Integrand size = 34, antiderivative size = 79 \[ \int \cot ^2(c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=-2 a^2 (A-i B) x+\frac {a^2 B \log (\cos (c+d x))}{d}+\frac {a^2 (2 i A+B) \log (\sin (c+d x))}{d}-\frac {A \cot (c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{d} \]
-2*a^2*(A-I*B)*x+a^2*B*ln(cos(d*x+c))/d+a^2*(2*I*A+B)*ln(sin(d*x+c))/d-A*c ot(d*x+c)*(a^2+I*a^2*tan(d*x+c))/d
Time = 0.75 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.68 \[ \int \cot ^2(c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=\frac {a^2 (-A \cot (c+d x)+(2 i A+B) \log (\tan (c+d x))-2 i (A-i B) \log (i+\tan (c+d x)))}{d} \]
(a^2*(-(A*Cot[c + d*x]) + ((2*I)*A + B)*Log[Tan[c + d*x]] - (2*I)*(A - I*B )*Log[I + Tan[c + d*x]]))/d
Time = 0.63 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.03, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {3042, 4076, 3042, 4072, 3042, 3956, 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.
\(\displaystyle \int \cot ^2(c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a+i a \tan (c+d x))^2 (A+B \tan (c+d x))}{\tan (c+d x)^2}dx\) |
\(\Big \downarrow \) 4076 |
\(\displaystyle \int \cot (c+d x) (i \tan (c+d x) a+a) (a (2 i A+B)+i a B \tan (c+d x))dx-\frac {A \cot (c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(i \tan (c+d x) a+a) (a (2 i A+B)+i a B \tan (c+d x))}{\tan (c+d x)}dx-\frac {A \cot (c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{d}\) |
\(\Big \downarrow \) 4072 |
\(\displaystyle \int \cot (c+d x) \left (a^2 (2 i A+B)-2 a^2 (A-i B) \tan (c+d x)\right )dx+a^2 (-B) \int \tan (c+d x)dx-\frac {A \cot (c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {a^2 (2 i A+B)-2 a^2 (A-i B) \tan (c+d x)}{\tan (c+d x)}dx+a^2 (-B) \int \tan (c+d x)dx-\frac {A \cot (c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{d}\) |
\(\Big \downarrow \) 3956 |
\(\displaystyle \int \frac {a^2 (2 i A+B)-2 a^2 (A-i B) \tan (c+d x)}{\tan (c+d x)}dx-\frac {A \cot (c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{d}+\frac {a^2 B \log (\cos (c+d x))}{d}\) |
\(\Big \downarrow \) 4014 |
\(\displaystyle a^2 (B+2 i A) \int \cot (c+d x)dx-2 a^2 x (A-i B)-\frac {A \cot (c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{d}+\frac {a^2 B \log (\cos (c+d x))}{d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a^2 (B+2 i A) \int -\tan \left (c+d x+\frac {\pi }{2}\right )dx-2 a^2 x (A-i B)-\frac {A \cot (c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{d}+\frac {a^2 B \log (\cos (c+d x))}{d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -a^2 (B+2 i A) \int \tan \left (\frac {1}{2} (2 c+\pi )+d x\right )dx-2 a^2 x (A-i B)-\frac {A \cot (c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{d}+\frac {a^2 B \log (\cos (c+d x))}{d}\) |
\(\Big \downarrow \) 3956 |
\(\displaystyle \frac {a^2 (B+2 i A) \log (-\sin (c+d x))}{d}-2 a^2 x (A-i B)-\frac {A \cot (c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{d}+\frac {a^2 B \log (\cos (c+d x))}{d}\) |
-2*a^2*(A - I*B)*x + (a^2*B*Log[Cos[c + d*x]])/d + (a^2*((2*I)*A + B)*Log[ -Sin[c + d*x]])/d - (A*Cot[c + d*x]*(a^2 + I*a^2*Tan[c + d*x]))/d
3.1.13.3.1 Defintions of rubi rules used
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_)])*((c_.) + (d_.)*tan[(e_.) + (f_ .)*(x_)]))/((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[B*(d/ b) Int[Tan[e + f*x], x], x] + Simp[1/b Int[Simp[A*b*c + (A*b*d + B*(b*c - a*d))*Tan[e + f*x], x]/(a + b*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d , e, f, A, B}, x] && NeQ[b*c - a*d, 0]
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^2)*(B*c - A*d)*(a + b*Tan[e + f*x])^(m - 1)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(b*c + a*d)*(n + 1))), x] - Simp[a/(d*(b*c + a*d)*(n + 1)) Int[ (a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^(n + 1)*Simp[A*b*d*(m - n - 2) - B*(b*c*(m - 1) + a*d*(n + 1)) + (a*A*d*(m + n) - B*(a*c*(m - 1) + b *d*(n + 1)))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && GtQ[m, 1] && LtQ[n, -1]
Time = 0.23 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.94
method | result | size |
parallelrisch | \(\frac {a^{2} \left (2 i B d x +2 i A \ln \left (\tan \left (d x +c \right )\right )-i A \ln \left (\sec ^{2}\left (d x +c \right )\right )-2 A d x -A \cot \left (d x +c \right )+B \ln \left (\tan \left (d x +c \right )\right )-B \ln \left (\sec ^{2}\left (d x +c \right )\right )\right )}{d}\) | \(74\) |
derivativedivides | \(\frac {-A \,a^{2} \left (d x +c \right )+B \,a^{2} \ln \left (\cos \left (d x +c \right )\right )+2 i A \,a^{2} \ln \left (\sin \left (d x +c \right )\right )+2 i B \,a^{2} \left (d x +c \right )+A \,a^{2} \left (-\cot \left (d x +c \right )-d x -c \right )+B \,a^{2} \ln \left (\sin \left (d x +c \right )\right )}{d}\) | \(88\) |
