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

Optimal. Leaf size=113 \[ -\frac {\sqrt {\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )}{2 x^2}-\sqrt {\pi } c^2 \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )-\frac {1}{2} \sqrt {\pi } b c^2 \text {Li}_2\left (-e^{\sinh ^{-1}(c x)}\right )+\frac {1}{2} \sqrt {\pi } b c^2 \text {Li}_2\left (e^{\sinh ^{-1}(c x)}\right )-\frac {\sqrt {\pi } b c}{2 x} \]

[Out]

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

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Rubi [A]  time = 0.20, antiderivative size = 201, normalized size of antiderivative = 1.78, 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 = {5737, 30, 5760, 4182, 2279, 2391} \[ -\frac {b c^2 \sqrt {\pi c^2 x^2+\pi } \text {PolyLog}\left (2,-e^{\sinh ^{-1}(c x)}\right )}{2 \sqrt {c^2 x^2+1}}+\frac {b c^2 \sqrt {\pi c^2 x^2+\pi } \text {PolyLog}\left (2,e^{\sinh ^{-1}(c x)}\right )}{2 \sqrt {c^2 x^2+1}}-\frac {\sqrt {\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )}{2 x^2}-\frac {c^2 \sqrt {\pi c^2 x^2+\pi } \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt {c^2 x^2+1}}-\frac {b c \sqrt {\pi c^2 x^2+\pi }}{2 x \sqrt {c^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

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 5737

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(
(f*x)^(m + 1)*Sqrt[d + e*x^2]*(a + b*ArcSinh[c*x])^n)/(f*(m + 1)), x] + (-Dist[(b*c*n*Sqrt[d + e*x^2])/(f*(m +
1)*Sqrt[1 + c^2*x^2]), Int[(f*x)^(m + 1)*(a + b*ArcSinh[c*x])^(n - 1), x], x] - Dist[(c^2*Sqrt[d + e*x^2])/(f
^2*(m + 1)*Sqrt[1 + c^2*x^2]), Int[((f*x)^(m + 2)*(a + b*ArcSinh[c*x])^n)/Sqrt[1 + c^2*x^2], x], x]) /; FreeQ[
{a, b, c, d, e, f}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && LtQ[m, -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 {\sqrt {\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )}{x^3} \, dx &=-\frac {\sqrt {\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )}{2 x^2}+\frac {\left (b c \sqrt {\pi +c^2 \pi x^2}\right ) \int \frac {1}{x^2} \, dx}{2 \sqrt {1+c^2 x^2}}+\frac {\left (c^2 \sqrt {\pi +c^2 \pi x^2}\right ) \int \frac {a+b \sinh ^{-1}(c x)}{x \sqrt {1+c^2 x^2}} \, dx}{2 \sqrt {1+c^2 x^2}}\\ &=-\frac {b c \sqrt {\pi +c^2 \pi x^2}}{2 x \sqrt {1+c^2 x^2}}-\frac {\sqrt {\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )}{2 x^2}+\frac {\left (c^2 \sqrt {\pi +c^2 \pi x^2}\right ) \operatorname {Subst}\left (\int (a+b x) \text {csch}(x) \, dx,x,\sinh ^{-1}(c x)\right )}{2 \sqrt {1+c^2 x^2}}\\ &=-\frac {b c \sqrt {\pi +c^2 \pi x^2}}{2 x \sqrt {1+c^2 x^2}}-\frac {\sqrt {\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )}{2 x^2}-\frac {c^2 \sqrt {\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{\sqrt {1+c^2 x^2}}-\frac {\left (b c^2 \sqrt {\pi +c^2 \pi x^2}\right ) \operatorname {Subst}\left (\int \log \left (1-e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{2 \sqrt {1+c^2 x^2}}+\frac {\left (b c^2 \sqrt {\pi +c^2 \pi x^2}\right ) \operatorname {Subst}\left (\int \log \left (1+e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{2 \sqrt {1+c^2 x^2}}\\ &=-\frac {b c \sqrt {\pi +c^2 \pi x^2}}{2 x \sqrt {1+c^2 x^2}}-\frac {\sqrt {\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )}{2 x^2}-\frac {c^2 \sqrt {\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{\sqrt {1+c^2 x^2}}-\frac {\left (b c^2 \sqrt {\pi +c^2 \pi x^2}\right ) \operatorname {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{2 \sqrt {1+c^2 x^2}}+\frac {\left (b c^2 \sqrt {\pi +c^2 \pi x^2}\right ) \operatorname {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{2 \sqrt {1+c^2 x^2}}\\ &=-\frac {b c \sqrt {\pi +c^2 \pi x^2}}{2 x \sqrt {1+c^2 x^2}}-\frac {\sqrt {\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )}{2 x^2}-\frac {c^2 \sqrt {\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{\sqrt {1+c^2 x^2}}-\frac {b c^2 \sqrt {\pi +c^2 \pi x^2} \text {Li}_2\left (-e^{\sinh ^{-1}(c x)}\right )}{2 \sqrt {1+c^2 x^2}}+\frac {b c^2 \sqrt {\pi +c^2 \pi x^2} \text {Li}_2\left (e^{\sinh ^{-1}(c x)}\right )}{2 \sqrt {1+c^2 x^2}}\\ \end {align*}

