3.668 \(\int \frac {e^{-\tanh ^{-1}(a x)}}{(c-\frac {c}{a^2 x^2})^3} \, dx\)

Optimal. Leaf size=130 \[ -\frac {x (5-8 a x)}{5 c^3 \sqrt {1-a^2 x^2}}+\frac {16 \sqrt {1-a^2 x^2}}{5 a c^3}+\frac {a^2 x^3 (5-6 a x)}{15 c^3 \left (1-a^2 x^2\right )^{3/2}}-\frac {a^4 x^5 (1-a x)}{5 c^3 \left (1-a^2 x^2\right )^{5/2}}+\frac {\sin ^{-1}(a x)}{a c^3} \]

[Out]

-1/5*a^4*x^5*(-a*x+1)/c^3/(-a^2*x^2+1)^(5/2)+1/15*a^2*x^3*(-6*a*x+5)/c^3/(-a^2*x^2+1)^(3/2)+arcsin(a*x)/a/c^3-
1/5*x*(-8*a*x+5)/c^3/(-a^2*x^2+1)^(1/2)+16/5*(-a^2*x^2+1)^(1/2)/a/c^3

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Rubi [A]  time = 0.20, antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {6157, 6149, 819, 641, 216} \[ -\frac {a^4 x^5 (1-a x)}{5 c^3 \left (1-a^2 x^2\right )^{5/2}}+\frac {a^2 x^3 (5-6 a x)}{15 c^3 \left (1-a^2 x^2\right )^{3/2}}-\frac {x (5-8 a x)}{5 c^3 \sqrt {1-a^2 x^2}}+\frac {16 \sqrt {1-a^2 x^2}}{5 a c^3}+\frac {\sin ^{-1}(a x)}{a c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^4*x^5*(1 - a*x))/(5*c^3*(1 - a^2*x^2)^(5/2)) + (a^2*x^3*(5 - 6*a*x))/(15*c^3*(1 - a^2*x^2)^(3/2)) - (x*(5
- 8*a*x))/(5*c^3*Sqrt[1 - a^2*x^2]) + (16*Sqrt[1 - a^2*x^2])/(5*a*c^3) + ArcSin[a*x]/(a*c^3)

Rule 216

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

Rule 641

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(e*(a + c*x^2)^(p + 1))/(2*c*(p + 1)),
x] + Dist[d, Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[p, -1]

Rule 819

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(
m - 1)*(a + c*x^2)^(p + 1)*(a*(e*f + d*g) - (c*d*f - a*e*g)*x))/(2*a*c*(p + 1)), x] - Dist[1/(2*a*c*(p + 1)),
Int[(d + e*x)^(m - 2)*(a + c*x^2)^(p + 1)*Simp[a*e*(e*f*(m - 1) + d*g*m) - c*d^2*f*(2*p + 3) + e*(a*e*g*m - c*
d*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && GtQ
[m, 1] && (EqQ[d, 0] || (EqQ[m, 2] && EqQ[p, -3] && RationalQ[a, c, d, e, f, g]) ||  !ILtQ[m + 2*p + 3, 0])

Rule 6149

Int[E^(ArcTanh[(a_.)*(x_)]*(n_))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[c^p, Int[(x^m*(1 -
a^2*x^2)^(p + n/2))/(1 - a*x)^n, x], x] /; FreeQ[{a, c, d, m, p}, x] && EqQ[a^2*c + d, 0] && (IntegerQ[p] || G
tQ[c, 0]) && ILtQ[(n - 1)/2, 0] &&  !IntegerQ[p - n/2]

Rule 6157

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_)^2)^(p_.), x_Symbol] :> Dist[d^p, Int[(u*(1 - a^2*x^
2)^p*E^(n*ArcTanh[a*x]))/x^(2*p), x], x] /; FreeQ[{a, c, d, n}, x] && EqQ[c + a^2*d, 0] && IntegerQ[p]

