3.361 \(\int \frac{e^{n \tan ^{-1}(a x)}}{\sqrt [3]{c+a^2 c x^2}} \, dx\)

Optimal. Leaf size=120 \[ -\frac{3\ 2^{\frac{2}{3}-\frac{i n}{2}} \sqrt [3]{a^2 x^2+1} (1-i a x)^{\frac{1}{6} (4+3 i n)} \, _2F_1\left (\frac{1}{6} (3 i n+2),\frac{1}{6} (3 i n+4);\frac{1}{6} (3 i n+10);\frac{1}{2} (1-i a x)\right )}{a (-3 n+4 i) \sqrt [3]{a^2 c x^2+c}} \]

[Out]

(-3*2^(2/3 - (I/2)*n)*(1 - I*a*x)^((4 + (3*I)*n)/6)*(1 + a^2*x^2)^(1/3)*Hypergeometric2F1[(2 + (3*I)*n)/6, (4
+ (3*I)*n)/6, (10 + (3*I)*n)/6, (1 - I*a*x)/2])/(a*(4*I - 3*n)*(c + a^2*c*x^2)^(1/3))

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Rubi [A]  time = 0.108826, antiderivative size = 120, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {5076, 5073, 69} \[ -\frac{3\ 2^{\frac{2}{3}-\frac{i n}{2}} \sqrt [3]{a^2 x^2+1} (1-i a x)^{\frac{1}{6} (4+3 i n)} \, _2F_1\left (\frac{1}{6} (3 i n+2),\frac{1}{6} (3 i n+4);\frac{1}{6} (3 i n+10);\frac{1}{2} (1-i a x)\right )}{a (-3 n+4 i) \sqrt [3]{a^2 c x^2+c}} \]

Antiderivative was successfully verified.

[In]

Int[E^(n*ArcTan[a*x])/(c + a^2*c*x^2)^(1/3),x]

[Out]

(-3*2^(2/3 - (I/2)*n)*(1 - I*a*x)^((4 + (3*I)*n)/6)*(1 + a^2*x^2)^(1/3)*Hypergeometric2F1[(2 + (3*I)*n)/6, (4
+ (3*I)*n)/6, (10 + (3*I)*n)/6, (1 - I*a*x)/2])/(a*(4*I - 3*n)*(c + a^2*c*x^2)^(1/3))

Rule 5076

Int[E^(ArcTan[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(c^IntPart[p]*(c + d*x^2)^FracP
art[p])/(1 + a^2*x^2)^FracPart[p], Int[(1 + a^2*x^2)^p*E^(n*ArcTan[a*x]), x], x] /; FreeQ[{a, c, d, n, p}, x]
&& EqQ[d, a^2*c] &&  !(IntegerQ[p] || GtQ[c, 0])

Rule 5073

Int[E^(ArcTan[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[c^p, Int[(1 - I*a*x)^(p + (I*n
)/2)*(1 + I*a*x)^(p - (I*n)/2), x], x] /; FreeQ[{a, c, d, n, p}, x] && EqQ[d, a^2*c] && (IntegerQ[p] || GtQ[c,
0])

Rule 69

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Hypergeometric2F1[
-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b*(m + 1)*(b/(b*c - a*d))^n), x] /; FreeQ[{a, b, c, d, m, n}
, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] ||  !(Ra
tionalQ[n] && GtQ[-(d/(b*c - a*d)), 0]))

Rubi steps

\begin{align*} \int \frac{e^{n \tan ^{-1}(a x)}}{\sqrt [3]{c+a^2 c x^2}} \, dx &=\frac{\sqrt [3]{1+a^2 x^2} \int \frac{e^{n \tan ^{-1}(a x)}}{\sqrt [3]{1+a^2 x^2}} \, dx}{\sqrt [3]{c+a^2 c x^2}}\\ &=\frac{\sqrt [3]{1+a^2 x^2} \int (1-i a x)^{-\frac{1}{3}+\frac{i n}{2}} (1+i a x)^{-\frac{1}{3}-\frac{i n}{2}} \, dx}{\sqrt [3]{c+a^2 c x^2}}\\ &=-\frac{3\ 2^{\frac{2}{3}-\frac{i n}{2}} (1-i a x)^{\frac{1}{6} (4+3 i n)} \sqrt [3]{1+a^2 x^2} \, _2F_1\left (\frac{1}{6} (2+3 i n),\frac{1}{6} (4+3 i n);\frac{1}{6} (10+3 i n);\frac{1}{2} (1-i a x)\right )}{a (4 i-3 n) \sqrt [3]{c+a^2 c x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0461303, size = 120, normalized size = 1. \[ \frac{3\ 2^{\frac{2}{3}-\frac{i n}{2}} \sqrt [3]{a^2 x^2+1} (1-i a x)^{\frac{2}{3}+\frac{i n}{2}} \, _2F_1\left (\frac{i n}{2}+\frac{1}{3},\frac{i n}{2}+\frac{2}{3};\frac{i n}{2}+\frac{5}{3};\frac{1}{2}-\frac{i a x}{2}\right )}{a (3 n-4 i) \sqrt [3]{a^2 c x^2+c}} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(n*ArcTan[a*x])/(c + a^2*c*x^2)^(1/3),x]

[Out]

(3*2^(2/3 - (I/2)*n)*(1 - I*a*x)^(2/3 + (I/2)*n)*(1 + a^2*x^2)^(1/3)*Hypergeometric2F1[1/3 + (I/2)*n, 2/3 + (I
/2)*n, 5/3 + (I/2)*n, 1/2 - (I/2)*a*x])/(a*(-4*I + 3*n)*(c + a^2*c*x^2)^(1/3))

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Maple [F]  time = 0.271, size = 0, normalized size = 0. \begin{align*} \int{{{\rm e}^{n\arctan \left ( ax \right ) }}{\frac{1}{\sqrt [3]{{a}^{2}c{x}^{2}+c}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(n*arctan(a*x))/(a^2*c*x^2+c)^(1/3),x)

[Out]

int(exp(n*arctan(a*x))/(a^2*c*x^2+c)^(1/3),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*arctan(a*x))/(a^2*c*x^2+c)^(1/3),x, algorithm="maxima")

[Out]

integrate(e^(n*arctan(a*x))/(a^2*c*x^2 + c)^(1/3), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{e^{\left (n \arctan \left (a x\right )\right )}}{{\left (a^{2} c x^{2} + c\right )}^{\frac{1}{3}}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*arctan(a*x))/(a^2*c*x^2+c)^(1/3),x, algorithm="fricas")

[Out]

integral(e^(n*arctan(a*x))/(a^2*c*x^2 + c)^(1/3), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{e^{n \operatorname{atan}{\left (a x \right )}}}{\sqrt [3]{c \left (a^{2} x^{2} + 1\right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*atan(a*x))/(a**2*c*x**2+c)**(1/3),x)

[Out]

Integral(exp(n*atan(a*x))/(c*(a**2*x**2 + 1))**(1/3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*arctan(a*x))/(a^2*c*x^2+c)^(1/3),x, algorithm="giac")

[Out]

integrate(e^(n*arctan(a*x))/(a^2*c*x^2 + c)^(1/3), x)