3.271 \(\int \frac{e^{2 \tan ^{-1}(a x)}}{(c+a^2 c x^2)^{5/2}} \, dx\)

Optimal. Leaf size=76 \[ \frac{6 (a x+2) e^{2 \tan ^{-1}(a x)}}{65 a c^2 \sqrt{a^2 c x^2+c}}+\frac{(3 a x+2) e^{2 \tan ^{-1}(a x)}}{13 a c \left (a^2 c x^2+c\right )^{3/2}} \]

[Out]

(E^(2*ArcTan[a*x])*(2 + 3*a*x))/(13*a*c*(c + a^2*c*x^2)^(3/2)) + (6*E^(2*ArcTan[a*x])*(2 + a*x))/(65*a*c^2*Sqr
t[c + a^2*c*x^2])

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Rubi [A]  time = 0.0811724, antiderivative size = 76, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.087, Rules used = {5070, 5069} \[ \frac{6 (a x+2) e^{2 \tan ^{-1}(a x)}}{65 a c^2 \sqrt{a^2 c x^2+c}}+\frac{(3 a x+2) e^{2 \tan ^{-1}(a x)}}{13 a c \left (a^2 c x^2+c\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[E^(2*ArcTan[a*x])/(c + a^2*c*x^2)^(5/2),x]

[Out]

(E^(2*ArcTan[a*x])*(2 + 3*a*x))/(13*a*c*(c + a^2*c*x^2)^(3/2)) + (6*E^(2*ArcTan[a*x])*(2 + a*x))/(65*a*c^2*Sqr
t[c + a^2*c*x^2])

Rule 5070

Int[E^(ArcTan[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((n - 2*a*(p + 1)*x)*(c + d*x^2
)^(p + 1)*E^(n*ArcTan[a*x]))/(a*c*(n^2 + 4*(p + 1)^2)), x] + Dist[(2*(p + 1)*(2*p + 3))/(c*(n^2 + 4*(p + 1)^2)
), Int[(c + d*x^2)^(p + 1)*E^(n*ArcTan[a*x]), x], x] /; FreeQ[{a, c, d, n}, x] && EqQ[d, a^2*c] && LtQ[p, -1]
&&  !IntegerQ[I*n] && NeQ[n^2 + 4*(p + 1)^2, 0] && IntegerQ[2*p]

Rule 5069

Int[E^(ArcTan[(a_.)*(x_)]*(n_.))/((c_) + (d_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[((n + a*x)*E^(n*ArcTan[a*x]))/
(a*c*(n^2 + 1)*Sqrt[c + d*x^2]), x] /; FreeQ[{a, c, d, n}, x] && EqQ[d, a^2*c] &&  !IntegerQ[I*n]

Rubi steps

\begin{align*} \int \frac{e^{2 \tan ^{-1}(a x)}}{\left (c+a^2 c x^2\right )^{5/2}} \, dx &=\frac{e^{2 \tan ^{-1}(a x)} (2+3 a x)}{13 a c \left (c+a^2 c x^2\right )^{3/2}}+\frac{6 \int \frac{e^{2 \tan ^{-1}(a x)}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx}{13 c}\\ &=\frac{e^{2 \tan ^{-1}(a x)} (2+3 a x)}{13 a c \left (c+a^2 c x^2\right )^{3/2}}+\frac{6 e^{2 \tan ^{-1}(a x)} (2+a x)}{65 a c^2 \sqrt{c+a^2 c x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0340057, size = 62, normalized size = 0.82 \[ \frac{\left (6 a^3 x^3+12 a^2 x^2+21 a x+22\right ) e^{2 \tan ^{-1}(a x)}}{65 c^2 \left (a^3 x^2+a\right ) \sqrt{a^2 c x^2+c}} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(2*ArcTan[a*x])/(c + a^2*c*x^2)^(5/2),x]

[Out]

(E^(2*ArcTan[a*x])*(22 + 21*a*x + 12*a^2*x^2 + 6*a^3*x^3))/(65*c^2*(a + a^3*x^2)*Sqrt[c + a^2*c*x^2])

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Maple [A]  time = 0.04, size = 56, normalized size = 0.7 \begin{align*}{\frac{ \left ({a}^{2}{x}^{2}+1 \right ) \left ( 6\,{a}^{3}{x}^{3}+12\,{a}^{2}{x}^{2}+21\,ax+22 \right ){{\rm e}^{2\,\arctan \left ( ax \right ) }}}{65\,a} \left ({a}^{2}c{x}^{2}+c \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(2*arctan(a*x))/(a^2*c*x^2+c)^(5/2),x)

[Out]

1/65*(a^2*x^2+1)*(6*a^3*x^3+12*a^2*x^2+21*a*x+22)*exp(2*arctan(a*x))/a/(a^2*c*x^2+c)^(5/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{e^{\left (2 \, \arctan \left (a x\right )\right )}}{{\left (a^{2} c x^{2} + c\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(2*arctan(a*x))/(a^2*c*x^2+c)^(5/2),x, algorithm="maxima")

[Out]

integrate(e^(2*arctan(a*x))/(a^2*c*x^2 + c)^(5/2), x)

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Fricas [A]  time = 2.01751, size = 163, normalized size = 2.14 \begin{align*} \frac{{\left (6 \, a^{3} x^{3} + 12 \, a^{2} x^{2} + 21 \, a x + 22\right )} \sqrt{a^{2} c x^{2} + c} e^{\left (2 \, \arctan \left (a x\right )\right )}}{65 \,{\left (a^{5} c^{3} x^{4} + 2 \, a^{3} c^{3} x^{2} + a c^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(2*arctan(a*x))/(a^2*c*x^2+c)^(5/2),x, algorithm="fricas")

[Out]

1/65*(6*a^3*x^3 + 12*a^2*x^2 + 21*a*x + 22)*sqrt(a^2*c*x^2 + c)*e^(2*arctan(a*x))/(a^5*c^3*x^4 + 2*a^3*c^3*x^2
+ a*c^3)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(2*atan(a*x))/(a**2*c*x**2+c)**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{e^{\left (2 \, \arctan \left (a x\right )\right )}}{{\left (a^{2} c x^{2} + c\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(2*arctan(a*x))/(a^2*c*x^2+c)^(5/2),x, algorithm="giac")

[Out]

integrate(e^(2*arctan(a*x))/(a^2*c*x^2 + c)^(5/2), x)