Integrand size = 17, antiderivative size = 71 \[ \int (d+c d x)^2 (a+b \text {arctanh}(c x)) \, dx=\frac {2}{3} b d^2 x+\frac {b d^2 (1+c x)^2}{6 c}+\frac {d^2 (1+c x)^3 (a+b \text {arctanh}(c x))}{3 c}+\frac {4 b d^2 \log (1-c x)}{3 c} \]
2/3*b*d^2*x+1/6*b*d^2*(c*x+1)^2/c+1/3*d^2*(c*x+1)^3*(a+b*arctanh(c*x))/c+4 /3*b*d^2*ln(-c*x+1)/c
Time = 0.06 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.30 \[ \int (d+c d x)^2 (a+b \text {arctanh}(c x)) \, dx=\frac {d^2 \left (6 a c x+6 b c x+6 a c^2 x^2+b c^2 x^2+2 a c^3 x^3+2 b c x \left (3+3 c x+c^2 x^2\right ) \text {arctanh}(c x)+6 b \log (1-c x)+b \log \left (1-c^2 x^2\right )\right )}{6 c} \]
(d^2*(6*a*c*x + 6*b*c*x + 6*a*c^2*x^2 + b*c^2*x^2 + 2*a*c^3*x^3 + 2*b*c*x* (3 + 3*c*x + c^2*x^2)*ArcTanh[c*x] + 6*b*Log[1 - c*x] + b*Log[1 - c^2*x^2] ))/(6*c)
Time = 0.24 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.82, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {6478, 27, 456, 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.
\(\displaystyle \int (c d x+d)^2 (a+b \text {arctanh}(c x)) \, dx\) |
\(\Big \downarrow \) 6478 |
\(\displaystyle \frac {d^2 (c x+1)^3 (a+b \text {arctanh}(c x))}{3 c}-\frac {b \int \frac {d^3 (c x+1)^3}{1-c^2 x^2}dx}{3 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {d^2 (c x+1)^3 (a+b \text {arctanh}(c x))}{3 c}-\frac {1}{3} b d^2 \int \frac {(c x+1)^3}{1-c^2 x^2}dx\) |
\(\Big \downarrow \) 456 |
\(\displaystyle \frac {d^2 (c x+1)^3 (a+b \text {arctanh}(c x))}{3 c}-\frac {1}{3} b d^2 \int \frac {(c x+1)^2}{1-c x}dx\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {d^2 (c x+1)^3 (a+b \text {arctanh}(c x))}{3 c}-\frac {1}{3} b d^2 \int \left (-c x+\frac {4}{1-c x}-3\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {d^2 (c x+1)^3 (a+b \text {arctanh}(c x))}{3 c}-\frac {1}{3} b d^2 \left (-\frac {c x^2}{2}-\frac {4 \log (1-c x)}{c}-3 x\right )\) |
(d^2*(1 + c*x)^3*(a + b*ArcTanh[c*x]))/(3*c) - (b*d^2*(-3*x - (c*x^2)/2 - (4*Log[1 - c*x])/c))/3
3.1.13.3.1 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[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ (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] && !Integ erQ[n]))
Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))*((d_) + (e_.)*(x_))^(q_.), x_Symbol ] :> Simp[(d + e*x)^(q + 1)*((a + b*ArcTanh[c*x])/(e*(q + 1))), x] - Simp[b *(c/(e*(q + 1))) Int[(d + e*x)^(q + 1)/(1 - c^2*x^2), x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[q, -1]
Time = 0.96 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.13
method | result | size |
derivativedivides | \(\frac {\frac {a \,d^{2} \left (c x +1\right )^{3}}{3}+d^{2} b \left (\frac {c^{3} x^{3} \operatorname {arctanh}\left (c x \right )}{3}+c^{2} x^{2} \operatorname {arctanh}\left (c x \right )+c x \,\operatorname {arctanh}\left (c x \right )+\frac {\operatorname {arctanh}\left (c x \right )}{3}+\frac {c^{2} x^{2}}{6}+c x +\frac {4 \ln \left (c x -1\right )}{3}\right )}{c}\) | \(80\) |
