Integrand size = 31, antiderivative size = 107 \[ \int \frac {x^2 (a+b \text {arctanh}(c x))}{(1+c x)^2 \left (1-c^2 x^2\right )} \, dx=-\frac {b}{16 c^3 (1+c x)^2}+\frac {5 b}{16 c^3 (1+c x)}-\frac {5 b \text {arctanh}(c x)}{16 c^3}-\frac {a+b \text {arctanh}(c x)}{4 c^3 (1+c x)^2}+\frac {3 (a+b \text {arctanh}(c x))}{4 c^3 (1+c x)}+\frac {(a+b \text {arctanh}(c x))^2}{8 b c^3} \] Output:
-1/16*b/c^3/(c*x+1)^2+5/16*b/c^3/(c*x+1)-5/16*b*arctanh(c*x)/c^3-1/4*(a+b* arctanh(c*x))/c^3/(c*x+1)^2+3/4*(a+b*arctanh(c*x))/c^3/(c*x+1)+1/8*(a+b*ar ctanh(c*x))^2/b/c^3
Time = 0.08 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.90 \[ \int \frac {x^2 (a+b \text {arctanh}(c x))}{(1+c x)^2 \left (1-c^2 x^2\right )} \, dx=-\frac {\frac {2 (4 a+b)}{(1+c x)^2}-\frac {2 (12 a+5 b)}{1+c x}-\frac {8 b (2+3 c x) \text {arctanh}(c x)}{(1+c x)^2}-4 b \text {arctanh}(c x)^2+(4 a-5 b) \log (1-c x)+(-4 a+5 b) \log (1+c x)}{32 c^3} \] Input:
Integrate[(x^2*(a + b*ArcTanh[c*x]))/((1 + c*x)^2*(1 - c^2*x^2)),x]
Output:
-1/32*((2*(4*a + b))/(1 + c*x)^2 - (2*(12*a + 5*b))/(1 + c*x) - (8*b*(2 + 3*c*x)*ArcTanh[c*x])/(1 + c*x)^2 - 4*b*ArcTanh[c*x]^2 + (4*a - 5*b)*Log[1 - c*x] + (-4*a + 5*b)*Log[1 + c*x])/c^3
Time = 0.94 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.28, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {2003, 7293, 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.
| \(\displaystyle \int \frac {x^2 (a+b \text {arctanh}(c x))}{(c x+1)^2 \left (1-c^2 x^2\right )} \, dx\) |
| \(\Big \downarrow \) 2003 |
| \(\displaystyle \int \frac {x^2 (a+b \text {arctanh}(c x))}{(1-c x) (c x+1)^3}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (-\frac {a x^2}{(c x-1) (c x+1)^3}-\frac {b x^2 \text {arctanh}(c x)}{(c x-1) (c x+1)^3}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {a \text {arctanh}(c x)}{4 c^3}+\frac {3 a}{4 c^3 (c x+1)}-\frac {a}{4 c^3 (c x+1)^2}+\frac {b \text {arctanh}(c x)^2}{8 c^3}+\frac {3 b \text {arctanh}(c x)}{4 c^3 (c x+1)}-\frac {b \text {arctanh}(c x)}{4 c^3 (c x+1)^2}-\frac {5 b \text {arctanh}(c x)}{16 c^3}+\frac {5 b}{16 c^3 (c x+1)}-\frac {b}{16 c^3 (c x+1)^2}\) |
Input:
Int[(x^2*(a + b*ArcTanh[c*x]))/((1 + c*x)^2*(1 - c^2*x^2)),x]
Output:
-1/4*a/(c^3*(1 + c*x)^2) - b/(16*c^3*(1 + c*x)^2) + (3*a)/(4*c^3*(1 + c*x) ) + (5*b)/(16*c^3*(1 + c*x)) + (a*ArcTanh[c*x])/(4*c^3) - (5*b*ArcTanh[c*x ])/(16*c^3) - (b*ArcTanh[c*x])/(4*c^3*(1 + c*x)^2) + (3*b*ArcTanh[c*x])/(4 *c^3*(1 + c*x)) + (b*ArcTanh[c*x]^2)/(8*c^3)
Int[(u_)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] : > Int[u*(c + d*x)^(n + p)*(a/c + (b/d)*x)^p, x] /; FreeQ[{a, b, c, d, n, p} , x] && EqQ[b*c^2 + a*d^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[c, 0] && !IntegerQ[n]))
Time = 0.67 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.26
| method | result | size |
| parallelrisch | \(-\frac {8 a \,c^{2} x^{2}+3 b c x -2 \,\operatorname {arctanh}\left (c x \right ) b c x +5 \,\operatorname {arctanh}\left (c x \right ) b \,c^{2} x^{2}+4 b \,c^{2} x^{2}+4 a c x -3 b \,\operatorname {arctanh}\left (c x \right )-4 a \,\operatorname {arctanh}\left (c x \right )-2 b \operatorname {arctanh}\left (c x \right )^{2}-4 x^{2} \operatorname {arctanh}\left (c x \right ) a \,c^{2}-8 x \,\operatorname {arctanh}\left (c x \right ) a c -2 b \operatorname {arctanh}\left (c x \right )^{2} x^{2} c^{2}-4 b \operatorname {arctanh}\left (c x \right )^{2} x c}{16 \left (c x +1\right )^{2} c^{3}}\) | \(135\) |
| derivativedivides | \(\frac {-a \left (\frac {1}{4 \left (c x +1\right )^{2}}-\frac {3}{4 \left (c x +1\right )}-\frac {\ln \left (c x +1\right )}{8}+\frac {\ln \left (c x -1\right )}{8}\right )-b \left (\frac {\operatorname {arctanh}\left (c x \right )}{4 \left (c x +1\right )^{2}}-\frac {3 \,\operatorname {arctanh}\left (c x \right )}{4 \left (c x +1\right )}-\frac {\operatorname {arctanh}\left (c x \right ) \ln \left (c x +1\right )}{8}+\frac {\operatorname {arctanh}\left (c x \right ) \ln \left (c x -1\right )}{8}-\frac {\left (\ln \left (c x +1\right )-\ln \left (\frac {c x}{2}+\frac {1}{2}\right )\right ) \ln \left (-\frac {c x}{2}+\frac {1}{2}\right )}{16}+\frac {\ln \left (c x +1\right )^{2}}{32}+\frac {\ln \left (c x -1\right )^{2}}{32}-\frac {\ln \left (c x -1\right ) \ln \left (\frac {c x}{2}+\frac {1}{2}\right )}{16}+\frac {1}{16 \left (c x +1\right )^{2}}-\frac {5}{16 \left (c x +1\right )}+\frac {5 \ln \left (c x +1\right )}{32}-\frac {5 \ln \left (c x -1\right )}{32}\right )}{c^{3}}\) | \(192\) |
