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

Optimal. Leaf size=268 \[ -\frac{x \sqrt{1-a^2 x^2}}{a^5 c^2 \sqrt{c-a^2 c x^2}}-\frac{\sqrt{1-a^2 x^2}}{a^6 c^2 (1-a x) \sqrt{c-a^2 c x^2}}+\frac{\sqrt{1-a^2 x^2}}{8 a^6 c^2 (a x+1) \sqrt{c-a^2 c x^2}}+\frac{\sqrt{1-a^2 x^2}}{8 a^6 c^2 (1-a x)^2 \sqrt{c-a^2 c x^2}}-\frac{23 \sqrt{1-a^2 x^2} \log (1-a x)}{16 a^6 c^2 \sqrt{c-a^2 c x^2}}+\frac{7 \sqrt{1-a^2 x^2} \log (a x+1)}{16 a^6 c^2 \sqrt{c-a^2 c x^2}} \]

[Out]

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

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Rubi [A]  time = 0.246031, antiderivative size = 268, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {6153, 6150, 88} \[ -\frac{x \sqrt{1-a^2 x^2}}{a^5 c^2 \sqrt{c-a^2 c x^2}}-\frac{\sqrt{1-a^2 x^2}}{a^6 c^2 (1-a x) \sqrt{c-a^2 c x^2}}+\frac{\sqrt{1-a^2 x^2}}{8 a^6 c^2 (a x+1) \sqrt{c-a^2 c x^2}}+\frac{\sqrt{1-a^2 x^2}}{8 a^6 c^2 (1-a x)^2 \sqrt{c-a^2 c x^2}}-\frac{23 \sqrt{1-a^2 x^2} \log (1-a x)}{16 a^6 c^2 \sqrt{c-a^2 c x^2}}+\frac{7 \sqrt{1-a^2 x^2} \log (a x+1)}{16 a^6 c^2 \sqrt{c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6153

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(c^IntPart[p]*(c +
d*x^2)^FracPart[p])/(1 - a^2*x^2)^FracPart[p], Int[x^m*(1 - a^2*x^2)^p*E^(n*ArcTanh[a*x]), x], x] /; FreeQ[{a,
c, d, m, n, p}, x] && EqQ[a^2*c + d, 0] &&  !(IntegerQ[p] || GtQ[c, 0]) &&  !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)} x^5}{\left (c-a^2 c x^2\right )^{5/2}} \, dx &=\frac{\sqrt{1-a^2 x^2} \int \frac{e^{\tanh ^{-1}(a x)} x^5}{\left (1-a^2 x^2\right )^{5/2}} \, dx}{c^2 \sqrt{c-a^2 c x^2}}\\ &=\frac{\sqrt{1-a^2 x^2} \int \frac{x^5}{(1-a x)^3 (1+a x)^2} \, dx}{c^2 \sqrt{c-a^2 c x^2}}\\ &=\frac{\sqrt{1-a^2 x^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}{c^2 \sqrt{c-a^2 c x^2}}\\ &=-\frac{x \sqrt{1-a^2 x^2}}{a^5 c^2 \sqrt{c-a^2 c x^2}}+\frac{\sqrt{1-a^2 x^2}}{8 a^6 c^2 (1-a x)^2 \sqrt{c-a^2 c x^2}}-\frac{\sqrt{1-a^2 x^2}}{a^6 c^2 (1-a x) \sqrt{c-a^2 c x^2}}+\frac{\sqrt{1-a^2 x^2}}{8 a^6 c^2 (1+a x) \sqrt{c-a^2 c x^2}}-\frac{23 \sqrt{1-a^2 x^2} \log (1-a x)}{16 a^6 c^2 \sqrt{c-a^2 c x^2}}+\frac{7 \sqrt{1-a^2 x^2} \log (1+a x)}{16 a^6 c^2 \sqrt{c-a^2 c x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0984278, size = 87, normalized size = 0.32 \[ \frac{\sqrt{1-a^2 x^2} \left (2 \left (-8 a x+\frac{8}{a x-1}+\frac{1}{a x+1}+\frac{1}{(a x-1)^2}\right )-23 \log (1-a x)+7 \log (a x+1)\right )}{16 a^6 c^2 \sqrt{c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[1 - a^2*x^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^2*Sqrt[c - a^2*c*x^2])

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Maple [A]  time = 0.095, size = 182, normalized size = 0.7 \begin{align*} -{\frac{-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}{ \left ( 16\,{a}^{2}{x}^{2}-16 \right ){c}^{3}{a}^{6} \left ( ax+1 \right ) \left ( ax-1 \right ) ^{2}}\sqrt{-{a}^{2}{x}^{2}+1}\sqrt{-c \left ({a}^{2}{x}^{2}-1 \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(-a^2*x^2+1)^(1/2)*(-c*(a^2*x^2-1))^(1/2)*(-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^2*x^2-1)/c^3/a^6/(a*x+1)/(a*x-1)^2

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a \int -\frac{x^{6}}{{\left (a^{4} c^{\frac{5}{2}} x^{4} - 2 \, a^{2} c^{\frac{5}{2}} x^{2} + c^{\frac{5}{2}}\right )}{\left (a x + 1\right )}{\left (a x - 1\right )}}\,{d x} + \frac{1}{4 \,{\left (a^{10} c^{\frac{5}{2}} x^{4} - 2 \, a^{8} c^{\frac{5}{2}} x^{2} + a^{6} c^{\frac{5}{2}}\right )}} + \frac{1}{a^{8} c^{\frac{5}{2}} x^{2} - a^{6} c^{\frac{5}{2}}} - \frac{\log \left (-a^{2} c x^{2} + c\right )}{2 \, a^{6} c^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*integrate(-x^6/((a^4*c^(5/2)*x^4 - 2*a^2*c^(5/2)*x^2 + c^(5/2))*(a*x + 1)*(a*x - 1)), x) + 1/4/(a^10*c^(5/2)
*x^4 - 2*a^8*c^(5/2)*x^2 + a^6*c^(5/2)) + 1/(a^8*c^(5/2)*x^2 - a^6*c^(5/2)) - 1/2*log(-a^2*c*x^2 + c)/(a^6*c^(
5/2))

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{-a^{2} c x^{2} + c} \sqrt{-a^{2} x^{2} + 1} x^{5}}{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)*x^5/(-a^2*c*x^2+c)^(5/2),x, algorithm="fricas")

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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