3.3.87 \(\int \frac {\sinh ^{-1}(a x^2)}{x} \, dx\) [287]

Optimal. Leaf size=54 \[ -\frac {1}{4} \sinh ^{-1}\left (a x^2\right )^2+\frac {1}{2} \sinh ^{-1}\left (a x^2\right ) \log \left (1-e^{2 \sinh ^{-1}\left (a x^2\right )}\right )+\frac {1}{4} \text {PolyLog}\left (2,e^{2 \sinh ^{-1}\left (a x^2\right )}\right ) \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 54, 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 = {5869, 3797, 2221, 2317, 2438} \begin {gather*} \frac {1}{4} \text {Li}_2\left (e^{2 \sinh ^{-1}\left (a x^2\right )}\right )-\frac {1}{4} \sinh ^{-1}\left (a x^2\right )^2+\frac {1}{2} \sinh ^{-1}\left (a x^2\right ) \log \left (1-e^{2 \sinh ^{-1}\left (a x^2\right )}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 5869

Int[ArcSinh[(a_.)*(x_)^(p_)]^(n_.)/(x_), x_Symbol] :> Dist[1/p, Subst[Int[x^n*Coth[x], x], x, ArcSinh[a*x^p]],
x] /; FreeQ[{a, p}, x] && IGtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {\sinh ^{-1}\left (a x^2\right )}{x} \, dx &=\frac {1}{2} \text {Subst}\left (\int x \coth (x) \, dx,x,\sinh ^{-1}\left (a x^2\right )\right )\\ &=-\frac {1}{4} \sinh ^{-1}\left (a x^2\right )^2-\text {Subst}\left (\int \frac {e^{2 x} x}{1-e^{2 x}} \, dx,x,\sinh ^{-1}\left (a x^2\right )\right )\\ &=-\frac {1}{4} \sinh ^{-1}\left (a x^2\right )^2+\frac {1}{2} \sinh ^{-1}\left (a x^2\right ) \log \left (1-e^{2 \sinh ^{-1}\left (a x^2\right )}\right )-\frac {1}{2} \text {Subst}\left (\int \log \left (1-e^{2 x}\right ) \, dx,x,\sinh ^{-1}\left (a x^2\right )\right )\\ &=-\frac {1}{4} \sinh ^{-1}\left (a x^2\right )^2+\frac {1}{2} \sinh ^{-1}\left (a x^2\right ) \log \left (1-e^{2 \sinh ^{-1}\left (a x^2\right )}\right )-\frac {1}{4} \text {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{2 \sinh ^{-1}\left (a x^2\right )}\right )\\ &=-\frac {1}{4} \sinh ^{-1}\left (a x^2\right )^2+\frac {1}{2} \sinh ^{-1}\left (a x^2\right ) \log \left (1-e^{2 \sinh ^{-1}\left (a x^2\right )}\right )+\frac {1}{4} \text {Li}_2\left (e^{2 \sinh ^{-1}\left (a x^2\right )}\right )\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 54, normalized size = 1.00 \begin {gather*} -\frac {1}{4} \sinh ^{-1}\left (a x^2\right )^2+\frac {1}{2} \sinh ^{-1}\left (a x^2\right ) \log \left (1-e^{2 \sinh ^{-1}\left (a x^2\right )}\right )+\frac {1}{4} \text {PolyLog}\left (2,e^{2 \sinh ^{-1}\left (a x^2\right )}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.41, size = 0, normalized size = 0.00 \[\int \frac {\arcsinh \left (a \,x^{2}\right )}{x}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arcsinh(a*x^2)/x,x)

[Out]

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

[Out]

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

[Out]

integral(arcsinh(a*x^2)/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(asinh(a*x**2)/x, x)

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

[Out]

integrate(arcsinh(a*x^2)/x, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {\mathrm {asinh}\left (a\,x^2\right )}{x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(asinh(a*x^2)/x, x)

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