Integrand size = 18, antiderivative size = 77 \[ \int \frac {x (a+b \text {arctanh}(c x))}{(d+c d x)^3} \, dx=\frac {b}{8 c^2 d^3 (1+c x)^2}-\frac {3 b}{8 c^2 d^3 (1+c x)}-\frac {b \text {arctanh}(c x)}{8 c^2 d^3}+\frac {x^2 (a+b \text {arctanh}(c x))}{2 d^3 (1+c x)^2} \] Output:
1/8*b/c^2/d^3/(c*x+1)^2-3/8*b/c^2/d^3/(c*x+1)-1/8*b*arctanh(c*x)/c^2/d^3+1 /2*x^2*(a+b*arctanh(c*x))/d^3/(c*x+1)^2
Time = 0.09 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.29 \[ \int \frac {x (a+b \text {arctanh}(c x))}{(d+c d x)^3} \, dx=-\frac {8 a+4 b+16 a c x+6 b c x+8 (b+2 b c x) \text {arctanh}(c x)+3 b (1+c x)^2 \log (1-c x)-3 b \log (1+c x)-6 b c x \log (1+c x)-3 b c^2 x^2 \log (1+c x)}{16 c^2 d^3 (1+c x)^2} \] Input:
Integrate[(x*(a + b*ArcTanh[c*x]))/(d + c*d*x)^3,x]
Output:
-1/16*(8*a + 4*b + 16*a*c*x + 6*b*c*x + 8*(b + 2*b*c*x)*ArcTanh[c*x] + 3*b *(1 + c*x)^2*Log[1 - c*x] - 3*b*Log[1 + c*x] - 6*b*c*x*Log[1 + c*x] - 3*b* c^2*x^2*Log[1 + c*x])/(c^2*d^3*(1 + c*x)^2)
Time = 0.34 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.97, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {6498, 27, 99, 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.
\(\displaystyle \int \frac {x (a+b \text {arctanh}(c x))}{(c d x+d)^3} \, dx\) |
\(\Big \downarrow \) 6498 |
\(\displaystyle \frac {x^2 (a+b \text {arctanh}(c x))}{2 d^3 (c x+1)^2}-b c \int \frac {x^2}{2 d^3 (1-c x) (c x+1)^3}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {x^2 (a+b \text {arctanh}(c x))}{2 d^3 (c x+1)^2}-\frac {b c \int \frac {x^2}{(1-c x) (c x+1)^3}dx}{2 d^3}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \frac {x^2 (a+b \text {arctanh}(c x))}{2 d^3 (c x+1)^2}-\frac {b c \int \left (-\frac {3}{4 c^2 (c x+1)^2}+\frac {1}{2 c^2 (c x+1)^3}-\frac {1}{4 c^2 \left (c^2 x^2-1\right )}\right )dx}{2 d^3}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {x^2 (a+b \text {arctanh}(c x))}{2 d^3 (c x+1)^2}-\frac {b c \left (\frac {\text {arctanh}(c x)}{4 c^3}+\frac {3}{4 c^3 (c x+1)}-\frac {1}{4 c^3 (c x+1)^2}\right )}{2 d^3}\) |
Input:
Int[(x*(a + b*ArcTanh[c*x]))/(d + c*d*x)^3,x]
Output:
(x^2*(a + b*ArcTanh[c*x]))/(2*d^3*(1 + c*x)^2) - (b*c*(-1/4*1/(c^3*(1 + c* x)^2) + 3/(4*c^3*(1 + c*x)) + ArcTanh[c*x]/(4*c^3)))/(2*d^3)
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_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | | (GtQ[m, 0] && GeQ[n, -1]))
Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_.)*((d_.) + (e_.)*( x_))^(q_.), x_Symbol] :> With[{u = IntHide[(f*x)^m*(d + e*x)^q, x]}, Simp[( a + b*ArcTanh[c*x]) u, x] - Simp[b*c Int[SimplifyIntegrand[u/(1 - c^2*x ^2), x], x], x]] /; FreeQ[{a, b, c, d, e, f, q}, x] && NeQ[q, -1] && Intege rQ[2*m] && ((IGtQ[m, 0] && IGtQ[q, 0]) || (ILtQ[m + q + 1, 0] && LtQ[m*q, 0 ]))
Time = 0.34 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.88
method | result | size |
parallelrisch | \(\frac {3 \,\operatorname {arctanh}\left (c x \right ) b \,c^{2} x^{2}+4 a \,c^{2} x^{2}+2 b \,c^{2} x^{2}-2 \,\operatorname {arctanh}\left (c x \right ) b c x +b c x -b \,\operatorname {arctanh}\left (c x \right )}{8 d^{3} \left (c x +1\right )^{2} c^{2}}\) | \(68\) |
