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

Optimal. Leaf size=216 \[ -\frac {1}{2} a e \text {Li}_2\left (c^2 x^2\right )+a d \log (x)+\frac {1}{2} b e \text {Li}_2(-c x) \left (-\log \left (1-c^2 x^2\right )+\log (1-c x)+\log (c x+1)\right )-\frac {1}{2} b e \text {Li}_2(c x) \left (-\log \left (1-c^2 x^2\right )+\log (1-c x)+\log (c x+1)\right )-\frac {1}{2} b d \text {Li}_2(-c x)+\frac {1}{2} b d \text {Li}_2(c x)+b e \text {Li}_3(1-c x)-b e \text {Li}_3(c x+1)-b e \text {Li}_2(1-c x) \log (1-c x)+b e \text {Li}_2(c x+1) \log (c x+1)-\frac {1}{2} b e \log (c x) \log ^2(1-c x)+\frac {1}{2} b e \log (-c x) \log ^2(c x+1) \]

[Out]

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

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Rubi [A]  time = 0.28, antiderivative size = 216, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 9, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6079, 5912, 6077, 2391, 6075, 2396, 2433, 2374, 6589} \[ -\frac {1}{2} a e \text {PolyLog}\left (2,c^2 x^2\right )+\frac {1}{2} b e \left (-\log \left (1-c^2 x^2\right )+\log (1-c x)+\log (c x+1)\right ) \text {PolyLog}(2,-c x)-\frac {1}{2} b e \left (-\log \left (1-c^2 x^2\right )+\log (1-c x)+\log (c x+1)\right ) \text {PolyLog}(2,c x)-\frac {1}{2} b d \text {PolyLog}(2,-c x)+\frac {1}{2} b d \text {PolyLog}(2,c x)+b e \text {PolyLog}(3,1-c x)-b e \text {PolyLog}(3,c x+1)-b e \log (1-c x) \text {PolyLog}(2,1-c x)+b e \log (c x+1) \text {PolyLog}(2,c x+1)+a d \log (x)-\frac {1}{2} b e \log (c x) \log ^2(1-c x)+\frac {1}{2} b e \log (-c x) \log ^2(c x+1) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2374

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

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]

Rule 2396

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

Rule 2433

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_.) + Log[(h_.)*((i_.) + (j_.)*(x_))^(m_.)]*
(g_.))*((k_.) + (l_.)*(x_))^(r_.), x_Symbol] :> Dist[1/e, Subst[Int[((k*x)/d)^r*(a + b*Log[c*x^n])^p*(f + g*Lo
g[h*((e*i - d*j)/e + (j*x)/e)^m]), x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, h, i, j, k, l, n, p, r},
x] && EqQ[e*k - d*l, 0]

Rule 5912

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

Rule 6075

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

Rule 6077

Int[(Log[(f_.) + (g_.)*(x_)^2]*(ArcTanh[(c_.)*(x_)]*(b_.) + (a_)))/(x_), x_Symbol] :> Dist[a, Int[Log[f + g*x^
2]/x, x], x] + Dist[b, Int[(Log[f + g*x^2]*ArcTanh[c*x])/x, x], x] /; FreeQ[{a, b, c, f, g}, x]

Rule 6079

Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))*(Log[(f_.) + (g_.)*(x_)^2]*(e_.) + (d_)))/(x_), x_Symbol] :> Dist[d,
Int[(a + b*ArcTanh[c*x])/x, x], x] + Dist[e, Int[(Log[f + g*x^2]*(a + b*ArcTanh[c*x]))/x, x], x] /; FreeQ[{a,
b, c, d, e, f, g}, x]

