3.693 \(\int \frac{e^{\tanh ^{-1}(a x)}}{(c-\frac{c}{a^2 x^2})^{5/2}} \, dx\)

Optimal. Leaf size=267 \[ -\frac{\left (1-a^2 x^2\right )^{5/2}}{a^6 x^5 (1-a x) \left (c-\frac{c}{a^2 x^2}\right )^{5/2}}+\frac{\left (1-a^2 x^2\right )^{5/2}}{8 a^6 x^5 (a x+1) \left (c-\frac{c}{a^2 x^2}\right )^{5/2}}+\frac{\left (1-a^2 x^2\right )^{5/2}}{8 a^6 x^5 (1-a x)^2 \left (c-\frac{c}{a^2 x^2}\right )^{5/2}}-\frac{\left (1-a^2 x^2\right )^{5/2}}{a^5 x^4 \left (c-\frac{c}{a^2 x^2}\right )^{5/2}}-\frac{23 \left (1-a^2 x^2\right )^{5/2} \log (1-a x)}{16 a^6 x^5 \left (c-\frac{c}{a^2 x^2}\right )^{5/2}}+\frac{7 \left (1-a^2 x^2\right )^{5/2} \log (a x+1)}{16 a^6 x^5 \left (c-\frac{c}{a^2 x^2}\right )^{5/2}} \]

[Out]

-((1 - a^2*x^2)^(5/2)/(a^5*(c - c/(a^2*x^2))^(5/2)*x^4)) + (1 - a^2*x^2)^(5/2)/(8*a^6*(c - c/(a^2*x^2))^(5/2)*
x^5*(1 - a*x)^2) - (1 - a^2*x^2)^(5/2)/(a^6*(c - c/(a^2*x^2))^(5/2)*x^5*(1 - a*x)) + (1 - a^2*x^2)^(5/2)/(8*a^
6*(c - c/(a^2*x^2))^(5/2)*x^5*(1 + a*x)) - (23*(1 - a^2*x^2)^(5/2)*Log[1 - a*x])/(16*a^6*(c - c/(a^2*x^2))^(5/
2)*x^5) + (7*(1 - a^2*x^2)^(5/2)*Log[1 + a*x])/(16*a^6*(c - c/(a^2*x^2))^(5/2)*x^5)

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Rubi [A]  time = 0.204935, antiderivative size = 267, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {6160, 6150, 88} \[ -\frac{\left (1-a^2 x^2\right )^{5/2}}{a^6 x^5 (1-a x) \left (c-\frac{c}{a^2 x^2}\right )^{5/2}}+\frac{\left (1-a^2 x^2\right )^{5/2}}{8 a^6 x^5 (a x+1) \left (c-\frac{c}{a^2 x^2}\right )^{5/2}}+\frac{\left (1-a^2 x^2\right )^{5/2}}{8 a^6 x^5 (1-a x)^2 \left (c-\frac{c}{a^2 x^2}\right )^{5/2}}-\frac{\left (1-a^2 x^2\right )^{5/2}}{a^5 x^4 \left (c-\frac{c}{a^2 x^2}\right )^{5/2}}-\frac{23 \left (1-a^2 x^2\right )^{5/2} \log (1-a x)}{16 a^6 x^5 \left (c-\frac{c}{a^2 x^2}\right )^{5/2}}+\frac{7 \left (1-a^2 x^2\right )^{5/2} \log (a x+1)}{16 a^6 x^5 \left (c-\frac{c}{a^2 x^2}\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcTanh[a*x]/(c - c/(a^2*x^2))^(5/2),x]

[Out]

-((1 - a^2*x^2)^(5/2)/(a^5*(c - c/(a^2*x^2))^(5/2)*x^4)) + (1 - a^2*x^2)^(5/2)/(8*a^6*(c - c/(a^2*x^2))^(5/2)*
x^5*(1 - a*x)^2) - (1 - a^2*x^2)^(5/2)/(a^6*(c - c/(a^2*x^2))^(5/2)*x^5*(1 - a*x)) + (1 - a^2*x^2)^(5/2)/(8*a^
6*(c - c/(a^2*x^2))^(5/2)*x^5*(1 + a*x)) - (23*(1 - a^2*x^2)^(5/2)*Log[1 - a*x])/(16*a^6*(c - c/(a^2*x^2))^(5/
2)*x^5) + (7*(1 - a^2*x^2)^(5/2)*Log[1 + a*x])/(16*a^6*(c - c/(a^2*x^2))^(5/2)*x^5)