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Mathematica [A]  time = 3.32, size = 185, normalized size = 1.64 \[ \frac {1}{8} \sqrt {\pi } \left (-\frac {4 a \sqrt {c^2 x^2+1}}{x^2}-4 a c^2 \log \left (\pi \left (\sqrt {c^2 x^2+1}+1\right )\right )+4 a c^2 \log (x)+b c^2 \left (4 \text {Li}_2\left (-e^{-\sinh ^{-1}(c x)}\right )-4 \text {Li}_2\left (e^{-\sinh ^{-1}(c x)}\right )+4 \sinh ^{-1}(c x) \log \left (1-e^{-\sinh ^{-1}(c x)}\right )-4 \sinh ^{-1}(c x) \log \left (e^{-\sinh ^{-1}(c x)}+1\right )+2 \tanh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )-2 \coth \left (\frac {1}{2} \sinh ^{-1}(c x)\right )-\sinh ^{-1}(c x) \text {csch}^2\left (\frac {1}{2} \sinh ^{-1}(c x)\right )-\sinh ^{-1}(c x) \text {sech}^2\left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Sqrt[Pi]*((-4*a*Sqrt[1 + c^2*x^2])/x^2 + 4*a*c^2*Log[x] - 4*a*c^2*Log[Pi*(1 + Sqrt[1 + c^2*x^2])] + b*c^2*(-2
*Coth[ArcSinh[c*x]/2] - ArcSinh[c*x]*Csch[ArcSinh[c*x]/2]^2 + 4*ArcSinh[c*x]*Log[1 - E^(-ArcSinh[c*x])] - 4*Ar
cSinh[c*x]*Log[1 + E^(-ArcSinh[c*x])] + 4*PolyLog[2, -E^(-ArcSinh[c*x])] - 4*PolyLog[2, E^(-ArcSinh[c*x])] - A
rcSinh[c*x]*Sech[ArcSinh[c*x]/2]^2 + 2*Tanh[ArcSinh[c*x]/2])))/8

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fricas [F]  time = 0.66, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {\pi + \pi c^{2} x^{2}} {\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}}{x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(c*x))*(pi*c^2*x^2+pi)^(1/2)/x^3,x, algorithm="fricas")

[Out]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(c*x))*(pi*c^2*x^2+pi)^(1/2)/x^3,x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [A]  time = 0.30, size = 243, normalized size = 2.15 \[ -\frac {a \left (\pi \,c^{2} x^{2}+\pi \right )^{\frac {3}{2}}}{2 \pi \,x^{2}}-\frac {a \sqrt {\pi }\, \arctanh \left (\frac {\sqrt {\pi }}{\sqrt {\pi \,c^{2} x^{2}+\pi }}\right ) c^{2}}{2}+\frac {a \sqrt {\pi \,c^{2} x^{2}+\pi }\, c^{2}}{2}-\frac {b \sqrt {\pi }\, \arcsinh \left (c x \right ) c^{2}}{2 \sqrt {c^{2} x^{2}+1}}-\frac {b c \sqrt {\pi }}{2 x}-\frac {b \sqrt {\pi }\, \arcsinh \left (c x \right )}{2 \sqrt {c^{2} x^{2}+1}\, x^{2}}-\frac {b \,c^{2} \sqrt {\pi }\, \arcsinh \left (c x \right ) \ln \left (1+c x +\sqrt {c^{2} x^{2}+1}\right )}{2}-\frac {b \,c^{2} \polylog \left (2, -c x -\sqrt {c^{2} x^{2}+1}\right ) \sqrt {\pi }}{2}+\frac {b \,c^{2} \sqrt {\pi }\, \arcsinh \left (c x \right ) \ln \left (1-c x -\sqrt {c^{2} x^{2}+1}\right )}{2}+\frac {b \,c^{2} \polylog \left (2, c x +\sqrt {c^{2} x^{2}+1}\right ) \sqrt {\pi }}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsinh(c*x))*(Pi*c^2*x^2+Pi)^(1/2)/x^3,x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(c*x))*(pi*c^2*x^2+pi)^(1/2)/x^3,x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*asinh(c*x))*(Pi + Pi*c^2*x^2)^(1/2))/x^3,x)

[Out]

int(((a + b*asinh(c*x))*(Pi + Pi*c^2*x^2)^(1/2))/x^3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asinh(c*x))*(pi*c**2*x**2+pi)**(1/2)/x**3,x)

[Out]

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

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