Integrand size = 22, antiderivative size = 71 \[ \int \frac {\cot (c+d x)}{(a+i a \tan (c+d x))^2} \, dx=-\frac {3 i x}{4 a^2}+\frac {\log (\sin (c+d x))}{a^2 d}+\frac {3}{4 a^2 d (1+i \tan (c+d x))}+\frac {1}{4 d (a+i a \tan (c+d x))^2} \]
Time = 0.87 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.15 \[ \int \frac {\cot (c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\frac {\frac {2}{(a+i a \tan (c+d x))^2}-\frac {7 \log (i-\tan (c+d x))-8 \log (\tan (c+d x))+\log (i+\tan (c+d x))+\frac {6 i}{-i+\tan (c+d x)}}{a^2}}{8 d} \]
(2/(a + I*a*Tan[c + d*x])^2 - (7*Log[I - Tan[c + d*x]] - 8*Log[Tan[c + d*x ]] + Log[I + Tan[c + d*x]] + (6*I)/(-I + Tan[c + d*x]))/a^2)/(8*d)
Time = 0.54 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.20, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.455, Rules used = {3042, 4042, 27, 3042, 4079, 3042, 4014, 3042, 25, 3956}
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.
\(\displaystyle \int \frac {\cot (c+d x)}{(a+i a \tan (c+d x))^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\tan (c+d x) (a+i a \tan (c+d x))^2}dx\) |
\(\Big \downarrow \) 4042 |
\(\displaystyle \frac {\int \frac {2 \cot (c+d x) (2 a-i a \tan (c+d x))}{i \tan (c+d x) a+a}dx}{4 a^2}+\frac {1}{4 d (a+i a \tan (c+d x))^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {\cot (c+d x) (2 a-i a \tan (c+d x))}{i \tan (c+d x) a+a}dx}{2 a^2}+\frac {1}{4 d (a+i a \tan (c+d x))^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {2 a-i a \tan (c+d x)}{\tan (c+d x) (i \tan (c+d x) a+a)}dx}{2 a^2}+\frac {1}{4 d (a+i a \tan (c+d x))^2}\) |
\(\Big \downarrow \) 4079 |
\(\displaystyle \frac {\frac {\int \cot (c+d x) \left (4 a^2-3 i a^2 \tan (c+d x)\right )dx}{2 a^2}+\frac {3}{2 d (1+i \tan (c+d x))}}{2 a^2}+\frac {1}{4 d (a+i a \tan (c+d x))^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\int \frac {4 a^2-3 i a^2 \tan (c+d x)}{\tan (c+d x)}dx}{2 a^2}+\frac {3}{2 d (1+i \tan (c+d x))}}{2 a^2}+\frac {1}{4 d (a+i a \tan (c+d x))^2}\) |
\(\Big \downarrow \) 4014 |
\(\displaystyle \frac {\frac {4 a^2 \int \cot (c+d x)dx-3 i a^2 x}{2 a^2}+\frac {3}{2 d (1+i \tan (c+d x))}}{2 a^2}+\frac {1}{4 d (a+i a \tan (c+d x))^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {4 a^2 \int -\tan \left (c+d x+\frac {\pi }{2}\right )dx-3 i a^2 x}{2 a^2}+\frac {3}{2 d (1+i \tan (c+d x))}}{2 a^2}+\frac {1}{4 d (a+i a \tan (c+d x))^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\frac {-4 a^2 \int \tan \left (\frac {1}{2} (2 c+\pi )+d x\right )dx-3 i a^2 x}{2 a^2}+\frac {3}{2 d (1+i \tan (c+d x))}}{2 a^2}+\frac {1}{4 d (a+i a \tan (c+d x))^2}\) |
\(\Big \downarrow \) 3956 |
\(\displaystyle \frac {\frac {\frac {4 a^2 \log (-\sin (c+d x))}{d}-3 i a^2 x}{2 a^2}+\frac {3}{2 d (1+i \tan (c+d x))}}{2 a^2}+\frac {1}{4 d (a+i a \tan (c+d x))^2}\) |
(((-3*I)*a^2*x + (4*a^2*Log[-Sin[c + d*x]])/d)/(2*a^2) + 3/(2*d*(1 + I*Tan [c + d*x])))/(2*a^2) + 1/(4*d*(a + I*a*Tan[c + d*x])^2)
3.1.63.3.1 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[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Log[RemoveContent[Cos[c + d *x], x]]/d, x] /; FreeQ[{c, d}, x]
Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_.) + (b_.)*tan[(e_.) + (f_. )*(x_)]), x_Symbol] :> Simp[(a*c + b*d)*(x/(a^2 + b^2)), x] + Simp[(b*c - a *d)/(a^2 + b^2) Int[(b - a*Tan[e + f*x])/(a + b*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && N eQ[a*c + b*d, 0]
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[a*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(2*f*m*(b*c - a*d))), x] + Simp[1/(2*a*m*(b*c - a*d)) In t[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[b*c*m - a*d*(2*m + n + 1) + b*d*(m + n + 1)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && LtQ[m, 0] && (IntegerQ[m] || IntegersQ[2*m, 2*n])
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[(a*A + b*B)*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(2*f*m*( b*c - a*d))), x] + Simp[1/(2*a*m*(b*c - a*d)) Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[A*(b*c*m - a*d*(2*m + n + 1)) + B*(a*c*m - b*d*(n + 1)) + d*(A*b - a*B)*(m + n + 1)*Tan[e + f*x], x], x], x] /; Free Q[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[m, 0] && !GtQ[n, 0]
Time = 0.38 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.01
method | result | size |
