Integrand size = 21, antiderivative size = 63 \[ \int \frac {a+b \text {arctanh}(c+d x)}{(c e+d e x)^2} \, dx=-\frac {a+b \text {arctanh}(c+d x)}{d e^2 (c+d x)}+\frac {b \log (c+d x)}{d e^2}-\frac {b \log \left (1-(c+d x)^2\right )}{2 d e^2} \] Output:
-(a+b*arctanh(d*x+c))/d/e^2/(d*x+c)+b*ln(d*x+c)/d/e^2-1/2*b*ln(1-(d*x+c)^2 )/d/e^2
Time = 0.07 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.10 \[ \int \frac {a+b \text {arctanh}(c+d x)}{(c e+d e x)^2} \, dx=-\frac {\frac {2 a}{c+d x}+\frac {2 b \text {arctanh}(c+d x)}{c+d x}-2 b \log (c+d x)+b \log \left (1-c^2-2 c d x-d^2 x^2\right )}{2 d e^2} \] Input:
Integrate[(a + b*ArcTanh[c + d*x])/(c*e + d*e*x)^2,x]
Output:
-1/2*((2*a)/(c + d*x) + (2*b*ArcTanh[c + d*x])/(c + d*x) - 2*b*Log[c + d*x ] + b*Log[1 - c^2 - 2*c*d*x - d^2*x^2])/(d*e^2)
Time = 0.29 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.84, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6657, 27, 6452, 243, 47, 14, 16}
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.
\(\displaystyle \int \frac {a+b \text {arctanh}(c+d x)}{(c e+d e x)^2} \, dx\) |
\(\Big \downarrow \) 6657 |
\(\displaystyle \frac {\int \frac {a+b \text {arctanh}(c+d x)}{e^2 (c+d x)^2}d(c+d x)}{d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {a+b \text {arctanh}(c+d x)}{(c+d x)^2}d(c+d x)}{d e^2}\) |
\(\Big \downarrow \) 6452 |
\(\displaystyle \frac {b \int \frac {1}{(c+d x) \left (1-(c+d x)^2\right )}d(c+d x)-\frac {a+b \text {arctanh}(c+d x)}{c+d x}}{d e^2}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {\frac {1}{2} b \int \frac {1}{(-c-d x+1) (c+d x)^2}d(c+d x)^2-\frac {a+b \text {arctanh}(c+d x)}{c+d x}}{d e^2}\) |
\(\Big \downarrow \) 47 |
\(\displaystyle \frac {\frac {1}{2} b \left (\int \frac {1}{-c-d x+1}d(c+d x)^2+\int \frac {1}{(c+d x)^2}d(c+d x)^2\right )-\frac {a+b \text {arctanh}(c+d x)}{c+d x}}{d e^2}\) |
\(\Big \downarrow \) 14 |
\(\displaystyle \frac {\frac {1}{2} b \left (\int \frac {1}{-c-d x+1}d(c+d x)^2+\log \left ((c+d x)^2\right )\right )-\frac {a+b \text {arctanh}(c+d x)}{c+d x}}{d e^2}\) |
\(\Big \downarrow \) 16 |
\(\displaystyle \frac {\frac {1}{2} b \left (\log \left ((c+d x)^2\right )-\log (-c-d x+1)\right )-\frac {a+b \text {arctanh}(c+d x)}{c+d x}}{d e^2}\) |
Input:
Int[(a + b*ArcTanh[c + d*x])/(c*e + d*e*x)^2,x]
Output:
(-((a + b*ArcTanh[c + d*x])/(c + d*x)) + (b*(-Log[1 - c - d*x] + Log[(c + d*x)^2]))/2)/(d*e^2)
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Simp[b/(b*c - a*d) Int[1/(a + b*x), x], x] - Simp[d/(b*c - a*d) Int[1/(c + d*x), x ], x] /; FreeQ[{a, b, c, d}, x]
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 = 0.53 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.02
