Integrand size = 21, antiderivative size = 69 \[ \int (c e+d e x)^2 (a+b \text {arctanh}(c+d x)) \, dx=\frac {b e^2 (c+d x)^2}{6 d}+\frac {e^2 (c+d x)^3 (a+b \text {arctanh}(c+d x))}{3 d}+\frac {b e^2 \log \left (1-(c+d x)^2\right )}{6 d} \]
Time = 0.07 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.86 \[ \int (c e+d e x)^2 (a+b \text {arctanh}(c+d x)) \, dx=\frac {e^2 \left ((c+d x)^2 (b+2 a (c+d x))+2 b (c+d x)^3 \text {arctanh}(c+d x)+b \log \left (1-(c+d x)^2\right )\right )}{6 d} \]
(e^2*((c + d*x)^2*(b + 2*a*(c + d*x)) + 2*b*(c + d*x)^3*ArcTanh[c + d*x] + b*Log[1 - (c + d*x)^2]))/(6*d)
Time = 0.30 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.78, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {6657, 27, 6452, 243, 49, 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.
\(\displaystyle \int (c e+d e x)^2 (a+b \text {arctanh}(c+d x)) \, dx\) |
\(\Big \downarrow \) 6657 |
\(\displaystyle \frac {\int e^2 (c+d x)^2 (a+b \text {arctanh}(c+d x))d(c+d x)}{d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {e^2 \int (c+d x)^2 (a+b \text {arctanh}(c+d x))d(c+d x)}{d}\) |
\(\Big \downarrow \) 6452 |
\(\displaystyle \frac {e^2 \left (\frac {1}{3} (c+d x)^3 (a+b \text {arctanh}(c+d x))-\frac {1}{3} b \int \frac {(c+d x)^3}{1-(c+d x)^2}d(c+d x)\right )}{d}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {e^2 \left (\frac {1}{3} (c+d x)^3 (a+b \text {arctanh}(c+d x))-\frac {1}{6} b \int \frac {(c+d x)^2}{-c-d x+1}d(c+d x)^2\right )}{d}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {e^2 \left (\frac {1}{3} (c+d x)^3 (a+b \text {arctanh}(c+d x))-\frac {1}{6} b \int \left (\frac {1}{-c-d x+1}-1\right )d(c+d x)^2\right )}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {e^2 \left (\frac {1}{3} (c+d x)^3 (a+b \text {arctanh}(c+d x))-\frac {1}{6} b (-\log (-c-d x+1)-c-d x)\right )}{d}\) |
3.1.10.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] : > Simp[x^(m + 1)*((a + b*ArcTanh[c*x^n])^p/(m + 1)), x] - Simp[b*c*n*(p/(m + 1)) Int[x^(m + n)*((a + b*ArcTanh[c*x^n])^(p - 1)/(1 - c^2*x^(2*n))), x ], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0] && (EqQ[p, 1] || (EqQ[n, 1 ] && IntegerQ[m])) && NeQ[m, -1]
Int[((a_.) + ArcTanh[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^( m_.), x_Symbol] :> Simp[1/d Subst[Int[(f*(x/d))^m*(a + b*ArcTanh[x])^p, x ], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[d*e - c*f, 0] && IGtQ[p, 0]
Time = 0.13 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.97
method | result | size |
derivativedivides | \(\frac {\frac {e^{2} a \left (d x +c \right )^{3}}{3}+e^{2} b \left (\frac {\left (d x +c \right )^{3} \operatorname {arctanh}\left (d x +c \right )}{3}+\frac {\left (d x +c \right )^{2}}{6}+\frac {\ln \left (d x +c -1\right )}{6}+\frac {\ln \left (d x +c +1\right )}{6}\right )}{d}\) | \(67\) |
