3.706 \(\int \frac {e^{\coth ^{-1}(a x)}}{x^2 (c-a^2 c x^2)^{5/2}} \, dx\)

Optimal. Leaf size=307 \[ -\frac {3 a^6 x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}{4 (1-a x) \left (c-a^2 c x^2\right )^{5/2}}+\frac {a^6 x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}{8 (a x+1) \left (c-a^2 c x^2\right )^{5/2}}-\frac {a^6 x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}{8 (1-a x)^2 \left (c-a^2 c x^2\right )^{5/2}}-\frac {a^6 x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} \log (x)}{\left (c-a^2 c x^2\right )^{5/2}}+\frac {23 a^6 x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} \log (1-a x)}{16 \left (c-a^2 c x^2\right )^{5/2}}-\frac {7 a^6 x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} \log (a x+1)}{16 \left (c-a^2 c x^2\right )^{5/2}}+\frac {a^5 x^4 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}{\left (c-a^2 c x^2\right )^{5/2}} \]

[Out]

a^5*(1-1/a^2/x^2)^(5/2)*x^4/(-a^2*c*x^2+c)^(5/2)-1/8*a^6*(1-1/a^2/x^2)^(5/2)*x^5/(-a*x+1)^2/(-a^2*c*x^2+c)^(5/
2)-3/4*a^6*(1-1/a^2/x^2)^(5/2)*x^5/(-a*x+1)/(-a^2*c*x^2+c)^(5/2)+1/8*a^6*(1-1/a^2/x^2)^(5/2)*x^5/(a*x+1)/(-a^2
*c*x^2+c)^(5/2)-a^6*(1-1/a^2/x^2)^(5/2)*x^5*ln(x)/(-a^2*c*x^2+c)^(5/2)+23/16*a^6*(1-1/a^2/x^2)^(5/2)*x^5*ln(-a
*x+1)/(-a^2*c*x^2+c)^(5/2)-7/16*a^6*(1-1/a^2/x^2)^(5/2)*x^5*ln(a*x+1)/(-a^2*c*x^2+c)^(5/2)

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Rubi [A]  time = 0.29, antiderivative size = 307, 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} \[ -\frac {3 a^6 x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}{4 (1-a x) \left (c-a^2 c x^2\right )^{5/2}}+\frac {a^6 x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}{8 (a x+1) \left (c-a^2 c x^2\right )^{5/2}}-\frac {a^6 x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}{8 (1-a x)^2 \left (c-a^2 c x^2\right )^{5/2}}+\frac {a^5 x^4 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}{\left (c-a^2 c x^2\right )^{5/2}}-\frac {a^6 x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} \log (x)}{\left (c-a^2 c x^2\right )^{5/2}}+\frac {23 a^6 x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} \log (1-a x)}{16 \left (c-a^2 c x^2\right )^{5/2}}-\frac {7 a^6 x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} \log (a x+1)}{16 \left (c-a^2 c x^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcCoth[a*x]/(x^2*(c - a^2*c*x^2)^(5/2)),x]

[Out]

(a^5*(1 - 1/(a^2*x^2))^(5/2)*x^4)/(c - a^2*c*x^2)^(5/2) - (a^6*(1 - 1/(a^2*x^2))^(5/2)*x^5)/(8*(1 - a*x)^2*(c
- a^2*c*x^2)^(5/2)) - (3*a^6*(1 - 1/(a^2*x^2))^(5/2)*x^5)/(4*(1 - a*x)*(c - a^2*c*x^2)^(5/2)) + (a^6*(1 - 1/(a
^2*x^2))^(5/2)*x^5)/(8*(1 + a*x)*(c - a^2*c*x^2)^(5/2)) - (a^6*(1 - 1/(a^2*x^2))^(5/2)*x^5*Log[x])/(c - a^2*c*
x^2)^(5/2) + (23*a^6*(1 - 1/(a^2*x^2))^(5/2)*x^5*Log[1 - a*x])/(16*(c - a^2*c*x^2)^(5/2)) - (7*a^6*(1 - 1/(a^2
*x^2))^(5/2)*x^5*Log[1 + a*x])/(16*(c - a^2*c*x^2)^(5/2))

