3.853 \(\int \frac {e^{3 \coth ^{-1}(a x)}}{(c-\frac {c}{a^2 x^2})^{5/2}} \, dx\)

Optimal. Leaf size=267 \[ \frac {x \sqrt {1-\frac {1}{a^2 x^2}}}{c^2 \sqrt {c-\frac {c}{a^2 x^2}}}+\frac {31 \sqrt {1-\frac {1}{a^2 x^2}}}{8 a c^2 (1-a x) \sqrt {c-\frac {c}{a^2 x^2}}}-\frac {9 \sqrt {1-\frac {1}{a^2 x^2}}}{8 a c^2 (1-a x)^2 \sqrt {c-\frac {c}{a^2 x^2}}}+\frac {\sqrt {1-\frac {1}{a^2 x^2}}}{6 a c^2 (1-a x)^3 \sqrt {c-\frac {c}{a^2 x^2}}}+\frac {49 \sqrt {1-\frac {1}{a^2 x^2}} \log (1-a x)}{16 a c^2 \sqrt {c-\frac {c}{a^2 x^2}}}-\frac {\sqrt {1-\frac {1}{a^2 x^2}} \log (a x+1)}{16 a c^2 \sqrt {c-\frac {c}{a^2 x^2}}} \]

[Out]

x*(1-1/a^2/x^2)^(1/2)/c^2/(c-c/a^2/x^2)^(1/2)+1/6*(1-1/a^2/x^2)^(1/2)/a/c^2/(-a*x+1)^3/(c-c/a^2/x^2)^(1/2)-9/8
*(1-1/a^2/x^2)^(1/2)/a/c^2/(-a*x+1)^2/(c-c/a^2/x^2)^(1/2)+31/8*(1-1/a^2/x^2)^(1/2)/a/c^2/(-a*x+1)/(c-c/a^2/x^2
)^(1/2)+49/16*ln(-a*x+1)*(1-1/a^2/x^2)^(1/2)/a/c^2/(c-c/a^2/x^2)^(1/2)-1/16*ln(a*x+1)*(1-1/a^2/x^2)^(1/2)/a/c^
2/(c-c/a^2/x^2)^(1/2)

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Rubi [A]  time = 0.18, antiderivative size = 267, 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, 88} \[ \frac {x \sqrt {1-\frac {1}{a^2 x^2}}}{c^2 \sqrt {c-\frac {c}{a^2 x^2}}}+\frac {31 \sqrt {1-\frac {1}{a^2 x^2}}}{8 a c^2 (1-a x) \sqrt {c-\frac {c}{a^2 x^2}}}-\frac {9 \sqrt {1-\frac {1}{a^2 x^2}}}{8 a c^2 (1-a x)^2 \sqrt {c-\frac {c}{a^2 x^2}}}+\frac {\sqrt {1-\frac {1}{a^2 x^2}}}{6 a c^2 (1-a x)^3 \sqrt {c-\frac {c}{a^2 x^2}}}+\frac {49 \sqrt {1-\frac {1}{a^2 x^2}} \log (1-a x)}{16 a c^2 \sqrt {c-\frac {c}{a^2 x^2}}}-\frac {\sqrt {1-\frac {1}{a^2 x^2}} \log (a x+1)}{16 a c^2 \sqrt {c-\frac {c}{a^2 x^2}}} \]

Antiderivative was successfully verified.

[In]

Int[E^(3*ArcCoth[a*x])/(c - c/(a^2*x^2))^(5/2),x]

[Out]

(Sqrt[1 - 1/(a^2*x^2)]*x)/(c^2*Sqrt[c - c/(a^2*x^2)]) + Sqrt[1 - 1/(a^2*x^2)]/(6*a*c^2*Sqrt[c - c/(a^2*x^2)]*(
1 - a*x)^3) - (9*Sqrt[1 - 1/(a^2*x^2)])/(8*a*c^2*Sqrt[c - c/(a^2*x^2)]*(1 - a*x)^2) + (31*Sqrt[1 - 1/(a^2*x^2)
])/(8*a*c^2*Sqrt[c - c/(a^2*x^2)]*(1 - a*x)) + (49*Sqrt[1 - 1/(a^2*x^2)]*Log[1 - a*x])/(16*a*c^2*Sqrt[c - c/(a
^2*x^2)]) - (Sqrt[1 - 1/(a^2*x^2)]*Log[1 + a*x])/(16*a*c^2*Sqrt[c - c/(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)}}{\left (c-\frac {c}{a^2 x^2}\right )^{5/2}} \, 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 )^{5/2}} \, dx}{c^2 \sqrt {c-\frac {c}{a^2 x^2}}}\\ &=\frac {\left (a^5 \sqrt {1-\frac {1}{a^2 x^2}}\right ) \int \frac {x^5}{(-1+a x)^4 (1+a x)} \, dx}{c^2 \sqrt {c-\frac {c}{a^2 x^2}}}\\ &=\frac {\left (a^5 \sqrt {1-\frac {1}{a^2 x^2}}\right ) \int \left (\frac {1}{a^5}+\frac {1}{2 a^5 (-1+a x)^4}+\frac {9}{4 a^5 (-1+a x)^3}+\frac {31}{8 a^5 (-1+a x)^2}+\frac {49}{16 a^5 (-1+a x)}-\frac {1}{16 a^5 (1+a x)}\right ) \, dx}{c^2 \sqrt {c-\frac {c}{a^2 x^2}}}\\ &=\frac {\sqrt {1-\frac {1}{a^2 x^2}} x}{c^2 \sqrt {c-\frac {c}{a^2 x^2}}}+\frac {\sqrt {1-\frac {1}{a^2 x^2}}}{6 a c^2 \sqrt {c-\frac {c}{a^2 x^2}} (1-a x)^3}-\frac {9 \sqrt {1-\frac {1}{a^2 x^2}}}{8 a c^2 \sqrt {c-\frac {c}{a^2 x^2}} (1-a x)^2}+\frac {31 \sqrt {1-\frac {1}{a^2 x^2}}}{8 a c^2 \sqrt {c-\frac {c}{a^2 x^2}} (1-a x)}+\frac {49 \sqrt {1-\frac {1}{a^2 x^2}} \log (1-a x)}{16 a c^2 \sqrt {c-\frac {c}{a^2 x^2}}}-\frac {\sqrt {1-\frac {1}{a^2 x^2}} \log (1+a x)}{16 a c^2 \sqrt {c-\frac {c}{a^2 x^2}}}\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 86, normalized size = 0.32 \[ \frac {\left (1-\frac {1}{a^2 x^2}\right )^{5/2} \left (48 a x+\frac {186}{1-a x}-\frac {54}{(a x-1)^2}-\frac {8}{(a x-1)^3}+147 \log (1-a x)-3 \log (a x+1)\right )}{48 a \left (c-\frac {c}{a^2 x^2}\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(3*ArcCoth[a*x])/(c - c/(a^2*x^2))^(5/2),x]

