3.1.34 \(\int e^{-\tanh ^{-1}(a x)} x^3 \, dx\) [34]

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

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6259, 847, 794, 222} \begin {gather*} -\frac {3 \text {ArcSin}(a x)}{8 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} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*(x^2*Sqrt[1 - a^2*x^2])/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 222

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]

Rule 794

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 847

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 6259

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]

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*}

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Mathematica [A]
time = 0.03, size = 51, normalized size = 0.59 \begin {gather*} \frac {\sqrt {1-a^2 x^2} \left (-16+9 a x-8 a^2 x^2+6 a^3 x^3\right )-9 \text {ArcSin}(a x)}{24 a^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^3/E^ArcTanh[a*x],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 [B] Leaf count of result is larger than twice the leaf count of optimal. \(203\) vs. \(2(73)=146\).
time = 1.20, size = 204, normalized size = 2.34

method	result	size
risch	\(-\frac {\left (6 a^{3} x^{3}-8 a^{2} x^{2}+9 a x -16\right ) \left (a^{2} x^{2}-1\right )}{24 a^{4} \sqrt {-a^{2} x^{2}+1}}-\frac {3 \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{8 a^{3} \sqrt {a^{2}}}\)	\(80\)
default	\(\frac {-\frac {x \left (-a^{2} x^{2}+1\right )^{\frac {3}{2}}}{4 a^{2}}+\frac {\frac {x \sqrt {-a^{2} x^{2}+1}}{2}+\frac {\arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{2 \sqrt {a^{2}}}}{4 a^{2}}}{a}+\frac {\left (-a^{2} x^{2}+1\right )^{\frac {3}{2}}}{3 a^{4}}+\frac {\frac {x \sqrt {-a^{2} x^{2}+1}}{2}+\frac {\arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{2 \sqrt {a^{2}}}}{a^{3}}-\frac {\sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}+\frac {a \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}\right )}{\sqrt {a^{2}}}}{a^{4}}\)	\(204\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(a*x+1)*(-a^2*x^2+1)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.47, size = 80, normalized size = 0.92 \begin {gather*} -\frac {{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}} x}{4 \, a^{3}} + \frac {5 \, \sqrt {-a^{2} x^{2} + 1} x}{8 \, a^{3}} + \frac {{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}{3 \, a^{4}} - \frac {3 \, \arcsin \left (a x\right )}{8 \, a^{4}} - \frac {\sqrt {-a^{2} x^{2} + 1}}{a^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(-a^2*x^2 + 1)^(3/2)*x/a^3 + 5/8*sqrt(-a^2*x^2 + 1)*x/a^3 + 1/3*(-a^2*x^2 + 1)^(3/2)/a^4 - 3/8*arcsin(a*x
)/a^4 - sqrt(-a^2*x^2 + 1)/a^4

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Fricas [A]
time = 0.38, size = 65, normalized size = 0.75 \begin {gather*} \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 {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(a*x+1)*(-a^2*x^2+1)^(1/2),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 [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{3} \sqrt {- \left (a x - 1\right ) \left (a x + 1\right )}}{a x + 1}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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Mupad [B]
time = 0.81, size = 97, normalized size = 1.11 \begin {gather*} -\frac {3\,\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{8\,a^3\,\sqrt {-a^2}}-\frac {\sqrt {1-a^2\,x^2}\,\left (\frac {2}{3\,{\left (-a^2\right )}^{3/2}}-\frac {3\,x\,\sqrt {-a^2}}{8\,a^3}+\frac {a^2\,x^2}{3\,{\left (-a^2\right )}^{3/2}}+\frac {x^3\,{\left (-a^2\right )}^{3/2}}{4\,a^3}\right )}{\sqrt {-a^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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