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

Optimal. Leaf size=125 \[ \frac{(a x+1)^3}{5 a c^2 \left (1-a^2 x^2\right )^{5/2}}-\frac{6 (a x+1)^2}{5 a c^2 \left (1-a^2 x^2\right )^{3/2}}+\frac{24 (a x+1)}{5 a c^2 \sqrt{1-a^2 x^2}}+\frac{\sqrt{1-a^2 x^2}}{a c^2}-\frac{3 \sin ^{-1}(a x)}{a c^2} \]

[Out]

(1 + a*x)^3/(5*a*c^2*(1 - a^2*x^2)^(5/2)) - (6*(1 + a*x)^2)/(5*a*c^2*(1 - a^2*x^2)^(3/2)) + (24*(1 + a*x))/(5*
a*c^2*Sqrt[1 - a^2*x^2]) + Sqrt[1 - a^2*x^2]/(a*c^2) - (3*ArcSin[a*x])/(a*c^2)

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Rubi [A]  time = 0.334767, antiderivative size = 125, normalized size of antiderivative = 1., 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, 6148, 1635, 641, 216} \[ \frac{(a x+1)^3}{5 a c^2 \left (1-a^2 x^2\right )^{5/2}}-\frac{6 (a x+1)^2}{5 a c^2 \left (1-a^2 x^2\right )^{3/2}}+\frac{24 (a x+1)}{5 a c^2 \sqrt{1-a^2 x^2}}+\frac{\sqrt{1-a^2 x^2}}{a c^2}-\frac{3 \sin ^{-1}(a x)}{a c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(1 + a*x)^3/(5*a*c^2*(1 - a^2*x^2)^(5/2)) - (6*(1 + a*x)^2)/(5*a*c^2*(1 - a^2*x^2)^(3/2)) + (24*(1 + a*x))/(5*
a*c^2*Sqrt[1 - a^2*x^2]) + Sqrt[1 - a^2*x^2]/(a*c^2) - (3*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 1635

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq,
a*e + c*d*x, x], f = PolynomialRemainder[Pq, a*e + c*d*x, x]}, -Simp[(d*f*(d + e*x)^m*(a + c*x^2)^(p + 1))/(2*
a*e*(p + 1)), x] + Dist[d/(2*a*(p + 1)), Int[(d + e*x)^(m - 1)*(a + c*x^2)^(p + 1)*ExpandToSum[2*a*e*(p + 1)*Q
+ f*(m + 2*p + 2), x], x], x]] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && EqQ[c*d^2 + a*e^2, 0] && ILtQ[p +
1/2, 0] && GtQ[m, 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^2 x^2}\right )^2} \, dx &=\frac{a^4 \int \frac{e^{3 \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)^3}{\left (1-a^2 x^2\right )^{7/2}} \, dx}{c^2}\\ &=\frac{(1+a x)^3}{5 a c^2 \left (1-a^2 x^2\right )^{5/2}}-\frac{a^4 \int \frac{(1+a x)^2 \left (\frac{3}{a^4}+\frac{5 x}{a^3}+\frac{5 x^2}{a^2}+\frac{5 x^3}{a}\right )}{\left (1-a^2 x^2\right )^{5/2}} \, dx}{5 c^2}\\ &=\frac{(1+a x)^3}{5 a c^2 \left (1-a^2 x^2\right )^{5/2}}-\frac{6 (1+a x)^2}{5 a c^2 \left (1-a^2 x^2\right )^{3/2}}+\frac{a^4 \int \frac{(1+a x) \left (\frac{27}{a^4}+\frac{30 x}{a^3}+\frac{15 x^2}{a^2}\right )}{\left (1-a^2 x^2\right )^{3/2}} \, dx}{15 c^2}\\ &=\frac{(1+a x)^3}{5 a c^2 \left (1-a^2 x^2\right )^{5/2}}-\frac{6 (1+a x)^2}{5 a c^2 \left (1-a^2 x^2\right )^{3/2}}+\frac{24 (1+a x)}{5 a c^2 \sqrt{1-a^2 x^2}}-\frac{a^4 \int \frac{\frac{45}{a^4}+\frac{15 x}{a^3}}{\sqrt{1-a^2 x^2}} \, dx}{15 c^2}\\ &=\frac{(1+a x)^3}{5 a c^2 \left (1-a^2 x^2\right )^{5/2}}-\frac{6 (1+a x)^2}{5 a c^2 \left (1-a^2 x^2\right )^{3/2}}+\frac{24 (1+a x)}{5 a c^2 \sqrt{1-a^2 x^2}}+\frac{\sqrt{1-a^2 x^2}}{a c^2}-\frac{3 \int \frac{1}{\sqrt{1-a^2 x^2}} \, dx}{c^2}\\ &=\frac{(1+a x)^3}{5 a c^2 \left (1-a^2 x^2\right )^{5/2}}-\frac{6 (1+a x)^2}{5 a c^2 \left (1-a^2 x^2\right )^{3/2}}+\frac{24 (1+a x)}{5 a c^2 \sqrt{1-a^2 x^2}}+\frac{\sqrt{1-a^2 x^2}}{a c^2}-\frac{3 \sin ^{-1}(a x)}{a c^2}\\ \end{align*}

Mathematica [A]  time = 0.0989422, size = 88, normalized size = 0.7 \[ \frac{-5 a^4 x^4+34 a^3 x^3-18 a^2 x^2-15 (a x-1)^2 \sqrt{1-a^2 x^2} \sin ^{-1}(a x)-33 a x+24}{5 a c^2 (a x-1)^2 \sqrt{1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(24 - 33*a*x - 18*a^2*x^2 + 34*a^3*x^3 - 5*a^4*x^4 - 15*(-1 + a*x)^2*Sqrt[1 - a^2*x^2]*ArcSin[a*x])/(5*a*c^2*(
-1 + a*x)^2*Sqrt[1 - a^2*x^2])

