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

Optimal. Leaf size=148 \[ \frac {c x \sqrt {c-\frac {c}{a^2 x^2}}}{\sqrt {1-\frac {1}{a^2 x^2}}}-\frac {3 c \sqrt {c-\frac {c}{a^2 x^2}}}{a^2 x \sqrt {1-\frac {1}{a^2 x^2}}}-\frac {3 c \log (x) \sqrt {c-\frac {c}{a^2 x^2}}}{a \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {c \sqrt {c-\frac {c}{a^2 x^2}}}{2 a^3 x^2 \sqrt {1-\frac {1}{a^2 x^2}}} \]

[Out]

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

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Rubi [A]  time = 0.12, antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {6197, 6193, 43} \[ \frac {c x \sqrt {c-\frac {c}{a^2 x^2}}}{\sqrt {1-\frac {1}{a^2 x^2}}}-\frac {3 c \sqrt {c-\frac {c}{a^2 x^2}}}{a^2 x \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {c \sqrt {c-\frac {c}{a^2 x^2}}}{2 a^3 x^2 \sqrt {1-\frac {1}{a^2 x^2}}}-\frac {3 c \log (x) \sqrt {c-\frac {c}{a^2 x^2}}}{a \sqrt {1-\frac {1}{a^2 x^2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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

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Mathematica [A]  time = 0.07, size = 59, normalized size = 0.40 \[ \frac {\left (c-\frac {c}{a^2 x^2}\right )^{3/2} \left (a^3 x-3 a^2 \log (x)-\frac {3 a}{x}+\frac {1}{2 x^2}\right )}{a^3 \left (1-\frac {1}{a^2 x^2}\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.63, size = 42, normalized size = 0.28 \[ \frac {{\left (2 \, a^{3} c x^{3} - 6 \, a^{2} c x^{2} \log \relax (x) - 6 \, a c x + c\right )} \sqrt {a^{2} c}}{2 \, a^{4} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*a^3*c*x^3 - 6*a^2*c*x^2*log(x) - 6*a*c*x + c)*sqrt(a^2*c)/(a^4*x^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.06, size = 69, normalized size = 0.47 \[ -\frac {\left (-2 x^{3} a^{3}+6 a^{2} \ln \relax (x ) x^{2}+6 a x -1\right ) x \left (\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}\right )^{\frac {3}{2}} \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}}}{2 \left (a x -1\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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