3.3.82 \(\int \frac {x^4 \sinh ^{-1}(a x)^2}{\sqrt {1+a^2 x^2}} \, dx\) [282]

Optimal. Leaf size=153 \[ -\frac {15 x \sqrt {1+a^2 x^2}}{64 a^4}+\frac {x^3 \sqrt {1+a^2 x^2}}{32 a^2}+\frac {15 \sinh ^{-1}(a x)}{64 a^5}+\frac {3 x^2 \sinh ^{-1}(a x)}{8 a^3}-\frac {x^4 \sinh ^{-1}(a x)}{8 a}-\frac {3 x \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^2}{8 a^4}+\frac {x^3 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^2}{4 a^2}+\frac {\sinh ^{-1}(a x)^3}{8 a^5} \]

[Out]

15/64*arcsinh(a*x)/a^5+3/8*x^2*arcsinh(a*x)/a^3-1/8*x^4*arcsinh(a*x)/a+1/8*arcsinh(a*x)^3/a^5-15/64*x*(a^2*x^2
+1)^(1/2)/a^4+1/32*x^3*(a^2*x^2+1)^(1/2)/a^2-3/8*x*arcsinh(a*x)^2*(a^2*x^2+1)^(1/2)/a^4+1/4*x^3*arcsinh(a*x)^2
*(a^2*x^2+1)^(1/2)/a^2

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Rubi [A]
time = 0.20, antiderivative size = 153, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {5812, 5783, 5776, 327, 221} \begin {gather*} \frac {\sinh ^{-1}(a x)^3}{8 a^5}+\frac {15 \sinh ^{-1}(a x)}{64 a^5}+\frac {3 x^2 \sinh ^{-1}(a x)}{8 a^3}+\frac {x^3 \sqrt {a^2 x^2+1}}{32 a^2}+\frac {x^3 \sqrt {a^2 x^2+1} \sinh ^{-1}(a x)^2}{4 a^2}-\frac {15 x \sqrt {a^2 x^2+1}}{64 a^4}-\frac {3 x \sqrt {a^2 x^2+1} \sinh ^{-1}(a x)^2}{8 a^4}-\frac {x^4 \sinh ^{-1}(a x)}{8 a} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^4*ArcSinh[a*x]^2)/Sqrt[1 + a^2*x^2],x]

[Out]

(-15*x*Sqrt[1 + a^2*x^2])/(64*a^4) + (x^3*Sqrt[1 + a^2*x^2])/(32*a^2) + (15*ArcSinh[a*x])/(64*a^5) + (3*x^2*Ar
cSinh[a*x])/(8*a^3) - (x^4*ArcSinh[a*x])/(8*a) - (3*x*Sqrt[1 + a^2*x^2]*ArcSinh[a*x]^2)/(8*a^4) + (x^3*Sqrt[1
+ a^2*x^2]*ArcSinh[a*x]^2)/(4*a^2) + ArcSinh[a*x]^3/(8*a^5)

Rule 221

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[Rt[b, 2]*(x/Sqrt[a])]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

Rule 327

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

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 5783

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(1/(b*c*(n + 1)))*S
imp[Sqrt[1 + c^2*x^2]/Sqrt[d + e*x^2]]*(a + b*ArcSinh[c*x])^(n + 1), x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ
[e, c^2*d] && NeQ[n, -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^4 \sinh ^{-1}(a x)^2}{\sqrt {1+a^2 x^2}} \, dx &=\frac {x^3 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^2}{4 a^2}-\frac {3 \int \frac {x^2 \sinh ^{-1}(a x)^2}{\sqrt {1+a^2 x^2}} \, dx}{4 a^2}-\frac {\int x^3 \sinh ^{-1}(a x) \, dx}{2 a}\\ &=-\frac {x^4 \sinh ^{-1}(a x)}{8 a}-\frac {3 x \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^2}{8 a^4}+\frac {x^3 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^2}{4 a^2}+\frac {1}{8} \int \frac {x^4}{\sqrt {1+a^2 x^2}} \, dx+\frac {3 \int \frac {\sinh ^{-1}(a x)^2}{\sqrt {1+a^2 x^2}} \, dx}{8 a^4}+\frac {3 \int x \sinh ^{-1}(a x) \, dx}{4 a^3}\\ &=\frac {x^3 \sqrt {1+a^2 x^2}}{32 a^2}+\frac {3 x^2 \sinh ^{-1}(a x)}{8 a^3}-\frac {x^4 \sinh ^{-1}(a x)}{8 a}-\frac {3 x \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^2}{8 a^4}+\frac {x^3 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^2}{4 a^2}+\frac {\sinh ^{-1}(a x)^3}{8 a^5}-\frac {3 \int \frac {x^2}{\sqrt {1+a^2 x^2}} \, dx}{32 a^2}-\frac {3 \int \frac {x^2}{\sqrt {1+a^2 x^2}} \, dx}{8 a^2}\\ &=-\frac {15 x \sqrt {1+a^2 x^2}}{64 a^4}+\frac {x^3 \sqrt {1+a^2 x^2}}{32 a^2}+\frac {3 x^2 \sinh ^{-1}(a x)}{8 a^3}-\frac {x^4 \sinh ^{-1}(a x)}{8 a}-\frac {3 x \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^2}{8 a^4}+\frac {x^3 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^2}{4 a^2}+\frac {\sinh ^{-1}(a x)^3}{8 a^5}+\frac {3 \int \frac {1}{\sqrt {1+a^2 x^2}} \, dx}{64 a^4}+\frac {3 \int \frac {1}{\sqrt {1+a^2 x^2}} \, dx}{16 a^4}\\ &=-\frac {15 x \sqrt {1+a^2 x^2}}{64 a^4}+\frac {x^3 \sqrt {1+a^2 x^2}}{32 a^2}+\frac {15 \sinh ^{-1}(a x)}{64 a^5}+\frac {3 x^2 \sinh ^{-1}(a x)}{8 a^3}-\frac {x^4 \sinh ^{-1}(a x)}{8 a}-\frac {3 x \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^2}{8 a^4}+\frac {x^3 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^2}{4 a^2}+\frac {\sinh ^{-1}(a x)^3}{8 a^5}\\ \end {align*}

