3.22 \(\int x^4 \sinh ^{-1}(a x)^3 \, dx\)

Optimal. Leaf size=195 \[ \frac {16 x \sinh ^{-1}(a x)}{25 a^4}-\frac {8 x^3 \sinh ^{-1}(a x)}{75 a^2}-\frac {3 x^4 \sqrt {a^2 x^2+1} \sinh ^{-1}(a x)^2}{25 a}-\frac {6 \left (a^2 x^2+1\right )^{5/2}}{625 a^5}+\frac {76 \left (a^2 x^2+1\right )^{3/2}}{1125 a^5}-\frac {298 \sqrt {a^2 x^2+1}}{375 a^5}-\frac {8 \sqrt {a^2 x^2+1} \sinh ^{-1}(a x)^2}{25 a^5}+\frac {4 x^2 \sqrt {a^2 x^2+1} \sinh ^{-1}(a x)^2}{25 a^3}+\frac {1}{5} x^5 \sinh ^{-1}(a x)^3+\frac {6}{125} x^5 \sinh ^{-1}(a x) \]

[Out]

76/1125*(a^2*x^2+1)^(3/2)/a^5-6/625*(a^2*x^2+1)^(5/2)/a^5+16/25*x*arcsinh(a*x)/a^4-8/75*x^3*arcsinh(a*x)/a^2+6
/125*x^5*arcsinh(a*x)+1/5*x^5*arcsinh(a*x)^3-298/375*(a^2*x^2+1)^(1/2)/a^5-8/25*arcsinh(a*x)^2*(a^2*x^2+1)^(1/
2)/a^5+4/25*x^2*arcsinh(a*x)^2*(a^2*x^2+1)^(1/2)/a^3-3/25*x^4*arcsinh(a*x)^2*(a^2*x^2+1)^(1/2)/a

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Rubi [A]  time = 0.37, antiderivative size = 195, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 7, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.700, Rules used = {5661, 5758, 5717, 5653, 261, 266, 43} \[ -\frac {6 \left (a^2 x^2+1\right )^{5/2}}{625 a^5}+\frac {76 \left (a^2 x^2+1\right )^{3/2}}{1125 a^5}-\frac {298 \sqrt {a^2 x^2+1}}{375 a^5}-\frac {3 x^4 \sqrt {a^2 x^2+1} \sinh ^{-1}(a x)^2}{25 a}-\frac {8 x^3 \sinh ^{-1}(a x)}{75 a^2}+\frac {4 x^2 \sqrt {a^2 x^2+1} \sinh ^{-1}(a x)^2}{25 a^3}-\frac {8 \sqrt {a^2 x^2+1} \sinh ^{-1}(a x)^2}{25 a^5}+\frac {16 x \sinh ^{-1}(a x)}{25 a^4}+\frac {1}{5} x^5 \sinh ^{-1}(a x)^3+\frac {6}{125} x^5 \sinh ^{-1}(a x) \]

Antiderivative was successfully verified.

[In]

Int[x^4*ArcSinh[a*x]^3,x]

[Out]

(-298*Sqrt[1 + a^2*x^2])/(375*a^5) + (76*(1 + a^2*x^2)^(3/2))/(1125*a^5) - (6*(1 + a^2*x^2)^(5/2))/(625*a^5) +
(16*x*ArcSinh[a*x])/(25*a^4) - (8*x^3*ArcSinh[a*x])/(75*a^2) + (6*x^5*ArcSinh[a*x])/125 - (8*Sqrt[1 + a^2*x^2
]*ArcSinh[a*x]^2)/(25*a^5) + (4*x^2*Sqrt[1 + a^2*x^2]*ArcSinh[a*x]^2)/(25*a^3) - (3*x^4*Sqrt[1 + a^2*x^2]*ArcS
inh[a*x]^2)/(25*a) + (x^5*ArcSinh[a*x]^3)/5

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 261

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 5653

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 5661

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 5717

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*d^IntPart[p]*(d + e*x^2)^FracPart[p])/(2*c*(p +
1)*(1 + c^2*x^2)^FracPart[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 5758