[Out]

((1 - 1/(a^2*x^2))^(5/2)*(48*a*x + 186/(1 - a*x) - 8/(-1 + a*x)^3 - 54/(-1 + a*x)^2 + 147*Log[1 - a*x] - 3*Log
[1 + a*x]))/(48*a*(c - c/(a^2*x^2))^(5/2))

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fricas [A]  time = 0.63, size = 138, normalized size = 0.52 \[ \frac {{\left (48 \, a^{4} x^{4} - 144 \, a^{3} x^{3} - 42 \, a^{2} x^{2} + 270 \, a x - 3 \, {\left (a^{3} x^{3} - 3 \, a^{2} x^{2} + 3 \, a x - 1\right )} \log \left (a x + 1\right ) + 147 \, {\left (a^{3} x^{3} - 3 \, a^{2} x^{2} + 3 \, a x - 1\right )} \log \left (a x - 1\right ) - 140\right )} \sqrt {a^{2} c}}{48 \, {\left (a^{5} c^{3} x^{3} - 3 \, a^{4} c^{3} x^{2} + 3 \, a^{3} c^{3} x - a^{2} c^{3}\right )}} \]

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)^(5/2),x, algorithm="fricas")

[Out]

1/48*(48*a^4*x^4 - 144*a^3*x^3 - 42*a^2*x^2 + 270*a*x - 3*(a^3*x^3 - 3*a^2*x^2 + 3*a*x - 1)*log(a*x + 1) + 147
*(a^3*x^3 - 3*a^2*x^2 + 3*a*x - 1)*log(a*x - 1) - 140)*sqrt(a^2*c)/(a^5*c^3*x^3 - 3*a^4*c^3*x^2 + 3*a^3*c^3*x
- a^2*c^3)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (c - \frac {c}{a^{2} x^{2}}\right )}^{\frac {5}{2}} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}}}\,{d x} \]

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)^(5/2),x, algorithm="giac")

[Out]

integrate(1/((c - c/(a^2*x^2))^(5/2)*((a*x - 1)/(a*x + 1))^(3/2)), x)

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maple [A]  time = 0.07, size = 175, normalized size = 0.66 \[ \frac {\left (a x -1\right ) \left (a x +1\right ) \left (48 x^{4} a^{4}+147 \ln \left (a x -1\right ) x^{3} a^{3}-3 a^{3} x^{3} \ln \left (a x +1\right )-144 x^{3} a^{3}-441 \ln \left (a x -1\right ) x^{2} a^{2}+9 \ln \left (a x +1\right ) x^{2} a^{2}-42 a^{2} x^{2}+441 \ln \left (a x -1\right ) x a -9 a x \ln \left (a x +1\right )+270 a x -147 \ln \left (a x -1\right )+3 \ln \left (a x +1\right )-140\right )}{48 \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}} a^{6} x^{5} \left (\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}\right )^{\frac {5}{2}}} \]

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)^(5/2),x)

[Out]

1/48/((a*x-1)/(a*x+1))^(3/2)*(a*x-1)*(a*x+1)*(48*x^4*a^4+147*ln(a*x-1)*x^3*a^3-3*a^3*x^3*ln(a*x+1)-144*x^3*a^3
-441*ln(a*x-1)*x^2*a^2+9*ln(a*x+1)*x^2*a^2-42*a^2*x^2+441*ln(a*x-1)*x*a-9*a*x*ln(a*x+1)+270*a*x-147*ln(a*x-1)+
3*ln(a*x+1)-140)/a^6/x^5/(c*(a^2*x^2-1)/a^2/x^2)^(5/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (c - \frac {c}{a^{2} x^{2}}\right )}^{\frac {5}{2}} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}}}\,{d x} \]

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)^(5/2),x, algorithm="maxima")

[Out]

integrate(1/((c - c/(a^2*x^2))^(5/2)*((a*x - 1)/(a*x + 1))^(3/2)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {1}{{\left (c-\frac {c}{a^2\,x^2}\right )}^{5/2}\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((c - c/(a^2*x^2))^(5/2)*((a*x - 1)/(a*x + 1))^(3/2)),x)

[Out]

int(1/((c - c/(a^2*x^2))^(5/2)*((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)**(5/2),x)

[Out]

Timed out

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