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

Optimal. Leaf size=252 \[ \frac {a^5 x^3 \left (1-\frac {1}{a^2 x^2}\right )^{3/2}}{2 (1-a x) \left (c-a^2 c x^2\right )^{3/2}}+\frac {2 a^5 x^3 \left (1-\frac {1}{a^2 x^2}\right )^{3/2} \log (x)}{\left (c-a^2 c x^2\right )^{3/2}}-\frac {7 a^5 x^3 \left (1-\frac {1}{a^2 x^2}\right )^{3/2} \log (1-a x)}{4 \left (c-a^2 c x^2\right )^{3/2}}-\frac {a^5 x^3 \left (1-\frac {1}{a^2 x^2}\right )^{3/2} \log (a x+1)}{4 \left (c-a^2 c x^2\right )^{3/2}}-\frac {a^4 x^2 \left (1-\frac {1}{a^2 x^2}\right )^{3/2}}{\left (c-a^2 c x^2\right )^{3/2}}-\frac {a^3 x \left (1-\frac {1}{a^2 x^2}\right )^{3/2}}{2 \left (c-a^2 c x^2\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.27, antiderivative size = 252, 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} \[ \frac {a^5 x^3 \left (1-\frac {1}{a^2 x^2}\right )^{3/2}}{2 (1-a x) \left (c-a^2 c x^2\right )^{3/2}}-\frac {a^4 x^2 \left (1-\frac {1}{a^2 x^2}\right )^{3/2}}{\left (c-a^2 c x^2\right )^{3/2}}-\frac {a^3 x \left (1-\frac {1}{a^2 x^2}\right )^{3/2}}{2 \left (c-a^2 c x^2\right )^{3/2}}+\frac {2 a^5 x^3 \left (1-\frac {1}{a^2 x^2}\right )^{3/2} \log (x)}{\left (c-a^2 c x^2\right )^{3/2}}-\frac {7 a^5 x^3 \left (1-\frac {1}{a^2 x^2}\right )^{3/2} \log (1-a x)}{4 \left (c-a^2 c x^2\right )^{3/2}}-\frac {a^5 x^3 \left (1-\frac {1}{a^2 x^2}\right )^{3/2} \log (a x+1)}{4 \left (c-a^2 c x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^3*(1 - 1/(a^2*x^2))^(3/2)*x)/(2*(c - a^2*c*x^2)^(3/2)) - (a^4*(1 - 1/(a^2*x^2))^(3/2)*x^2)/(c - a^2*c*x^2)
^(3/2) + (a^5*(1 - 1/(a^2*x^2))^(3/2)*x^3)/(2*(1 - a*x)*(c - a^2*c*x^2)^(3/2)) + (2*a^5*(1 - 1/(a^2*x^2))^(3/2
)*x^3*Log[x])/(c - a^2*c*x^2)^(3/2) - (7*a^5*(1 - 1/(a^2*x^2))^(3/2)*x^3*Log[1 - a*x])/(4*(c - a^2*c*x^2)^(3/2
)) - (a^5*(1 - 1/(a^2*x^2))^(3/2)*x^3*Log[1 + a*x])/(4*(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^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} x^6} \, 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 {1}{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}{x^3}+\frac {a}{x^2}+\frac {2 a^2}{x}+\frac {a^3}{2 (-1+a x)^2}-\frac {7 a^3}{4 (-1+a x)}-\frac {a^3}{4 (1+a x)}\right ) \, dx}{\left (c-a^2 c x^2\right )^{3/2}}\\ &=-\frac {a^3 \left (1-\frac {1}{a^2 x^2}\right )^{3/2} x}{2 \left (c-a^2 c x^2\right )^{3/2}}-\frac {a^4 \left (1-\frac {1}{a^2 x^2}\right )^{3/2} x^2}{\left (c-a^2 c x^2\right )^{3/2}}+\frac {a^5 \left (1-\frac {1}{a^2 x^2}\right )^{3/2} x^3}{2 (1-a x) \left (c-a^2 c x^2\right )^{3/2}}+\frac {2 a^5 \left (1-\frac {1}{a^2 x^2}\right )^{3/2} x^3 \log (x)}{\left (c-a^2 c x^2\right )^{3/2}}-\frac {7 a^5 \left (1-\frac {1}{a^2 x^2}\right )^{3/2} x^3 \log (1-a x)}{4 \left (c-a^2 c x^2\right )^{3/2}}-\frac {a^5 \left (1-\frac {1}{a^2 x^2}\right )^{3/2} x^3 \log (1+a x)}{4 \left (c-a^2 c x^2\right )^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 94, normalized size = 0.37 \[ \frac {a^3 x^3 \left (1-\frac {1}{a^2 x^2}\right )^{3/2} \left (\frac {2 a^2}{1-a x}+8 a^2 \log (x)-7 a^2 \log (1-a x)-a^2 \log (a x+1)-\frac {4 a}{x}-\frac {2}{x^2}\right )}{4 \left (c-a^2 c x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.54, size = 113, normalized size = 0.45 \[ -\frac {{\left (6 \, a^{2} x^{2} - 2 \, a x + {\left (a^{3} x^{3} - a^{2} x^{2}\right )} \log \left (a x + 1\right ) + 7 \, {\left (a^{3} x^{3} - a^{2} x^{2}\right )} \log \left (a x - 1\right ) - 8 \, {\left (a^{3} x^{3} - a^{2} x^{2}\right )} \log \relax (x) - 2\right )} \sqrt {-a^{2} c}}{4 \, {\left (a^{2} c^{2} x^{3} - a c^{2} x^{2}\right )}} \]

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*(6*a^2*x^2 - 2*a*x + (a^3*x^3 - a^2*x^2)*log(a*x + 1) + 7*(a^3*x^3 - a^2*x^2)*log(a*x - 1) - 8*(a^3*x^3 -
a^2*x^2)*log(x) - 2)*sqrt(-a^2*c)/(a^2*c^2*x^3 - a*c^2*x^2)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

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]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [A]  time = 0.07, size = 138, normalized size = 0.55 \[ \frac {\sqrt {-c \left (a^{2} x^{2}-1\right )}\, \left (8 a^{3} \ln \relax (x ) x^{3}-7 \ln \left (a x -1\right ) x^{3} a^{3}-a^{3} x^{3} \ln \left (a x +1\right )-8 a^{2} \ln \relax (x ) x^{2}+7 \ln \left (a x -1\right ) x^{2} a^{2}+\ln \left (a x +1\right ) x^{2} a^{2}-6 a^{2} x^{2}+2 a x +2\right )}{4 \sqrt {\frac {a x -1}{a x +1}}\, \left (a^{2} x^{2}-1\right ) c^{2} x^{2}} \]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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