3.2 \(\int e^{\tanh ^{-1}(a x)} x^3 \, dx\)

Optimal. Leaf size=87 \[ -\frac{x^3 \sqrt{1-a^2 x^2}}{4 a}-\frac{x^2 \sqrt{1-a^2 x^2}}{3 a^2}-\frac{(9 a x+16) \sqrt{1-a^2 x^2}}{24 a^4}+\frac{3 \sin ^{-1}(a x)}{8 a^4} \]

[Out]

-(x^2*Sqrt[1 - a^2*x^2])/(3*a^2) - (x^3*Sqrt[1 - a^2*x^2])/(4*a) - ((16 + 9*a*x)*Sqrt[1 - a^2*x^2])/(24*a^4) +
(3*ArcSin[a*x])/(8*a^4)

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Rubi [A]  time = 0.0754742, antiderivative size = 87, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {6124, 833, 780, 216} \[ -\frac{x^3 \sqrt{1-a^2 x^2}}{4 a}-\frac{x^2 \sqrt{1-a^2 x^2}}{3 a^2}-\frac{(9 a x+16) \sqrt{1-a^2 x^2}}{24 a^4}+\frac{3 \sin ^{-1}(a x)}{8 a^4} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcTanh[a*x]*x^3,x]

[Out]

-(x^2*Sqrt[1 - a^2*x^2])/(3*a^2) - (x^3*Sqrt[1 - a^2*x^2])/(4*a) - ((16 + 9*a*x)*Sqrt[1 - a^2*x^2])/(24*a^4) +
(3*ArcSin[a*x])/(8*a^4)

Rule 6124

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(x_)^(m_.), x_Symbol] :> Int[x^m*((1 + a*x)^((n + 1)/2)/((1 - a*x)^((n - 1)/
2)*Sqrt[1 - a^2*x^2])), x] /; FreeQ[{a, m}, x] && IntegerQ[(n - 1)/2]

Rule 833

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(g*(d + e*x)
^m*(a + c*x^2)^(p + 1))/(c*(m + 2*p + 2)), x] + Dist[1/(c*(m + 2*p + 2)), Int[(d + e*x)^(m - 1)*(a + c*x^2)^p*
Simp[c*d*f*(m + 2*p + 2) - a*e*g*m + c*(e*f*(m + 2*p + 2) + d*g*m)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, p
}, x] && NeQ[c*d^2 + a*e^2, 0] && GtQ[m, 0] && NeQ[m + 2*p + 2, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ
[2*m, 2*p]) &&  !(IGtQ[m, 0] && EqQ[f, 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 e^{\tanh ^{-1}(a x)} x^3 \, dx &=\int \frac{x^3 (1+a x)}{\sqrt{1-a^2 x^2}} \, dx\\ &=-\frac{x^3 \sqrt{1-a^2 x^2}}{4 a}-\frac{\int \frac{x^2 \left (-3 a-4 a^2 x\right )}{\sqrt{1-a^2 x^2}} \, dx}{4 a^2}\\ &=-\frac{x^2 \sqrt{1-a^2 x^2}}{3 a^2}-\frac{x^3 \sqrt{1-a^2 x^2}}{4 a}+\frac{\int \frac{x \left (8 a^2+9 a^3 x\right )}{\sqrt{1-a^2 x^2}} \, dx}{12 a^4}\\ &=-\frac{x^2 \sqrt{1-a^2 x^2}}{3 a^2}-\frac{x^3 \sqrt{1-a^2 x^2}}{4 a}-\frac{(16+9 a x) \sqrt{1-a^2 x^2}}{24 a^4}+\frac{3 \int \frac{1}{\sqrt{1-a^2 x^2}} \, dx}{8 a^3}\\ &=-\frac{x^2 \sqrt{1-a^2 x^2}}{3 a^2}-\frac{x^3 \sqrt{1-a^2 x^2}}{4 a}-\frac{(16+9 a x) \sqrt{1-a^2 x^2}}{24 a^4}+\frac{3 \sin ^{-1}(a x)}{8 a^4}\\ \end{align*}

Mathematica [A]  time = 0.0371607, size = 52, normalized size = 0.6 \[ \frac{9 \sin ^{-1}(a x)-\sqrt{1-a^2 x^2} \left (6 a^3 x^3+8 a^2 x^2+9 a x+16\right )}{24 a^4} \]

Antiderivative was successfully verified.

[In]

Integrate[E^ArcTanh[a*x]*x^3,x]

[Out]

(-(Sqrt[1 - a^2*x^2]*(16 + 9*a*x + 8*a^2*x^2 + 6*a^3*x^3)) + 9*ArcSin[a*x])/(24*a^4)

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Maple [A]  time = 0.037, size = 107, normalized size = 1.2 \begin{align*} -{\frac{{x}^{3}}{4\,a}\sqrt{-{a}^{2}{x}^{2}+1}}-{\frac{3\,x}{8\,{a}^{3}}\sqrt{-{a}^{2}{x}^{2}+1}}+{\frac{3}{8\,{a}^{3}}\arctan \left ({x\sqrt{{a}^{2}}{\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}} \right ){\frac{1}{\sqrt{{a}^{2}}}}}-{\frac{{x}^{2}}{3\,{a}^{2}}\sqrt{-{a}^{2}{x}^{2}+1}}-{\frac{2}{3\,{a}^{4}}\sqrt{-{a}^{2}{x}^{2}+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,x)

