Integrand size = 27, antiderivative size = 225 \[ \int x^3 (a+b \text {arctanh}(c x)) \left (d+e \log \left (1-c^2 x^2\right )\right ) \, dx=\frac {b (2 d-3 e) x}{8 c^3}-\frac {2 b e x}{3 c^3}+\frac {b (2 d-e) x^3}{24 c}-\frac {b e x^3}{18 c}-\frac {b (2 d-3 e) \text {arctanh}(c x)}{8 c^4}+\frac {2 b e \text {arctanh}(c x)}{3 c^4}-\frac {e x^2 (a+b \text {arctanh}(c x))}{4 c^2}-\frac {1}{8} e x^4 (a+b \text {arctanh}(c x))+\frac {b e x \log \left (1-c^2 x^2\right )}{4 c^3}+\frac {b e x^3 \log \left (1-c^2 x^2\right )}{12 c}-\frac {e (a+b \text {arctanh}(c x)) \log \left (1-c^2 x^2\right )}{4 c^4}+\frac {1}{4} x^4 (a+b \text {arctanh}(c x)) \left (d+e \log \left (1-c^2 x^2\right )\right ) \]
1/8*b*(2*d-3*e)*x/c^3-2/3*b*e*x/c^3+1/24*b*(2*d-e)*x^3/c-1/18*b*e*x^3/c-1/ 8*b*(2*d-3*e)*arctanh(c*x)/c^4+2/3*b*e*arctanh(c*x)/c^4-1/4*e*x^2*(a+b*arc tanh(c*x))/c^2-1/8*e*x^4*(a+b*arctanh(c*x))+1/4*b*e*x*ln(-c^2*x^2+1)/c^3+1 /12*b*e*x^3*ln(-c^2*x^2+1)/c-1/4*e*(a+b*arctanh(c*x))*ln(-c^2*x^2+1)/c^4+1 /4*x^4*(a+b*arctanh(c*x))*(d+e*ln(-c^2*x^2+1))
Time = 0.09 (sec) , antiderivative size = 192, normalized size of antiderivative = 0.85 \[ \int x^3 (a+b \text {arctanh}(c x)) \left (d+e \log \left (1-c^2 x^2\right )\right ) \, dx=\frac {6 b c (6 d-25 e) x-36 a c^2 e x^2+2 b c^3 (6 d-7 e) x^3+18 a c^4 (2 d-e) x^4-18 b c^2 x^2 \left (-2 c^2 d x^2+e \left (2+c^2 x^2\right )\right ) \text {arctanh}(c x)+3 (6 b d-12 a e-25 b e) \log (1-c x)-3 (6 b d+12 a e-25 b e) \log (1+c x)+12 e \left (3 a c^4 x^4+b c x \left (3+c^2 x^2\right )+3 b \left (-1+c^4 x^4\right ) \text {arctanh}(c x)\right ) \log \left (1-c^2 x^2\right )}{144 c^4} \]
(6*b*c*(6*d - 25*e)*x - 36*a*c^2*e*x^2 + 2*b*c^3*(6*d - 7*e)*x^3 + 18*a*c^ 4*(2*d - e)*x^4 - 18*b*c^2*x^2*(-2*c^2*d*x^2 + e*(2 + c^2*x^2))*ArcTanh[c* x] + 3*(6*b*d - 12*a*e - 25*b*e)*Log[1 - c*x] - 3*(6*b*d + 12*a*e - 25*b*e )*Log[1 + c*x] + 12*e*(3*a*c^4*x^4 + b*c*x*(3 + c^2*x^2) + 3*b*(-1 + c^4*x ^4)*ArcTanh[c*x])*Log[1 - c^2*x^2])/(144*c^4)
Time = 0.50 (sec) , antiderivative size = 222, normalized size of antiderivative = 0.99, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {6645, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int x^3 (a+b \text {arctanh}(c x)) \left (e \log \left (1-c^2 x^2\right )+d\right ) \, dx\) |
| \(\Big \downarrow \) 6645 |
