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

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

[Out]

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

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Rubi [A]  time = 0.150325, antiderivative size = 96, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {6157, 6148, 819, 641, 216} \[ \frac{a^2 x^3 (a x+1)}{3 c^2 \left (1-a^2 x^2\right )^{3/2}}-\frac{x (4 a x+3)}{3 c^2 \sqrt{1-a^2 x^2}}-\frac{8 \sqrt{1-a^2 x^2}}{3 a c^2}+\frac{\sin ^{-1}(a x)}{a c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6148

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] || Gt
Q[c, 0]) && IGtQ[(n + 1)/2, 0] &&  !IntegerQ[p - 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^{\tanh ^{-1}(a x)}}{\left (c-\frac{c}{a^2 x^2}\right )^2} \, dx &=\frac{a^4 \int \frac{e^{\tanh ^{-1}(a x)} x^4}{\left (1-a^2 x^2\right )^2} \, dx}{c^2}\\ &=\frac{a^4 \int \frac{x^4 (1+a x)}{\left (1-a^2 x^2\right )^{5/2}} \, dx}{c^2}\\ &=\frac{a^2 x^3 (1+a x)}{3 c^2 \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^2}\\ &=\frac{a^2 x^3 (1+a x)}{3 c^2 \left (1-a^2 x^2\right )^{3/2}}-\frac{x (3+4 a x)}{3 c^2 \sqrt{1-a^2 x^2}}+\frac{\int \frac{3+8 a x}{\sqrt{1-a^2 x^2}} \, dx}{3 c^2}\\ &=\frac{a^2 x^3 (1+a x)}{3 c^2 \left (1-a^2 x^2\right )^{3/2}}-\frac{x (3+4 a x)}{3 c^2 \sqrt{1-a^2 x^2}}-\frac{8 \sqrt{1-a^2 x^2}}{3 a c^2}+\frac{\int \frac{1}{\sqrt{1-a^2 x^2}} \, dx}{c^2}\\ &=\frac{a^2 x^3 (1+a x)}{3 c^2 \left (1-a^2 x^2\right )^{3/2}}-\frac{x (3+4 a x)}{3 c^2 \sqrt{1-a^2 x^2}}-\frac{8 \sqrt{1-a^2 x^2}}{3 a c^2}+\frac{\sin ^{-1}(a x)}{a c^2}\\ \end{align*}

Mathematica [A]  time = 0.0503823, size = 78, normalized size = 0.81 \[ \frac{3 a^3 x^3-7 a^2 x^2+3 (a x-1) \sqrt{1-a^2 x^2} \sin ^{-1}(a x)-5 a x+8}{3 a c^2 (a x-1) \sqrt{1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.048, size = 177, normalized size = 1.8 \begin{align*} -{\frac{1}{a{c}^{2}}\sqrt{-{a}^{2}{x}^{2}+1}}+{\frac{1}{{c}^{2}}\arctan \left ({x\sqrt{{a}^{2}}{\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}} \right ){\frac{1}{\sqrt{{a}^{2}}}}}+{\frac{1}{6\,{a}^{3}{c}^{2}}\sqrt{-{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,a \left ( x-{a}^{-1} \right ) } \left ( x-{a}^{-1} \right ) ^{-2}}+{\frac{19}{12\,{a}^{2}{c}^{2}}\sqrt{-{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,a \left ( x-{a}^{-1} \right ) } \left ( x-{a}^{-1} \right ) ^{-1}}-{\frac{1}{4\,{a}^{2}{c}^{2} \left ( x+{a}^{-1} \right ) }\sqrt{-{a}^{2} \left ( x+{a}^{-1} \right ) ^{2}+2\,a \left ( x+{a}^{-1} \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{a x + 1}{\sqrt{-a^{2} x^{2} + 1}{\left (c - \frac{c}{a^{2} x^{2}}\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.118, 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^{2} x^{3} - a^{3} c^{2} x^{2} - a^{2} c^{2} x + a c^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{a x + 1}{\sqrt{-a^{2} x^{2} + 1}{\left (c - \frac{c}{a^{2} x^{2}}\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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