3.6.96 \(\int \frac {e^{-\coth ^{-1}(a x)}}{(c-a^2 c x^2)^3} \, dx\) [596]

Optimal. Leaf size=91 \[ -\frac {8 e^{-\coth ^{-1}(a x)}}{15 a c^3}+\frac {e^{-\coth ^{-1}(a x)} (1+4 a x)}{15 a c^3 \left (1-a^2 x^2\right )^2}+\frac {4 e^{-\coth ^{-1}(a x)} (1+2 a x)}{15 a c^3 \left (1-a^2 x^2\right )} \]

[Out]

-8/15/a/c^3*((a*x-1)/(a*x+1))^(1/2)+1/15*(4*a*x+1)/a/c^3*((a*x-1)/(a*x+1))^(1/2)/(-a^2*x^2+1)^2+4/15*(2*a*x+1)
/a/c^3*((a*x-1)/(a*x+1))^(1/2)/(-a^2*x^2+1)

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Rubi [A]
time = 0.07, antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {6320, 6318} \begin {gather*} \frac {4 (2 a x+1) e^{-\coth ^{-1}(a x)}}{15 a c^3 \left (1-a^2 x^2\right )}+\frac {(4 a x+1) e^{-\coth ^{-1}(a x)}}{15 a c^3 \left (1-a^2 x^2\right )^2}-\frac {8 e^{-\coth ^{-1}(a x)}}{15 a c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-8/(15*a*c^3*E^ArcCoth[a*x]) + (1 + 4*a*x)/(15*a*c^3*E^ArcCoth[a*x]*(1 - a^2*x^2)^2) + (4*(1 + 2*a*x))/(15*a*c
^3*E^ArcCoth[a*x]*(1 - a^2*x^2))

Rule 6318

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))/((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[E^(n*ArcCoth[a*x])/(a*c*n), x] /; F
reeQ[{a, c, d, n}, x] && EqQ[a^2*c + d, 0] &&  !IntegerQ[n/2]

Rule 6320

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(n + 2*a*(p + 1)*x)*(c + d*x^2
)^(p + 1)*(E^(n*ArcCoth[a*x])/(a*c*(n^2 - 4*(p + 1)^2))), x] - Dist[2*(p + 1)*((2*p + 3)/(c*(n^2 - 4*(p + 1)^2
))), Int[(c + d*x^2)^(p + 1)*E^(n*ArcCoth[a*x]), x], x] /; FreeQ[{a, c, d, n}, x] && EqQ[a^2*c + d, 0] &&  !In
tegerQ[n/2] && LtQ[p, -1] && NeQ[p, -3/2] && NeQ[n^2 - 4*(p + 1)^2, 0] && (IntegerQ[p] ||  !IntegerQ[n])

Rubi steps

\begin {align*} \int \frac {e^{-\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx &=\frac {e^{-\coth ^{-1}(a x)} (1+4 a x)}{15 a c^3 \left (1-a^2 x^2\right )^2}+\frac {4 \int \frac {e^{-\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^2} \, dx}{5 c}\\ &=\frac {e^{-\coth ^{-1}(a x)} (1+4 a x)}{15 a c^3 \left (1-a^2 x^2\right )^2}+\frac {4 e^{-\coth ^{-1}(a x)} (1+2 a x)}{15 a c^3 \left (1-a^2 x^2\right )}+\frac {8 \int \frac {e^{-\coth ^{-1}(a x)}}{c-a^2 c x^2} \, dx}{15 c^2}\\ &=-\frac {8 e^{-\coth ^{-1}(a x)}}{15 a c^3}+\frac {e^{-\coth ^{-1}(a x)} (1+4 a x)}{15 a c^3 \left (1-a^2 x^2\right )^2}+\frac {4 e^{-\coth ^{-1}(a x)} (1+2 a x)}{15 a c^3 \left (1-a^2 x^2\right )}\\ \end {align*}

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Mathematica [A]
time = 0.07, size = 64, normalized size = 0.70 \begin {gather*} -\frac {\sqrt {1-\frac {1}{a^2 x^2}} x \left (3-12 a x-12 a^2 x^2+8 a^3 x^3+8 a^4 x^4\right )}{15 (-1+a x)^2 (c+a c x)^3} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-1/15*(Sqrt[1 - 1/(a^2*x^2)]*x*(3 - 12*a*x - 12*a^2*x^2 + 8*a^3*x^3 + 8*a^4*x^4))/((-1 + a*x)^2*(c + a*c*x)^3)

