\(\int \frac {e^{n \arctan (a x)}}{(c+a^2 c x^2)^{4/3}} \, dx\) [363]

Optimal result

Integrand size = 23, antiderivative size = 123 \[ \int \frac {e^{n \arctan (a x)}}{\left (c+a^2 c x^2\right )^{4/3}} \, dx=\frac {3\ 2^{-\frac {1}{3}-\frac {i n}{2}} (1-i a x)^{\frac {1}{6} (-2+3 i n)} \sqrt [3]{1+a^2 x^2} \operatorname {Hypergeometric2F1}\left (\frac {1}{6} (-2+3 i n),\frac {1}{6} (8+3 i n),\frac {1}{6} (4+3 i n),\frac {1}{2} (1-i a x)\right )}{a c (2 i+3 n) \sqrt [3]{c+a^2 c x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {5184, 5181, 71} \[ \int \frac {e^{n \arctan (a x)}}{\left (c+a^2 c x^2\right )^{4/3}} \, dx=\frac {3\ 2^{-\frac {1}{3}-\frac {i n}{2}} \sqrt [3]{a^2 x^2+1} (1-i a x)^{\frac {1}{6} (-2+3 i n)} \operatorname {Hypergeometric2F1}\left (\frac {1}{6} (3 i n-2),\frac {1}{6} (3 i n+8),\frac {1}{6} (3 i n+4),\frac {1}{2} (1-i a x)\right )}{a c (3 n+2 i) \sqrt [3]{a^2 c x^2+c}} \]

[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 71

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c
- a*d))^n))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], 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 5181

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 5184

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*} \text {integral}& = \frac {\sqrt [3]{1+a^2 x^2} \int \frac {e^{n \arctan (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} \operatorname {Hypergeometric2F1}\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] (verified)

Time = 0.04 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.00 \[ \int \frac {e^{n \arctan (a x)}}{\left (c+a^2 c x^2\right )^{4/3}} \, dx=\frac {3\ 2^{-\frac {1}{3}-\frac {i n}{2}} (1-i a x)^{-\frac {1}{3}+\frac {i n}{2}} \sqrt [3]{1+a^2 x^2} \operatorname {Hypergeometric2F1}\left (-\frac {1}{3}+\frac {i n}{2},\frac {4}{3}+\frac {i n}{2},\frac {2}{3}+\frac {i n}{2},\frac {1}{2}-\frac {i a x}{2}\right )}{a c (2 i+3 n) \sqrt [3]{c+a^2 c x^2}} \]

[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))

Maple [F]

\[\int \frac {{\mathrm e}^{n \arctan \left (a x \right )}}{\left (a^{2} c \,x^{2}+c \right )^{\frac {4}{3}}}d x\]

[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)

Fricas [F]

\[ \int \frac {e^{n \arctan (a x)}}{\left (c+a^2 c x^2\right )^{4/3}} \, dx=\int { \frac {e^{\left (n \arctan \left (a x\right )\right )}}{{\left (a^{2} c x^{2} + c\right )}^{\frac {4}{3}}} \,d x } \]

[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)

Sympy [F]

\[ \int \frac {e^{n \arctan (a x)}}{\left (c+a^2 c x^2\right )^{4/3}} \, dx=\int \frac {e^{n \operatorname {atan}{\left (a x \right )}}}{\left (c \left (a^{2} x^{2} + 1\right )\right )^{\frac {4}{3}}}\, dx \]

[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)

Maxima [F]

\[ \int \frac {e^{n \arctan (a x)}}{\left (c+a^2 c x^2\right )^{4/3}} \, dx=\int { \frac {e^{\left (n \arctan \left (a x\right )\right )}}{{\left (a^{2} c x^{2} + c\right )}^{\frac {4}{3}}} \,d x } \]

[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)

Giac [F]

\[ \int \frac {e^{n \arctan (a x)}}{\left (c+a^2 c x^2\right )^{4/3}} \, dx=\int { \frac {e^{\left (n \arctan \left (a x\right )\right )}}{{\left (a^{2} c x^{2} + c\right )}^{\frac {4}{3}}} \,d x } \]

[In]

integrate(exp(n*arctan(a*x))/(a^2*c*x^2+c)^(4/3),x, algorithm="giac")

[Out]

sage0*x

Mupad [F(-1)]

Timed out. \[ \int \frac {e^{n \arctan (a x)}}{\left (c+a^2 c x^2\right )^{4/3}} \, dx=\int \frac {{\mathrm {e}}^{n\,\mathrm {atan}\left (a\,x\right )}}{{\left (c\,a^2\,x^2+c\right )}^{4/3}} \,d x \]

[In]

int(exp(n*atan(a*x))/(c + a^2*c*x^2)^(4/3),x)

[Out]

int(exp(n*atan(a*x))/(c + a^2*c*x^2)^(4/3), x)