default | \(\frac {\frac {e^{2} a \left (d x +c \right )^{3}}{3}+e^{2} b \left (\frac {\left (d x +c \right )^{3} \operatorname {arctanh}\left (d x +c \right )}{3}+\frac {\left (d x +c \right )^{2}}{6}+\frac {\ln \left (d x +c -1\right )}{6}+\frac {\ln \left (d x +c +1\right )}{6}\right )}{d}\) | \(67\) |
parts | \(\frac {e^{2} a \left (d x +c \right )^{3}}{3 d}+\frac {e^{2} b \left (\frac {\left (d x +c \right )^{3} \operatorname {arctanh}\left (d x +c \right )}{3}+\frac {\left (d x +c \right )^{2}}{6}+\frac {\ln \left (d x +c -1\right )}{6}+\frac {\ln \left (d x +c +1\right )}{6}\right )}{d}\) | \(69\) |
risch | \(\frac {e^{2} \left (d x +c \right )^{3} b \ln \left (d x +c +1\right )}{6 d}-\frac {e^{2} d^{2} b \,x^{3} \ln \left (-d x -c +1\right )}{6}-\frac {e^{2} d b c \,x^{2} \ln \left (-d x -c +1\right )}{2}+\frac {e^{2} d^{2} a \,x^{3}}{3}-\frac {e^{2} b \,c^{2} x \ln \left (-d x -c +1\right )}{2}+e^{2} d a c \,x^{2}-\frac {e^{2} b \,c^{3} \ln \left (-d x -c +1\right )}{6 d}+e^{2} a \,c^{2} x +\frac {e^{2} d b \,x^{2}}{6}+\frac {e^{2} b c x}{3}+\frac {e^{2} b \ln \left (d^{2} x^{2}+2 c d x +c^{2}-1\right )}{6 d}\) | \(186\) |
parallelrisch | \(\frac {2 b \,d^{4} e^{2} \operatorname {arctanh}\left (d x +c \right ) x^{3}+2 x^{3} a \,d^{4} e^{2}+6 x^{2} \operatorname {arctanh}\left (d x +c \right ) b c \,d^{3} e^{2}+6 x^{2} a c \,d^{3} e^{2}+6 x \,\operatorname {arctanh}\left (d x +c \right ) b \,c^{2} d^{2} e^{2}+x^{2} b \,d^{3} e^{2}+6 x a \,c^{2} d^{2} e^{2}+2 \,\operatorname {arctanh}\left (d x +c \right ) b \,c^{3} d \,e^{2}+2 x b c \,d^{2} e^{2}-18 a \,c^{3} d \,e^{2}-5 b \,c^{2} d \,e^{2}+2 \ln \left (d x +c -1\right ) b d \,e^{2}+2 \,\operatorname {arctanh}\left (d x +c \right ) b d \,e^{2}+6 d \,e^{2} c a +b d \,e^{2}}{6 d^{2}}\) | \(200\) |
1/d*(1/3*e^2*a*(d*x+c)^3+e^2*b*(1/3*(d*x+c)^3*arctanh(d*x+c)+1/6*(d*x+c)^2 +1/6*ln(d*x+c-1)+1/6*ln(d*x+c+1)))
Leaf count of result is larger than twice the leaf count of optimal. 144 vs. \(2 (63) = 126\).
Time = 0.27 (sec) , antiderivative size = 144, normalized size of antiderivative = 2.09 \[ \int (c e+d e x)^2 (a+b \text {arctanh}(c+d x)) \, dx=\frac {2 \, a d^{3} e^{2} x^{3} + {\left (6 \, a c + b\right )} d^{2} e^{2} x^{2} + 2 \, {\left (3 \, a c^{2} + b c\right )} d e^{2} x + {\left (b c^{3} + b\right )} e^{2} \log \left (d x + c + 1\right ) - {\left (b c^{3} - b\right )} e^{2} \log \left (d x + c - 1\right ) + {\left (b d^{3} e^{2} x^{3} + 3 \, b c d^{2} e^{2} x^{2} + 3 \, b c^{2} d e^{2} x\right )} \log \left (-\frac {d x + c + 1}{d x + c - 1}\right )}{6 \, d} \]
1/6*(2*a*d^3*e^2*x^3 + (6*a*c + b)*d^2*e^2*x^2 + 2*(3*a*c^2 + b*c)*d*e^2*x + (b*c^3 + b)*e^2*log(d*x + c + 1) - (b*c^3 - b)*e^2*log(d*x + c - 1) + ( b*d^3*e^2*x^3 + 3*b*c*d^2*e^2*x^2 + 3*b*c^2*d*e^2*x)*log(-(d*x + c + 1)/(d *x + c - 1)))/d
Leaf count of result is larger than twice the leaf count of optimal. 180 vs. \(2 (56) = 112\).
