### 3.699 $$\int \frac{e^{\coth ^{-1}(a x)} x^5}{(c-a^2 c x^2)^{5/2}} \, dx$$

Optimal. Leaf size=262 $\frac{x^6 \left (1-\frac{1}{a^2 x^2}\right )^{5/2}}{\left (c-a^2 c x^2\right )^{5/2}}+\frac{x^5 \left (1-\frac{1}{a^2 x^2}\right )^{5/2}}{a (1-a x) \left (c-a^2 c x^2\right )^{5/2}}-\frac{x^5 \left (1-\frac{1}{a^2 x^2}\right )^{5/2}}{8 a (a x+1) \left (c-a^2 c x^2\right )^{5/2}}-\frac{x^5 \left (1-\frac{1}{a^2 x^2}\right )^{5/2}}{8 a (1-a x)^2 \left (c-a^2 c x^2\right )^{5/2}}+\frac{23 x^5 \left (1-\frac{1}{a^2 x^2}\right )^{5/2} \log (1-a x)}{16 a \left (c-a^2 c x^2\right )^{5/2}}-\frac{7 x^5 \left (1-\frac{1}{a^2 x^2}\right )^{5/2} \log (a x+1)}{16 a \left (c-a^2 c x^2\right )^{5/2}}$

[Out]

((1 - 1/(a^2*x^2))^(5/2)*x^6)/(c - a^2*c*x^2)^(5/2) - ((1 - 1/(a^2*x^2))^(5/2)*x^5)/(8*a*(1 - a*x)^2*(c - a^2*
c*x^2)^(5/2)) + ((1 - 1/(a^2*x^2))^(5/2)*x^5)/(a*(1 - a*x)*(c - a^2*c*x^2)^(5/2)) - ((1 - 1/(a^2*x^2))^(5/2)*x
^5)/(8*a*(1 + a*x)*(c - a^2*c*x^2)^(5/2)) + (23*(1 - 1/(a^2*x^2))^(5/2)*x^5*Log[1 - a*x])/(16*a*(c - a^2*c*x^2
)^(5/2)) - (7*(1 - 1/(a^2*x^2))^(5/2)*x^5*Log[1 + a*x])/(16*a*(c - a^2*c*x^2)^(5/2))

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Rubi [A]  time = 0.242256, antiderivative size = 262, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 25, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.12, Rules used = {6192, 6193, 88} $\frac{x^6 \left (1-\frac{1}{a^2 x^2}\right )^{5/2}}{\left (c-a^2 c x^2\right )^{5/2}}+\frac{x^5 \left (1-\frac{1}{a^2 x^2}\right )^{5/2}}{a (1-a x) \left (c-a^2 c x^2\right )^{5/2}}-\frac{x^5 \left (1-\frac{1}{a^2 x^2}\right )^{5/2}}{8 a (a x+1) \left (c-a^2 c x^2\right )^{5/2}}-\frac{x^5 \left (1-\frac{1}{a^2 x^2}\right )^{5/2}}{8 a (1-a x)^2 \left (c-a^2 c x^2\right )^{5/2}}+\frac{23 x^5 \left (1-\frac{1}{a^2 x^2}\right )^{5/2} \log (1-a x)}{16 a \left (c-a^2 c x^2\right )^{5/2}}-\frac{7 x^5 \left (1-\frac{1}{a^2 x^2}\right )^{5/2} \log (a x+1)}{16 a \left (c-a^2 c x^2\right )^{5/2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

((1 - 1/(a^2*x^2))^(5/2)*x^6)/(c - a^2*c*x^2)^(5/2) - ((1 - 1/(a^2*x^2))^(5/2)*x^5)/(8*a*(1 - a*x)^2*(c - a^2*
c*x^2)^(5/2)) + ((1 - 1/(a^2*x^2))^(5/2)*x^5)/(a*(1 - a*x)*(c - a^2*c*x^2)^(5/2)) - ((1 - 1/(a^2*x^2))^(5/2)*x
^5)/(8*a*(1 + a*x)*(c - a^2*c*x^2)^(5/2)) + (23*(1 - 1/(a^2*x^2))^(5/2)*x^5*Log[1 - a*x])/(16*a*(c - a^2*c*x^2
)^(5/2)) - (7*(1 - 1/(a^2*x^2))^(5/2)*x^5*Log[1 + a*x])/(16*a*(c - a^2*c*x^2)^(5/2))

Rule 6192

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 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 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]))