method | result | size |
derivativedivides | \(\frac {-\frac {a}{e^{2} \left (d x +c \right )}+\frac {b \left (-\frac {\operatorname {arctanh}\left (d x +c \right )}{d x +c}-\frac {\ln \left (d x +c -1\right )}{2}+\ln \left (d x +c \right )-\frac {\ln \left (d x +c +1\right )}{2}\right )}{e^{2}}}{d}\) | \(64\) |
default | \(\frac {-\frac {a}{e^{2} \left (d x +c \right )}+\frac {b \left (-\frac {\operatorname {arctanh}\left (d x +c \right )}{d x +c}-\frac {\ln \left (d x +c -1\right )}{2}+\ln \left (d x +c \right )-\frac {\ln \left (d x +c +1\right )}{2}\right )}{e^{2}}}{d}\) | \(64\) |
parts | \(-\frac {a}{d \left (d x +c \right ) e^{2}}+\frac {b \left (-\frac {\operatorname {arctanh}\left (d x +c \right )}{d x +c}-\frac {\ln \left (d x +c -1\right )}{2}+\ln \left (d x +c \right )-\frac {\ln \left (d x +c +1\right )}{2}\right )}{e^{2} d}\) | \(66\) |
parallelrisch | \(-\frac {3 \ln \left (d x +c -1\right ) x b c \,d^{2}-3 \ln \left (d x +c \right ) x b c \,d^{2}+3 x \,\operatorname {arctanh}\left (d x +c \right ) b c \,d^{2}+3 \ln \left (d x +c -1\right ) b \,c^{2} d -3 \ln \left (d x +c \right ) b \,c^{2} d +3 \,\operatorname {arctanh}\left (d x +c \right ) b \,c^{2} d -a \,d^{2} x +3 \,\operatorname {arctanh}\left (d x +c \right ) b c d +2 a c d}{3 \left (d x +c \right ) d^{2} e^{2} c}\) | \(126\) |
risch | \(-\frac {b \ln \left (d x +c +1\right )}{2 d \left (d x +c \right ) e^{2}}-\frac {\ln \left (d^{2} x^{2}+2 c d x +c^{2}-1\right ) b d x -2 \ln \left (-d x -c \right ) b d x +\ln \left (d^{2} x^{2}+2 c d x +c^{2}-1\right ) b c -2 \ln \left (-d x -c \right ) b c -b \ln \left (-d x -c +1\right )+2 a}{2 e^{2} \left (d x +c \right ) d}\) | \(127\) |
Input:
int((a+b*arctanh(d*x+c))/(d*e*x+c*e)^2,x,method=_RETURNVERBOSE)
Output:
1/d*(-a/e^2/(d*x+c)+b/e^2*(-1/(d*x+c)*arctanh(d*x+c)-1/2*ln(d*x+c-1)+ln(d* x+c)-1/2*ln(d*x+c+1)))
Time = 0.09 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.35 \[ \int \frac {a+b \text {arctanh}(c+d x)}{(c e+d e x)^2} \, dx=-\frac {{\left (b d x + b c\right )} \log \left (d^{2} x^{2} + 2 \, c d x + c^{2} - 1\right ) - 2 \, {\left (b d x + b c\right )} \log \left (d x + c\right ) + b \log \left (-\frac {d x + c + 1}{d x + c - 1}\right ) + 2 \, a}{2 \, {\left (d^{2} e^{2} x + c d e^{2}\right )}} \] Input:
integrate((a+b*arctanh(d*x+c))/(d*e*x+c*e)^2,x, algorithm="fricas")
Output:
-1/2*((b*d*x + b*c)*log(d^2*x^2 + 2*c*d*x + c^2 - 1) - 2*(b*d*x + b*c)*log (d*x + c) + b*log(-(d*x + c + 1)/(d*x + c - 1)) + 2*a)/(d^2*e^2*x + c*d*e^ 2)
Leaf count of result is larger than twice the leaf count of optimal. 219 vs. \(2 (53) = 106\).
