3.1.31 \(\int \frac {\sinh ^{-1}(a x)^3}{x^5} \, dx\) [31]

Optimal. Leaf size=159 \[ -\frac {a^3 \sqrt {1+a^2 x^2}}{4 x}-\frac {a^2 \sinh ^{-1}(a x)}{4 x^2}+\frac {1}{2} a^4 \sinh ^{-1}(a x)^2-\frac {a \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^2}{4 x^3}+\frac {a^3 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^2}{2 x}-\frac {\sinh ^{-1}(a x)^3}{4 x^4}-a^4 \sinh ^{-1}(a x) \log \left (1-e^{2 \sinh ^{-1}(a x)}\right )-\frac {1}{2} a^4 \text {PolyLog}\left (2,e^{2 \sinh ^{-1}(a x)}\right ) \]

[Out]

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

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Rubi [A]
time = 0.19, antiderivative size = 159, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.900, Rules used = {5776, 5809, 5800, 5775, 3797, 2221, 2317, 2438, 270} \begin {gather*} -\frac {1}{2} a^4 \text {Li}_2\left (e^{2 \sinh ^{-1}(a x)}\right )+\frac {1}{2} a^4 \sinh ^{-1}(a x)^2-a^4 \sinh ^{-1}(a x) \log \left (1-e^{2 \sinh ^{-1}(a x)}\right )-\frac {a^2 \sinh ^{-1}(a x)}{4 x^2}-\frac {a \sqrt {a^2 x^2+1} \sinh ^{-1}(a x)^2}{4 x^3}-\frac {a^3 \sqrt {a^2 x^2+1}}{4 x}+\frac {a^3 \sqrt {a^2 x^2+1} \sinh ^{-1}(a x)^2}{2 x}-\frac {\sinh ^{-1}(a x)^3}{4 x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[ArcSinh[a*x]^3/x^5,x]

[Out]

-1/4*(a^3*Sqrt[1 + a^2*x^2])/x - (a^2*ArcSinh[a*x])/(4*x^2) + (a^4*ArcSinh[a*x]^2)/2 - (a*Sqrt[1 + a^2*x^2]*Ar
cSinh[a*x]^2)/(4*x^3) + (a^3*Sqrt[1 + a^2*x^2]*ArcSinh[a*x]^2)/(2*x) - ArcSinh[a*x]^3/(4*x^4) - a^4*ArcSinh[a*
x]*Log[1 - E^(2*ArcSinh[a*x])] - (a^4*PolyLog[2, E^(2*ArcSinh[a*x])])/2

Rule 270

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

Rule 2221

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x]
- Dist[d*(m/(b*f*g*n*Log[F])), Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2317

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2438

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2, (-c)*e*x^n]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 3797

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (Complex[0, fz_])*(f_.)*(x_)], x_Symbol] :> Simp[(-I)*((
c + d*x)^(m + 1)/(d*(m + 1))), x] + Dist[2*I, Int[((c + d*x)^m*(E^(2*((-I)*e + f*fz*x))/(1 + E^(2*((-I)*e + f*
fz*x))/E^(2*I*k*Pi))))/E^(2*I*k*Pi), x], x] /; FreeQ[{c, d, e, f, fz}, x] && IntegerQ[4*k] && IGtQ[m, 0]