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^2 \left (c-a^2 c x^2\right )^{5/2}} \, dx &=\frac {\left (\left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5\right ) \int \frac {e^{\coth ^{-1}(a x)}}{\left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^7} \, dx}{\left (c-a^2 c x^2\right )^{5/2}}\\ &=\frac {\left (a^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5\right ) \int \frac {1}{x^2 (-1+a x)^3 (1+a x)^2} \, dx}{\left (c-a^2 c x^2\right )^{5/2}}\\ &=\frac {\left (a^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5\right ) \int \left (-\frac {1}{x^2}-\frac {a}{x}+\frac {a^2}{4 (-1+a x)^3}-\frac {3 a^2}{4 (-1+a x)^2}+\frac {23 a^2}{16 (-1+a x)}-\frac {a^2}{8 (1+a x)^2}-\frac {7 a^2}{16 (1+a x)}\right ) \, dx}{\left (c-a^2 c x^2\right )^{5/2}}\\ &=\frac {a^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^4}{\left (c-a^2 c x^2\right )^{5/2}}-\frac {a^6 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5}{8 (1-a x)^2 \left (c-a^2 c x^2\right )^{5/2}}-\frac {3 a^6 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5}{4 (1-a x) \left (c-a^2 c x^2\right )^{5/2}}+\frac {a^6 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5}{8 (1+a x) \left (c-a^2 c x^2\right )^{5/2}}-\frac {a^6 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5 \log (x)}{\left (c-a^2 c x^2\right )^{5/2}}+\frac {23 a^6 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5 \log (1-a x)}{16 \left (c-a^2 c x^2\right )^{5/2}}-\frac {7 a^6 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5 \log (1+a x)}{16 \left (c-a^2 c x^2\right )^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 99, normalized size = 0.32 \[ \frac {a^5 x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} \left (\frac {12 a}{a x-1}+\frac {2 a}{a x+1}-\frac {2 a}{(a x-1)^2}-16 a \log (x)+23 a \log (1-a x)-7 a \log (a x+1)+\frac {16}{x}\right )}{16 \left (c-a^2 c x^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[E^ArcCoth[a*x]/(x^2*(c - a^2*c*x^2)^(5/2)),x]

[Out]

(a^5*(1 - 1/(a^2*x^2))^(5/2)*x^5*(16/x - (2*a)/(-1 + a*x)^2 + (12*a)/(-1 + a*x) + (2*a)/(1 + a*x) - 16*a*Log[x
] + 23*a*Log[1 - a*x] - 7*a*Log[1 + a*x]))/(16*(c - a^2*c*x^2)^(5/2))

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fricas [A]  time = 0.69, size = 174, normalized size = 0.57 \[ -\frac {{\left (30 \, a^{3} x^{3} - 22 \, a^{2} x^{2} - 28 \, a x - 7 \, {\left (a^{4} x^{4} - a^{3} x^{3} - a^{2} x^{2} + a x\right )} \log \left (a x + 1\right ) + 23 \, {\left (a^{4} x^{4} - a^{3} x^{3} - a^{2} x^{2} + a x\right )} \log \left (a x - 1\right ) - 16 \, {\left (a^{4} x^{4} - a^{3} x^{3} - a^{2} x^{2} + a x\right )} \log \relax (x) + 16\right )} \sqrt {-a^{2} c}}{16 \, {\left (a^{4} c^{3} x^{4} - a^{3} c^{3} x^{3} - a^{2} c^{3} x^{2} + a c^{3} x\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((a*x-1)/(a*x+1))^(1/2)/x^2/(-a^2*c*x^2+c)^(5/2),x, algorithm="fricas")

[Out]

