Integrand size = 20, antiderivative size = 150 \[ \int \frac {x^2 (a+b \text {arctanh}(c x))}{(d+c d x)^3} \, dx=-\frac {b}{8 c^3 d^3 (1+c x)^2}+\frac {7 b}{8 c^3 d^3 (1+c x)}-\frac {7 b \text {arctanh}(c x)}{8 c^3 d^3}-\frac {a+b \text {arctanh}(c x)}{2 c^3 d^3 (1+c x)^2}+\frac {2 (a+b \text {arctanh}(c x))}{c^3 d^3 (1+c x)}-\frac {(a+b \text {arctanh}(c x)) \log \left (\frac {2}{1+c x}\right )}{c^3 d^3}+\frac {b \operatorname {PolyLog}\left (2,1-\frac {2}{1+c x}\right )}{2 c^3 d^3} \] Output:
-1/8*b/c^3/d^3/(c*x+1)^2+7/8*b/c^3/d^3/(c*x+1)-7/8*b*arctanh(c*x)/c^3/d^3- 1/2*(a+b*arctanh(c*x))/c^3/d^3/(c*x+1)^2+2*(a+b*arctanh(c*x))/c^3/d^3/(c*x +1)-(a+b*arctanh(c*x))*ln(2/(c*x+1))/c^3/d^3+1/2*b*polylog(2,1-2/(c*x+1))/ c^3/d^3
Time = 0.39 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.97 \[ \int \frac {x^2 (a+b \text {arctanh}(c x))}{(d+c d x)^3} \, dx=\frac {-\frac {16 a}{(1+c x)^2}+\frac {64 a}{1+c x}+32 a \log (1+c x)+b \left (12 \cosh (2 \text {arctanh}(c x))-\cosh (4 \text {arctanh}(c x))+16 \operatorname {PolyLog}\left (2,-e^{-2 \text {arctanh}(c x)}\right )-12 \sinh (2 \text {arctanh}(c x))+\sinh (4 \text {arctanh}(c x))+4 \text {arctanh}(c x) \left (6 \cosh (2 \text {arctanh}(c x))-\cosh (4 \text {arctanh}(c x))-8 \log \left (1+e^{-2 \text {arctanh}(c x)}\right )-6 \sinh (2 \text {arctanh}(c x))+\sinh (4 \text {arctanh}(c x))\right )\right )}{32 c^3 d^3} \] Input:
Integrate[(x^2*(a + b*ArcTanh[c*x]))/(d + c*d*x)^3,x]
Output:
((-16*a)/(1 + c*x)^2 + (64*a)/(1 + c*x) + 32*a*Log[1 + c*x] + b*(12*Cosh[2 *ArcTanh[c*x]] - Cosh[4*ArcTanh[c*x]] + 16*PolyLog[2, -E^(-2*ArcTanh[c*x]) ] - 12*Sinh[2*ArcTanh[c*x]] + Sinh[4*ArcTanh[c*x]] + 4*ArcTanh[c*x]*(6*Cos h[2*ArcTanh[c*x]] - Cosh[4*ArcTanh[c*x]] - 8*Log[1 + E^(-2*ArcTanh[c*x])] - 6*Sinh[2*ArcTanh[c*x]] + Sinh[4*ArcTanh[c*x]])))/(32*c^3*d^3)
Time = 0.49 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {6502, 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.
