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

Optimal. Leaf size=60 \[ \frac{\text{PolyLog}\left (2,e^{2 \sinh ^{-1}(a+b x)}\right )}{2 d}-\frac{\sinh ^{-1}(a+b x)^2}{2 d}+\frac{\sinh ^{-1}(a+b x) \log \left (1-e^{2 \sinh ^{-1}(a+b x)}\right )}{d} \]

[Out]

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

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Rubi [A]  time = 0.095525, antiderivative size = 60, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.368, Rules used = {5865, 12, 5659, 3716, 2190, 2279, 2391} \[ \frac{\text{PolyLog}\left (2,e^{2 \sinh ^{-1}(a+b x)}\right )}{2 d}-\frac{\sinh ^{-1}(a+b x)^2}{2 d}+\frac{\sinh ^{-1}(a+b x) \log \left (1-e^{2 \sinh ^{-1}(a+b x)}\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 5865

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 5659

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

Rule 3716

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (Complex[0, fz_])*(f_.)*(x_)], x_Symbol] :> -Simp[(I*(c
+ d*x)^(m + 1))/(d*(m + 1)), x] + Dist[2*I, Int[((c + d*x)^m*E^(2*(-(I*e) + f*fz*x)))/(E^(2*I*k*Pi)*(1 + E^(2*
(-(I*e) + f*fz*x))/E^(2*I*k*Pi))), x], x] /; FreeQ[{c, d, e, f, fz}, x] && IntegerQ[4*k] && IGtQ[m, 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}(a+b x)}{\frac{a d}{b}+d x} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{b \sinh ^{-1}(x)}{d x} \, dx,x,a+b x\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int \frac{\sinh ^{-1}(x)}{x} \, dx,x,a+b x\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int x \coth (x) \, dx,x,\sinh ^{-1}(a+b x)\right )}{d}\\ &=-\frac{\sinh ^{-1}(a+b x)^2}{2 d}-\frac{2 \operatorname{Subst}\left (\int \frac{e^{2 x} x}{1-e^{2 x}} \, dx,x,\sinh ^{-1}(a+b x)\right )}{d}\\ &=-\frac{\sinh ^{-1}(a+b x)^2}{2 d}+\frac{\sinh ^{-1}(a+b x) \log \left (1-e^{2 \sinh ^{-1}(a+b x)}\right )}{d}-\frac{\operatorname{Subst}\left (\int \log \left (1-e^{2 x}\right ) \, dx,x,\sinh ^{-1}(a+b x)\right )}{d}\\ &=-\frac{\sinh ^{-1}(a+b x)^2}{2 d}+\frac{\sinh ^{-1}(a+b x) \log \left (1-e^{2 \sinh ^{-1}(a+b x)}\right )}{d}-\frac{\operatorname{Subst}\left (\int \frac{\log (1-x)}{x} \, dx,x,e^{2 \sinh ^{-1}(a+b x)}\right )}{2 d}\\ &=-\frac{\sinh ^{-1}(a+b x)^2}{2 d}+\frac{\sinh ^{-1}(a+b x) \log \left (1-e^{2 \sinh ^{-1}(a+b x)}\right )}{d}+\frac{\text{Li}_2\left (e^{2 \sinh ^{-1}(a+b x)}\right )}{2 d}\\ \end{align*}

Mathematica [A]  time = 0.0165718, size = 52, normalized size = 0.87 \[ \frac{\text{PolyLog}\left (2,e^{2 \sinh ^{-1}(a+b x)}\right )-\sinh ^{-1}(a+b x) \left (\sinh ^{-1}(a+b x)-2 \log \left (1-e^{2 \sinh ^{-1}(a+b x)}\right )\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.037, size = 125, normalized size = 2.1 \begin{align*} -{\frac{ \left ({\it Arcsinh} \left ( bx+a \right ) \right ) ^{2}}{2\,d}}+{\frac{{\it Arcsinh} \left ( bx+a \right ) }{d}\ln \left ( 1+bx+a+\sqrt{1+ \left ( bx+a \right ) ^{2}} \right ) }+{\frac{1}{d}{\it polylog} \left ( 2,-bx-a-\sqrt{1+ \left ( bx+a \right ) ^{2}} \right ) }+{\frac{{\it Arcsinh} \left ( bx+a \right ) }{d}\ln \left ( 1-bx-a-\sqrt{1+ \left ( bx+a \right ) ^{2}} \right ) }+{\frac{1}{d}{\it polylog} \left ( 2,bx+a+\sqrt{1+ \left ( bx+a \right ) ^{2}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(b*arcsinh(b*x + a)/(b*d*x + a*d), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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