\(\int x (a+b \coth ^{-1}(c x)) (d+e \log (1-c^2 x^2)) \, dx\) [268]

Optimal result

Integrand size = 25, antiderivative size = 140 \[ \int x \left (a+b \coth ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right ) \, dx=\frac {b (d-e) x}{2 c}-\frac {b e x}{c}+\frac {1}{2} d x^2 \left (a+b \coth ^{-1}(c x)\right )-\frac {1}{2} e x^2 \left (a+b \coth ^{-1}(c x)\right )-\frac {b (d-e) \text {arctanh}(c x)}{2 c^2}+\frac {b e \text {arctanh}(c x)}{c^2}+\frac {b e x \log \left (1-c^2 x^2\right )}{2 c}-\frac {e \left (1-c^2 x^2\right ) \left (a+b \coth ^{-1}(c x)\right ) \log \left (1-c^2 x^2\right )}{2 c^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {2504, 2436, 2332, 6231, 327, 213, 2498, 212} \[ \int x \left (a+b \coth ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right ) \, dx=-\frac {e \left (1-c^2 x^2\right ) \log \left (1-c^2 x^2\right ) \left (a+b \coth ^{-1}(c x)\right )}{2 c^2}+\frac {1}{2} d x^2 \left (a+b \coth ^{-1}(c x)\right )-\frac {1}{2} e x^2 \left (a+b \coth ^{-1}(c x)\right )-\frac {b (d-e) \text {arctanh}(c x)}{2 c^2}+\frac {b e \text {arctanh}(c x)}{c^2}+\frac {b e x \log \left (1-c^2 x^2\right )}{2 c}+\frac {b x (d-e)}{2 c}-\frac {b e x}{c} \]

[In]

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

[Out]

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

Rule 212

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

Rule 213

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

Rule 327

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

Rule 2332

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 2436

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

Rule 2498

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)], x_Symbol] :> Simp[x*Log[c*(d + e*x^n)^p], x] - Dist[e*n*p, Int[
x^n/(d + e*x^n), x], x] /; FreeQ[{c, d, e, n, p}, x]

Rule 2504

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[I
nt[x^(Simplify[(m + 1)/n] - 1)*(a + b*Log[c*(d + e*x)^p])^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n, p,
q}, x] && IntegerQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0] || IGtQ[q, 0]) &&  !(EqQ[q, 1] && ILtQ[n, 0] &&
IGtQ[m, 0])

Rule 6231

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

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} d x^2 \left (a+b \coth ^{-1}(c x)\right )-\frac {1}{2} e x^2 \left (a+b \coth ^{-1}(c x)\right )-\frac {e \left (1-c^2 x^2\right ) \left (a+b \coth ^{-1}(c x)\right ) \log \left (1-c^2 x^2\right )}{2 c^2}-(b c) \int \left (-\frac {(d-e) x^2}{2 \left (-1+c^2 x^2\right )}-\frac {e \log \left (1-c^2 x^2\right )}{2 c^2}\right ) \, dx \\ & = \frac {1}{2} d x^2 \left (a+b \coth ^{-1}(c x)\right )-\frac {1}{2} e x^2 \left (a+b \coth ^{-1}(c x)\right )-\frac {e \left (1-c^2 x^2\right ) \left (a+b \coth ^{-1}(c x)\right ) \log \left (1-c^2 x^2\right )}{2 c^2}+\frac {1}{2} (b c (d-e)) \int \frac {x^2}{-1+c^2 x^2} \, dx+\frac {(b e) \int \log \left (1-c^2 x^2\right ) \, dx}{2 c} \\ & = \frac {b (d-e) x}{2 c}+\frac {1}{2} d x^2 \left (a+b \coth ^{-1}(c x)\right )-\frac {1}{2} e x^2 \left (a+b \coth ^{-1}(c x)\right )+\frac {b e x \log \left (1-c^2 x^2\right )}{2 c}-\frac {e \left (1-c^2 x^2\right ) \left (a+b \coth ^{-1}(c x)\right ) \log \left (1-c^2 x^2\right )}{2 c^2}+\frac {(b (d-e)) \int \frac {1}{-1+c^2 x^2} \, dx}{2 c}+(b c e) \int \frac {x^2}{1-c^2 x^2} \, dx \\ & = \frac {b (d-e) x}{2 c}-\frac {b e x}{c}+\frac {1}{2} d x^2 \left (a+b \coth ^{-1}(c x)\right )-\frac {1}{2} e x^2 \left (a+b \coth ^{-1}(c x)\right )-\frac {b (d-e) \text {arctanh}(c x)}{2 c^2}+\frac {b e x \log \left (1-c^2 x^2\right )}{2 c}-\frac {e \left (1-c^2 x^2\right ) \left (a+b \coth ^{-1}(c x)\right ) \log \left (1-c^2 x^2\right )}{2 c^2}+\frac {(b e) \int \frac {1}{1-c^2 x^2} \, dx}{c} \\ & = \frac {b (d-e) x}{2 c}-\frac {b e x}{c}+\frac {1}{2} d x^2 \left (a+b \coth ^{-1}(c x)\right )-\frac {1}{2} e x^2 \left (a+b \coth ^{-1}(c x)\right )-\frac {b (d-e) \text {arctanh}(c x)}{2 c^2}+\frac {b e \text {arctanh}(c x)}{c^2}+\frac {b e x \log \left (1-c^2 x^2\right )}{2 c}-\frac {e \left (1-c^2 x^2\right ) \left (a+b \coth ^{-1}(c x)\right ) \log \left (1-c^2 x^2\right )}{2 c^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.92 \[ \int x \left (a+b \coth ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right ) \, dx=\frac {2 b c (d-3 e) x+2 a c^2 (d-e) x^2+2 b c^2 (d-e) x^2 \coth ^{-1}(c x)+(b (d-3 e)-2 a e) \log (1-c x)-(b (d-3 e)+2 a e) \log (1+c x)+2 e \left (c x (b+a c x)+b \left (-1+c^2 x^2\right ) \coth ^{-1}(c x)\right ) \log \left (1-c^2 x^2\right )}{4 c^2} \]

