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

Optimal. Leaf size=171 \[ \frac {x \sqrt {1-\frac {1}{a^2 x^2}}}{c \sqrt {c-\frac {c}{a^2 x^2}}}+\frac {3 \sqrt {1-\frac {1}{a^2 x^2}}}{a c (1-a x) \sqrt {c-\frac {c}{a^2 x^2}}}-\frac {\sqrt {1-\frac {1}{a^2 x^2}}}{2 a c (1-a x)^2 \sqrt {c-\frac {c}{a^2 x^2}}}+\frac {3 \sqrt {1-\frac {1}{a^2 x^2}} \log (1-a x)}{a c \sqrt {c-\frac {c}{a^2 x^2}}} \]

[Out]

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

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Rubi [A]  time = 0.15, antiderivative size = 171, 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} \[ \frac {x \sqrt {1-\frac {1}{a^2 x^2}}}{c \sqrt {c-\frac {c}{a^2 x^2}}}+\frac {3 \sqrt {1-\frac {1}{a^2 x^2}}}{a c (1-a x) \sqrt {c-\frac {c}{a^2 x^2}}}-\frac {\sqrt {1-\frac {1}{a^2 x^2}}}{2 a c (1-a x)^2 \sqrt {c-\frac {c}{a^2 x^2}}}+\frac {3 \sqrt {1-\frac {1}{a^2 x^2}} \log (1-a x)}{a c \sqrt {c-\frac {c}{a^2 x^2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.08, size = 64, normalized size = 0.37 \[ \frac {\left (1-\frac {1}{a^2 x^2}\right )^{3/2} \left (2 a x+\frac {5-6 a x}{(a x-1)^2}+6 \log (1-a x)\right )}{2 a \left (c-\frac {c}{a^2 x^2}\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((1 - 1/(a^2*x^2))^(3/2)*(2*a*x + (5 - 6*a*x)/(-1 + a*x)^2 + 6*Log[1 - a*x]))/(2*a*(c - c/(a^2*x^2))^(3/2))

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fricas [A]  time = 0.50, size = 81, normalized size = 0.47 \[ \frac {{\left (2 \, a^{3} x^{3} - 4 \, a^{2} x^{2} - 4 \, a x + 6 \, {\left (a^{2} x^{2} - 2 \, a x + 1\right )} \log \left (a x - 1\right ) + 5\right )} \sqrt {a^{2} c}}{2 \, {\left (a^{4} c^{2} x^{2} - 2 \, a^{3} c^{2} x + a^{2} c^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*a^3*x^3 - 4*a^2*x^2 - 4*a*x + 6*(a^2*x^2 - 2*a*x + 1)*log(a*x - 1) + 5)*sqrt(a^2*c)/(a^4*c^2*x^2 - 2*a^
3*c^2*x + a^2*c^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.07, size = 102, normalized size = 0.60 \[ \frac {\left (a x -1\right ) \left (2 x^{3} a^{3}+6 \ln \left (a x -1\right ) x^{2} a^{2}-4 a^{2} x^{2}-12 \ln \left (a x -1\right ) x a -4 a x +6 \ln \left (a x -1\right )+5\right )}{2 \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}} a^{4} x^{3} \left (\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}\right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/((a*x-1)/(a*x+1))^(3/2)*(a*x-1)*(2*x^3*a^3+6*ln(a*x-1)*x^2*a^2-4*a^2*x^2-12*ln(a*x-1)*x*a-4*a*x+6*ln(a*x-1
)+5)/a^4/x^3/(c*(a^2*x^2-1)/a^2/x^2)^(3/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/((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 \[ \int \frac {1}{{\left (c-\frac {c}{a^2\,x^2}\right )}^{3/2}\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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