3.98 \(\int \frac{a+b \cosh ^{-1}(c+d x)}{c e+d e x} \, dx\)

Optimal. Leaf size=81 \[ -\frac{b \text{PolyLog}\left (2,-e^{-2 \cosh ^{-1}(c+d x)}\right )}{2 d e}+\frac{\left (a+b \cosh ^{-1}(c+d x)\right )^2}{2 b d e}+\frac{\log \left (e^{-2 \cosh ^{-1}(c+d x)}+1\right ) \left (a+b \cosh ^{-1}(c+d x)\right )}{d e} \]

[Out]

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

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Rubi [A]  time = 0.107032, antiderivative size = 81, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {5866, 12, 5660, 3718, 2190, 2279, 2391} \[ \frac{b \text{PolyLog}\left (2,-e^{2 \cosh ^{-1}(c+d x)}\right )}{2 d e}-\frac{\left (a+b \cosh ^{-1}(c+d x)\right )^2}{2 b d e}+\frac{\log \left (e^{2 \cosh ^{-1}(c+d x)}+1\right ) \left (a+b \cosh ^{-1}(c+d x)\right )}{d e} \]

Warning: Unable to verify antiderivative.

[In]

Int[(a + b*ArcCosh[c + d*x])/(c*e + d*e*x),x]

[Out]

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

Rule 5866

Int[((a_.) + ArcCosh[(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*ArcCosh[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 5660

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

Rule 3718

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (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)))/(1 + E^(2*(-(I*e) + f*fz*x))), x],
x] /; FreeQ[{c, d, e, f, fz}, x] && 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{a+b \cosh ^{-1}(c+d x)}{c e+d e x} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{a+b \cosh ^{-1}(x)}{e x} \, dx,x,c+d x\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \frac{a+b \cosh ^{-1}(x)}{x} \, dx,x,c+d x\right )}{d e}\\ &=\frac{\operatorname{Subst}\left (\int (a+b x) \tanh (x) \, dx,x,\cosh ^{-1}(c+d x)\right )}{d e}\\ &=-\frac{\left (a+b \cosh ^{-1}(c+d x)\right )^2}{2 b d e}+\frac{2 \operatorname{Subst}\left (\int \frac{e^{2 x} (a+b x)}{1+e^{2 x}} \, dx,x,\cosh ^{-1}(c+d x)\right )}{d e}\\ &=-\frac{\left (a+b \cosh ^{-1}(c+d x)\right )^2}{2 b d e}+\frac{\left (a+b \cosh ^{-1}(c+d x)\right ) \log \left (1+e^{2 \cosh ^{-1}(c+d x)}\right )}{d e}-\frac{b \operatorname{Subst}\left (\int \log \left (1+e^{2 x}\right ) \, dx,x,\cosh ^{-1}(c+d x)\right )}{d e}\\ &=-\frac{\left (a+b \cosh ^{-1}(c+d x)\right )^2}{2 b d e}+\frac{\left (a+b \cosh ^{-1}(c+d x)\right ) \log \left (1+e^{2 \cosh ^{-1}(c+d x)}\right )}{d e}-\frac{b \operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{2 \cosh ^{-1}(c+d x)}\right )}{2 d e}\\ &=-\frac{\left (a+b \cosh ^{-1}(c+d x)\right )^2}{2 b d e}+\frac{\left (a+b \cosh ^{-1}(c+d x)\right ) \log \left (1+e^{2 \cosh ^{-1}(c+d x)}\right )}{d e}+\frac{b \text{Li}_2\left (-e^{2 \cosh ^{-1}(c+d x)}\right )}{2 d e}\\ \end{align*}

Mathematica [A]  time = 0.0615755, size = 69, normalized size = 0.85 \[ \frac{-b \text{PolyLog}\left (2,-e^{-2 \cosh ^{-1}(c+d x)}\right )+2 a \log (c+d x)+b \cosh ^{-1}(c+d x)^2+2 b \cosh ^{-1}(c+d x) \log \left (e^{-2 \cosh ^{-1}(c+d x)}+1\right )}{2 d e} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*ArcCosh[c + d*x])/(c*e + d*e*x),x]

[Out]

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

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Maple [A]  time = 0.031, size = 111, normalized size = 1.4 \begin{align*}{\frac{a\ln \left ( dx+c \right ) }{de}}-{\frac{b \left ({\rm arccosh} \left (dx+c\right ) \right ) ^{2}}{2\,de}}+{\frac{b{\rm arccosh} \left (dx+c\right )}{de}\ln \left ( \left ( dx+c+\sqrt{dx+c-1}\sqrt{dx+c+1} \right ) ^{2}+1 \right ) }+{\frac{b}{2\,de}{\it polylog} \left ( 2,- \left ( dx+c+\sqrt{dx+c-1}\sqrt{dx+c+1} \right ) ^{2} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arccosh(d*x+c))/(d*e*x+c*e),x)

[Out]

1/d*a/e*ln(d*x+c)-1/2/d*b/e*arccosh(d*x+c)^2+1/d*b/e*arccosh(d*x+c)*ln((d*x+c+(d*x+c-1)^(1/2)*(d*x+c+1)^(1/2))
^2+1)+1/2/d*b/e*polylog(2,-(d*x+c+(d*x+c-1)^(1/2)*(d*x+c+1)^(1/2))^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((a+b*arccosh(d*x+c))/(d*e*x+c*e),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{arcosh}\left (d x + c\right ) + a}{d e x + c e}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccosh(d*x+c))/(d*e*x+c*e),x, algorithm="fricas")

[Out]

integral((b*arccosh(d*x + c) + a)/(d*e*x + c*e), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*acosh(d*x+c))/(d*e*x+c*e),x)

[Out]

(Integral(a/(c + d*x), x) + Integral(b*acosh(c + d*x)/(c + d*x), x))/e

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccosh(d*x+c))/(d*e*x+c*e),x, algorithm="giac")

[Out]

integrate((b*arccosh(d*x + c) + a)/(d*e*x + c*e), x)