[In]

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

[Out]

(2*b*c*(d - 3*e)*x + 2*a*c^2*(d - e)*x^2 + 2*b*c^2*(d - e)*x^2*ArcCoth[c*x] + (b*(d - 3*e) - 2*a*e)*Log[1 - c*
x] - (b*(d - 3*e) + 2*a*e)*Log[1 + c*x] + 2*e*(c*x*(b + a*c*x) + b*(-1 + c^2*x^2)*ArcCoth[c*x])*Log[1 - c^2*x^
2])/(4*c^2)

Maple [A] (verified)

Time = 1.63 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.24

method	result	size
parallelrisch	\(\frac {\ln \left (-c^{2} x^{2}+1\right ) \operatorname {arccoth}\left (c x \right ) b \,c^{2} e \,x^{2}+\operatorname {arccoth}\left (c x \right ) b \,c^{2} d \,x^{2}-\operatorname {arccoth}\left (c x \right ) b \,c^{2} e \,x^{2}+\ln \left (-c^{2} x^{2}+1\right ) a \,c^{2} e \,x^{2}+a \,c^{2} d \,x^{2}-a \,c^{2} e \,x^{2}+\ln \left (-c^{2} x^{2}+1\right ) b c e x +b c d x -3 b x e c -\operatorname {arccoth}\left (c x \right ) \ln \left (-c^{2} x^{2}+1\right ) b e -\operatorname {arccoth}\left (c x \right ) b d +3 \,\operatorname {arccoth}\left (c x \right ) b e -\ln \left (-c^{2} x^{2}+1\right ) a e}{2 c^{2}}\)	\(174\)
default	\(\text {Expression too large to display}\)	\(2474\)
parts	\(\text {Expression too large to display}\)	\(2474\)
risch	\(\text {Expression too large to display}\)	\(6696\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.99 \[ \int x \left (a+b \coth ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right ) \, dx=\frac {2 \, {\left (a c^{2} d - a c^{2} e\right )} x^{2} + 2 \, {\left (b c d - 3 \, b c e\right )} x + 2 \, {\left (a c^{2} e x^{2} + b c e x - a e\right )} \log \left (-c^{2} x^{2} + 1\right ) + {\left ({\left (b c^{2} d - b c^{2} e\right )} x^{2} - b d + 3 \, b e + {\left (b c^{2} e x^{2} - b e\right )} \log \left (-c^{2} x^{2} + 1\right )\right )} \log \left (\frac {c x + 1}{c x - 1}\right )}{4 \, c^{2}} \]

[In]

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

[Out]

1/4*(2*(a*c^2*d - a*c^2*e)*x^2 + 2*(b*c*d - 3*b*c*e)*x + 2*(a*c^2*e*x^2 + b*c*e*x - a*e)*log(-c^2*x^2 + 1) + (
(b*c^2*d - b*c^2*e)*x^2 - b*d + 3*b*e + (b*c^2*e*x^2 - b*e)*log(-c^2*x^2 + 1))*log((c*x + 1)/(c*x - 1)))/c^2