Time = 0.37 (sec) , antiderivative size = 180, normalized size of antiderivative = 2.61 \[ \int (c e+d e x)^2 (a+b \text {arctanh}(c+d x)) \, dx=\begin {cases} a c^{2} e^{2} x + a c d e^{2} x^{2} + \frac {a d^{2} e^{2} x^{3}}{3} + \frac {b c^{3} e^{2} \operatorname {atanh}{\left (c + d x \right )}}{3 d} + b c^{2} e^{2} x \operatorname {atanh}{\left (c + d x \right )} + b c d e^{2} x^{2} \operatorname {atanh}{\left (c + d x \right )} + \frac {b c e^{2} x}{3} + \frac {b d^{2} e^{2} x^{3} \operatorname {atanh}{\left (c + d x \right )}}{3} + \frac {b d e^{2} x^{2}}{6} + \frac {b e^{2} \log {\left (\frac {c}{d} + x + \frac {1}{d} \right )}}{3 d} - \frac {b e^{2} \operatorname {atanh}{\left (c + d x \right )}}{3 d} & \text {for}\: d \neq 0 \\c^{2} e^{2} x \left (a + b \operatorname {atanh}{\left (c \right )}\right ) & \text {otherwise} \end {cases} \]
Piecewise((a*c**2*e**2*x + a*c*d*e**2*x**2 + a*d**2*e**2*x**3/3 + b*c**3*e **2*atanh(c + d*x)/(3*d) + b*c**2*e**2*x*atanh(c + d*x) + b*c*d*e**2*x**2* atanh(c + d*x) + b*c*e**2*x/3 + b*d**2*e**2*x**3*atanh(c + d*x)/3 + b*d*e* *2*x**2/6 + b*e**2*log(c/d + x + 1/d)/(3*d) - b*e**2*atanh(c + d*x)/(3*d), Ne(d, 0)), (c**2*e**2*x*(a + b*atanh(c)), True))
Leaf count of result is larger than twice the leaf count of optimal. 225 vs. \(2 (63) = 126\).
Time = 0.19 (sec) , antiderivative size = 225, normalized size of antiderivative = 3.26 \[ \int (c e+d e x)^2 (a+b \text {arctanh}(c+d x)) \, dx=\frac {1}{3} \, a d^{2} e^{2} x^{3} + a c d e^{2} x^{2} + \frac {1}{2} \, {\left (2 \, x^{2} \operatorname {artanh}\left (d x + c\right ) + d {\left (\frac {2 \, x}{d^{2}} - \frac {{\left (c^{2} + 2 \, c + 1\right )} \log \left (d x + c + 1\right )}{d^{3}} + \frac {{\left (c^{2} - 2 \, c + 1\right )} \log \left (d x + c - 1\right )}{d^{3}}\right )}\right )} b c d e^{2} + \frac {1}{6} \, {\left (2 \, x^{3} \operatorname {artanh}\left (d x + c\right ) + d {\left (\frac {d x^{2} - 4 \, c x}{d^{3}} + \frac {{\left (c^{3} + 3 \, c^{2} + 3 \, c + 1\right )} \log \left (d x + c + 1\right )}{d^{4}} - \frac {{\left (c^{3} - 3 \, c^{2} + 3 \, c - 1\right )} \log \left (d x + c - 1\right )}{d^{4}}\right )}\right )} b d^{2} e^{2} + a c^{2} e^{2} x + \frac {{\left (2 \, {\left (d x + c\right )} \operatorname {artanh}\left (d x + c\right ) + \log \left (-{\left (d x + c\right )}^{2} + 1\right )\right )} b c^{2} e^{2}}{2 \, d} \]
1/3*a*d^2*e^2*x^3 + a*c*d*e^2*x^2 + 1/2*(2*x^2*arctanh(d*x + c) + d*(2*x/d ^2 - (c^2 + 2*c + 1)*log(d*x + c + 1)/d^3 + (c^2 - 2*c + 1)*log(d*x + c - 1)/d^3))*b*c*d*e^2 + 1/6*(2*x^3*arctanh(d*x + c) + d*((d*x^2 - 4*c*x)/d^3 + (c^3 + 3*c^2 + 3*c + 1)*log(d*x + c + 1)/d^4 - (c^3 - 3*c^2 + 3*c - 1)*l og(d*x + c - 1)/d^4))*b*d^2*e^2 + a*c^2*e^2*x + 1/2*(2*(d*x + c)*arctanh(d *x + c) + log(-(d*x + c)^2 + 1))*b*c^2*e^2/d
Leaf count of result is larger than twice the leaf count of optimal. 322 vs. \(2 (63) = 126\).