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Maple [A]
time = 0.21, size = 68, normalized size = 0.75

method	result	size
gosper	\(-\frac {\sqrt {\frac {a x -1}{a x +1}}\, \left (8 a^{4} x^{4}+8 a^{3} x^{3}-12 a^{2} x^{2}-12 a x +3\right )}{15 \left (a^{2} x^{2}-1\right )^{2} c^{3} a}\)	\(65\)
default	\(-\frac {\sqrt {\frac {a x -1}{a x +1}}\, \left (8 a^{4} x^{4}+8 a^{3} x^{3}-12 a^{2} x^{2}-12 a x +3\right )}{15 c^{3} \left (a x +1\right )^{2} a \left (a x -1\right )^{2}}\)	\(68\)
trager	\(-\frac {\left (8 a^{4} x^{4}+8 a^{3} x^{3}-12 a^{2} x^{2}-12 a x +3\right ) \sqrt {-\frac {-a x +1}{a x +1}}}{15 a \,c^{3} \left (a x -1\right )^{2} \left (a x +1\right )^{2}}\)	\(70\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a*x-1)/(a*x+1))^(1/2)/(-a^2*c*x^2+c)^3,x,method=_RETURNVERBOSE)

[Out]

-1/15*((a*x-1)/(a*x+1))^(1/2)*(8*a^4*x^4+8*a^3*x^3-12*a^2*x^2-12*a*x+3)/c^3/(a*x+1)^2/a/(a*x-1)^2

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Maxima [A]
time = 0.27, size = 102, normalized size = 1.12 \begin {gather*} -\frac {1}{240} \, a {\left (\frac {3 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {5}{2}} - 20 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}} + 90 \, \sqrt {\frac {a x - 1}{a x + 1}}}{a^{2} c^{3}} + \frac {5 \, {\left (\frac {12 \, {\left (a x - 1\right )}}{a x + 1} - 1\right )}}{a^{2} c^{3} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/240*a*((3*((a*x - 1)/(a*x + 1))^(5/2) - 20*((a*x - 1)/(a*x + 1))^(3/2) + 90*sqrt((a*x - 1)/(a*x + 1)))/(a^2
*c^3) + 5*(12*(a*x - 1)/(a*x + 1) - 1)/(a^2*c^3*((a*x - 1)/(a*x + 1))^(3/2)))

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Fricas [A]
time = 0.35, size = 76, normalized size = 0.84 \begin {gather*} -\frac {{\left (8 \, a^{4} x^{4} + 8 \, a^{3} x^{3} - 12 \, a^{2} x^{2} - 12 \, a x + 3\right )} \sqrt {\frac {a x - 1}{a x + 1}}}{15 \, {\left (a^{5} c^{3} x^{4} - 2 \, a^{3} c^{3} x^{2} + a c^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*(8*a^4*x^4 + 8*a^3*x^3 - 12*a^2*x^2 - 12*a*x + 3)*sqrt((a*x - 1)/(a*x + 1))/(a^5*c^3*x^4 - 2*a^3*c^3*x^2
+ a*c^3)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \frac {\int \frac {\sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{a^{6} x^{6} - 3 a^{4} x^{4} + 3 a^{2} x^{2} - 1}\, dx}{c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral(sqrt(a*x/(a*x + 1) - 1/(a*x + 1))/(a**6*x**6 - 3*a**4*x**4 + 3*a**2*x**2 - 1), x)/c**3

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

[Out]

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

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Mupad [B]
time = 1.24, size = 109, normalized size = 1.20 \begin {gather*} \frac {{\left (\frac {a\,x-1}{a\,x+1}\right )}^{3/2}}{12\,a\,c^3}-\frac {3\,\sqrt {\frac {a\,x-1}{a\,x+1}}}{8\,a\,c^3}-\frac {{\left (\frac {a\,x-1}{a\,x+1}\right )}^{5/2}}{80\,a\,c^3}-\frac {\frac {4\,\left (a\,x-1\right )}{a\,x+1}-\frac {1}{3}}{16\,a\,c^3\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((a*x - 1)/(a*x + 1))^(3/2)/(12*a*c^3) - (3*((a*x - 1)/(a*x + 1))^(1/2))/(8*a*c^3) - ((a*x - 1)/(a*x + 1))^(5/
2)/(80*a*c^3) - ((4*(a*x - 1))/(a*x + 1) - 1/3)/(16*a*c^3*((a*x - 1)/(a*x + 1))^(3/2))

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