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

Optimal. Leaf size=101 \[ \frac{i b c \text{PolyLog}\left (2,-i e^{\sinh ^{-1}(c x)}\right )}{d}-\frac{i b c \text{PolyLog}\left (2,i e^{\sinh ^{-1}(c x)}\right )}{d}-\frac{a+b \sinh ^{-1}(c x)}{d x}-\frac{2 c \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{d}-\frac{b c \tanh ^{-1}\left (\sqrt{c^2 x^2+1}\right )}{d} \]

[Out]

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

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Rubi [A]  time = 0.149639, antiderivative size = 101, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 8, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {5747, 5693, 4180, 2279, 2391, 266, 63, 208} \[ \frac{i b c \text{PolyLog}\left (2,-i e^{\sinh ^{-1}(c x)}\right )}{d}-\frac{i b c \text{PolyLog}\left (2,i e^{\sinh ^{-1}(c x)}\right )}{d}-\frac{a+b \sinh ^{-1}(c x)}{d x}-\frac{2 c \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{d}-\frac{b c \tanh ^{-1}\left (\sqrt{c^2 x^2+1}\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 5747

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*d^IntPart[p]*(d + e*x^
2)^FracPart[p])/(f*(m + 1)*(1 + c^2*x^2)^FracPart[p]), Int[(f*x)^(m + 1)*(1 + c^2*x^2)^(p + 1/2)*(a + b*ArcSin
h[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e, f, p}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && LtQ[m, -1] && Int
egerQ[m]

Rule 5693

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

Rule 4180

Int[csc[(e_.) + Pi*(k_.) + (Complex[0, fz_])*(f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c
+ d*x)^m*ArcTanh[E^(-(I*e) + f*fz*x)/E^(I*k*Pi)])/(f*fz*I), x] + (-Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*
Log[1 - E^(-(I*e) + f*fz*x)/E^(I*k*Pi)], x], x] + Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*Log[1 + E^(-(I*e)
+ f*fz*x)/E^(I*k*Pi)], x], x]) /; FreeQ[{c, d, e, f, fz}, x] && IntegerQ[2*k] && 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 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

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

Rubi steps

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

Mathematica [A]  time = 0.162546, size = 182, normalized size = 1.8 \[ -\frac{-b \sqrt{-c^2} x \text{PolyLog}\left (2,\frac{c e^{\sinh ^{-1}(c x)}}{\sqrt{-c^2}}\right )+b \sqrt{-c^2} x \text{PolyLog}\left (2,\frac{\sqrt{-c^2} e^{\sinh ^{-1}(c x)}}{c}\right )+a c x \tan ^{-1}(c x)+a+b c x \tanh ^{-1}\left (\sqrt{c^2 x^2+1}\right )+b \sqrt{-c^2} x \sinh ^{-1}(c x) \log \left (\frac{c e^{\sinh ^{-1}(c x)}}{\sqrt{-c^2}}+1\right )-b \sqrt{-c^2} x \sinh ^{-1}(c x) \log \left (\frac{\sqrt{-c^2} e^{\sinh ^{-1}(c x)}}{c}+1\right )+b \sinh ^{-1}(c x)}{d x} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [A]  time = 0.013, size = 202, normalized size = 2. \begin{align*} -{\frac{a}{dx}}-{\frac{ca\arctan \left ( cx \right ) }{d}}-{\frac{b{\it Arcsinh} \left ( cx \right ) }{dx}}-{\frac{bc{\it Arcsinh} \left ( cx \right ) \arctan \left ( cx \right ) }{d}}-{\frac{bc}{d}{\it Artanh} \left ({\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}} \right ) }-{\frac{bc\arctan \left ( cx \right ) }{d}\ln \left ( 1+{i \left ( 1+icx \right ){\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}} \right ) }+{\frac{bc\arctan \left ( cx \right ) }{d}\ln \left ( 1-{i \left ( 1+icx \right ){\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}} \right ) }+{\frac{icb}{d}{\it dilog} \left ( 1+{i \left ( 1+icx \right ){\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}} \right ) }-{\frac{icb}{d}{\it dilog} \left ( 1-{i \left ( 1+icx \right ){\frac{1}{\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^2/(c^2*d*x^2+d),x)

[Out]

-a/d/x-c*a/d*arctan(c*x)-b/d*arcsinh(c*x)/x-c*b/d*arcsinh(c*x)*arctan(c*x)-c*b/d*arctanh(1/(c^2*x^2+1)^(1/2))-
c*b/d*arctan(c*x)*ln(1+I*(1+I*c*x)/(c^2*x^2+1)^(1/2))+c*b/d*arctan(c*x)*ln(1-I*(1+I*c*x)/(c^2*x^2+1)^(1/2))+I*
c*b/d*dilog(1+I*(1+I*c*x)/(c^2*x^2+1)^(1/2))-I*c*b/d*dilog(1-I*(1+I*c*x)/(c^2*x^2+1)^(1/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a*(c*arctan(c*x)/d + 1/(d*x)) + b*integrate(log(c*x + sqrt(c^2*x^2 + 1))/(c^2*d*x^4 + d*x^2), 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^{2} d x^{4} + d x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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 )} x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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