\(\int e^{-\coth ^{-1}(a x)} (c-\frac {c}{a^2 x^2})^{5/2} \, dx\) [856]

Optimal result

Integrand size = 24, antiderivative size = 238 \[ \int e^{-\coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{5/2} \, dx=\frac {c^2 \sqrt {c-\frac {c}{a^2 x^2}}}{4 a^5 \sqrt {1-\frac {1}{a^2 x^2}} x^4}-\frac {c^2 \sqrt {c-\frac {c}{a^2 x^2}}}{3 a^4 \sqrt {1-\frac {1}{a^2 x^2}} x^3}-\frac {c^2 \sqrt {c-\frac {c}{a^2 x^2}}}{a^3 \sqrt {1-\frac {1}{a^2 x^2}} x^2}+\frac {2 c^2 \sqrt {c-\frac {c}{a^2 x^2}}}{a^2 \sqrt {1-\frac {1}{a^2 x^2}} x}+\frac {c^2 \sqrt {c-\frac {c}{a^2 x^2}} x}{\sqrt {1-\frac {1}{a^2 x^2}}}-\frac {c^2 \sqrt {c-\frac {c}{a^2 x^2}} \log (x)}{a \sqrt {1-\frac {1}{a^2 x^2}}} \]

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

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {6332, 6328, 90} \[ \int e^{-\coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{5/2} \, dx=\frac {c^2 x \sqrt {c-\frac {c}{a^2 x^2}}}{\sqrt {1-\frac {1}{a^2 x^2}}}+\frac {2 c^2 \sqrt {c-\frac {c}{a^2 x^2}}}{a^2 x \sqrt {1-\frac {1}{a^2 x^2}}}-\frac {c^2 \log (x) \sqrt {c-\frac {c}{a^2 x^2}}}{a \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {c^2 \sqrt {c-\frac {c}{a^2 x^2}}}{4 a^5 x^4 \sqrt {1-\frac {1}{a^2 x^2}}}-\frac {c^2 \sqrt {c-\frac {c}{a^2 x^2}}}{3 a^4 x^3 \sqrt {1-\frac {1}{a^2 x^2}}}-\frac {c^2 \sqrt {c-\frac {c}{a^2 x^2}}}{a^3 x^2 \sqrt {1-\frac {1}{a^2 x^2}}} \]

[In]

Int[(c - c/(a^2*x^2))^(5/2)/E^ArcCoth[a*x],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)])/(3*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)] - (c^2*Sqrt[c - c/(a^2*x^2
)]*Log[x])/(a*Sqrt[1 - 1/(a^2*x^2)])

Rule 90

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 6328

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_)^2)^(p_.), x_Symbol] :> Dist[c^p/a^(2*p), Int[(u/x^(
2*p))*(-1 + a*x)^(p - n/2)*(1 + a*x)^(p + n/2), 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 6332

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*} \text {integral}& = \frac {\left (c^2 \sqrt {c-\frac {c}{a^2 x^2}}\right ) \int e^{-\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)^3 (1+a x)^2}{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 {a}{x^4}+\frac {2 a^2}{x^3}-\frac {2 a^3}{x^2}-\frac {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}}}{3 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 {c^2 \sqrt {c-\frac {c}{a^2 x^2}} \log (x)}{a \sqrt {1-\frac {1}{a^2 x^2}}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.32 \[ \int e^{-\coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{5/2} \, dx=\frac {\left (c-\frac {c}{a^2 x^2}\right )^{5/2} \left (\frac {1}{4 a^5 x^4}-\frac {1}{3 a^4 x^3}-\frac {1}{a^3 x^2}+\frac {2}{a^2 x}+x-\frac {\log (x)}{a}\right )}{\left (1-\frac {1}{a^2 x^2}\right )^{5/2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.40

method	result	size
default	\(-\frac {{\left (\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}\right )}^{\frac {5}{2}} \sqrt {\frac {a x -1}{a x +1}}\, x \left (-12 a^{5} x^{5}+12 \ln \left (x \right ) x^{4} a^{4}-24 a^{3} x^{3}+12 a^{2} x^{2}+4 a x -3\right )}{12 \left (a x -1\right ) \left (a^{2} x^{2}-1\right )^{2}}\)	\(96\)

[In]

int((c-c/a^2/x^2)^(5/2)*((a*x-1)/(a*x+1))^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/12*(c*(a^2*x^2-1)/a^2/x^2)^(5/2)*((a*x-1)/(a*x+1))^(1/2)*x*(-12*a^5*x^5+12*ln(x)*x^4*a^4-24*a^3*x^3+12*a^2*
x^2+4*a*x-3)/(a*x-1)/(a^2*x^2-1)^2

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.31 \[ \int e^{-\coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{5/2} \, dx=\frac {{\left (12 \, a^{5} c^{2} x^{5} - 12 \, a^{4} c^{2} x^{4} \log \left (x\right ) + 24 \, a^{3} c^{2} x^{3} - 12 \, a^{2} c^{2} x^{2} - 4 \, a c^{2} x + 3 \, c^{2}\right )} \sqrt {a^{2} c}}{12 \, a^{6} x^{4}} \]

[In]

integrate((c-c/a^2/x^2)^(5/2)*((a*x-1)/(a*x+1))^(1/2),x, algorithm="fricas")

[Out]

1/12*(12*a^5*c^2*x^5 - 12*a^4*c^2*x^4*log(x) + 24*a^3*c^2*x^3 - 12*a^2*c^2*x^2 - 4*a*c^2*x + 3*c^2)*sqrt(a^2*c
)/(a^6*x^4)

Sympy [F(-1)]

Timed out. \[ \int e^{-\coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{5/2} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

\[ \int e^{-\coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{5/2} \, dx=\int { {\left (c - \frac {c}{a^{2} x^{2}}\right )}^{\frac {5}{2}} \sqrt {\frac {a x - 1}{a x + 1}} \,d x } \]

[In]

integrate((c-c/a^2/x^2)^(5/2)*((a*x-1)/(a*x+1))^(1/2),x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int e^{-\coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{5/2} \, dx=\int { {\left (c - \frac {c}{a^{2} x^{2}}\right )}^{\frac {5}{2}} \sqrt {\frac {a x - 1}{a x + 1}} \,d x } \]

[In]

integrate((c-c/a^2/x^2)^(5/2)*((a*x-1)/(a*x+1))^(1/2),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int e^{-\coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{5/2} \, dx=\int {\left (c-\frac {c}{a^2\,x^2}\right )}^{5/2}\,\sqrt {\frac {a\,x-1}{a\,x+1}} \,d x \]

[In]

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

[Out]

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