Integrand size = 21, antiderivative size = 61 \[ \int \frac {a+b \arctan (c+d x)}{(c e+d e x)^2} \, dx=-\frac {a+b \arctan (c+d x)}{d e^2 (c+d x)}+\frac {b \log (c+d x)}{d e^2}-\frac {b \log \left (1+(c+d x)^2\right )}{2 d e^2} \]
Time = 0.02 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.85 \[ \int \frac {a+b \arctan (c+d x)}{(c e+d e x)^2} \, dx=\frac {\frac {-a-b \arctan (c+d x)}{c+d x}+b \left (\log (c+d x)-\frac {1}{2} \log \left (1+(c+d x)^2\right )\right )}{d e^2} \]
Time = 0.24 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.87, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {5566, 27, 5361, 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.
\(\displaystyle \int \frac {a+b \arctan (c+d x)}{(c e+d e x)^2} \, dx\) |
\(\Big \downarrow \) 5566 |
\(\displaystyle \frac {\int \frac {a+b \arctan (c+d x)}{e^2 (c+d x)^2}d(c+d x)}{d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {a+b \arctan (c+d x)}{(c+d x)^2}d(c+d x)}{d e^2}\) |
\(\Big \downarrow \) 5361 |
\(\displaystyle \frac {b \int \frac {1}{(c+d x) \left ((c+d x)^2+1\right )}d(c+d x)-\frac {a+b \arctan (c+d x)}{c+d x}}{d e^2}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {\frac {1}{2} b \int \frac {1}{(c+d x)^2 \left ((c+d x)^2+1\right )}d(c+d x)^2-\frac {a+b \arctan (c+d x)}{c+d x}}{d e^2}\) |
\(\Big \downarrow \) 47 |
\(\displaystyle \frac {\frac {1}{2} b \left (\int \frac {1}{(c+d x)^2}d(c+d x)^2-\int \frac {1}{(c+d x)^2+1}d(c+d x)^2\right )-\frac {a+b \arctan (c+d x)}{c+d x}}{d e^2}\) |
\(\Big \downarrow \) 14 |
\(\displaystyle \frac {\frac {1}{2} b \left (\log \left ((c+d x)^2\right )-\int \frac {1}{(c+d x)^2+1}d(c+d x)^2\right )-\frac {a+b \arctan (c+d x)}{c+d x}}{d e^2}\) |
\(\Big \downarrow \) 16 |
\(\displaystyle \frac {\frac {1}{2} b \left (\log \left ((c+d x)^2\right )-\log \left ((c+d x)^2+1\right )\right )-\frac {a+b \arctan (c+d x)}{c+d x}}{d e^2}\) |
(-((a + b*ArcTan[c + d*x])/(c + d*x)) + (b*(Log[(c + d*x)^2] - Log[1 + (c + d*x)^2]))/2)/(d*e^2)
3.1.5.3.1 Defintions of rubi rules used
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_.) + ArcTan[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)*((a + b*ArcTan[c*x^n])^p/(m + 1)), x] - Simp[b*c*n*(p/(m + 1)) Int[x^(m + n)*((a + b*ArcTan[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_.) + ArcTan[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m _.), x_Symbol] :> Simp[1/d Subst[Int[(f*(x/d))^m*(a + b*ArcTan[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 = 0.28 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.95
method | result | size |
derivativedivides | \(\frac {-\frac {a}{e^{2} \left (d x +c \right )}+\frac {b \left (-\frac {\arctan \left (d x +c \right )}{d x +c}+\ln \left (d x +c \right )-\frac {\ln \left (1+\left (d x +c \right )^{2}\right )}{2}\right )}{e^{2}}}{d}\) | \(58\) |
default | \(\frac {-\frac {a}{e^{2} \left (d x +c \right )}+\frac {b \left (-\frac {\arctan \left (d x +c \right )}{d x +c}+\ln \left (d x +c \right )-\frac {\ln \left (1+\left (d x +c \right )^{2}\right )}{2}\right )}{e^{2}}}{d}\) | \(58\) |
parts | \(-\frac {a}{d \,e^{2} \left (d x +c \right )}+\frac {b \left (-\frac {\arctan \left (d x +c \right )}{d x +c}+\ln \left (d x +c \right )-\frac {\ln \left (1+\left (d x +c \right )^{2}\right )}{2}\right )}{e^{2} d}\) | \(60\) |
parallelrisch | \(\frac {6 \ln \left (d x +c \right ) x b c \,d^{2}-3 \ln \left (d^{2} x^{2}+2 c d x +c^{2}+1\right ) x b c \,d^{2}+6 \ln \left (d x +c \right ) b \,c^{2} d -3 \ln \left (d^{2} x^{2}+2 c d x +c^{2}+1\right ) b \,c^{2} d +2 x a \,d^{2}-6 b \arctan \left (d x +c \right ) c d -4 a c d}{6 \left (d x +c \right ) c \,d^{2} e^{2}}\) | \(121\) |
risch | \(\frac {i b \ln \left (1+i \left (d x +c \right )\right )}{2 d \,e^{2} \left (d x +c \right )}-\frac {-2 \ln \left (-d x -c \right ) b d x +\ln \left (-d^{2} x^{2}-2 c d x -c^{2}-1\right ) b d x -2 \ln \left (-d x -c \right ) b c +\ln \left (-d^{2} x^{2}-2 c d x -c^{2}-1\right ) b c +i b \ln \left (1-i \left (d x +c \right )\right )+2 a}{2 e^{2} \left (d x +c \right ) d}\) | \(140\) |
Time = 0.28 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.23 \[ \int \frac {a+b \arctan (c+d x)}{(c e+d e x)^2} \, dx=-\frac {2 \, b \arctan \left (d x + c\right ) + {\left (b d x + b c\right )} \log \left (d^{2} x^{2} + 2 \, c d x + c^{2} + 1\right ) - 2 \, {\left (b d x + b c\right )} \log \left (d x + c\right ) + 2 \, a}{2 \, {\left (d^{2} e^{2} x + c d e^{2}\right )}} \]
-1/2*(2*b*arctan(d*x + c) + (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) + 2*a)/(d^2*e^2*x + c*d*e^2)
Result contains complex when optimal does not.
