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/(-a^2*c*x^2+c)^(3/2)+1/2*(1-1/a^2/x^2)^(3/2)*x^5/(-a^2*c*x^2+c)^(3/2)+1/2*(1-1/a^2/x
^2)^(3/2)*x^3/a^2/(-a*x+1)/(-a^2*c*x^2+c)^(3/2)+7/4*(1-1/a^2/x^2)^(3/2)*x^3*ln(-a*x+1)/a^2/(-a^2*c*x^2+c)^(3/2
)+1/4*(1-1/a^2/x^2)^(3/2)*x^3*ln(a*x+1)/a^2/(-a^2*c*x^2+c)^(3/2)

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Rubi [A]  time = 0.23, antiderivative size = 211, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, 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 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 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]

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*}

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Mathematica [A]  time = 0.09, 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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fricas [A]  time = 0.61, size = 76, normalized size = 0.36 \[ \frac {{\left (2 \, a^{3} x^{3} + 2 \, a^{2} x^{2} - 4 \, a x + {\left (a x - 1\right )} \log \left (a x + 1\right ) + 7 \, {\left (a x - 1\right )} \log \left (a x - 1\right ) - 2\right )} \sqrt {-a^{2} c}}{4 \, {\left (a^{7} c^{2} x - a^{6} c^{2}\right )}} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{4}}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}} \sqrt {\frac {a x - 1}{a x + 1}}}\,{d x} \]

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)

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maple [A]  time = 0.06, size = 106, normalized size = 0.50 \[ \frac {\sqrt {-c \left (a^{2} x^{2}-1\right )}\, \left (2 x^{3} a^{3}+2 a^{2} x^{2}+7 \ln \left (a x -1\right ) x a +a x \ln \left (a x +1\right )-4 a x -7 \ln \left (a x -1\right )-\ln \left (a x +1\right )-2\right )}{4 \sqrt {\frac {a x -1}{a x +1}}\, \left (a^{2} x^{2}-1\right ) c^{2} a^{5}} \]

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+7*ln(a*x-1)*x*a+a*x*ln(a*x+1)-4*a*x-7*
ln(a*x-1)-ln(a*x+1)-2)/(a^2*x^2-1)/c^2/a^5

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{4}}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}} \sqrt {\frac {a x - 1}{a x + 1}}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^4}{{\left (c-a^2\,c\,x^2\right )}^{3/2}\,\sqrt {\frac {a\,x-1}{a\,x+1}}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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