3.96 \(\int \frac {a+b \sinh ^{-1}(c x)}{x (\pi +c^2 \pi x^2)^{3/2}} \, dx\)

Optimal. Leaf size=94 \[ \frac {a+b \sinh ^{-1}(c x)}{\pi \sqrt {\pi c^2 x^2+\pi }}-\frac {2 \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{\pi ^{3/2}}-\frac {b \text {Li}_2\left (-e^{\sinh ^{-1}(c x)}\right )}{\pi ^{3/2}}+\frac {b \text {Li}_2\left (e^{\sinh ^{-1}(c x)}\right )}{\pi ^{3/2}}-\frac {b \tan ^{-1}(c x)}{\pi ^{3/2}} \]

[Out]

-b*arctan(c*x)/Pi^(3/2)-2*(a+b*arcsinh(c*x))*arctanh(c*x+(c^2*x^2+1)^(1/2))/Pi^(3/2)-b*polylog(2,-c*x-(c^2*x^2
+1)^(1/2))/Pi^(3/2)+b*polylog(2,c*x+(c^2*x^2+1)^(1/2))/Pi^(3/2)+(a+b*arcsinh(c*x))/Pi/(Pi*c^2*x^2+Pi)^(1/2)

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Rubi [A]  time = 0.22, antiderivative size = 119, normalized size of antiderivative = 1.27, number of steps used = 8, number of rules used = 6, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {5755, 5760, 4182, 2279, 2391, 203} \[ -\frac {b \text {PolyLog}\left (2,-e^{\sinh ^{-1}(c x)}\right )}{\pi ^{3/2}}+\frac {b \text {PolyLog}\left (2,e^{\sinh ^{-1}(c x)}\right )}{\pi ^{3/2}}+\frac {a+b \sinh ^{-1}(c x)}{\pi \sqrt {\pi c^2 x^2+\pi }}-\frac {2 \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{\pi ^{3/2}}-\frac {b \sqrt {c^2 x^2+1} \tan ^{-1}(c x)}{\pi \sqrt {\pi c^2 x^2+\pi }} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSinh[c*x])/(x*(Pi + c^2*Pi*x^2)^(3/2)),x]

[Out]

(a + b*ArcSinh[c*x])/(Pi*Sqrt[Pi + c^2*Pi*x^2]) - (b*Sqrt[1 + c^2*x^2]*ArcTan[c*x])/(Pi*Sqrt[Pi + c^2*Pi*x^2])
- (2*(a + b*ArcSinh[c*x])*ArcTanh[E^ArcSinh[c*x]])/Pi^(3/2) - (b*PolyLog[2, -E^ArcSinh[c*x]])/Pi^(3/2) + (b*P
olyLog[2, E^ArcSinh[c*x]])/Pi^(3/2)

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

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 5755

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)/(2*d*f*(p + 1)), x] + (Dist[(m + 2*p + 3)/(2*d*(p
+ 1)), Int[(f*x)^m*(d + e*x^2)^(p + 1)*(a + b*ArcSinh[c*x])^n, x], x] + Dist[(b*c*n*d^IntPart[p]*(d + e*x^2)^F
racPart[p])/(2*f*(p + 1)*(1 + c^2*x^2)^FracPart[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}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && LtQ[p, -1] &&  !GtQ
[m, 1] && (IntegerQ[m] || IntegerQ[p] || EqQ[n, 1])