derivativedivides | \(\frac {\frac {a \left (\frac {1}{2 \left (c x +1\right )^{2}}-\frac {1}{c x +1}\right )}{d^{3}}+\frac {b \left (\frac {\operatorname {arctanh}\left (c x \right )}{2 \left (c x +1\right )^{2}}-\frac {\operatorname {arctanh}\left (c x \right )}{c x +1}-\frac {3 \ln \left (c x -1\right )}{16}+\frac {1}{8 \left (c x +1\right )^{2}}-\frac {3}{8 \left (c x +1\right )}+\frac {3 \ln \left (c x +1\right )}{16}\right )}{d^{3}}}{c^{2}}\) | \(96\) |
default | \(\frac {\frac {a \left (\frac {1}{2 \left (c x +1\right )^{2}}-\frac {1}{c x +1}\right )}{d^{3}}+\frac {b \left (\frac {\operatorname {arctanh}\left (c x \right )}{2 \left (c x +1\right )^{2}}-\frac {\operatorname {arctanh}\left (c x \right )}{c x +1}-\frac {3 \ln \left (c x -1\right )}{16}+\frac {1}{8 \left (c x +1\right )^{2}}-\frac {3}{8 \left (c x +1\right )}+\frac {3 \ln \left (c x +1\right )}{16}\right )}{d^{3}}}{c^{2}}\) | \(96\) |
parts | \(\frac {a \left (-\frac {1}{c^{2} \left (c x +1\right )}+\frac {1}{2 c^{2} \left (c x +1\right )^{2}}\right )}{d^{3}}+\frac {b \left (\frac {\operatorname {arctanh}\left (c x \right )}{2 \left (c x +1\right )^{2}}-\frac {\operatorname {arctanh}\left (c x \right )}{c x +1}-\frac {3 \ln \left (c x -1\right )}{16}+\frac {1}{8 \left (c x +1\right )^{2}}-\frac {3}{8 \left (c x +1\right )}+\frac {3 \ln \left (c x +1\right )}{16}\right )}{d^{3} c^{2}}\) | \(101\) |
orering | \(-\frac {\left (c x +1\right ) \left (4 x^{3} c^{3}-7 c^{2} x^{2}+c x +2\right ) \left (a +b \,\operatorname {arctanh}\left (c x \right )\right )}{8 c^{2} \left (c d x +d \right )^{3}}-\frac {\left (2 c x +1\right ) \left (c x -1\right ) \left (c x +1\right )^{2} \left (\frac {a +b \,\operatorname {arctanh}\left (c x \right )}{\left (c d x +d \right )^{3}}+\frac {x b c}{\left (-c^{2} x^{2}+1\right ) \left (c d x +d \right )^{3}}-\frac {3 x \left (a +b \,\operatorname {arctanh}\left (c x \right )\right ) c d}{\left (c d x +d \right )^{4}}\right )}{8 c^{2}}\) | \(135\) |
risch | \(-\frac {b \left (2 c x +1\right ) \ln \left (c x +1\right )}{4 c^{2} d^{3} \left (c x +1\right )^{2}}-\frac {3 \ln \left (c x -1\right ) b \,c^{2} x^{2}-3 \ln \left (-c x -1\right ) b \,c^{2} x^{2}+6 \ln \left (c x -1\right ) b c x -6 \ln \left (-c x -1\right ) b c x -8 b c x \ln \left (-c x +1\right )+16 a c x +6 b c x +3 b \ln \left (c x -1\right )-3 b \ln \left (-c x -1\right )-4 b \ln \left (-c x +1\right )+8 a +4 b}{16 c^{2} d^{3} \left (c x +1\right )^{2}}\) | \(157\) |
Input:
int(x*(a+b*arctanh(c*x))/(c*d*x+d)^3,x,method=_RETURNVERBOSE)
Output:
1/8*(3*arctanh(c*x)*b*c^2*x^2+4*a*c^2*x^2+2*b*c^2*x^2-2*arctanh(c*x)*b*c*x +b*c*x-b*arctanh(c*x))/d^3/(c*x+1)^2/c^2
Time = 0.08 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.09 \[ \int \frac {x (a+b \text {arctanh}(c x))}{(d+c d x)^3} \, dx=-\frac {2 \, {\left (8 \, a + 3 \, b\right )} c x - {\left (3 \, b c^{2} x^{2} - 2 \, b c x - b\right )} \log \left (-\frac {c x + 1}{c x - 1}\right ) + 8 \, a + 4 \, b}{16 \, {\left (c^{4} d^{3} x^{2} + 2 \, c^{3} d^{3} x + c^{2} d^{3}\right )}} \] Input:
integrate(x*(a+b*arctanh(c*x))/(c*d*x+d)^3,x, algorithm="fricas")
Output:
-1/16*(2*(8*a + 3*b)*c*x - (3*b*c^2*x^2 - 2*b*c*x - b)*log(-(c*x + 1)/(c*x - 1)) + 8*a + 4*b)/(c^4*d^3*x^2 + 2*c^3*d^3*x + c^2*d^3)
Leaf count of result is larger than twice the leaf count of optimal. 277 vs. \(2 (71) = 142\).
