3.275 \(\int \frac{(a+b \coth ^{-1}(c x)) (d+e \log (1-c^2 x^2))}{x^2} \, dx\)

Optimal. Leaf size=105 \[ -\frac{1}{2} b c e \text{PolyLog}\left (2,\frac{1}{1-c^2 x^2}\right )-\frac{\left (a+b \coth ^{-1}(c x)\right ) \left (e \log \left (1-c^2 x^2\right )+d\right )}{x}-\frac{c e \left (a+b \coth ^{-1}(c x)\right )^2}{b}+\frac{1}{2} b c \log \left (1-\frac{1}{1-c^2 x^2}\right ) \left (e \log \left (1-c^2 x^2\right )+d\right ) \]

[Out]

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

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Rubi [A]  time = 0.269076, antiderivative size = 94, normalized size of antiderivative = 0.9, number of steps used = 8, number of rules used = 8, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.296, Rules used = {6082, 2475, 2411, 2344, 2301, 2316, 2315, 5949} \[ -\frac{1}{2} b c e \text{PolyLog}\left (2,c^2 x^2\right )-\frac{\left (a+b \coth ^{-1}(c x)\right ) \left (e \log \left (1-c^2 x^2\right )+d\right )}{x}-\frac{c e \left (a+b \coth ^{-1}(c x)\right )^2}{b}-\frac{b c \left (e \log \left (1-c^2 x^2\right )+d\right )^2}{4 e}+b c d \log (x) \]

Antiderivative was successfully verified.

[In]

Int[((a + b*ArcCoth[c*x])*(d + e*Log[1 - c^2*x^2]))/x^2,x]

[Out]

-((c*e*(a + b*ArcCoth[c*x])^2)/b) + b*c*d*Log[x] - ((a + b*ArcCoth[c*x])*(d + e*Log[1 - c^2*x^2]))/x - (b*c*(d
+ e*Log[1 - c^2*x^2])^2)/(4*e) - (b*c*e*PolyLog[2, c^2*x^2])/2

Rule 6082

Int[((a_.) + ArcCoth[(c_.)*(x_)]*(b_.))*((d_.) + Log[(f_.) + (g_.)*(x_)^2]*(e_.))*(x_)^(m_.), x_Symbol] :> Sim
p[(x^(m + 1)*(d + e*Log[f + g*x^2])*(a + b*ArcCoth[c*x]))/(m + 1), x] + (-Dist[(b*c)/(m + 1), Int[(x^(m + 1)*(
d + e*Log[f + g*x^2]))/(1 - c^2*x^2), x], x] - Dist[(2*e*g)/(m + 1), Int[(x^(m + 2)*(a + b*ArcCoth[c*x]))/(f +
g*x^2), x], x]) /; FreeQ[{a, b, c, d, e, f, g}, x] && ILtQ[m/2, 0]

Rule 2475

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m_.)*((f_) + (g_.)*(x_)^(s_))^(r_.),
x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(f + g*x^(s/n))^r*(a + b*Log[c*(d + e*x)^p])^q,
x], x, x^n], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p, q, r, s}, x] && IntegerQ[r] && IntegerQ[s/n] && Intege
rQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0] || IGtQ[q, 0])

Rule 2411

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_.) + (g_.)*(x_))^(q_.)*((h_.) + (i_.)*(x_))
^(r_.), x_Symbol] :> Dist[1/e, Subst[Int[((g*x)/e)^q*((e*h - d*i)/e + (i*x)/e)^r*(a + b*Log[c*x^n])^p, x], x,
d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, h, i, n, p, q, r}, x] && EqQ[e*f - d*g, 0] && (IGtQ[p, 0] || IGtQ[
r, 0]) && IntegerQ[2*r]

Rule 2344

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

Rule 2301

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

Rule 2316

Int[((a_.) + Log[(c_.)*(x_)]*(b_.))/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[((a + b*Log[-((c*d)/e)])*Log[d + e*
x])/e, x] + Dist[b, Int[Log[-((e*x)/d)]/(d + e*x), x], x] /; FreeQ[{a, b, c, d, e}, x] && GtQ[-((c*d)/e), 0]

Rule 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 0]

