3.270 \(\int \frac {(a+b \coth ^{-1}(c x)) (d+e \log (1-c^2 x^2))}{x^3} \, dx\)

Optimal. Leaf size=247 \[ -\frac {\left (a+b \coth ^{-1}(c x)\right ) \left (e \log \left (1-c^2 x^2\right )+d\right )}{2 x^2}+\frac {1}{2} c^2 e (a+b) \log (1-c x)+\frac {1}{2} c^2 e (a-b) \log (c x+1)-a c^2 e \log (x)-\frac {b c \left (e \log \left (1-c^2 x^2\right )+d\right )}{2 x}+\frac {1}{2} b c^2 \tanh ^{-1}(c x) \left (e \log \left (1-c^2 x^2\right )+d\right )+\frac {1}{2} b c^2 e \text {Li}_2\left (1-\frac {2}{1-c x}\right )+\frac {1}{2} b c^2 e \text {Li}_2\left (\frac {2}{c x+1}-1\right )-\frac {1}{2} b c^2 e \tanh ^{-1}(c x)^2-\frac {1}{2} b c^2 e \coth ^{-1}(c x)^2+b c^2 e \log \left (\frac {2}{1-c x}\right ) \tanh ^{-1}(c x)-b c^2 e \log \left (2-\frac {2}{c x+1}\right ) \coth ^{-1}(c x) \]

[Out]

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

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Rubi [A]  time = 0.49, antiderivative size = 247, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 13, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.482, Rules used = {5917, 325, 206, 6086, 6725, 801, 5989, 5933, 2447, 5984, 5918, 2402, 2315} \[ \frac {1}{2} b c^2 e \text {PolyLog}\left (2,1-\frac {2}{1-c x}\right )+\frac {1}{2} b c^2 e \text {PolyLog}\left (2,\frac {2}{c x+1}-1\right )-\frac {\left (a+b \coth ^{-1}(c x)\right ) \left (e \log \left (1-c^2 x^2\right )+d\right )}{2 x^2}+\frac {1}{2} c^2 e (a+b) \log (1-c x)+\frac {1}{2} c^2 e (a-b) \log (c x+1)-a c^2 e \log (x)-\frac {b c \left (e \log \left (1-c^2 x^2\right )+d\right )}{2 x}+\frac {1}{2} b c^2 \tanh ^{-1}(c x) \left (e \log \left (1-c^2 x^2\right )+d\right )-\frac {1}{2} b c^2 e \tanh ^{-1}(c x)^2-\frac {1}{2} b c^2 e \coth ^{-1}(c x)^2+b c^2 e \log \left (\frac {2}{1-c x}\right ) \tanh ^{-1}(c x)-b c^2 e \log \left (2-\frac {2}{c x+1}\right ) \coth ^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

Int[((a + b*ArcCoth[c*x])*(d + e*Log[1 - c^2*x^2]))/x^3,x]

[Out]

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

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 801

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
(d + e*x)^m*(f + g*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

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 2402

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> -Dist[e/g, Subst[Int[Log[2*d*x]/(1 - 2*
d*x), x], x, 1/(d + e*x)], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]

Rule 2447

Int[Log[u_]*(Pq_)^(m_.), x_Symbol] :> With[{C = FullSimplify[(Pq^m*(1 - u))/D[u, x]]}, Simp[C*PolyLog[2, 1 - u
], x] /; FreeQ[C, x]] /; IntegerQ[m] && PolyQ[Pq, x] && RationalFunctionQ[u, x] && LeQ[RationalFunctionExponen
ts[u, x][[2]], Expon[Pq, x]]

Rule 5917

Int[((a_.) + ArcCoth[(c_.)*(x_)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcC
oth[c*x])^p)/(d*(m + 1)), x] - Dist[(b*c*p)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcCoth[c*x])^(p - 1))/(1 -
c^2*x^2), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[p, 0] && (EqQ[p, 1] || IntegerQ[m]) && NeQ[m, -1]

Rule 5918

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

Rule 5933

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

Rule 5984

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

Rule 5989

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

Rule 6086

Int[((a_.) + ArcCoth[(c_.)*(x_)]*(b_.))*((d_.) + Log[(f_.) + (g_.)*(x_)^2]*(e_.))*(x_)^(m_.), x_Symbol] :> Wit
h[{u = IntHide[x^m*(a + b*ArcCoth[c*x]), x]}, Dist[d + e*Log[f + g*x^2], u, x] - Dist[2*e*g, Int[ExpandIntegra
nd[(x*u)/(f + g*x^2), x], x], x]] /; FreeQ[{a, b, c, d, e, f, g}, x] && IntegerQ[m] && NeQ[m, -1]

