3.35 \(\int e^{-\tanh ^{-1}(a x)} x^2 \, dx\)

Optimal. Leaf size=73 \[ \frac {\sin ^{-1}(a x)}{2 a^3}-\frac {x \sqrt {1-a^2 x^2}}{2 a^2}-\frac {\left (1-a^2 x^2\right )^{3/2}}{3 a^3}+\frac {\sqrt {1-a^2 x^2}}{a^3} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {6124, 797, 641, 195, 216} \[ -\frac {\left (1-a^2 x^2\right )^{3/2}}{3 a^3}-\frac {x \sqrt {1-a^2 x^2}}{2 a^2}+\frac {\sqrt {1-a^2 x^2}}{a^3}+\frac {\sin ^{-1}(a x)}{2 a^3} \]

Antiderivative was successfully verified.

[In]

Int[x^2/E^ArcTanh[a*x],x]

[Out]

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

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

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]

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 797

Int[(x_)^2*((f_) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[1/c, Int[(f + g*x)*(a + c*x^2)^(p
+ 1), x], x] - Dist[a/c, Int[(f + g*x)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, f, g, p}, x] && EqQ[a*g^2 + f^2*
c, 0]

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]

Rubi steps

\begin {align*} \int e^{-\tanh ^{-1}(a x)} x^2 \, dx &=\int \frac {x^2 (1-a x)}{\sqrt {1-a^2 x^2}} \, dx\\ &=\frac {\int \frac {1-a x}{\sqrt {1-a^2 x^2}} \, dx}{a^2}-\frac {\int (1-a x) \sqrt {1-a^2 x^2} \, dx}{a^2}\\ &=\frac {\sqrt {1-a^2 x^2}}{a^3}-\frac {\left (1-a^2 x^2\right )^{3/2}}{3 a^3}+\frac {\int \frac {1}{\sqrt {1-a^2 x^2}} \, dx}{a^2}-\frac {\int \sqrt {1-a^2 x^2} \, dx}{a^2}\\ &=\frac {\sqrt {1-a^2 x^2}}{a^3}-\frac {x \sqrt {1-a^2 x^2}}{2 a^2}-\frac {\left (1-a^2 x^2\right )^{3/2}}{3 a^3}+\frac {\sin ^{-1}(a x)}{a^3}-\frac {\int \frac {1}{\sqrt {1-a^2 x^2}} \, dx}{2 a^2}\\ &=\frac {\sqrt {1-a^2 x^2}}{a^3}-\frac {x \sqrt {1-a^2 x^2}}{2 a^2}-\frac {\left (1-a^2 x^2\right )^{3/2}}{3 a^3}+\frac {\sin ^{-1}(a x)}{2 a^3}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 43, normalized size = 0.59 \[ \frac {\sqrt {1-a^2 x^2} \left (2 a^2 x^2-3 a x+4\right )+3 \sin ^{-1}(a x)}{6 a^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2/E^ArcTanh[a*x],x]

[Out]

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

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fricas [A]  time = 0.43, size = 57, normalized size = 0.78 \[ \frac {{\left (2 \, a^{2} x^{2} - 3 \, a x + 4\right )} \sqrt {-a^{2} x^{2} + 1} - 6 \, \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right )}{6 \, a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*((2*a^2*x^2 - 3*a*x + 4)*sqrt(-a^2*x^2 + 1) - 6*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)))/a^3

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(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,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [B]  time = 0.04, size = 134, normalized size = 1.84 \[ -\frac {\left (-a^{2} x^{2}+1\right )^{\frac {3}{2}}}{3 a^{3}}-\frac {x \sqrt {-a^{2} x^{2}+1}}{2 a^{2}}-\frac {\arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{2 a^{2} \sqrt {a^{2}}}+\frac {\sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}{a^{3}}+\frac {\arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}\right )}{a^{2} \sqrt {a^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(-a^2*x^2+1)^(3/2)/a^3-1/2*x*(-a^2*x^2+1)^(1/2)/a^2-1/2/a^2/(a^2)^(1/2)*arctan((a^2)^(1/2)*x/(-a^2*x^2+1)
^(1/2))+1/a^3*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(1/2)+1/a^2/(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 = 0.41, size = 61, normalized size = 0.84 \[ -\frac {\sqrt {-a^{2} x^{2} + 1} x}{2 \, a^{2}} - \frac {{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}{3 \, a^{3}} + \frac {\arcsin \left (a x\right )}{2 \, a^{3}} + \frac {\sqrt {-a^{2} x^{2} + 1}}{a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.83, size = 82, normalized size = 1.12 \[ \frac {\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{2\,a^2\,\sqrt {-a^2}}+\frac {\sqrt {1-a^2\,x^2}\,\left (\frac {2\,a}{3\,{\left (-a^2\right )}^{3/2}}-\frac {x\,\sqrt {-a^2}}{2\,a^2}+\frac {a^3\,x^2}{3\,{\left (-a^2\right )}^{3/2}}\right )}{\sqrt {-a^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2} \sqrt {- \left (a x - 1\right ) \left (a x + 1\right )}}{a x + 1}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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