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

Optimal. Leaf size=74 \[ \frac{x^2 (a x+1)}{a^2 c \sqrt{1-a^2 x^2}}+\frac{(3 a x+4) \sqrt{1-a^2 x^2}}{2 a^4 c}-\frac{3 \sin ^{-1}(a x)}{2 a^4 c} \]

[Out]

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

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Rubi [A]  time = 0.107834, antiderivative size = 74, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {6148, 819, 780, 216} \[ \frac{x^2 (a x+1)}{a^2 c \sqrt{1-a^2 x^2}}+\frac{(3 a x+4) \sqrt{1-a^2 x^2}}{2 a^4 c}-\frac{3 \sin ^{-1}(a x)}{2 a^4 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 780

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(((e*f + d*g)*(2*p
+ 3) + 2*e*g*(p + 1)*x)*(a + c*x^2)^(p + 1))/(2*c*(p + 1)*(2*p + 3)), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(c*
(2*p + 3)), Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, p}, x] &&  !LeQ[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)} x^3}{c-a^2 c x^2} \, dx &=\frac{\int \frac{x^3 (1+a x)}{\left (1-a^2 x^2\right )^{3/2}} \, dx}{c}\\ &=\frac{x^2 (1+a x)}{a^2 c \sqrt{1-a^2 x^2}}-\frac{\int \frac{x (2+3 a x)}{\sqrt{1-a^2 x^2}} \, dx}{a^2 c}\\ &=\frac{x^2 (1+a x)}{a^2 c \sqrt{1-a^2 x^2}}+\frac{(4+3 a x) \sqrt{1-a^2 x^2}}{2 a^4 c}-\frac{3 \int \frac{1}{\sqrt{1-a^2 x^2}} \, dx}{2 a^3 c}\\ &=\frac{x^2 (1+a x)}{a^2 c \sqrt{1-a^2 x^2}}+\frac{(4+3 a x) \sqrt{1-a^2 x^2}}{2 a^4 c}-\frac{3 \sin ^{-1}(a x)}{2 a^4 c}\\ \end{align*}

Mathematica [A]  time = 0.0390005, size = 65, normalized size = 0.88 \[ -\frac{a^3 x^3+2 a^2 x^2+3 \sqrt{1-a^2 x^2} \sin ^{-1}(a x)-3 a x-4}{2 a^4 c \sqrt{1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.039, size = 119, normalized size = 1.6 \begin{align*}{\frac{x}{2\,{a}^{3}c}\sqrt{-{a}^{2}{x}^{2}+1}}-{\frac{3}{2\,{a}^{3}c}\arctan \left ({x\sqrt{{a}^{2}}{\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}} \right ){\frac{1}{\sqrt{{a}^{2}}}}}+{\frac{1}{{a}^{4}c}\sqrt{-{a}^{2}{x}^{2}+1}}-{\frac{1}{c{a}^{5}}\sqrt{-{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,a \left ( x-{a}^{-1} \right ) } \left ( x-{a}^{-1} \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.71213, size = 414, normalized size = 5.59 \begin{align*} -\frac{a^{2} c{\left (\frac{\sqrt{-a^{2} x^{2} + 1} c}{\sqrt{a^{2} c^{2}} a^{5} c x + a^{5} c^{2}} + \frac{\sqrt{-a^{2} x^{2} + 1} c}{\sqrt{a^{2} c^{2}} a^{5} c x - a^{5} c^{2}} - \frac{\sqrt{-a^{2} x^{2} + 1}}{a^{6} c x + \sqrt{a^{2} c^{2}} a^{4}} + \frac{\sqrt{-a^{2} x^{2} + 1}}{a^{6} c x - \sqrt{a^{2} c^{2}} a^{4}} - \frac{\sqrt{a^{2} c^{2}} \sqrt{-a^{2} x^{2} + 1} x}{a^{5} c^{2}} - \frac{2 \, \sqrt{a^{2} c^{2}} \sqrt{-a^{2} x^{2} + 1}}{a^{6} c^{2}} + \frac{\sqrt{a^{2} c^{2}} \arcsin \left (\frac{x}{c \sqrt{\frac{1}{a^{2} c^{2}}}}\right )}{a^{7} c^{3} \sqrt{\frac{1}{a^{2} c^{2}}}} + \frac{2 \, \left (a^{2} c^{2}\right )^{\frac{3}{2}} \arcsin \left (\frac{x}{c \sqrt{\frac{1}{a^{2} c^{2}}}}\right )}{a^{9} c^{5} \sqrt{\frac{1}{a^{2} c^{2}}}}\right )}}{2 \, \sqrt{a^{2} 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)*x^3/(-a^2*c*x^2+c),x, algorithm="maxima")

[Out]

-1/2*a^2*c*(sqrt(-a^2*x^2 + 1)*c/(sqrt(a^2*c^2)*a^5*c*x + a^5*c^2) + sqrt(-a^2*x^2 + 1)*c/(sqrt(a^2*c^2)*a^5*c
*x - a^5*c^2) - sqrt(-a^2*x^2 + 1)/(a^6*c*x + sqrt(a^2*c^2)*a^4) + sqrt(-a^2*x^2 + 1)/(a^6*c*x - sqrt(a^2*c^2)
*a^4) - sqrt(a^2*c^2)*sqrt(-a^2*x^2 + 1)*x/(a^5*c^2) - 2*sqrt(a^2*c^2)*sqrt(-a^2*x^2 + 1)/(a^6*c^2) + sqrt(a^2
*c^2)*arcsin(x/(c*sqrt(1/(a^2*c^2))))/(a^7*c^3*sqrt(1/(a^2*c^2))) + 2*(a^2*c^2)^(3/2)*arcsin(x/(c*sqrt(1/(a^2*
c^2))))/(a^9*c^5*sqrt(1/(a^2*c^2))))/sqrt(a^2*c^2)

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Fricas [A]  time = 1.59515, size = 174, normalized size = 2.35 \begin{align*} \frac{4 \, a x + 6 \,{\left (a x - 1\right )} \arctan \left (\frac{\sqrt{-a^{2} x^{2} + 1} - 1}{a x}\right ) +{\left (a^{2} x^{2} + a x - 4\right )} \sqrt{-a^{2} x^{2} + 1} - 4}{2 \,{\left (a^{5} c x - a^{4} c\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)*x^3/(-a^2*c*x^2+c),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.20777, size = 122, normalized size = 1.65 \begin{align*} \frac{1}{2} \, \sqrt{-a^{2} x^{2} + 1}{\left (\frac{x}{a^{3} c} + \frac{2}{a^{4} c}\right )} - \frac{3 \, \arcsin \left (a x\right ) \mathrm{sgn}\left (a\right )}{2 \, a^{3} c{\left | a \right |}} + \frac{2}{a^{3} c{\left (\frac{\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a}{a^{2} x} - 1\right )}{\left | a \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)*x^3/(-a^2*c*x^2+c),x, algorithm="giac")

[Out]

1/2*sqrt(-a^2*x^2 + 1)*(x/(a^3*c) + 2/(a^4*c)) - 3/2*arcsin(a*x)*sgn(a)/(a^3*c*abs(a)) + 2/(a^3*c*((sqrt(-a^2*
x^2 + 1)*abs(a) + a)/(a^2*x) - 1)*abs(a))