Time = 0.84 (sec) , antiderivative size = 277, normalized size of antiderivative = 3.60 \[ \int \frac {x (a+b \text {arctanh}(c x))}{(d+c d x)^3} \, dx=\begin {cases} - \frac {8 a c x}{8 c^{4} d^{3} x^{2} + 16 c^{3} d^{3} x + 8 c^{2} d^{3}} - \frac {4 a}{8 c^{4} d^{3} x^{2} + 16 c^{3} d^{3} x + 8 c^{2} d^{3}} + \frac {3 b c^{2} x^{2} \operatorname {atanh}{\left (c x \right )}}{8 c^{4} d^{3} x^{2} + 16 c^{3} d^{3} x + 8 c^{2} d^{3}} - \frac {2 b c x \operatorname {atanh}{\left (c x \right )}}{8 c^{4} d^{3} x^{2} + 16 c^{3} d^{3} x + 8 c^{2} d^{3}} - \frac {3 b c x}{8 c^{4} d^{3} x^{2} + 16 c^{3} d^{3} x + 8 c^{2} d^{3}} - \frac {b \operatorname {atanh}{\left (c x \right )}}{8 c^{4} d^{3} x^{2} + 16 c^{3} d^{3} x + 8 c^{2} d^{3}} - \frac {2 b}{8 c^{4} d^{3} x^{2} + 16 c^{3} d^{3} x + 8 c^{2} d^{3}} & \text {for}\: c \neq 0 \\\frac {a x^{2}}{2 d^{3}} & \text {otherwise} \end {cases} \] Input:
integrate(x*(a+b*atanh(c*x))/(c*d*x+d)**3,x)
Output:
Piecewise((-8*a*c*x/(8*c**4*d**3*x**2 + 16*c**3*d**3*x + 8*c**2*d**3) - 4* a/(8*c**4*d**3*x**2 + 16*c**3*d**3*x + 8*c**2*d**3) + 3*b*c**2*x**2*atanh( c*x)/(8*c**4*d**3*x**2 + 16*c**3*d**3*x + 8*c**2*d**3) - 2*b*c*x*atanh(c*x )/(8*c**4*d**3*x**2 + 16*c**3*d**3*x + 8*c**2*d**3) - 3*b*c*x/(8*c**4*d**3 *x**2 + 16*c**3*d**3*x + 8*c**2*d**3) - b*atanh(c*x)/(8*c**4*d**3*x**2 + 1 6*c**3*d**3*x + 8*c**2*d**3) - 2*b/(8*c**4*d**3*x**2 + 16*c**3*d**3*x + 8* c**2*d**3), Ne(c, 0)), (a*x**2/(2*d**3), True))
Leaf count of result is larger than twice the leaf count of optimal. 152 vs. \(2 (69) = 138\).
