3.508 \(\int \frac {e^{-3 \tanh ^{-1}(a x)}}{(c-\frac {c}{a x})^4} \, dx\)

Optimal. Leaf size=96 \[ -\frac {x (4 a x+3)}{3 c^4 \sqrt {1-a^2 x^2}}-\frac {8 \sqrt {1-a^2 x^2}}{3 a c^4}+\frac {a^2 x^3 (a x+1)}{3 c^4 \left (1-a^2 x^2\right )^{3/2}}+\frac {\sin ^{-1}(a x)}{a c^4} \]

[Out]

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

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Rubi [A]  time = 0.16, antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {6131, 6128, 850, 819, 641, 216} \[ \frac {a^2 x^3 (a x+1)}{3 c^4 \left (1-a^2 x^2\right )^{3/2}}-\frac {x (4 a x+3)}{3 c^4 \sqrt {1-a^2 x^2}}-\frac {8 \sqrt {1-a^2 x^2}}{3 a c^4}+\frac {\sin ^{-1}(a x)}{a c^4} \]

Antiderivative was successfully verified.

[In]

Int[1/(E^(3*ArcTanh[a*x])*(c - c/(a*x))^4),x]

[Out]

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

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 850

Int[((x_)^(n_.)*((a_) + (c_.)*(x_)^2)^(p_))/((d_) + (e_.)*(x_)), x_Symbol] :> Int[x^n*(a/d + (c*x)/e)*(a + c*x
^2)^(p - 1), x] /; FreeQ[{a, c, d, e, n, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && ( !IntegerQ[n] ||
!IntegerQ[2*p] || IGtQ[n, 2] || (GtQ[p, 0] && NeQ[n, 2]))

Rule 6128

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

Rule 6131

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

Rubi steps

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

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Mathematica [A]  time = 0.19, size = 68, normalized size = 0.71 \[ \frac {\frac {\sqrt {1-a^2 x^2} \left (-3 a^3 x^3+7 a^2 x^2+5 a x-8\right )}{(a x-1)^2 (a x+1)}+3 \sin ^{-1}(a x)}{3 a c^4} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(E^(3*ArcTanh[a*x])*(c - c/(a*x))^4),x]

[Out]

((Sqrt[1 - a^2*x^2]*(-8 + 5*a*x + 7*a^2*x^2 - 3*a^3*x^3))/((-1 + a*x)^2*(1 + a*x)) + 3*ArcSin[a*x])/(3*a*c^4)

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fricas [A]  time = 0.85, size = 142, normalized size = 1.48 \[ -\frac {8 \, a^{3} x^{3} - 8 \, a^{2} x^{2} - 8 \, a x + 6 \, {\left (a^{3} x^{3} - a^{2} x^{2} - a x + 1\right )} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) + {\left (3 \, a^{3} x^{3} - 7 \, a^{2} x^{2} - 5 \, a x + 8\right )} \sqrt {-a^{2} x^{2} + 1} + 8}{3 \, {\left (a^{4} c^{4} x^{3} - a^{3} c^{4} x^{2} - a^{2} c^{4} x + a c^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/(c-c/a/x)^4,x, algorithm="fricas")

[Out]

-1/3*(8*a^3*x^3 - 8*a^2*x^2 - 8*a*x + 6*(a^3*x^3 - a^2*x^2 - a*x + 1)*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) +
(3*a^3*x^3 - 7*a^2*x^2 - 5*a*x + 8)*sqrt(-a^2*x^2 + 1) + 8)/(a^4*c^4*x^3 - a^3*c^4*x^2 - a^2*c^4*x + a*c^4)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}{{\left (a x + 1\right )}^{3} {\left (c - \frac {c}{a x}\right )}^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/(c-c/a/x)^4,x, algorithm="giac")

[Out]

integrate((-a^2*x^2 + 1)^(3/2)/((a*x + 1)^3*(c - c/(a*x))^4), x)