Time = 0.28 (sec) , antiderivative size = 322, normalized size of antiderivative = 4.67 \[ \int (c e+d e x)^2 (a+b \text {arctanh}(c+d x)) \, dx=-\frac {1}{6} \, {\left ({\left (c + 1\right )} d - {\left (c - 1\right )} d\right )} {\left (\frac {b e^{2} \log \left (-\frac {d x + c + 1}{d x + c - 1} + 1\right )}{d^{2}} - \frac {b e^{2} \log \left (-\frac {d x + c + 1}{d x + c - 1}\right )}{d^{2}} - \frac {{\left (\frac {3 \, {\left (d x + c + 1\right )}^{2} b e^{2}}{{\left (d x + c - 1\right )}^{2}} + b e^{2}\right )} \log \left (-\frac {d x + c + 1}{d x + c - 1}\right )}{\frac {{\left (d x + c + 1\right )}^{3} d^{2}}{{\left (d x + c - 1\right )}^{3}} - \frac {3 \, {\left (d x + c + 1\right )}^{2} d^{2}}{{\left (d x + c - 1\right )}^{2}} + \frac {3 \, {\left (d x + c + 1\right )} d^{2}}{d x + c - 1} - d^{2}} - \frac {2 \, {\left (\frac {3 \, {\left (d x + c + 1\right )}^{2} a e^{2}}{{\left (d x + c - 1\right )}^{2}} + a e^{2} + \frac {{\left (d x + c + 1\right )}^{2} b e^{2}}{{\left (d x + c - 1\right )}^{2}} - \frac {{\left (d x + c + 1\right )} b e^{2}}{d x + c - 1}\right )}}{\frac {{\left (d x + c + 1\right )}^{3} d^{2}}{{\left (d x + c - 1\right )}^{3}} - \frac {3 \, {\left (d x + c + 1\right )}^{2} d^{2}}{{\left (d x + c - 1\right )}^{2}} + \frac {3 \, {\left (d x + c + 1\right )} d^{2}}{d x + c - 1} - d^{2}}\right )} \]
-1/6*((c + 1)*d - (c - 1)*d)*(b*e^2*log(-(d*x + c + 1)/(d*x + c - 1) + 1)/ d^2 - b*e^2*log(-(d*x + c + 1)/(d*x + c - 1))/d^2 - (3*(d*x + c + 1)^2*b*e ^2/(d*x + c - 1)^2 + b*e^2)*log(-(d*x + c + 1)/(d*x + c - 1))/((d*x + c + 1)^3*d^2/(d*x + c - 1)^3 - 3*(d*x + c + 1)^2*d^2/(d*x + c - 1)^2 + 3*(d*x + c + 1)*d^2/(d*x + c - 1) - d^2) - 2*(3*(d*x + c + 1)^2*a*e^2/(d*x + c - 1)^2 + a*e^2 + (d*x + c + 1)^2*b*e^2/(d*x + c - 1)^2 - (d*x + c + 1)*b*e^2 /(d*x + c - 1))/((d*x + c + 1)^3*d^2/(d*x + c - 1)^3 - 3*(d*x + c + 1)^2*d ^2/(d*x + c - 1)^2 + 3*(d*x + c + 1)*d^2/(d*x + c - 1) - d^2))
Time = 0.68 (sec) , antiderivative size = 237, normalized size of antiderivative = 3.43 \[ \int (c e+d e x)^2 (a+b \text {arctanh}(c+d x)) \, dx=\frac {a\,d^2\,e^2\,x^3}{3}+\frac {b\,c\,e^2\,x}{3}+\frac {b\,e^2\,\ln \left (c+d\,x-1\right )}{6\,d}+\frac {b\,e^2\,\ln \left (c+d\,x+1\right )}{6\,d}+a\,c^2\,e^2\,x+\frac {b\,d\,e^2\,x^2}{6}+a\,c\,d\,e^2\,x^2+\frac {b\,c^2\,e^2\,x\,\ln \left (c+d\,x+1\right )}{2}-\frac {b\,c^3\,e^2\,\ln \left (c+d\,x-1\right )}{6\,d}+\frac {b\,c^3\,e^2\,\ln \left (c+d\,x+1\right )}{6\,d}-\frac {b\,c^2\,e^2\,x\,\ln \left (1-d\,x-c\right )}{2}+\frac {b\,d^2\,e^2\,x^3\,\ln \left (c+d\,x+1\right )}{6}-\frac {b\,d^2\,e^2\,x^3\,\ln \left (1-d\,x-c\right )}{6}+\frac {b\,c\,d\,e^2\,x^2\,\ln \left (c+d\,x+1\right )}{2}-\frac {b\,c\,d\,e^2\,x^2\,\ln \left (1-d\,x-c\right )}{2} \]
(a*d^2*e^2*x^3)/3 + (b*c*e^2*x)/3 + (b*e^2*log(c + d*x - 1))/(6*d) + (b*e^ 2*log(c + d*x + 1))/(6*d) + a*c^2*e^2*x + (b*d*e^2*x^2)/6 + a*c*d*e^2*x^2 + (b*c^2*e^2*x*log(c + d*x + 1))/2 - (b*c^3*e^2*log(c + d*x - 1))/(6*d) + (b*c^3*e^2*log(c + d*x + 1))/(6*d) - (b*c^2*e^2*x*log(1 - d*x - c))/2 + (b *d^2*e^2*x^3*log(c + d*x + 1))/6 - (b*d^2*e^2*x^3*log(1 - d*x - c))/6 + (b *c*d*e^2*x^2*log(c + d*x + 1))/2 - (b*c*d*e^2*x^2*log(1 - d*x - c))/2