Rule 5949

Int[((a_.) + ArcCoth[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(a + b*ArcCoth[c*x])^(p
+ 1)/(b*c*d*(p + 1)), x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[c^2*d + e, 0] && NeQ[p, -1]

Rubi steps

\begin{align*} \int \frac{\left (a+b \coth ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right )}{x^2} \, dx &=-\frac{\left (a+b \coth ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right )}{x}+(b c) \int \frac{d+e \log \left (1-c^2 x^2\right )}{x \left (1-c^2 x^2\right )} \, dx-\left (2 c^2 e\right ) \int \frac{a+b \coth ^{-1}(c x)}{1-c^2 x^2} \, dx\\ &=-\frac{c e \left (a+b \coth ^{-1}(c x)\right )^2}{b}-\frac{\left (a+b \coth ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right )}{x}+\frac{1}{2} (b c) \operatorname{Subst}\left (\int \frac{d+e \log \left (1-c^2 x\right )}{x \left (1-c^2 x\right )} \, dx,x,x^2\right )\\ &=-\frac{c e \left (a+b \coth ^{-1}(c x)\right )^2}{b}-\frac{\left (a+b \coth ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right )}{x}-\frac{b \operatorname{Subst}\left (\int \frac{d+e \log (x)}{x \left (\frac{1}{c^2}-\frac{x}{c^2}\right )} \, dx,x,1-c^2 x^2\right )}{2 c}\\ &=-\frac{c e \left (a+b \coth ^{-1}(c x)\right )^2}{b}-\frac{\left (a+b \coth ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right )}{x}-\frac{b \operatorname{Subst}\left (\int \frac{d+e \log (x)}{\frac{1}{c^2}-\frac{x}{c^2}} \, dx,x,1-c^2 x^2\right )}{2 c}-\frac{1}{2} (b c) \operatorname{Subst}\left (\int \frac{d+e \log (x)}{x} \, dx,x,1-c^2 x^2\right )\\ &=-\frac{c e \left (a+b \coth ^{-1}(c x)\right )^2}{b}+b c d \log (x)-\frac{\left (a+b \coth ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right )}{x}-\frac{b c \left (d+e \log \left (1-c^2 x^2\right )\right )^2}{4 e}-\frac{(b e) \operatorname{Subst}\left (\int \frac{\log (x)}{\frac{1}{c^2}-\frac{x}{c^2}} \, dx,x,1-c^2 x^2\right )}{2 c}\\ &=-\frac{c e \left (a+b \coth ^{-1}(c x)\right )^2}{b}+b c d \log (x)-\frac{\left (a+b \coth ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right )}{x}-\frac{b c \left (d+e \log \left (1-c^2 x^2\right )\right )^2}{4 e}-\frac{1}{2} b c e \text{Li}_2\left (c^2 x^2\right )\\ \end{align*}

Mathematica [B]  time = 0.203594, size = 332, normalized size = 3.16 \[ -\frac{4 b c e x \text{PolyLog}(2,-c x)+4 b c e x \text{PolyLog}(2,c x)-2 b c e x \text{PolyLog}\left (2,\frac{1}{2}-\frac{c x}{2}\right )-2 b c e x \text{PolyLog}\left (2,\frac{1}{2} (c x+1)\right )+4 a e \log \left (1-c^2 x^2\right )+8 a c e x \tanh ^{-1}(c x)+4 a d+2 b c d x \log \left (1-c^2 x^2\right )-4 b c e x \log (x) \log \left (1-c^2 x^2\right )+2 b c e x \log \left (x-\frac{1}{c}\right ) \log \left (1-c^2 x^2\right )+2 b c e x \log \left (\frac{1}{c}+x\right ) \log \left (1-c^2 x^2\right )+4 b e \log \left (1-c^2 x^2\right ) \coth ^{-1}(c x)-4 b c d x \log (x)+4 b d \coth ^{-1}(c x)-b c e x \log ^2\left (x-\frac{1}{c}\right )-b c e x \log ^2\left (\frac{1}{c}+x\right )-2 b c e x \log \left (\frac{1}{c}+x\right ) \log \left (\frac{1}{2} (1-c x)\right )+4 b c e x \log (x) \log (1-c x)-2 b c e x \log \left (x-\frac{1}{c}\right ) \log \left (\frac{1}{2} (c x+1)\right )+4 b c e x \log (x) \log (c x+1)+4 b c e x \coth ^{-1}(c x)^2}{4 x} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*ArcCoth[c*x])*(d + e*Log[1 - c^2*x^2]))/x^2,x]