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 \left (\pi +c^2 \pi x^2\right )^{3/2}} \, dx &=\frac {a+b \sinh ^{-1}(c x)}{\pi \sqrt {\pi +c^2 \pi x^2}}+\frac {\int \frac {a+b \sinh ^{-1}(c x)}{x \sqrt {\pi +c^2 \pi x^2}} \, dx}{\pi }-\frac {\left (b c \sqrt {1+c^2 x^2}\right ) \int \frac {1}{1+c^2 x^2} \, dx}{\pi \sqrt {\pi +c^2 \pi x^2}}\\ &=\frac {a+b \sinh ^{-1}(c x)}{\pi \sqrt {\pi +c^2 \pi x^2}}-\frac {b \sqrt {1+c^2 x^2} \tan ^{-1}(c x)}{\pi \sqrt {\pi +c^2 \pi x^2}}+\frac {\operatorname {Subst}\left (\int (a+b x) \text {csch}(x) \, dx,x,\sinh ^{-1}(c x)\right )}{\pi ^{3/2}}\\ &=\frac {a+b \sinh ^{-1}(c x)}{\pi \sqrt {\pi +c^2 \pi x^2}}-\frac {b \sqrt {1+c^2 x^2} \tan ^{-1}(c x)}{\pi \sqrt {\pi +c^2 \pi x^2}}-\frac {2 \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{\pi ^{3/2}}-\frac {b \operatorname {Subst}\left (\int \log \left (1-e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{\pi ^{3/2}}+\frac {b \operatorname {Subst}\left (\int \log \left (1+e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{\pi ^{3/2}}\\ &=\frac {a+b \sinh ^{-1}(c x)}{\pi \sqrt {\pi +c^2 \pi x^2}}-\frac {b \sqrt {1+c^2 x^2} \tan ^{-1}(c x)}{\pi \sqrt {\pi +c^2 \pi x^2}}-\frac {2 \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{\pi ^{3/2}}-\frac {b \operatorname {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{\pi ^{3/2}}+\frac {b \operatorname {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{\pi ^{3/2}}\\ &=\frac {a+b \sinh ^{-1}(c x)}{\pi \sqrt {\pi +c^2 \pi x^2}}-\frac {b \sqrt {1+c^2 x^2} \tan ^{-1}(c x)}{\pi \sqrt {\pi +c^2 \pi x^2}}-\frac {2 \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{\pi ^{3/2}}-\frac {b \text {Li}_2\left (-e^{\sinh ^{-1}(c x)}\right )}{\pi ^{3/2}}+\frac {b \text {Li}_2\left (e^{\sinh ^{-1}(c x)}\right )}{\pi ^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.32, size = 143, normalized size = 1.52 \[ \frac {\frac {a}{\sqrt {c^2 x^2+1}}-a \log \left (\pi \left (\sqrt {c^2 x^2+1}+1\right )\right )+a \log (x)+\frac {b \sinh ^{-1}(c x)}{\sqrt {c^2 x^2+1}}+b \text {Li}_2\left (-e^{-\sinh ^{-1}(c x)}\right )-b \text {Li}_2\left (e^{-\sinh ^{-1}(c x)}\right )+b \sinh ^{-1}(c x) \log \left (1-e^{-\sinh ^{-1}(c x)}\right )-b \sinh ^{-1}(c x) \log \left (e^{-\sinh ^{-1}(c x)}+1\right )-2 b \tan ^{-1}\left (\tanh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )}{\pi ^{3/2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*ArcSinh[c*x])/(x*(Pi + c^2*Pi*x^2)^(3/2)),x]

[Out]

(a/Sqrt[1 + c^2*x^2] + (b*ArcSinh[c*x])/Sqrt[1 + c^2*x^2] - 2*b*ArcTan[Tanh[ArcSinh[c*x]/2]] + 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])])/Pi^(3/2)