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

Rubi steps

\begin {align*} \int x^4 \sinh ^{-1}(a x)^3 \, dx &=\frac {1}{5} x^5 \sinh ^{-1}(a x)^3-\frac {1}{5} (3 a) \int \frac {x^5 \sinh ^{-1}(a x)^2}{\sqrt {1+a^2 x^2}} \, dx\\ &=-\frac {3 x^4 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^2}{25 a}+\frac {1}{5} x^5 \sinh ^{-1}(a x)^3+\frac {6}{25} \int x^4 \sinh ^{-1}(a x) \, dx+\frac {12 \int \frac {x^3 \sinh ^{-1}(a x)^2}{\sqrt {1+a^2 x^2}} \, dx}{25 a}\\ &=\frac {6}{125} x^5 \sinh ^{-1}(a x)+\frac {4 x^2 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^2}{25 a^3}-\frac {3 x^4 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^2}{25 a}+\frac {1}{5} x^5 \sinh ^{-1}(a x)^3-\frac {8 \int \frac {x \sinh ^{-1}(a x)^2}{\sqrt {1+a^2 x^2}} \, dx}{25 a^3}-\frac {8 \int x^2 \sinh ^{-1}(a x) \, dx}{25 a^2}-\frac {1}{125} (6 a) \int \frac {x^5}{\sqrt {1+a^2 x^2}} \, dx\\ &=-\frac {8 x^3 \sinh ^{-1}(a x)}{75 a^2}+\frac {6}{125} x^5 \sinh ^{-1}(a x)-\frac {8 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^2}{25 a^5}+\frac {4 x^2 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^2}{25 a^3}-\frac {3 x^4 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^2}{25 a}+\frac {1}{5} x^5 \sinh ^{-1}(a x)^3+\frac {16 \int \sinh ^{-1}(a x) \, dx}{25 a^4}+\frac {8 \int \frac {x^3}{\sqrt {1+a^2 x^2}} \, dx}{75 a}-\frac {1}{125} (3 a) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {1+a^2 x}} \, dx,x,x^2\right )\\ &=\frac {16 x \sinh ^{-1}(a x)}{25 a^4}-\frac {8 x^3 \sinh ^{-1}(a x)}{75 a^2}+\frac {6}{125} x^5 \sinh ^{-1}(a x)-\frac {8 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^2}{25 a^5}+\frac {4 x^2 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^2}{25 a^3}-\frac {3 x^4 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^2}{25 a}+\frac {1}{5} x^5 \sinh ^{-1}(a x)^3-\frac {16 \int \frac {x}{\sqrt {1+a^2 x^2}} \, dx}{25 a^3}+\frac {4 \operatorname {Subst}\left (\int \frac {x}{\sqrt {1+a^2 x}} \, dx,x,x^2\right )}{75 a}-\frac {1}{125} (3 a) \operatorname {Subst}\left (\int \left (\frac {1}{a^4 \sqrt {1+a^2 x}}-\frac {2 \sqrt {1+a^2 x}}{a^4}+\frac {\left (1+a^2 x\right )^{3/2}}{a^4}\right ) \, dx,x,x^2\right )\\ &=-\frac {86 \sqrt {1+a^2 x^2}}{125 a^5}+\frac {4 \left (1+a^2 x^2\right )^{3/2}}{125 a^5}-\frac {6 \left (1+a^2 x^2\right )^{5/2}}{625 a^5}+\frac {16 x \sinh ^{-1}(a x)}{25 a^4}-\frac {8 x^3 \sinh ^{-1}(a x)}{75 a^2}+\frac {6}{125} x^5 \sinh ^{-1}(a x)-\frac {8 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^2}{25 a^5}+\frac {4 x^2 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^2}{25 a^3}-\frac {3 x^4 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^2}{25 a}+\frac {1}{5} x^5 \sinh ^{-1}(a x)^3+\frac {4 \operatorname {Subst}\left (\int \left (-\frac {1}{a^2 \sqrt {1+a^2 x}}+\frac {\sqrt {1+a^2 x}}{a^2}\right ) \, dx,x,x^2\right )}{75 a}\\ &=-\frac {298 \sqrt {1+a^2 x^2}}{375 a^5}+\frac {76 \left (1+a^2 x^2\right )^{3/2}}{1125 a^5}-\frac {6 \left (1+a^2 x^2\right )^{5/2}}{625 a^5}+\frac {16 x \sinh ^{-1}(a x)}{25 a^4}-\frac {8 x^3 \sinh ^{-1}(a x)}{75 a^2}+\frac {6}{125} x^5 \sinh ^{-1}(a x)-\frac {8 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^2}{25 a^5}+\frac {4 x^2 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^2}{25 a^3}-\frac {3 x^4 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^2}{25 a}+\frac {1}{5} x^5 \sinh ^{-1}(a x)^3\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 120, normalized size = 0.62 \[ \frac {1125 a^5 x^5 \sinh ^{-1}(a x)^3-2 \sqrt {a^2 x^2+1} \left (27 a^4 x^4-136 a^2 x^2+2072\right )+30 a x \left (9 a^4 x^4-20 a^2 x^2+120\right ) \sinh ^{-1}(a x)-225 \sqrt {a^2 x^2+1} \left (3 a^4 x^4-4 a^2 x^2+8\right ) \sinh ^{-1}(a x)^2}{5625 a^5} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4*ArcSinh[a*x]^3,x]