Time = 0.04 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.97 \[ \int \frac {x (a+b \text {arctanh}(c x))}{(d+c d x)^3} \, dx=-\frac {1}{16} \, {\left (c {\left (\frac {2 \, {\left (3 \, c x + 2\right )}}{c^{5} d^{3} x^{2} + 2 \, c^{4} d^{3} x + c^{3} d^{3}} - \frac {3 \, \log \left (c x + 1\right )}{c^{3} d^{3}} + \frac {3 \, \log \left (c x - 1\right )}{c^{3} d^{3}}\right )} + \frac {8 \, {\left (2 \, c x + 1\right )} \operatorname {artanh}\left (c x\right )}{c^{4} d^{3} x^{2} + 2 \, c^{3} d^{3} x + c^{2} d^{3}}\right )} b - \frac {{\left (2 \, c x + 1\right )} a}{2 \, {\left (c^{4} d^{3} x^{2} + 2 \, c^{3} d^{3} x + c^{2} d^{3}\right )}} \] Input:
integrate(x*(a+b*arctanh(c*x))/(c*d*x+d)^3,x, algorithm="maxima")
Output:
-1/16*(c*(2*(3*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)) + 8*(2*c*x + 1)*arctanh(c*x)/(c ^4*d^3*x^2 + 2*c^3*d^3*x + c^2*d^3))*b - 1/2*(2*c*x + 1)*a/(c^4*d^3*x^2 + 2*c^3*d^3*x + c^2*d^3)
Time = 0.12 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.48 \[ \int \frac {x (a+b \text {arctanh}(c x))}{(d+c d x)^3} \, dx=\frac {1}{32} \, c {\left (\frac {2 \, {\left (c x - 1\right )}^{2} {\left (\frac {2 \, {\left (c x + 1\right )} b}{c x - 1} + b\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{{\left (c x + 1\right )}^{2} c^{3} d^{3}} + \frac {{\left (c x - 1\right )}^{2} {\left (\frac {8 \, {\left (c x + 1\right )} a}{c x - 1} + 4 \, a + \frac {4 \, {\left (c x + 1\right )} b}{c x - 1} + b\right )}}{{\left (c x + 1\right )}^{2} c^{3} d^{3}}\right )} \] Input:
integrate(x*(a+b*arctanh(c*x))/(c*d*x+d)^3,x, algorithm="giac")
Output:
1/32*c*(2*(c*x - 1)^2*(2*(c*x + 1)*b/(c*x - 1) + b)*log(-(c*x + 1)/(c*x - 1))/((c*x + 1)^2*c^3*d^3) + (c*x - 1)^2*(8*(c*x + 1)*a/(c*x - 1) + 4*a + 4 *(c*x + 1)*b/(c*x - 1) + b)/((c*x + 1)^2*c^3*d^3))
Time = 4.40 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.05 \[ \int \frac {x (a+b \text {arctanh}(c x))}{(d+c d x)^3} \, dx=\frac {c\,\left (b\,x-2\,b\,x\,\mathrm {atanh}\left (c\,x\right )\right )-b\,\mathrm {atanh}\left (c\,x\right )+c^2\,\left (4\,a\,x^2+2\,b\,x^2+3\,b\,x^2\,\mathrm {atanh}\left (c\,x\right )\right )}{8\,c^4\,d^3\,x^2+16\,c^3\,d^3\,x+8\,c^2\,d^3} \] Input:
int((x*(a + b*atanh(c*x)))/(d + c*d*x)^3,x)
Output:
(c*(b*x - 2*b*x*atanh(c*x)) - b*atanh(c*x) + c^2*(4*a*x^2 + 2*b*x^2 + 3*b* x^2*atanh(c*x)))/(8*c^2*d^3 + 16*c^3*d^3*x + 8*c^4*d^3*x^2)
Time = 0.17 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.64 \[ \int \frac {x (a+b \text {arctanh}(c x))}{(d+c d x)^3} \, dx=\frac {8 \mathit {atanh} \left (c x \right ) b \,c^{2} x^{2}+\mathrm {log}\left (c x -1\right ) b \,c^{2} x^{2}+2 \,\mathrm {log}\left (c x -1\right ) b c x +\mathrm {log}\left (c x -1\right ) b -\mathrm {log}\left (c x +1\right ) b \,c^{2} x^{2}-2 \,\mathrm {log}\left (c x +1\right ) b c x -\mathrm {log}\left (c x +1\right ) b +8 a \,c^{2} x^{2}+3 b \,c^{2} x^{2}-b}{16 c^{2} d^{3} \left (c^{2} x^{2}+2 c x +1\right )} \] Input:
int(x*(a+b*atanh(c*x))/(c*d*x+d)^3,x)
Output:
(8*atanh(c*x)*b*c**2*x**2 + log(c*x - 1)*b*c**2*x**2 + 2*log(c*x - 1)*b*c* x + log(c*x - 1)*b - log(c*x + 1)*b*c**2*x**2 - 2*log(c*x + 1)*b*c*x - log (c*x + 1)*b + 8*a*c**2*x**2 + 3*b*c**2*x**2 - b)/(16*c**2*d**3*(c**2*x**2 + 2*c*x + 1))