3.926 \(\int \frac {e^{-3 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}}}{x^4} \, dx\)

Optimal. Leaf size=221 \[ \frac {4 a^2 \sqrt {c-\frac {c}{a^2 x^2}}}{x \sqrt {1-\frac {1}{a^2 x^2}}}-\frac {2 a \sqrt {c-\frac {c}{a^2 x^2}}}{x^2 \sqrt {1-\frac {1}{a^2 x^2}}}-\frac {\sqrt {c-\frac {c}{a^2 x^2}}}{4 a x^4 \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {\sqrt {c-\frac {c}{a^2 x^2}}}{x^3 \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {4 a^3 \log (x) \sqrt {c-\frac {c}{a^2 x^2}}}{\sqrt {1-\frac {1}{a^2 x^2}}}-\frac {4 a^3 \sqrt {c-\frac {c}{a^2 x^2}} \log (a x+1)}{\sqrt {1-\frac {1}{a^2 x^2}}} \]

[Out]

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

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Rubi [A]  time = 0.29, antiderivative size = 221, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {6197, 6193, 88} \[ \frac {4 a^2 \sqrt {c-\frac {c}{a^2 x^2}}}{x \sqrt {1-\frac {1}{a^2 x^2}}}-\frac {2 a \sqrt {c-\frac {c}{a^2 x^2}}}{x^2 \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {\sqrt {c-\frac {c}{a^2 x^2}}}{x^3 \sqrt {1-\frac {1}{a^2 x^2}}}-\frac {\sqrt {c-\frac {c}{a^2 x^2}}}{4 a x^4 \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {4 a^3 \log (x) \sqrt {c-\frac {c}{a^2 x^2}}}{\sqrt {1-\frac {1}{a^2 x^2}}}-\frac {4 a^3 \sqrt {c-\frac {c}{a^2 x^2}} \log (a x+1)}{\sqrt {1-\frac {1}{a^2 x^2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-Sqrt[c - c/(a^2*x^2)]/(4*a*Sqrt[1 - 1/(a^2*x^2)]*x^4) + Sqrt[c - c/(a^2*x^2)]/(Sqrt[1 - 1/(a^2*x^2)]*x^3) - (
2*a*Sqrt[c - c/(a^2*x^2)])/(Sqrt[1 - 1/(a^2*x^2)]*x^2) + (4*a^2*Sqrt[c - c/(a^2*x^2)])/(Sqrt[1 - 1/(a^2*x^2)]*
x) + (4*a^3*Sqrt[c - c/(a^2*x^2)]*Log[x])/Sqrt[1 - 1/(a^2*x^2)] - (4*a^3*Sqrt[c - c/(a^2*x^2)]*Log[1 + a*x])/S
qrt[1 - 1/(a^2*x^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 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 6197

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

Rubi steps

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

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Mathematica [A]  time = 0.06, size = 80, normalized size = 0.36 \[ \frac {\sqrt {c-\frac {c}{a^2 x^2}} \left (4 a^4 \log (x)-4 a^4 \log (a x+1)+\frac {4 a^3}{x}-\frac {2 a^2}{x^2}+\frac {a}{x^3}-\frac {1}{4 x^4}\right )}{a \sqrt {1-\frac {1}{a^2 x^2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.54, size = 99, normalized size = 0.45 \[ \frac {16 \, a^{5} \sqrt {c} x^{4} \log \left (\frac {2 \, a^{3} c x^{2} + 2 \, a^{2} c x - \sqrt {a^{2} c} {\left (2 \, a x + 1\right )} \sqrt {c} + a c}{a x^{2} + x}\right ) + {\left (16 \, a^{3} x^{3} - 8 \, a^{2} x^{2} + 4 \, a x - 1\right )} \sqrt {a^{2} c}}{4 \, a^{2} x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c/a^2/x^2)^(1/2)*((a*x-1)/(a*x+1))^(3/2)/x^4,x, algorithm="fricas")

[Out]

1/4*(16*a^5*sqrt(c)*x^4*log((2*a^3*c*x^2 + 2*a^2*c*x - sqrt(a^2*c)*(2*a*x + 1)*sqrt(c) + a*c)/(a*x^2 + x)) + (
16*a^3*x^3 - 8*a^2*x^2 + 4*a*x - 1)*sqrt(a^2*c))/(a^2*x^4)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c/a^2/x^2)^(1/2)*((a*x-1)/(a*x+1))^(3/2)/x^4,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.06, size = 98, normalized size = 0.44 \[ \frac {\left (16 a^{4} \ln \relax (x ) x^{4}-16 \ln \left (a x +1\right ) x^{4} a^{4}+16 x^{3} a^{3}-8 a^{2} x^{2}+4 a x -1\right ) \sqrt {\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}}\, \left (a x +1\right ) \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}}}{4 \left (a x -1\right )^{2} x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c/a^2/x^2)^(1/2)*((a*x-1)/(a*x+1))^(3/2)/x^4,x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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