3.363 \(\int \frac{e^{n \tan ^{-1}(a x)}}{(c+a^2 c x^2)^{4/3}} \, dx\)

Optimal. Leaf size=123 \[ \frac{3\ 2^{-\frac{1}{3}-\frac{i n}{2}} \sqrt [3]{a^2 x^2+1} (1-i a x)^{\frac{1}{6} (-2+3 i n)} \, _2F_1\left (\frac{1}{6} (3 i n-2),\frac{1}{6} (3 i n+8);\frac{1}{6} (3 i n+4);\frac{1}{2} (1-i a x)\right )}{a c (3 n+2 i) \sqrt [3]{a^2 c x^2+c}} \]

[Out]

(3*2^(-1/3 - (I/2)*n)*(1 - I*a*x)^((-2 + (3*I)*n)/6)*(1 + a^2*x^2)^(1/3)*Hypergeometric2F1[(-2 + (3*I)*n)/6, (
8 + (3*I)*n)/6, (4 + (3*I)*n)/6, (1 - I*a*x)/2])/(a*c*(2*I + 3*n)*(c + a^2*c*x^2)^(1/3))

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Rubi [A]  time = 0.119131, antiderivative size = 123, 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} \[ \frac{3\ 2^{-\frac{1}{3}-\frac{i n}{2}} \sqrt [3]{a^2 x^2+1} (1-i a x)^{\frac{1}{6} (-2+3 i n)} \, _2F_1\left (\frac{1}{6} (3 i n-2),\frac{1}{6} (3 i n+8);\frac{1}{6} (3 i n+4);\frac{1}{2} (1-i a x)\right )}{a c (3 n+2 i) \sqrt [3]{a^2 c x^2+c}} \]

Antiderivative was successfully verified.

[In]

Int[E^(n*ArcTan[a*x])/(c + a^2*c*x^2)^(4/3),x]

[Out]

(3*2^(-1/3 - (I/2)*n)*(1 - I*a*x)^((-2 + (3*I)*n)/6)*(1 + a^2*x^2)^(1/3)*Hypergeometric2F1[(-2 + (3*I)*n)/6, (
8 + (3*I)*n)/6, (4 + (3*I)*n)/6, (1 - I*a*x)/2])/(a*c*(2*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)}}{\left (c+a^2 c x^2\right )^{4/3}} \, dx &=\frac{\sqrt [3]{1+a^2 x^2} \int \frac{e^{n \tan ^{-1}(a x)}}{\left (1+a^2 x^2\right )^{4/3}} \, dx}{c \sqrt [3]{c+a^2 c x^2}}\\ &=\frac{\sqrt [3]{1+a^2 x^2} \int (1-i a x)^{-\frac{4}{3}+\frac{i n}{2}} (1+i a x)^{-\frac{4}{3}-\frac{i n}{2}} \, dx}{c \sqrt [3]{c+a^2 c x^2}}\\ &=\frac{3\ 2^{-\frac{1}{3}-\frac{i n}{2}} (1-i a x)^{\frac{1}{6} (-2+3 i n)} \sqrt [3]{1+a^2 x^2} \, _2F_1\left (\frac{1}{6} (-2+3 i n),\frac{1}{6} (8+3 i n);\frac{1}{6} (4+3 i n);\frac{1}{2} (1-i a x)\right )}{a c (2 i+3 n) \sqrt [3]{c+a^2 c x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0533024, size = 123, normalized size = 1. \[ \frac{3\ 2^{-\frac{1}{3}-\frac{i n}{2}} \sqrt [3]{a^2 x^2+1} (1-i a x)^{-\frac{1}{3}+\frac{i n}{2}} \, _2F_1\left (\frac{i n}{2}-\frac{1}{3},\frac{i n}{2}+\frac{4}{3};\frac{i n}{2}+\frac{2}{3};\frac{1}{2}-\frac{i a x}{2}\right )}{a c (3 n+2 i) \sqrt [3]{a^2 c x^2+c}} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(n*ArcTan[a*x])/(c + a^2*c*x^2)^(4/3),x]

[Out]

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

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Maple [F]  time = 0.266, size = 0, normalized size = 0. \begin{align*} \int{{{\rm e}^{n\arctan \left ( ax \right ) }} \left ({a}^{2}c{x}^{2}+c \right ) ^{-{\frac{4}{3}}}}\, 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)^(4/3),x)

[Out]

int(exp(n*arctan(a*x))/(a^2*c*x^2+c)^(4/3),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{4}{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)^(4/3),x, algorithm="maxima")

[Out]

integrate(e^(n*arctan(a*x))/(a^2*c*x^2 + c)^(4/3), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (a^{2} c x^{2} + c\right )}^{\frac{2}{3}} e^{\left (n \arctan \left (a x\right )\right )}}{a^{4} c^{2} x^{4} + 2 \, a^{2} c^{2} x^{2} + c^{2}}, 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)^(4/3),x, algorithm="fricas")

[Out]

integral((a^2*c*x^2 + c)^(2/3)*e^(n*arctan(a*x))/(a^4*c^2*x^4 + 2*a^2*c^2*x^2 + c^2), 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 )}}}{\left (c \left (a^{2} x^{2} + 1\right )\right )^{\frac{4}{3}}}\, 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)**(4/3),x)

[Out]

Integral(exp(n*atan(a*x))/(c*(a**2*x**2 + 1))**(4/3), 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{4}{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)^(4/3),x, algorithm="giac")

[Out]

integrate(e^(n*arctan(a*x))/(a^2*c*x^2 + c)^(4/3), x)