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

Optimal. Leaf size=129 \[ -\frac{a^4 x^5 (a x+1)}{5 c^3 \left (1-a^2 x^2\right )^{5/2}}+\frac{a^2 x^3 (6 a x+5)}{15 c^3 \left (1-a^2 x^2\right )^{3/2}}-\frac{x (8 a x+5)}{5 c^3 \sqrt{1-a^2 x^2}}-\frac{16 \sqrt{1-a^2 x^2}}{5 a c^3}+\frac{\sin ^{-1}(a x)}{a c^3} \]

[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)

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Rubi [A]  time = 0.196946, antiderivative size = 129, normalized size of antiderivative = 1., number of steps used = 7, 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} \[ -\frac{a^4 x^5 (a x+1)}{5 c^3 \left (1-a^2 x^2\right )^{5/2}}+\frac{a^2 x^3 (6 a x+5)}{15 c^3 \left (1-a^2 x^2\right )^{3/2}}-\frac{x (8 a x+5)}{5 c^3 \sqrt{1-a^2 x^2}}-\frac{16 \sqrt{1-a^2 x^2}}{5 a c^3}+\frac{\sin ^{-1}(a x)}{a c^3} \]

Antiderivative was successfully verified.

[In]

Int[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 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 )^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*}

Mathematica [A]  time = 0.0761693, size = 108, normalized size = 0.84 \[ \frac{15 a^5 x^5-38 a^4 x^4-52 a^3 x^3+87 a^2 x^2+15 (a x-1)^2 (a x+1) \sqrt{1-a^2 x^2} \sin ^{-1}(a x)+33 a x-48}{15 a c^3 (a x-1)^2 (a x+1) \sqrt{1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[E^ArcTanh[a*x]/(c - c/(a^2*x^2))^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)^2*(1 + a*x)*Sqrt[1 - a^2*x^2
]*ArcSin[a*x])/(15*a*c^3*(-1 + a*x)^2*(1 + a*x)*Sqrt[1 - a^2*x^2])

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Maple [B]  time = 0.053, size = 259, normalized size = 2. \begin{align*} -{\frac{1}{a{c}^{3}}\sqrt{-{a}^{2}{x}^{2}+1}}+{\frac{1}{{c}^{3}}\arctan \left ({x\sqrt{{a}^{2}}{\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}} \right ){\frac{1}{\sqrt{{a}^{2}}}}}+{\frac{1}{20\,{a}^{4}{c}^{3}}\sqrt{-{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,a \left ( x-{a}^{-1} \right ) } \left ( x-{a}^{-1} \right ) ^{-3}}+{\frac{23}{60\,{a}^{3}{c}^{3}}\sqrt{-{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,a \left ( x-{a}^{-1} \right ) } \left ( x-{a}^{-1} \right ) ^{-2}}+{\frac{493}{240\,{a}^{2}{c}^{3}}\sqrt{-{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,a \left ( x-{a}^{-1} \right ) } \left ( x-{a}^{-1} \right ) ^{-1}}+{\frac{1}{24\,{a}^{3}{c}^{3} \left ( x+{a}^{-1} \right ) ^{2}}\sqrt{-{a}^{2} \left ( x+{a}^{-1} \right ) ^{2}+2\,a \left ( x+{a}^{-1} \right ) }}-{\frac{25}{48\,{a}^{2}{c}^{3} \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)^3,x)

[Out]

-(-a^2*x^2+1)^(1/2)/a/c^3+1/c^3/(a^2)^(1/2)*arctan((a^2)^(1/2)*x/(-a^2*x^2+1)^(1/2))+1/20/a^4/c^3/(x-1/a)^3*(-
a^2*(x-1/a)^2-2*a*(x-1/a))^(1/2)+23/60/a^3/c^3/(x-1/a)^2*(-a^2*(x-1/a)^2-2*a*(x-1/a))^(1/2)+493/240/a^2/c^3/(x
-1/a)*(-a^2*(x-1/a)^2-2*a*(x-1/a))^(1/2)+1/24/a^3/c^3/(x+1/a)^2*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(1/2)-25/48/a^2/c
^3/(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 )}^{3}}\,{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)^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.22016, size = 456, normalized size = 3.53 \begin{align*} -\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 )}} \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)^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 + 3
3*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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{a^{6} \int \frac{x^{6}}{a^{5} x^{5} \sqrt{- a^{2} x^{2} + 1} - a^{4} x^{4} \sqrt{- a^{2} x^{2} + 1} - 2 a^{3} x^{3} \sqrt{- a^{2} x^{2} + 1} + 2 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^{3}} \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)**3,x)

[Out]

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

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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 )}^{3}}\,{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)^3,x, algorithm="giac")

[Out]

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