3.8.3 \(\int \frac {e^{\coth ^{-1}(a x)} x}{(c-a^2 c x^2)^{5/2}} \, dx\) [703]

Optimal. Leaf size=137 \[ -\frac {a^3 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5}{8 (1-a x)^2 \left (c-a^2 c x^2\right )^{5/2}}-\frac {a^3 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5}{8 (1+a x) \left (c-a^2 c x^2\right )^{5/2}}+\frac {a^3 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5 \tanh ^{-1}(a x)}{8 \left (c-a^2 c x^2\right )^{5/2}} \]

[Out]

-1/8*a^3*(1-1/a^2/x^2)^(5/2)*x^5/(-a*x+1)^2/(-a^2*c*x^2+c)^(5/2)-1/8*a^3*(1-1/a^2/x^2)^(5/2)*x^5/(a*x+1)/(-a^2
*c*x^2+c)^(5/2)+1/8*a^3*(1-1/a^2/x^2)^(5/2)*x^5*arctanh(a*x)/(-a^2*c*x^2+c)^(5/2)

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Rubi [A]
time = 0.15, antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {6327, 6328, 78, 213} \begin {gather*} -\frac {a^3 x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}{8 (a x+1) \left (c-a^2 c x^2\right )^{5/2}}-\frac {a^3 x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}{8 (1-a x)^2 \left (c-a^2 c x^2\right )^{5/2}}+\frac {a^3 x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} \tanh ^{-1}(a x)}{8 \left (c-a^2 c x^2\right )^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/8*(a^3*(1 - 1/(a^2*x^2))^(5/2)*x^5)/((1 - a*x)^2*(c - a^2*c*x^2)^(5/2)) - (a^3*(1 - 1/(a^2*x^2))^(5/2)*x^5)
/(8*(1 + a*x)*(c - a^2*c*x^2)^(5/2)) + (a^3*(1 - 1/(a^2*x^2))^(5/2)*x^5*ArcTanh[a*x])/(8*(c - a^2*c*x^2)^(5/2)
)

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 213

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(-1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])]
, x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 6327

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(c + d*x^2)^p/(x^(2*p)*(
1 - 1/(a^2*x^2))^p), Int[u*x^(2*p)*(1 - 1/(a^2*x^2))^p*E^(n*ArcCoth[a*x]), x], x] /; FreeQ[{a, c, d, n, p}, x]
&& EqQ[a^2*c + d, 0] &&  !IntegerQ[n/2] &&  !IntegerQ[p]

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]

Rubi steps

\begin {align*} \int \frac {e^{\coth ^{-1}(a x)} x}{\left (c-a^2 c x^2\right )^{5/2}} \, dx &=\frac {\left (\left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5\right ) \int \frac {e^{\coth ^{-1}(a x)}}{\left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^4} \, dx}{\left (c-a^2 c x^2\right )^{5/2}}\\ &=\frac {\left (a^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5\right ) \int \frac {x}{(-1+a x)^3 (1+a x)^2} \, dx}{\left (c-a^2 c x^2\right )^{5/2}}\\ &=\frac {\left (a^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5\right ) \int \left (\frac {1}{4 a (-1+a x)^3}+\frac {1}{8 a (1+a x)^2}-\frac {1}{8 a \left (-1+a^2 x^2\right )}\right ) \, dx}{\left (c-a^2 c x^2\right )^{5/2}}\\ &=-\frac {a^3 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5}{8 (1-a x)^2 \left (c-a^2 c x^2\right )^{5/2}}-\frac {a^3 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5}{8 (1+a x) \left (c-a^2 c x^2\right )^{5/2}}-\frac {\left (a^4 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5\right ) \int \frac {1}{-1+a^2 x^2} \, dx}{8 \left (c-a^2 c x^2\right )^{5/2}}\\ &=-\frac {a^3 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5}{8 (1-a x)^2 \left (c-a^2 c x^2\right )^{5/2}}-\frac {a^3 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5}{8 (1+a x) \left (c-a^2 c x^2\right )^{5/2}}+\frac {a^3 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5 \tanh ^{-1}(a x)}{8 \left (c-a^2 c x^2\right )^{5/2}}\\ \end {align*}

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Mathematica [A]
time = 0.10, size = 60, normalized size = 0.44 \begin {gather*} -\frac {a^3 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5 \left (\frac {1}{(-1+a x)^2}+\frac {1}{1+a x}-\tanh ^{-1}(a x)\right )}{8 \left (c-a^2 c x^2\right )^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/8*(a^3*(1 - 1/(a^2*x^2))^(5/2)*x^5*((-1 + a*x)^(-2) + (1 + a*x)^(-1) - ArcTanh[a*x]))/(c - a^2*c*x^2)^(5/2)

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Maple [A]
time = 0.09, size = 164, normalized size = 1.20

method	result	size
default	\(-\frac {\sqrt {-c \left (a^{2} x^{2}-1\right )}\, \left (\ln \left (a x +1\right ) a^{3} x^{3}-x^{3} \ln \left (a x -1\right ) a^{3}-\ln \left (a x +1\right ) a^{2} x^{2}+x^{2} \ln \left (a x -1\right ) a^{2}-2 a^{2} x^{2}-\ln \left (a x +1\right ) a x +x \ln \left (a x -1\right ) a +2 a x +\ln \left (a x +1\right )-\ln \left (a x -1\right )-4\right )}{16 \sqrt {\frac {a x -1}{a x +1}}\, \left (a x -1\right ) \left (a^{2} x^{2}-1\right ) c^{3} a^{2} \left (a x +1\right )}\)	\(164\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16/((a*x-1)/(a*x+1))^(1/2)/(a*x-1)*(-c*(a^2*x^2-1))^(1/2)*(ln(a*x+1)*a^3*x^3-x^3*ln(a*x-1)*a^3-ln(a*x+1)*a^
2*x^2+x^2*ln(a*x-1)*a^2-2*a^2*x^2-ln(a*x+1)*a*x+x*ln(a*x-1)*a+2*a*x+ln(a*x+1)-ln(a*x-1)-4)/(a^2*x^2-1)/c^3/a^2
/(a*x+1)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.34, size = 134, normalized size = 0.98 \begin {gather*} -\frac {{\left (a^{4} x^{3} - a^{3} x^{2} - a^{2} x + a\right )} \sqrt {-c} \log \left (\frac {a^{2} c x^{2} - 2 \, \sqrt {-a^{2} c} \sqrt {-c} x + c}{a^{2} x^{2} - 1}\right ) - 2 \, {\left (a^{2} x^{2} - a x + 2\right )} \sqrt {-a^{2} c}}{16 \, {\left (a^{6} c^{3} x^{3} - a^{5} c^{3} x^{2} - a^{4} c^{3} x + a^{3} c^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*((a^4*x^3 - a^3*x^2 - a^2*x + a)*sqrt(-c)*log((a^2*c*x^2 - 2*sqrt(-a^2*c)*sqrt(-c)*x + c)/(a^2*x^2 - 1))
- 2*(a^2*x^2 - a*x + 2)*sqrt(-a^2*c))/(a^6*c^3*x^3 - a^5*c^3*x^2 - a^4*c^3*x + a^3*c^3)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x}{{\left (c-a^2\,c\,x^2\right )}^{5/2}\,\sqrt {\frac {a\,x-1}{a\,x+1}}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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