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

Optimal. Leaf size=186 \[ \frac {2 x^2 \sqrt {c-\frac {c}{a^2 x^2}}}{a^2 \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {x^4 \sqrt {c-\frac {c}{a^2 x^2}}}{4 \sqrt {1-\frac {1}{a^2 x^2}}}-\frac {x^3 \sqrt {c-\frac {c}{a^2 x^2}}}{a \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {4 \sqrt {c-\frac {c}{a^2 x^2}} \log (a x+1)}{a^4 \sqrt {1-\frac {1}{a^2 x^2}}}-\frac {4 x \sqrt {c-\frac {c}{a^2 x^2}}}{a^3 \sqrt {1-\frac {1}{a^2 x^2}}} \]

[Out]

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

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Rubi [A]  time = 0.29, antiderivative size = 186, 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} \[ \frac {x^4 \sqrt {c-\frac {c}{a^2 x^2}}}{4 \sqrt {1-\frac {1}{a^2 x^2}}}-\frac {x^3 \sqrt {c-\frac {c}{a^2 x^2}}}{a \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {2 x^2 \sqrt {c-\frac {c}{a^2 x^2}}}{a^2 \sqrt {1-\frac {1}{a^2 x^2}}}-\frac {4 x \sqrt {c-\frac {c}{a^2 x^2}}}{a^3 \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {4 \sqrt {c-\frac {c}{a^2 x^2}} \log (a x+1)}{a^4 \sqrt {1-\frac {1}{a^2 x^2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*Sqrt[c - c/(a^2*x^2)]*x)/(a^3*Sqrt[1 - 1/(a^2*x^2)]) + (2*Sqrt[c - c/(a^2*x^2)]*x^2)/(a^2*Sqrt[1 - 1/(a^2*
x^2)]) - (Sqrt[c - c/(a^2*x^2)]*x^3)/(a*Sqrt[1 - 1/(a^2*x^2)]) + (Sqrt[c - c/(a^2*x^2)]*x^4)/(4*Sqrt[1 - 1/(a^
2*x^2)]) + (4*Sqrt[c - c/(a^2*x^2)]*Log[1 + a*x])/(a^4*Sqrt[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 e^{-3 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}} x^3 \, dx &=\frac {\sqrt {c-\frac {c}{a^2 x^2}} \int e^{-3 \coth ^{-1}(a x)} \sqrt {1-\frac {1}{a^2 x^2}} x^3 \, dx}{\sqrt {1-\frac {1}{a^2 x^2}}}\\ &=\frac {\sqrt {c-\frac {c}{a^2 x^2}} \int \frac {x^2 (-1+a x)^2}{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 {4}{a^2}+\frac {4 x}{a}-3 x^2+a x^3+\frac {4}{a^2 (1+a x)}\right ) \, dx}{a \sqrt {1-\frac {1}{a^2 x^2}}}\\ &=-\frac {4 \sqrt {c-\frac {c}{a^2 x^2}} x}{a^3 \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {2 \sqrt {c-\frac {c}{a^2 x^2}} x^2}{a^2 \sqrt {1-\frac {1}{a^2 x^2}}}-\frac {\sqrt {c-\frac {c}{a^2 x^2}} x^3}{a \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {\sqrt {c-\frac {c}{a^2 x^2}} x^4}{4 \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {4 \sqrt {c-\frac {c}{a^2 x^2}} \log (1+a x)}{a^4 \sqrt {1-\frac {1}{a^2 x^2}}}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 69, normalized size = 0.37 \[ \frac {\sqrt {c-\frac {c}{a^2 x^2}} \left (a x \left (a^3 x^3-4 a^2 x^2+8 a x-16\right )+16 \log (a x+1)\right )}{4 a^4 \sqrt {1-\frac {1}{a^2 x^2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c - c/(a^2*x^2)]*(a*x*(-16 + 8*a*x - 4*a^2*x^2 + a^3*x^3) + 16*Log[1 + a*x]))/(4*a^4*Sqrt[1 - 1/(a^2*x^2
)])

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fricas [A]  time = 0.43, size = 48, normalized size = 0.26 \[ \frac {{\left (a^{4} x^{4} - 4 \, a^{3} x^{3} + 8 \, a^{2} x^{2} - 16 \, a x + 16 \, \log \left (a x + 1\right )\right )} \sqrt {a^{2} c}}{4 \, a^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.06, size = 89, normalized size = 0.48 \[ \frac {\left (x^{4} a^{4}-4 x^{3} a^{3}+8 a^{2} x^{2}-16 a x +16 \ln \left (a x +1\right )\right ) x \sqrt {\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}}\, \left (a x +1\right ) \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}}}{4 \left (a x -1\right )^{2} a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^3\,\sqrt {c-\frac {c}{a^2\,x^2}}\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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