Integrand size = 27, antiderivative size = 206 \[ \int x^2 (a+b \text {arctanh}(c x)) \left (d+e \log \left (1-c^2 x^2\right )\right ) \, dx=-\frac {2 a e x}{3 c^2}-\frac {5 b e x^2}{18 c}-\frac {2 b e x \text {arctanh}(c x)}{3 c^2}-\frac {2}{9} e x^3 (a+b \text {arctanh}(c x))+\frac {e (a+b \text {arctanh}(c x))^2}{3 b c^3}-\frac {11 b e \log \left (1-c^2 x^2\right )}{18 c^3}-\frac {b e \log ^2\left (1-c^2 x^2\right )}{12 c^3}+\frac {b x^2 \left (d+e \log \left (1-c^2 x^2\right )\right )}{6 c}+\frac {1}{3} x^3 (a+b \text {arctanh}(c x)) \left (d+e \log \left (1-c^2 x^2\right )\right )+\frac {b \log \left (1-c^2 x^2\right ) \left (d+e \log \left (1-c^2 x^2\right )\right )}{6 c^3} \] Output:
-2/3*a*e*x/c^2-5/18*b*e*x^2/c-2/3*b*e*x*arctanh(c*x)/c^2-2/9*e*x^3*(a+b*ar ctanh(c*x))+1/3*e*(a+b*arctanh(c*x))^2/b/c^3-11/18*b*e*ln(-c^2*x^2+1)/c^3- 1/12*b*e*ln(-c^2*x^2+1)^2/c^3+1/6*b*x^2*(d+e*ln(-c^2*x^2+1))/c+1/3*x^3*(a+ b*arctanh(c*x))*(d+e*ln(-c^2*x^2+1))+1/6*b*ln(-c^2*x^2+1)*(d+e*ln(-c^2*x^2 +1))/c^3
Time = 0.07 (sec) , antiderivative size = 183, normalized size of antiderivative = 0.89 \[ \int x^2 (a+b \text {arctanh}(c x)) \left (d+e \log \left (1-c^2 x^2\right )\right ) \, dx=\frac {-24 a c e x+2 b c^2 (3 d-5 e) x^2+4 a c^3 (3 d-2 e) x^3+4 b c x \left (3 c^2 d x^2-2 e \left (3+c^2 x^2\right )\right ) \text {arctanh}(c x)+12 b e \text {arctanh}(c x)^2+2 (3 b d-6 a e-11 b e) \log (1-c x)+2 (3 b d+6 a e-11 b e) \log (1+c x)+6 c^2 e x^2 (b+2 a c x+2 b c x \text {arctanh}(c x)) \log \left (1-c^2 x^2\right )+3 b e \log ^2\left (1-c^2 x^2\right )}{36 c^3} \] Input:
Integrate[x^2*(a + b*ArcTanh[c*x])*(d + e*Log[1 - c^2*x^2]),x]
Output:
(-24*a*c*e*x + 2*b*c^2*(3*d - 5*e)*x^2 + 4*a*c^3*(3*d - 2*e)*x^3 + 4*b*c*x *(3*c^2*d*x^2 - 2*e*(3 + c^2*x^2))*ArcTanh[c*x] + 12*b*e*ArcTanh[c*x]^2 + 2*(3*b*d - 6*a*e - 11*b*e)*Log[1 - c*x] + 2*(3*b*d + 6*a*e - 11*b*e)*Log[1 + c*x] + 6*c^2*e*x^2*(b + 2*a*c*x + 2*b*c*x*ArcTanh[c*x])*Log[1 - c^2*x^2 ] + 3*b*e*Log[1 - c^2*x^2]^2)/(36*c^3)
Time = 0.92 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.08, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {6647, 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.