[Out]

-(4*a*d + 4*b*d*ArcCoth[c*x] + 4*b*c*e*x*ArcCoth[c*x]^2 + 8*a*c*e*x*ArcTanh[c*x] - 4*b*c*d*x*Log[x] - b*c*e*x*
Log[-c^(-1) + x]^2 - b*c*e*x*Log[c^(-1) + x]^2 - 2*b*c*e*x*Log[c^(-1) + x]*Log[(1 - c*x)/2] + 4*b*c*e*x*Log[x]
*Log[1 - c*x] - 2*b*c*e*x*Log[-c^(-1) + x]*Log[(1 + c*x)/2] + 4*b*c*e*x*Log[x]*Log[1 + c*x] + 4*a*e*Log[1 - c^
2*x^2] + 2*b*c*d*x*Log[1 - c^2*x^2] + 4*b*e*ArcCoth[c*x]*Log[1 - c^2*x^2] - 4*b*c*e*x*Log[x]*Log[1 - c^2*x^2]
+ 2*b*c*e*x*Log[-c^(-1) + x]*Log[1 - c^2*x^2] + 2*b*c*e*x*Log[c^(-1) + x]*Log[1 - c^2*x^2] + 4*b*c*e*x*PolyLog
[2, -(c*x)] + 4*b*c*e*x*PolyLog[2, c*x] - 2*b*c*e*x*PolyLog[2, 1/2 - (c*x)/2] - 2*b*c*e*x*PolyLog[2, (1 + c*x)
/2])/(4*x)

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Maple [F]  time = 1.642, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( a+b{\rm arccoth} \left (cx\right ) \right ) \left ( d+e\ln \left ( -{c}^{2}{x}^{2}+1 \right ) \right ) }{{x}^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arccoth(c*x))*(d+e*ln(-c^2*x^2+1))/x^2,x)

[Out]

int((a+b*arccoth(c*x))*(d+e*ln(-c^2*x^2+1))/x^2,x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccoth(c*x))*(d+e*log(-c^2*x^2+1))/x^2,x, algorithm="maxima")

[Out]

-1/2*(c*(log(c^2*x^2 - 1) - log(x^2)) + 2*arccoth(c*x)/x)*b*d - (c^2*(log(c*x + 1)/c - log(c*x - 1)/c) + log(-
c^2*x^2 + 1)/x)*a*e - 1/2*b*e*(log(c*x + 1)^2/x - integrate(-((c*x + 1)*log(c*x - 1)^2 - (I*pi + (I*pi*c + 2*c
)*x)*log(c*x + 1) - (-I*pi - I*pi*c*x)*log(c*x - 1))/(c*x^3 + x^2), x)) - a*d/x

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b d \operatorname{arcoth}\left (c x\right ) + a d +{\left (b e \operatorname{arcoth}\left (c x\right ) + a e\right )} \log \left (-c^{2} x^{2} + 1\right )}{x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccoth(c*x))*(d+e*log(-c^2*x^2+1))/x^2,x, algorithm="fricas")

[Out]

integral((b*d*arccoth(c*x) + a*d + (b*e*arccoth(c*x) + a*e)*log(-c^2*x^2 + 1))/x^2, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b \operatorname{acoth}{\left (c x \right )}\right ) \left (d + e \log{\left (- c^{2} x^{2} + 1 \right )}\right )}{x^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*acoth(c*x))*(d+e*ln(-c**2*x**2+1))/x**2,x)

[Out]

Integral((a + b*acoth(c*x))*(d + e*log(-c**2*x**2 + 1))/x**2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccoth(c*x))*(d+e*log(-c^2*x^2+1))/x^2,x, algorithm="giac")

[Out]

integrate((b*arccoth(c*x) + a)*(e*log(-c^2*x^2 + 1) + d)/x^2, x)