Rule 6589

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

Rubi steps

\begin {align*} \int \frac {\left (a+b \tanh ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right )}{x} \, dx &=d \int \frac {a+b \tanh ^{-1}(c x)}{x} \, dx+e \int \frac {\left (a+b \tanh ^{-1}(c x)\right ) \log \left (1-c^2 x^2\right )}{x} \, dx\\ &=a d \log (x)-\frac {1}{2} b d \text {Li}_2(-c x)+\frac {1}{2} b d \text {Li}_2(c x)+(a e) \int \frac {\log \left (1-c^2 x^2\right )}{x} \, dx+(b e) \int \frac {\tanh ^{-1}(c x) \log \left (1-c^2 x^2\right )}{x} \, dx\\ &=a d \log (x)-\frac {1}{2} b d \text {Li}_2(-c x)+\frac {1}{2} b d \text {Li}_2(c x)-\frac {1}{2} a e \text {Li}_2\left (c^2 x^2\right )-\frac {1}{2} (b e) \int \frac {\log ^2(1-c x)}{x} \, dx+\frac {1}{2} (b e) \int \frac {\log ^2(1+c x)}{x} \, dx+\left (b e \left (-\log (1-c x)-\log (1+c x)+\log \left (1-c^2 x^2\right )\right )\right ) \int \frac {\tanh ^{-1}(c x)}{x} \, dx\\ &=a d \log (x)-\frac {1}{2} b e \log (c x) \log ^2(1-c x)+\frac {1}{2} b e \log (-c x) \log ^2(1+c x)-\frac {1}{2} b d \text {Li}_2(-c x)+\frac {1}{2} b e \left (\log (1-c x)+\log (1+c x)-\log \left (1-c^2 x^2\right )\right ) \text {Li}_2(-c x)+\frac {1}{2} b d \text {Li}_2(c x)-\frac {1}{2} b e \left (\log (1-c x)+\log (1+c x)-\log \left (1-c^2 x^2\right )\right ) \text {Li}_2(c x)-\frac {1}{2} a e \text {Li}_2\left (c^2 x^2\right )-(b c e) \int \frac {\log (c x) \log (1-c x)}{1-c x} \, dx-(b c e) \int \frac {\log (-c x) \log (1+c x)}{1+c x} \, dx\\ &=a d \log (x)-\frac {1}{2} b e \log (c x) \log ^2(1-c x)+\frac {1}{2} b e \log (-c x) \log ^2(1+c x)-\frac {1}{2} b d \text {Li}_2(-c x)+\frac {1}{2} b e \left (\log (1-c x)+\log (1+c x)-\log \left (1-c^2 x^2\right )\right ) \text {Li}_2(-c x)+\frac {1}{2} b d \text {Li}_2(c x)-\frac {1}{2} b e \left (\log (1-c x)+\log (1+c x)-\log \left (1-c^2 x^2\right )\right ) \text {Li}_2(c x)-\frac {1}{2} a e \text {Li}_2\left (c^2 x^2\right )+(b e) \operatorname {Subst}\left (\int \frac {\log (x) \log \left (c \left (\frac {1}{c}-\frac {x}{c}\right )\right )}{x} \, dx,x,1-c x\right )-(b e) \operatorname {Subst}\left (\int \frac {\log (x) \log \left (-c \left (-\frac {1}{c}+\frac {x}{c}\right )\right )}{x} \, dx,x,1+c x\right )\\ &=a d \log (x)-\frac {1}{2} b e \log (c x) \log ^2(1-c x)+\frac {1}{2} b e \log (-c x) \log ^2(1+c x)-\frac {1}{2} b d \text {Li}_2(-c x)+\frac {1}{2} b e \left (\log (1-c x)+\log (1+c x)-\log \left (1-c^2 x^2\right )\right ) \text {Li}_2(-c x)+\frac {1}{2} b d \text {Li}_2(c x)-\frac {1}{2} b e \left (\log (1-c x)+\log (1+c x)-\log \left (1-c^2 x^2\right )\right ) \text {Li}_2(c x)-\frac {1}{2} a e \text {Li}_2\left (c^2 x^2\right )-b e \log (1-c x) \text {Li}_2(1-c x)+b e \log (1+c x) \text {Li}_2(1+c x)+(b e) \operatorname {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,1-c x\right )-(b e) \operatorname {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,1+c x\right )\\ &=a d \log (x)-\frac {1}{2} b e \log (c x) \log ^2(1-c x)+\frac {1}{2} b e \log (-c x) \log ^2(1+c x)-\frac {1}{2} b d \text {Li}_2(-c x)+\frac {1}{2} b e \left (\log (1-c x)+\log (1+c x)-\log \left (1-c^2 x^2\right )\right ) \text {Li}_2(-c x)+\frac {1}{2} b d \text {Li}_2(c x)-\frac {1}{2} b e \left (\log (1-c x)+\log (1+c x)-\log \left (1-c^2 x^2\right )\right ) \text {Li}_2(c x)-\frac {1}{2} a e \text {Li}_2\left (c^2 x^2\right )-b e \log (1-c x) \text {Li}_2(1-c x)+b e \log (1+c x) \text {Li}_2(1+c x)+b e \text {Li}_3(1-c x)-b e \text {Li}_3(1+c x)\\ \end {align*}