-1/16*(30*a^3*x^3 - 22*a^2*x^2 - 28*a*x - 7*(a^4*x^4 - a^3*x^3 - a^2*x^2 + a*x)*log(a*x + 1) + 23*(a^4*x^4 - a
^3*x^3 - a^2*x^2 + a*x)*log(a*x - 1) - 16*(a^4*x^4 - a^3*x^3 - a^2*x^2 + a*x)*log(x) + 16)*sqrt(-a^2*c)/(a^4*c
^3*x^4 - a^3*c^3*x^3 - a^2*c^3*x^2 + a*c^3*x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {5}{2}} x^{2} \sqrt {\frac {a x - 1}{a x + 1}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((a*x-1)/(a*x+1))^(1/2)/x^2/(-a^2*c*x^2+c)^(5/2),x, algorithm="giac")

[Out]

integrate(1/((-a^2*c*x^2 + c)^(5/2)*x^2*sqrt((a*x - 1)/(a*x + 1))), x)

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maple [A]  time = 0.06, size = 225, normalized size = 0.73 \[ \frac {\sqrt {-c \left (a^{2} x^{2}-1\right )}\, \left (16 a^{4} \ln \relax (x ) x^{4}-23 \ln \left (a x -1\right ) x^{4} a^{4}+7 \ln \left (a x +1\right ) x^{4} a^{4}-16 a^{3} \ln \relax (x ) x^{3}+23 \ln \left (a x -1\right ) x^{3} a^{3}-7 a^{3} x^{3} \ln \left (a x +1\right )-30 x^{3} a^{3}-16 a^{2} \ln \relax (x ) x^{2}+23 \ln \left (a x -1\right ) x^{2} a^{2}-7 \ln \left (a x +1\right ) x^{2} a^{2}+22 a^{2} x^{2}+16 a \ln \relax (x ) x -23 \ln \left (a x -1\right ) x a +7 a x \ln \left (a x +1\right )+28 a x -16\right )}{16 \sqrt {\frac {a x -1}{a x +1}}\, \left (a x -1\right ) \left (a^{2} x^{2}-1\right ) c^{3} x \left (a x +1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a*x-1)/(a*x+1))^(1/2)/x^2/(-a^2*c*x^2+c)^(5/2),x)

[Out]

1/16/((a*x-1)/(a*x+1))^(1/2)/(a*x-1)*(-c*(a^2*x^2-1))^(1/2)*(16*a^4*ln(x)*x^4-23*ln(a*x-1)*x^4*a^4+7*ln(a*x+1)
*x^4*a^4-16*a^3*ln(x)*x^3+23*ln(a*x-1)*x^3*a^3-7*a^3*x^3*ln(a*x+1)-30*x^3*a^3-16*a^2*ln(x)*x^2+23*ln(a*x-1)*x^
2*a^2-7*ln(a*x+1)*x^2*a^2+22*a^2*x^2+16*a*ln(x)*x-23*ln(a*x-1)*x*a+7*a*x*ln(a*x+1)+28*a*x-16)/(a^2*x^2-1)/c^3/
x/(a*x+1)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {5}{2}} x^{2} \sqrt {\frac {a x - 1}{a x + 1}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((a*x-1)/(a*x+1))^(1/2)/x^2/(-a^2*c*x^2+c)^(5/2),x, algorithm="maxima")

[Out]

integrate(1/((-a^2*c*x^2 + c)^(5/2)*x^2*sqrt((a*x - 1)/(a*x + 1))), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {1}{x^2\,{\left (c-a^2\,c\,x^2\right )}^{5/2}\,\sqrt {\frac {a\,x-1}{a\,x+1}}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^2*(c - a^2*c*x^2)^(5/2)*((a*x - 1)/(a*x + 1))^(1/2)),x)

[Out]

int(1/(x^2*(c - a^2*c*x^2)^(5/2)*((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**2/(-a**2*c*x**2+c)**(5/2),x)

[Out]

Timed out

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