3.1 \(\int \frac{\sinh ^{-1}(c x)}{d+e x} \, dx\)

Optimal. Leaf size=170 \[ \frac{\text{PolyLog}\left (2,-\frac{e e^{\sinh ^{-1}(c x)}}{c d-\sqrt{c^2 d^2+e^2}}\right )}{e}+\frac{\text{PolyLog}\left (2,-\frac{e e^{\sinh ^{-1}(c x)}}{\sqrt{c^2 d^2+e^2}+c d}\right )}{e}+\frac{\sinh ^{-1}(c x) \log \left (\frac{e e^{\sinh ^{-1}(c x)}}{c d-\sqrt{c^2 d^2+e^2}}+1\right )}{e}+\frac{\sinh ^{-1}(c x) \log \left (\frac{e e^{\sinh ^{-1}(c x)}}{\sqrt{c^2 d^2+e^2}+c d}+1\right )}{e}-\frac{\sinh ^{-1}(c x)^2}{2 e} \]

[Out]

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

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Rubi [A]  time = 0.264292, antiderivative size = 170, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.417, Rules used = {5799, 5561, 2190, 2279, 2391} \[ \frac{\text{PolyLog}\left (2,-\frac{e e^{\sinh ^{-1}(c x)}}{c d-\sqrt{c^2 d^2+e^2}}\right )}{e}+\frac{\text{PolyLog}\left (2,-\frac{e e^{\sinh ^{-1}(c x)}}{\sqrt{c^2 d^2+e^2}+c d}\right )}{e}+\frac{\sinh ^{-1}(c x) \log \left (\frac{e e^{\sinh ^{-1}(c x)}}{c d-\sqrt{c^2 d^2+e^2}}+1\right )}{e}+\frac{\sinh ^{-1}(c x) \log \left (\frac{e e^{\sinh ^{-1}(c x)}}{\sqrt{c^2 d^2+e^2}+c d}+1\right )}{e}-\frac{\sinh ^{-1}(c x)^2}{2 e} \]

Antiderivative was successfully verified.

[In]

Int[ArcSinh[c*x]/(d + e*x),x]

[Out]

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

Rule 5799

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

Rule 5561

Int[(Cosh[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))^(m_.))/((a_) + (b_.)*Sinh[(c_.) + (d_.)*(x_)]), x_Symbol] :
> -Simp[(e + f*x)^(m + 1)/(b*f*(m + 1)), x] + (Int[((e + f*x)^m*E^(c + d*x))/(a - Rt[a^2 + b^2, 2] + b*E^(c +
d*x)), x] + Int[((e + f*x)^m*E^(c + d*x))/(a + Rt[a^2 + b^2, 2] + b*E^(c + d*x)), x]) /; FreeQ[{a, b, c, d, e,
f}, x] && IGtQ[m, 0] && NeQ[a^2 + b^2, 0]

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, 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]

Rubi steps

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

Mathematica [A]  time = 0.009979, size = 168, normalized size = 0.99 \[ \frac{\text{PolyLog}\left (2,\frac{e e^{\sinh ^{-1}(c x)}}{\sqrt{c^2 d^2+e^2}-c d}\right )}{e}+\frac{\text{PolyLog}\left (2,-\frac{e e^{\sinh ^{-1}(c x)}}{\sqrt{c^2 d^2+e^2}+c d}\right )}{e}+\frac{\sinh ^{-1}(c x) \log \left (\frac{e e^{\sinh ^{-1}(c x)}}{c d-\sqrt{c^2 d^2+e^2}}+1\right )}{e}+\frac{\sinh ^{-1}(c x) \log \left (\frac{e e^{\sinh ^{-1}(c x)}}{\sqrt{c^2 d^2+e^2}+c d}+1\right )}{e}-\frac{\sinh ^{-1}(c x)^2}{2 e} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcSinh[c*x]/(d + e*x),x]

[Out]

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

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Maple [A]  time = 0.076, size = 263, normalized size = 1.6 \begin{align*} -{\frac{ \left ({\it Arcsinh} \left ( cx \right ) \right ) ^{2}}{2\,e}}+{\frac{{\it Arcsinh} \left ( cx \right ) }{e}\ln \left ({ \left ( - \left ( cx+\sqrt{{c}^{2}{x}^{2}+1} \right ) e-cd+\sqrt{{c}^{2}{d}^{2}+{e}^{2}} \right ) \left ( -cd+\sqrt{{c}^{2}{d}^{2}+{e}^{2}} \right ) ^{-1}} \right ) }+{\frac{{\it Arcsinh} \left ( cx \right ) }{e}\ln \left ({ \left ( \left ( cx+\sqrt{{c}^{2}{x}^{2}+1} \right ) e+cd+\sqrt{{c}^{2}{d}^{2}+{e}^{2}} \right ) \left ( cd+\sqrt{{c}^{2}{d}^{2}+{e}^{2}} \right ) ^{-1}} \right ) }+{\frac{1}{e}{\it dilog} \left ({ \left ( \left ( cx+\sqrt{{c}^{2}{x}^{2}+1} \right ) e+cd+\sqrt{{c}^{2}{d}^{2}+{e}^{2}} \right ) \left ( cd+\sqrt{{c}^{2}{d}^{2}+{e}^{2}} \right ) ^{-1}} \right ) }+{\frac{1}{e}{\it dilog} \left ({ \left ( - \left ( cx+\sqrt{{c}^{2}{x}^{2}+1} \right ) e-cd+\sqrt{{c}^{2}{d}^{2}+{e}^{2}} \right ) \left ( -cd+\sqrt{{c}^{2}{d}^{2}+{e}^{2}} \right ) ^{-1}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arcsinh(c*x)/(e*x+d),x)

[Out]

-1/2*arcsinh(c*x)^2/e+1/e*arcsinh(c*x)*ln((-(c*x+(c^2*x^2+1)^(1/2))*e-c*d+(c^2*d^2+e^2)^(1/2))/(-c*d+(c^2*d^2+
e^2)^(1/2)))+1/e*arcsinh(c*x)*ln(((c*x+(c^2*x^2+1)^(1/2))*e+c*d+(c^2*d^2+e^2)^(1/2))/(c*d+(c^2*d^2+e^2)^(1/2))
)+1/e*dilog(((c*x+(c^2*x^2+1)^(1/2))*e+c*d+(c^2*d^2+e^2)^(1/2))/(c*d+(c^2*d^2+e^2)^(1/2)))+1/e*dilog((-(c*x+(c
^2*x^2+1)^(1/2))*e-c*d+(c^2*d^2+e^2)^(1/2))/(-c*d+(c^2*d^2+e^2)^(1/2)))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsinh(c*x)/(e*x+d),x, algorithm="maxima")

[Out]

integrate(arcsinh(c*x)/(e*x + d), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\operatorname{arsinh}\left (c x\right )}{e x + d}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsinh(c*x)/(e*x+d),x, algorithm="fricas")

[Out]

integral(arcsinh(c*x)/(e*x + d), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(asinh(c*x)/(e*x+d),x)

[Out]

Integral(asinh(c*x)/(d + e*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsinh(c*x)/(e*x+d),x, algorithm="giac")

[Out]

integrate(arcsinh(c*x)/(e*x + d), x)