3.4.43 \(\int \frac {x^3 \sinh ^{-1}(a x)^3}{\sqrt {1+a^2 x^2}} \, dx\) [343]

Optimal. Leaf size=153 \[ \frac {40 x}{9 a^3}-\frac {2 x^3}{27 a}-\frac {40 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)}{9 a^4}+\frac {2 x^2 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)}{9 a^2}+\frac {2 x \sinh ^{-1}(a x)^2}{a^3}-\frac {x^3 \sinh ^{-1}(a x)^2}{3 a}-\frac {2 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^3}{3 a^4}+\frac {x^2 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^3}{3 a^2} \]

[Out]

40/9*x/a^3-2/27*x^3/a+2*x*arcsinh(a*x)^2/a^3-1/3*x^3*arcsinh(a*x)^2/a-40/9*arcsinh(a*x)*(a^2*x^2+1)^(1/2)/a^4+
2/9*x^2*arcsinh(a*x)*(a^2*x^2+1)^(1/2)/a^2-2/3*arcsinh(a*x)^3*(a^2*x^2+1)^(1/2)/a^4+1/3*x^2*arcsinh(a*x)^3*(a^
2*x^2+1)^(1/2)/a^2

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Rubi [A]
time = 0.22, antiderivative size = 153, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {5812, 5798, 5772, 8, 5776, 30} \begin {gather*} \frac {40 x}{9 a^3}+\frac {2 x \sinh ^{-1}(a x)^2}{a^3}+\frac {x^2 \sqrt {a^2 x^2+1} \sinh ^{-1}(a x)^3}{3 a^2}+\frac {2 x^2 \sqrt {a^2 x^2+1} \sinh ^{-1}(a x)}{9 a^2}-\frac {2 \sqrt {a^2 x^2+1} \sinh ^{-1}(a x)^3}{3 a^4}-\frac {40 \sqrt {a^2 x^2+1} \sinh ^{-1}(a x)}{9 a^4}-\frac {2 x^3}{27 a}-\frac {x^3 \sinh ^{-1}(a x)^2}{3 a} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^3*ArcSinh[a*x]^3)/Sqrt[1 + a^2*x^2],x]

[Out]

(40*x)/(9*a^3) - (2*x^3)/(27*a) - (40*Sqrt[1 + a^2*x^2]*ArcSinh[a*x])/(9*a^4) + (2*x^2*Sqrt[1 + a^2*x^2]*ArcSi
nh[a*x])/(9*a^2) + (2*x*ArcSinh[a*x]^2)/a^3 - (x^3*ArcSinh[a*x]^2)/(3*a) - (2*Sqrt[1 + a^2*x^2]*ArcSinh[a*x]^3
)/(3*a^4) + (x^2*Sqrt[1 + a^2*x^2]*ArcSinh[a*x]^3)/(3*a^2)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 5772

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.), x_Symbol] :> Simp[x*(a + b*ArcSinh[c*x])^n, x] - Dist[b*c*n, In
t[x*((a + b*ArcSinh[c*x])^(n - 1)/Sqrt[1 + c^2*x^2]), x], x] /; FreeQ[{a, b, c}, x] && GtQ[n, 0]

Rule 5776

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*ArcS
inh[c*x])^n/(d*(m + 1))), x] - Dist[b*c*(n/(d*(m + 1))), Int[(d*x)^(m + 1)*((a + b*ArcSinh[c*x])^(n - 1)/Sqrt[
1 + c^2*x^2]), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && NeQ[m, -1]

Rule 5798

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*(x_)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(d + e*x^2)^
(p + 1)*((a + b*ArcSinh[c*x])^n/(2*e*(p + 1))), x] - Dist[b*(n/(2*c*(p + 1)))*Simp[(d + e*x^2)^p/(1 + c^2*x^2)
^p], Int[(1 + c^2*x^2)^(p + 1/2)*(a + b*ArcSinh[c*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[e
, c^2*d] && GtQ[n, 0] && NeQ[p, -1]

