### 3.691 $$\int \frac{e^{\coth ^{-1}(a x)} x^4}{(c-a^2 c x^2)^{3/2}} \, dx$$

Optimal. Leaf size=211 $\frac{x^5 \left (1-\frac{1}{a^2 x^2}\right )^{3/2}}{2 \left (c-a^2 c x^2\right )^{3/2}}+\frac{x^4 \left (1-\frac{1}{a^2 x^2}\right )^{3/2}}{a \left (c-a^2 c x^2\right )^{3/2}}+\frac{x^3 \left (1-\frac{1}{a^2 x^2}\right )^{3/2}}{2 a^2 (1-a x) \left (c-a^2 c x^2\right )^{3/2}}+\frac{7 x^3 \left (1-\frac{1}{a^2 x^2}\right )^{3/2} \log (1-a x)}{4 a^2 \left (c-a^2 c x^2\right )^{3/2}}+\frac{x^3 \left (1-\frac{1}{a^2 x^2}\right )^{3/2} \log (a x+1)}{4 a^2 \left (c-a^2 c x^2\right )^{3/2}}$

[Out]

((1 - 1/(a^2*x^2))^(3/2)*x^4)/(a*(c - a^2*c*x^2)^(3/2)) + ((1 - 1/(a^2*x^2))^(3/2)*x^5)/(2*(c - a^2*c*x^2)^(3/
2)) + ((1 - 1/(a^2*x^2))^(3/2)*x^3)/(2*a^2*(1 - a*x)*(c - a^2*c*x^2)^(3/2)) + (7*(1 - 1/(a^2*x^2))^(3/2)*x^3*L
og[1 - a*x])/(4*a^2*(c - a^2*c*x^2)^(3/2)) + ((1 - 1/(a^2*x^2))^(3/2)*x^3*Log[1 + a*x])/(4*a^2*(c - a^2*c*x^2)
^(3/2))

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Rubi [A]  time = 0.230426, antiderivative size = 211, 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} $\frac{x^5 \left (1-\frac{1}{a^2 x^2}\right )^{3/2}}{2 \left (c-a^2 c x^2\right )^{3/2}}+\frac{x^4 \left (1-\frac{1}{a^2 x^2}\right )^{3/2}}{a \left (c-a^2 c x^2\right )^{3/2}}+\frac{x^3 \left (1-\frac{1}{a^2 x^2}\right )^{3/2}}{2 a^2 (1-a x) \left (c-a^2 c x^2\right )^{3/2}}+\frac{7 x^3 \left (1-\frac{1}{a^2 x^2}\right )^{3/2} \log (1-a x)}{4 a^2 \left (c-a^2 c x^2\right )^{3/2}}+\frac{x^3 \left (1-\frac{1}{a^2 x^2}\right )^{3/2} \log (a x+1)}{4 a^2 \left (c-a^2 c x^2\right )^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

Int[(E^ArcCoth[a*x]*x^4)/(c - a^2*c*x^2)^(3/2),x]

[Out]

((1 - 1/(a^2*x^2))^(3/2)*x^4)/(a*(c - a^2*c*x^2)^(3/2)) + ((1 - 1/(a^2*x^2))^(3/2)*x^5)/(2*(c - a^2*c*x^2)^(3/
2)) + ((1 - 1/(a^2*x^2))^(3/2)*x^3)/(2*a^2*(1 - a*x)*(c - a^2*c*x^2)^(3/2)) + (7*(1 - 1/(a^2*x^2))^(3/2)*x^3*L
og[1 - a*x])/(4*a^2*(c - a^2*c*x^2)^(3/2)) + ((1 - 1/(a^2*x^2))^(3/2)*x^3*Log[1 + a*x])/(4*a^2*(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^4}{\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)} 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^4}{(-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^4}+\frac{x}{a^3}+\frac{1}{2 a^4 (-1+a x)^2}+\frac{7}{4 a^4 (-1+a x)}+\frac{1}{4 a^4 (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}{a \left (c-a^2 c x^2\right )^{3/2}}+\frac{\left (1-\frac{1}{a^2 x^2}\right )^{3/2} x^5}{2 \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^2 (1-a x) \left (c-a^2 c x^2\right )^{3/2}}+\frac{7 \left (1-\frac{1}{a^2 x^2}\right )^{3/2} x^3 \log (1-a x)}{4 a^2 \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^2 \left (c-a^2 c x^2\right )^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.0815126, size = 77, normalized size = 0.36 $\frac{x^3 \left (1-\frac{1}{a^2 x^2}\right )^{3/2} \left (2 \left (a^2 x^2+2 a x+\frac{1}{1-a x}\right )+7 \log (1-a x)+\log (a x+1)\right )}{4 a^2 \left (c-a^2 c x^2\right )^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(E^ArcCoth[a*x]*x^4)/(c - a^2*c*x^2)^(3/2),x]

[Out]

((1 - 1/(a^2*x^2))^(3/2)*x^3*(2*(2*a*x + a^2*x^2 + (1 - a*x)^(-1)) + 7*Log[1 - a*x] + Log[1 + a*x]))/(4*a^2*(c
- a^2*c*x^2)^(3/2))

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Maple [A]  time = 0.16, size = 106, normalized size = 0.5 \begin{align*}{\frac{2\,{x}^{3}{a}^{3}+2\,{a}^{2}{x}^{2}+ax\ln \left ( ax+1 \right ) +7\,\ln \left ( ax-1 \right ) xa-4\,ax-\ln \left ( ax+1 \right ) -7\,\ln \left ( ax-1 \right ) -2}{ \left ( 4\,{a}^{2}{x}^{2}-4 \right ){c}^{2}{a}^{5}}\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^4/(-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)*(2*x^3*a^3+2*a^2*x^2+a*x*ln(a*x+1)+7*ln(a*x-1)*x*a-4*a*x-ln
(a*x+1)-7*ln(a*x-1)-2)/(a^2*x^2-1)/c^2/a^5

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{4}}{{\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^4/(-a^2*c*x^2+c)^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.73703, size = 173, normalized size = 0.82 \begin{align*} \frac{{\left (2 \, a^{3} x^{3} + 2 \, a^{2} x^{2} - 4 \, a x +{\left (a x - 1\right )} \log \left (a x + 1\right ) + 7 \,{\left (a x - 1\right )} \log \left (a x - 1\right ) - 2\right )} \sqrt{-a^{2} c}}{4 \,{\left (a^{7} c^{2} x - a^{6} 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^4/(-a^2*c*x^2+c)^(3/2),x, algorithm="fricas")

[Out]

1/4*(2*a^3*x^3 + 2*a^2*x^2 - 4*a*x + (a*x - 1)*log(a*x + 1) + 7*(a*x - 1)*log(a*x - 1) - 2)*sqrt(-a^2*c)/(a^7*
c^2*x - a^6*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**4/(-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^{4}}{{\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^4/(-a^2*c*x^2+c)^(3/2),x, algorithm="giac")

[Out]

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