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

Optimal. Leaf size=98 \[ \frac{3 i b \sqrt [3]{x} \text{PolyLog}\left (2,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}-\frac{3 b \text{PolyLog}\left (3,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^3}+a x-\frac{3 b x^{2/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+i b x \]

[Out]

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

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Rubi [A]  time = 0.163829, antiderivative size = 98, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.429, Rules used = {3739, 3719, 2190, 2531, 2282, 6589} \[ a x+\frac{3 i b \sqrt [3]{x} \text{Li}_2\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}-\frac{3 b \text{Li}_3\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^3}-\frac{3 b x^{2/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+i b x \]

Antiderivative was successfully verified.

[In]

Int[a + b*Tan[c + d*x^(1/3)],x]

[Out]

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

Mathematica [A]  time = 0.0269151, size = 98, normalized size = 1. \[ \frac{3 i b \sqrt [3]{x} \text{PolyLog}\left (2,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}-\frac{3 b \text{PolyLog}\left (3,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^3}+a x-\frac{3 b x^{2/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+i b x \]

Antiderivative was successfully verified.

[In]

Integrate[a + b*Tan[c + d*x^(1/3)],x]

[Out]

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

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Maple [F]  time = 0.141, size = 0, normalized size = 0. \begin{align*} \int a+b\tan \left ( c+d\sqrt [3]{x} \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a x + 2 \, b \int \frac{\sin \left (2 \, d x^{\frac{1}{3}} + 2 \, c\right )}{\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}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [C]  time = 1.42031, size = 738, normalized size = 7.53 \begin{align*} \frac{4 \, a d^{3} x - 6 \, b d^{2} x^{\frac{2}{3}} \log \left (-\frac{2 \,{\left (i \, \tan \left (d x^{\frac{1}{3}} + c\right ) - 1\right )}}{\tan \left (d x^{\frac{1}{3}} + c\right )^{2} + 1}\right ) - 6 \, b d^{2} x^{\frac{2}{3}} \log \left (-\frac{2 \,{\left (-i \, \tan \left (d x^{\frac{1}{3}} + c\right ) - 1\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 (i \, \tan \left (d x^{\frac{1}{3}} + c\right ) - 1\right )}}{\tan \left (d x^{\frac{1}{3}} + c\right )^{2} + 1} + 1\right ) + 6 i \, b d x^{\frac{1}{3}}{\rm Li}_2\left (\frac{2 \,{\left (-i \, \tan \left (d x^{\frac{1}{3}} + c\right ) - 1\right )}}{\tan \left (d x^{\frac{1}{3}} + c\right )^{2} + 1} + 1\right ) - 3 \, b{\rm polylog}\left (3, \frac{\tan \left (d x^{\frac{1}{3}} + c\right )^{2} + 2 i \, \tan \left (d x^{\frac{1}{3}} + c\right ) - 1}{\tan \left (d x^{\frac{1}{3}} + c\right )^{2} + 1}\right ) - 3 \, b{\rm polylog}\left (3, \frac{\tan \left (d x^{\frac{1}{3}} + c\right )^{2} - 2 i \, \tan \left (d x^{\frac{1}{3}} + c\right ) - 1}{\tan \left (d x^{\frac{1}{3}} + c\right )^{2} + 1}\right )}{4 \, d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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