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

Optimal. Leaf size=35 \[ \frac{(a x+1) e^{\tan ^{-1}(a x)}}{2 a c \sqrt{a^2 c x^2+c}} \]

[Out]

(E^ArcTan[a*x]*(1 + a*x))/(2*a*c*Sqrt[c + a^2*c*x^2])

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Rubi [A]  time = 0.0361105, antiderivative size = 35, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.048, Rules used = {5069} \[ \frac{(a x+1) e^{\tan ^{-1}(a x)}}{2 a c \sqrt{a^2 c x^2+c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(E^ArcTan[a*x]*(1 + a*x))/(2*a*c*Sqrt[c + a^2*c*x^2])

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

Mathematica [A]  time = 0.0148304, size = 35, normalized size = 1. \[ \frac{(a x+1) e^{\tan ^{-1}(a x)}}{2 a c \sqrt{a^2 c x^2+c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(E^ArcTan[a*x]*(1 + a*x))/(2*a*c*Sqrt[c + a^2*c*x^2])

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Maple [A]  time = 0.038, size = 37, normalized size = 1.1 \begin{align*}{\frac{ \left ({a}^{2}{x}^{2}+1 \right ) \left ( ax+1 \right ){{\rm e}^{\arctan \left ( ax \right ) }}}{2\,a} \left ({a}^{2}c{x}^{2}+c \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(a^2*x^2+1)*(a*x+1)*exp(arctan(a*x))/a/(a^2*c*x^2+c)^(3/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{e^{\left (\arctan \left (a x\right )\right )}}{{\left (a^{2} c x^{2} + c\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.87468, size = 99, normalized size = 2.83 \begin{align*} \frac{\sqrt{a^{2} c x^{2} + c}{\left (a x + 1\right )} e^{\left (\arctan \left (a x\right )\right )}}{2 \,{\left (a^{3} c^{2} x^{2} + a c^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(a^2*c*x^2 + c)*(a*x + 1)*e^(arctan(a*x))/(a^3*c^2*x^2 + a*c^2)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{e^{\operatorname{atan}{\left (a x \right )}}}{\left (c \left (a^{2} x^{2} + 1\right )\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{e^{\left (\arctan \left (a x\right )\right )}}{{\left (a^{2} c x^{2} + c\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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