default | \(\frac {-A \,a^{2} \left (d x +c \right )+B \,a^{2} \ln \left (\cos \left (d x +c \right )\right )+2 i A \,a^{2} \ln \left (\sin \left (d x +c \right )\right )+2 i B \,a^{2} \left (d x +c \right )+A \,a^{2} \left (-\cot \left (d x +c \right )-d x -c \right )+B \,a^{2} \ln \left (\sin \left (d x +c \right )\right )}{d}\) | \(88\) |
norman | \(\frac {\left (2 i B \,a^{2}-2 A \,a^{2}\right ) x \tan \left (d x +c \right )-\frac {A \,a^{2}}{d}}{\tan \left (d x +c \right )}+\frac {\left (2 i A \,a^{2}+B \,a^{2}\right ) \ln \left (\tan \left (d x +c \right )\right )}{d}-\frac {\left (i A \,a^{2}+B \,a^{2}\right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{d}\) | \(96\) |
risch | \(-\frac {4 i a^{2} B c}{d}+\frac {4 a^{2} A c}{d}-\frac {2 i A \,a^{2}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}+\frac {a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) B}{d}+\frac {2 i a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) A}{d}+\frac {a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) B}{d}\) | \(108\) |
a^2*(2*I*B*d*x+2*I*A*ln(tan(d*x+c))-I*A*ln(sec(d*x+c)^2)-2*A*d*x-A*cot(d*x +c)+B*ln(tan(d*x+c))-B*ln(sec(d*x+c)^2))/d
Time = 0.26 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.29 \[ \int \cot ^2(c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=\frac {-2 i \, A a^{2} + {\left (B a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - B a^{2}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + {\left ({\left (2 i \, A + B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (-2 i \, A - B\right )} a^{2}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right )}{d e^{\left (2 i \, d x + 2 i \, c\right )} - d} \]
(-2*I*A*a^2 + (B*a^2*e^(2*I*d*x + 2*I*c) - B*a^2)*log(e^(2*I*d*x + 2*I*c) + 1) + ((2*I*A + B)*a^2*e^(2*I*d*x + 2*I*c) + (-2*I*A - B)*a^2)*log(e^(2*I *d*x + 2*I*c) - 1))/(d*e^(2*I*d*x + 2*I*c) - d)
Time = 1.32 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.38 \[ \int \cot ^2(c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=- \frac {2 i A a^{2}}{d e^{2 i c} e^{2 i d x} - d} + \frac {B a^{2} \log {\left (e^{2 i d x} + e^{- 2 i c} \right )}}{d} + \frac {i a^{2} \cdot \left (2 A - i B\right ) \log {\left (e^{2 i d x} + \frac {\left (A a^{2} - i B a^{2} - a^{2} \cdot \left (2 A - i B\right )\right ) e^{- 2 i c}}{A a^{2}} \right )}}{d} \]
-2*I*A*a**2/(d*exp(2*I*c)*exp(2*I*d*x) - d) + B*a**2*log(exp(2*I*d*x) + ex p(-2*I*c))/d + I*a**2*(2*A - I*B)*log(exp(2*I*d*x) + (A*a**2 - I*B*a**2 - a**2*(2*A - I*B))*exp(-2*I*c)/(A*a**2))/d
Time = 0.29 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.94 \[ \int \cot ^2(c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=-\frac {2 \, {\left (d x + c\right )} {\left (A - i \, B\right )} a^{2} - {\left (-i \, A - B\right )} a^{2} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - {\left (2 i \, A + B\right )} a^{2} \log \left (\tan \left (d x + c\right )\right ) + \frac {A a^{2}}{\tan \left (d x + c\right )}}{d} \]
-(2*(d*x + c)*(A - I*B)*a^2 - (-I*A - B)*a^2*log(tan(d*x + c)^2 + 1) - (2* I*A + B)*a^2*log(tan(d*x + c)) + A*a^2/tan(d*x + c))/d
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 155 vs. \(2 (73) = 146\).
Time = 0.86 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.96 \[ \int \cot ^2(c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=\frac {2 \, B a^{2} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right ) + 2 \, B a^{2} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right ) + A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 8 \, {\left (i \, A a^{2} + B a^{2}\right )} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + i\right ) + 2 \, {\left (2 i \, A a^{2} + B a^{2}\right )} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + \frac {-4 i \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - A a^{2}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}}{2 \, d} \]
1/2*(2*B*a^2*log(tan(1/2*d*x + 1/2*c) + 1) + 2*B*a^2*log(tan(1/2*d*x + 1/2 *c) - 1) + A*a^2*tan(1/2*d*x + 1/2*c) - 8*(I*A*a^2 + B*a^2)*log(tan(1/2*d* x + 1/2*c) + I) + 2*(2*I*A*a^2 + B*a^2)*log(tan(1/2*d*x + 1/2*c)) + (-4*I* A*a^2*tan(1/2*d*x + 1/2*c) - 2*B*a^2*tan(1/2*d*x + 1/2*c) - A*a^2)/tan(1/2 *d*x + 1/2*c))/d
Time = 7.69 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.10 \[ \int \cot ^2(c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=\frac {B\,a^2\,\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )}{d}-\frac {2\,B\,a^2\,\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )}{d}-\frac {A\,a^2\,\mathrm {cot}\left (c+d\,x\right )}{d}+\frac {A\,a^2\,\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )\,2{}\mathrm {i}}{d}-\frac {A\,a^2\,\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,2{}\mathrm {i}}{d} \]