default | \(\frac {\frac {a \,d^{2} \left (c x +1\right )^{3}}{3}+d^{2} b \left (\frac {c^{3} x^{3} \operatorname {arctanh}\left (c x \right )}{3}+c^{2} x^{2} \operatorname {arctanh}\left (c x \right )+c x \,\operatorname {arctanh}\left (c x \right )+\frac {\operatorname {arctanh}\left (c x \right )}{3}+\frac {c^{2} x^{2}}{6}+c x +\frac {4 \ln \left (c x -1\right )}{3}\right )}{c}\) | \(80\) |
parts | \(\frac {a \,d^{2} \left (c x +1\right )^{3}}{3 c}+\frac {d^{2} b \left (\frac {c^{3} x^{3} \operatorname {arctanh}\left (c x \right )}{3}+c^{2} x^{2} \operatorname {arctanh}\left (c x \right )+c x \,\operatorname {arctanh}\left (c x \right )+\frac {\operatorname {arctanh}\left (c x \right )}{3}+\frac {c^{2} x^{2}}{6}+c x +\frac {4 \ln \left (c x -1\right )}{3}\right )}{c}\) | \(82\) |
parallelrisch | \(\frac {2 x^{3} \operatorname {arctanh}\left (c x \right ) b \,d^{2} c^{3}+2 a \,c^{3} d^{2} x^{3}+6 x^{2} \operatorname {arctanh}\left (c x \right ) b \,c^{2} d^{2}+6 a \,c^{2} d^{2} x^{2}+b \,c^{2} d^{2} x^{2}+6 b c \,d^{2} x \,\operatorname {arctanh}\left (c x \right )+6 a \,d^{2} c x +6 b c \,d^{2} x +8 \ln \left (c x -1\right ) b \,d^{2}+2 b \,d^{2} \operatorname {arctanh}\left (c x \right )}{6 c}\) | \(124\) |
risch | \(\frac {d^{2} \left (c x +1\right )^{3} b \ln \left (c x +1\right )}{6 c}-\frac {d^{2} c^{2} b \,x^{3} \ln \left (-c x +1\right )}{6}+\frac {a \,c^{2} d^{2} x^{3}}{3}-\frac {d^{2} c b \,x^{2} \ln \left (-c x +1\right )}{2}+a c \,d^{2} x^{2}+\frac {b c \,d^{2} x^{2}}{6}-\frac {b \,d^{2} x \ln \left (-c x +1\right )}{2}+a \,d^{2} x +d^{2} b x -\frac {\ln \left (-c x +1\right ) b \,d^{2}}{6 c}+\frac {4 d^{2} b \ln \left (c x -1\right )}{3 c}\) | \(148\) |
1/c*(1/3*a*d^2*(c*x+1)^3+d^2*b*(1/3*c^3*x^3*arctanh(c*x)+c^2*x^2*arctanh(c *x)+c*x*arctanh(c*x)+1/3*arctanh(c*x)+1/6*c^2*x^2+c*x+4/3*ln(c*x-1)))
Time = 0.26 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.61 \[ \int (d+c d x)^2 (a+b \text {arctanh}(c x)) \, dx=\frac {2 \, a c^{3} d^{2} x^{3} + {\left (6 \, a + b\right )} c^{2} d^{2} x^{2} + 6 \, {\left (a + b\right )} c d^{2} x + b d^{2} \log \left (c x + 1\right ) + 7 \, b d^{2} \log \left (c x - 1\right ) + {\left (b c^{3} d^{2} x^{3} + 3 \, b c^{2} d^{2} x^{2} + 3 \, b c d^{2} x\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{6 \, c} \]
1/6*(2*a*c^3*d^2*x^3 + (6*a + b)*c^2*d^2*x^2 + 6*(a + b)*c*d^2*x + b*d^2*l og(c*x + 1) + 7*b*d^2*log(c*x - 1) + (b*c^3*d^2*x^3 + 3*b*c^2*d^2*x^2 + 3* b*c*d^2*x)*log(-(c*x + 1)/(c*x - 1)))/c
Leaf count of result is larger than twice the leaf count of optimal. 131 vs. \(2 (63) = 126\).