Rule 6725

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 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^3} \, dx &=-\frac {b c \left (d+e \log \left (1-c^2 x^2\right )\right )}{2 x}-\frac {\left (a+b \coth ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right )}{2 x^2}+\frac {1}{2} b c^2 \tanh ^{-1}(c x) \left (d+e \log \left (1-c^2 x^2\right )\right )+\left (2 c^2 e\right ) \int \left (\frac {a+b c x+b \coth ^{-1}(c x)}{2 x \left (-1+c^2 x^2\right )}-\frac {b c^2 x \tanh ^{-1}(c x)}{2 \left (-1+c^2 x^2\right )}\right ) \, dx\\ &=-\frac {b c \left (d+e \log \left (1-c^2 x^2\right )\right )}{2 x}-\frac {\left (a+b \coth ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right )}{2 x^2}+\frac {1}{2} b c^2 \tanh ^{-1}(c x) \left (d+e \log \left (1-c^2 x^2\right )\right )+\left (c^2 e\right ) \int \frac {a+b c x+b \coth ^{-1}(c x)}{x \left (-1+c^2 x^2\right )} \, dx-\left (b c^4 e\right ) \int \frac {x \tanh ^{-1}(c x)}{-1+c^2 x^2} \, dx\\ &=-\frac {1}{2} b c^2 e \tanh ^{-1}(c x)^2-\frac {b c \left (d+e \log \left (1-c^2 x^2\right )\right )}{2 x}-\frac {\left (a+b \coth ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right )}{2 x^2}+\frac {1}{2} b c^2 \tanh ^{-1}(c x) \left (d+e \log \left (1-c^2 x^2\right )\right )+\left (c^2 e\right ) \int \left (\frac {a+b c x}{x \left (-1+c^2 x^2\right )}+\frac {b \coth ^{-1}(c x)}{x \left (-1+c^2 x^2\right )}\right ) \, dx+\left (b c^3 e\right ) \int \frac {\tanh ^{-1}(c x)}{1-c x} \, dx\\ &=-\frac {1}{2} b c^2 e \tanh ^{-1}(c x)^2+b c^2 e \tanh ^{-1}(c x) \log \left (\frac {2}{1-c x}\right )-\frac {b c \left (d+e \log \left (1-c^2 x^2\right )\right )}{2 x}-\frac {\left (a+b \coth ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right )}{2 x^2}+\frac {1}{2} b c^2 \tanh ^{-1}(c x) \left (d+e \log \left (1-c^2 x^2\right )\right )+\left (c^2 e\right ) \int \frac {a+b c x}{x \left (-1+c^2 x^2\right )} \, dx+\left (b c^2 e\right ) \int \frac {\coth ^{-1}(c x)}{x \left (-1+c^2 x^2\right )} \, dx-\left (b c^3 e\right ) \int \frac {\log \left (\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx\\ &=-\frac {1}{2} b c^2 e \coth ^{-1}(c x)^2-\frac {1}{2} b c^2 e \tanh ^{-1}(c x)^2+b c^2 e \tanh ^{-1}(c x) \log \left (\frac {2}{1-c x}\right )-\frac {b c \left (d+e \log \left (1-c^2 x^2\right )\right )}{2 x}-\frac {\left (a+b \coth ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right )}{2 x^2}+\frac {1}{2} b c^2 \tanh ^{-1}(c x) \left (d+e \log \left (1-c^2 x^2\right )\right )+\left (c^2 e\right ) \int \left (-\frac {a}{x}+\frac {(a+b) c}{2 (-1+c x)}+\frac {(a-b) c}{2 (1+c x)}\right ) \, dx-\left (b c^2 e\right ) \int \frac {\coth ^{-1}(c x)}{x (1+c x)} \, dx+\left (b c^2 e\right ) \operatorname {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1-c x}\right )\\ &=-\frac {1}{2} b c^2 e \coth ^{-1}(c x)^2-\frac {1}{2} b c^2 e \tanh ^{-1}(c x)^2-a c^2 e \log (x)+b c^2 e \tanh ^{-1}(c x) \log \left (\frac {2}{1-c x}\right )+\frac {1}{2} (a+b) c^2 e \log (1-c x)+\frac {1}{2} (a-b) c^2 e \log (1+c x)-\frac {b c \left (d+e \log \left (1-c^2 x^2\right )\right )}{2 x}-\frac {\left (a+b \coth ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right )}{2 x^2}+\frac {1}{2} b c^2 \tanh ^{-1}(c x) \left (d+e \log \left (1-c^2 x^2\right )\right )-b c^2 e \coth ^{-1}(c x) \log \left (2-\frac {2}{1+c x}\right )+\frac {1}{2} b c^2 e \text {Li}_2\left (1-\frac {2}{1-c x}\right )+\left (b c^3 e\right ) \int \frac {\log \left (2-\frac {2}{1+c x}\right )}{1-c^2 x^2} \, dx\\ &=-\frac {1}{2} b c^2 e \coth ^{-1}(c x)^2-\frac {1}{2} b c^2 e \tanh ^{-1}(c x)^2-a c^2 e \log (x)+b c^2 e \tanh ^{-1}(c x) \log \left (\frac {2}{1-c x}\right )+\frac {1}{2} (a+b) c^2 e \log (1-c x)+\frac {1}{2} (a-b) c^2 e \log (1+c x)-\frac {b c \left (d+e \log \left (1-c^2 x^2\right )\right )}{2 x}-\frac {\left (a+b \coth ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right )}{2 x^2}+\frac {1}{2} b c^2 \tanh ^{-1}(c x) \left (d+e \log \left (1-c^2 x^2\right )\right )-b c^2 e \coth ^{-1}(c x) \log \left (2-\frac {2}{1+c x}\right )+\frac {1}{2} b c^2 e \text {Li}_2\left (1-\frac {2}{1-c x}\right )+\frac {1}{2} b c^2 e \text {Li}_2\left (-1+\frac {2}{1+c x}\right )\\ \end {align*}