| \(\displaystyle \int x^2 (a+b \text {arctanh}(c x)) \left (e \log \left (1-c^2 x^2\right )+d\right ) \, dx\) |
| \(\Big \downarrow \) 6647 |
| \(\displaystyle 2 c^2 e \int \left (\frac {(2 c x \text {arctanh}(c x) b+b+2 a c x) x^3}{6 c \left (1-c^2 x^2\right )}+\frac {b \log \left (1-c^2 x^2\right ) x}{6 c^3 \left (1-c^2 x^2\right )}\right )dx+\frac {1}{3} x^3 (a+b \text {arctanh}(c x)) \left (e \log \left (1-c^2 x^2\right )+d\right )+\frac {b x^2 \left (e \log \left (1-c^2 x^2\right )+d\right )}{6 c}+\frac {b \log \left (1-c^2 x^2\right ) \left (e \log \left (1-c^2 x^2\right )+d\right )}{6 c^3}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{3} x^3 (a+b \text {arctanh}(c x)) \left (e \log \left (1-c^2 x^2\right )+d\right )+2 c^2 e \left (\frac {a \text {arctanh}(c x)}{3 c^5}-\frac {a x}{3 c^4}-\frac {a x^3}{9 c^2}+\frac {b \text {arctanh}(c x)^2}{6 c^5}-\frac {b x \text {arctanh}(c x)}{3 c^4}-\frac {b x^3 \text {arctanh}(c x)}{9 c^2}-\frac {5 b x^2}{36 c^3}-\frac {b \log ^2\left (1-c^2 x^2\right )}{24 c^5}-\frac {11 b \log \left (1-c^2 x^2\right )}{36 c^5}\right )+\frac {b x^2 \left (e \log \left (1-c^2 x^2\right )+d\right )}{6 c}+\frac {b \log \left (1-c^2 x^2\right ) \left (e \log \left (1-c^2 x^2\right )+d\right )}{6 c^3}\) |
Input:
Int[x^2*(a + b*ArcTanh[c*x])*(d + e*Log[1 - c^2*x^2]),x]
Output:
(b*x^2*(d + e*Log[1 - c^2*x^2]))/(6*c) + (x^3*(a + b*ArcTanh[c*x])*(d + e* Log[1 - c^2*x^2]))/3 + (b*Log[1 - c^2*x^2]*(d + e*Log[1 - c^2*x^2]))/(6*c^ 3) + 2*c^2*e*(-1/3*(a*x)/c^4 - (5*b*x^2)/(36*c^3) - (a*x^3)/(9*c^2) + (a*A rcTanh[c*x])/(3*c^5) - (b*x*ArcTanh[c*x])/(3*c^4) - (b*x^3*ArcTanh[c*x])/( 9*c^2) + (b*ArcTanh[c*x]^2)/(6*c^5) - (11*b*Log[1 - c^2*x^2])/(36*c^5) - ( b*Log[1 - c^2*x^2]^2)/(24*c^5))
Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))*((d_.) + Log[(f_.) + (g_.)*(x_)^2]* (e_.))*(x_)^(m_.), x_Symbol] :> With[{u = IntHide[x^m*(a + b*ArcTanh[c*x]), x]}, Simp[(d + e*Log[f + g*x^2]) u, x] - Simp[2*e*g Int[ExpandIntegran d[x*(u/(f + g*x^2)), x], x], x]] /; FreeQ[{a, b, c, d, e, f, g}, x] && Inte gerQ[m] && NeQ[m, -1]
Time = 3.06 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.11
| method | result | size |
