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 \[ -\frac {\left (a+b \coth ^{-1}(c x)\right ) \left (e \log \left (1-c^2 x^2\right )+d\right )}{x}-\frac {c e \left (a+b \coth ^{-1}(c x)\right )^2}{b}+\frac {1}{2} b c \log \left (1-\frac {1}{1-c^2 x^2}\right ) \left (e \log \left (1-c^2 x^2\right )+d\right )-\frac {1}{2} b c e \text {Li}_2\left (\frac {1}{1-c^2 x^2}\right ) \]

[Out]

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

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Rubi [A]  time = 0.27, antiderivative size = 94, normalized size of antiderivative = 0.90, 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 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 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 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 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 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 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 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]

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]

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*}

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Mathematica [B]  time = 0.21, size = 332, normalized size = 3.16 \[ -\frac {4 a e \log \left (1-c^2 x^2\right )+8 a c e x \tanh ^{-1}(c x)+4 a d+2 b c d x \log \left (1-c^2 x^2\right )-4 b c e x \log (x) \log \left (1-c^2 x^2\right )+2 b c e x \log \left (x-\frac {1}{c}\right ) \log \left (1-c^2 x^2\right )+2 b c e x \log \left (\frac {1}{c}+x\right ) \log \left (1-c^2 x^2\right )+4 b e \log \left (1-c^2 x^2\right ) \coth ^{-1}(c x)-4 b c d x \log (x)+4 b d \coth ^{-1}(c x)+4 b c e x \text {Li}_2(-c x)+4 b c e x \text {Li}_2(c x)-2 b c e x \text {Li}_2\left (\frac {1}{2}-\frac {c x}{2}\right )-2 b c e x \text {Li}_2\left (\frac {1}{2} (c x+1)\right )-b c e x \log ^2\left (x-\frac {1}{c}\right )-b c e x \log ^2\left (\frac {1}{c}+x\right )-2 b c e x \log \left (\frac {1}{c}+x\right ) \log \left (\frac {1}{2} (1-c x)\right )+4 b c e x \log (x) \log (1-c x)-2 b c e x \log \left (x-\frac {1}{c}\right ) \log \left (\frac {1}{2} (c x+1)\right )+4 b c e x \log (x) \log (c x+1)+4 b c e x \coth ^{-1}(c x)^2}{4 x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*(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*Lo
g[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*Pol
yLog[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])/x

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fricas [F]  time = 0.57, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b d \operatorname {arcoth}\left (c x\right ) + a d + {\left (b e \operatorname {arcoth}\left (c x\right ) + a e\right )} \log \left (-c^{2} x^{2} + 1\right )}{x^{2}}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \operatorname {arcoth}\left (c x\right ) + a\right )} {\left (e \log \left (-c^{2} x^{2} + 1\right ) + d\right )}}{x^{2}}\,{d x} \]

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)

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maple [F]  time = 5.65, size = 0, normalized size = 0.00 \[ \int \frac {\left (a +b \,\mathrm {arccoth}\left (c x \right )\right ) \left (d +e \ln \left (-c^{2} x^{2}+1\right )\right )}{x^{2}}\, dx \]

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 = 0.00, size = 0, normalized size = 0.00 \[ -\frac {1}{2} \, {\left (c {\left (\log \left (c^{2} x^{2} - 1\right ) - \log \left (x^{2}\right )\right )} + \frac {2 \, \operatorname {arcoth}\left (c x\right )}{x}\right )} b d - {\left (c^{2} {\left (\frac {\log \left (c x + 1\right )}{c} - \frac {\log \left (c x - 1\right )}{c}\right )} + \frac {\log \left (-c^{2} x^{2} + 1\right )}{x}\right )} a e - \frac {1}{2} \, b e {\left (\frac {\log \left (c x + 1\right )^{2}}{x} - \int -\frac {{\left (c x + 1\right )} \log \left (c x - 1\right )^{2} - {\left (i \, \pi + {\left (i \, \pi c + 2 \, c\right )} x\right )} \log \left (c x + 1\right ) - {\left (-i \, \pi - i \, \pi c x\right )} \log \left (c x - 1\right )}{c x^{3} + x^{2}}\,{d x}\right )} - \frac {a d}{x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\left (a+b\,\mathrm {acoth}\left (c\,x\right )\right )\,\left (d+e\,\ln \left (1-c^2\,x^2\right )\right )}{x^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b \operatorname {acoth}{\left (c x \right )}\right ) \left (d + e \log {\left (- c^{2} x^{2} + 1 \right )}\right )}{x^{2}}\, dx \]

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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