| \(\displaystyle -b c \int \left (-\frac {\left (2 e-c^2 (2 d-e) x^2\right ) x^2}{8 c^2 \left (1-c^2 x^2\right )}-\frac {e \left (c^2 x^2+1\right ) \log \left (1-c^2 x^2\right )}{4 c^4}\right )dx+\frac {1}{4} x^4 (a+b \text {arctanh}(c x)) \left (e \log \left (1-c^2 x^2\right )+d\right )-\frac {e x^2 (a+b \text {arctanh}(c x))}{4 c^2}-\frac {e \log \left (1-c^2 x^2\right ) (a+b \text {arctanh}(c x))}{4 c^4}-\frac {1}{8} e x^4 (a+b \text {arctanh}(c x))\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{4} x^4 (a+b \text {arctanh}(c x)) \left (e \log \left (1-c^2 x^2\right )+d\right )-\frac {e x^2 (a+b \text {arctanh}(c x))}{4 c^2}-\frac {e \log \left (1-c^2 x^2\right ) (a+b \text {arctanh}(c x))}{4 c^4}-\frac {1}{8} e x^4 (a+b \text {arctanh}(c x))-b c \left (\frac {(2 d-3 e) \text {arctanh}(c x)}{8 c^5}-\frac {2 e \text {arctanh}(c x)}{3 c^5}-\frac {x (2 d-3 e)}{8 c^4}+\frac {2 e x}{3 c^4}-\frac {x^3 (2 d-e)}{24 c^2}+\frac {e x^3}{18 c^2}-\frac {e x^3 \log \left (1-c^2 x^2\right )}{12 c^2}-\frac {e x \log \left (1-c^2 x^2\right )}{4 c^4}\right )\) |
-1/4*(e*x^2*(a + b*ArcTanh[c*x]))/c^2 - (e*x^4*(a + b*ArcTanh[c*x]))/8 - ( e*(a + b*ArcTanh[c*x])*Log[1 - c^2*x^2])/(4*c^4) + (x^4*(a + b*ArcTanh[c*x ])*(d + e*Log[1 - c^2*x^2]))/4 - b*c*(-1/8*((2*d - 3*e)*x)/c^4 + (2*e*x)/( 3*c^4) - ((2*d - e)*x^3)/(24*c^2) + (e*x^3)/(18*c^2) + ((2*d - 3*e)*ArcTan h[c*x])/(8*c^5) - (2*e*ArcTanh[c*x])/(3*c^5) - (e*x*Log[1 - c^2*x^2])/(4*c ^4) - (e*x^3*Log[1 - c^2*x^2])/(12*c^2))
3.6.23.3.1 Defintions of rubi rules used
Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))*((d_.) + Log[(f_.) + (g_.)*(x_)^2]* (e_.))*(x_)^(m_.), x_Symbol] :> With[{u = IntHide[x^m*(d + e*Log[f + g*x^2] ), x]}, Simp[(a + b*ArcTanh[c*x]) u, x] - Simp[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]
Time = 1.38 (sec) , antiderivative size = 249, normalized size of antiderivative = 1.11
| method | result | size |
| parallelrisch | \(\frac {-18 \,\operatorname {arctanh}\left (c x \right ) b d -18 \ln \left (-c^{2} x^{2}+1\right ) a e -18 \,\operatorname {arctanh}\left (c x \right ) \ln \left (-c^{2} x^{2}+1\right ) b e -18 a \,c^{2} e \,x^{2}+18 b c d x +18 a e \ln \left (-c^{2} x^{2}+1\right ) x^{4} c^{4}+18 b \,\operatorname {arctanh}\left (c x \right ) x^{4} c^{4} d -9 b \,\operatorname {arctanh}\left (c x \right ) x^{4} c^{4} e +6 b e \ln \left (-c^{2} x^{2}+1\right ) x^{3} c^{3}-75 x b e c -18 \,\operatorname {arctanh}\left (c x \right ) b \,c^{2} e \,x^{2}+18 \ln \left (-c^{2} x^{2}+1\right ) b c e x +75 \,\operatorname {arctanh}\left (c x \right ) b e -9 a \,c^{4} e \,x^{4}-7 b \,c^{3} e \,x^{3}-18 a e +6 b \,c^{3} d \,x^{3}+18 a \,c^{4} d \,x^{4}+18 b e \ln \left (-c^{2} x^{2}+1\right ) \operatorname {arctanh}\left (c x \right ) x^{4} c^{4}}{72 c^{4}}\) | \(249\) |