\(\displaystyle \int \frac {x^2 (a+b \text {arctanh}(c x))}{(c d x+d)^3} \, dx\) |
\(\Big \downarrow \) 6502 |
\(\displaystyle \int \left (\frac {a+b \text {arctanh}(c x)}{c^2 d^3 (c x+1)}-\frac {2 (a+b \text {arctanh}(c x))}{c^2 d^3 (c x+1)^2}+\frac {a+b \text {arctanh}(c x)}{c^2 d^3 (c x+1)^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2 (a+b \text {arctanh}(c x))}{c^3 d^3 (c x+1)}-\frac {a+b \text {arctanh}(c x)}{2 c^3 d^3 (c x+1)^2}-\frac {\log \left (\frac {2}{c x+1}\right ) (a+b \text {arctanh}(c x))}{c^3 d^3}-\frac {7 b \text {arctanh}(c x)}{8 c^3 d^3}+\frac {b \operatorname {PolyLog}\left (2,1-\frac {2}{c x+1}\right )}{2 c^3 d^3}+\frac {7 b}{8 c^3 d^3 (c x+1)}-\frac {b}{8 c^3 d^3 (c x+1)^2}\) |
Input:
Int[(x^2*(a + b*ArcTanh[c*x]))/(d + c*d*x)^3,x]
Output:
-1/8*b/(c^3*d^3*(1 + c*x)^2) + (7*b)/(8*c^3*d^3*(1 + c*x)) - (7*b*ArcTanh[ c*x])/(8*c^3*d^3) - (a + b*ArcTanh[c*x])/(2*c^3*d^3*(1 + c*x)^2) + (2*(a + b*ArcTanh[c*x]))/(c^3*d^3*(1 + c*x)) - ((a + b*ArcTanh[c*x])*Log[2/(1 + c *x)])/(c^3*d^3) + (b*PolyLog[2, 1 - 2/(1 + c*x)])/(2*c^3*d^3)
Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e _.)*(x_))^(q_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*ArcTanh[c*x])^p, ( f*x)^m*(d + e*x)^q, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[p, 0] && IntegerQ[q] && (GtQ[q, 0] || NeQ[a, 0] || IntegerQ[m])
Time = 0.41 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.05
method | result | size |
derivativedivides | \(\frac {\frac {a \left (-\frac {1}{2 \left (c x +1\right )^{2}}+\frac {2}{c x +1}+\ln \left (c x +1\right )\right )}{d^{3}}+\frac {b \left (-\frac {\operatorname {arctanh}\left (c x \right )}{2 \left (c x +1\right )^{2}}+\frac {2 \,\operatorname {arctanh}\left (c x \right )}{c x +1}+\operatorname {arctanh}\left (c x \right ) \ln \left (c x +1\right )-\frac {\ln \left (c x +1\right )^{2}}{4}+\frac {\left (\ln \left (c x +1\right )-\ln \left (\frac {c x}{2}+\frac {1}{2}\right )\right ) \ln \left (-\frac {c x}{2}+\frac {1}{2}\right )}{2}-\frac {\operatorname {dilog}\left (\frac {c x}{2}+\frac {1}{2}\right )}{2}-\frac {1}{8 \left (c x +1\right )^{2}}+\frac {7}{8 \left (c x +1\right )}-\frac {7 \ln \left (c x +1\right )}{16}+\frac {7 \ln \left (c x -1\right )}{16}\right )}{d^{3}}}{c^{3}}\) | \(157\) |
default | \(\frac {\frac {a \left (-\frac {1}{2 \left (c x +1\right )^{2}}+\frac {2}{c x +1}+\ln \left (c x +1\right )\right )}{d^{3}}+\frac {b \left (-\frac {\operatorname {arctanh}\left (c x \right )}{2 \left (c x +1\right )^{2}}+\frac {2 \,\operatorname {arctanh}\left (c x \right )}{c x +1}+\operatorname {arctanh}\left (c x \right ) \ln \left (c x +1\right )-\frac {\ln \left (c x +1\right )^{2}}{4}+\frac {\left (\ln \left (c x +1\right )-\ln \left (\frac {c x}{2}+\frac {1}{2}\right )\right ) \ln \left (-\frac {c x}{2}+\frac {1}{2}\right )}{2}-\frac {\operatorname {dilog}\left (\frac {c x}{2}+\frac {1}{2}\right )}{2}-\frac {1}{8 \left (c x +1\right )^{2}}+\frac {7}{8 \left (c x +1\right )}-\frac {7 \ln \left (c x +1\right )}{16}+\frac {7 \ln \left (c x -1\right )}{16}\right )}{d^{3}}}{c^{3}}\) | \(157\) |