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Mathematica [A]
time = 0.05, size = 98, normalized size = 0.64 \begin {gather*} \frac {a x \sqrt {1+a^2 x^2} \left (-15+2 a^2 x^2\right )+\left (15+24 a^2 x^2-8 a^4 x^4\right ) \sinh ^{-1}(a x)+8 a x \sqrt {1+a^2 x^2} \left (-3+2 a^2 x^2\right ) \sinh ^{-1}(a x)^2+8 \sinh ^{-1}(a x)^3}{64 a^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^4*ArcSinh[a*x]^2)/Sqrt[1 + a^2*x^2],x]

[Out]

(a*x*Sqrt[1 + a^2*x^2]*(-15 + 2*a^2*x^2) + (15 + 24*a^2*x^2 - 8*a^4*x^4)*ArcSinh[a*x] + 8*a*x*Sqrt[1 + a^2*x^2
]*(-3 + 2*a^2*x^2)*ArcSinh[a*x]^2 + 8*ArcSinh[a*x]^3)/(64*a^5)

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Maple [F]
time = 180.00, size = 0, normalized size = 0.00 \[\int \frac {x^{4} \arcsinh \left (a x \right )^{2}}{\sqrt {a^{2} x^{2}+1}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*arcsinh(a*x)^2/(a^2*x^2+1)^(1/2),x, algorithm="maxima")

[Out]

integrate(x^4*arcsinh(a*x)^2/sqrt(a^2*x^2 + 1), x)

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Fricas [A]
time = 0.38, size = 131, normalized size = 0.86 \begin {gather*} \frac {8 \, {\left (2 \, a^{3} x^{3} - 3 \, a x\right )} \sqrt {a^{2} x^{2} + 1} \log \left (a x + \sqrt {a^{2} x^{2} + 1}\right )^{2} + 8 \, \log \left (a x + \sqrt {a^{2} x^{2} + 1}\right )^{3} - {\left (8 \, a^{4} x^{4} - 24 \, a^{2} x^{2} - 15\right )} \log \left (a x + \sqrt {a^{2} x^{2} + 1}\right ) + {\left (2 \, a^{3} x^{3} - 15 \, a x\right )} \sqrt {a^{2} x^{2} + 1}}{64 \, a^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*arcsinh(a*x)^2/(a^2*x^2+1)^(1/2),x, algorithm="fricas")

[Out]

1/64*(8*(2*a^3*x^3 - 3*a*x)*sqrt(a^2*x^2 + 1)*log(a*x + sqrt(a^2*x^2 + 1))^2 + 8*log(a*x + sqrt(a^2*x^2 + 1))^
3 - (8*a^4*x^4 - 24*a^2*x^2 - 15)*log(a*x + sqrt(a^2*x^2 + 1)) + (2*a^3*x^3 - 15*a*x)*sqrt(a^2*x^2 + 1))/a^5

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Sympy [A]
time = 0.86, size = 146, normalized size = 0.95 \begin {gather*} \begin {cases} - \frac {x^{4} \operatorname {asinh}{\left (a x \right )}}{8 a} + \frac {x^{3} \sqrt {a^{2} x^{2} + 1} \operatorname {asinh}^{2}{\left (a x \right )}}{4 a^{2}} + \frac {x^{3} \sqrt {a^{2} x^{2} + 1}}{32 a^{2}} + \frac {3 x^{2} \operatorname {asinh}{\left (a x \right )}}{8 a^{3}} - \frac {3 x \sqrt {a^{2} x^{2} + 1} \operatorname {asinh}^{2}{\left (a x \right )}}{8 a^{4}} - \frac {15 x \sqrt {a^{2} x^{2} + 1}}{64 a^{4}} + \frac {\operatorname {asinh}^{3}{\left (a x \right )}}{8 a^{5}} + \frac {15 \operatorname {asinh}{\left (a x \right )}}{64 a^{5}} & \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**4*asinh(a*x)**2/(a**2*x**2+1)**(1/2),x)

[Out]

Piecewise((-x**4*asinh(a*x)/(8*a) + x**3*sqrt(a**2*x**2 + 1)*asinh(a*x)**2/(4*a**2) + x**3*sqrt(a**2*x**2 + 1)
/(32*a**2) + 3*x**2*asinh(a*x)/(8*a**3) - 3*x*sqrt(a**2*x**2 + 1)*asinh(a*x)**2/(8*a**4) - 15*x*sqrt(a**2*x**2
+ 1)/(64*a**4) + asinh(a*x)**3/(8*a**5) + 15*asinh(a*x)/(64*a**5), Ne(a, 0)), (0, True))

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*arcsinh(a*x)^2/(a^2*x^2+1)^(1/2),x, algorithm="giac")

[Out]

integrate(x^4*arcsinh(a*x)^2/sqrt(a^2*x^2 + 1), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^4\,{\mathrm {asinh}\left (a\,x\right )}^2}{\sqrt {a^2\,x^2+1}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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