Rule 6160

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_)^2)^(p_), x_Symbol] :> Dist[(x^(2*p)*(c + d/x^2)^p)/
(1 + (c*x^2)/d)^p, Int[(u*(1 + (c*x^2)/d)^p*E^(n*ArcTanh[a*x]))/x^(2*p), x], x] /; FreeQ[{a, c, d, n, p}, x] &
& EqQ[c + a^2*d, 0] &&  !IntegerQ[p] &&  !IntegerQ[n/2]

Rule 6150

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[c^p, Int[x^m*(1 -
a*x)^(p - n/2)*(1 + a*x)^(p + n/2), x], x] /; FreeQ[{a, c, d, m, n, p}, x] && EqQ[a^2*c + d, 0] && (IntegerQ[p
] || GtQ[c, 0])

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rubi steps

\begin{align*} \int \frac{e^{\tanh ^{-1}(a x)}}{\left (c-\frac{c}{a^2 x^2}\right )^{5/2}} \, dx &=\frac{\left (1-a^2 x^2\right )^{5/2} \int \frac{e^{\tanh ^{-1}(a x)} x^5}{\left (1-a^2 x^2\right )^{5/2}} \, dx}{\left (c-\frac{c}{a^2 x^2}\right )^{5/2} x^5}\\ &=\frac{\left (1-a^2 x^2\right )^{5/2} \int \frac{x^5}{(1-a x)^3 (1+a x)^2} \, dx}{\left (c-\frac{c}{a^2 x^2}\right )^{5/2} x^5}\\ &=\frac{\left (1-a^2 x^2\right )^{5/2} \int \left (-\frac{1}{a^5}-\frac{1}{4 a^5 (-1+a x)^3}-\frac{1}{a^5 (-1+a x)^2}-\frac{23}{16 a^5 (-1+a x)}-\frac{1}{8 a^5 (1+a x)^2}+\frac{7}{16 a^5 (1+a x)}\right ) \, dx}{\left (c-\frac{c}{a^2 x^2}\right )^{5/2} x^5}\\ &=-\frac{\left (1-a^2 x^2\right )^{5/2}}{a^5 \left (c-\frac{c}{a^2 x^2}\right )^{5/2} x^4}+\frac{\left (1-a^2 x^2\right )^{5/2}}{8 a^6 \left (c-\frac{c}{a^2 x^2}\right )^{5/2} x^5 (1-a x)^2}-\frac{\left (1-a^2 x^2\right )^{5/2}}{a^6 \left (c-\frac{c}{a^2 x^2}\right )^{5/2} x^5 (1-a x)}+\frac{\left (1-a^2 x^2\right )^{5/2}}{8 a^6 \left (c-\frac{c}{a^2 x^2}\right )^{5/2} x^5 (1+a x)}-\frac{23 \left (1-a^2 x^2\right )^{5/2} \log (1-a x)}{16 a^6 \left (c-\frac{c}{a^2 x^2}\right )^{5/2} x^5}+\frac{7 \left (1-a^2 x^2\right )^{5/2} \log (1+a x)}{16 a^6 \left (c-\frac{c}{a^2 x^2}\right )^{5/2} x^5}\\ \end{align*}

Mathematica [A]  time = 0.132773, size = 87, normalized size = 0.33 \[ \frac{\left (1-a^2 x^2\right )^{5/2} \left (2 \left (-8 a x+\frac{8}{a x-1}+\frac{1}{a x+1}+\frac{1}{(a x-1)^2}\right )-23 \log (1-a x)+7 \log (a x+1)\right )}{16 a^6 x^5 \left (c-\frac{c}{a^2 x^2}\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[E^ArcTanh[a*x]/(c - c/(a^2*x^2))^(5/2),x]

[Out]

