\(\int e^{2 \arctan (a x)} (c+a^2 c x^2)^{3/2} \, dx\) [267]

Optimal result

Integrand size = 23, antiderivative size = 88 \[ \int e^{2 \arctan (a x)} \left (c+a^2 c x^2\right )^{3/2} \, dx=\frac {\left (\frac {2}{29}+\frac {5 i}{29}\right ) 2^{\frac {5}{2}-i} c (1-i a x)^{\frac {5}{2}+i} \sqrt {c+a^2 c x^2} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2}+i,\frac {5}{2}+i,\frac {7}{2}+i,\frac {1}{2} (1-i a x)\right )}{a \sqrt {1+a^2 x^2}} \]

[Out]

(2/29+5/29*I)*2^(5/2-I)*c*(1-I*a*x)^(5/2+I)*hypergeom([5/2+I, -3/2+I],[7/2+I],1/2-1/2*I*a*x)*(a^2*c*x^2+c)^(1/
2)/a/(a^2*x^2+1)^(1/2)

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 88, 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} \[ \int e^{2 \arctan (a x)} \left (c+a^2 c x^2\right )^{3/2} \, dx=\frac {\left (\frac {2}{29}+\frac {5 i}{29}\right ) 2^{\frac {5}{2}-i} c (1-i a x)^{\frac {5}{2}+i} \sqrt {a^2 c x^2+c} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2}+i,\frac {5}{2}+i,\frac {7}{2}+i,\frac {1}{2} (1-i a x)\right )}{a \sqrt {a^2 x^2+1}} \]

[In]

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

[Out]

((2/29 + (5*I)/29)*2^(5/2 - I)*c*(1 - I*a*x)^(5/2 + I)*Sqrt[c + a^2*c*x^2]*Hypergeometric2F1[-3/2 + I, 5/2 + I
, 7/2 + I, (1 - I*a*x)/2])/(a*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^{2 \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}+i} (1+i a x)^{\frac {3}{2}-i} \, dx}{\sqrt {1+a^2 x^2}} \\ & = \frac {\left (\frac {2}{29}+\frac {5 i}{29}\right ) 2^{\frac {5}{2}-i} c (1-i a x)^{\frac {5}{2}+i} \sqrt {c+a^2 c x^2} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2}+i,\frac {5}{2}+i,\frac {7}{2}+i,\frac {1}{2} (1-i a x)\right )}{a \sqrt {1+a^2 x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00 \[ \int e^{2 \arctan (a x)} \left (c+a^2 c x^2\right )^{3/2} \, dx=\frac {\left (\frac {2}{29}+\frac {5 i}{29}\right ) 2^{\frac {5}{2}-i} c (1-i a x)^{\frac {5}{2}+i} \sqrt {c+a^2 c x^2} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2}+i,\frac {5}{2}+i,\frac {7}{2}+i,\frac {1}{2} (1-i a x)\right )}{a \sqrt {1+a^2 x^2}} \]

[In]

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

[Out]

((2/29 + (5*I)/29)*2^(5/2 - I)*c*(1 - I*a*x)^(5/2 + I)*Sqrt[c + a^2*c*x^2]*Hypergeometric2F1[-3/2 + I, 5/2 + I
, 7/2 + I, (1 - I*a*x)/2])/(a*Sqrt[1 + a^2*x^2])

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

integrate(exp(2*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^(2*arctan(a*x)), x)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

integrate(exp(2*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^(2*arctan(a*x)), x)

Giac [F(-2)]

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

[In]

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

[In]

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

[Out]

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