Time = 1.10 (sec) , antiderivative size = 219, normalized size of antiderivative = 3.48 \[ \int \frac {a+b \text {arctanh}(c+d x)}{(c e+d e x)^2} \, dx=\begin {cases} - \frac {a}{c d e^{2} + d^{2} e^{2} x} + \frac {b c \log {\left (\frac {c}{d} + x \right )}}{c d e^{2} + d^{2} e^{2} x} - \frac {b c \log {\left (\frac {c}{d} + x + \frac {1}{d} \right )}}{c d e^{2} + d^{2} e^{2} x} + \frac {b c \operatorname {atanh}{\left (c + d x \right )}}{c d e^{2} + d^{2} e^{2} x} + \frac {b d x \log {\left (\frac {c}{d} + x \right )}}{c d e^{2} + d^{2} e^{2} x} - \frac {b d x \log {\left (\frac {c}{d} + x + \frac {1}{d} \right )}}{c d e^{2} + d^{2} e^{2} x} + \frac {b d x \operatorname {atanh}{\left (c + d x \right )}}{c d e^{2} + d^{2} e^{2} x} - \frac {b \operatorname {atanh}{\left (c + d x \right )}}{c d e^{2} + d^{2} e^{2} x} & \text {for}\: d \neq 0 \\\frac {x \left (a + b \operatorname {atanh}{\left (c \right )}\right )}{c^{2} e^{2}} & \text {otherwise} \end {cases} \] Input:
integrate((a+b*atanh(d*x+c))/(d*e*x+c*e)**2,x)
Output:
Piecewise((-a/(c*d*e**2 + d**2*e**2*x) + b*c*log(c/d + x)/(c*d*e**2 + d**2 *e**2*x) - b*c*log(c/d + x + 1/d)/(c*d*e**2 + d**2*e**2*x) + b*c*atanh(c + d*x)/(c*d*e**2 + d**2*e**2*x) + b*d*x*log(c/d + x)/(c*d*e**2 + d**2*e**2* x) - b*d*x*log(c/d + x + 1/d)/(c*d*e**2 + d**2*e**2*x) + b*d*x*atanh(c + d *x)/(c*d*e**2 + d**2*e**2*x) - b*atanh(c + d*x)/(c*d*e**2 + d**2*e**2*x), Ne(d, 0)), (x*(a + b*atanh(c))/(c**2*e**2), True))
Time = 0.03 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.51 \[ \int \frac {a+b \text {arctanh}(c+d x)}{(c e+d e x)^2} \, dx=-\frac {1}{2} \, {\left (d {\left (\frac {\log \left (d x + c + 1\right )}{d^{2} e^{2}} - \frac {2 \, \log \left (d x + c\right )}{d^{2} e^{2}} + \frac {\log \left (d x + c - 1\right )}{d^{2} e^{2}}\right )} + \frac {2 \, \operatorname {artanh}\left (d x + c\right )}{d^{2} e^{2} x + c d e^{2}}\right )} b - \frac {a}{d^{2} e^{2} x + c d e^{2}} \] Input:
integrate((a+b*arctanh(d*x+c))/(d*e*x+c*e)^2,x, algorithm="maxima")
Output:
-1/2*(d*(log(d*x + c + 1)/(d^2*e^2) - 2*log(d*x + c)/(d^2*e^2) + log(d*x + c - 1)/(d^2*e^2)) + 2*arctanh(d*x + c)/(d^2*e^2*x + c*d*e^2))*b - a/(d^2* e^2*x + c*d*e^2)
Leaf count of result is larger than twice the leaf count of optimal. 152 vs. \(2 (61) = 122\).
