Integrand size = 17, antiderivative size = 57 \[ \int \frac {a+b \text {arctanh}(c x)}{(d+c d x)^2} \, dx=-\frac {b}{2 c d^2 (1+c x)}+\frac {b \text {arctanh}(c x)}{2 c d^2}-\frac {a+b \text {arctanh}(c x)}{c d^2 (1+c x)} \] Output:
-1/2*b/c/d^2/(c*x+1)+1/2*b*arctanh(c*x)/c/d^2-(a+b*arctanh(c*x))/c/d^2/(c* x+1)
Time = 0.09 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.12 \[ \int \frac {a+b \text {arctanh}(c x)}{(d+c d x)^2} \, dx=\frac {-4 a-2 b-4 b \text {arctanh}(c x)-(b+b c x) \log (1-c x)+b \log (1+c x)+b c x \log (1+c x)}{4 c d^2 (1+c x)} \] Input:
Integrate[(a + b*ArcTanh[c*x])/(d + c*d*x)^2,x]
Output:
(-4*a - 2*b - 4*b*ArcTanh[c*x] - (b + b*c*x)*Log[1 - c*x] + b*Log[1 + c*x] + b*c*x*Log[1 + c*x])/(4*c*d^2*(1 + c*x))
Time = 0.25 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.96, 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, 54, 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.
\(\displaystyle \int \frac {a+b \text {arctanh}(c x)}{(c d x+d)^2} \, dx\) |
\(\Big \downarrow \) 6478 |
\(\displaystyle \frac {b \int \frac {1}{d (c x+1) \left (1-c^2 x^2\right )}dx}{d}-\frac {a+b \text {arctanh}(c x)}{c d^2 (c x+1)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {b \int \frac {1}{(c x+1) \left (1-c^2 x^2\right )}dx}{d^2}-\frac {a+b \text {arctanh}(c x)}{c d^2 (c x+1)}\) |
\(\Big \downarrow \) 456 |
\(\displaystyle \frac {b \int \frac {1}{(1-c x) (c x+1)^2}dx}{d^2}-\frac {a+b \text {arctanh}(c x)}{c d^2 (c x+1)}\) |
\(\Big \downarrow \) 54 |
\(\displaystyle \frac {b \int \left (\frac {1}{2 (c x+1)^2}-\frac {1}{2 \left (c^2 x^2-1\right )}\right )dx}{d^2}-\frac {a+b \text {arctanh}(c x)}{c d^2 (c x+1)}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {b \left (\frac {\text {arctanh}(c x)}{2 c}-\frac {1}{2 c (c x+1)}\right )}{d^2}-\frac {a+b \text {arctanh}(c x)}{c d^2 (c x+1)}\) |
Input:
Int[(a + b*ArcTanh[c*x])/(d + c*d*x)^2,x]
Output:
-((a + b*ArcTanh[c*x])/(c*d^2*(1 + c*x))) + (b*(-1/2*1/(c*(1 + c*x)) + Arc Tanh[c*x]/(2*c)))/d^2
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[E xpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && ILtQ[m, 0] && IntegerQ[n] && !(IGtQ[n, 0] && LtQ[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 = 0.21 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.72
method | result | size |
parallelrisch | \(\frac {\operatorname {arctanh}\left (c x \right ) b c x +2 a c x +b c x -b \,\operatorname {arctanh}\left (c x \right )}{2 d^{2} \left (c x +1\right ) c}\) | \(41\) |
derivativedivides | \(\frac {-\frac {a}{d^{2} \left (c x +1\right )}+\frac {b \left (-\frac {\operatorname {arctanh}\left (c x \right )}{c x +1}-\frac {\ln \left (c x -1\right )}{4}-\frac {1}{2 \left (c x +1\right )}+\frac {\ln \left (c x +1\right )}{4}\right )}{d^{2}}}{c}\) | \(63\) |
default | \(\frac {-\frac {a}{d^{2} \left (c x +1\right )}+\frac {b \left (-\frac {\operatorname {arctanh}\left (c x \right )}{c x +1}-\frac {\ln \left (c x -1\right )}{4}-\frac {1}{2 \left (c x +1\right )}+\frac {\ln \left (c x +1\right )}{4}\right )}{d^{2}}}{c}\) | \(63\) |
parts | \(-\frac {a}{d^{2} c \left (c x +1\right )}+\frac {b \left (-\frac {\operatorname {arctanh}\left (c x \right )}{c x +1}-\frac {\ln \left (c x -1\right )}{4}-\frac {1}{2 \left (c x +1\right )}+\frac {\ln \left (c x +1\right )}{4}\right )}{d^{2} c}\) | \(65\) |
risch | \(-\frac {b \ln \left (c x +1\right )}{2 c \,d^{2} \left (c x +1\right )}+\frac {\ln \left (-c x -1\right ) b c x -\ln \left (c x -1\right ) b c x +b \ln \left (-c x -1\right )-b \ln \left (c x -1\right )+2 b \ln \left (-c x +1\right )-4 a -2 b}{4 d^{2} \left (c x +1\right ) c}\) | \(96\) |
orering | \(-\frac {\left (c x +1\right ) \left (2 c^{2} x^{2}-3 c x +1\right ) \left (a +b \,\operatorname {arctanh}\left (c x \right )\right )}{2 c \left (c d x +d \right )^{2}}-\frac {x \left (c x -1\right ) \left (c x +1\right )^{2} \left (\frac {b c}{\left (-c^{2} x^{2}+1\right ) \left (c d x +d \right )^{2}}-\frac {2 \left (a +b \,\operatorname {arctanh}\left (c x \right )\right ) c d}{\left (c d x +d \right )^{3}}\right )}{2 c}\) | \(104\) |
Input:
int((a+b*arctanh(c*x))/(c*d*x+d)^2,x,method=_RETURNVERBOSE)
Output:
1/2*(arctanh(c*x)*b*c*x+2*a*c*x+b*c*x-b*arctanh(c*x))/d^2/(c*x+1)/c
Time = 0.08 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.86 \[ \int \frac {a+b \text {arctanh}(c x)}{(d+c d x)^2} \, dx=\frac {{\left (b c x - b\right )} \log \left (-\frac {c x + 1}{c x - 1}\right ) - 4 \, a - 2 \, b}{4 \, {\left (c^{2} d^{2} x + c d^{2}\right )}} \] Input:
integrate((a+b*arctanh(c*x))/(c*d*x+d)^2,x, algorithm="fricas")
Output:
1/4*((b*c*x - b)*log(-(c*x + 1)/(c*x - 1)) - 4*a - 2*b)/(c^2*d^2*x + c*d^2 )
Leaf count of result is larger than twice the leaf count of optimal. 95 vs. \(2 (44) = 88\).