| parallelrisch | \(\frac {-44 \,\operatorname {arctanh}\left (c x \right ) b e +12 a \,c^{3} d \,x^{3}+6 b \,c^{2} d \,x^{2}+12 \ln \left (c x -1\right ) b d +12 \,\operatorname {arctanh}\left (c x \right ) b d +12 d b \,\operatorname {arctanh}\left (c x \right ) x^{3} c^{3}-24 a c e x -8 a \,c^{3} e \,x^{3}-10 b \,c^{2} e \,x^{2}+24 \,\operatorname {arctanh}\left (c x \right ) a e +12 e b \operatorname {arctanh}\left (c x \right )^{2}+3 e b \ln \left (-c^{2} x^{2}+1\right )^{2}-8 e b \,\operatorname {arctanh}\left (c x \right ) x^{3} c^{3}-24 e b \,\operatorname {arctanh}\left (c x \right ) x c +6 x^{2} \ln \left (-c^{2} x^{2}+1\right ) b \,c^{2} e +12 x^{3} \operatorname {arctanh}\left (c x \right ) \ln \left (-c^{2} x^{2}+1\right ) b e \,c^{3}-44 \ln \left (c x -1\right ) b e +12 x^{3} \ln \left (-c^{2} x^{2}+1\right ) a e \,c^{3}}{36 c^{3}}\) | \(229\) |
| default | \(\text {Expression too large to display}\) | \(3856\) |
| parts | \(\text {Expression too large to display}\) | \(3856\) |
| risch | \(\text {Expression too large to display}\) | \(4015\) |
Input:
int(x^2*(a+b*arctanh(c*x))*(d+e*ln(-c^2*x^2+1)),x,method=_RETURNVERBOSE)
Output:
1/36*(-44*arctanh(c*x)*b*e+12*a*c^3*d*x^3+6*b*c^2*d*x^2+12*ln(c*x-1)*b*d+1 2*arctanh(c*x)*b*d+12*d*b*arctanh(c*x)*x^3*c^3-24*a*c*e*x-8*a*c^3*e*x^3-10 *b*c^2*e*x^2+24*arctanh(c*x)*a*e+12*e*b*arctanh(c*x)^2+3*e*b*ln(-c^2*x^2+1 )^2-8*e*b*arctanh(c*x)*x^3*c^3-24*e*b*arctanh(c*x)*x*c+6*x^2*ln(-c^2*x^2+1 )*b*c^2*e+12*x^3*arctanh(c*x)*ln(-c^2*x^2+1)*b*e*c^3-44*ln(c*x-1)*b*e+12*x ^3*ln(-c^2*x^2+1)*a*e*c^3)/c^3
Time = 0.11 (sec) , antiderivative size = 200, normalized size of antiderivative = 0.97 \[ \int x^2 (a+b \text {arctanh}(c x)) \left (d+e \log \left (1-c^2 x^2\right )\right ) \, dx=-\frac {24 \, a c e x - 4 \, {\left (3 \, a c^{3} d - 2 \, a c^{3} e\right )} x^{3} - 3 \, b e \log \left (-c^{2} x^{2} + 1\right )^{2} - 3 \, b e \log \left (-\frac {c x + 1}{c x - 1}\right )^{2} - 2 \, {\left (3 \, b c^{2} d - 5 \, b c^{2} e\right )} x^{2} - 2 \, {\left (6 \, a c^{3} e x^{3} + 3 \, b c^{2} e x^{2} + 3 \, b d - 11 \, b e\right )} \log \left (-c^{2} x^{2} + 1\right ) - 2 \, {\left (3 \, b c^{3} e x^{3} \log \left (-c^{2} x^{2} + 1\right ) - 6 \, b c e x + {\left (3 \, b c^{3} d - 2 \, b c^{3} e\right )} x^{3} + 6 \, a e\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{36 \, c^{3}} \] Input:
integrate(x^2*(a+b*arctanh(c*x))*(d+e*log(-c^2*x^2+1)),x, algorithm="frica s")
Output:
-1/36*(24*a*c*e*x - 4*(3*a*c^3*d - 2*a*c^3*e)*x^3 - 3*b*e*log(-c^2*x^2 + 1 )^2 - 3*b*e*log(-(c*x + 1)/(c*x - 1))^2 - 2*(3*b*c^2*d - 5*b*c^2*e)*x^2 - 2*(6*a*c^3*e*x^3 + 3*b*c^2*e*x^2 + 3*b*d - 11*b*e)*log(-c^2*x^2 + 1) - 2*( 3*b*c^3*e*x^3*log(-c^2*x^2 + 1) - 6*b*c*e*x + (3*b*c^3*d - 2*b*c^3*e)*x^3 + 6*a*e)*log(-(c*x + 1)/(c*x - 1)))/c^3