| default | \(\text {Expression too large to display}\) | \(3609\) |
| parts | \(\text {Expression too large to display}\) | \(3609\) |
| risch | \(\text {Expression too large to display}\) | \(8859\) |
1/72*(-18*arctanh(c*x)*b*d-18*ln(-c^2*x^2+1)*a*e-18*arctanh(c*x)*ln(-c^2*x ^2+1)*b*e-18*a*c^2*e*x^2+18*b*c*d*x+18*a*e*ln(-c^2*x^2+1)*x^4*c^4+18*b*arc tanh(c*x)*x^4*c^4*d-9*b*arctanh(c*x)*x^4*c^4*e+6*b*e*ln(-c^2*x^2+1)*x^3*c^ 3-75*x*b*e*c-18*arctanh(c*x)*b*c^2*e*x^2+18*ln(-c^2*x^2+1)*b*c*e*x+75*arct anh(c*x)*b*e-9*a*c^4*e*x^4-7*b*c^3*e*x^3-18*a*e+6*b*c^3*d*x^3+18*a*c^4*d*x ^4+18*b*e*ln(-c^2*x^2+1)*arctanh(c*x)*x^4*c^4)/c^4
Time = 0.25 (sec) , antiderivative size = 196, normalized size of antiderivative = 0.87 \[ \int x^3 (a+b \text {arctanh}(c x)) \left (d+e \log \left (1-c^2 x^2\right )\right ) \, dx=-\frac {36 \, a c^{2} e x^{2} - 18 \, {\left (2 \, a c^{4} d - a c^{4} e\right )} x^{4} - 2 \, {\left (6 \, b c^{3} d - 7 \, b c^{3} e\right )} x^{3} - 6 \, {\left (6 \, b c d - 25 \, b c e\right )} x - 12 \, {\left (3 \, a c^{4} e x^{4} + b c^{3} e x^{3} + 3 \, b c e x - 3 \, a e\right )} \log \left (-c^{2} x^{2} + 1\right ) + 3 \, {\left (6 \, b c^{2} e x^{2} - 3 \, {\left (2 \, b c^{4} d - b c^{4} e\right )} x^{4} + 6 \, b d - 25 \, b e - 6 \, {\left (b c^{4} e x^{4} - b e\right )} \log \left (-c^{2} x^{2} + 1\right )\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{144 \, c^{4}} \]
-1/144*(36*a*c^2*e*x^2 - 18*(2*a*c^4*d - a*c^4*e)*x^4 - 2*(6*b*c^3*d - 7*b *c^3*e)*x^3 - 6*(6*b*c*d - 25*b*c*e)*x - 12*(3*a*c^4*e*x^4 + b*c^3*e*x^3 + 3*b*c*e*x - 3*a*e)*log(-c^2*x^2 + 1) + 3*(6*b*c^2*e*x^2 - 3*(2*b*c^4*d - b*c^4*e)*x^4 + 6*b*d - 25*b*e - 6*(b*c^4*e*x^4 - b*e)*log(-c^2*x^2 + 1))*l og(-(c*x + 1)/(c*x - 1)))/c^4
Time = 1.41 (sec) , antiderivative size = 279, normalized size of antiderivative = 1.24 \[ \int x^3 (a+b \text {arctanh}(c x)) \left (d+e \log \left (1-c^2 x^2\right )\right ) \, dx=\begin {cases} \frac {a d x^{4}}{4} + \frac {a e x^{4} \log {\left (- c^{2} x^{2} + 1 \right )}}{4} - \frac {a e x^{4}}{8} - \frac {a e x^{2}}{4 c^{2}} - \frac {a e \log {\left (- c^{2} x^{2} + 1 \right )}}{4 c^{4}} + \frac {b d x^{4} \operatorname {atanh}{\left (c x \right )}}{4} + \frac {b e x^{4} \log {\left (- c^{2} x^{2} + 1 \right )} \operatorname {atanh}{\left (c x \right )}}{4} - \frac {b e x^{4} \operatorname {atanh}{\left (c x \right )}}{8} + \frac {b d x^{3}}{12 c} + \frac {b e x^{3} \log {\left (- c^{2} x^{2} + 1 \right )}}{12 c} - \frac {7 b e x^{3}}{72 