3.1.88 \(\int \frac {a+b \sinh ^{-1}(c x)}{x^3 \sqrt {\pi +c^2 \pi x^2}} \, dx\) [88]

Optimal. Leaf size=115 \[ -\frac {b c}{2 \sqrt {\pi } x}-\frac {\sqrt {\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )}{2 \pi x^2}+\frac {c^2 \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{\sqrt {\pi }}+\frac {b c^2 \text {PolyLog}\left (2,-e^{\sinh ^{-1}(c x)}\right )}{2 \sqrt {\pi }}-\frac {b c^2 \text {PolyLog}\left (2,e^{\sinh ^{-1}(c x)}\right )}{2 \sqrt {\pi }} \]

[Out]

-1/2*b*c/x/Pi^(1/2)+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)/Pi/x^2

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Rubi [A]
time = 0.14, antiderivative size = 115, normalized size of antiderivative = 1.00, 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 = {5809, 5816, 4267, 2317, 2438, 30} \begin {gather*} -\frac {\sqrt {\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )}{2 \pi x^2}+\frac {c^2 \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt {\pi }}+\frac {b c^2 \text {Li}_2\left (-e^{\sinh ^{-1}(c x)}\right )}{2 \sqrt {\pi }}-\frac {b c^2 \text {Li}_2\left (e^{\sinh ^{-1}(c x)}\right )}{2 \sqrt {\pi }}-\frac {b c}{2 \sqrt {\pi } x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 30

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

Rule 2317

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 2438

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 4267

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 5809

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/(d*f*(m + 1))), x] + (-Dist[c^2*((m + 2*p + 3)/(f^2*
(m + 1))), Int[(f*x)^(m + 2)*(d + e*x^2)^p*(a + b*ArcSinh[c*x])^n, x], x] - Dist[b*c*(n/(f*(m + 1)))*Simp[(d +
e*x^2)^p/(1 + c^2*x^2)^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, p}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && ILtQ[m, -1]

Rule 5816

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

Rubi steps

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

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Mathematica [A]
time = 1.80, size = 185, normalized size = 1.61 \begin {gather*} \frac {-\frac {4 a \sqrt {1+c^2 x^2}}{x^2}-4 a c^2 \log (x)+4 a c^2 \log \left (\pi \left (1+\sqrt {1+c^2 x^2}\right )\right )+b c^2 \left (-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 )-4 \sinh ^{-1}(c x) \log \left (1-e^{-\sinh ^{-1}(c x)}\right )+4 \sinh ^{-1}(c x) \log \left (1+e^{-\sinh ^{-1}(c x)}\right )-4 \text {PolyLog}\left (2,-e^{-\sinh ^{-1}(c x)}\right )+4 \text {PolyLog}\left (2,e^{-\sinh ^{-1}(c x)}\right )-\sinh ^{-1}(c x) \text {sech}^2\left (\frac {1}{2} \sinh ^{-1}(c x)\right )+2 \tanh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )}{8 \sqrt {\pi }} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-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[ArcS
inh[c*x]/2] - ArcSinh[c*x]*Csch[ArcSinh[c*x]/2]^2 - 4*ArcSinh[c*x]*Log[1 - E^(-ArcSinh[c*x])] + 4*ArcSinh[c*x]
*Log[1 + E^(-ArcSinh[c*x])] - 4*PolyLog[2, -E^(-ArcSinh[c*x])] + 4*PolyLog[2, E^(-ArcSinh[c*x])] - ArcSinh[c*x
]*Sech[ArcSinh[c*x]/2]^2 + 2*Tanh[ArcSinh[c*x]/2]))/(8*Sqrt[Pi])

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Maple [A]
time = 4.29, size = 226, normalized size = 1.97

method	result	size
default	\(a \left (-\frac {\sqrt {\pi \,c^{2} x^{2}+\pi }}{2 \pi \,x^{2}}+\frac {c^{2} \arctanh \left (\frac {\sqrt {\pi }}{\sqrt {\pi \,c^{2} x^{2}+\pi }}\right )}{2 \sqrt {\pi }}\right )-\frac {b \arcsinh \left (c x \right ) c^{2}}{2 \sqrt {\pi }\, \sqrt {c^{2} x^{2}+1}}-\frac {b c}{2 x \sqrt {\pi }}-\frac {b \arcsinh \left (c x \right )}{2 \sqrt {\pi }\, \sqrt {c^{2} x^{2}+1}\, x^{2}}+\frac {b \,c^{2} \arcsinh \left (c x \right ) \ln \left (1+c x +\sqrt {c^{2} x^{2}+1}\right )}{2 \sqrt {\pi }}+\frac {b \,c^{2} \polylog \left (2, -c x -\sqrt {c^{2} x^{2}+1}\right )}{2 \sqrt {\pi }}-\frac {b \,c^{2} \arcsinh \left (c x \right ) \ln \left (1-c x -\sqrt {c^{2} x^{2}+1}\right )}{2 \sqrt {\pi }}-\frac {b \,c^{2} \polylog \left (2, c x +\sqrt {c^{2} x^{2}+1}\right )}{2 \sqrt {\pi }}\)	\(226\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*(-1/2/Pi/x^2*(Pi*c^2*x^2+Pi)^(1/2)+1/2/Pi^(1/2)*c^2*arctanh(Pi^(1/2)/(Pi*c^2*x^2+Pi)^(1/2)))-1/2*b/Pi^(1/2)/
(c^2*x^2+1)^(1/2)*arcsinh(c*x)*c^2-1/2*b*c/x/Pi^(1/2)-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 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(c^2*arcsinh(1/(c*abs(x)))/sqrt(pi) - sqrt(pi + pi*c^2*x^2)/(pi*x^2))*a + b*integrate(log(c*x + sqrt(c^2*x
^2 + 1))/(sqrt(pi + pi*c^2*x^2)*x^3), 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((a+b*arcsinh(c*x))/x^3/(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^5 + pi*x^3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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((a+b*arcsinh(c*x))/x^3/(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^3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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