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maple [B]  time = 0.06, size = 424, normalized size = 4.42 \[ -\frac {23 \left (-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )\right )^{\frac {3}{2}}}{96 a \,c^{4}}-\frac {29 \left (-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )\right )^{\frac {3}{2}}}{32 a \,c^{4}}+\frac {\left (-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )\right )^{\frac {5}{2}}}{24 a^{5} c^{4} \left (x -\frac {1}{a}\right )^{4}}+\frac {17 \left (-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )\right )^{\frac {5}{2}}}{48 a^{4} c^{4} \left (x -\frac {1}{a}\right )^{3}}-\frac {\left (-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )\right )^{\frac {5}{2}}}{16 a^{4} c^{4} \left (x +\frac {1}{a}\right )^{3}}-\frac {\left (-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )\right )^{\frac {5}{2}}}{4 a^{3} c^{4} \left (x +\frac {1}{a}\right )^{2}}-\frac {23 \sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}\, x}{64 c^{4}}-\frac {23 \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}\right )}{64 c^{4} \sqrt {a^{2}}}-\frac {43 \left (-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )\right )^{\frac {5}{2}}}{48 a^{3} c^{4} \left (x -\frac {1}{a}\right )^{2}}+\frac {87 \sqrt {-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )}\, x}{64 c^{4}}+\frac {87 \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )}}\right )}{64 c^{4} \sqrt {a^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-23/96/a/c^4*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(3/2)-29/32/a/c^4*(-a^2*(x-1/a)^2-2*a*(x-1/a))^(3/2)+1/24/a^5/c^4/(x
-1/a)^4*(-a^2*(x-1/a)^2-2*a*(x-1/a))^(5/2)+17/48/a^4/c^4/(x-1/a)^3*(-a^2*(x-1/a)^2-2*a*(x-1/a))^(5/2)-1/16/a^4
/c^4/(x+1/a)^3*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(5/2)-1/4/a^3/c^4/(x+1/a)^2*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(5/2)-23/
64/c^4*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(1/2)*x-23/64/c^4/(a^2)^(1/2)*arctan((a^2)^(1/2)*x/(-a^2*(x+1/a)^2+2*a*(x+
1/a))^(1/2))-43/48/a^3/c^4/(x-1/a)^2*(-a^2*(x-1/a)^2-2*a*(x-1/a))^(5/2)+87/64/c^4*(-a^2*(x-1/a)^2-2*a*(x-1/a))
^(1/2)*x+87/64/c^4/(a^2)^(1/2)*arctan((a^2)^(1/2)*x/(-a^2*(x-1/a)^2-2*a*(x-1/a))^(1/2))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}{{\left (a x + 1\right )}^{3} {\left (c - \frac {c}{a x}\right )}^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/(c-c/a/x)^4,x, algorithm="maxima")

[Out]

integrate((-a^2*x^2 + 1)^(3/2)/((a*x + 1)^3*(c - c/(a*x))^4), x)

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mupad [B]  time = 0.07, size = 188, normalized size = 1.96 \[ \frac {a\,\sqrt {1-a^2\,x^2}}{6\,\left (a^4\,c^4\,x^2-2\,a^3\,c^4\,x+a^2\,c^4\right )}+\frac {\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{c^4\,\sqrt {-a^2}}-\frac {\sqrt {1-a^2\,x^2}}{a\,c^4}+\frac {\sqrt {1-a^2\,x^2}}{4\,\sqrt {-a^2}\,\left (c^4\,x\,\sqrt {-a^2}+\frac {c^4\,\sqrt {-a^2}}{a}\right )}-\frac {19\,\sqrt {1-a^2\,x^2}}{12\,\sqrt {-a^2}\,\left (c^4\,x\,\sqrt {-a^2}-\frac {c^4\,\sqrt {-a^2}}{a}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a*(1 - a^2*x^2)^(1/2))/(6*(a^2*c^4 - 2*a^3*c^4*x + a^4*c^4*x^2)) + asinh(x*(-a^2)^(1/2))/(c^4*(-a^2)^(1/2)) -
(1 - a^2*x^2)^(1/2)/(a*c^4) + (1 - a^2*x^2)^(1/2)/(4*(-a^2)^(1/2)*(c^4*x*(-a^2)^(1/2) + (c^4*(-a^2)^(1/2))/a)
) - (19*(1 - a^2*x^2)^(1/2))/(12*(-a^2)^(1/2)*(c^4*x*(-a^2)^(1/2) - (c^4*(-a^2)^(1/2))/a))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x+1)**3*(-a**2*x**2+1)**(3/2)/(c-c/a/x)**4,x)

[Out]

a**4*(Integral(x**4*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) + Integral(-a**2*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**4

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