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.56 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.49 \[ \int x \left (a+b \coth ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right ) \, dx=\begin {cases} \frac {a d x^{2}}{2} + \frac {a e x^{2} \log {\left (- c^{2} x^{2} + 1 \right )}}{2} - \frac {a e x^{2}}{2} - \frac {a e \log {\left (- c^{2} x^{2} + 1 \right )}}{2 c^{2}} + \frac {b d x^{2} \operatorname {acoth}{\left (c x \right )}}{2} + \frac {b e x^{2} \log {\left (- c^{2} x^{2} + 1 \right )} \operatorname {acoth}{\left (c x \right )}}{2} - \frac {b e x^{2} \operatorname {acoth}{\left (c x \right )}}{2} + \frac {b d x}{2 c} + \frac {b e x \log {\left (- c^{2} x^{2} + 1 \right )}}{2 c} - \frac {3 b e x}{2 c} - \frac {b d \operatorname {acoth}{\left (c x \right )}}{2 c^{2}} - \frac {b e \log {\left (- c^{2} x^{2} + 1 \right )} \operatorname {acoth}{\left (c x \right )}}{2 c^{2}} + \frac {3 b e \operatorname {acoth}{\left (c x \right )}}{2 c^{2}} & \text {for}\: c \neq 0 \\\frac {d x^{2} \left (a + \frac {i \pi b}{2}\right )}{2} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((a*d*x**2/2 + a*e*x**2*log(-c**2*x**2 + 1)/2 - a*e*x**2/2 - a*e*log(-c**2*x**2 + 1)/(2*c**2) + b*d*x
**2*acoth(c*x)/2 + b*e*x**2*log(-c**2*x**2 + 1)*acoth(c*x)/2 - b*e*x**2*acoth(c*x)/2 + b*d*x/(2*c) + b*e*x*log
(-c**2*x**2 + 1)/(2*c) - 3*b*e*x/(2*c) - b*d*acoth(c*x)/(2*c**2) - b*e*log(-c**2*x**2 + 1)*acoth(c*x)/(2*c**2)
+ 3*b*e*acoth(c*x)/(2*c**2), Ne(c, 0)), (d*x**2*(a + I*pi*b/2)/2, True))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.22 \[ \int x \left (a+b \coth ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right ) \, dx=\frac {1}{2} \, a d x^{2} + \frac {1}{4} \, {\left (2 \, x^{2} \operatorname {arcoth}\left (c x\right ) + c {\left (\frac {2 \, x}{c^{2}} - \frac {\log \left (c x + 1\right )}{c^{3}} + \frac {\log \left (c x - 1\right )}{c^{3}}\right )}\right )} b d - \frac {{\left (c^{2} x^{2} - {\left (c^{2} x^{2} - 1\right )} \log \left (-c^{2} x^{2} + 1\right ) - 1\right )} b e \operatorname {arcoth}\left (c x\right )}{2 \, c^{2}} - \frac {{\left (c^{2} x^{2} - {\left (c^{2} x^{2} - 1\right )} \log \left (-c^{2} x^{2} + 1\right ) - 1\right )} a e}{2 \, c^{2}} - \frac {{\left (3 \, c x - {\left (c x + 1\right )} \log \left (c x + 1\right ) - {\left (c x - 1\right )} \log \left (-c x + 1\right )\right )} b e}{2 \, c^{2}} \]

[In]

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

[Out]

1/2*a*d*x^2 + 1/4*(2*x^2*arccoth(c*x) + c*(2*x/c^2 - log(c*x + 1)/c^3 + log(c*x - 1)/c^3))*b*d - 1/2*(c^2*x^2
- (c^2*x^2 - 1)*log(-c^2*x^2 + 1) - 1)*b*e*arccoth(c*x)/c^2 - 1/2*(c^2*x^2 - (c^2*x^2 - 1)*log(-c^2*x^2 + 1) -
1)*a*e/c^2 - 1/2*(3*c*x - (c*x + 1)*log(c*x + 1) - (c*x - 1)*log(-c*x + 1))*b*e/c^2