Time = 0.12 (sec) , antiderivative size = 152, normalized size of antiderivative = 2.41 \[ \int \frac {a+b \text {arctanh}(c+d x)}{(c e+d e x)^2} \, dx=\frac {1}{2} \, {\left ({\left (c + 1\right )} d - {\left (c - 1\right )} d\right )} {\left (\frac {b \log \left (-\frac {d x + c + 1}{d x + c - 1}\right )}{\frac {{\left (d x + c + 1\right )} d^{2} e^{2}}{d x + c - 1} + d^{2} e^{2}} + \frac {2 \, a}{\frac {{\left (d x + c + 1\right )} d^{2} e^{2}}{d x + c - 1} + d^{2} e^{2}} + \frac {b \log \left (-\frac {d x + c + 1}{d x + c - 1} - 1\right )}{d^{2} e^{2}} - \frac {b \log \left (-\frac {d x + c + 1}{d x + c - 1}\right )}{d^{2} e^{2}}\right )} \] Input:
integrate((a+b*arctanh(d*x+c))/(d*e*x+c*e)^2,x, algorithm="giac")
Output:
1/2*((c + 1)*d - (c - 1)*d)*(b*log(-(d*x + c + 1)/(d*x + c - 1))/((d*x + c + 1)*d^2*e^2/(d*x + c - 1) + d^2*e^2) + 2*a/((d*x + c + 1)*d^2*e^2/(d*x + c - 1) + d^2*e^2) + b*log(-(d*x + c + 1)/(d*x + c - 1) - 1)/(d^2*e^2) - b *log(-(d*x + c + 1)/(d*x + c - 1))/(d^2*e^2))
Time = 4.68 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.94 \[ \int \frac {a+b \text {arctanh}(c+d x)}{(c e+d e x)^2} \, dx=\frac {b\,\ln \left (1-d\,x-c\right )}{2\,x\,d^2\,e^2+2\,c\,d\,e^2}-\frac {b\,\ln \left (c+d\,x+1\right )}{2\,\left (x\,d^2\,e^2+c\,d\,e^2\right )}-\frac {a}{x\,d^2\,e^2+c\,d\,e^2}-\frac {b\,\ln \left (c^2+2\,c\,d\,x+d^2\,x^2-1\right )}{2\,d\,e^2}+\frac {b\,\ln \left (c+d\,x\right )}{d\,e^2} \] Input:
int((a + b*atanh(c + d*x))/(c*e + d*e*x)^2,x)
Output:
(b*log(1 - d*x - c))/(2*d^2*e^2*x + 2*c*d*e^2) - (b*log(c + d*x + 1))/(2*( d^2*e^2*x + c*d*e^2)) - a/(d^2*e^2*x + c*d*e^2) - (b*log(c^2 + d^2*x^2 + 2 *c*d*x - 1))/(2*d*e^2) + (b*log(c + d*x))/(d*e^2)
Time = 0.18 (sec) , antiderivative size = 155, normalized size of antiderivative = 2.46 \[ \int \frac {a+b \text {arctanh}(c+d x)}{(c e+d e x)^2} \, dx=\frac {2 \mathit {atanh} \left (d x +c \right ) b d x -\mathrm {log}\left (d x +c -1\right ) b \,c^{2}-\mathrm {log}\left (d x +c -1\right ) b c d x +\mathrm {log}\left (d x +c -1\right ) b c +\mathrm {log}\left (d x +c -1\right ) b d x -\mathrm {log}\left (d x +c +1\right ) b \,c^{2}-\mathrm {log}\left (d x +c +1\right ) b c d x -\mathrm {log}\left (d x +c +1\right ) b c -\mathrm {log}\left (d x +c +1\right ) b d x +2 \,\mathrm {log}\left (d x +c \right ) b \,c^{2}+2 \,\mathrm {log}\left (d x +c \right ) b c d x +2 a d x}{2 c d \,e^{2} \left (d x +c \right )} \] Input:
int((a+b*atanh(d*x+c))/(d*e*x+c*e)^2,x)
Output:
(2*atanh(c + d*x)*b*d*x - log(c + d*x - 1)*b*c**2 - log(c + d*x - 1)*b*c*d *x + log(c + d*x - 1)*b*c + log(c + d*x - 1)*b*d*x - log(c + d*x + 1)*b*c* *2 - log(c + d*x + 1)*b*c*d*x - log(c + d*x + 1)*b*c - log(c + d*x + 1)*b* d*x + 2*log(c + d*x)*b*c**2 + 2*log(c + d*x)*b*c*d*x + 2*a*d*x)/(2*c*d*e** 2*(c + d*x))