| default | \(\frac {-a \left (\frac {1}{4 \left (c x +1\right )^{2}}-\frac {3}{4 \left (c x +1\right )}-\frac {\ln \left (c x +1\right )}{8}+\frac {\ln \left (c x -1\right )}{8}\right )-b \left (\frac {\operatorname {arctanh}\left (c x \right )}{4 \left (c x +1\right )^{2}}-\frac {3 \,\operatorname {arctanh}\left (c x \right )}{4 \left (c x +1\right )}-\frac {\operatorname {arctanh}\left (c x \right ) \ln \left (c x +1\right )}{8}+\frac {\operatorname {arctanh}\left (c x \right ) \ln \left (c x -1\right )}{8}-\frac {\left (\ln \left (c x +1\right )-\ln \left (\frac {c x}{2}+\frac {1}{2}\right )\right ) \ln \left (-\frac {c x}{2}+\frac {1}{2}\right )}{16}+\frac {\ln \left (c x +1\right )^{2}}{32}+\frac {\ln \left (c x -1\right )^{2}}{32}-\frac {\ln \left (c x -1\right ) \ln \left (\frac {c x}{2}+\frac {1}{2}\right )}{16}+\frac {1}{16 \left (c x +1\right )^{2}}-\frac {5}{16 \left (c x +1\right )}+\frac {5 \ln \left (c x +1\right )}{32}-\frac {5 \ln \left (c x -1\right )}{32}\right )}{c^{3}}\) | \(192\) |
| parts | \(-a \left (\frac {1}{4 c^{3} \left (c x +1\right )^{2}}-\frac {3}{4 c^{3} \left (c x +1\right )}-\frac {\ln \left (c x +1\right )}{8 c^{3}}+\frac {\ln \left (c x -1\right )}{8 c^{3}}\right )-\frac {b \left (\frac {\operatorname {arctanh}\left (c x \right )}{4 \left (c x +1\right )^{2}}-\frac {3 \,\operatorname {arctanh}\left (c x \right )}{4 \left (c x +1\right )}-\frac {\operatorname {arctanh}\left (c x \right ) \ln \left (c x +1\right )}{8}+\frac {\operatorname {arctanh}\left (c x \right ) \ln \left (c x -1\right )}{8}-\frac {\left (\ln \left (c x +1\right )-\ln \left (\frac {c x}{2}+\frac {1}{2}\right )\right ) \ln \left (-\frac {c x}{2}+\frac {1}{2}\right )}{16}+\frac {\ln \left (c x +1\right )^{2}}{32}+\frac {\ln \left (c x -1\right )^{2}}{32}-\frac {\ln \left (c x -1\right ) \ln \left (\frac {c x}{2}+\frac {1}{2}\right )}{16}+\frac {1}{16 \left (c x +1\right )^{2}}-\frac {5}{16 \left (c x +1\right )}+\frac {5 \ln \left (c x +1\right )}{32}-\frac {5 \ln \left (c x -1\right )}{32}\right )}{c^{3}}\) | \(203\) |
| risch | \(\frac {b \ln \left (c x +1\right )^{2}}{32 c^{3}}-\frac {b \left (c^{2} x^{2} \ln \left (-c x +1\right )+2 c x \ln \left (-c x +1\right )-6 c x +\ln \left (-c x +1\right )-4\right ) \ln \left (c x +1\right )}{16 c^{3} \left (c x +1\right )^{2}}-\frac {-b \,c^{2} x^{2} \ln \left (-c x +1\right )^{2}+4 \ln \left (c x -1\right ) a \,c^{2} x^{2}-5 \ln \left (c x -1\right ) b \,c^{2} x^{2}-4 \ln \left (-c x -1\right ) a \,c^{2} x^{2}+5 \ln \left (-c x -1\right ) b \,c^{2} x^{2}-2 b c x \ln \left (-c x +1\right )^{2}+8 \ln \left (c x -1\right ) a c x -10 \ln \left (c x -1\right ) b c x -8 \ln \left (-c x -1\right ) a c x +10 \ln \left (-c x -1\right ) b c x +12 b c x \ln \left (-c x +1\right )-24 a c x -10 b c x -b \ln \left (-c x +1\right )^{2}+4 \ln \left (c x -1\right ) a -5 b \ln \left (c x -1\right )-4 a \ln \left (-c x -1\right )+5 b \ln \left (-c x -1\right )+8 b \ln \left (-c x +1\right )-16 a -8 b}{32 c^{3} \left (c x +1\right )^{2}}\) | \(314\) |
Input:
int(x^2*(a+b*arctanh(c*x))/(c*x+1)^2/(-c^2*x^2+1),x,method=_RETURNVERBOSE)
Output:
-1/16*(8*a*c^2*x^2+3*b*c*x-2*arctanh(c*x)*b*c*x+5*arctanh(c*x)*b*c^2*x^2+4 *b*c^2*x^2+4*a*c*x-3*b*arctanh(c*x)-4*a*arctanh(c*x)-2*b*arctanh(c*x)^2-4* x^2*arctanh(c*x)*a*c^2-8*x*arctanh(c*x)*a*c-2*b*arctanh(c*x)^2*x^2*c^2-4*b *arctanh(c*x)^2*x*c)/(c*x+1)^2/c^3
Time = 0.09 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.10 \[ \int \frac {x^2 (a+b \text {arctanh}(c x))}{(1+c x)^2 \left (1-c^2 x^2\right )} \, dx=\frac {2 \, {\left (12 \, a + 5 \, b\right )} c x + {\left (b c^{2} x^{2} + 2 \, b c x + b\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )^{2} + {\left ({\left (4 \, a - 5 \, b\right )} c^{2} x^{2} + 2 \, {\left (4 \, a + b\right )} c x + 4 \, a + 3 \, b\right )} \log \left (-\frac {c x + 1}{c x - 1}\right ) + 16 \, a + 8 \, b}{32 \, {\left (c^{5} x^{2} + 2 \, c^{4} x + c^{3}\right )}} \] Input:
integrate(x^2*(a+b*arctanh(c*x))/(c*x+1)^2/(-c^2*x^2+1),x, algorithm="fric as")
Output:
1/32*(2*(12*a + 5*b)*c*x + (b*c^2*x^2 + 2*b*c*x + b)*log(-(c*x + 1)/(c*x - 1))^2 + ((4*a - 5*b)*c^2*x^2 + 2*(4*a + b)*c*x + 4*a + 3*b)*log(-(c*x + 1 )/(c*x - 1)) + 16*a + 8*b)/(c^5*x^2 + 2*c^4*x + c^3)
\[ \int \frac {x^2 (a+b \text {arctanh}(c x))}{(1+c x)^2 \left (1-c^2 x^2\right )} \, dx=- \int \frac {a x^{2}}{c^{4} x^{4} + 2 c^{3} x^{3} - 2 c x - 1}\, dx - \int \frac {b x^{2} \operatorname {atanh}{\left (c x \right )}}{c^{4} x^{4} + 2 c^{3} x^{3} - 2 c x - 1}\, dx \] Input:
integrate(x**2*(a+b*atanh(c*x))/(c*x+1)**2/(-c**2*x**2+1),x)
Output:
-Integral(a*x**2/(c**4*x**4 + 2*c**3*x**3 - 2*c*x - 1), x) - Integral(b*x* *2*atanh(c*x)/(c**4*x**4 + 2*c**3*x**3 - 2*c*x - 1), x)
Leaf count of result is larger than twice the leaf count of optimal. 245 vs. \(2 (95) = 190\).