c} - \frac {b e x^{2} \operatorname {atanh}{\left (c x \right )}}{4 c^{2}} + \frac {b d x}{4 c^{3}} + \frac {b e x \log {\left (- c^{2} x^{2} + 1 \right )}}{4 c^{3}} - \frac {25 b e x}{24 c^{3}} - \frac {b d \operatorname {atanh}{\left (c x \right )}}{4 c^{4}} - \frac {b e \log {\left (- c^{2} x^{2} + 1 \right )} \operatorname {atanh}{\left (c x \right )}}{4 c^{4}} + \frac {25 b e \operatorname {atanh}{\left (c x \right )}}{24 c^{4}} & \text {for}\: c \neq 0 \\\frac {a d x^{4}}{4} & \text {otherwise} \end {cases} \]
Piecewise((a*d*x**4/4 + a*e*x**4*log(-c**2*x**2 + 1)/4 - a*e*x**4/8 - a*e* x**2/(4*c**2) - a*e*log(-c**2*x**2 + 1)/(4*c**4) + b*d*x**4*atanh(c*x)/4 + b*e*x**4*log(-c**2*x**2 + 1)*atanh(c*x)/4 - b*e*x**4*atanh(c*x)/8 + b*d*x **3/(12*c) + b*e*x**3*log(-c**2*x**2 + 1)/(12*c) - 7*b*e*x**3/(72*c) - b*e *x**2*atanh(c*x)/(4*c**2) + b*d*x/(4*c**3) + b*e*x*log(-c**2*x**2 + 1)/(4* c**3) - 25*b*e*x/(24*c**3) - b*d*atanh(c*x)/(4*c**4) - b*e*log(-c**2*x**2 + 1)*atanh(c*x)/(4*c**4) + 25*b*e*atanh(c*x)/(24*c**4), Ne(c, 0)), (a*d*x* *4/4, True))
Result contains complex when optimal does not.
Time = 0.19 (sec) , antiderivative size = 271, normalized size of antiderivative = 1.20 \[ \int x^3 (a+b \text {arctanh}(c x)) \left (d+e \log \left (1-c^2 x^2\right )\right ) \, dx=\frac {1}{4} \, a d x^{4} + \frac {1}{8} \, {\left (2 \, x^{4} \log \left (-c^{2} x^{2} + 1\right ) - c^{2} {\left (\frac {c^{2} x^{4} + 2 \, x^{2}}{c^{4}} + \frac {2 \, \log \left (c^{2} x^{2} - 1\right )}{c^{6}}\right )}\right )} b e \operatorname {artanh}\left (c x\right ) + \frac {1}{24} \, {\left (6 \, x^{4} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {2 \, {\left (c^{2} x^{3} + 3 \, x\right )}}{c^{4}} - \frac {3 \, \log \left (c x + 1\right )}{c^{5}} + \frac {3 \, \log \left (c x - 1\right )}{c^{5}}\right )}\right )} b d + \frac {1}{8} \, {\left (2 \, x^{4} \log \left (-c^{2} x^{2} + 1\right ) - c^{2} {\left (\frac {c^{2} x^{4} + 2 \, x^{2}}{c^{4}} + \frac {2 \, \log \left (c^{2} x^{2} - 1\right )}{c^{6}}\right )}\right )} a e - \frac {{\left (2 \, {\left (-6 i \, \pi c^{3} + 7 \, c^{3}\right )} x^{3} + 6 \, {\left (-6 i \, \pi c + 25 \, c\right )} x + 3 \, {\left (6 i \, \pi - 4 \, c^{3} x^{3} - 12 \, c x - 25\right )} \log \left (c x + 1\right ) + 3 \, {\left (-6 i \, \pi - 4 \, c^{3} x^{3} - 12 \, c x + 25\right )} \log \left (c x - 1\right )\right )} b e}{144 \, c^{4}} \]
