3.86 \(\int \frac {a+b \sinh ^{-1}(c x)}{x \sqrt {\pi +c^2 \pi x^2}} \, dx\)

Optimal. Leaf size=56 \[ -\frac {2 \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt {\pi }}-\frac {b \text {Li}_2\left (-e^{\sinh ^{-1}(c x)}\right )}{\sqrt {\pi }}+\frac {b \text {Li}_2\left (e^{\sinh ^{-1}(c x)}\right )}{\sqrt {\pi }} \]

[Out]

-2*(a+b*arcsinh(c*x))*arctanh(c*x+(c^2*x^2+1)^(1/2))/Pi^(1/2)-b*polylog(2,-c*x-(c^2*x^2+1)^(1/2))/Pi^(1/2)+b*p
olylog(2,c*x+(c^2*x^2+1)^(1/2))/Pi^(1/2)

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Rubi [A]  time = 0.12, antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {5760, 4182, 2279, 2391} \[ -\frac {b \text {PolyLog}\left (2,-e^{\sinh ^{-1}(c x)}\right )}{\sqrt {\pi }}+\frac {b \text {PolyLog}\left (2,e^{\sinh ^{-1}(c x)}\right )}{\sqrt {\pi }}-\frac {2 \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt {\pi }} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSinh[c*x])/(x*Sqrt[Pi + c^2*Pi*x^2]),x]

[Out]

(-2*(a + b*ArcSinh[c*x])*ArcTanh[E^ArcSinh[c*x]])/Sqrt[Pi] - (b*PolyLog[2, -E^ArcSinh[c*x]])/Sqrt[Pi] + (b*Pol
yLog[2, E^ArcSinh[c*x]])/Sqrt[Pi]

Rule 2279

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 2391

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 4182

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

Rule 5760

Int[(((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Dist[1/(c^(m
+ 1)*Sqrt[d]), Subst[Int[(a + b*x)^n*Sinh[x]^m, x], x, ArcSinh[c*x]], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[
e, c^2*d] && GtQ[d, 0] && IGtQ[n, 0] && IntegerQ[m]

Rubi steps

\begin {align*} \int \frac {a+b \sinh ^{-1}(c x)}{x \sqrt {\pi +c^2 \pi x^2}} \, dx &=\frac {\operatorname {Subst}\left (\int (a+b x) \text {csch}(x) \, dx,x,\sinh ^{-1}(c x)\right )}{\sqrt {\pi }}\\ &=-\frac {2 \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{\sqrt {\pi }}-\frac {b \operatorname {Subst}\left (\int \log \left (1-e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{\sqrt {\pi }}+\frac {b \operatorname {Subst}\left (\int \log \left (1+e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{\sqrt {\pi }}\\ &=-\frac {2 \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{\sqrt {\pi }}-\frac {b \operatorname {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{\sqrt {\pi }}+\frac {b \operatorname {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{\sqrt {\pi }}\\ &=-\frac {2 \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{\sqrt {\pi }}-\frac {b \text {Li}_2\left (-e^{\sinh ^{-1}(c x)}\right )}{\sqrt {\pi }}+\frac {b \text {Li}_2\left (e^{\sinh ^{-1}(c x)}\right )}{\sqrt {\pi }}\\ \end {align*}

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Mathematica [A]  time = 0.16, size = 96, normalized size = 1.71 \[ \frac {-a \log \left (\pi \left (\sqrt {c^2 x^2+1}+1\right )\right )+a \log (x)+b \text {Li}_2\left (-e^{-\sinh ^{-1}(c x)}\right )-b \text {Li}_2\left (e^{-\sinh ^{-1}(c x)}\right )+b \sinh ^{-1}(c x) \log \left (1-e^{-\sinh ^{-1}(c x)}\right )-b \sinh ^{-1}(c x) \log \left (e^{-\sinh ^{-1}(c x)}+1\right )}{\sqrt {\pi }} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*ArcSinh[c*x])/(x*Sqrt[Pi + c^2*Pi*x^2]),x]

[Out]

(b*ArcSinh[c*x]*Log[1 - E^(-ArcSinh[c*x])] - b*ArcSinh[c*x]*Log[1 + E^(-ArcSinh[c*x])] + a*Log[x] - a*Log[Pi*(
1 + Sqrt[1 + c^2*x^2])] + b*PolyLog[2, -E^(-ArcSinh[c*x])] - b*PolyLog[2, E^(-ArcSinh[c*x])])/Sqrt[Pi]

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fricas [F]  time = 0.49, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {\pi + \pi c^{2} x^{2}} {\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}}{\pi c^{2} x^{3} + \pi x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(pi + pi*c^2*x^2)*(b*arcsinh(c*x) + a)/(pi*c^2*x^3 + pi*x), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {b \operatorname {arsinh}\left (c x\right ) + a}{\sqrt {\pi + \pi c^{2} x^{2}} x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*arcsinh(c*x) + a)/(sqrt(pi + pi*c^2*x^2)*x), x)

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maple [A]  time = 0.04, size = 72, normalized size = 1.29 \[ -\frac {a \arctanh \left (\frac {\sqrt {\pi }}{\sqrt {\pi \,c^{2} x^{2}+\pi }}\right )}{\sqrt {\pi }}+\frac {b \left (4 \dilog \left (\frac {1}{c x +\sqrt {c^{2} x^{2}+1}}\right )-\dilog \left (\frac {1}{\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{2}}\right )\right )}{2 \sqrt {\pi }} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsinh(c*x))/x/(Pi*c^2*x^2+Pi)^(1/2),x)

[Out]

-a/Pi^(1/2)*arctanh(Pi^(1/2)/(Pi*c^2*x^2+Pi)^(1/2))+1/2*b*(4*dilog(1/(c*x+(c^2*x^2+1)^(1/2)))-dilog(1/(c*x+(c^
2*x^2+1)^(1/2))^2))/Pi^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ b \int \frac {\log \left (c x + \sqrt {c^{2} x^{2} + 1}\right )}{\sqrt {\pi + \pi c^{2} x^{2}} x}\,{d x} - \frac {a \operatorname {arsinh}\left (\frac {1}{c {\left | x \right |}}\right )}{\sqrt {\pi }} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*integrate(log(c*x + sqrt(c^2*x^2 + 1))/(sqrt(pi + pi*c^2*x^2)*x), x) - a*arcsinh(1/(c*abs(x)))/sqrt(pi)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {a+b\,\mathrm {asinh}\left (c\,x\right )}{x\,\sqrt {\Pi \,c^2\,x^2+\Pi }} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asinh(c*x))/(x*(Pi + Pi*c^2*x^2)^(1/2)),x)

[Out]

int((a + b*asinh(c*x))/(x*(Pi + Pi*c^2*x^2)^(1/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {a}{x \sqrt {c^{2} x^{2} + 1}}\, dx + \int \frac {b \operatorname {asinh}{\left (c x \right )}}{x \sqrt {c^{2} x^{2} + 1}}\, dx}{\sqrt {\pi }} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asinh(c*x))/x/(pi*c**2*x**2+pi)**(1/2),x)

[Out]

(Integral(a/(x*sqrt(c**2*x**2 + 1)), x) + Integral(b*asinh(c*x)/(x*sqrt(c**2*x**2 + 1)), x))/sqrt(pi)

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