Time = 0.90 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.25 \[ \int x^2 (a+b \text {arctanh}(c x)) \left (d+e \log \left (1-c^2 x^2\right )\right ) \, dx=\begin {cases} \frac {a d x^{3}}{3} + \frac {a e x^{3} \log {\left (- c^{2} x^{2} + 1 \right )}}{3} - \frac {2 a e x^{3}}{9} - \frac {2 a e x}{3 c^{2}} + \frac {2 a e \operatorname {atanh}{\left (c x \right )}}{3 c^{3}} + \frac {b d x^{3} \operatorname {atanh}{\left (c x \right )}}{3} + \frac {b e x^{3} \log {\left (- c^{2} x^{2} + 1 \right )} \operatorname {atanh}{\left (c x \right )}}{3} - \frac {2 b e x^{3} \operatorname {atanh}{\left (c x \right )}}{9} + \frac {b d x^{2}}{6 c} + \frac {b e x^{2} \log {\left (- c^{2} x^{2} + 1 \right )}}{6 c} - \frac {5 b e x^{2}}{18 c} - \frac {2 b e x \operatorname {atanh}{\left (c x \right )}}{3 c^{2}} + \frac {b d \log {\left (- c^{2} x^{2} + 1 \right )}}{6 c^{3}} + \frac {b e \log {\left (- c^{2} x^{2} + 1 \right )}^{2}}{12 c^{3}} - \frac {11 b e \log {\left (- c^{2} x^{2} + 1 \right )}}{18 c^{3}} + \frac {b e \operatorname {atanh}^{2}{\left (c x \right )}}{3 c^{3}} & \text {for}\: c \neq 0 \\\frac {a d x^{3}}{3} & \text {otherwise} \end {cases} \] Input:
integrate(x**2*(a+b*atanh(c*x))*(d+e*ln(-c**2*x**2+1)),x)
Output:
Piecewise((a*d*x**3/3 + a*e*x**3*log(-c**2*x**2 + 1)/3 - 2*a*e*x**3/9 - 2* a*e*x/(3*c**2) + 2*a*e*atanh(c*x)/(3*c**3) + b*d*x**3*atanh(c*x)/3 + b*e*x **3*log(-c**2*x**2 + 1)*atanh(c*x)/3 - 2*b*e*x**3*atanh(c*x)/9 + b*d*x**2/ (6*c) + b*e*x**2*log(-c**2*x**2 + 1)/(6*c) - 5*b*e*x**2/(18*c) - 2*b*e*x*a tanh(c*x)/(3*c**2) + b*d*log(-c**2*x**2 + 1)/(6*c**3) + b*e*log(-c**2*x**2 + 1)**2/(12*c**3) - 11*b*e*log(-c**2*x**2 + 1)/(18*c**3) + b*e*atanh(c*x) **2/(3*c**3), Ne(c, 0)), (a*d*x**3/3, True))
Result contains complex when optimal does not.
Time = 0.04 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.22 \[ \int x^2 (a+b \text {arctanh}(c x)) \left (d+e \log \left (1-c^2 x^2\right )\right ) \, dx=\frac {1}{3} \, a d x^{3} + \frac {1}{9} \, {\left (3 \, x^{3} \log \left (-c^{2} x^{2} + 1\right ) - c^{2} {\left (\frac {2 \, {\left (c^{2} x^{3} + 3 \, x\right )}}{c^{4}} - \frac {3 \, \log \left (c x + 1\right )}{c^{5}} + \frac {3 \, \log \left (c x - 1\right )}{c^{5}}\right )}\right )} b e \operatorname {artanh}\left (c x\right ) + \frac {1}{6} \, {\left (2 \, x^{3} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {x^{2}}{c^{2}} + \frac {\log \left (c^{2} x^{2} - 1\right )}{c^{4}}\right )}\right )} b d + \frac {1}{9} \, {\left (3 \, x^{3} \log \left (-c^{2} x^{2} + 1\right ) - c^{2} {\left (\frac {2 \, {\left (c^{2} x^{3} + 3 \, x\right )}}{c^{4}} - \frac {3 \, \log \left (c x + 1\right )}{c^{5}} + \frac {3 \, \log \left (c x - 1\right )}{c^{5}}\right )}\right )} a e + \frac {{\left ({\left (3 i \, \pi c^{2} - 5 \, c^{2}\right )} x^{2} + {\left (3 i \, \pi + 3 \, c^{2} x^{2} + 6 \, \log \left (c x - 1\right ) - 11\right )} \log \left (c x + 1\right ) + {\left (3 i \, \pi + 3 \, c^{2} x^{2} - 11\right )} \log \left (c x - 1\right )\right )} b e}{18 \, c^{3}} \] Input:
integrate(x^2*(a+b*arctanh(c*x))*(d+e*log(-c^2*x^2+1)),x, algorithm="maxim a")
Output:
1/3*a*d*x^3 + 1/9*(3*x^3*log(-c^2*x^2 + 1) - c^2*(2*(c^2*x^3 + 3*x)/c^4 - 3*log(c*x + 1)/c^5 + 3*log(c*x - 1)/c^5))*b*e*arctanh(c*x) + 1/6*(2*x^3*ar ctanh(c*x) + c*(x^2/c^2 + log(c^2*x^2 - 1)/c^4))*b*d + 1/9*(3*x^3*log(-c^2 *x^2 + 1) - c^2*(2*(c^2*x^3 + 3*x)/c^4 - 3*log(c*x + 1)/c^5 + 3*log(c*x - 1)/c^5))*a*e + 1/18*((3*I*pi*c^2 - 5*c^2)*x^2 + (3*I*pi + 3*c^2*x^2 + 6*lo g(c*x - 1) - 11)*log(c*x + 1) + (3*I*pi + 3*c^2*x^2 - 11)*log(c*x - 1))*b* e/c^3
Time = 0.25 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.18 \[ \int x^2 (a+b \text {arctanh}(c x)) \left (d+e \log \left (1-c^2 x^2\right )\right ) \, dx=-\frac {1}{6} \, b e x^{3} \log \left (-c x + 1\right )^{2} + \frac {1}{9} \, {\left (3 \, a d - 2 \, a e\right )} x^{3} + \frac {1}{6} \, {\left (b e x^{3} + \frac {b e}{c^{3}}\right )} \log \left (c x + 1\right )^{2} + \frac {{\left (3 \, b d - 5 \, b e\right )} x^{2}}{18 \, c} + \frac {1}{18} \, {\left ({\left (3 \, b d + 6 \, a e - 2 \, b e\right )} x^{3} + \frac {3 \, b e x^{2}}{c} - \frac {6 \, b e x}{c^{2}}\right )} \log \left (c x + 1\right ) - \frac {1}{18} \, {\left ({\left (3 \, b d - 6 \, a e - 2 \, b e\right )} x^{3} - \frac {3 \, b e x^{2}}{c} - \frac {6 \, b e x}{c^{2}} - \frac {6 \, b e \log \left (c x - 1\right )}{c^{3}}\right )} \log \left (-c x + 1\right ) - \frac {2 \, a e x}{3 \, c^{2}} - \frac {b e \log \left (c x - 1\right )^{2}}{6 \, c^{3}} + \frac {{\left (3 \, b d + 6 \, a e - 11 \, b e\right )} \log \left (c x + 1\right )}{18 \, c^{3}} + \frac {{\left (3 \, b d - 6 \, a e - 11 \, b e\right )} \log \left (c x - 1\right )}{18 \, c^{3}} \] Input:
integrate(x^2*(a+b*arctanh(c*x))*(d+e*log(-c^2*x^2+1)),x, algorithm="giac" )
Output:
-1/6*b*e*x^3*log(-c*x + 1)^2 + 1/9*(3*a*d - 2*a*e)*x^3 + 1/6*(b*e*x^3 + b* e/c^3)*log(c*x + 1)^2 + 1/18*(3*b*d - 5*b*e)*x^2/c + 1/18*((3*b*d + 6*a*e - 2*b*e)*x^3 + 3*b*e*x^2/c - 6*b*e*x/c^2)*log(c*x + 1) - 1/18*((3*b*d - 6* a*e - 2*b*e)*x^3 - 3*b*e*x^2/c - 6*b*e*x/c^2 - 6*b*e*log(c*x - 1)/c^3)*log (-c*x + 1) - 2/3*a*e*x/c^2 - 1/6*b*e*log(c*x - 1)^2/c^3 + 1/18*(3*b*d + 6* a*e - 11*b*e)*log(c*x + 1)/c^3 + 1/18*(3*b*d - 6*a*e - 11*b*e)*log(c*x - 1 )/c^3
Time = 5.40 (sec) , antiderivative size = 515, normalized size of antiderivative = 2.50 \[ \int x^2 (a+b \text {arctanh}(c x)) \left (d+e \log \left (1-c^2 x^2\right )\right ) \, dx=\frac {a\,d\,x^3}{3}-\frac {2\,a\,e\,x^3}{9}+\frac {b\,d\,x^3\,\ln \left (c\,x+1\right )}{6}-\frac {b\,d\,x^3\,\ln \left (1-c\,x\right )}{6}-\frac {b\,e\,x^3\,\ln \left (c\,x+1\right )}{9}+\frac {b\,e\,x^3\,\ln \left (1-c\,x\right )}{9}+\frac {b\,e\,{\ln \left (c\,x+1\right )}^2}{6\,c^3}+\frac {b\,e\,{\ln \left (1-c\,x\right )}^2}{6\,c^3}-\frac {2\,a\,e\,x}{3\,c^2}+\frac {b\,d\,x^2}{6\,c}-\frac {5\,b\,e\,x^2}{18\,c}+\frac {a\,e\,x^3\,\ln \left (1-c^2\,x^2\right )}{3}-\frac {a\,e\,\ln \left (c\,x-1\right )}{3\,c^3}+\frac {a\,e\,\ln \left (c\,x+1\right )}{3\,c^3}+\frac {b\,d\,\ln \left (c\,x-1\right )}{6\,c^3}+\frac {b\,d\,\ln \left (c\,x+1\right )}{6\,c^3}-\frac {11\,b\,e\,\ln \left (c\,x-1\right )}{18\,c^3}-\frac {11\,b\,e\,\ln \left (c\,x+1\right )}{18\,c^3}-\frac {b\,e\,\ln \left (c\,x+1\right )\,\ln \left (-\frac {2\,a\,e-2\,a\,c\,e\,x}{3\,c^2}\right )}{6\,c^3}-\frac {b\,e\,\ln \left (c\,x+1\right )\,\ln \left (-\frac {2\,a\,e+2\,a\,c\,e\,x}{3\,c^2}\right )}{6\,c^3}-\frac {b\,e\,\ln \left (1-c\,x\right )\,\ln \left (-\frac {2\,a\,e-2\,a\,c\,e\,x}{3\,c^2}\right )}{6\,c^3}-\frac {b\,e\,\ln \left (1-c\,x\right )\,\ln \left (-\frac {2\,a\,e+2\,a\,c\,e\,x}{3\,c^2}\right )}{6\,c^3}-\frac {b\,e\,x\,\ln \left (c\,x+1\right )}{3\,c^2}+\frac {b\,e\,x\,\ln \left (1-c\,x\right )}{3\,c^2}+\frac {b\,e\,x^2\,\ln \left (1-c^2\,x^2\right )}{6\,c}+\frac {b\,e\,\ln \left (-\frac {2\,a\,e-2\,a\,c\,e\,x}{3\,c^2}\right )\,\ln \left (1-c^2\,x^2\right )}{6\,c^3}+\frac {b\,e\,\ln \left (-\frac {2\,a\,e+2\,a\,c\,e\,x}{3\,c^2}\right )\,\ln \left (1-c^2\,x^2\right )}{6\,c^3}+\frac {b\,e\,x^3\,\ln \left (c\,x+1\right )\,\ln \left (1-c^2\,x^2\right )}{6}-\frac {b\,e\,x^3\,\ln \left (1-c\,x\right )\,\ln \left (1-c^2\,x^2\right )}{6} \] Input:
int(x^2*(a + b*atanh(c*x))*(d + e*log(1 - c^2*x^2)),x)
Output:
