3.1376 \(\int \frac {e^{-\tanh ^{-1}(a x)} x^2}{(c-a^2 c x^2)^{3/2}} \, dx\)

Optimal. Leaf size=138 \[ -\frac {\sqrt {1-a^2 x^2}}{2 a^3 c (a x+1) \sqrt {c-a^2 c x^2}}-\frac {\sqrt {1-a^2 x^2} \log (1-a x)}{4 a^3 c \sqrt {c-a^2 c x^2}}-\frac {3 \sqrt {1-a^2 x^2} \log (a x+1)}{4 a^3 c \sqrt {c-a^2 c x^2}} \]

[Out]

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

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Rubi [A]  time = 0.26, antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {6153, 6150, 88} \[ -\frac {\sqrt {1-a^2 x^2}}{2 a^3 c (a x+1) \sqrt {c-a^2 c x^2}}-\frac {\sqrt {1-a^2 x^2} \log (1-a x)}{4 a^3 c \sqrt {c-a^2 c x^2}}-\frac {3 \sqrt {1-a^2 x^2} \log (a x+1)}{4 a^3 c \sqrt {c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

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

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]))

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 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]

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) (1+a x)^2} \, dx}{c \sqrt {c-a^2 c x^2}}\\ &=\frac {\sqrt {1-a^2 x^2} \int \left (-\frac {1}{4 a^2 (-1+a x)}+\frac {1}{2 a^2 (1+a x)^2}-\frac {3}{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 {\sqrt {1-a^2 x^2} \log (1-a x)}{4 a^3 c \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}}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 72, normalized size = 0.52 \[ -\frac {\sqrt {1-a^2 x^2} ((a x+1) \log (1-a x)+3 (a x+1) \log (a x+1)+2)}{4 a^3 (a c x+c) \sqrt {c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.76, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1} x^{2}}{a^{5} c^{2} x^{5} + a^{4} c^{2} x^{4} - 2 \, a^{3} c^{2} x^{3} - 2 \, a^{2} c^{2} x^{2} + a c^{2} x + c^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(a*x+1)*(-a^2*x^2+1)^(1/2)/(-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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {-a^{2} x^{2} + 1} x^{2}}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}} {\left (a x + 1\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(a*x+1)*(-a^2*x^2+1)^(1/2)/(-a^2*c*x^2+c)^(3/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.05, size = 88, normalized size = 0.64 \[ \frac {\sqrt {-a^{2} x^{2}+1}\, \sqrt {-c \left (a^{2} x^{2}-1\right )}\, \left (\ln \left (a x -1\right ) x a +3 a x \ln \left (a x +1\right )+\ln \left (a x -1\right )+3 \ln \left (a x +1\right )+2\right )}{4 \left (a^{2} x^{2}-1\right ) c^{2} a^{3} \left (a x +1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(a*x+1)*(-a^2*x^2+1)^(1/2)/(-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)*(ln(a*x-1)*x*a+3*a*x*ln(a*x+1)+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 [A]  time = 0.35, size = 52, normalized size = 0.38 \[ -\frac {\sqrt {c}}{2 \, {\left (a^{4} c^{2} x + a^{3} c^{2}\right )}} - \frac {3 \, \log \left (a x + 1\right )}{4 \, a^{3} c^{\frac {3}{2}}} - \frac {\log \left (a x - 1\right )}{4 \, a^{3} c^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(a*x+1)*(-a^2*x^2+1)^(1/2)/(-a^2*c*x^2+c)^(3/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^2\,\sqrt {1-a^2\,x^2}}{{\left (c-a^2\,c\,x^2\right )}^{3/2}\,\left (a\,x+1\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2} \sqrt {- \left (a x - 1\right ) \left (a x + 1\right )}}{\left (- c \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {3}{2}} \left (a x + 1\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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