((1 - a^2*x^2)^(5/2)*(2*(-8*a*x + (-1 + a*x)^(-2) + 8/(-1 + a*x) + (1 + a*x)^(-1)) - 23*Log[1 - a*x] + 7*Log[1
+ a*x]))/(16*a^6*(c - c/(a^2*x^2))^(5/2)*x^5)

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Maple [A]  time = 0.178, size = 167, normalized size = 0.6 \begin{align*}{\frac{ \left ( ax+1 \right ) \left ( -16\,{x}^{4}{a}^{4}+7\,{a}^{3}{x}^{3}\ln \left ( ax+1 \right ) -23\,\ln \left ( ax-1 \right ){x}^{3}{a}^{3}+16\,{x}^{3}{a}^{3}-7\,\ln \left ( ax+1 \right ){a}^{2}{x}^{2}+23\,\ln \left ( ax-1 \right ){a}^{2}{x}^{2}+34\,{a}^{2}{x}^{2}-7\,ax\ln \left ( ax+1 \right ) +23\,\ln \left ( ax-1 \right ) xa-18\,ax+7\,\ln \left ( ax+1 \right ) -23\,\ln \left ( ax-1 \right ) -12 \right ) }{16\,{a}^{6}{x}^{5}}\sqrt{-{a}^{2}{x}^{2}+1} \left ({\frac{c \left ({a}^{2}{x}^{2}-1 \right ) }{{a}^{2}{x}^{2}}} \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x+1)/(-a^2*x^2+1)^(1/2)/(c-c/a^2/x^2)^(5/2),x)

[Out]

1/16*(-a^2*x^2+1)^(1/2)*(a*x+1)*(-16*x^4*a^4+7*a^3*x^3*ln(a*x+1)-23*ln(a*x-1)*x^3*a^3+16*x^3*a^3-7*ln(a*x+1)*a
^2*x^2+23*ln(a*x-1)*a^2*x^2+34*a^2*x^2-7*a*x*ln(a*x+1)+23*ln(a*x-1)*x*a-18*a*x+7*ln(a*x+1)-23*ln(a*x-1)-12)/a^
6/x^5/(c*(a^2*x^2-1)/a^2/x^2)^(5/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{a x + 1}{\sqrt{-a^{2} x^{2} + 1}{\left (c - \frac{c}{a^{2} x^{2}}\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a^2*x^2+1)^(1/2)/(c-c/a^2/x^2)^(5/2),x, algorithm="maxima")

[Out]

integrate((a*x + 1)/(sqrt(-a^2*x^2 + 1)*(c - c/(a^2*x^2))^(5/2)), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{\sqrt{-a^{2} x^{2} + 1} a^{6} x^{6} \sqrt{\frac{a^{2} c x^{2} - c}{a^{2} x^{2}}}}{a^{7} c^{3} x^{7} - a^{6} c^{3} x^{6} - 3 \, a^{5} c^{3} x^{5} + 3 \, a^{4} c^{3} x^{4} + 3 \, a^{3} c^{3} x^{3} - 3 \, a^{2} c^{3} x^{2} - a c^{3} x + c^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a^2*x^2+1)^(1/2)/(c-c/a^2/x^2)^(5/2),x, algorithm="fricas")

[Out]

integral(-sqrt(-a^2*x^2 + 1)*a^6*x^6*sqrt((a^2*c*x^2 - c)/(a^2*x^2))/(a^7*c^3*x^7 - a^6*c^3*x^6 - 3*a^5*c^3*x^
5 + 3*a^4*c^3*x^4 + 3*a^3*c^3*x^3 - 3*a^2*c^3*x^2 - a*c^3*x + c^3), x)

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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((a*x+1)/(-a**2*x**2+1)**(1/2)/(c-c/a**2/x**2)**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{a x + 1}{\sqrt{-a^{2} x^{2} + 1}{\left (c - \frac{c}{a^{2} x^{2}}\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a^2*x^2+1)^(1/2)/(c-c/a^2/x^2)^(5/2),x, algorithm="giac")

[Out]

integrate((a*x + 1)/(sqrt(-a^2*x^2 + 1)*(c - c/(a^2*x^2))^(5/2)), x)