3.1.21 \(\int \frac {\sinh ^{-1}(a x)^2}{x^5} \, dx\) [21]

Optimal. Leaf size=85 \[ -\frac {a^2}{12 x^2}-\frac {a \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)}{6 x^3}+\frac {a^3 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)}{3 x}-\frac {\sinh ^{-1}(a x)^2}{4 x^4}-\frac {1}{3} a^4 \log (x) \]

[Out]

-1/12*a^2/x^2-1/4*arcsinh(a*x)^2/x^4-1/3*a^4*ln(x)-1/6*a*arcsinh(a*x)*(a^2*x^2+1)^(1/2)/x^3+1/3*a^3*arcsinh(a*
x)*(a^2*x^2+1)^(1/2)/x

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Rubi [A]
time = 0.10, antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5776, 5809, 5800, 29, 30} \begin {gather*} -\frac {1}{3} a^4 \log (x)-\frac {a^2}{12 x^2}-\frac {a \sqrt {a^2 x^2+1} \sinh ^{-1}(a x)}{6 x^3}+\frac {a^3 \sqrt {a^2 x^2+1} \sinh ^{-1}(a x)}{3 x}-\frac {\sinh ^{-1}(a x)^2}{4 x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[ArcSinh[a*x]^2/x^5,x]

[Out]

-1/12*a^2/x^2 - (a*Sqrt[1 + a^2*x^2]*ArcSinh[a*x])/(6*x^3) + (a^3*Sqrt[1 + a^2*x^2]*ArcSinh[a*x])/(3*x) - ArcS
inh[a*x]^2/(4*x^4) - (a^4*Log[x])/3

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 30

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

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 5800

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Simp[
(f*x)^(m + 1)*(d + e*x^2)^(p + 1)*((a + b*ArcSinh[c*x])^n/(d*f*(m + 1))), x] - Dist[b*c*(n/(f*(m + 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, m, p}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && EqQ[m + 2*p + 3, 0] && NeQ[m, -1]

Rule 5809

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Simp[
(f*x)^(m + 1)*(d + e*x^2)^(p + 1)*((a + b*ArcSinh[c*x])^n/(d*f*(m + 1))), x] + (-Dist[c^2*((m + 2*p + 3)/(f^2*
(m + 1))), Int[(f*x)^(m + 2)*(d + e*x^2)^p*(a + b*ArcSinh[c*x])^n, x], x] - Dist[b*c*(n/(f*(m + 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] && ILtQ[m, -1]

Rubi steps

\begin {align*} \int \frac {\sinh ^{-1}(a x)^2}{x^5} \, dx &=-\frac {\sinh ^{-1}(a x)^2}{4 x^4}+\frac {1}{2} a \int \frac {\sinh ^{-1}(a x)}{x^4 \sqrt {1+a^2 x^2}} \, dx\\ &=-\frac {a \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)}{6 x^3}-\frac {\sinh ^{-1}(a x)^2}{4 x^4}+\frac {1}{6} a^2 \int \frac {1}{x^3} \, dx-\frac {1}{3} a^3 \int \frac {\sinh ^{-1}(a x)}{x^2 \sqrt {1+a^2 x^2}} \, dx\\ &=-\frac {a^2}{12 x^2}-\frac {a \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)}{6 x^3}+\frac {a^3 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)}{3 x}-\frac {\sinh ^{-1}(a x)^2}{4 x^4}-\frac {1}{3} a^4 \int \frac {1}{x} \, dx\\ &=-\frac {a^2}{12 x^2}-\frac {a \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)}{6 x^3}+\frac {a^3 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)}{3 x}-\frac {\sinh ^{-1}(a x)^2}{4 x^4}-\frac {1}{3} a^4 \log (x)\\ \end {align*}

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Mathematica [A]
time = 0.04, size = 64, normalized size = 0.75 \begin {gather*} -\frac {a^2 x^2-2 a x \sqrt {1+a^2 x^2} \left (-1+2 a^2 x^2\right ) \sinh ^{-1}(a x)+3 \sinh ^{-1}(a x)^2+4 a^4 x^4 \log (x)}{12 x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[ArcSinh[a*x]^2/x^5,x]

[Out]

-1/12*(a^2*x^2 - 2*a*x*Sqrt[1 + a^2*x^2]*(-1 + 2*a^2*x^2)*ArcSinh[a*x] + 3*ArcSinh[a*x]^2 + 4*a^4*x^4*Log[x])/
x^4