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Mathematica [A]  time = 0.16, size = 161, normalized size = 0.65 \[ \frac {1}{2} \left (-\frac {e \log \left (1-c^2 x^2\right ) \left (a+\left (b-b c^2 x^2\right ) \coth ^{-1}(c x)+b c x\right )}{x^2}+c^2 e (a+b) \log (1-c x)+c^2 e (a-b) \log (c x+1)-2 a c^2 e \log (x)-\frac {a d}{x^2}-b c^2 e \left (\text {Li}_2\left (-\frac {1}{c x}\right )-\text {Li}_2\left (\frac {1}{c x}\right )\right )-\frac {b d \left (c x (c x \log (1-c x)-c x \log (c x+1)+2)+2 \coth ^{-1}(c x)\right )}{2 x^2}\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[((a + b*ArcCoth[c*x])*(d + e*Log[1 - c^2*x^2]))/x^3,x]

[Out]

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

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fricas [F]  time = 0.87, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b d \operatorname {arcoth}\left (c x\right ) + a d + {\left (b e \operatorname {arcoth}\left (c x\right ) + a e\right )} \log \left (-c^{2} x^{2} + 1\right )}{x^{3}}, x\right ) \]

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^3,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^3, x)

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

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^3,x, algorithm="giac")

[Out]

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

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maple [F]  time = 11.44, size = 0, normalized size = 0.00 \[ \int \frac {\left (a +b \,\mathrm {arccoth}\left (c x \right )\right ) \left (d +e \ln \left (-c^{2} x^{2}+1\right )\right )}{x^{3}}\, dx \]

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^3,x)

[Out]

int((a+b*arccoth(c*x))*(d+e*ln(-c^2*x^2+1))/x^3,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{4} \, {\left ({\left (c \log \left (c x + 1\right ) - c \log \left (c x - 1\right ) - \frac {2}{x}\right )} c - \frac {2 \, \operatorname {arcoth}\left (c x\right )}{x^{2}}\right )} b d + \frac {1}{2} \, {\left (c^{2} {\left (\log \left (c^{2} x^{2} - 1\right ) - \log \left (x^{2}\right )\right )} - \frac {\log \left (-c^{2} x^{2} + 1\right )}{x^{2}}\right )} a e - \frac {1}{4} \, b e {\left (\frac {\log \left (c x + 1\right )^{2}}{x^{2}} - 2 \, \int -\frac {{\left (c x + 1\right )} \log \left (c x - 1\right )^{2} - {\left (i \, \pi + {\left (i \, \pi c + c\right )} x\right )} \log \left (c x + 1\right ) - {\left (-i \, \pi - i \, \pi c x\right )} \log \left (c x - 1\right )}{c x^{4} + x^{3}}\,{d x}\right )} - \frac {a d}{2 \, x^{2}} \]

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^3,x, algorithm="maxima")

[Out]

1/4*((c*log(c*x + 1) - c*log(c*x - 1) - 2/x)*c - 2*arccoth(c*x)/x^2)*b*d + 1/2*(c^2*(log(c^2*x^2 - 1) - log(x^
2)) - log(-c^2*x^2 + 1)/x^2)*a*e - 1/4*b*e*(log(c*x + 1)^2/x^2 - 2*integrate(-((c*x + 1)*log(c*x - 1)^2 - (I*p
i + (I*pi*c + c)*x)*log(c*x + 1) - (-I*pi - I*pi*c*x)*log(c*x - 1))/(c*x^4 + x^3), x)) - 1/2*a*d/x^2

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\left (a+b\,\mathrm {acoth}\left (c\,x\right )\right )\,\left (d+e\,\ln \left (1-c^2\,x^2\right )\right )}{x^3} \,d x \]

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^3,x)

[Out]

int(((a + b*acoth(c*x))*(d + e*log(1 - c^2*x^2)))/x^3, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b \operatorname {acoth}{\left (c x \right )}\right ) \left (d + e \log {\left (- c^{2} x^{2} + 1 \right )}\right )}{x^{3}}\, dx \]

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**3,x)

[Out]

Integral((a + b*acoth(c*x))*(d + e*log(-c**2*x**2 + 1))/x**3, x)

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