3.42 \(\int \frac{a+b \sinh ^{-1}(c x)}{x (d+c^2 d x^2)^2} \, dx\)

Optimal. Leaf size=110 \[ -\frac{b \text{PolyLog}\left (2,-e^{2 \sinh ^{-1}(c x)}\right )}{2 d^2}+\frac{b \text{PolyLog}\left (2,e^{2 \sinh ^{-1}(c x)}\right )}{2 d^2}+\frac{a+b \sinh ^{-1}(c x)}{2 d^2 \left (c^2 x^2+1\right )}-\frac{2 \tanh ^{-1}\left (e^{2 \sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{d^2}-\frac{b c x}{2 d^2 \sqrt{c^2 x^2+1}} \]

[Out]

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

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Rubi [A]  time = 0.176965, antiderivative size = 110, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.292, Rules used = {5755, 5720, 5461, 4182, 2279, 2391, 191} \[ -\frac{b \text{PolyLog}\left (2,-e^{2 \sinh ^{-1}(c x)}\right )}{2 d^2}+\frac{b \text{PolyLog}\left (2,e^{2 \sinh ^{-1}(c x)}\right )}{2 d^2}+\frac{a+b \sinh ^{-1}(c x)}{2 d^2 \left (c^2 x^2+1\right )}-\frac{2 \tanh ^{-1}\left (e^{2 \sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{d^2}-\frac{b c x}{2 d^2 \sqrt{c^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 5720

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

Rule 5461

Int[Csch[(a_.) + (b_.)*(x_)]^(n_.)*((c_.) + (d_.)*(x_))^(m_.)*Sech[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Dis
t[2^n, Int[(c + d*x)^m*Csch[2*a + 2*b*x]^n, x], x] /; FreeQ[{a, b, c, d}, x] && RationalQ[m] && IntegerQ[n]

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 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 191

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^(p + 1))/a, x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rubi steps

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

Mathematica [B]  time = 0.444966, size = 234, normalized size = 2.13 \[ -\frac{2 b \text{PolyLog}\left (2,\frac{c e^{\sinh ^{-1}(c x)}}{\sqrt{-c^2}}\right )+2 b \text{PolyLog}\left (2,\frac{\sqrt{-c^2} e^{\sinh ^{-1}(c x)}}{c}\right )-b \text{PolyLog}\left (2,e^{2 \sinh ^{-1}(c x)}\right )+\frac{a^2}{b}-\frac{a}{c^2 x^2+1}+a \log \left (c^2 x^2+1\right )+2 a \sinh ^{-1}(c x)-2 a \log \left (1-e^{2 \sinh ^{-1}(c x)}\right )+\frac{b c x}{\sqrt{c^2 x^2+1}}-\frac{b \sinh ^{-1}(c x)}{c^2 x^2+1}+2 b \sinh ^{-1}(c x) \log \left (\frac{c e^{\sinh ^{-1}(c x)}}{\sqrt{-c^2}}+1\right )+2 b \sinh ^{-1}(c x) \log \left (\frac{\sqrt{-c^2} e^{\sinh ^{-1}(c x)}}{c}+1\right )-2 b \sinh ^{-1}(c x) \log \left (1-e^{2 \sinh ^{-1}(c x)}\right )}{2 d^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [B]  time = 0.085, size = 283, normalized size = 2.6 \begin{align*}{\frac{a\ln \left ( cx \right ) }{{d}^{2}}}+{\frac{a}{2\,{d}^{2} \left ({c}^{2}{x}^{2}+1 \right ) }}-{\frac{a\ln \left ({c}^{2}{x}^{2}+1 \right ) }{2\,{d}^{2}}}-{\frac{bcx}{2\,{d}^{2}}{\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}}+{\frac{b{c}^{2}{x}^{2}}{2\,{d}^{2} \left ({c}^{2}{x}^{2}+1 \right ) }}+{\frac{b{\it Arcsinh} \left ( cx \right ) }{2\,{d}^{2} \left ({c}^{2}{x}^{2}+1 \right ) }}+{\frac{b}{2\,{d}^{2} \left ({c}^{2}{x}^{2}+1 \right ) }}-{\frac{b{\it Arcsinh} \left ( cx \right ) }{{d}^{2}}\ln \left ( 1+ \left ( cx+\sqrt{{c}^{2}{x}^{2}+1} \right ) ^{2} \right ) }-{\frac{b}{2\,{d}^{2}}{\it polylog} \left ( 2,- \left ( cx+\sqrt{{c}^{2}{x}^{2}+1} \right ) ^{2} \right ) }+{\frac{b{\it Arcsinh} \left ( cx \right ) }{{d}^{2}}\ln \left ( 1+cx+\sqrt{{c}^{2}{x}^{2}+1} \right ) }+{\frac{b}{{d}^{2}}{\it polylog} \left ( 2,-cx-\sqrt{{c}^{2}{x}^{2}+1} \right ) }+{\frac{b{\it Arcsinh} \left ( cx \right ) }{{d}^{2}}\ln \left ( 1-cx-\sqrt{{c}^{2}{x}^{2}+1} \right ) }+{\frac{b}{{d}^{2}}{\it polylog} \left ( 2,cx+\sqrt{{c}^{2}{x}^{2}+1} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{2} \, a{\left (\frac{1}{c^{2} d^{2} x^{2} + d^{2}} - \frac{\log \left (c^{2} x^{2} + 1\right )}{d^{2}} + \frac{2 \, \log \left (x\right )}{d^{2}}\right )} + b \int \frac{\log \left (c x + \sqrt{c^{2} x^{2} + 1}\right )}{c^{4} d^{2} x^{5} + 2 \, c^{2} d^{2} x^{3} + d^{2} x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a*(1/(c^2*d^2*x^2 + d^2) - log(c^2*x^2 + 1)/d^2 + 2*log(x)/d^2) + b*integrate(log(c*x + sqrt(c^2*x^2 + 1))
/(c^4*d^2*x^5 + 2*c^2*d^2*x^3 + d^2*x), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b \operatorname{arsinh}\left (c x\right ) + a}{c^{4} d^{2} x^{5} + 2 \, c^{2} d^{2} x^{3} + d^{2} x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{a}{c^{4} x^{5} + 2 c^{2} x^{3} + x}\, dx + \int \frac{b \operatorname{asinh}{\left (c x \right )}}{c^{4} x^{5} + 2 c^{2} x^{3} + x}\, dx}{d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(a/(c**4*x**5 + 2*c**2*x**3 + x), x) + Integral(b*asinh(c*x)/(c**4*x**5 + 2*c**2*x**3 + x), x))/d**2

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{b \operatorname{arsinh}\left (c x\right ) + a}{{\left (c^{2} d x^{2} + d\right )}^{2} x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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