3.1.56 \(\int \frac {x}{\sinh ^{-1}(a x)^2} \, dx\) [56]

Optimal. Leaf size=37 \[ -\frac {x \sqrt {1+a^2 x^2}}{a \sinh ^{-1}(a x)}+\frac {\text {Chi}\left (2 \sinh ^{-1}(a x)\right )}{a^2} \]

[Out]

Chi(2*arcsinh(a*x))/a^2-x*(a^2*x^2+1)^(1/2)/a/arcsinh(a*x)

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Rubi [A]
time = 0.02, antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {5778, 3382} \begin {gather*} \frac {\text {Chi}\left (2 \sinh ^{-1}(a x)\right )}{a^2}-\frac {x \sqrt {a^2 x^2+1}}{a \sinh ^{-1}(a x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((x*Sqrt[1 + a^2*x^2])/(a*ArcSinh[a*x])) + CoshIntegral[2*ArcSinh[a*x]]/a^2

Rule 3382

Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CoshIntegral[c*f*(fz/d)
+ f*fz*x]/d, x] /; FreeQ[{c, d, e, f, fz}, x] && EqQ[d*(e - Pi/2) - c*f*fz*I, 0]

Rule 5778

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Simp[x^m*Sqrt[1 + c^2*x^2]*((a + b*ArcSi
nh[c*x])^(n + 1)/(b*c*(n + 1))), x] - Dist[1/(b^2*c^(m + 1)*(n + 1)), Subst[Int[ExpandTrigReduce[x^(n + 1), Si
nh[-a/b + x/b]^(m - 1)*(m + (m + 1)*Sinh[-a/b + x/b]^2), x], x], x, a + b*ArcSinh[c*x]], x] /; FreeQ[{a, b, c}
, x] && IGtQ[m, 0] && GeQ[n, -2] && LtQ[n, -1]

Rubi steps

\begin {align*} \int \frac {x}{\sinh ^{-1}(a x)^2} \, dx &=-\frac {x \sqrt {1+a^2 x^2}}{a \sinh ^{-1}(a x)}+\frac {\text {Subst}\left (\int \frac {\cosh (2 x)}{x} \, dx,x,\sinh ^{-1}(a x)\right )}{a^2}\\ &=-\frac {x \sqrt {1+a^2 x^2}}{a \sinh ^{-1}(a x)}+\frac {\text {Chi}\left (2 \sinh ^{-1}(a x)\right )}{a^2}\\ \end {align*}

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Mathematica [A]
time = 0.00, size = 32, normalized size = 0.86 \begin {gather*} \frac {\text {Chi}\left (2 \sinh ^{-1}(a x)\right )}{a^2}-\frac {\sinh \left (2 \sinh ^{-1}(a x)\right )}{2 a^2 \sinh ^{-1}(a x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

CoshIntegral[2*ArcSinh[a*x]]/a^2 - Sinh[2*ArcSinh[a*x]]/(2*a^2*ArcSinh[a*x])

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Maple [A]
time = 1.57, size = 28, normalized size = 0.76

method	result	size
derivativedivides	\(\frac {-\frac {\sinh \left (2 \arcsinh \left (a x \right )\right )}{2 \arcsinh \left (a x \right )}+\hyperbolicCosineIntegral \left (2 \arcsinh \left (a x \right )\right )}{a^{2}}\)	\(28\)
default	\(\frac {-\frac {\sinh \left (2 \arcsinh \left (a x \right )\right )}{2 \arcsinh \left (a x \right )}+\hyperbolicCosineIntegral \left (2 \arcsinh \left (a x \right )\right )}{a^{2}}\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/a^2*(-1/2/arcsinh(a*x)*sinh(2*arcsinh(a*x))+Chi(2*arcsinh(a*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/arcsinh(a*x)^2,x, algorithm="maxima")

[Out]

-(a^3*x^4 + a*x^2 + (a^2*x^3 + x)*sqrt(a^2*x^2 + 1))/((a^3*x^2 + sqrt(a^2*x^2 + 1)*a^2*x + a)*log(a*x + sqrt(a
^2*x^2 + 1))) + integrate((2*a^5*x^5 + 2*(a^2*x^2 + 1)*a^3*x^3 + 4*a^3*x^3 + 2*a*x + (4*a^4*x^4 + 4*a^2*x^2 +
1)*sqrt(a^2*x^2 + 1))/((a^5*x^4 + (a^2*x^2 + 1)*a^3*x^2 + 2*a^3*x^2 + 2*(a^4*x^3 + a^2*x)*sqrt(a^2*x^2 + 1) +
a)*log(a*x + 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(x/arcsinh(a*x)^2,x, algorithm="fricas")

[Out]

integral(x/arcsinh(a*x)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

integrate(x/arcsinh(a*x)^2, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {x}{{\mathrm {asinh}\left (a\,x\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x/asinh(a*x)^2, x)

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