3.597 \(\int \frac {e^{-\coth ^{-1}(a x)}}{(c-a^2 c x^2)^4} \, dx\)

Optimal. Leaf size=127 \[ \frac {8 (2 a x+1) e^{-\coth ^{-1}(a x)}}{35 a c^4 \left (1-a^2 x^2\right )}+\frac {2 (4 a x+1) e^{-\coth ^{-1}(a x)}}{35 a c^4 \left (1-a^2 x^2\right )^2}+\frac {(6 a x+1) e^{-\coth ^{-1}(a x)}}{35 a c^4 \left (1-a^2 x^2\right )^3}-\frac {16 e^{-\coth ^{-1}(a x)}}{35 a c^4} \]

[Out]

-16/35/a/c^4*((a*x-1)/(a*x+1))^(1/2)+1/35*(6*a*x+1)/a/c^4*((a*x-1)/(a*x+1))^(1/2)/(-a^2*x^2+1)^3+2/35*(4*a*x+1
)/a/c^4*((a*x-1)/(a*x+1))^(1/2)/(-a^2*x^2+1)^2+8/35*(2*a*x+1)/a/c^4*((a*x-1)/(a*x+1))^(1/2)/(-a^2*x^2+1)

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Rubi [A]  time = 0.14, antiderivative size = 127, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {6185, 6183} \[ \frac {8 (2 a x+1) e^{-\coth ^{-1}(a x)}}{35 a c^4 \left (1-a^2 x^2\right )}+\frac {2 (4 a x+1) e^{-\coth ^{-1}(a x)}}{35 a c^4 \left (1-a^2 x^2\right )^2}+\frac {(6 a x+1) e^{-\coth ^{-1}(a x)}}{35 a c^4 \left (1-a^2 x^2\right )^3}-\frac {16 e^{-\coth ^{-1}(a x)}}{35 a c^4} \]

Antiderivative was successfully verified.

[In]

Int[1/(E^ArcCoth[a*x]*(c - a^2*c*x^2)^4),x]

[Out]

-16/(35*a*c^4*E^ArcCoth[a*x]) + (1 + 6*a*x)/(35*a*c^4*E^ArcCoth[a*x]*(1 - a^2*x^2)^3) + (2*(1 + 4*a*x))/(35*a*
c^4*E^ArcCoth[a*x]*(1 - a^2*x^2)^2) + (8*(1 + 2*a*x))/(35*a*c^4*E^ArcCoth[a*x]*(1 - a^2*x^2))

Rule 6183

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 6185

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 )^4} \, dx &=\frac {e^{-\coth ^{-1}(a x)} (1+6 a x)}{35 a c^4 \left (1-a^2 x^2\right )^3}+\frac {6 \int \frac {e^{-\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx}{7 c}\\ &=\frac {e^{-\coth ^{-1}(a x)} (1+6 a x)}{35 a c^4 \left (1-a^2 x^2\right )^3}+\frac {2 e^{-\coth ^{-1}(a x)} (1+4 a x)}{35 a c^4 \left (1-a^2 x^2\right )^2}+\frac {24 \int \frac {e^{-\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^2} \, dx}{35 c^2}\\ &=\frac {e^{-\coth ^{-1}(a x)} (1+6 a x)}{35 a c^4 \left (1-a^2 x^2\right )^3}+\frac {2 e^{-\coth ^{-1}(a x)} (1+4 a x)}{35 a c^4 \left (1-a^2 x^2\right )^2}+\frac {8 e^{-\coth ^{-1}(a x)} (1+2 a x)}{35 a c^4 \left (1-a^2 x^2\right )}+\frac {16 \int \frac {e^{-\coth ^{-1}(a x)}}{c-a^2 c x^2} \, dx}{35 c^3}\\ &=-\frac {16 e^{-\coth ^{-1}(a x)}}{35 a c^4}+\frac {e^{-\coth ^{-1}(a x)} (1+6 a x)}{35 a c^4 \left (1-a^2 x^2\right )^3}+\frac {2 e^{-\coth ^{-1}(a x)} (1+4 a x)}{35 a c^4 \left (1-a^2 x^2\right )^2}+\frac {8 e^{-\coth ^{-1}(a x)} (1+2 a x)}{35 a c^4 \left (1-a^2 x^2\right )}\\ \end {align*}

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Mathematica [A]  time = 0.29, size = 80, normalized size = 0.63 \[ -\frac {x \sqrt {1-\frac {1}{a^2 x^2}} \left (16 a^6 x^6+16 a^5 x^5-40 a^4 x^4-40 a^3 x^3+30 a^2 x^2+30 a x-5\right )}{35 (a x-1)^3 (a c x+c)^4} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/(E^ArcCoth[a*x]*(c - a^2*c*x^2)^4),x]

[Out]