Rule 5812

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Simp[
f*(f*x)^(m - 1)*(d + e*x^2)^(p + 1)*((a + b*ArcSinh[c*x])^n/(e*(m + 2*p + 1))), x] + (-Dist[f^2*((m - 1)/(c^2*
(m + 2*p + 1))), Int[(f*x)^(m - 2)*(d + e*x^2)^p*(a + b*ArcSinh[c*x])^n, x], x] - Dist[b*f*(n/(c*(m + 2*p + 1)
))*Simp[(d + e*x^2)^p/(1 + c^2*x^2)^p], Int[(f*x)^(m - 1)*(1 + c^2*x^2)^(p + 1/2)*(a + b*ArcSinh[c*x])^(n - 1)
, x], x]) /; FreeQ[{a, b, c, d, e, f, p}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && IGtQ[m, 1] && NeQ[m + 2*p + 1, 0
]

Rubi steps

\begin {align*} \int \frac {x^3 \sinh ^{-1}(a x)^3}{\sqrt {1+a^2 x^2}} \, dx &=\frac {x^2 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^3}{3 a^2}-\frac {2 \int \frac {x \sinh ^{-1}(a x)^3}{\sqrt {1+a^2 x^2}} \, dx}{3 a^2}-\frac {\int x^2 \sinh ^{-1}(a x)^2 \, dx}{a}\\ &=-\frac {x^3 \sinh ^{-1}(a x)^2}{3 a}-\frac {2 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^3}{3 a^4}+\frac {x^2 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^3}{3 a^2}+\frac {2}{3} \int \frac {x^3 \sinh ^{-1}(a x)}{\sqrt {1+a^2 x^2}} \, dx+\frac {2 \int \sinh ^{-1}(a x)^2 \, dx}{a^3}\\ &=\frac {2 x^2 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)}{9 a^2}+\frac {2 x \sinh ^{-1}(a x)^2}{a^3}-\frac {x^3 \sinh ^{-1}(a x)^2}{3 a}-\frac {2 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^3}{3 a^4}+\frac {x^2 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^3}{3 a^2}-\frac {4 \int \frac {x \sinh ^{-1}(a x)}{\sqrt {1+a^2 x^2}} \, dx}{9 a^2}-\frac {4 \int \frac {x \sinh ^{-1}(a x)}{\sqrt {1+a^2 x^2}} \, dx}{a^2}-\frac {2 \int x^2 \, dx}{9 a}\\ &=-\frac {2 x^3}{27 a}-\frac {40 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)}{9 a^4}+\frac {2 x^2 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)}{9 a^2}+\frac {2 x \sinh ^{-1}(a x)^2}{a^3}-\frac {x^3 \sinh ^{-1}(a x)^2}{3 a}-\frac {2 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^3}{3 a^4}+\frac {x^2 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^3}{3 a^2}+\frac {4 \int 1 \, dx}{9 a^3}+\frac {4 \int 1 \, dx}{a^3}\\ &=\frac {40 x}{9 a^3}-\frac {2 x^3}{27 a}-\frac {40 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)}{9 a^4}+\frac {2 x^2 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)}{9 a^2}+\frac {2 x \sinh ^{-1}(a x)^2}{a^3}-\frac {x^3 \sinh ^{-1}(a x)^2}{3 a}-\frac {2 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^3}{3 a^4}+\frac {x^2 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^3}{3 a^2}\\ \end {align*}

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Mathematica [A]
time = 0.05, size = 98, normalized size = 0.64 \begin {gather*} \frac {-2 a x \left (-60+a^2 x^2\right )+6 \left (-20+a^2 x^2\right ) \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)-9 a x \left (-6+a^2 x^2\right ) \sinh ^{-1}(a x)^2+9 \left (-2+a^2 x^2\right ) \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^3}{27 a^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^3*ArcSinh[a*x]^3)/Sqrt[1 + a^2*x^2],x]

[Out]

(-2*a*x*(-60 + a^2*x^2) + 6*(-20 + a^2*x^2)*Sqrt[1 + a^2*x^2]*ArcSinh[a*x] - 9*a*x*(-6 + a^2*x^2)*ArcSinh[a*x]
^2 + 9*(-2 + a^2*x^2)*Sqrt[1 + a^2*x^2]*ArcSinh[a*x]^3)/(27*a^4)

