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

Optimal. Leaf size=234 \[ \frac {c^2 x \sqrt {c-\frac {c}{a^2 x^2}}}{\sqrt {1-\frac {1}{a^2 x^2}}}-\frac {2 c^2 \sqrt {c-\frac {c}{a^2 x^2}}}{a^2 x \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {3 c^2 \log (x) \sqrt {c-\frac {c}{a^2 x^2}}}{a \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {c^2 \sqrt {c-\frac {c}{a^2 x^2}}}{4 a^5 x^4 \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {c^2 \sqrt {c-\frac {c}{a^2 x^2}}}{a^4 x^3 \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {c^2 \sqrt {c-\frac {c}{a^2 x^2}}}{a^3 x^2 \sqrt {1-\frac {1}{a^2 x^2}}} \]

[Out]

1/4*c^2*(c-c/a^2/x^2)^(1/2)/a^5/x^4/(1-1/a^2/x^2)^(1/2)+c^2*(c-c/a^2/x^2)^(1/2)/a^4/x^3/(1-1/a^2/x^2)^(1/2)+c^
2*(c-c/a^2/x^2)^(1/2)/a^3/x^2/(1-1/a^2/x^2)^(1/2)-2*c^2*(c-c/a^2/x^2)^(1/2)/a^2/x/(1-1/a^2/x^2)^(1/2)+c^2*x*(c
-c/a^2/x^2)^(1/2)/(1-1/a^2/x^2)^(1/2)+3*c^2*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 = 0.14, antiderivative size = 234, 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, 75} \[ \frac {c^2 x \sqrt {c-\frac {c}{a^2 x^2}}}{\sqrt {1-\frac {1}{a^2 x^2}}}-\frac {2 c^2 \sqrt {c-\frac {c}{a^2 x^2}}}{a^2 x \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {c^2 \sqrt {c-\frac {c}{a^2 x^2}}}{a^3 x^2 \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {c^2 \sqrt {c-\frac {c}{a^2 x^2}}}{a^4 x^3 \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {c^2 \sqrt {c-\frac {c}{a^2 x^2}}}{4 a^5 x^4 \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {3 c^2 \log (x) \sqrt {c-\frac {c}{a^2 x^2}}}{a \sqrt {1-\frac {1}{a^2 x^2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c^2*Sqrt[c - c/(a^2*x^2)])/(4*a^5*Sqrt[1 - 1/(a^2*x^2)]*x^4) + (c^2*Sqrt[c - c/(a^2*x^2)])/(a^4*Sqrt[1 - 1/(a
^2*x^2)]*x^3) + (c^2*Sqrt[c - c/(a^2*x^2)])/(a^3*Sqrt[1 - 1/(a^2*x^2)]*x^2) - (2*c^2*Sqrt[c - c/(a^2*x^2)])/(a
^2*Sqrt[1 - 1/(a^2*x^2)]*x) + (c^2*Sqrt[c - c/(a^2*x^2)]*x)/Sqrt[1 - 1/(a^2*x^2)] + (3*c^2*Sqrt[c - c/(a^2*x^2
)]*Log[x])/(a*Sqrt[1 - 1/(a^2*x^2)])

Rule 75

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && EqQ[b*e + a*f, 0] &&  !(ILtQ[n
+ p + 2, 0] && GtQ[n + 2*p, 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 )^{5/2} \, dx &=\frac {\left (c^2 \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 )^{5/2} \, dx}{\sqrt {1-\frac {1}{a^2 x^2}}}\\ &=\frac {\left (c^2 \sqrt {c-\frac {c}{a^2 x^2}}\right ) \int \frac {(-1+a x) (1+a x)^4}{x^5} \, dx}{a^5 \sqrt {1-\frac {1}{a^2 x^2}}}\\ &=\frac {\left (c^2 \sqrt {c-\frac {c}{a^2 x^2}}\right ) \int \left (a^5-\frac {1}{x^5}-\frac {3 a}{x^4}-\frac {2 a^2}{x^3}+\frac {2 a^3}{x^2}+\frac {3 a^4}{x}\right ) \, dx}{a^5 \sqrt {1-\frac {1}{a^2 x^2}}}\\ &=\frac {c^2 \sqrt {c-\frac {c}{a^2 x^2}}}{4 a^5 \sqrt {1-\frac {1}{a^2 x^2}} x^4}+\frac {c^2 \sqrt {c-\frac {c}{a^2 x^2}}}{a^4 \sqrt {1-\frac {1}{a^2 x^2}} x^3}+\frac {c^2 \sqrt {c-\frac {c}{a^2 x^2}}}{a^3 \sqrt {1-\frac {1}{a^2 x^2}} x^2}-\frac {2 c^2 \sqrt {c-\frac {c}{a^2 x^2}}}{a^2 \sqrt {1-\frac {1}{a^2 x^2}} x}+\frac {c^2 \sqrt {c-\frac {c}{a^2 x^2}} x}{\sqrt {1-\frac {1}{a^2 x^2}}}+\frac {3 c^2 \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 = 0.06, size = 87, normalized size = 0.37 \[ \frac {\left (c-\frac {c}{a^2 x^2}\right )^{5/2} \left (\frac {3}{4} a \left (a^4 x+4 a^3 \log (x)-\frac {6 a^2}{x}-\frac {2 a}{x^2}-\frac {1}{3 x^3}\right )+\frac {(a x+1)^5}{4 x^4}\right )}{a^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((c - c/(a^2*x^2))^(5/2)*((1 + a*x)^5/(4*x^4) + (3*a*(-1/3*1/x^3 - (2*a)/x^2 - (6*a^2)/x + a^4*x + 4*a^3*Log[x
]))/4))/(a^5*(1 - 1/(a^2*x^2))^(5/2))

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fricas [A]  time = 0.60, size = 72, normalized size = 0.31 \[ \frac {{\left (4 \, a^{5} c^{2} x^{5} + 12 \, a^{4} c^{2} x^{4} \log \relax (x) - 8 \, a^{3} c^{2} x^{3} + 4 \, a^{2} c^{2} x^{2} + 4 \, a c^{2} x + c^{2}\right )} \sqrt {a^{2} c}}{4 \, a^{6} x^{4}} \]

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]

1/4*(4*a^5*c^2*x^5 + 12*a^4*c^2*x^4*log(x) - 8*a^3*c^2*x^3 + 4*a^2*c^2*x^2 + 4*a*c^2*x + c^2)*sqrt(a^2*c)/(a^6
*x^4)

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

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

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maple [A]  time = 0.06, size = 96, normalized size = 0.41 \[ \frac {\left (4 x^{5} a^{5}+12 a^{4} \ln \relax (x ) x^{4}-8 x^{3} a^{3}+4 a^{2} x^{2}+4 a x +1\right ) \left (\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}\right )^{\frac {5}{2}} x}{4 \left (a x +1\right )^{3} \left (a^{2} x^{2}-1\right ) \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}}} \]

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)^(5/2),x)

[Out]

1/4*(4*x^5*a^5+12*a^4*ln(x)*x^4-8*x^3*a^3+4*a^2*x^2+4*a*x+1)*(c*(a^2*x^2-1)/a^2/x^2)^(5/2)*x/(a*x+1)^3/(a^2*x^
2-1)/((a*x-1)/(a*x+1))^(3/2)

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

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((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 \[ \int \frac {{\left (c-\frac {c}{a^2\,x^2}\right )}^{5/2}}{{\left (\frac {a\,x-1}{a\,x+1}\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((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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