1/4*a*d*x^4 + 1/8*(2*x^4*log(-c^2*x^2 + 1) - c^2*((c^2*x^4 + 2*x^2)/c^4 + 2*log(c^2*x^2 - 1)/c^6))*b*e*arctanh(c*x) + 1/24*(6*x^4*arctanh(c*x) + c*( 2*(c^2*x^3 + 3*x)/c^4 - 3*log(c*x + 1)/c^5 + 3*log(c*x - 1)/c^5))*b*d + 1/ 8*(2*x^4*log(-c^2*x^2 + 1) - c^2*((c^2*x^4 + 2*x^2)/c^4 + 2*log(c^2*x^2 - 1)/c^6))*a*e - 1/144*(2*(-6*I*pi*c^3 + 7*c^3)*x^3 + 6*(-6*I*pi*c + 25*c)*x + 3*(6*I*pi - 4*c^3*x^3 - 12*c*x - 25)*log(c*x + 1) + 3*(-6*I*pi - 4*c^3* x^3 - 12*c*x + 25)*log(c*x - 1))*b*e/c^4
Exception generated. \[ \int x^3 (a+b \text {arctanh}(c x)) \left (d+e \log \left (1-c^2 x^2\right )\right ) \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value
Time = 4.57 (sec) , antiderivative size = 851, normalized size of antiderivative = 3.78 \[ \int x^3 (a+b \text {arctanh}(c x)) \left (d+e \log \left (1-c^2 x^2\right )\right ) \, dx={\ln \left (1-c\,x\right )}^2\,\left (\frac {b\,e}{8\,c^4}-\frac {b\,e\,x^4}{8}\right )-{\ln \left (c\,x+1\right )}^2\,\left (\frac {b\,e}{8\,c^4}-\frac {b\,e\,x^4}{8}\right )+\ln \left (1-c\,x\right )\,\left (\frac {x^4\,\left (a\,e-\frac {b\,d}{2}+\frac {b\,e}{4}+\frac {b\,e\,\left (\ln \left (c\,x+1\right )+\ln \left (1-c\,x\right )-\ln \left (1-c^2\,x^2\right )\right )}{2}\right )}{4}-\frac {x^2\,\left (\frac {16\,a\,e-8\,b\,d+8\,b\,e\,\left (\ln \left (c\,x+1\right )+\ln \left (1-c\,x\right )-\ln \left (1-c^2\,x^2\right )\right )}{c}-\frac {16\,a\,e-8\,b\,d+4\,b\,e+8\,b\,e\,\left (\ln \left (c\,x+1\right )+\ln \left (1-c\,x\right )-\ln \left (1-c^2\,x^2\right )\right )}{c}\right )}{32\,c}+\frac {b\,e\,x}{4\,c^3}+\frac {b\,e\,x^3}{12\,c}\right )-x^2\,\left (\frac {a\,\left (e-2\,d+2\,e\,\left (\ln \left (c\,x+1\right )+\ln \left (1-c\,x\right )-\ln \left (1-c^2\,x^2\right )\right )\right )}{4\,c^2}+\frac {a\,\left (d-e\,\left (\ln \left (c\,x+1\right )+\ln \left (1-c\,x\right )-\ln \left (1-c^2\,x^2\right )\right )\right )}{2\,c^2}\right )-x\,\left (\frac {b\,\left (7\,e-6\,d+6\,e\,\left (\ln \left (c\,x+1\right )+\ln \left (1-c\,x\right )-\ln \left (1-c^2\,x^2\right )\right )\right )}{24\,c^3}+\frac {3\,b\,e}{4\,c^3}\right )-\frac {a\,x^4\,\left (e-2\,d+2\,e\,\left (\ln \left (c\,x+1\right )+\ln \left (1-c\,x\right )-\ln \left (1-c^2\,x^2\right )\right )\right )}{8}-\frac {\ln \left (\frac {x\,\left (12\,a\,e-6\,b\,d+25\,b\,e+6\,b\,e\,\left (\ln \left (c\,x+1\right )+\ln \left (1-c\,x\right )-\ln \left (1-c^2\,x^2\right )\right )\right )}{24\,c^2}-\frac {25\,b\,e-6\,b\,d+6\,b\,e\,\left (\ln \left (c\,x+1\right )+\ln \left (1-c\,x\right )-\ln \left (1-c^2\,x^2\right )\right )}{24\,c^3}-\frac {a\,e\,x}{2\,c^2}\right )\,\left (12\,a\,e-6\,b\,d+25\,b\,e+6\,b\,e\,\left (\ln \left (c\,x+1\right )+\ln \left (1-c\,x\right )-\ln \left (1-c^2\,x^2\right )\right )\right )}{48\,c^4}-\frac {\ln \left (\frac {x\,\left (12\,a\,e+6\,b\,d-25\,b\,e-6\,b\,e\,\left (\ln \left (c\,x+1\right )+\ln \left (1-c\,x\right )-\ln \left (1-c^2\,x^2\right )\right )\right )}{24\,c^2}-\frac {25\,b\,e-6\,b\,d+6\,b\,e\,\left (\ln \left (c\,x+1\right )+\ln \left (1-c\,x\right )-\ln \left (1-c^2\,x^2\right )\right )}{24\,c^3}-\frac {a\,e\,x}{2\,c^2}\right )\,\left (12\,a\,e+6\,b\,d-25\,b\,e-6\,b\,e\,\left (\ln \left (c\,x+1\right )+\ln \left (1-c\,x\right )-\ln \left (1-c^2\,x^2\right )\right )\right )}{48\,c^4}+c\,\ln \left (c\,x+1\right )\,\left (\frac {x^4\,\left (4\,a\,e+2\,b\,d-b\,e-2\,b\,e\,\left (\ln \left (c\,x+1\right )+\ln \left (1-c\,x\right )-\ln \left (1-c^2\,x^2\right )\right )\right )}{16\,c}+\frac {b\,e\,x}{4\,c^4}+\frac {b\,e\,x^3}{12\,c^2}-\frac {b\,e\,x^2}{8\,c^3}\right )-\frac {b\,x^3\,\left (7\,e-6\,d+6\,e\,\left (\ln \left (c\,x+1\right )+\ln \left (1-c\,x\right )-\ln \left (1-c^2\,x^2\right )\right )\right )}{72\,c} \]
log(1 - c*x)^2*((b*e)/(8*c^4) - (b*e*x^4)/8) - log(c*x + 1)^2*((b*e)/(8*c^ 4) - (b*e*x^4)/8) + log(1 - c*x)*((x^4*(a*e - (b*d)/2 + (b*e)/4 + (b*e*(lo g(c*x + 1) + log(1 - c*x) - log(1 - c^2*x^2)))/2))/4 - (x^2*((16*a*e - 8*b *d + 8*b*e*(log(c*x + 1) + log(1 - c*x) - log(1 - c^2*x^2)))/c - (16*a*e - 8*b*d + 4*b*e + 8*b*e*(log(c*x + 1) + log(1 - c*x) - log(1 - c^2*x^2)))/c ))/(32*c) + (b*e*x)/(4*c^3) + (b*e*x^3)/(12*c)) - x^2*((a*(e - 2*d + 2*e*( log(c*x + 1) + log(1 - c*x) - log(1 - c^2*x^2))))/(4*c^2) + (a*(d - e*(log (c*x + 1) + log(1 - c*x) - log(1 - c^2*x^2))))/(2*c^2)) - x*((b*(7*e - 6*d + 6*e*(log(c*x + 1) + log(1 - c*x) - log(1 - c^2*x^2))))/(24*c^3) + (3*b* e)/(4*c^3)) - (a*x^4*(e - 2*d + 2*e*(log(c*x + 1) + log(1 - c*x) - log(1 - c^2*x^2))))/8 - (log((x*(12*a*e - 6*b*d + 25*b*e + 6*b*e*(log(c*x + 1) + log(1 - c*x) - log(1 - c^2*x^2))))/(24*c^2) - (25*b*e - 6*b*d + 6*b*e*(log (c*x + 1) + log(1 - c*x) - log(1 - c^2*x^2)))/(24*c^3) - (a*e*x)/(2*c^2))* (12*a*e - 6*b*d + 25*b*e + 6*b*e*(log(c*x + 1) + log(1 - c*x) - log(1 - c^ 2*x^2))))/(48*c^4) - (log((x*(12*a*e + 6*b*d - 25*b*e - 6*b*e*(log(c*x + 1 ) + log(1 - c*x) - log(1 - c^2*x^2))))/(24*c^2) - (25*b*e - 6*b*d + 6*b*e* (log(c*x + 1) + log(1 - c*x) - log(1 - c^2*x^2)))/(24*c^3) - (a*e*x)/(2*c^ 2))*(12*a*e + 6*b*d - 25*b*e - 6*b*e*(log(c*x + 1) + log(1 - c*x) - log(1 - c^2*x^2))))/(48*c^4) + c*log(c*x + 1)*((x^4*(4*a*e + 2*b*d - b*e - 2*b*e *(log(c*x + 1) + log(1 - c*x) - log(1 - c^2*x^2))))/(16*c) + (b*e*x)/(4...