Rubi steps

\begin{align*} \int \frac{e^{\coth ^{-1}(a x)} x^5}{\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}} \, 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^5}{(-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}{a^5}+\frac{1}{4 a^5 (-1+a x)^3}+\frac{1}{a^5 (-1+a x)^2}+\frac{23}{16 a^5 (-1+a x)}+\frac{1}{8 a^5 (1+a x)^2}-\frac{7}{16 a^5 (1+a x)}\right ) \, dx}{\left (c-a^2 c x^2\right )^{5/2}}\\ &=\frac{\left (1-\frac{1}{a^2 x^2}\right )^{5/2} x^6}{\left (c-a^2 c x^2\right )^{5/2}}-\frac{\left (1-\frac{1}{a^2 x^2}\right )^{5/2} x^5}{8 a (1-a x)^2 \left (c-a^2 c x^2\right )^{5/2}}+\frac{\left (1-\frac{1}{a^2 x^2}\right )^{5/2} x^5}{a (1-a x) \left (c-a^2 c x^2\right )^{5/2}}-\frac{\left (1-\frac{1}{a^2 x^2}\right )^{5/2} x^5}{8 a (1+a x) \left (c-a^2 c x^2\right )^{5/2}}+\frac{23 \left (1-\frac{1}{a^2 x^2}\right )^{5/2} x^5 \log (1-a x)}{16 a \left (c-a^2 c x^2\right )^{5/2}}-\frac{7 \left (1-\frac{1}{a^2 x^2}\right )^{5/2} x^5 \log (1+a x)}{16 a \left (c-a^2 c x^2\right )^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.109289, size = 89, normalized size = 0.34 $\frac{x^5 \left (1-\frac{1}{a^2 x^2}\right )^{5/2} \left (16 a x+\frac{16}{1-a x}-\frac{2}{a x+1}-\frac{2}{(a x-1)^2}+23 \log (1-a x)-7 \log (a x+1)\right )}{16 a \left (c-a^2 c x^2\right )^{5/2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

((1 - 1/(a^2*x^2))^(5/2)*x^5*(16*a*x + 16/(1 - a*x) - 2/(-1 + a*x)^2 - 2/(1 + a*x) + 23*Log[1 - a*x] - 7*Log[1
+ a*x]))/(16*a*(c - a^2*c*x^2)^(5/2))

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Maple [A]  time = 0.139, size = 185, normalized size = 0.7 \begin{align*}{\frac{-16\,{x}^{4}{a}^{4}+7\,{a}^{3}{x}^{3}\ln \left ( ax+1 \right ) -23\,\ln \left ( ax-1 \right ){x}^{3}{a}^{3}+16\,{x}^{3}{a}^{3}-7\,\ln \left ( ax+1 \right ){a}^{2}{x}^{2}+23\,\ln \left ( ax-1 \right ){a}^{2}{x}^{2}+34\,{a}^{2}{x}^{2}-7\,ax\ln \left ( ax+1 \right ) +23\,\ln \left ( ax-1 \right ) xa-18\,ax+7\,\ln \left ( ax+1 \right ) -23\,\ln \left ( ax-1 \right ) -12}{ \left ( 16\,ax-16 \right ) \left ({a}^{2}{x}^{2}-1 \right ){c}^{3}{a}^{6} \left ( ax+1 \right ) }\sqrt{-c \left ({a}^{2}{x}^{2}-1 \right ) }{\frac{1}{\sqrt{{\frac{ax-1}{ax+1}}}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16/((a*x-1)/(a*x+1))^(1/2)/(a*x-1)*(-c*(a^2*x^2-1))^(1/2)*(-16*x^4*a^4+7*a^3*x^3*ln(a*x+1)-23*ln(a*x-1)*x^3*
a^3+16*x^3*a^3-7*ln(a*x+1)*a^2*x^2+23*ln(a*x-1)*a^2*x^2+34*a^2*x^2-7*a*x*ln(a*x+1)+23*ln(a*x-1)*x*a-18*a*x+7*l
n(a*x+1)-23*ln(a*x-1)-12)/(a^2*x^2-1)/c^3/a^6/(a*x+1)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{5}}{{\left (-a^{2} c x^{2} + c\right )}^{\frac{5}{2}} \sqrt{\frac{a x - 1}{a x + 1}}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.59674, size = 296, normalized size = 1.13 \begin{align*} -\frac{{\left (16 \, a^{4} x^{4} - 16 \, a^{3} x^{3} - 34 \, a^{2} x^{2} + 18 \, a x - 7 \,{\left (a^{3} x^{3} - a^{2} x^{2} - a x + 1\right )} \log \left (a x + 1\right ) + 23 \,{\left (a^{3} x^{3} - a^{2} x^{2} - a x + 1\right )} \log \left (a x - 1\right ) + 12\right )} \sqrt{-a^{2} c}}{16 \,{\left (a^{10} c^{3} x^{3} - a^{9} c^{3} x^{2} - a^{8} c^{3} x + a^{7} c^{3}\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(16*a^4*x^4 - 16*a^3*x^3 - 34*a^2*x^2 + 18*a*x - 7*(a^3*x^3 - a^2*x^2 - a*x + 1)*log(a*x + 1) + 23*(a^3*
x^3 - a^2*x^2 - a*x + 1)*log(a*x - 1) + 12)*sqrt(-a^2*c)/(a^10*c^3*x^3 - a^9*c^3*x^2 - a^8*c^3*x + a^7*c^3)

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{5}}{{\left (-a^{2} c x^{2} + c\right )}^{\frac{5}{2}} \sqrt{\frac{a x - 1}{a x + 1}}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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