Rule 5775

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/(x_), x_Symbol] :> Dist[1/b, Subst[Int[x^n*Coth[-a/b + x/b], x],
x, a + b*ArcSinh[c*x]], x] /; FreeQ[{a, b, c}, x] && IGtQ[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 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)^3}{x^5} \, dx &=-\frac {\sinh ^{-1}(a x)^3}{4 x^4}+\frac {1}{4} (3 a) \int \frac {\sinh ^{-1}(a x)^2}{x^4 \sqrt {1+a^2 x^2}} \, dx\\ &=-\frac {a \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^2}{4 x^3}-\frac {\sinh ^{-1}(a x)^3}{4 x^4}+\frac {1}{2} a^2 \int \frac {\sinh ^{-1}(a x)}{x^3} \, dx-\frac {1}{2} a^3 \int \frac {\sinh ^{-1}(a x)^2}{x^2 \sqrt {1+a^2 x^2}} \, dx\\ &=-\frac {a^2 \sinh ^{-1}(a x)}{4 x^2}-\frac {a \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^2}{4 x^3}+\frac {a^3 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^2}{2 x}-\frac {\sinh ^{-1}(a x)^3}{4 x^4}+\frac {1}{4} a^3 \int \frac {1}{x^2 \sqrt {1+a^2 x^2}} \, dx-a^4 \int \frac {\sinh ^{-1}(a x)}{x} \, dx\\ &=-\frac {a^3 \sqrt {1+a^2 x^2}}{4 x}-\frac {a^2 \sinh ^{-1}(a x)}{4 x^2}-\frac {a \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^2}{4 x^3}+\frac {a^3 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^2}{2 x}-\frac {\sinh ^{-1}(a x)^3}{4 x^4}-a^4 \text {Subst}\left (\int x \coth (x) \, dx,x,\sinh ^{-1}(a x)\right )\\ &=-\frac {a^3 \sqrt {1+a^2 x^2}}{4 x}-\frac {a^2 \sinh ^{-1}(a x)}{4 x^2}+\frac {1}{2} a^4 \sinh ^{-1}(a x)^2-\frac {a \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^2}{4 x^3}+\frac {a^3 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^2}{2 x}-\frac {\sinh ^{-1}(a x)^3}{4 x^4}+\left (2 a^4\right ) \text {Subst}\left (\int \frac {e^{2 x} x}{1-e^{2 x}} \, dx,x,\sinh ^{-1}(a x)\right )\\ &=-\frac {a^3 \sqrt {1+a^2 x^2}}{4 x}-\frac {a^2 \sinh ^{-1}(a x)}{4 x^2}+\frac {1}{2} a^4 \sinh ^{-1}(a x)^2-\frac {a \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^2}{4 x^3}+\frac {a^3 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^2}{2 x}-\frac {\sinh ^{-1}(a x)^3}{4 x^4}-a^4 \sinh ^{-1}(a x) \log \left (1-e^{2 \sinh ^{-1}(a x)}\right )+a^4 \text {Subst}\left (\int \log \left (1-e^{2 x}\right ) \, dx,x,\sinh ^{-1}(a x)\right )\\ &=-\frac {a^3 \sqrt {1+a^2 x^2}}{4 x}-\frac {a^2 \sinh ^{-1}(a x)}{4 x^2}+\frac {1}{2} a^4 \sinh ^{-1}(a x)^2-\frac {a \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^2}{4 x^3}+\frac {a^3 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^2}{2 x}-\frac {\sinh ^{-1}(a x)^3}{4 x^4}-a^4 \sinh ^{-1}(a x) \log \left (1-e^{2 \sinh ^{-1}(a x)}\right )+\frac {1}{2} a^4 \text {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{2 \sinh ^{-1}(a x)}\right )\\ &=-\frac {a^3 \sqrt {1+a^2 x^2}}{4 x}-\frac {a^2 \sinh ^{-1}(a x)}{4 x^2}+\frac {1}{2} a^4 \sinh ^{-1}(a x)^2-\frac {a \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^2}{4 x^3}+\frac {a^3 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^2}{2 x}-\frac {\sinh ^{-1}(a x)^3}{4 x^4}-a^4 \sinh ^{-1}(a x) \log \left (1-e^{2 \sinh ^{-1}(a x)}\right )-\frac {1}{2} a^4 \text {Li}_2\left (e^{2 \sinh ^{-1}(a x)}\right )\\ \end {align*}

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Mathematica [A]
time = 0.44, size = 107, normalized size = 0.67 \begin {gather*} \frac {1}{4} \left (-\frac {\sinh ^{-1}(a x)^3}{x^4}+a^4 \left (-\frac {\sqrt {1+a^2 x^2} \left (1+\left (-2+\frac {1}{a^2 x^2}\right ) \sinh ^{-1}(a x)^2\right )}{a x}-\sinh ^{-1}(a x) \left (\frac {1}{a^2 x^2}+2 \sinh ^{-1}(a x)+4 \log \left (1-e^{-2 \sinh ^{-1}(a x)}\right )\right )+2 \text {PolyLog}\left (2,e^{-2 \sinh ^{-1}(a x)}\right )\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[ArcSinh[a*x]^3/x^5,x]

[Out]

