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

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

[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)

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Rubi [A]  time = 0.164805, antiderivative size = 96, normalized size of antiderivative = 1., 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 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]

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 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 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 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 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]

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*}

Mathematica [A]  time = 0.174464, 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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Maple [B]  time = 0.064, size = 424, normalized size = 4.4 \begin{align*} -{\frac{43}{48\,{a}^{3}{c}^{4}} \left ( -{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,a \left ( x-{a}^{-1} \right ) \right ) ^{{\frac{5}{2}}} \left ( x-{a}^{-1} \right ) ^{-2}}+{\frac{87\,x}{64\,{c}^{4}}\sqrt{-{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,a \left ( x-{a}^{-1} \right ) }}+{\frac{87}{64\,{c}^{4}}\arctan \left ({x\sqrt{{a}^{2}}{\frac{1}{\sqrt{-{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,a \left ( x-{a}^{-1} \right ) }}}} \right ){\frac{1}{\sqrt{{a}^{2}}}}}+{\frac{1}{24\,{a}^{5}{c}^{4}} \left ( -{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,a \left ( x-{a}^{-1} \right ) \right ) ^{{\frac{5}{2}}} \left ( x-{a}^{-1} \right ) ^{-4}}+{\frac{17}{48\,{a}^{4}{c}^{4}} \left ( -{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,a \left ( x-{a}^{-1} \right ) \right ) ^{{\frac{5}{2}}} \left ( x-{a}^{-1} \right ) ^{-3}}-{\frac{1}{16\,{a}^{4}{c}^{4} \left ( x+{a}^{-1} \right ) ^{3}} \left ( -{a}^{2} \left ( x+{a}^{-1} \right ) ^{2}+2\,a \left ( x+{a}^{-1} \right ) \right ) ^{{\frac{5}{2}}}}-{\frac{1}{4\,{a}^{3}{c}^{4} \left ( x+{a}^{-1} \right ) ^{2}} \left ( -{a}^{2} \left ( x+{a}^{-1} \right ) ^{2}+2\,a \left ( x+{a}^{-1} \right ) \right ) ^{{\frac{5}{2}}}}-{\frac{23\,x}{64\,{c}^{4}}\sqrt{-{a}^{2} \left ( x+{a}^{-1} \right ) ^{2}+2\,a \left ( x+{a}^{-1} \right ) }}-{\frac{23}{64\,{c}^{4}}\arctan \left ({x\sqrt{{a}^{2}}{\frac{1}{\sqrt{-{a}^{2} \left ( x+{a}^{-1} \right ) ^{2}+2\,a \left ( x+{a}^{-1} \right ) }}}} \right ){\frac{1}{\sqrt{{a}^{2}}}}}-{\frac{29}{32\,a{c}^{4}} \left ( -{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,a \left ( x-{a}^{-1} \right ) \right ) ^{{\frac{3}{2}}}}-{\frac{23}{96\,a{c}^{4}} \left ( -{a}^{2} \left ( x+{a}^{-1} \right ) ^{2}+2\,a \left ( x+{a}^{-1} \right ) \right ) ^{{\frac{3}{2}}}} \end{align*}

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]

-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))+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))-29/32/
a/c^4*(-a^2*(x-1/a)^2-2*a*(x-1/a))^(3/2)-23/96/a/c^4*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(3/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

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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Fricas [A]  time = 2.47331, size = 297, normalized size = 3.09 \begin{align*} -\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 )}} \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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 - \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}\, dx\right )}{c^{4}} \end{align*}

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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

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)