Time = 0.22 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.85 \[ \int (d+c d x)^2 (a+b \text {arctanh}(c x)) \, dx=\begin {cases} \frac {a c^{2} d^{2} x^{3}}{3} + a c d^{2} x^{2} + a d^{2} x + \frac {b c^{2} d^{2} x^{3} \operatorname {atanh}{\left (c x \right )}}{3} + b c d^{2} x^{2} \operatorname {atanh}{\left (c x \right )} + \frac {b c d^{2} x^{2}}{6} + b d^{2} x \operatorname {atanh}{\left (c x \right )} + b d^{2} x + \frac {4 b d^{2} \log {\left (x - \frac {1}{c} \right )}}{3 c} + \frac {b d^{2} \operatorname {atanh}{\left (c x \right )}}{3 c} & \text {for}\: c \neq 0 \\a d^{2} x & \text {otherwise} \end {cases} \]
Piecewise((a*c**2*d**2*x**3/3 + a*c*d**2*x**2 + a*d**2*x + b*c**2*d**2*x** 3*atanh(c*x)/3 + b*c*d**2*x**2*atanh(c*x) + b*c*d**2*x**2/6 + b*d**2*x*ata nh(c*x) + b*d**2*x + 4*b*d**2*log(x - 1/c)/(3*c) + b*d**2*atanh(c*x)/(3*c) , Ne(c, 0)), (a*d**2*x, True))
Leaf count of result is larger than twice the leaf count of optimal. 147 vs. \(2 (63) = 126\).
Time = 0.20 (sec) , antiderivative size = 147, normalized size of antiderivative = 2.07 \[ \int (d+c d x)^2 (a+b \text {arctanh}(c x)) \, dx=\frac {1}{3} \, a c^{2} d^{2} x^{3} + \frac {1}{6} \, {\left (2 \, x^{3} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {x^{2}}{c^{2}} + \frac {\log \left (c^{2} x^{2} - 1\right )}{c^{4}}\right )}\right )} b c^{2} d^{2} + a c d^{2} x^{2} + \frac {1}{2} \, {\left (2 \, x^{2} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {2 \, x}{c^{2}} - \frac {\log \left (c x + 1\right )}{c^{3}} + \frac {\log \left (c x - 1\right )}{c^{3}}\right )}\right )} b c d^{2} + a d^{2} x + \frac {{\left (2 \, c x \operatorname {artanh}\left (c x\right ) + \log \left (-c^{2} x^{2} + 1\right )\right )} b d^{2}}{2 \, c} \]
1/3*a*c^2*d^2*x^3 + 1/6*(2*x^3*arctanh(c*x) + c*(x^2/c^2 + log(c^2*x^2 - 1 )/c^4))*b*c^2*d^2 + a*c*d^2*x^2 + 1/2*(2*x^2*arctanh(c*x) + c*(2*x/c^2 - l og(c*x + 1)/c^3 + log(c*x - 1)/c^3))*b*c*d^2 + a*d^2*x + 1/2*(2*c*x*arctan h(c*x) + log(-c^2*x^2 + 1))*b*d^2/c
Leaf count of result is larger than twice the leaf count of optimal. 330 vs. \(2 (63) = 126\).