(a*d*x^3)/3 - (2*a*e*x^3)/9 + (b*d*x^3*log(c*x + 1))/6 - (b*d*x^3*log(1 - c*x))/6 - (b*e*x^3*log(c*x + 1))/9 + (b*e*x^3*log(1 - c*x))/9 + (b*e*log(c *x + 1)^2)/(6*c^3) + (b*e*log(1 - c*x)^2)/(6*c^3) - (2*a*e*x)/(3*c^2) + (b *d*x^2)/(6*c) - (5*b*e*x^2)/(18*c) + (a*e*x^3*log(1 - c^2*x^2))/3 - (a*e*l og(c*x - 1))/(3*c^3) + (a*e*log(c*x + 1))/(3*c^3) + (b*d*log(c*x - 1))/(6* c^3) + (b*d*log(c*x + 1))/(6*c^3) - (11*b*e*log(c*x - 1))/(18*c^3) - (11*b *e*log(c*x + 1))/(18*c^3) - (b*e*log(c*x + 1)*log(-(2*a*e - 2*a*c*e*x)/(3* c^2)))/(6*c^3) - (b*e*log(c*x + 1)*log(-(2*a*e + 2*a*c*e*x)/(3*c^2)))/(6*c ^3) - (b*e*log(1 - c*x)*log(-(2*a*e - 2*a*c*e*x)/(3*c^2)))/(6*c^3) - (b*e* log(1 - c*x)*log(-(2*a*e + 2*a*c*e*x)/(3*c^2)))/(6*c^3) - (b*e*x*log(c*x + 1))/(3*c^2) + (b*e*x*log(1 - c*x))/(3*c^2) + (b*e*x^2*log(1 - c^2*x^2))/( 6*c) + (b*e*log(-(2*a*e - 2*a*c*e*x)/(3*c^2))*log(1 - c^2*x^2))/(6*c^3) + (b*e*log(-(2*a*e + 2*a*c*e*x)/(3*c^2))*log(1 - c^2*x^2))/(6*c^3) + (b*e*x^ 3*log(c*x + 1)*log(1 - c^2*x^2))/6 - (b*e*x^3*log(1 - c*x)*log(1 - c^2*x^2 ))/6
Time = 0.17 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.18 \[ \int x^2 (a+b \text {arctanh}(c x)) \left (d+e \log \left (1-c^2 x^2\right )\right ) \, dx=\frac {12 \mathit {atanh} \left (c x \right )^{2} b e +12 \mathit {atanh} \left (c x \right ) \mathrm {log}\left (-c^{2} x^{2}+1\right ) b \,c^{3} e \,x^{3}+12 \mathit {atanh} \left (c x \right ) b \,c^{3} d \,x^{3}-8 \mathit {atanh} \left (c x \right ) b \,c^{3} e \,x^{3}-24 \mathit {atanh} \left (c x \right ) b c e x +3 \mathrm {log}\left (-c^{2} x^{2}+1\right )^{2} b e +12 \,\mathrm {log}\left (-c^{2} x^{2}+1\right ) a \,c^{3} e \,x^{3}+12 \,\mathrm {log}\left (-c^{2} x^{2}+1\right ) a e +6 \,\mathrm {log}\left (-c^{2} x^{2}+1\right ) b \,c^{2} e \,x^{2}+6 \,\mathrm {log}\left (-c^{2} x^{2}+1\right ) b d -22 \,\mathrm {log}\left (-c^{2} x^{2}+1\right ) b e -24 \,\mathrm {log}\left (c^{2} x -c \right ) a e +12 a \,c^{3} d \,x^{3}-8 a \,c^{3} e \,x^{3}-24 a c e x +6 b \,c^{2} d \,x^{2}-10 b \,c^{2} e \,x^{2}}{36 c^{3}} \] Input:
int(x^2*(a+b*atanh(c*x))*(d+e*log(-c^2*x^2+1)),x)
Output:
(12*atanh(c*x)**2*b*e + 12*atanh(c*x)*log( - c**2*x**2 + 1)*b*c**3*e*x**3 + 12*atanh(c*x)*b*c**3*d*x**3 - 8*atanh(c*x)*b*c**3*e*x**3 - 24*atanh(c*x) *b*c*e*x + 3*log( - c**2*x**2 + 1)**2*b*e + 12*log( - c**2*x**2 + 1)*a*c** 3*e*x**3 + 12*log( - c**2*x**2 + 1)*a*e + 6*log( - c**2*x**2 + 1)*b*c**2*e *x**2 + 6*log( - c**2*x**2 + 1)*b*d - 22*log( - c**2*x**2 + 1)*b*e - 24*lo g(c**2*x - c)*a*e + 12*a*c**3*d*x**3 - 8*a*c**3*e*x**3 - 24*a*c*e*x + 6*b* c**2*d*x**2 - 10*b*c**2*e*x**2)/(36*c**3)