Rubi steps

\begin {align*} \int \frac {e^{-\tanh ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^3} \, dx &=-\frac {a^6 \int \frac {e^{-\tanh ^{-1}(a x)} x^6}{\left (1-a^2 x^2\right )^3} \, dx}{c^3}\\ &=-\frac {a^6 \int \frac {x^6 (1-a x)}{\left (1-a^2 x^2\right )^{7/2}} \, dx}{c^3}\\ &=-\frac {a^4 x^5 (1-a x)}{5 c^3 \left (1-a^2 x^2\right )^{5/2}}+\frac {a^4 \int \frac {x^4 (5-6 a x)}{\left (1-a^2 x^2\right )^{5/2}} \, dx}{5 c^3}\\ &=-\frac {a^4 x^5 (1-a x)}{5 c^3 \left (1-a^2 x^2\right )^{5/2}}+\frac {a^2 x^3 (5-6 a x)}{15 c^3 \left (1-a^2 x^2\right )^{3/2}}-\frac {a^2 \int \frac {x^2 (15-24 a x)}{\left (1-a^2 x^2\right )^{3/2}} \, dx}{15 c^3}\\ &=-\frac {a^4 x^5 (1-a x)}{5 c^3 \left (1-a^2 x^2\right )^{5/2}}+\frac {a^2 x^3 (5-6 a x)}{15 c^3 \left (1-a^2 x^2\right )^{3/2}}-\frac {x (5-8 a x)}{5 c^3 \sqrt {1-a^2 x^2}}+\frac {\int \frac {15-48 a x}{\sqrt {1-a^2 x^2}} \, dx}{15 c^3}\\ &=-\frac {a^4 x^5 (1-a x)}{5 c^3 \left (1-a^2 x^2\right )^{5/2}}+\frac {a^2 x^3 (5-6 a x)}{15 c^3 \left (1-a^2 x^2\right )^{3/2}}-\frac {x (5-8 a x)}{5 c^3 \sqrt {1-a^2 x^2}}+\frac {16 \sqrt {1-a^2 x^2}}{5 a c^3}+\frac {\int \frac {1}{\sqrt {1-a^2 x^2}} \, dx}{c^3}\\ &=-\frac {a^4 x^5 (1-a x)}{5 c^3 \left (1-a^2 x^2\right )^{5/2}}+\frac {a^2 x^3 (5-6 a x)}{15 c^3 \left (1-a^2 x^2\right )^{3/2}}-\frac {x (5-8 a x)}{5 c^3 \sqrt {1-a^2 x^2}}+\frac {16 \sqrt {1-a^2 x^2}}{5 a c^3}+\frac {\sin ^{-1}(a x)}{a c^3}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 108, normalized size = 0.83 \[ \frac {-15 a^5 x^5-38 a^4 x^4+52 a^3 x^3+87 a^2 x^2+15 (a x-1) (a x+1)^2 \sqrt {1-a^2 x^2} \sin ^{-1}(a x)-33 a x-48}{15 a c^3 (a x-1) (a x+1)^2 \sqrt {1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-48 - 33*a*x + 87*a^2*x^2 + 52*a^3*x^3 - 38*a^4*x^4 - 15*a^5*x^5 + 15*(-1 + a*x)*(1 + a*x)^2*Sqrt[1 - a^2*x^2
]*ArcSin[a*x])/(15*a*c^3*(-1 + a*x)*(1 + a*x)^2*Sqrt[1 - a^2*x^2])

