3.1.1 \(\int e^{\tanh ^{-1}(a x)} x^4 \, dx\) [1]

Optimal. Leaf size=111 \[ -\frac {4 x^2 \sqrt {1-a^2 x^2}}{15 a^3}-\frac {x^3 \sqrt {1-a^2 x^2}}{4 a^2}-\frac {x^4 \sqrt {1-a^2 x^2}}{5 a}-\frac {(64+45 a x) \sqrt {1-a^2 x^2}}{120 a^5}+\frac {3 \text {ArcSin}(a x)}{8 a^5} \]

[Out]

3/8*arcsin(a*x)/a^5-4/15*x^2*(-a^2*x^2+1)^(1/2)/a^3-1/4*x^3*(-a^2*x^2+1)^(1/2)/a^2-1/5*x^4*(-a^2*x^2+1)^(1/2)/
a-1/120*(45*a*x+64)*(-a^2*x^2+1)^(1/2)/a^5

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Rubi [A]
time = 0.07, antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {6259, 847, 794, 222} \begin {gather*} \frac {3 \text {ArcSin}(a x)}{8 a^5}-\frac {x^4 \sqrt {1-a^2 x^2}}{5 a}-\frac {x^3 \sqrt {1-a^2 x^2}}{4 a^2}-\frac {(45 a x+64) \sqrt {1-a^2 x^2}}{120 a^5}-\frac {4 x^2 \sqrt {1-a^2 x^2}}{15 a^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^ArcTanh[a*x]*x^4,x]

[Out]

(-4*x^2*Sqrt[1 - a^2*x^2])/(15*a^3) - (x^3*Sqrt[1 - a^2*x^2])/(4*a^2) - (x^4*Sqrt[1 - a^2*x^2])/(5*a) - ((64 +
45*a*x)*Sqrt[1 - a^2*x^2])/(120*a^5) + (3*ArcSin[a*x])/(8*a^5)

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^4 \, dx &=\int \frac {x^4 (1+a x)}{\sqrt {1-a^2 x^2}} \, dx\\ &=-\frac {x^4 \sqrt {1-a^2 x^2}}{5 a}-\frac {\int \frac {x^3 \left (-4 a-5 a^2 x\right )}{\sqrt {1-a^2 x^2}} \, dx}{5 a^2}\\ &=-\frac {x^3 \sqrt {1-a^2 x^2}}{4 a^2}-\frac {x^4 \sqrt {1-a^2 x^2}}{5 a}+\frac {\int \frac {x^2 \left (15 a^2+16 a^3 x\right )}{\sqrt {1-a^2 x^2}} \, dx}{20 a^4}\\ &=-\frac {4 x^2 \sqrt {1-a^2 x^2}}{15 a^3}-\frac {x^3 \sqrt {1-a^2 x^2}}{4 a^2}-\frac {x^4 \sqrt {1-a^2 x^2}}{5 a}-\frac {\int \frac {x \left (-32 a^3-45 a^4 x\right )}{\sqrt {1-a^2 x^2}} \, dx}{60 a^6}\\ &=-\frac {4 x^2 \sqrt {1-a^2 x^2}}{15 a^3}-\frac {x^3 \sqrt {1-a^2 x^2}}{4 a^2}-\frac {x^4 \sqrt {1-a^2 x^2}}{5 a}-\frac {(64+45 a x) \sqrt {1-a^2 x^2}}{120 a^5}+\frac {3 \int \frac {1}{\sqrt {1-a^2 x^2}} \, dx}{8 a^4}\\ &=-\frac {4 x^2 \sqrt {1-a^2 x^2}}{15 a^3}-\frac {x^3 \sqrt {1-a^2 x^2}}{4 a^2}-\frac {x^4 \sqrt {1-a^2 x^2}}{5 a}-\frac {(64+45 a x) \sqrt {1-a^2 x^2}}{120 a^5}+\frac {3 \sin ^{-1}(a x)}{8 a^5}\\ \end {align*}

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Mathematica [A]
time = 0.06, size = 60, normalized size = 0.54 \begin {gather*} \frac {-\sqrt {1-a^2 x^2} \left (64+45 a x+32 a^2 x^2+30 a^3 x^3+24 a^4 x^4\right )+45 \text {ArcSin}(a x)}{120 a^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^ArcTanh[a*x]*x^4,x]

[Out]

(-(Sqrt[1 - a^2*x^2]*(64 + 45*a*x + 32*a^2*x^2 + 30*a^3*x^3 + 24*a^4*x^4)) + 45*ArcSin[a*x])/(120*a^5)

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Maple [A]
time = 0.62, size = 142, normalized size = 1.28