Time = 0.28 (sec) , antiderivative size = 330, normalized size of antiderivative = 4.65 \[ \int (d+c d x)^2 (a+b \text {arctanh}(c x)) \, dx=-\frac {2}{3} \, {\left (\frac {2 \, b d^{2} \log \left (-\frac {c x + 1}{c x - 1} + 1\right )}{c^{2}} - \frac {2 \, b d^{2} \log \left (-\frac {c x + 1}{c x - 1}\right )}{c^{2}} - \frac {2 \, {\left (\frac {3 \, {\left (c x + 1\right )}^{2} b d^{2}}{{\left (c x - 1\right )}^{2}} - \frac {3 \, {\left (c x + 1\right )} b d^{2}}{c x - 1} + b d^{2}\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{\frac {{\left (c x + 1\right )}^{3} c^{2}}{{\left (c x - 1\right )}^{3}} - \frac {3 \, {\left (c x + 1\right )}^{2} c^{2}}{{\left (c x - 1\right )}^{2}} + \frac {3 \, {\left (c x + 1\right )} c^{2}}{c x - 1} - c^{2}} - \frac {\frac {12 \, {\left (c x + 1\right )}^{2} a d^{2}}{{\left (c x - 1\right )}^{2}} - \frac {12 \, {\left (c x + 1\right )} a d^{2}}{c x - 1} + 4 \, a d^{2} + \frac {4 \, {\left (c x + 1\right )}^{2} b d^{2}}{{\left (c x - 1\right )}^{2}} - \frac {7 \, {\left (c x + 1\right )} b d^{2}}{c x - 1} + 3 \, b d^{2}}{\frac {{\left (c x + 1\right )}^{3} c^{2}}{{\left (c x - 1\right )}^{3}} - \frac {3 \, {\left (c x + 1\right )}^{2} c^{2}}{{\left (c x - 1\right )}^{2}} + \frac {3 \, {\left (c x + 1\right )} c^{2}}{c x - 1} - c^{2}}\right )} c \]
-2/3*(2*b*d^2*log(-(c*x + 1)/(c*x - 1) + 1)/c^2 - 2*b*d^2*log(-(c*x + 1)/( c*x - 1))/c^2 - 2*(3*(c*x + 1)^2*b*d^2/(c*x - 1)^2 - 3*(c*x + 1)*b*d^2/(c* x - 1) + b*d^2)*log(-(c*x + 1)/(c*x - 1))/((c*x + 1)^3*c^2/(c*x - 1)^3 - 3 *(c*x + 1)^2*c^2/(c*x - 1)^2 + 3*(c*x + 1)*c^2/(c*x - 1) - c^2) - (12*(c*x + 1)^2*a*d^2/(c*x - 1)^2 - 12*(c*x + 1)*a*d^2/(c*x - 1) + 4*a*d^2 + 4*(c* x + 1)^2*b*d^2/(c*x - 1)^2 - 7*(c*x + 1)*b*d^2/(c*x - 1) + 3*b*d^2)/((c*x + 1)^3*c^2/(c*x - 1)^3 - 3*(c*x + 1)^2*c^2/(c*x - 1)^2 + 3*(c*x + 1)*c^2/( c*x - 1) - c^2))*c
Time = 3.73 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.48 \[ \int (d+c d x)^2 (a+b \text {arctanh}(c x)) \, dx=\frac {d^2\,\left (6\,a\,x+6\,b\,x+6\,b\,x\,\mathrm {atanh}\left (c\,x\right )\right )}{6}+\frac {c^2\,d^2\,\left (2\,a\,x^3+2\,b\,x^3\,\mathrm {atanh}\left (c\,x\right )\right )}{6}-\frac {d^2\,\left (6\,b\,\mathrm {atanh}\left (c\,x\right )-4\,b\,\ln \left (c^2\,x^2-1\right )\right )}{6\,c}+\frac {c\,d^2\,\left (6\,a\,x^2+b\,x^2+6\,b\,x^2\,\mathrm {atanh}\left (c\,x\right )\right )}{6} \]