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Maple [A]
time = 2.64, size = 112, normalized size = 1.32

method	result	size
derivativedivides	\(a^{4} \left (\frac {2 \arcsinh \left (a x \right )}{3}-\frac {-4 \arcsinh \left (a x \right ) \sqrt {a^{2} x^{2}+1}\, a^{3} x^{3}+4 a^{4} x^{4} \arcsinh \left (a x \right )+2 a x \arcsinh \left (a x \right ) \sqrt {a^{2} x^{2}+1}+3 \arcsinh \left (a x \right )^{2}+a^{2} x^{2}}{12 a^{4} x^{4}}-\frac {\ln \left (\left (a x +\sqrt {a^{2} x^{2}+1}\right )^{2}-1\right )}{3}\right )\)	\(112\)
default	\(a^{4} \left (\frac {2 \arcsinh \left (a x \right )}{3}-\frac {-4 \arcsinh \left (a x \right ) \sqrt {a^{2} x^{2}+1}\, a^{3} x^{3}+4 a^{4} x^{4} \arcsinh \left (a x \right )+2 a x \arcsinh \left (a x \right ) \sqrt {a^{2} x^{2}+1}+3 \arcsinh \left (a x \right )^{2}+a^{2} x^{2}}{12 a^{4} x^{4}}-\frac {\ln \left (\left (a x +\sqrt {a^{2} x^{2}+1}\right )^{2}-1\right )}{3}\right )\)	\(112\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arcsinh(a*x)^2/x^5,x,method=_RETURNVERBOSE)

[Out]

a^4*(2/3*arcsinh(a*x)-1/12*(-4*arcsinh(a*x)*(a^2*x^2+1)^(1/2)*a^3*x^3+4*a^4*x^4*arcsinh(a*x)+2*a*x*arcsinh(a*x
)*(a^2*x^2+1)^(1/2)+3*arcsinh(a*x)^2+a^2*x^2)/a^4/x^4-1/3*ln((a*x+(a^2*x^2+1)^(1/2))^2-1))

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Maxima [A]
time = 0.26, size = 71, normalized size = 0.84 \begin {gather*} -\frac {1}{12} \, {\left (4 \, a^{2} \log \left (x\right ) + \frac {1}{x^{2}}\right )} a^{2} + \frac {1}{6} \, {\left (\frac {2 \, \sqrt {a^{2} x^{2} + 1} a^{2}}{x} - \frac {\sqrt {a^{2} x^{2} + 1}}{x^{3}}\right )} a \operatorname {arsinh}\left (a x\right ) - \frac {\operatorname {arsinh}\left (a x\right )^{2}}{4 \, x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsinh(a*x)^2/x^5,x, algorithm="maxima")

[Out]

-1/12*(4*a^2*log(x) + 1/x^2)*a^2 + 1/6*(2*sqrt(a^2*x^2 + 1)*a^2/x - sqrt(a^2*x^2 + 1)/x^3)*a*arcsinh(a*x) - 1/
4*arcsinh(a*x)^2/x^4

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Fricas [A]
time = 0.35, size = 85, normalized size = 1.00 \begin {gather*} -\frac {4 \, a^{4} x^{4} \log \left (x\right ) + a^{2} x^{2} - 2 \, {\left (2 \, a^{3} x^{3} - a x\right )} \sqrt {a^{2} x^{2} + 1} \log \left (a x + \sqrt {a^{2} x^{2} + 1}\right ) + 3 \, \log \left (a x + \sqrt {a^{2} x^{2} + 1}\right )^{2}}{12 \, x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsinh(a*x)^2/x^5,x, algorithm="fricas")

[Out]

-1/12*(4*a^4*x^4*log(x) + a^2*x^2 - 2*(2*a^3*x^3 - a*x)*sqrt(a^2*x^2 + 1)*log(a*x + sqrt(a^2*x^2 + 1)) + 3*log
(a*x + sqrt(a^2*x^2 + 1))^2)/x^4

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\operatorname {asinh}^{2}{\left (a x \right )}}{x^{5}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(asinh(a*x)**2/x**5,x)

[Out]

Integral(asinh(a*x)**2/x**5, x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 148 vs. \(2 (71) = 142\).
time = 0.46, size = 148, normalized size = 1.74 \begin {gather*} -\frac {1}{12} \, {\left (2 \, a^{3} \log \left (x^{2}\right ) - 4 \, a^{3} \log \left (-x {\left | a \right |} + \sqrt {a^{2} x^{2} + 1}\right ) - \frac {8 \, {\left (3 \, {\left (x {\left | a \right |} - \sqrt {a^{2} x^{2} + 1}\right )}^{2} - 1\right )} a^{2} {\left | a \right |} \log \left (a x + \sqrt {a^{2} x^{2} + 1}\right )}{{\left ({\left (x {\left | a \right |} - \sqrt {a^{2} x^{2} + 1}\right )}^{2} - 1\right )}^{3}} - \frac {2 \, a^{3} x^{2} - a}{x^{2}}\right )} a - \frac {\log \left (a x + \sqrt {a^{2} x^{2} + 1}\right )^{2}}{4 \, x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsinh(a*x)^2/x^5,x, algorithm="giac")

[Out]

-1/12*(2*a^3*log(x^2) - 4*a^3*log(-x*abs(a) + sqrt(a^2*x^2 + 1)) - 8*(3*(x*abs(a) - sqrt(a^2*x^2 + 1))^2 - 1)*
a^2*abs(a)*log(a*x + sqrt(a^2*x^2 + 1))/((x*abs(a) - sqrt(a^2*x^2 + 1))^2 - 1)^3 - (2*a^3*x^2 - a)/x^2)*a - 1/
4*log(a*x + sqrt(a^2*x^2 + 1))^2/x^4

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\mathrm {asinh}\left (a\,x\right )}^2}{x^5} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(asinh(a*x)^2/x^5,x)

[Out]

int(asinh(a*x)^2/x^5, x)

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