-1/35*(Sqrt[1 - 1/(a^2*x^2)]*x*(-5 + 30*a*x + 30*a^2*x^2 - 40*a^3*x^3 - 40*a^4*x^4 + 16*a^5*x^5 + 16*a^6*x^6))
/((-1 + a*x)^3*(c + a*c*x)^4)

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fricas [A]  time = 0.63, size = 104, normalized size = 0.82 \[ -\frac {{\left (16 \, a^{6} x^{6} + 16 \, a^{5} x^{5} - 40 \, a^{4} x^{4} - 40 \, a^{3} x^{3} + 30 \, a^{2} x^{2} + 30 \, a x - 5\right )} \sqrt {\frac {a x - 1}{a x + 1}}}{35 \, {\left (a^{7} c^{4} x^{6} - 3 \, a^{5} c^{4} x^{4} + 3 \, a^{3} c^{4} x^{2} - a c^{4}\right )}} \]

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)^4,x, algorithm="fricas")

[Out]

-1/35*(16*a^6*x^6 + 16*a^5*x^5 - 40*a^4*x^4 - 40*a^3*x^3 + 30*a^2*x^2 + 30*a*x - 5)*sqrt((a*x - 1)/(a*x + 1))/
(a^7*c^4*x^6 - 3*a^5*c^4*x^4 + 3*a^3*c^4*x^2 - a*c^4)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {\frac {a x - 1}{a x + 1}}}{{\left (a^{2} c x^{2} - c\right )}^{4}}\,{d x} \]

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)^4,x, algorithm="giac")

[Out]

integrate(sqrt((a*x - 1)/(a*x + 1))/(a^2*c*x^2 - c)^4, x)

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maple [A]  time = 0.04, size = 81, normalized size = 0.64 \[ -\frac {\sqrt {\frac {a x -1}{a x +1}}\, \left (16 x^{6} a^{6}+16 x^{5} a^{5}-40 x^{4} a^{4}-40 x^{3} a^{3}+30 a^{2} x^{2}+30 a x -5\right )}{35 \left (a^{2} x^{2}-1\right )^{3} c^{4} a} \]

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)^4,x)

[Out]

-1/35*((a*x-1)/(a*x+1))^(1/2)*(16*a^6*x^6+16*a^5*x^5-40*a^4*x^4-40*a^3*x^3+30*a^2*x^2+30*a*x-5)/(a^2*x^2-1)^3/
c^4/a

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maxima [A]  time = 0.31, size = 135, normalized size = 1.06 \[ \frac {1}{2240} \, a {\left (\frac {5 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {7}{2}} - 42 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {5}{2}} + 175 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}} - 700 \, \sqrt {\frac {a x - 1}{a x + 1}}}{a^{2} c^{4}} + \frac {7 \, {\left (\frac {10 \, {\left (a x - 1\right )}}{a x + 1} - \frac {75 \, {\left (a x - 1\right )}^{2}}{{\left (a x + 1\right )}^{2}} - 1\right )}}{a^{2} c^{4} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {5}{2}}}\right )} \]

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)^4,x, algorithm="maxima")

[Out]

1/2240*a*((5*((a*x - 1)/(a*x + 1))^(7/2) - 42*((a*x - 1)/(a*x + 1))^(5/2) + 175*((a*x - 1)/(a*x + 1))^(3/2) -
700*sqrt((a*x - 1)/(a*x + 1)))/(a^2*c^4) + 7*(10*(a*x - 1)/(a*x + 1) - 75*(a*x - 1)^2/(a*x + 1)^2 - 1)/(a^2*c^
4*((a*x - 1)/(a*x + 1))^(5/2)))

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mupad [B]  time = 0.04, size = 148, normalized size = 1.17 \[ \frac {5\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{3/2}}{64\,a\,c^4}-\frac {5\,\sqrt {\frac {a\,x-1}{a\,x+1}}}{16\,a\,c^4}-\frac {3\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{5/2}}{160\,a\,c^4}+\frac {{\left (\frac {a\,x-1}{a\,x+1}\right )}^{7/2}}{448\,a\,c^4}-\frac {\frac {15\,{\left (a\,x-1\right )}^2}{{\left (a\,x+1\right )}^2}-\frac {2\,\left (a\,x-1\right )}{a\,x+1}+\frac {1}{5}}{64\,a\,c^4\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{5/2}} \]

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)^4,x)

[Out]

(5*((a*x - 1)/(a*x + 1))^(3/2))/(64*a*c^4) - (5*((a*x - 1)/(a*x + 1))^(1/2))/(16*a*c^4) - (3*((a*x - 1)/(a*x +
1))^(5/2))/(160*a*c^4) + ((a*x - 1)/(a*x + 1))^(7/2)/(448*a*c^4) - ((15*(a*x - 1)^2)/(a*x + 1)^2 - (2*(a*x -
1))/(a*x + 1) + 1/5)/(64*a*c^4*((a*x - 1)/(a*x + 1))^(5/2))

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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))**(1/2)/(-a**2*c*x**2+c)**4,x)

[Out]

Timed out

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