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

Optimal. Leaf size=120 \[ -\frac {3\ 2^{\frac {1}{3}-\frac {i n}{2}} \left (a^2 x^2+1\right )^{2/3} (1-i a x)^{\frac {1}{6} (2+3 i n)} \, _2F_1\left (\frac {1}{6} (3 i n+2),\frac {1}{6} (3 i n+4);\frac {1}{6} (3 i n+8);\frac {1}{2} (1-i a x)\right )}{a (-3 n+2 i) \left (a^2 c x^2+c\right )^{2/3}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.11, antiderivative size = 120, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {5076, 5073, 69} \[ -\frac {3\ 2^{\frac {1}{3}-\frac {i n}{2}} \left (a^2 x^2+1\right )^{2/3} (1-i a x)^{\frac {1}{6} (2+3 i n)} \, _2F_1\left (\frac {1}{6} (3 i n+2),\frac {1}{6} (3 i n+4);\frac {1}{6} (3 i n+8);\frac {1}{2} (1-i a x)\right )}{a (-3 n+2 i) \left (a^2 c x^2+c\right )^{2/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]))

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 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])

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.05, size = 120, normalized size = 1.00 \[ \frac {3\ 2^{\frac {1}{3}-\frac {i n}{2}} \left (a^2 x^2+1\right )^{2/3} (1-i a x)^{\frac {1}{3}+\frac {i n}{2}} \, _2F_1\left (\frac {i n}{2}+\frac {1}{3},\frac {i n}{2}+\frac {2}{3};\frac {i n}{2}+\frac {4}{3};\frac {1}{2}-\frac {i a x}{2}\right )}{a (3 n-2 i) \left (a^2 c x^2+c\right )^{2/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [F]  time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {e^{\left (n \arctan \left (a x\right )\right )}}{{\left (a^{2} c x^{2} + c\right )}^{\frac {2}{3}}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

________________________________________________________________________________________

maple [F]  time = 0.29, size = 0, normalized size = 0.00 \[ \int \frac {{\mathrm e}^{n \arctan \left (a x \right )}}{\left (a^{2} c \,x^{2}+c \right )^{\frac {2}{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {e^{\left (n \arctan \left (a x\right )\right )}}{{\left (a^{2} c x^{2} + c\right )}^{\frac {2}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\mathrm {e}}^{n\,\mathrm {atan}\left (a\,x\right )}}{{\left (c\,a^2\,x^2+c\right )}^{2/3}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {e^{n \operatorname {atan}{\left (a x \right )}}}{\left (c \left (a^{2} x^{2} + 1\right )\right )^{\frac {2}{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________