3.1.3 \(\int e^{\tanh ^{-1}(a x)} x^2 \, dx\) [3]

Optimal. Leaf size=74 \[ -\frac {\sqrt {1-a^2 x^2}}{a^3}-\frac {x \sqrt {1-a^2 x^2}}{2 a^2}+\frac {\left (1-a^2 x^2\right )^{3/2}}{3 a^3}+\frac {\text {ArcSin}(a x)}{2 a^3} \]

[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 = 0.04, antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6259, 811, 655, 201, 222} \begin {gather*} \frac {\text {ArcSin}(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} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^ArcTanh[a*x]*x^2,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 201

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 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 655

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 811

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 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^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 = 0.05, size = 44, normalized size = 0.59 \begin {gather*} \frac {-\sqrt {1-a^2 x^2} \left (4+3 a x+2 a^2 x^2\right )+3 \text {ArcSin}(a x)}{6 a^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^ArcTanh[a*x]*x^2,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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Maple [A]
time = 0.40, size = 90, normalized size = 1.22

method	result	size
risch	\(\frac {\left (2 a^{2} x^{2}+3 a x +4\right ) \left (a^{2} x^{2}-1\right )}{6 a^{3} \sqrt {-a^{2} x^{2}+1}}+\frac {\arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{2 a^{2} \sqrt {a^{2}}}\)	\(72\)
default	\(a \left (-\frac {x^{2} \sqrt {-a^{2} x^{2}+1}}{3 a^{2}}-\frac {2 \sqrt {-a^{2} x^{2}+1}}{3 a^{4}}\right )-\frac {x \sqrt {-a^{2} x^{2}+1}}{2 a^{2}}+\frac {\arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{2 a^{2} \sqrt {a^{2}}}\)	\(90\)
meijerg	\(\frac {\frac {4 \sqrt {\pi }}{3}-\frac {\sqrt {\pi }\, \left (4 a^{2} x^{2}+8\right ) \sqrt {-a^{2} x^{2}+1}}{6}}{2 a^{3} \sqrt {\pi }}-\frac {-\frac {\sqrt {\pi }\, x \left (-a^{2}\right )^{\frac {3}{2}} \sqrt {-a^{2} x^{2}+1}}{a^{2}}+\frac {\sqrt {\pi }\, \left (-a^{2}\right )^{\frac {3}{2}} \arcsin \left (a x \right )}{a^{3}}}{2 a^{2} \sqrt {\pi }\, \sqrt {-a^{2}}}\)	\(105\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.46, size = 65, normalized size = 0.88 \begin {gather*} -\frac {\sqrt {-a^{2} x^{2} + 1} x^{2}}{3 \, a} - \frac {\sqrt {-a^{2} x^{2} + 1} x}{2 \, a^{2}} + \frac {\arcsin \left (a x\right )}{2 \, a^{3}} - \frac {2 \, \sqrt {-a^{2} x^{2} + 1}}{3 \, a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.38, size = 57, normalized size = 0.77 \begin {gather*} -\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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*x^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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Sympy [A]
time = 2.02, size = 133, normalized size = 1.80 \begin {gather*} a \left (\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}\right ) + \begin {cases} - \frac {i x \sqrt {a^{2} x^{2} - 1}}{2 a^{2}} - \frac {i \operatorname {acosh}{\left (a x \right )}}{2 a^{3}} & \text {for}\: \left |{a^{2} x^{2}}\right | > 1 \\\frac {x^{3}}{2 \sqrt {- a^{2} x^{2} + 1}} - \frac {x}{2 a^{2} \sqrt {- a^{2} x^{2} + 1}} + \frac {\operatorname {asin}{\left (a x \right )}}{2 a^{3}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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((a*x+1)/(-a^2*x^2+1)^(1/2)*x^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 = 0.05, size = 82, normalized size = 1.11 \begin {gather*} \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}{3\,a\,\sqrt {-a^2}}+\frac {a\,x^2}{3\,\sqrt {-a^2}}-\frac {x\,\sqrt {-a^2}}{2\,a^2}\right )}{\sqrt {-a^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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