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Maple [F]
time = 180.00, size = 0, normalized size = 0.00 \[\int \frac {x^{3} \arcsinh \left (a x \right )^{3}}{\sqrt {a^{2} x^{2}+1}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.26, size = 127, normalized size = 0.83 \begin {gather*} \frac {1}{3} \, {\left (\frac {\sqrt {a^{2} x^{2} + 1} x^{2}}{a^{2}} - \frac {2 \, \sqrt {a^{2} x^{2} + 1}}{a^{4}}\right )} \operatorname {arsinh}\left (a x\right )^{3} + \frac {2}{27} \, a {\left (\frac {3 \, {\left (\sqrt {a^{2} x^{2} + 1} x^{2} - \frac {20 \, \sqrt {a^{2} x^{2} + 1}}{a^{2}}\right )} \operatorname {arsinh}\left (a x\right )}{a^{3}} - \frac {a^{2} x^{3} - 60 \, x}{a^{4}}\right )} - \frac {{\left (a^{2} x^{3} - 6 \, x\right )} \operatorname {arsinh}\left (a x\right )^{2}}{3 \, a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(sqrt(a^2*x^2 + 1)*x^2/a^2 - 2*sqrt(a^2*x^2 + 1)/a^4)*arcsinh(a*x)^3 + 2/27*a*(3*(sqrt(a^2*x^2 + 1)*x^2 -
20*sqrt(a^2*x^2 + 1)/a^2)*arcsinh(a*x)/a^3 - (a^2*x^3 - 60*x)/a^4) - 1/3*(a^2*x^3 - 6*x)*arcsinh(a*x)^2/a^3

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Fricas [A]
time = 0.38, size = 128, normalized size = 0.84 \begin {gather*} -\frac {2 \, a^{3} x^{3} - 9 \, \sqrt {a^{2} x^{2} + 1} {\left (a^{2} x^{2} - 2\right )} \log \left (a x + \sqrt {a^{2} x^{2} + 1}\right )^{3} + 9 \, {\left (a^{3} x^{3} - 6 \, a x\right )} \log \left (a x + \sqrt {a^{2} x^{2} + 1}\right )^{2} - 6 \, \sqrt {a^{2} x^{2} + 1} {\left (a^{2} x^{2} - 20\right )} \log \left (a x + \sqrt {a^{2} x^{2} + 1}\right ) - 120 \, a x}{27 \, a^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/27*(2*a^3*x^3 - 9*sqrt(a^2*x^2 + 1)*(a^2*x^2 - 2)*log(a*x + sqrt(a^2*x^2 + 1))^3 + 9*(a^3*x^3 - 6*a*x)*log(
a*x + sqrt(a^2*x^2 + 1))^2 - 6*sqrt(a^2*x^2 + 1)*(a^2*x^2 - 20)*log(a*x + sqrt(a^2*x^2 + 1)) - 120*a*x)/a^4

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Sympy [A]
time = 0.60, size = 148, normalized size = 0.97 \begin {gather*} \begin {cases} - \frac {x^{3} \operatorname {asinh}^{2}{\left (a x \right )}}{3 a} - \frac {2 x^{3}}{27 a} + \frac {x^{2} \sqrt {a^{2} x^{2} + 1} \operatorname {asinh}^{3}{\left (a x \right )}}{3 a^{2}} + \frac {2 x^{2} \sqrt {a^{2} x^{2} + 1} \operatorname {asinh}{\left (a x \right )}}{9 a^{2}} + \frac {2 x \operatorname {asinh}^{2}{\left (a x \right )}}{a^{3}} + \frac {40 x}{9 a^{3}} - \frac {2 \sqrt {a^{2} x^{2} + 1} \operatorname {asinh}^{3}{\left (a x \right )}}{3 a^{4}} - \frac {40 \sqrt {a^{2} x^{2} + 1} \operatorname {asinh}{\left (a x \right )}}{9 a^{4}} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-x**3*asinh(a*x)**2/(3*a) - 2*x**3/(27*a) + x**2*sqrt(a**2*x**2 + 1)*asinh(a*x)**3/(3*a**2) + 2*x**
2*sqrt(a**2*x**2 + 1)*asinh(a*x)/(9*a**2) + 2*x*asinh(a*x)**2/a**3 + 40*x/(9*a**3) - 2*sqrt(a**2*x**2 + 1)*asi
nh(a*x)**3/(3*a**4) - 40*sqrt(a**2*x**2 + 1)*asinh(a*x)/(9*a**4), Ne(a, 0)), (0, 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(x^3*arcsinh(a*x)^3/(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 [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^3\,{\mathrm {asinh}\left (a\,x\right )}^3}{\sqrt {a^2\,x^2+1}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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