method	result	size
risch	\(\frac {\left (24 a^{4} x^{4}+30 a^{3} x^{3}+32 a^{2} x^{2}+45 a x +64\right ) \left (a^{2} x^{2}-1\right )}{120 a^{5} \sqrt {-a^{2} x^{2}+1}}+\frac {3 \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{8 a^{4} \sqrt {a^{2}}}\)	\(88\)
meijerg	\(-\frac {-\frac {16 \sqrt {\pi }}{15}+\frac {\sqrt {\pi }\, \left (6 a^{4} x^{4}+8 a^{2} x^{2}+16\right ) \sqrt {-a^{2} x^{2}+1}}{15}}{2 a^{5} \sqrt {\pi }}+\frac {-\frac {\sqrt {\pi }\, x \left (-a^{2}\right )^{\frac {5}{2}} \left (10 a^{2} x^{2}+15\right ) \sqrt {-a^{2} x^{2}+1}}{20 a^{4}}+\frac {3 \sqrt {\pi }\, \left (-a^{2}\right )^{\frac {5}{2}} \arcsin \left (a x \right )}{4 a^{5}}}{2 a^{4} \sqrt {\pi }\, \sqrt {-a^{2}}}\)	\(124\)
default	\(a \left (-\frac {x^{4} \sqrt {-a^{2} x^{2}+1}}{5 a^{2}}+\frac {-\frac {4 x^{2} \sqrt {-a^{2} x^{2}+1}}{15 a^{2}}-\frac {8 \sqrt {-a^{2} x^{2}+1}}{15 a^{4}}}{a^{2}}\right )-\frac {x^{3} \sqrt {-a^{2} x^{2}+1}}{4 a^{2}}+\frac {-\frac {3 x \sqrt {-a^{2} x^{2}+1}}{8 a^{2}}+\frac {3 \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{8 a^{2} \sqrt {a^{2}}}}{a^{2}}\)	\(142\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x+1)/(-a^2*x^2+1)^(1/2)*x^4,x,method=_RETURNVERBOSE)

[Out]

a*(-1/5*x^4/a^2*(-a^2*x^2+1)^(1/2)+4/5/a^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))-1/4*x
^3*(-a^2*x^2+1)^(1/2)/a^2+3/4/a^2*(-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 = 105, normalized size = 0.95 \begin {gather*} -\frac {\sqrt {-a^{2} x^{2} + 1} x^{4}}{5 \, a} - \frac {\sqrt {-a^{2} x^{2} + 1} x^{3}}{4 \, a^{2}} - \frac {4 \, \sqrt {-a^{2} x^{2} + 1} x^{2}}{15 \, a^{3}} - \frac {3 \, \sqrt {-a^{2} x^{2} + 1} x}{8 \, a^{4}} + \frac {3 \, \arcsin \left (a x\right )}{8 \, a^{5}} - \frac {8 \, \sqrt {-a^{2} x^{2} + 1}}{15 \, a^{5}} \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^4,x, algorithm="maxima")

[Out]

-1/5*sqrt(-a^2*x^2 + 1)*x^4/a - 1/4*sqrt(-a^2*x^2 + 1)*x^3/a^2 - 4/15*sqrt(-a^2*x^2 + 1)*x^2/a^3 - 3/8*sqrt(-a
^2*x^2 + 1)*x/a^4 + 3/8*arcsin(a*x)/a^5 - 8/15*sqrt(-a^2*x^2 + 1)/a^5

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Fricas [A]
time = 0.45, size = 73, normalized size = 0.66 \begin {gather*} -\frac {{\left (24 \, a^{4} x^{4} + 30 \, a^{3} x^{3} + 32 \, a^{2} x^{2} + 45 \, a x + 64\right )} \sqrt {-a^{2} x^{2} + 1} + 90 \, \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right )}{120 \, a^{5}} \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^4,x, algorithm="fricas")

[Out]

-1/120*((24*a^4*x^4 + 30*a^3*x^3 + 32*a^2*x^2 + 45*a*x + 64)*sqrt(-a^2*x^2 + 1) + 90*arctan((sqrt(-a^2*x^2 + 1
) - 1)/(a*x)))/a^5

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Sympy [A]
time = 3.42, size = 221, normalized size = 1.99 \begin {gather*} a \left (\begin {cases} - \frac {x^{4} \sqrt {- a^{2} x^{2} + 1}}{5 a^{2}} - \frac {4 x^{2} \sqrt {- a^{2} x^{2} + 1}}{15 a^{4}} - \frac {8 \sqrt {- a^{2} x^{2} + 1}}{15 a^{6}} & \text {for}\: a \neq 0 \\\frac {x^{6}}{6} & \text {otherwise} \end {cases}\right ) + \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} \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**4,x)

[Out]

a*Piecewise((-x**4*sqrt(-a**2*x**2 + 1)/(5*a**2) - 4*x**2*sqrt(-a**2*x**2 + 1)/(15*a**4) - 8*sqrt(-a**2*x**2 +
1)/(15*a**6), Ne(a, 0)), (x**6/6, True)) + 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*s
qrt(-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))

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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^4,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 = 0.08, size = 112, normalized size = 1.01 \begin {gather*} \frac {3\,\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{8\,a^4\,\sqrt {-a^2}}+\frac {\sqrt {1-a^2\,x^2}\,\left (\frac {8}{15\,a^3\,\sqrt {-a^2}}+\frac {a\,x^4}{5\,\sqrt {-a^2}}-\frac {3\,x\,\sqrt {-a^2}}{8\,a^4}+\frac {4\,x^2}{15\,a\,\sqrt {-a^2}}+\frac {x^3\,{\left (-a^2\right )}^{3/2}}{4\,a^4}\right )}{\sqrt {-a^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(3*asinh(x*(-a^2)^(1/2)))/(8*a^4*(-a^2)^(1/2)) + ((1 - a^2*x^2)^(1/2)*(8/(15*a^3*(-a^2)^(1/2)) + (a*x^4)/(5*(-
a^2)^(1/2)) - (3*x*(-a^2)^(1/2))/(8*a^4) + (4*x^2)/(15*a*(-a^2)^(1/2)) + (x^3*(-a^2)^(3/2))/(4*a^4)))/(-a^2)^(
1/2)

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