3.969 \(\int \frac{e^{\tanh ^{-1}(a x)} x^2}{(c-a^2 c x^2)^{3/2}} \, dx\)

Optimal. Leaf size=139 \[ \frac{\sqrt{1-a^2 x^2}}{2 a^3 c (1-a x) \sqrt{c-a^2 c x^2}}+\frac{3 \sqrt{1-a^2 x^2} \log (1-a x)}{4 a^3 c \sqrt{c-a^2 c x^2}}+\frac{\sqrt{1-a^2 x^2} \log (a x+1)}{4 a^3 c \sqrt{c-a^2 c x^2}} \]

[Out]

Sqrt[1 - a^2*x^2]/(2*a^3*c*(1 - a*x)*Sqrt[c - a^2*c*x^2]) + (3*Sqrt[1 - a^2*x^2]*Log[1 - a*x])/(4*a^3*c*Sqrt[c
- a^2*c*x^2]) + (Sqrt[1 - a^2*x^2]*Log[1 + a*x])/(4*a^3*c*Sqrt[c - a^2*c*x^2])

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Rubi [A]  time = 0.221601, antiderivative size = 139, 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} \[ \frac{\sqrt{1-a^2 x^2}}{2 a^3 c (1-a x) \sqrt{c-a^2 c x^2}}+\frac{3 \sqrt{1-a^2 x^2} \log (1-a x)}{4 a^3 c \sqrt{c-a^2 c x^2}}+\frac{\sqrt{1-a^2 x^2} \log (a x+1)}{4 a^3 c \sqrt{c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(E^ArcTanh[a*x]*x^2)/(c - a^2*c*x^2)^(3/2),x]

[Out]

Sqrt[1 - a^2*x^2]/(2*a^3*c*(1 - a*x)*Sqrt[c - a^2*c*x^2]) + (3*Sqrt[1 - a^2*x^2]*Log[1 - a*x])/(4*a^3*c*Sqrt[c
- a^2*c*x^2]) + (Sqrt[1 - a^2*x^2]*Log[1 + a*x])/(4*a^3*c*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^2}{\left (c-a^2 c x^2\right )^{3/2}} \, dx &=\frac{\sqrt{1-a^2 x^2} \int \frac{e^{\tanh ^{-1}(a x)} x^2}{\left (1-a^2 x^2\right )^{3/2}} \, dx}{c \sqrt{c-a^2 c x^2}}\\ &=\frac{\sqrt{1-a^2 x^2} \int \frac{x^2}{(1-a x)^2 (1+a x)} \, dx}{c \sqrt{c-a^2 c x^2}}\\ &=\frac{\sqrt{1-a^2 x^2} \int \left (\frac{1}{2 a^2 (-1+a x)^2}+\frac{3}{4 a^2 (-1+a x)}+\frac{1}{4 a^2 (1+a x)}\right ) \, dx}{c \sqrt{c-a^2 c x^2}}\\ &=\frac{\sqrt{1-a^2 x^2}}{2 a^3 c (1-a x) \sqrt{c-a^2 c x^2}}+\frac{3 \sqrt{1-a^2 x^2} \log (1-a x)}{4 a^3 c \sqrt{c-a^2 c x^2}}+\frac{\sqrt{1-a^2 x^2} \log (1+a x)}{4 a^3 c \sqrt{c-a^2 c x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0405422, size = 76, normalized size = 0.55 \[ \frac{\sqrt{1-a^2 x^2} \left (\frac{1}{2 a^3 (1-a x)}+\frac{3 \log (1-a x)}{4 a^3}+\frac{\log (a x+1)}{4 a^3}\right )}{c \sqrt{c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(E^ArcTanh[a*x]*x^2)/(c - a^2*c*x^2)^(3/2),x]

[Out]

(Sqrt[1 - a^2*x^2]*(1/(2*a^3*(1 - a*x)) + (3*Log[1 - a*x])/(4*a^3) + Log[1 + a*x]/(4*a^3)))/(c*Sqrt[c - a^2*c*
x^2])

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Maple [A]  time = 0.095, size = 90, normalized size = 0.7 \begin{align*} -{\frac{ax\ln \left ( ax+1 \right ) +3\,\ln \left ( ax-1 \right ) xa-\ln \left ( ax+1 \right ) -3\,\ln \left ( ax-1 \right ) -2}{ \left ( 4\,{a}^{2}{x}^{2}-4 \right ){c}^{2}{a}^{3} \left ( ax-1 \right ) }\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^2/(-a^2*c*x^2+c)^(3/2),x)

[Out]

-1/4*(-a^2*x^2+1)^(1/2)*(-c*(a^2*x^2-1))^(1/2)*(a*x*ln(a*x+1)+3*ln(a*x-1)*x*a-ln(a*x+1)-3*ln(a*x-1)-2)/(a^2*x^
2-1)/c^2/a^3/(a*x-1)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (a x + 1\right )} x^{2}}{{\left (-a^{2} c x^{2} + c\right )}^{\frac{3}{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^2/(-a^2*c*x^2+c)^(3/2),x, algorithm="maxima")

[Out]

integrate((a*x + 1)*x^2/((-a^2*c*x^2 + c)^(3/2)*sqrt(-a^2*x^2 + 1)), x)

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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^{2}}{a^{5} c^{2} x^{5} - a^{4} c^{2} x^{4} - 2 \, a^{3} c^{2} x^{3} + 2 \, a^{2} c^{2} x^{2} + a c^{2} x - c^{2}}, 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^2/(-a^2*c*x^2+c)^(3/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{2} \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{3}{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**2/(-a**2*c*x**2+c)**(3/2),x)

[Out]

Integral(x**2*(a*x + 1)/(sqrt(-(a*x - 1)*(a*x + 1))*(-c*(a*x - 1)*(a*x + 1))**(3/2)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (a x + 1\right )} x^{2}}{{\left (-a^{2} c x^{2} + c\right )}^{\frac{3}{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^2/(-a^2*c*x^2+c)^(3/2),x, algorithm="giac")

[Out]

integrate((a*x + 1)*x^2/((-a^2*c*x^2 + c)^(3/2)*sqrt(-a^2*x^2 + 1)), x)