risch | \(-\frac {7 i x}{4 a^{2}}+\frac {{\mathrm e}^{-2 i \left (d x +c \right )}}{2 a^{2} d}+\frac {{\mathrm e}^{-4 i \left (d x +c \right )}}{16 a^{2} d}-\frac {2 i c}{a^{2} d}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{a^{2} d}\) | \(72\) |
derivativedivides | \(-\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d \,a^{2}}-\frac {3 i \arctan \left (\tan \left (d x +c \right )\right )}{4 d \,a^{2}}+\frac {\ln \left (\tan \left (d x +c \right )\right )}{a^{2} d}-\frac {3 i}{4 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )}-\frac {1}{4 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )^{2}}\) | \(90\) |
default | \(-\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d \,a^{2}}-\frac {3 i \arctan \left (\tan \left (d x +c \right )\right )}{4 d \,a^{2}}+\frac {\ln \left (\tan \left (d x +c \right )\right )}{a^{2} d}-\frac {3 i}{4 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )}-\frac {1}{4 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )^{2}}\) | \(90\) |
norman | \(\frac {\frac {1}{a d}-\frac {3 i x}{4 a}-\frac {5 i \tan \left (d x +c \right )}{4 d a}-\frac {3 i \left (\tan ^{3}\left (d x +c \right )\right )}{4 d a}-\frac {3 i x \left (\tan ^{2}\left (d x +c \right )\right )}{2 a}-\frac {3 i x \left (\tan ^{4}\left (d x +c \right )\right )}{4 a}+\frac {\tan ^{2}\left (d x +c \right )}{2 a d}}{a \left (1+\tan ^{2}\left (d x +c \right )\right )^{2}}+\frac {\ln \left (\tan \left (d x +c \right )\right )}{a^{2} d}-\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d \,a^{2}}\) | \(144\) |
-7/4*I*x/a^2+1/2/a^2/d*exp(-2*I*(d*x+c))+1/16/a^2/d*exp(-4*I*(d*x+c))-2*I/ a^2/d*c+1/a^2/d*ln(exp(2*I*(d*x+c))-1)
Time = 0.24 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.93 \[ \int \frac {\cot (c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\frac {{\left (-28 i \, d x e^{\left (4 i \, d x + 4 i \, c\right )} + 16 \, e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right ) + 8 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} e^{\left (-4 i \, d x - 4 i \, c\right )}}{16 \, a^{2} d} \]
1/16*(-28*I*d*x*e^(4*I*d*x + 4*I*c) + 16*e^(4*I*d*x + 4*I*c)*log(e^(2*I*d* x + 2*I*c) - 1) + 8*e^(2*I*d*x + 2*I*c) + 1)*e^(-4*I*d*x - 4*I*c)/(a^2*d)
Time = 0.22 (sec) , antiderivative size = 150, normalized size of antiderivative = 2.11 \[ \int \frac {\cot (c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\begin {cases} \frac {\left (16 a^{2} d e^{4 i c} e^{- 2 i d x} + 2 a^{2} d e^{2 i c} e^{- 4 i d x}\right ) e^{- 6 i c}}{32 a^{4} d^{2}} & \text {for}\: a^{4} d^{2} e^{6 i c} \neq 0 \\x \left (\frac {\left (- 7 i e^{4 i c} - 4 i e^{2 i c} - i\right ) e^{- 4 i c}}{4 a^{2}} + \frac {7 i}{4 a^{2}}\right ) & \text {otherwise} \end {cases} - \frac {7 i x}{4 a^{2}} + \frac {\log {\left (e^{2 i d x} - e^{- 2 i c} \right )}}{a^{2} d} \]
Piecewise(((16*a**2*d*exp(4*I*c)*exp(-2*I*d*x) + 2*a**2*d*exp(2*I*c)*exp(- 4*I*d*x))*exp(-6*I*c)/(32*a**4*d**2), Ne(a**4*d**2*exp(6*I*c), 0)), (x*((- 7*I*exp(4*I*c) - 4*I*exp(2*I*c) - I)*exp(-4*I*c)/(4*a**2) + 7*I/(4*a**2)), True)) - 7*I*x/(4*a**2) + log(exp(2*I*d*x) - exp(-2*I*c))/(a**2*d)
Exception generated. \[ \int \frac {\cot (c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\text {Exception raised: RuntimeError} \]
Time = 0.60 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.14 \[ \int \frac {\cot (c+d x)}{(a+i a \tan (c+d x))^2} \, dx=-\frac {\frac {2 \, \log \left (\tan \left (d x + c\right ) + i\right )}{a^{2}} + \frac {14 \, \log \left (\tan \left (d x + c\right ) - i\right )}{a^{2}} - \frac {16 \, \log \left (\tan \left (d x + c\right )\right )}{a^{2}} - \frac {21 \, \tan \left (d x + c\right )^{2} - 54 i \, \tan \left (d x + c\right ) - 37}{a^{2} {\left (\tan \left (d x + c\right ) - i\right )}^{2}}}{16 \, d} \]
-1/16*(2*log(tan(d*x + c) + I)/a^2 + 14*log(tan(d*x + c) - I)/a^2 - 16*log (tan(d*x + c))/a^2 - (21*tan(d*x + c)^2 - 54*I*tan(d*x + c) - 37)/(a^2*(ta n(d*x + c) - I)^2))/d
Time = 4.86 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.37 \[ \int \frac {\cot (c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )}{a^2\,d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )}{8\,a^2\,d}-\frac {7\,\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )}{8\,a^2\,d}+\frac {\frac {3\,\mathrm {tan}\left (c+d\,x\right )}{4\,a^2}-\frac {1{}\mathrm {i}}{a^2}}{d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^2\,1{}\mathrm {i}+2\,\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )} \]