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fricas [F]  time = 0.62, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {\pi + \pi c^{2} x^{2}} {\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}}{\pi ^{2} c^{4} x^{5} + 2 \, \pi ^{2} c^{2} x^{3} + \pi ^{2} x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(c*x))/x/(pi*c^2*x^2+pi)^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(pi + pi*c^2*x^2)*(b*arcsinh(c*x) + a)/(pi^2*c^4*x^5 + 2*pi^2*c^2*x^3 + pi^2*x), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {b \operatorname {arsinh}\left (c x\right ) + a}{{\left (\pi + \pi c^{2} x^{2}\right )}^{\frac {3}{2}} x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(c*x))/x/(pi*c^2*x^2+pi)^(3/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.23, size = 156, normalized size = 1.66 \[ \frac {a}{\pi \sqrt {\pi \,c^{2} x^{2}+\pi }}-\frac {a \arctanh \left (\frac {\sqrt {\pi }}{\sqrt {\pi \,c^{2} x^{2}+\pi }}\right )}{\pi ^{\frac {3}{2}}}+\frac {b \arcsinh \left (c x \right )}{\pi ^{\frac {3}{2}} \sqrt {c^{2} x^{2}+1}}-\frac {2 b \arctan \left (c x +\sqrt {c^{2} x^{2}+1}\right )}{\pi ^{\frac {3}{2}}}-\frac {b \dilog \left (c x +\sqrt {c^{2} x^{2}+1}\right )}{\pi ^{\frac {3}{2}}}-\frac {b \dilog \left (1+c x +\sqrt {c^{2} x^{2}+1}\right )}{\pi ^{\frac {3}{2}}}-\frac {b \arcsinh \left (c x \right ) \ln \left (1+c x +\sqrt {c^{2} x^{2}+1}\right )}{\pi ^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsinh(c*x))/x/(Pi*c^2*x^2+Pi)^(3/2),x)

[Out]

a/Pi/(Pi*c^2*x^2+Pi)^(1/2)-a/Pi^(3/2)*arctanh(Pi^(1/2)/(Pi*c^2*x^2+Pi)^(1/2))+b/Pi^(3/2)/(c^2*x^2+1)^(1/2)*arc
sinh(c*x)-2*b/Pi^(3/2)*arctan(c*x+(c^2*x^2+1)^(1/2))-b/Pi^(3/2)*dilog(c*x+(c^2*x^2+1)^(1/2))-b/Pi^(3/2)*dilog(
1+c*x+(c^2*x^2+1)^(1/2))-b/Pi^(3/2)*arcsinh(c*x)*ln(1+c*x+(c^2*x^2+1)^(1/2))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -a {\left (\frac {\operatorname {arsinh}\left (\frac {1}{c {\left | x \right |}}\right )}{\pi ^{\frac {3}{2}}} - \frac {1}{\pi \sqrt {\pi + \pi c^{2} x^{2}}}\right )} + b \int \frac {\log \left (c x + \sqrt {c^{2} x^{2} + 1}\right )}{{\left (\pi + \pi c^{2} x^{2}\right )}^{\frac {3}{2}} x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(c*x))/x/(pi*c^2*x^2+pi)^(3/2),x, algorithm="maxima")

[Out]

-a*(arcsinh(1/(c*abs(x)))/pi^(3/2) - 1/(pi*sqrt(pi + pi*c^2*x^2))) + b*integrate(log(c*x + sqrt(c^2*x^2 + 1))/
((pi + pi*c^2*x^2)^(3/2)*x), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {a+b\,\mathrm {asinh}\left (c\,x\right )}{x\,{\left (\Pi \,c^2\,x^2+\Pi \right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asinh(c*x))/(x*(Pi + Pi*c^2*x^2)^(3/2)),x)

[Out]

int((a + b*asinh(c*x))/(x*(Pi + Pi*c^2*x^2)^(3/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {a}{c^{2} x^{3} \sqrt {c^{2} x^{2} + 1} + x \sqrt {c^{2} x^{2} + 1}}\, dx + \int \frac {b \operatorname {asinh}{\left (c x \right )}}{c^{2} x^{3} \sqrt {c^{2} x^{2} + 1} + x \sqrt {c^{2} x^{2} + 1}}\, dx}{\pi ^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asinh(c*x))/x/(pi*c**2*x**2+pi)**(3/2),x)

[Out]

(Integral(a/(c**2*x**3*sqrt(c**2*x**2 + 1) + x*sqrt(c**2*x**2 + 1)), x) + Integral(b*asinh(c*x)/(c**2*x**3*sqr
t(c**2*x**2 + 1) + x*sqrt(c**2*x**2 + 1)), x))/pi**(3/2)

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