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

Optimal. Leaf size=89 \[ \sqrt {\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )-2 \sqrt {\pi } \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )-\sqrt {\pi } b \text {Li}_2\left (-e^{\sinh ^{-1}(c x)}\right )+\sqrt {\pi } b \text {Li}_2\left (e^{\sinh ^{-1}(c x)}\right )+\sqrt {\pi } (-b) c x \]

[Out]

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

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Rubi [A]  time = 0.19, antiderivative size = 177, normalized size of antiderivative = 1.99, 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 = {5742, 5760, 4182, 2279, 2391, 8} \[ -\frac {b \sqrt {\pi c^2 x^2+\pi } \text {PolyLog}\left (2,-e^{\sinh ^{-1}(c x)}\right )}{\sqrt {c^2 x^2+1}}+\frac {b \sqrt {\pi c^2 x^2+\pi } \text {PolyLog}\left (2,e^{\sinh ^{-1}(c x)}\right )}{\sqrt {c^2 x^2+1}}+\sqrt {\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )-\frac {2 \sqrt {\pi c^2 x^2+\pi } \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt {c^2 x^2+1}}-\frac {b c x \sqrt {\pi c^2 x^2+\pi }}{\sqrt {c^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

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 5742

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 + 2)), x] + (Dist[Sqrt[d + e*x^2]/((m + 2)*Sqrt[1
+ c^2*x^2]), Int[((f*x)^m*(a + b*ArcSinh[c*x])^n)/Sqrt[1 + c^2*x^2], x], x] - Dist[(b*c*n*Sqrt[d + e*x^2])/(f*
(m + 2)*Sqrt[1 + c^2*x^2]), Int[(f*x)^(m + 1)*(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[m, -1] && (RationalQ[m] || 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 {\sqrt {\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )}{x} \, dx &=\sqrt {\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {\sqrt {\pi +c^2 \pi x^2} \int \frac {a+b \sinh ^{-1}(c x)}{x \sqrt {1+c^2 x^2}} \, dx}{\sqrt {1+c^2 x^2}}-\frac {\left (b c \sqrt {\pi +c^2 \pi x^2}\right ) \int 1 \, dx}{\sqrt {1+c^2 x^2}}\\ &=-\frac {b c x \sqrt {\pi +c^2 \pi x^2}}{\sqrt {1+c^2 x^2}}+\sqrt {\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {\sqrt {\pi +c^2 \pi x^2} \operatorname {Subst}\left (\int (a+b x) \text {csch}(x) \, dx,x,\sinh ^{-1}(c x)\right )}{\sqrt {1+c^2 x^2}}\\ &=-\frac {b c x \sqrt {\pi +c^2 \pi x^2}}{\sqrt {1+c^2 x^2}}+\sqrt {\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )-\frac {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 \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 )}{\sqrt {1+c^2 x^2}}+\frac {\left (b \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 )}{\sqrt {1+c^2 x^2}}\\ &=-\frac {b c x \sqrt {\pi +c^2 \pi x^2}}{\sqrt {1+c^2 x^2}}+\sqrt {\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )-\frac {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 \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 )}{\sqrt {1+c^2 x^2}}+\frac {\left (b \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 )}{\sqrt {1+c^2 x^2}}\\ &=-\frac {b c x \sqrt {\pi +c^2 \pi x^2}}{\sqrt {1+c^2 x^2}}+\sqrt {\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )-\frac {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 \sqrt {\pi +c^2 \pi x^2} \text {Li}_2\left (-e^{\sinh ^{-1}(c x)}\right )}{\sqrt {1+c^2 x^2}}+\frac {b \sqrt {\pi +c^2 \pi x^2} \text {Li}_2\left (e^{\sinh ^{-1}(c x)}\right )}{\sqrt {1+c^2 x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.20, size = 131, normalized size = 1.47 \[ \sqrt {\pi } \left (a \sqrt {c^2 x^2+1}-a \log \left (\pi \left (\sqrt {c^2 x^2+1}+1\right )\right )+a \log (x)+b \left (\sqrt {c^2 x^2+1} \sinh ^{-1}(c x)+\text {Li}_2\left (-e^{-\sinh ^{-1}(c x)}\right )-\text {Li}_2\left (e^{-\sinh ^{-1}(c x)}\right )-c x+\sinh ^{-1}(c x) \log \left (1-e^{-\sinh ^{-1}(c x)}\right )-\sinh ^{-1}(c x) \log \left (e^{-\sinh ^{-1}(c x)}+1\right )\right )\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F]  time = 0.61, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {\pi + \pi c^{2} x^{2}} {\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}}{x}, x\right ) \]

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,x, algorithm="fricas")

[Out]

integral(sqrt(pi + pi*c^2*x^2)*(b*arcsinh(c*x) + a)/x, 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,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 = 0.22, size = 171, normalized size = 1.92 \[ -\sqrt {\pi }\, \arctanh \left (\frac {\sqrt {\pi }}{\sqrt {\pi \,c^{2} x^{2}+\pi }}\right ) a +a \sqrt {\pi \,c^{2} x^{2}+\pi }+\sqrt {\pi }\, \arcsinh \left (c x \right ) \ln \left (1-c x -\sqrt {c^{2} x^{2}+1}\right ) b +\sqrt {\pi }\, \arcsinh \left (c x \right ) \sqrt {c^{2} x^{2}+1}\, b -\sqrt {\pi }\, \arcsinh \left (c x \right ) \ln \left (1+c x +\sqrt {c^{2} x^{2}+1}\right ) b -b c x \sqrt {\pi }+b \polylog \left (2, c x +\sqrt {c^{2} x^{2}+1}\right ) \sqrt {\pi }-b \polylog \left (2, -c x -\sqrt {c^{2} x^{2}+1}\right ) \sqrt {\pi } \]

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,x)

[Out]

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

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

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,x, algorithm="maxima")

[Out]

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

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

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,x)

[Out]

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

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

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,x)

[Out]

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

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