3.272 \(\int \frac{e^{2 \tan ^{-1}(a x)}}{(c+a^2 c x^2)^{7/2}} \, dx\)

Optimal. Leaf size=114 \[ \frac{24 (a x+2) e^{2 \tan ^{-1}(a x)}}{377 a c^3 \sqrt{a^2 c x^2+c}}+\frac{20 (3 a x+2) e^{2 \tan ^{-1}(a x)}}{377 a c^2 \left (a^2 c x^2+c\right )^{3/2}}+\frac{(5 a x+2) e^{2 \tan ^{-1}(a x)}}{29 a c \left (a^2 c x^2+c\right )^{5/2}} \]

[Out]

(E^(2*ArcTan[a*x])*(2 + 5*a*x))/(29*a*c*(c + a^2*c*x^2)^(5/2)) + (20*E^(2*ArcTan[a*x])*(2 + 3*a*x))/(377*a*c^2
*(c + a^2*c*x^2)^(3/2)) + (24*E^(2*ArcTan[a*x])*(2 + a*x))/(377*a*c^3*Sqrt[c + a^2*c*x^2])

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Rubi [A]  time = 0.12535, antiderivative size = 114, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.087, Rules used = {5070, 5069} \[ \frac{24 (a x+2) e^{2 \tan ^{-1}(a x)}}{377 a c^3 \sqrt{a^2 c x^2+c}}+\frac{20 (3 a x+2) e^{2 \tan ^{-1}(a x)}}{377 a c^2 \left (a^2 c x^2+c\right )^{3/2}}+\frac{(5 a x+2) e^{2 \tan ^{-1}(a x)}}{29 a c \left (a^2 c x^2+c\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[E^(2*ArcTan[a*x])/(c + a^2*c*x^2)^(7/2),x]

[Out]

(E^(2*ArcTan[a*x])*(2 + 5*a*x))/(29*a*c*(c + a^2*c*x^2)^(5/2)) + (20*E^(2*ArcTan[a*x])*(2 + 3*a*x))/(377*a*c^2
*(c + a^2*c*x^2)^(3/2)) + (24*E^(2*ArcTan[a*x])*(2 + a*x))/(377*a*c^3*Sqrt[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 )^{7/2}} \, dx &=\frac{e^{2 \tan ^{-1}(a x)} (2+5 a x)}{29 a c \left (c+a^2 c x^2\right )^{5/2}}+\frac{20 \int \frac{e^{2 \tan ^{-1}(a x)}}{\left (c+a^2 c x^2\right )^{5/2}} \, dx}{29 c}\\ &=\frac{e^{2 \tan ^{-1}(a x)} (2+5 a x)}{29 a c \left (c+a^2 c x^2\right )^{5/2}}+\frac{20 e^{2 \tan ^{-1}(a x)} (2+3 a x)}{377 a c^2 \left (c+a^2 c x^2\right )^{3/2}}+\frac{120 \int \frac{e^{2 \tan ^{-1}(a x)}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx}{377 c^2}\\ &=\frac{e^{2 \tan ^{-1}(a x)} (2+5 a x)}{29 a c \left (c+a^2 c x^2\right )^{5/2}}+\frac{20 e^{2 \tan ^{-1}(a x)} (2+3 a x)}{377 a c^2 \left (c+a^2 c x^2\right )^{3/2}}+\frac{24 e^{2 \tan ^{-1}(a x)} (2+a x)}{377 a c^3 \sqrt{c+a^2 c x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0378807, size = 81, normalized size = 0.71 \[ \frac{\left (24 a^5 x^5+48 a^4 x^4+108 a^3 x^3+136 a^2 x^2+149 a x+114\right ) e^{2 \tan ^{-1}(a x)}}{377 a c^3 \left (a^2 x^2+1\right )^2 \sqrt{a^2 c x^2+c}} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(2*ArcTan[a*x])/(c + a^2*c*x^2)^(7/2),x]

[Out]

(E^(2*ArcTan[a*x])*(114 + 149*a*x + 136*a^2*x^2 + 108*a^3*x^3 + 48*a^4*x^4 + 24*a^5*x^5))/(377*a*c^3*(1 + a^2*
x^2)^2*Sqrt[c + a^2*c*x^2])

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Maple [A]  time = 0.037, size = 72, normalized size = 0.6 \begin{align*}{\frac{ \left ({a}^{2}{x}^{2}+1 \right ) \left ( 24\,{a}^{5}{x}^{5}+48\,{a}^{4}{x}^{4}+108\,{a}^{3}{x}^{3}+136\,{a}^{2}{x}^{2}+149\,ax+114 \right ){{\rm e}^{2\,\arctan \left ( ax \right ) }}}{377\,a} \left ({a}^{2}c{x}^{2}+c \right ) ^{-{\frac{7}{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)^(7/2),x)

[Out]

1/377*(a^2*x^2+1)*(24*a^5*x^5+48*a^4*x^4+108*a^3*x^3+136*a^2*x^2+149*a*x+114)*exp(2*arctan(a*x))/a/(a^2*c*x^2+
c)^(7/2)

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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{7}{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)^(7/2),x, algorithm="maxima")

[Out]

integrate(e^(2*arctan(a*x))/(a^2*c*x^2 + c)^(7/2), x)

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Fricas [A]  time = 1.86729, size = 228, normalized size = 2. \begin{align*} \frac{{\left (24 \, a^{5} x^{5} + 48 \, a^{4} x^{4} + 108 \, a^{3} x^{3} + 136 \, a^{2} x^{2} + 149 \, a x + 114\right )} \sqrt{a^{2} c x^{2} + c} e^{\left (2 \, \arctan \left (a x\right )\right )}}{377 \,{\left (a^{7} c^{4} x^{6} + 3 \, a^{5} c^{4} x^{4} + 3 \, a^{3} c^{4} x^{2} + a c^{4}\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)^(7/2),x, algorithm="fricas")

[Out]

1/377*(24*a^5*x^5 + 48*a^4*x^4 + 108*a^3*x^3 + 136*a^2*x^2 + 149*a*x + 114)*sqrt(a^2*c*x^2 + c)*e^(2*arctan(a*
x))/(a^7*c^4*x^6 + 3*a^5*c^4*x^4 + 3*a^3*c^4*x^2 + a*c^4)

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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)**(7/2),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{7}{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)^(7/2),x, algorithm="giac")

[Out]

integrate(e^(2*arctan(a*x))/(a^2*c*x^2 + c)^(7/2), x)