*a^2*b^3 + 3*a*b^4 - 3*I*b^5)*cos(2*d*x^(1/3) + 2*c) +
(3*I*a^5 + 3*a^4*b + 6*I*a^3*b^2 + 6*a^2*b^3 + 3*I*a*b^4 + 3*b^5)*sin(2*d*x^(1/3) + 2*c)))/d^3

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Fricas [C]  time = 1.75075, size = 2923, normalized size = 4.79 \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)))^2,x, algorithm="fricas")

[Out]

-1/2*(6*b^3*d^2*x^(2/3) - 2*(a^3 - a*b^2)*d^3*x + 2*(a^3 - a*b^2)*d^3 - (-6*I*a^2*b*d*x^(1/3) - 3*I*a*b^2 + (-
6*I*a*b^2*d*x^(1/3) - 3*I*b^3)*tan(d*x^(1/3) + c))*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*a^2*b*d*x^(1/3) + 3*I*a*b^2 + (6*I*a*b^2*d*x^(1/3) + 3*I*b^3)*tan(d*x^(1/3) + c))*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*(a^2*b*d^2*x^(2/3) - a^2*b*c^2 + a*b^2*d*x^(1/3) + a*b^2*c + (a*b^2*d^
2*x^(2/3) - a*b^2*c^2 + b^3*d*x^(1/3) + b^3*c)*tan(d*x^(1/3) + c))*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*(a^2*b*d^2*x^(2/3) - a^2*b*c^2 + a*b^2*d*x^(1/3) + a*b^2*c + (a*b^2*d^2*x^(2/3) - a*b^2*c^2 + b^3*d
*x^(1/3) + b^3*c)*tan(d*x^(1/3) + c))*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*(a^2*b*c^2 - a*b
^2*c + (a*b^2*c^2 - b^3*c)*tan(d*x^(1/3) + c))*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*(a^2*b*c^2 - a*b^2*c + (a*b^2*c^2 - b^3*c)*tan(d*
x^(1/3) + c))*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)) - 3*(a*b^2*tan(d*x^(1/3) + c) + a^2*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*(a*b^2*tan(d*x^(1/3) + c) + a^2*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)) - 2*(3*a*b^2*d^2*x^(2/3) + (a^2*b - b^3)*d^3*x - (a^2*b - b^3)*d^3)*tan(d*x^(1/3) + c))/((a^4*b +
2*a^2*b^3 + b^5)*d^3*tan(d*x^(1/3) + c) + (a^5 + 2*a^3*b^2 + a*b^4)*d^3)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\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(1/(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 \frac{1}{{\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(1/(a+b*tan(c+d*x^(1/3)))^2,x, algorithm="giac")

[Out]

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