### 3.692 $$\int \frac{e^{\coth ^{-1}(a x)} x^3}{(c-a^2 c x^2)^{3/2}} \, dx$$

Optimal. Leaf size=172 $\frac{x^4 \left (1-\frac{1}{a^2 x^2}\right )^{3/2}}{\left (c-a^2 c x^2\right )^{3/2}}+\frac{x^3 \left (1-\frac{1}{a^2 x^2}\right )^{3/2}}{2 a (1-a x) \left (c-a^2 c x^2\right )^{3/2}}+\frac{5 x^3 \left (1-\frac{1}{a^2 x^2}\right )^{3/2} \log (1-a x)}{4 a \left (c-a^2 c x^2\right )^{3/2}}-\frac{x^3 \left (1-\frac{1}{a^2 x^2}\right )^{3/2} \log (a x+1)}{4 a \left (c-a^2 c x^2\right )^{3/2}}$

[Out]

((1 - 1/(a^2*x^2))^(3/2)*x^4)/(c - a^2*c*x^2)^(3/2) + ((1 - 1/(a^2*x^2))^(3/2)*x^3)/(2*a*(1 - a*x)*(c - a^2*c*
x^2)^(3/2)) + (5*(1 - 1/(a^2*x^2))^(3/2)*x^3*Log[1 - a*x])/(4*a*(c - a^2*c*x^2)^(3/2)) - ((1 - 1/(a^2*x^2))^(3
/2)*x^3*Log[1 + a*x])/(4*a*(c - a^2*c*x^2)^(3/2))

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Rubi [A]  time = 0.199598, antiderivative size = 172, 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 = {6192, 6193, 88} $\frac{x^4 \left (1-\frac{1}{a^2 x^2}\right )^{3/2}}{\left (c-a^2 c x^2\right )^{3/2}}+\frac{x^3 \left (1-\frac{1}{a^2 x^2}\right )^{3/2}}{2 a (1-a x) \left (c-a^2 c x^2\right )^{3/2}}+\frac{5 x^3 \left (1-\frac{1}{a^2 x^2}\right )^{3/2} \log (1-a x)}{4 a \left (c-a^2 c x^2\right )^{3/2}}-\frac{x^3 \left (1-\frac{1}{a^2 x^2}\right )^{3/2} \log (a x+1)}{4 a \left (c-a^2 c x^2\right )^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

((1 - 1/(a^2*x^2))^(3/2)*x^4)/(c - a^2*c*x^2)^(3/2) + ((1 - 1/(a^2*x^2))^(3/2)*x^3)/(2*a*(1 - a*x)*(c - a^2*c*
x^2)^(3/2)) + (5*(1 - 1/(a^2*x^2))^(3/2)*x^3*Log[1 - a*x])/(4*a*(c - a^2*c*x^2)^(3/2)) - ((1 - 1/(a^2*x^2))^(3
/2)*x^3*Log[1 + a*x])/(4*a*(c - a^2*c*x^2)^(3/2))

Rule 6192

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

Rule 6193

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

Rubi steps

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

Mathematica [A]  time = 0.0550632, size = 71, normalized size = 0.41 $\frac{x^3 \left (1-\frac{1}{a^2 x^2}\right )^{3/2} \left (4 a x+\frac{2}{1-a x}+5 \log (1-a x)-\log (a x+1)\right )}{4 a \left (c-a^2 c x^2\right )^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

((1 - 1/(a^2*x^2))^(3/2)*x^3*(4*a*x + 2/(1 - a*x) + 5*Log[1 - a*x] - Log[1 + a*x]))/(4*a*(c - a^2*c*x^2)^(3/2)
)

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Maple [A]  time = 0.152, size = 98, normalized size = 0.6 \begin{align*} -{\frac{-4\,{a}^{2}{x}^{2}+ax\ln \left ( ax+1 \right ) -5\,\ln \left ( ax-1 \right ) xa+4\,ax-\ln \left ( ax+1 \right ) +5\,\ln \left ( ax-1 \right ) +2}{ \left ( 4\,{a}^{2}{x}^{2}-4 \right ){c}^{2}{a}^{4}}\sqrt{-c \left ({a}^{2}{x}^{2}-1 \right ) }{\frac{1}{\sqrt{{\frac{ax-1}{ax+1}}}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4/((a*x-1)/(a*x+1))^(1/2)*(-c*(a^2*x^2-1))^(1/2)*(-4*a^2*x^2+a*x*ln(a*x+1)-5*ln(a*x-1)*x*a+4*a*x-ln(a*x+1)+
5*ln(a*x-1)+2)/(a^2*x^2-1)/c^2/a^4

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.59521, size = 157, normalized size = 0.91 \begin{align*} \frac{{\left (4 \, a^{2} x^{2} - 4 \, a x -{\left (a x - 1\right )} \log \left (a x + 1\right ) + 5 \,{\left (a x - 1\right )} \log \left (a x - 1\right ) - 2\right )} \sqrt{-a^{2} c}}{4 \,{\left (a^{6} c^{2} x - a^{5} c^{2}\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(4*a^2*x^2 - 4*a*x - (a*x - 1)*log(a*x + 1) + 5*(a*x - 1)*log(a*x - 1) - 2)*sqrt(-a^2*c)/(a^6*c^2*x - a^5*
c^2)

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

[Out]

Timed out

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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