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

Optimal. Leaf size=182 \[ \frac {a^4 x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}{8 (a x+1) \left (c-a^2 c x^2\right )^{5/2}}+\frac {a^4 x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}{8 (a x+1)^2 \left (c-a^2 c x^2\right )^{5/2}}+\frac {a^4 x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}{6 (a x+1)^3 \left (c-a^2 c x^2\right )^{5/2}}-\frac {a^4 x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} \tanh ^{-1}(a x)}{8 \left (c-a^2 c x^2\right )^{5/2}} \]

[Out]

1/6*a^4*(1-1/a^2/x^2)^(5/2)*x^5/(a*x+1)^3/(-a^2*c*x^2+c)^(5/2)+1/8*a^4*(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^4*(1-1/a^2/x^2)^(5/2)*x^5/(a*x+1)/(-a^2*c*x^2+c)^(5/2)-1/8*a^4*(1-1/a^2/x^2)^(5/2)*x^5*a
rctanh(a*x)/(-a^2*c*x^2+c)^(5/2)

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Rubi [A]  time = 0.20, antiderivative size = 182, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {6192, 6193, 44, 207} \[ \frac {a^4 x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}{8 (a x+1) \left (c-a^2 c x^2\right )^{5/2}}+\frac {a^4 x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}{8 (a x+1)^2 \left (c-a^2 c x^2\right )^{5/2}}+\frac {a^4 x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}{6 (a x+1)^3 \left (c-a^2 c x^2\right )^{5/2}}-\frac {a^4 x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} \tanh ^{-1}(a x)}{8 \left (c-a^2 c x^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^4*(1 - 1/(a^2*x^2))^(5/2)*x^5)/(6*(1 + a*x)^3*(c - a^2*c*x^2)^(5/2)) + (a^4*(1 - 1/(a^2*x^2))^(5/2)*x^5)/(8
*(1 + a*x)^2*(c - a^2*c*x^2)^(5/2)) + (a^4*(1 - 1/(a^2*x^2))^(5/2)*x^5)/(8*(1 + a*x)*(c - a^2*c*x^2)^(5/2)) -
(a^4*(1 - 1/(a^2*x^2))^(5/2)*x^5*ArcTanh[a*x])/(8*(c - a^2*c*x^2)^(5/2))

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

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^{-3 \coth ^{-1}(a 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^{-3 \coth ^{-1}(a x)}}{\left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5} \, 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 {1}{(-1+a x) (1+a 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 \left (-\frac {1}{2 (1+a x)^4}-\frac {1}{4 (1+a x)^3}-\frac {1}{8 (1+a x)^2}+\frac {1}{8 \left (-1+a^2 x^2\right )}\right ) \, dx}{\left (c-a^2 c x^2\right )^{5/2}}\\ &=\frac {a^4 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5}{6 (1+a x)^3 \left (c-a^2 c x^2\right )^{5/2}}+\frac {a^4 \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^4 \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^5 \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^4 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5}{6 (1+a x)^3 \left (c-a^2 c x^2\right )^{5/2}}+\frac {a^4 \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^4 \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^4 \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 = 71, normalized size = 0.39 \[ -\frac {x \sqrt {1-\frac {1}{a^2 x^2}} \left (-3 a^2 x^2-9 a x+3 (a x+1)^3 \tanh ^{-1}(a x)-10\right )}{24 c^2 (a x+1)^3 \sqrt {c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/24*(Sqrt[1 - 1/(a^2*x^2)]*x*(-10 - 9*a*x - 3*a^2*x^2 + 3*(1 + a*x)^3*ArcTanh[a*x]))/(c^2*(1 + a*x)^3*Sqrt[c
- a^2*c*x^2])

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fricas [A]  time = 0.56, size = 136, normalized size = 0.75 \[ -\frac {3 \, {\left (a^{4} x^{3} + 3 \, a^{3} x^{2} + 3 \, a^{2} x + a\right )} \sqrt {-c} \log \left (\frac {a^{2} c x^{2} + 2 \, \sqrt {-a^{2} c} \sqrt {-c} x + c}{a^{2} x^{2} - 1}\right ) + 2 \, {\left (3 \, a^{2} x^{2} + 9 \, a x + 10\right )} \sqrt {-a^{2} c}}{48 \, {\left (a^{5} c^{3} x^{3} + 3 \, a^{4} c^{3} x^{2} + 3 \, a^{3} c^{3} x + a^{2} c^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/48*(3*(a^4*x^3 + 3*a^3*x^2 + 3*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*(3*a^2*x^2 + 9*a*x + 10)*sqrt(-a^2*c))/(a^5*c^3*x^3 + 3*a^4*c^3*x^2 + 3*a^3*c^3*x + a^2*c^3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.06, size = 169, normalized size = 0.93 \[ -\frac {\left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}} \sqrt {-c \left (a^{2} x^{2}-1\right )}\, \left (3 \ln \left (a x -1\right ) x^{3} a^{3}-3 a^{3} x^{3} \ln \left (a x +1\right )+9 \ln \left (a x -1\right ) x^{2} a^{2}-9 \ln \left (a x +1\right ) x^{2} a^{2}+6 a^{2} x^{2}+9 \ln \left (a x -1\right ) x a -9 a x \ln \left (a x +1\right )+18 a x +3 \ln \left (a x -1\right )-3 \ln \left (a x +1\right )+20\right )}{48 \left (a x +1\right ) \left (a x -1\right ) \left (a^{2} x^{2}-1\right ) c^{3} a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/48*((a*x-1)/(a*x+1))^(3/2)/(a*x+1)/(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)+
9*ln(a*x-1)*x^2*a^2-9*ln(a*x+1)*x^2*a^2+6*a^2*x^2+9*ln(a*x-1)*x*a-9*a*x*ln(a*x+1)+18*a*x+3*ln(a*x-1)-3*ln(a*x+
1)+20)/(a^2*x^2-1)/c^3/a

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (\frac {a\,x-1}{a\,x+1}\right )}^{3/2}}{{\left (c-a^2\,c\,x^2\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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