[Out]

(-2*Sqrt[1 + a^2*x^2]*(2072 - 136*a^2*x^2 + 27*a^4*x^4) + 30*a*x*(120 - 20*a^2*x^2 + 9*a^4*x^4)*ArcSinh[a*x] -
225*Sqrt[1 + a^2*x^2]*(8 - 4*a^2*x^2 + 3*a^4*x^4)*ArcSinh[a*x]^2 + 1125*a^5*x^5*ArcSinh[a*x]^3)/(5625*a^5)

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fricas [A]  time = 0.41, size = 151, normalized size = 0.77 \[ \frac {1125 \, a^{5} x^{5} \log \left (a x + \sqrt {a^{2} x^{2} + 1}\right )^{3} - 225 \, {\left (3 \, a^{4} x^{4} - 4 \, a^{2} x^{2} + 8\right )} \sqrt {a^{2} x^{2} + 1} \log \left (a x + \sqrt {a^{2} x^{2} + 1}\right )^{2} + 30 \, {\left (9 \, a^{5} x^{5} - 20 \, a^{3} x^{3} + 120 \, a x\right )} \log \left (a x + \sqrt {a^{2} x^{2} + 1}\right ) - 2 \, {\left (27 \, a^{4} x^{4} - 136 \, a^{2} x^{2} + 2072\right )} \sqrt {a^{2} x^{2} + 1}}{5625 \, a^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*arcsinh(a*x)^3,x, algorithm="fricas")

[Out]

1/5625*(1125*a^5*x^5*log(a*x + sqrt(a^2*x^2 + 1))^3 - 225*(3*a^4*x^4 - 4*a^2*x^2 + 8)*sqrt(a^2*x^2 + 1)*log(a*
x + sqrt(a^2*x^2 + 1))^2 + 30*(9*a^5*x^5 - 20*a^3*x^3 + 120*a*x)*log(a*x + sqrt(a^2*x^2 + 1)) - 2*(27*a^4*x^4
- 136*a^2*x^2 + 2072)*sqrt(a^2*x^2 + 1))/a^5

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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^4*arcsinh(a*x)^3,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 [A]  time = 0.48, size = 172, normalized size = 0.88 \[ \frac {\frac {a^{5} x^{5} \arcsinh \left (a x \right )^{3}}{5}-\frac {8 \arcsinh \left (a x \right )^{2} \sqrt {a^{2} x^{2}+1}}{25}-\frac {3 a^{4} x^{4} \arcsinh \left (a x \right )^{2} \sqrt {a^{2} x^{2}+1}}{25}+\frac {4 \arcsinh \left (a x \right )^{2} \sqrt {a^{2} x^{2}+1}\, a^{2} x^{2}}{25}+\frac {16 a x \arcsinh \left (a x \right )}{25}-\frac {4144 \sqrt {a^{2} x^{2}+1}}{5625}+\frac {6 a^{5} x^{5} \arcsinh \left (a x \right )}{125}-\frac {6 a^{4} x^{4} \sqrt {a^{2} x^{2}+1}}{625}+\frac {272 a^{2} x^{2} \sqrt {a^{2} x^{2}+1}}{5625}-\frac {8 a^{3} x^{3} \arcsinh \left (a x \right )}{75}}{a^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*arcsinh(a*x)^3,x)

[Out]

1/a^5*(1/5*a^5*x^5*arcsinh(a*x)^3-8/25*arcsinh(a*x)^2*(a^2*x^2+1)^(1/2)-3/25*a^4*x^4*arcsinh(a*x)^2*(a^2*x^2+1
)^(1/2)+4/25*arcsinh(a*x)^2*(a^2*x^2+1)^(1/2)*a^2*x^2+16/25*a*x*arcsinh(a*x)-4144/5625*(a^2*x^2+1)^(1/2)+6/125
*a^5*x^5*arcsinh(a*x)-6/625*a^4*x^4*(a^2*x^2+1)^(1/2)+272/5625*a^2*x^2*(a^2*x^2+1)^(1/2)-8/75*a^3*x^3*arcsinh(
a*x))