Time = 0.58 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.67 \[ \int \frac {a+b \text {arctanh}(c x)}{(d+c d x)^2} \, dx=\begin {cases} - \frac {2 a}{2 c^{2} d^{2} x + 2 c d^{2}} + \frac {b c x \operatorname {atanh}{\left (c x \right )}}{2 c^{2} d^{2} x + 2 c d^{2}} - \frac {b \operatorname {atanh}{\left (c x \right )}}{2 c^{2} d^{2} x + 2 c d^{2}} - \frac {b}{2 c^{2} d^{2} x + 2 c d^{2}} & \text {for}\: c \neq 0 \\\frac {a x}{d^{2}} & \text {otherwise} \end {cases} \] Input:
integrate((a+b*atanh(c*x))/(c*d*x+d)**2,x)
Output:
Piecewise((-2*a/(2*c**2*d**2*x + 2*c*d**2) + b*c*x*atanh(c*x)/(2*c**2*d**2 *x + 2*c*d**2) - b*atanh(c*x)/(2*c**2*d**2*x + 2*c*d**2) - b/(2*c**2*d**2* x + 2*c*d**2), Ne(c, 0)), (a*x/d**2, True))
Time = 0.03 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.68 \[ \int \frac {a+b \text {arctanh}(c x)}{(d+c d x)^2} \, dx=-\frac {1}{4} \, {\left (c {\left (\frac {2}{c^{3} d^{2} x + c^{2} d^{2}} - \frac {\log \left (c x + 1\right )}{c^{2} d^{2}} + \frac {\log \left (c x - 1\right )}{c^{2} d^{2}}\right )} + \frac {4 \, \operatorname {artanh}\left (c x\right )}{c^{2} d^{2} x + c d^{2}}\right )} b - \frac {a}{c^{2} d^{2} x + c d^{2}} \] Input:
integrate((a+b*arctanh(c*x))/(c*d*x+d)^2,x, algorithm="maxima")
Output:
-1/4*(c*(2/(c^3*d^2*x + c^2*d^2) - log(c*x + 1)/(c^2*d^2) + log(c*x - 1)/( c^2*d^2)) + 4*arctanh(c*x)/(c^2*d^2*x + c*d^2))*b - a/(c^2*d^2*x + c*d^2)
Time = 0.12 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.11 \[ \int \frac {a+b \text {arctanh}(c x)}{(d+c d x)^2} \, dx=\frac {1}{4} \, c {\left (\frac {{\left (c x - 1\right )} b \log \left (-\frac {c x + 1}{c x - 1}\right )}{{\left (c x + 1\right )} c^{2} d^{2}} + \frac {{\left (c x - 1\right )} {\left (2 \, a + b\right )}}{{\left (c x + 1\right )} c^{2} d^{2}}\right )} \] Input:
integrate((a+b*arctanh(c*x))/(c*d*x+d)^2,x, algorithm="giac")
Output:
1/4*c*((c*x - 1)*b*log(-(c*x + 1)/(c*x - 1))/((c*x + 1)*c^2*d^2) + (c*x - 1)*(2*a + b)/((c*x + 1)*c^2*d^2))
Time = 3.88 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.79 \[ \int \frac {a+b \text {arctanh}(c x)}{(d+c d x)^2} \, dx=-\frac {b\,\mathrm {atanh}\left (c\,x\right )-c\,\left (2\,a\,x+b\,x+b\,x\,\mathrm {atanh}\left (c\,x\right )\right )}{2\,x\,c^2\,d^2+2\,c\,d^2} \] Input:
int((a + b*atanh(c*x))/(d + c*d*x)^2,x)
Output:
-(b*atanh(c*x) - c*(2*a*x + b*x + b*x*atanh(c*x)))/(2*c*d^2 + 2*c^2*d^2*x)
Time = 0.16 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.28 \[ \int \frac {a+b \text {arctanh}(c x)}{(d+c d x)^2} \, dx=\frac {4 \mathit {atanh} \left (c x \right ) b c x +\mathrm {log}\left (c x -1\right ) b c x +\mathrm {log}\left (c x -1\right ) b -\mathrm {log}\left (c x +1\right ) b c x -\mathrm {log}\left (c x +1\right ) b +4 a c x +2 b c x}{4 c \,d^{2} \left (c x +1\right )} \] Input:
int((a+b*atanh(c*x))/(c*d*x+d)^2,x)
Output:
(4*atanh(c*x)*b*c*x + log(c*x - 1)*b*c*x + log(c*x - 1)*b - log(c*x + 1)*b *c*x - log(c*x + 1)*b + 4*a*c*x + 2*b*c*x)/(4*c*d**2*(c*x + 1))