\(\int e^{n \arctan (a x)} (c+a^2 c x^2)^{3/2} \, dx\) [348]

Optimal result

Integrand size = 23, antiderivative size = 121 \[ \int e^{n \arctan (a x)} \left (c+a^2 c x^2\right )^{3/2} \, dx=-\frac {2^{\frac {5}{2}-\frac {i n}{2}} c (1-i a x)^{\frac {1}{2} (5+i n)} \sqrt {c+a^2 c x^2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (-3+i n),\frac {1}{2} (5+i n),\frac {1}{2} (7+i n),\frac {1}{2} (1-i a x)\right )}{a (5 i-n) \sqrt {1+a^2 x^2}} \]

[Out]

-2^(5/2-1/2*I*n)*c*(1-I*a*x)^(5/2+1/2*I*n)*hypergeom([5/2+1/2*I*n, -3/2+1/2*I*n],[7/2+1/2*I*n],1/2-1/2*I*a*x)*
(a^2*c*x^2+c)^(1/2)/a/(5*I-n)/(a^2*x^2+1)^(1/2)

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 121, 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} \[ \int e^{n \arctan (a x)} \left (c+a^2 c x^2\right )^{3/2} \, dx=-\frac {c 2^{\frac {5}{2}-\frac {i n}{2}} \sqrt {a^2 c x^2+c} (1-i a x)^{\frac {1}{2} (5+i n)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (i n-3),\frac {1}{2} (i n+5),\frac {1}{2} (i n+7),\frac {1}{2} (1-i a x)\right )}{a (-n+5 i) \sqrt {a^2 x^2+1}} \]

[In]

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

[Out]

-((2^(5/2 - (I/2)*n)*c*(1 - I*a*x)^((5 + I*n)/2)*Sqrt[c + a^2*c*x^2]*Hypergeometric2F1[(-3 + I*n)/2, (5 + I*n)
/2, (7 + I*n)/2, (1 - I*a*x)/2])/(a*(5*I - n)*Sqrt[1 + a^2*x^2]))

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 {\left (c \sqrt {c+a^2 c x^2}\right ) \int e^{n \arctan (a x)} \left (1+a^2 x^2\right )^{3/2} \, dx}{\sqrt {1+a^2 x^2}} \\ & = \frac {\left (c \sqrt {c+a^2 c x^2}\right ) \int (1-i a x)^{\frac {3}{2}+\frac {i n}{2}} (1+i a x)^{\frac {3}{2}-\frac {i n}{2}} \, dx}{\sqrt {1+a^2 x^2}} \\ & = -\frac {2^{\frac {5}{2}-\frac {i n}{2}} c (1-i a x)^{\frac {1}{2} (5+i n)} \sqrt {c+a^2 c x^2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (-3+i n),\frac {1}{2} (5+i n),\frac {1}{2} (7+i n),\frac {1}{2} (1-i a x)\right )}{a (5 i-n) \sqrt {1+a^2 x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.98 \[ \int e^{n \arctan (a x)} \left (c+a^2 c x^2\right )^{3/2} \, dx=\frac {2^{\frac {5}{2}-\frac {i n}{2}} c (1-i a x)^{\frac {5}{2}+\frac {i n}{2}} \sqrt {c+a^2 c x^2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (5+i n),\frac {1}{2} i (3 i+n),\frac {1}{2} (7+i n),\frac {1}{2} (1-i a x)\right )}{a (-5 i+n) \sqrt {1+a^2 x^2}} \]

[In]

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

[Out]

(2^(5/2 - (I/2)*n)*c*(1 - I*a*x)^(5/2 + (I/2)*n)*Sqrt[c + a^2*c*x^2]*Hypergeometric2F1[(5 + I*n)/2, (I/2)*(3*I
+ n), (7 + I*n)/2, (1 - I*a*x)/2])/(a*(-5*I + n)*Sqrt[1 + a^2*x^2])

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F(-2)]

Exception generated. \[ \int e^{n \arctan (a x)} \left (c+a^2 c x^2\right )^{3/2} \, dx=\text {Exception raised: TypeError} \]

[In]

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

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

Mupad [F(-1)]

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

[In]

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

[Out]

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