3.3.75 \(\int \frac {(a+b \coth ^{-1}(c x)) (d+e \log (1-c^2 x^2))}{x^2} \, dx\) [275]

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

[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 = 0.16, antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {6229, 2525, 2458, 2379, 2438, 6096} \begin {gather*} -\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 ) \end {gather*}

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) - ((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

Rule 2379

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

Rule 2438

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]

Rule 2458

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 2525

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 6096

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 6229

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) \text {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 \text {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 \text {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) \text {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) \text {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] Leaf count is larger than twice the leaf count of optimal. \(332\) vs. \(2(105)=210\).
time = 0.13, size = 332, normalized size = 3.16 \begin {gather*} -\frac {4 a d+4 b d \coth ^{-1}(c x)+4 b c e x \coth ^{-1}(c x)^2+8 a c e x \tanh ^{-1}(c x)-4 b c d x \log (x)-b c e x \log ^2\left (-\frac {1}{c}+x\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 (-\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)+4 a e \log \left (1-c^2 x^2\right )+2 b c d x \log \left (1-c^2 x^2\right )+4 b e \coth ^{-1}(c 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 (-\frac {1}{c}+x\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 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} (1+c x)\right )}{4 x} \end {gather*}

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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Maple [F]
time = 1.08, 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 \begin {gather*} \text {Failed to integrate} \end {gather*}

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*(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))*e - a*d/x

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \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 \end {gather*}

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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