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.39 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.72 \[ \int x \left (a+b \coth ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right ) \, dx=-\frac {1}{4} \, b e x^{2} \log \left (-c x + 1\right )^{2} - \frac {1}{4} \, {\left (-i \, \pi b d + i \, \pi b e - 2 \, a d + 2 \, a e\right )} x^{2} + \frac {1}{4} \, {\left (b e x^{2} - \frac {b e}{c^{2}}\right )} \log \left (c x + 1\right )^{2} - \frac {1}{4} \, {\left ({\left (-i \, \pi b e - b d - 2 \, a e + b e\right )} x^{2} - \frac {2 \, b e x}{c}\right )} \log \left (c x + 1\right ) - \frac {b e \log \left (c x - 1\right )^{2}}{4 \, c^{2}} - \frac {1}{4} \, {\left ({\left (-i \, \pi b e + b d - 2 \, a e - b e\right )} x^{2} - \frac {2 \, b e x}{c} - \frac {2 \, b e \log \left (c x - 1\right )}{c^{2}}\right )} \log \left (-c x + 1\right ) + \frac {{\left (b d - 3 \, b e\right )} x}{2 \, c} + \frac {{\left (-i \, \pi b e - b d - 2 \, a e + 3 \, b e\right )} \log \left (c x + 1\right )}{4 \, c^{2}} + \frac {{\left (-i \, \pi b e + b d - 2 \, a e - 3 \, b e\right )} \log \left (c x - 1\right )}{4 \, c^{2}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.60 (sec) , antiderivative size = 329, normalized size of antiderivative = 2.35 \[ \int x \left (a+b \coth ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right ) \, dx=\ln \left (1-\frac {1}{c\,x}\right )\,\left (\frac {\frac {b\,d\,x^3}{2}-\frac {b\,c^2\,d\,x^5}{2}}{2\,\left (c\,x^2+x\right )\,\left (c\,x-1\right )}-\frac {\frac {b\,e\,x^3}{2}-\frac {b\,c^2\,e\,x^5}{2}}{2\,\left (c\,x^2+x\right )\,\left (c\,x-1\right )}+\frac {\ln \left (1-c^2\,x^2\right )\,\left (\frac {b\,e\,x^3}{2}-\frac {b\,c^2\,e\,x^5}{2}\right )}{2\,\left (c\,x^2+x\right )\,\left (c\,x-1\right )}-\frac {b\,e\,\ln \left (1-c^2\,x^2\right )\,\left (x-c^2\,x^3\right )}{4\,c^2\,\left (c\,x^2+x\right )\,\left (c\,x-1\right )}\right )+\ln \left (1-c^2\,x^2\right )\,\left (\frac {a\,e\,x^2}{2}+\frac {b\,e\,x}{2\,c}\right )-\ln \left (\frac {1}{c\,x}+1\right )\,\left (\ln \left (1-c^2\,x^2\right )\,\left (\frac {b\,e}{4\,c^2}-\frac {b\,e\,x^2}{4}\right )-\frac {b\,d\,x^2}{4}+\frac {b\,e\,x^2}{4}\right )+\frac {a\,x^2\,\left (d-e\right )}{2}-\frac {\ln \left (c\,x+1\right )\,\left (2\,a\,e+b\,d-3\,b\,e\right )}{4\,c^2}-\frac {\ln \left (c\,x-1\right )\,\left (2\,a\,e-b\,d+3\,b\,e\right )}{4\,c^2}+\frac {b\,x\,\left (d-3\,e\right )}{2\,c} \]

[In]

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

[Out]

log(1 - 1/(c*x))*(((b*d*x^3)/2 - (b*c^2*d*x^5)/2)/(2*(x + c*x^2)*(c*x - 1)) - ((b*e*x^3)/2 - (b*c^2*e*x^5)/2)/
(2*(x + c*x^2)*(c*x - 1)) + (log(1 - c^2*x^2)*((b*e*x^3)/2 - (b*c^2*e*x^5)/2))/(2*(x + c*x^2)*(c*x - 1)) - (b*
e*log(1 - c^2*x^2)*(x - c^2*x^3))/(4*c^2*(x + c*x^2)*(c*x - 1))) + log(1 - c^2*x^2)*((a*e*x^2)/2 + (b*e*x)/(2*
c)) - log(1/(c*x) + 1)*(log(1 - c^2*x^2)*((b*e)/(4*c^2) - (b*e*x^2)/4) - (b*d*x^2)/4 + (b*e*x^2)/4) + (a*x^2*(
d - e))/2 - (log(c*x + 1)*(2*a*e + b*d - 3*b*e))/(4*c^2) - (log(c*x - 1)*(2*a*e - b*d + 3*b*e))/(4*c^2) + (b*x
*(d - 3*e))/(2*c)