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maxima [A]  time = 0.35, size = 165, normalized size = 0.85 \[ \frac {1}{5} \, x^{5} \operatorname {arsinh}\left (a x\right )^{3} - \frac {1}{25} \, {\left (\frac {3 \, \sqrt {a^{2} x^{2} + 1} x^{4}}{a^{2}} - \frac {4 \, \sqrt {a^{2} x^{2} + 1} x^{2}}{a^{4}} + \frac {8 \, \sqrt {a^{2} x^{2} + 1}}{a^{6}}\right )} a \operatorname {arsinh}\left (a x\right )^{2} - \frac {2}{5625} \, a {\left (\frac {27 \, \sqrt {a^{2} x^{2} + 1} a^{2} x^{4} - 136 \, \sqrt {a^{2} x^{2} + 1} x^{2} + \frac {2072 \, \sqrt {a^{2} x^{2} + 1}}{a^{2}}}{a^{4}} - \frac {15 \, {\left (9 \, a^{4} x^{5} - 20 \, a^{2} x^{3} + 120 \, x\right )} \operatorname {arsinh}\left (a x\right )}{a^{5}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*arcsinh(a*x)^3,x, algorithm="maxima")

[Out]

1/5*x^5*arcsinh(a*x)^3 - 1/25*(3*sqrt(a^2*x^2 + 1)*x^4/a^2 - 4*sqrt(a^2*x^2 + 1)*x^2/a^4 + 8*sqrt(a^2*x^2 + 1)
/a^6)*a*arcsinh(a*x)^2 - 2/5625*a*((27*sqrt(a^2*x^2 + 1)*a^2*x^4 - 136*sqrt(a^2*x^2 + 1)*x^2 + 2072*sqrt(a^2*x
^2 + 1)/a^2)/a^4 - 15*(9*a^4*x^5 - 20*a^2*x^3 + 120*x)*arcsinh(a*x)/a^5)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^4\,{\mathrm {asinh}\left (a\,x\right )}^3 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*asinh(a*x)^3,x)

[Out]

int(x^4*asinh(a*x)^3, x)

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sympy [A]  time = 5.45, size = 196, normalized size = 1.01 \[ \begin {cases} \frac {x^{5} \operatorname {asinh}^{3}{\left (a x \right )}}{5} + \frac {6 x^{5} \operatorname {asinh}{\left (a x \right )}}{125} - \frac {3 x^{4} \sqrt {a^{2} x^{2} + 1} \operatorname {asinh}^{2}{\left (a x \right )}}{25 a} - \frac {6 x^{4} \sqrt {a^{2} x^{2} + 1}}{625 a} - \frac {8 x^{3} \operatorname {asinh}{\left (a x \right )}}{75 a^{2}} + \frac {4 x^{2} \sqrt {a^{2} x^{2} + 1} \operatorname {asinh}^{2}{\left (a x \right )}}{25 a^{3}} + \frac {272 x^{2} \sqrt {a^{2} x^{2} + 1}}{5625 a^{3}} + \frac {16 x \operatorname {asinh}{\left (a x \right )}}{25 a^{4}} - \frac {8 \sqrt {a^{2} x^{2} + 1} \operatorname {asinh}^{2}{\left (a x \right )}}{25 a^{5}} - \frac {4144 \sqrt {a^{2} x^{2} + 1}}{5625 a^{5}} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*asinh(a*x)**3,x)

[Out]

Piecewise((x**5*asinh(a*x)**3/5 + 6*x**5*asinh(a*x)/125 - 3*x**4*sqrt(a**2*x**2 + 1)*asinh(a*x)**2/(25*a) - 6*
x**4*sqrt(a**2*x**2 + 1)/(625*a) - 8*x**3*asinh(a*x)/(75*a**2) + 4*x**2*sqrt(a**2*x**2 + 1)*asinh(a*x)**2/(25*
a**3) + 272*x**2*sqrt(a**2*x**2 + 1)/(5625*a**3) + 16*x*asinh(a*x)/(25*a**4) - 8*sqrt(a**2*x**2 + 1)*asinh(a*x
)**2/(25*a**5) - 4144*sqrt(a**2*x**2 + 1)/(5625*a**5), Ne(a, 0)), (0, True))

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