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Maple [A]  time = 0.05, size = 212, normalized size = 1.7 \begin{align*} -{\frac{a{x}^{2}}{{c}^{2}}{\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}}+7\,{\frac{1}{a{c}^{2}\sqrt{-{a}^{2}{x}^{2}+1}}}+10\,{\frac{x}{{c}^{2}\sqrt{-{a}^{2}{x}^{2}+1}}}-3\,{\frac{1}{{c}^{2}\sqrt{{a}^{2}}}\arctan \left ({\frac{\sqrt{{a}^{2}}x}{\sqrt{-{a}^{2}{x}^{2}+1}}} \right ) }+{\frac{2}{5\,{a}^{3}{c}^{2}} \left ( x-{a}^{-1} \right ) ^{-2}{\frac{1}{\sqrt{-{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,a \left ( x-{a}^{-1} \right ) }}}}+{\frac{13}{5\,{a}^{2}{c}^{2}} \left ( x-{a}^{-1} \right ) ^{-1}{\frac{1}{\sqrt{-{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,a \left ( x-{a}^{-1} \right ) }}}}-{\frac{26\,x}{5\,{c}^{2}}{\frac{1}{\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)^3/(-a^2*x^2+1)^(3/2)/(c-c/a^2/x^2)^2,x)

[Out]

-a/c^2*x^2/(-a^2*x^2+1)^(1/2)+7/a/c^2/(-a^2*x^2+1)^(1/2)+10*x/c^2/(-a^2*x^2+1)^(1/2)-3/c^2/(a^2)^(1/2)*arctan(
(a^2)^(1/2)*x/(-a^2*x^2+1)^(1/2))+2/5/a^3/c^2/(x-1/a)^2/(-a^2*(x-1/a)^2-2*a*(x-1/a))^(1/2)+13/5/a^2/c^2/(x-1/a
)/(-a^2*(x-1/a)^2-2*a*(x-1/a))^(1/2)-26/5/c^2/(-a^2*(x-1/a)^2-2*a*(x-1/a))^(1/2)*x

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (a x + 1\right )}^{3}}{{\left (-a^{2} x^{2} + 1\right )}^{\frac{3}{2}}{\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)^3/(-a^2*x^2+1)^(3/2)/(c-c/a^2/x^2)^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.15279, size = 317, normalized size = 2.54 \begin{align*} \frac{24 \, a^{3} x^{3} - 72 \, a^{2} x^{2} + 72 \, a x + 30 \,{\left (a^{3} x^{3} - 3 \, a^{2} x^{2} + 3 \, a x - 1\right )} \arctan \left (\frac{\sqrt{-a^{2} x^{2} + 1} - 1}{a x}\right ) +{\left (5 \, a^{3} x^{3} - 39 \, a^{2} x^{2} + 57 \, a x - 24\right )} \sqrt{-a^{2} x^{2} + 1} - 24}{5 \,{\left (a^{4} c^{2} x^{3} - 3 \, a^{3} c^{2} x^{2} + 3 \, 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)^3/(-a^2*x^2+1)^(3/2)/(c-c/a^2/x^2)^2,x, algorithm="fricas")

[Out]

1/5*(24*a^3*x^3 - 72*a^2*x^2 + 72*a*x + 30*(a^3*x^3 - 3*a^2*x^2 + 3*a*x - 1)*arctan((sqrt(-a^2*x^2 + 1) - 1)/(
a*x)) + (5*a^3*x^3 - 39*a^2*x^2 + 57*a*x - 24)*sqrt(-a^2*x^2 + 1) - 24)/(a^4*c^2*x^3 - 3*a^3*c^2*x^2 + 3*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} \left (\int \frac{x^{4}}{- a^{4} x^{4} \sqrt{- a^{2} x^{2} + 1} + 2 a^{3} x^{3} \sqrt{- a^{2} x^{2} + 1} - 2 a x \sqrt{- a^{2} x^{2} + 1} + \sqrt{- a^{2} x^{2} + 1}}\, dx + \int \frac{a x^{5}}{- a^{4} x^{4} \sqrt{- a^{2} x^{2} + 1} + 2 a^{3} x^{3} \sqrt{- a^{2} x^{2} + 1} - 2 a x \sqrt{- a^{2} x^{2} + 1} + \sqrt{- a^{2} x^{2} + 1}}\, dx\right )}{c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.18864, size = 243, normalized size = 1.94 \begin{align*} -\frac{3 \, \arcsin \left (a x\right ) \mathrm{sgn}\left (a\right )}{c^{2}{\left | a \right |}} + \frac{\sqrt{-a^{2} x^{2} + 1}}{a c^{2}} - \frac{2 \,{\left (\frac{80 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}}{a^{2} x} - \frac{120 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{2}}{a^{4} x^{2}} + \frac{70 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{3}}{a^{6} x^{3}} - \frac{15 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{4}}{a^{8} x^{4}} - 19\right )}}{5 \, c^{2}{\left (\frac{\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a}{a^{2} x} - 1\right )}^{5}{\left | a \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3*arcsin(a*x)*sgn(a)/(c^2*abs(a)) + sqrt(-a^2*x^2 + 1)/(a*c^2) - 2/5*(80*(sqrt(-a^2*x^2 + 1)*abs(a) + a)/(a^2
*x) - 120*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^2/(a^4*x^2) + 70*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^3/(a^6*x^3) - 15*(s
qrt(-a^2*x^2 + 1)*abs(a) + a)^4/(a^8*x^4) - 19)/(c^2*((sqrt(-a^2*x^2 + 1)*abs(a) + a)/(a^2*x) - 1)^5*abs(a))