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Mathematica [F]  time = 0.26, size = 0, normalized size = 0.00 \[ \int \frac {\left (a+b \tanh ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right )}{x} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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fricas [F]  time = 0.49, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b d \operatorname {artanh}\left (c x\right ) + a d + {\left (b e \operatorname {artanh}\left (c x\right ) + a e\right )} \log \left (-c^{2} x^{2} + 1\right )}{x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 10.98, size = 1638, normalized size = 7.58 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*e*polylog(3,-c*x+1)-b*e*polylog(3,c*x+1)+ln(c*x)*ln(c*x-1)*a*e-1/2*ln(c*x)*ln(c*x-1)*b*d-1/2*ln(c*x)*ln(c*x-
1)^2*b*e-polylog(2,-c*x+1)*ln(c*x-1)*b*e+1/2*b*e*ln(-c*x)*ln(c*x+1)^2+b*e*ln(c*x+1)*polylog(2,c*x+1)-1/4*I*ln(
c*x)*ln(c*x-1)*Pi*b*e*csgn(I*(c*x+1))*csgn(I*(c*x-1)*(c*x+1))^2+1/4*I*dilog(c*x)*Pi*b*e*csgn(I*(c*x-1))*csgn(I
*(c*x+1))*csgn(I*(c*x-1)*(c*x+1))-1/2*I*ln(c*x)*Pi*a*e*csgn(I*(c*x-1))*csgn(I*(c*x+1))*csgn(I*(c*x-1)*(c*x+1))
-1/4*I*ln(c*x)*ln(c*x-1)*Pi*b*e*csgn(I*(c*x-1))*csgn(I*(c*x-1)*(c*x+1))^2+dilog(c*x)*a*e-1/2*dilog(c*x)*b*d+ln
(c*x)*a*d+1/2*ln(c*x)*Pi^2*b*e-1/4*I*dilog(c*x)*Pi*b*e*csgn(I*(c*x-1)*(c*x+1))^3+1/4*I*ln(c*x)*ln(c*x-1)*Pi*b*
e*csgn(I*(c*x-1))*csgn(I*(c*x+1))*csgn(I*(c*x-1)*(c*x+1))+1/4*ln(c*x)*Pi^2*b*e*csgn(I*(c*x-1))^3*csgn(I*(c*x+1
))*csgn(I*(c*x-1)*(c*x+1))^2+I*ln(c*x)*Pi*a*e-1/4*I*ln(c*x)*ln(c*x-1)*Pi*b*e*csgn(I*(c*x-1)*(c*x+1))^3+1/4*ln(
c*x)*Pi^2*b*e*csgn(I*(c*x-1))^3*csgn(I*(c*x-1)*(c*x+1))^3+1/4*ln(c*x)*Pi^2*b*e*csgn(I*(c*x-1))^3*csgn(I*(c*x+1
))*csgn(I*(c*x-1)*(c*x+1))-(1/2*I*Pi*b*e*csgn(I*(c*x-1))^2-1/2*I*Pi*b*e*csgn(I*(c*x-1)*(c*x+1))^2-1/2*I*Pi*b*e
*csgn(I*(c*x-1))^3-1/4*I*Pi*b*e*csgn(I*(c*x-1))*csgn(I*(c*x+1))*csgn(I*(c*x-1)*(c*x+1))+1/4*I*Pi*b*e*csgn(I*(c
*x-1))*csgn(I*(c*x-1)*(c*x+1))^2+1/4*I*Pi*b*e*csgn(I*(c*x+1))*csgn(I*(c*x-1)*(c*x+1))^2+1/4*I*Pi*b*e*csgn(I*(c
*x-1)*(c*x+1))^3+a*e+1/2*b*d)*dilog(c*x+1)-1/4*ln(c*x)*Pi^2*b*e*csgn(I*(c*x-1))*csgn(I*(c*x+1))*csgn(I*(c*x-1)
*(c*x+1))+1/2*I*ln(c*x)*ln(c*x-1)*Pi*b*e*csgn(I*(c*x-1))^2+1/2*I*ln(c*x)*ln(c*x-1)*Pi*b*e*csgn(I*(c*x-1)*(c*x+