Time = 0.04 (sec) , antiderivative size = 245, normalized size of antiderivative = 2.29 \[ \int \frac {x^2 (a+b \text {arctanh}(c x))}{(1+c x)^2 \left (1-c^2 x^2\right )} \, dx=\frac {1}{8} \, b {\left (\frac {2 \, {\left (3 \, c x + 2\right )}}{c^{5} x^{2} + 2 \, c^{4} x + c^{3}} + \frac {\log \left (c x + 1\right )}{c^{3}} - \frac {\log \left (c x - 1\right )}{c^{3}}\right )} \operatorname {artanh}\left (c x\right ) - \frac {{\left ({\left (c^{2} x^{2} + 2 \, c x + 1\right )} \log \left (c x + 1\right )^{2} + {\left (c^{2} x^{2} + 2 \, c x + 1\right )} \log \left (c x - 1\right )^{2} - 10 \, c x + {\left (5 \, c^{2} x^{2} + 10 \, c x - 2 \, {\left (c^{2} x^{2} + 2 \, c x + 1\right )} \log \left (c x - 1\right ) + 5\right )} \log \left (c x + 1\right ) - 5 \, {\left (c^{2} x^{2} + 2 \, c x + 1\right )} \log \left (c x - 1\right ) - 8\right )} b c}{32 \, {\left (c^{6} x^{2} + 2 \, c^{5} x + c^{4}\right )}} + \frac {1}{8} \, a {\left (\frac {2 \, {\left (3 \, c x + 2\right )}}{c^{5} x^{2} + 2 \, c^{4} x + c^{3}} + \frac {\log \left (c x + 1\right )}{c^{3}} - \frac {\log \left (c x - 1\right )}{c^{3}}\right )} \] Input:
integrate(x^2*(a+b*arctanh(c*x))/(c*x+1)^2/(-c^2*x^2+1),x, algorithm="maxi ma")
Output:
1/8*b*(2*(3*c*x + 2)/(c^5*x^2 + 2*c^4*x + c^3) + log(c*x + 1)/c^3 - log(c* x - 1)/c^3)*arctanh(c*x) - 1/32*((c^2*x^2 + 2*c*x + 1)*log(c*x + 1)^2 + (c ^2*x^2 + 2*c*x + 1)*log(c*x - 1)^2 - 10*c*x + (5*c^2*x^2 + 10*c*x - 2*(c^2 *x^2 + 2*c*x + 1)*log(c*x - 1) + 5)*log(c*x + 1) - 5*(c^2*x^2 + 2*c*x + 1) *log(c*x - 1) - 8)*b*c/(c^6*x^2 + 2*c^5*x + c^4) + 1/8*a*(2*(3*c*x + 2)/(c ^5*x^2 + 2*c^4*x + c^3) + log(c*x + 1)/c^3 - log(c*x - 1)/c^3)
\[ \int \frac {x^2 (a+b \text {arctanh}(c x))}{(1+c x)^2 \left (1-c^2 x^2\right )} \, dx=\int { -\frac {{\left (b \operatorname {artanh}\left (c x\right ) + a\right )} x^{2}}{{\left (c^{2} x^{2} - 1\right )} {\left (c x + 1\right )}^{2}} \,d x } \] Input:
integrate(x^2*(a+b*arctanh(c*x))/(c*x+1)^2/(-c^2*x^2+1),x, algorithm="giac ")
Output:
integrate(-(b*arctanh(c*x) + a)*x^2/((c^2*x^2 - 1)*(c*x + 1)^2), x)
Time = 3.95 (sec) , antiderivative size = 223, normalized size of antiderivative = 2.08 \[ \int \frac {x^2 (a+b \text {arctanh}(c x))}{(1+c x)^2 \left (1-c^2 x^2\right )} \, dx=\frac {\frac {4\,\left (2\,a+b\right )}{c}+x\,\left (12\,a+5\,b\right )}{16\,c^4\,x^2+32\,c^3\,x+16\,c^2}-\ln \left (1-c\,x\right )\,\left (\frac {\frac {b}{c^3}+\frac {b\,x}{c^2}}{2\,c^2\,x^2+4\,c\,x+2}+\frac {b\,\ln \left (c\,x+1\right )}{16\,c^3}-\frac {b\,\left (3\,c^2\,x^2+10\,c\,x+11\right )}{16\,c^3\,\left (2\,c^2\,x^2+4\,c\,x+2\right )}\right )+\frac {\ln \left (c\,x+1\right )\,\left (\frac {5\,b}{32\,c^4}-\frac {3\,b\,x^2}{32\,c^2}+\frac {3\,b\,x}{16\,c^3}\right )}{2\,x+c\,x^2+\frac {1}{c}}+\frac {b\,{\ln \left (c\,x+1\right )}^2}{32\,c^3}+\frac {b\,{\ln \left (1-c\,x\right )}^2}{32\,c^3}-\frac {\mathrm {atan}\left (c\,x\,1{}\mathrm {i}\right )\,\left (2\,a-b\right )\,1{}\mathrm {i}}{8\,c^3} \] Input:
int(-(x^2*(a + b*atanh(c*x)))/((c^2*x^2 - 1)*(c*x + 1)^2),x)
Output:
((4*(2*a + b))/c + x*(12*a + 5*b))/(32*c^3*x + 16*c^2 + 16*c^4*x^2) - log( 1 - c*x)*((b/c^3 + (b*x)/c^2)/(4*c*x + 2*c^2*x^2 + 2) + (b*log(c*x + 1))/( 16*c^3) - (b*(10*c*x + 3*c^2*x^2 + 11))/(16*c^3*(4*c*x + 2*c^2*x^2 + 2))) + (log(c*x + 1)*((5*b)/(32*c^4) - (3*b*x^2)/(32*c^2) + (3*b*x)/(16*c^3)))/ (2*x + c*x^2 + 1/c) - (atan(c*x*1i)*(2*a - b)*1i)/(8*c^3) + (b*log(c*x + 1 )^2)/(32*c^3) + (b*log(1 - c*x)^2)/(32*c^3)
Time = 0.17 (sec) , antiderivative size = 238, normalized size of antiderivative = 2.22 \[ \int \frac {x^2 (a+b \text {arctanh}(c x))}{(1+c x)^2 \left (1-c^2 x^2\right )} \, dx=\frac {4 \mathit {atanh} \left (c x \right )^{2} b \,c^{2} x^{2}+8 \mathit {atanh} \left (c x \right )^{2} b c x +4 \mathit {atanh} \left (c x \right )^{2} b -12 \mathit {atanh} \left (c x \right ) b \,c^{2} x^{2}+4 \mathit {atanh} \left (c x \right ) b -4 \,\mathrm {log}\left (c x -1\right ) a \,c^{2} x^{2}-8 \,\mathrm {log}\left (c x -1\right ) a c x -4 \,\mathrm {log}\left (c x -1\right ) a -\mathrm {log}\left (c x -1\right ) b \,c^{2} x^{2}-2 \,\mathrm {log}\left (c x -1\right ) b c x -\mathrm {log}\left (c x -1\right ) b +4 \,\mathrm {log}\left (c x +1\right ) a \,c^{2} x^{2}+8 \,\mathrm {log}\left (c x +1\right ) a c x +4 \,\mathrm {log}\left (c x +1\right ) a +\mathrm {log}\left (c x +1\right ) b \,c^{2} x^{2}+2 \,\mathrm {log}\left (c x +1\right ) b c x +\mathrm {log}\left (c x +1\right ) b -12 a \,c^{2} x^{2}+4 a -5 b \,c^{2} x^{2}+3 b}{32 c^{3} \left (c^{2} x^{2}+2 c x +1\right )} \] Input:
int(x^2*(a+b*atanh(c*x))/(c*x+1)^2/(-c^2*x^2+1),x)
Output:
(4*atanh(c*x)**2*b*c**2*x**2 + 8*atanh(c*x)**2*b*c*x + 4*atanh(c*x)**2*b - 12*atanh(c*x)*b*c**2*x**2 + 4*atanh(c*x)*b - 4*log(c*x - 1)*a*c**2*x**2 - 8*log(c*x - 1)*a*c*x - 4*log(c*x - 1)*a - log(c*x - 1)*b*c**2*x**2 - 2*lo g(c*x - 1)*b*c*x - log(c*x - 1)*b + 4*log(c*x + 1)*a*c**2*x**2 + 8*log(c*x + 1)*a*c*x + 4*log(c*x + 1)*a + log(c*x + 1)*b*c**2*x**2 + 2*log(c*x + 1) *b*c*x + log(c*x + 1)*b - 12*a*c**2*x**2 + 4*a - 5*b*c**2*x**2 + 3*b)/(32* c**3*(c**2*x**2 + 2*c*x + 1))