[Out]

-1/4*x^3*(-a^2*x^2+1)^(1/2)/a-3/8/a^3*x*(-a^2*x^2+1)^(1/2)+3/8/a^3/(a^2)^(1/2)*arctan((a^2)^(1/2)*x/(-a^2*x^2+
1)^(1/2))-1/3*x^2*(-a^2*x^2+1)^(1/2)/a^2-2/3*(-a^2*x^2+1)^(1/2)/a^4

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Maxima [A]  time = 1.4249, size = 131, normalized size = 1.51 \begin{align*} -\frac{\sqrt{-a^{2} x^{2} + 1} x^{3}}{4 \, a} - \frac{\sqrt{-a^{2} x^{2} + 1} x^{2}}{3 \, a^{2}} - \frac{3 \, \sqrt{-a^{2} x^{2} + 1} x}{8 \, a^{3}} + \frac{3 \, \arcsin \left (\frac{a^{2} x}{\sqrt{a^{2}}}\right )}{8 \, \sqrt{a^{2}} a^{3}} - \frac{2 \, \sqrt{-a^{2} x^{2} + 1}}{3 \, a^{4}} \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,x, algorithm="maxima")

[Out]

-1/4*sqrt(-a^2*x^2 + 1)*x^3/a - 1/3*sqrt(-a^2*x^2 + 1)*x^2/a^2 - 3/8*sqrt(-a^2*x^2 + 1)*x/a^3 + 3/8*arcsin(a^2
*x/sqrt(a^2))/(sqrt(a^2)*a^3) - 2/3*sqrt(-a^2*x^2 + 1)/a^4

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Fricas [A]  time = 2.12157, size = 153, normalized size = 1.76 \begin{align*} -\frac{{\left (6 \, a^{3} x^{3} + 8 \, a^{2} x^{2} + 9 \, a x + 16\right )} \sqrt{-a^{2} x^{2} + 1} + 18 \, \arctan \left (\frac{\sqrt{-a^{2} x^{2} + 1} - 1}{a x}\right )}{24 \, a^{4}} \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,x, algorithm="fricas")

[Out]

-1/24*((6*a^3*x^3 + 8*a^2*x^2 + 9*a*x + 16)*sqrt(-a^2*x^2 + 1) + 18*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)))/a^
4

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Sympy [C]  time = 5.7756, size = 199, normalized size = 2.29 \begin{align*} a \left (\begin{cases} - \frac{i x^{5}}{4 \sqrt{a^{2} x^{2} - 1}} - \frac{i x^{3}}{8 a^{2} \sqrt{a^{2} x^{2} - 1}} + \frac{3 i x}{8 a^{4} \sqrt{a^{2} x^{2} - 1}} - \frac{3 i \operatorname{acosh}{\left (a x \right )}}{8 a^{5}} & \text{for}\: \left |{a^{2} x^{2}}\right | > 1 \\\frac{x^{5}}{4 \sqrt{- a^{2} x^{2} + 1}} + \frac{x^{3}}{8 a^{2} \sqrt{- a^{2} x^{2} + 1}} - \frac{3 x}{8 a^{4} \sqrt{- a^{2} x^{2} + 1}} + \frac{3 \operatorname{asin}{\left (a x \right )}}{8 a^{5}} & \text{otherwise} \end{cases}\right ) + \begin{cases} - \frac{x^{2} \sqrt{- a^{2} x^{2} + 1}}{3 a^{2}} - \frac{2 \sqrt{- a^{2} x^{2} + 1}}{3 a^{4}} & \text{for}\: a \neq 0 \\\frac{x^{4}}{4} & \text{otherwise} \end{cases} \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,x)

[Out]

a*Piecewise((-I*x**5/(4*sqrt(a**2*x**2 - 1)) - I*x**3/(8*a**2*sqrt(a**2*x**2 - 1)) + 3*I*x/(8*a**4*sqrt(a**2*x
**2 - 1)) - 3*I*acosh(a*x)/(8*a**5), Abs(a**2*x**2) > 1), (x**5/(4*sqrt(-a**2*x**2 + 1)) + x**3/(8*a**2*sqrt(-
a**2*x**2 + 1)) - 3*x/(8*a**4*sqrt(-a**2*x**2 + 1)) + 3*asin(a*x)/(8*a**5), True)) + Piecewise((-x**2*sqrt(-a*
*2*x**2 + 1)/(3*a**2) - 2*sqrt(-a**2*x**2 + 1)/(3*a**4), Ne(a, 0)), (x**4/4, True))

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Giac [A]  time = 1.22701, size = 80, normalized size = 0.92 \begin{align*} -\frac{1}{24} \, \sqrt{-a^{2} x^{2} + 1}{\left ({\left (2 \, x{\left (\frac{3 \, x}{a} + \frac{4}{a^{2}}\right )} + \frac{9}{a^{3}}\right )} x + \frac{16}{a^{4}}\right )} + \frac{3 \, \arcsin \left (a x\right ) \mathrm{sgn}\left (a\right )}{8 \, a^{3}{\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,x, algorithm="giac")

[Out]

-1/24*sqrt(-a^2*x^2 + 1)*((2*x*(3*x/a + 4/a^2) + 9/a^3)*x + 16/a^4) + 3/8*arcsin(a*x)*sgn(a)/(a^3*abs(a))