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fricas [A]  time = 0.50, size = 208, normalized size = 1.60 \[ \frac {48 \, a^{5} x^{5} + 48 \, a^{4} x^{4} - 96 \, a^{3} x^{3} - 96 \, a^{2} x^{2} + 48 \, a x - 30 \, {\left (a^{5} x^{5} + a^{4} x^{4} - 2 \, a^{3} x^{3} - 2 \, a^{2} x^{2} + a x + 1\right )} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) + {\left (15 \, a^{5} x^{5} + 38 \, a^{4} x^{4} - 52 \, a^{3} x^{3} - 87 \, a^{2} x^{2} + 33 \, a x + 48\right )} \sqrt {-a^{2} x^{2} + 1} + 48}{15 \, {\left (a^{6} c^{3} x^{5} + a^{5} c^{3} x^{4} - 2 \, a^{4} c^{3} x^{3} - 2 \, a^{3} c^{3} x^{2} + a^{2} c^{3} x + a c^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(48*a^5*x^5 + 48*a^4*x^4 - 96*a^3*x^3 - 96*a^2*x^2 + 48*a*x - 30*(a^5*x^5 + a^4*x^4 - 2*a^3*x^3 - 2*a^2*x
^2 + a*x + 1)*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) + (15*a^5*x^5 + 38*a^4*x^4 - 52*a^3*x^3 - 87*a^2*x^2 + 33
*a*x + 48)*sqrt(-a^2*x^2 + 1) + 48)/(a^6*c^3*x^5 + a^5*c^3*x^4 - 2*a^4*c^3*x^3 - 2*a^3*c^3*x^2 + a^2*c^3*x + a
*c^3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.06, size = 356, normalized size = 2.74 \[ \frac {\left (-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )\right )^{\frac {3}{2}}}{4 a^{3} c^{3} \left (x -\frac {1}{a}\right )^{2}}+\frac {19 \sqrt {-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )}}{32 a \,c^{3}}-\frac {19 \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )}}\right )}{32 c^{3} \sqrt {a^{2}}}+\frac {\left (-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )\right )^{\frac {3}{2}}}{48 a^{4} c^{3} \left (x -\frac {1}{a}\right )^{3}}+\frac {15 \left (-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )\right )^{\frac {3}{2}}}{16 a^{3} c^{3} \left (x +\frac {1}{a}\right )^{2}}+\frac {51 \sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}{32 a \,c^{3}}+\frac {51 \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}\right )}{32 c^{3} \sqrt {a^{2}}}-\frac {43 \left (-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )\right )^{\frac {3}{2}}}{240 a^{4} c^{3} \left (x +\frac {1}{a}\right )^{3}}+\frac {\left (-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )\right )^{\frac {3}{2}}}{40 a^{5} c^{3} \left (x +\frac {1}{a}\right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4/a^3/c^3/(x-1/a)^2*(-a^2*(x-1/a)^2-2*a*(x-1/a))^(3/2)+19/32/a/c^3*(-a^2*(x-1/a)^2-2*a*(x-1/a))^(1/2)-19/32/
c^3/(a^2)^(1/2)*arctan((a^2)^(1/2)*x/(-a^2*(x-1/a)^2-2*a*(x-1/a))^(1/2))+1/48/a^4/c^3/(x-1/a)^3*(-a^2*(x-1/a)^
2-2*a*(x-1/a))^(3/2)+15/16/a^3/c^3/(x+1/a)^2*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(3/2)+51/32/a/c^3*(-a^2*(x+1/a)^2+2*
a*(x+1/a))^(1/2)+51/32/c^3/(a^2)^(1/2)*arctan((a^2)^(1/2)*x/(-a^2*(x+1/a)^2+2*a*(x+1/a))^(1/2))-43/240/a^4/c^3
/(x+1/a)^3*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(3/2)+1/40/a^5/c^3/(x+1/a)^4*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(3/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.28, size = 365, normalized size = 2.81 \[ \frac {\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{c^3\,\sqrt {-a^2}}-\frac {5\,a\,\sqrt {1-a^2\,x^2}}{12\,\left (a^4\,c^3\,x^2+2\,a^3\,c^3\,x+a^2\,c^3\right )}-\frac {a\,\sqrt {1-a^2\,x^2}}{24\,\left (a^4\,c^3\,x^2-2\,a^3\,c^3\,x+a^2\,c^3\right )}+\frac {a^6\,\sqrt {1-a^2\,x^2}}{30\,\left (a^9\,c^3\,x^2+2\,a^8\,c^3\,x+a^7\,c^3\right )}+\frac {\sqrt {1-a^2\,x^2}}{a\,c^3}-\frac {493\,\sqrt {1-a^2\,x^2}}{240\,\sqrt {-a^2}\,\left (c^3\,x\,\sqrt {-a^2}+\frac {c^3\,\sqrt {-a^2}}{a}\right )}+\frac {25\,\sqrt {1-a^2\,x^2}}{48\,\sqrt {-a^2}\,\left (c^3\,x\,\sqrt {-a^2}-\frac {c^3\,\sqrt {-a^2}}{a}\right )}-\frac {\sqrt {1-a^2\,x^2}}{20\,\sqrt {-a^2}\,\left (3\,c^3\,x\,\sqrt {-a^2}+\frac {c^3\,\sqrt {-a^2}}{a}+a^2\,c^3\,x^3\,\sqrt {-a^2}+3\,a\,c^3\,x^2\,\sqrt {-a^2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

asinh(x*(-a^2)^(1/2))/(c^3*(-a^2)^(1/2)) - (5*a*(1 - a^2*x^2)^(1/2))/(12*(a^2*c^3 + 2*a^3*c^3*x + a^4*c^3*x^2)
) - (a*(1 - a^2*x^2)^(1/2))/(24*(a^2*c^3 - 2*a^3*c^3*x + a^4*c^3*x^2)) + (a^6*(1 - a^2*x^2)^(1/2))/(30*(a^7*c^
3 + 2*a^8*c^3*x + a^9*c^3*x^2)) + (1 - a^2*x^2)^(1/2)/(a*c^3) - (493*(1 - a^2*x^2)^(1/2))/(240*(-a^2)^(1/2)*(c
^3*x*(-a^2)^(1/2) + (c^3*(-a^2)^(1/2))/a)) + (25*(1 - a^2*x^2)^(1/2))/(48*(-a^2)^(1/2)*(c^3*x*(-a^2)^(1/2) - (
c^3*(-a^2)^(1/2))/a)) - (1 - a^2*x^2)^(1/2)/(20*(-a^2)^(1/2)*(3*c^3*x*(-a^2)^(1/2) + (c^3*(-a^2)^(1/2))/a + a^
2*c^3*x^3*(-a^2)^(1/2) + 3*a*c^3*x^2*(-a^2)^(1/2)))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {a^{6} \int \frac {x^{6} \sqrt {- a^{2} x^{2} + 1}}{a^{7} x^{7} + a^{6} x^{6} - 3 a^{5} x^{5} - 3 a^{4} x^{4} + 3 a^{3} x^{3} + 3 a^{2} x^{2} - a x - 1}\, dx}{c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**6*Integral(x**6*sqrt(-a**2*x**2 + 1)/(a**7*x**7 + a**6*x**6 - 3*a**5*x**5 - 3*a**4*x**4 + 3*a**3*x**3 + 3*a
**2*x**2 - a*x - 1), x)/c**3

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