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

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Maple [A]
time = 2.79, size = 213, normalized size = 1.34

method	result	size
derivativedivides	\(a^{4} \left (-\frac {-2 \arcsinh \left (a x \right )^{2} \sqrt {a^{2} x^{2}+1}\, a^{3} x^{3}+2 \arcsinh \left (a x \right )^{2} a^{4} x^{4}+a x \arcsinh \left (a x \right )^{2} \sqrt {a^{2} x^{2}+1}+a^{3} x^{3} \sqrt {a^{2} x^{2}+1}-a^{4} x^{4}+\arcsinh \left (a x \right )^{3}+a^{2} x^{2} \arcsinh \left (a x \right )}{4 a^{4} x^{4}}+\arcsinh \left (a x \right )^{2}-\arcsinh \left (a x \right ) \ln \left (1+a x +\sqrt {a^{2} x^{2}+1}\right )-\polylog \left (2, -a x -\sqrt {a^{2} x^{2}+1}\right )-\arcsinh \left (a x \right ) \ln \left (1-a x -\sqrt {a^{2} x^{2}+1}\right )-\polylog \left (2, a x +\sqrt {a^{2} x^{2}+1}\right )\right )\)	\(213\)
default	\(a^{4} \left (-\frac {-2 \arcsinh \left (a x \right )^{2} \sqrt {a^{2} x^{2}+1}\, a^{3} x^{3}+2 \arcsinh \left (a x \right )^{2} a^{4} x^{4}+a x \arcsinh \left (a x \right )^{2} \sqrt {a^{2} x^{2}+1}+a^{3} x^{3} \sqrt {a^{2} x^{2}+1}-a^{4} x^{4}+\arcsinh \left (a x \right )^{3}+a^{2} x^{2} \arcsinh \left (a x \right )}{4 a^{4} x^{4}}+\arcsinh \left (a x \right )^{2}-\arcsinh \left (a x \right ) \ln \left (1+a x +\sqrt {a^{2} x^{2}+1}\right )-\polylog \left (2, -a x -\sqrt {a^{2} x^{2}+1}\right )-\arcsinh \left (a x \right ) \ln \left (1-a x -\sqrt {a^{2} x^{2}+1}\right )-\polylog \left (2, a x +\sqrt {a^{2} x^{2}+1}\right )\right )\)	\(213\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arcsinh(a*x)^3/x^5,x,method=_RETURNVERBOSE)

[Out]

a^4*(-1/4*(-2*arcsinh(a*x)^2*(a^2*x^2+1)^(1/2)*a^3*x^3+2*arcsinh(a*x)^2*a^4*x^4+a*x*arcsinh(a*x)^2*(a^2*x^2+1)
^(1/2)+a^3*x^3*(a^2*x^2+1)^(1/2)-a^4*x^4+arcsinh(a*x)^3+a^2*x^2*arcsinh(a*x))/a^4/x^4+arcsinh(a*x)^2-arcsinh(a
*x)*ln(1+a*x+(a^2*x^2+1)^(1/2))-polylog(2,-a*x-(a^2*x^2+1)^(1/2))-arcsinh(a*x)*ln(1-a*x-(a^2*x^2+1)^(1/2))-pol
ylog(2,a*x+(a^2*x^2+1)^(1/2)))

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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(arcsinh(a*x)^3/x^5,x, algorithm="maxima")

[Out]

-1/4*log(a*x + sqrt(a^2*x^2 + 1))^3/x^4 + integrate(3/4*(a^3*x^2 + sqrt(a^2*x^2 + 1)*a^2*x + a)*log(a*x + sqrt
(a^2*x^2 + 1))^2/(a^3*x^7 + a*x^5 + (a^2*x^6 + x^4)*sqrt(a^2*x^2 + 1)), x)

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Fricas [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(arcsinh(a*x)^3/x^5,x, algorithm="fricas")

[Out]

integral(arcsinh(a*x)^3/x^5, x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\operatorname {asinh}^{3}{\left (a x \right )}}{x^{5}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(asinh(a*x)**3/x**5,x)

[Out]

Integral(asinh(a*x)**3/x**5, x)

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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(arcsinh(a*x)^3/x^5,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 {{\mathrm {asinh}\left (a\,x\right )}^3}{x^5} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(asinh(a*x)^3/x^5,x)

[Out]

int(asinh(a*x)^3/x^5, x)

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