parts | \(\frac {a \left (\frac {2}{c^{3} \left (c x +1\right )}+\frac {\ln \left (c x +1\right )}{c^{3}}-\frac {1}{2 c^{3} \left (c x +1\right )^{2}}\right )}{d^{3}}+\frac {b \left (-\frac {\operatorname {arctanh}\left (c x \right )}{2 \left (c x +1\right )^{2}}+\frac {2 \,\operatorname {arctanh}\left (c x \right )}{c x +1}+\operatorname {arctanh}\left (c x \right ) \ln \left (c x +1\right )-\frac {\ln \left (c x +1\right )^{2}}{4}+\frac {\left (\ln \left (c x +1\right )-\ln \left (\frac {c x}{2}+\frac {1}{2}\right )\right ) \ln \left (-\frac {c x}{2}+\frac {1}{2}\right )}{2}-\frac {\operatorname {dilog}\left (\frac {c x}{2}+\frac {1}{2}\right )}{2}-\frac {1}{8 \left (c x +1\right )^{2}}+\frac {7}{8 \left (c x +1\right )}-\frac {7 \ln \left (c x +1\right )}{16}+\frac {7 \ln \left (c x -1\right )}{16}\right )}{d^{3} c^{3}}\) | \(166\) |
risch | \(\frac {b \ln \left (c x +1\right )^{2}}{4 c^{3} d^{3}}+\frac {\left (\frac {b x}{c^{2}}+\frac {3 b}{4 c^{3}}\right ) \ln \left (c x +1\right )}{d^{3} \left (c x +1\right )^{2}}-\frac {7 b \ln \left (-c x -1\right )}{16 d^{3} c^{3}}-\frac {b \ln \left (-c x +1\right ) x}{2 d^{3} c^{2} \left (-c x -1\right )}+\frac {b \ln \left (-c x +1\right )}{2 d^{3} c^{3} \left (-c x -1\right )}+\frac {b \ln \left (\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (-\frac {c x}{2}+\frac {1}{2}\right )}{2 d^{3} c^{3}}-\frac {b \ln \left (\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (-c x +1\right )}{2 d^{3} c^{3}}+\frac {b \operatorname {dilog}\left (-\frac {c x}{2}+\frac {1}{2}\right )}{2 d^{3} c^{3}}+\frac {b}{8 d^{3} c^{3} \left (-c x -1\right )}-\frac {b \ln \left (-c x +1\right ) x^{2}}{16 d^{3} c \left (-c x -1\right )^{2}}-\frac {b \ln \left (-c x +1\right ) x}{8 d^{3} c^{2} \left (-c x -1\right )^{2}}+\frac {3 b \ln \left (-c x +1\right )}{16 d^{3} c^{3} \left (-c x -1\right )^{2}}-\frac {2 a}{d^{3} c^{3} \left (-c x -1\right )}+\frac {a \ln \left (-c x -1\right )}{d^{3} c^{3}}-\frac {a}{2 d^{3} c^{3} \left (-c x -1\right )^{2}}+\frac {b}{c^{3} d^{3} \left (c x +1\right )}-\frac {b}{8 c^{3} d^{3} \left (c x +1\right )^{2}}\) | \(349\) |
Input:
int(x^2*(a+b*arctanh(c*x))/(c*d*x+d)^3,x,method=_RETURNVERBOSE)
Output:
1/c^3*(a/d^3*(-1/2/(c*x+1)^2+2/(c*x+1)+ln(c*x+1))+b/d^3*(-1/2/(c*x+1)^2*ar ctanh(c*x)+2/(c*x+1)*arctanh(c*x)+arctanh(c*x)*ln(c*x+1)-1/4*ln(c*x+1)^2+1 /2*(ln(c*x+1)-ln(1/2*c*x+1/2))*ln(-1/2*c*x+1/2)-1/2*dilog(1/2*c*x+1/2)-1/8 /(c*x+1)^2+7/8/(c*x+1)-7/16*ln(c*x+1)+7/16*ln(c*x-1)))
\[ \int \frac {x^2 (a+b \text {arctanh}(c x))}{(d+c d x)^3} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (c x\right ) + a\right )} x^{2}}{{\left (c d x + d\right )}^{3}} \,d x } \] Input:
integrate(x^2*(a+b*arctanh(c*x))/(c*d*x+d)^3,x, algorithm="fricas")
Output:
integral((b*x^2*arctanh(c*x) + a*x^2)/(c^3*d^3*x^3 + 3*c^2*d^3*x^2 + 3*c*d ^3*x + d^3), x)
\[ \int \frac {x^2 (a+b \text {arctanh}(c x))}{(d+c d x)^3} \, dx=\frac {\int \frac {a x^{2}}{c^{3} x^{3} + 3 c^{2} x^{2} + 3 c x + 1}\, dx + \int \frac {b x^{2} \operatorname {atanh}{\left (c x \right )}}{c^{3} x^{3} + 3 c^{2} x^{2} + 3 c x + 1}\, dx}{d^{3}} \] Input:
integrate(x**2*(a+b*atanh(c*x))/(c*d*x+d)**3,x)
Output:
(Integral(a*x**2/(c**3*x**3 + 3*c**2*x**2 + 3*c*x + 1), x) + Integral(b*x* *2*atanh(c*x)/(c**3*x**3 + 3*c**2*x**2 + 3*c*x + 1), x))/d**3
\[ \int \frac {x^2 (a+b \text {arctanh}(c x))}{(d+c d x)^3} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (c x\right ) + a\right )} x^{2}}{{\left (c d x + d\right )}^{3}} \,d x } \] Input:
integrate(x^2*(a+b*arctanh(c*x))/(c*d*x+d)^3,x, algorithm="maxima")
Output:
1/32*(64*c^3*integrate(1/2*x^3*log(c*x + 1)/(c^6*d^3*x^4 + 2*c^5*d^3*x^3 - 2*c^3*d^3*x - c^2*d^3), x) - 4*c^2*(2*(3*c*x + 2)/(c^7*d^3*x^2 + 2*c^6*d^ 3*x + c^5*d^3) + log(c*x + 1)/(c^5*d^3) - log(c*x - 1)/(c^5*d^3)) + 64*c^2 *integrate(1/2*x^2*log(c*x + 1)/(c^6*d^3*x^4 + 2*c^5*d^3*x^3 - 2*c^3*d^3*x - c^2*d^3), x) + 7*c*(2*x/(c^5*d^3*x^2 + 2*c^4*d^3*x + c^3*d^3) - log(c*x + 1)/(c^4*d^3) + log(c*x - 1)/(c^4*d^3)) + 96*c*integrate(1/2*x*log(c*x + 1)/(c^6*d^3*x^4 + 2*c^5*d^3*x^3 - 2*c^3*d^3*x - c^2*d^3), x) - 8*(4*c*x + 2*(c^2*x^2 + 2*c*x + 1)*log(c*x + 1) + 3)*log(-c*x + 1)/(c^5*d^3*x^2 + 2* c^4*d^3*x + c^3*d^3) + 6*(c*x + 2)/(c^5*d^3*x^2 + 2*c^4*d^3*x + c^3*d^3) - 3*log(c*x + 1)/(c^3*d^3) + 3*log(c*x - 1)/(c^3*d^3) + 32*integrate(1/2*lo g(c*x + 1)/(c^6*d^3*x^4 + 2*c^5*d^3*x^3 - 2*c^3*d^3*x - c^2*d^3), x))*b + 1/2*a*((4*c*x + 3)/(c^5*d^3*x^2 + 2*c^4*d^3*x + c^3*d^3) + 2*log(c*x + 1)/ (c^3*d^3))
\[ \int \frac {x^2 (a+b \text {arctanh}(c x))}{(d+c d x)^3} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (c x\right ) + a\right )} x^{2}}{{\left (c d x + d\right )}^{3}} \,d x } \] Input:
integrate(x^2*(a+b*arctanh(c*x))/(c*d*x+d)^3,x, algorithm="giac")
Output:
integrate((b*arctanh(c*x) + a)*x^2/(c*d*x + d)^3, x)
Timed out. \[ \int \frac {x^2 (a+b \text {arctanh}(c x))}{(d+c d x)^3} \, dx=\int \frac {x^2\,\left (a+b\,\mathrm {atanh}\left (c\,x\right )\right )}{{\left (d+c\,d\,x\right )}^3} \,d x \] Input:
int((x^2*(a + b*atanh(c*x)))/(d + c*d*x)^3,x)
Output:
int((x^2*(a + b*atanh(c*x)))/(d + c*d*x)^3, x)
\[ \int \frac {x^2 (a+b \text {arctanh}(c x))}{(d+c d x)^3} \, dx=\frac {2 \left (\int \frac {\mathit {atanh} \left (c x \right ) x^{2}}{c^{3} x^{3}+3 c^{2} x^{2}+3 c x +1}d x \right ) b \,c^{5} x^{2}+4 \left (\int \frac {\mathit {atanh} \left (c x \right ) x^{2}}{c^{3} x^{3}+3 c^{2} x^{2}+3 c x +1}d x \right ) b \,c^{4} x +2 \left (\int \frac {\mathit {atanh} \left (c x \right ) x^{2}}{c^{3} x^{3}+3 c^{2} x^{2}+3 c x +1}d x \right ) b \,c^{3}+2 \,\mathrm {log}\left (c x +1\right ) a \,c^{2} x^{2}+4 \,\mathrm {log}\left (c x +1\right ) a c x +2 \,\mathrm {log}\left (c x +1\right ) a -2 a \,c^{2} x^{2}+a}{2 c^{3} d^{3} \left (c^{2} x^{2}+2 c x +1\right )} \] Input:
int(x^2*(a+b*atanh(c*x))/(c*d*x+d)^3,x)
Output:
(2*int((atanh(c*x)*x**2)/(c**3*x**3 + 3*c**2*x**2 + 3*c*x + 1),x)*b*c**5*x **2 + 4*int((atanh(c*x)*x**2)/(c**3*x**3 + 3*c**2*x**2 + 3*c*x + 1),x)*b*c **4*x + 2*int((atanh(c*x)*x**2)/(c**3*x**3 + 3*c**2*x**2 + 3*c*x + 1),x)*b *c**3 + 2*log(c*x + 1)*a*c**2*x**2 + 4*log(c*x + 1)*a*c*x + 2*log(c*x + 1) *a - 2*a*c**2*x**2 + a)/(2*c**3*d**3*(c**2*x**2 + 2*c*x + 1))