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

Optimal. Leaf size=176 \[ \frac {a x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}{2 (1-a x) \left (c-a^2 c x^2\right )^{5/2}}-\frac {a x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}{8 (a x+1) \left (c-a^2 c x^2\right )^{5/2}}-\frac {a x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}{8 (1-a x)^2 \left (c-a^2 c x^2\right )^{5/2}}-\frac {3 a x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} \tanh ^{-1}(a x)}{8 \left (c-a^2 c x^2\right )^{5/2}} \]

[Out]

-1/8*a*(1-1/a^2/x^2)^(5/2)*x^5/(-a*x+1)^2/(-a^2*c*x^2+c)^(5/2)+1/2*a*(1-1/a^2/x^2)^(5/2)*x^5/(-a*x+1)/(-a^2*c*
x^2+c)^(5/2)-1/8*a*(1-1/a^2/x^2)^(5/2)*x^5/(a*x+1)/(-a^2*c*x^2+c)^(5/2)-3/8*a*(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 = 0.27, antiderivative size = 176, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {6192, 6193, 88, 207} \[ \frac {a x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}{2 (1-a x) \left (c-a^2 c x^2\right )^{5/2}}-\frac {a x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}{8 (a x+1) \left (c-a^2 c x^2\right )^{5/2}}-\frac {a x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}{8 (1-a x)^2 \left (c-a^2 c x^2\right )^{5/2}}-\frac {3 a x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} \tanh ^{-1}(a x)}{8 \left (c-a^2 c x^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*(1 - 1/(a^2*x^2))^(5/2)*x^5)/(8*(1 - a*x)^2*(c - a^2*c*x^2)^(5/2)) + (a*(1 - 1/(a^2*x^2))^(5/2)*x^5)/(2*(1
- a*x)*(c - a^2*c*x^2)^(5/2)) - (a*(1 - 1/(a^2*x^2))^(5/2)*x^5)/(8*(1 + a*x)*(c - a^2*c*x^2)^(5/2)) - (3*a*(1
- 1/(a^2*x^2))^(5/2)*x^5*ArcTanh[a*x])/(8*(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 207

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

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^3}{\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^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^3}{(-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^3 (-1+a x)^3}+\frac {1}{2 a^3 (-1+a x)^2}+\frac {1}{8 a^3 (1+a x)^2}+\frac {3}{8 a^3 \left (-1+a^2 x^2\right )}\right ) \, dx}{\left (c-a^2 c x^2\right )^{5/2}}\\ &=-\frac {a \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 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5}{2 (1-a x) \left (c-a^2 c x^2\right )^{5/2}}-\frac {a \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 (3 a^2 \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 \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 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5}{2 (1-a x) \left (c-a^2 c x^2\right )^{5/2}}-\frac {a \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 {3 a \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 = 0.07, size = 86, normalized size = 0.49 \[ -\frac {x \sqrt {1-\frac {1}{a^2 x^2}} \left (5 a^2 x^2-a x+3 (a x-1)^2 (a x+1) \tanh ^{-1}(a x)-2\right )}{8 a^3 c^2 (a x-1)^2 (a x+1) \sqrt {c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.62, size = 136, normalized size = 0.77 \[ -\frac {3 \, {\left (a^{4} x^{3} - a^{3} x^{2} - a^{2} x + a\right )} \sqrt {-c} \log \left (\frac {a^{2} c x^{2} + 2 \, \sqrt {-a^{2} c} \sqrt {-c} x + c}{a^{2} x^{2} - 1}\right ) - 2 \, {\left (5 \, a^{2} x^{2} - a x - 2\right )} \sqrt {-a^{2} c}}{16 \, {\left (a^{8} c^{3} x^{3} - a^{7} c^{3} x^{2} - a^{6} c^{3} x + a^{5} c^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((a*x-1)/(a*x+1))^(1/2)*x^3/(-a^2*c*x^2+c)^(5/2),x, algorithm="fricas")

[Out]

-1/16*(3*(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*(5*a^2*x^2 - a*x - 2)*sqrt(-a^2*c))/(a^8*c^3*x^3 - a^7*c^3*x^2 - a^6*c^3*x + a^5*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^3/(-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 = 0.06, size = 169, normalized size = 0.96 \[ -\frac {\sqrt {-c \left (a^{2} x^{2}-1\right )}\, \left (3 \ln \left (a x -1\right ) x^{3} a^{3}-3 a^{3} x^{3} \ln \left (a x +1\right )-3 \ln \left (a x -1\right ) x^{2} a^{2}+3 \ln \left (a x +1\right ) x^{2} a^{2}-10 a^{2} x^{2}-3 \ln \left (a x -1\right ) x a +3 a x \ln \left (a x +1\right )+2 a x +3 \ln \left (a x -1\right )-3 \ln \left (a x +1\right )+4\right )}{16 \sqrt {\frac {a x -1}{a x +1}}\, \left (a x -1\right ) \left (a^{2} x^{2}-1\right ) c^{3} a^{4} \left (a x +1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a*x-1)/(a*x+1))^(1/2)*x^3/(-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)*(3*ln(a*x-1)*x^3*a^3-3*a^3*x^3*ln(a*x+1)-3*ln(a*x
-1)*x^2*a^2+3*ln(a*x+1)*x^2*a^2-10*a^2*x^2-3*ln(a*x-1)*x*a+3*a*x*ln(a*x+1)+2*a*x+3*ln(a*x-1)-3*ln(a*x+1)+4)/(a
^2*x^2-1)/c^3/a^4/(a*x+1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((a*x-1)/(a*x+1))^(1/2)*x^3/(-a^2*c*x^2+c)^(5/2),x, algorithm="maxima")

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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