Time = 12.22 (sec) , antiderivative size = 223, normalized size of antiderivative = 3.66 \[ \int \frac {a+b \arctan (c+d x)}{(c e+d e x)^2} \, dx=\begin {cases} - \frac {a}{c d e^{2} + d^{2} e^{2} x} + \frac {b c \log {\left (\frac {c}{d} + x \right )}}{c d e^{2} + d^{2} e^{2} x} - \frac {b c \log {\left (\frac {c}{d} + x - \frac {i}{d} \right )}}{c d e^{2} + d^{2} e^{2} x} + \frac {i b c \operatorname {atan}{\left (c + d x \right )}}{c d e^{2} + d^{2} e^{2} x} + \frac {b d x \log {\left (\frac {c}{d} + x \right )}}{c d e^{2} + d^{2} e^{2} x} - \frac {b d x \log {\left (\frac {c}{d} + x - \frac {i}{d} \right )}}{c d e^{2} + d^{2} e^{2} x} + \frac {i b d x \operatorname {atan}{\left (c + d x \right )}}{c d e^{2} + d^{2} e^{2} x} - \frac {b \operatorname {atan}{\left (c + d x \right )}}{c d e^{2} + d^{2} e^{2} x} & \text {for}\: d \neq 0 \\\frac {x \left (a + b \operatorname {atan}{\left (c \right )}\right )}{c^{2} e^{2}} & \text {otherwise} \end {cases} \]
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 - I/d)/(c*d*e**2 + d**2*e**2*x) + I*b*c*atan(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 - I/d)/(c*d*e**2 + d**2*e**2*x) + I*b*d*x*atan(c + d*x)/(c*d*e**2 + d**2*e**2*x) - b*atan(c + d*x)/(c*d*e**2 + d**2*e**2*x), Ne(d, 0)), (x*(a + b*atan(c))/(c**2*e**2), True))
Time = 0.21 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.51 \[ \int \frac {a+b \arctan (c+d x)}{(c e+d e x)^2} \, dx=-\frac {1}{2} \, {\left (d {\left (\frac {\log \left (d^{2} x^{2} + 2 \, c d x + c^{2} + 1\right )}{d^{2} e^{2}} - \frac {2 \, \log \left (d x + c\right )}{d^{2} e^{2}}\right )} + \frac {2 \, \arctan \left (d x + c\right )}{d^{2} e^{2} x + c d e^{2}}\right )} b - \frac {a}{d^{2} e^{2} x + c d e^{2}} \]
-1/2*(d*(log(d^2*x^2 + 2*c*d*x + c^2 + 1)/(d^2*e^2) - 2*log(d*x + c)/(d^2* e^2)) + 2*arctan(d*x + c)/(d^2*e^2*x + c*d*e^2))*b - a/(d^2*e^2*x + c*d*e^ 2)
\[ \int \frac {a+b \arctan (c+d x)}{(c e+d e x)^2} \, dx=\int { \frac {b \arctan \left (d x + c\right ) + a}{{\left (d e x + c e\right )}^{2}} \,d x } \]
Time = 0.75 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.44 \[ \int \frac {a+b \arctan (c+d x)}{(c e+d e x)^2} \, dx=\frac {b\,\ln \left (c+d\,x\right )}{d\,e^2}-\frac {b\,\mathrm {atan}\left (c+d\,x\right )}{x\,d^2\,e^2+c\,d\,e^2}-\frac {b\,\ln \left (c^2+2\,c\,d\,x+d^2\,x^2+1\right )}{2\,d\,e^2}-\frac {a}{x\,d^2\,e^2+c\,d\,e^2} \]