1))^2-1/2*I*ln(c*x)*ln(c*x-1)*Pi*b*e*csgn(I*(c*x-1))^3-1/4*ln(c*x)*Pi^2*b*e*csgn(I*(c*x-1))^2*csgn(I*(c*x-1)*(
c*x+1))^3-3/4*ln(c*x)*Pi^2*b*e*csgn(I*(c*x-1))^3*csgn(I*(c*x-1)*(c*x+1))^2-1/2*I*ln(c*x)*Pi*b*d-I*dilog(c*x)*P
i*b*e-1/4*ln(c*x)*Pi^2*b*e*csgn(I*(c*x-1))^2*csgn(I*(c*x+1))*csgn(I*(c*x-1)*(c*x+1))^2-1/4*ln(c*x)*Pi^2*b*e*cs
gn(I*(c*x-1))^4*csgn(I*(c*x+1))*csgn(I*(c*x-1)*(c*x+1))-1/4*I*dilog(c*x)*Pi*b*e*csgn(I*(c*x-1))*csgn(I*(c*x-1)
*(c*x+1))^2-1/4*I*dilog(c*x)*Pi*b*e*csgn(I*(c*x+1))*csgn(I*(c*x-1)*(c*x+1))^2+1/2*I*ln(c*x)*Pi*a*e*csgn(I*(c*x
-1))*csgn(I*(c*x-1)*(c*x+1))^2+1/2*I*ln(c*x)*Pi*a*e*csgn(I*(c*x+1))*csgn(I*(c*x-1)*(c*x+1))^2+1/2*I*ln(c*x)*Pi
*a*e*csgn(I*(c*x-1)*(c*x+1))^3-I*ln(c*x)*ln(c*x-1)*Pi*b*e+1/4*ln(c*x)*Pi^2*b*e*csgn(I*(c*x-1))*csgn(I*(c*x-1)*
(c*x+1))^2+1/4*ln(c*x)*Pi^2*b*e*csgn(I*(c*x+1))*csgn(I*(c*x-1)*(c*x+1))^2+1/4*ln(c*x)*Pi^2*b*e*csgn(I*(c*x-1))
^4*csgn(I*(c*x-1)*(c*x+1))^2+1/2*I*dilog(c*x)*Pi*b*e*csgn(I*(c*x-1))^2+1/2*I*dilog(c*x)*Pi*b*e*csgn(I*(c*x-1)*
(c*x+1))^2-I*ln(c*x)*Pi*a*e*csgn(I*(c*x-1)*(c*x+1))^2+1/2*I*ln(c*x)*Pi*b*d*csgn(I*(c*x-1))^2-1/2*I*ln(c*x)*Pi*
b*d*csgn(I*(c*x-1))^3-1/2*I*dilog(c*x)*Pi*b*e*csgn(I*(c*x-1))^3+1/2*ln(c*x)*Pi^2*b*e*csgn(I*(c*x-1))^3+1/4*ln(
c*x)*Pi^2*b*e*csgn(I*(c*x-1)*(c*x+1))^3+1/2*ln(c*x)*Pi^2*b*e*csgn(I*(c*x-1))^2*csgn(I*(c*x-1)*(c*x+1))^2-1/2*l
n(c*x)*Pi^2*b*e*csgn(I*(c*x-1)*(c*x+1))^2-1/2*ln(c*x)*Pi^2*b*e*csgn(I*(c*x-1))^2

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maxima [A]  time = 0.47, size = 152, normalized size = 0.70 \[ -\frac {1}{2} \, {\left (\log \left (c x\right ) \log \left (-c x + 1\right )^{2} + 2 \, {\rm Li}_2\left (-c x + 1\right ) \log \left (-c x + 1\right ) - 2 \, {\rm Li}_{3}(-c x + 1)\right )} b e + \frac {1}{2} \, {\left (\log \left (c x + 1\right )^{2} \log \left (-c x\right ) + 2 \, {\rm Li}_2\left (c x + 1\right ) \log \left (c x + 1\right ) - 2 \, {\rm Li}_{3}(c x + 1)\right )} b e + a d \log \relax (x) - \frac {1}{2} \, {\left (b d - 2 \, a e\right )} {\left (\log \left (c x\right ) \log \left (-c x + 1\right ) + {\rm Li}_2\left (-c x + 1\right )\right )} + \frac {1}{2} \, {\left (b d + 2 \, a e\right )} {\left (\log \left (c x + 1\right ) \log \left (-c x\right ) + {\rm Li}_2\left (c x + 1\right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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