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/(-a^2*c*x^2+c)^(5/2)-1/8*(1-1/a^2/x^2)^(5/2)*x^5/a/(-a*x+1)^2/(-a^2*c*x^2+c)^(5/2)+(1-
1/a^2/x^2)^(5/2)*x^5/a/(-a*x+1)/(-a^2*c*x^2+c)^(5/2)-1/8*(1-1/a^2/x^2)^(5/2)*x^5/a/(a*x+1)/(-a^2*c*x^2+c)^(5/2
)+23/16*(1-1/a^2/x^2)^(5/2)*x^5*ln(-a*x+1)/a/(-a^2*c*x^2+c)^(5/2)-7/16*(1-1/a^2/x^2)^(5/2)*x^5*ln(a*x+1)/a/(-a
^2*c*x^2+c)^(5/2)

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Rubi [A]  time = 0.24, antiderivative size = 262, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, 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 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 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]

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*}

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Mathematica [A]  time = 0.11, 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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fricas [A]  time = 0.51, size = 138, normalized size = 0.53 \[ -\frac {{\left (16 \, a^{4} x^{4} - 16 \, a^{3} x^{3} - 34 \, a^{2} x^{2} + 18 \, a x - 7 \, {\left (a^{3} x^{3} - a^{2} x^{2} - a x + 1\right )} \log \left (a x + 1\right ) + 23 \, {\left (a^{3} x^{3} - a^{2} x^{2} - a x + 1\right )} \log \left (a x - 1\right ) + 12\right )} \sqrt {-a^{2} c}}{16 \, {\left (a^{10} c^{3} x^{3} - a^{9} c^{3} x^{2} - a^{8} c^{3} x + a^{7} c^{3}\right )}} \]

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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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

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]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [A]  time = 0.07, size = 185, normalized size = 0.71 \[ -\frac {\sqrt {-c \left (a^{2} x^{2}-1\right )}\, \left (16 x^{4} a^{4}+23 \ln \left (a x -1\right ) x^{3} a^{3}-7 a^{3} x^{3} \ln \left (a x +1\right )-16 x^{3} a^{3}-23 \ln \left (a x -1\right ) x^{2} a^{2}+7 \ln \left (a x +1\right ) x^{2} a^{2}-34 a^{2} x^{2}-23 \ln \left (a x -1\right ) x a +7 a x \ln \left (a x +1\right )+18 a x +23 \ln \left (a x -1\right )-7 \ln \left (a x +1\right )+12\right )}{16 \sqrt {\frac {a x -1}{a x +1}}\, \left (a x -1\right ) \left (a^{2} x^{2}-1\right ) c^{3} a^{6} \left (a x +1\right )} \]

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

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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