\(\int \cot (c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx\) [12]

Optimal result

Integrand size = 32, antiderivative size = 75 \[ \int \cot (c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=2 a^2 (i A+B) x+\frac {a^2 (A-2 i B) \log (\cos (c+d x))}{d}+\frac {a^2 A \log (\sin (c+d x))}{d}+\frac {i B \left (a^2+i a^2 \tan (c+d x)\right )}{d} \]

[Out]

2*a^2*(I*A+B)*x+a^2*(A-2*I*B)*ln(cos(d*x+c))/d+a^2*A*ln(sin(d*x+c))/d+I*B*(a^2+I*a^2*tan(d*x+c))/d

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {3675, 3670, 3556, 3612} \[ \int \cot (c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=\frac {a^2 (A-2 i B) \log (\cos (c+d x))}{d}+2 a^2 x (B+i A)+\frac {a^2 A \log (\sin (c+d x))}{d}+\frac {i B \left (a^2+i a^2 \tan (c+d x)\right )}{d} \]

[In]

Int[Cot[c + d*x]*(a + I*a*Tan[c + d*x])^2*(A + B*Tan[c + d*x]),x]

[Out]

2*a^2*(I*A + B)*x + (a^2*(A - (2*I)*B)*Log[Cos[c + d*x]])/d + (a^2*A*Log[Sin[c + d*x]])/d + (I*B*(a^2 + I*a^2*
Tan[c + d*x]))/d

Rule 3556

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3612

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] + Dist[(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] && NeQ[a*c + b*d, 0]

Rule 3670

Int[(((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]))/((a_.) + (b_.)*tan[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> Dist[B*(d/b), Int[Tan[e + f*x], x], x] + Dist[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]

Rule 3675

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*B*(a + b*Tan[e + f*x])^(m - 1)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*
(m + n))), x] + Dist[1/(d*(m + n)), Int[(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^n*Simp[a*A*d*(m + n)
+ B*(a*c*(m - 1) - b*d*(n + 1)) - (B*(b*c - a*d)*(m - 1) - d*(A*b + a*B)*(m + n))*Tan[e + f*x], x], x], x] /;
FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && GtQ[m, 1] &&  !LtQ[n, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {i B \left (a^2+i a^2 \tan (c+d x)\right )}{d}+\int \cot (c+d x) (a+i a \tan (c+d x)) (a A+a (i A+2 B) \tan (c+d x)) \, dx \\ & = \frac {i B \left (a^2+i a^2 \tan (c+d x)\right )}{d}-\left (a^2 (A-2 i B)\right ) \int \tan (c+d x) \, dx+\int \cot (c+d x) \left (a^2 A+2 a^2 (i A+B) \tan (c+d x)\right ) \, dx \\ & = 2 a^2 (i A+B) x+\frac {a^2 (A-2 i B) \log (\cos (c+d x))}{d}+\frac {i B \left (a^2+i a^2 \tan (c+d x)\right )}{d}+\left (a^2 A\right ) \int \cot (c+d x) \, dx \\ & = 2 a^2 (i A+B) x+\frac {a^2 (A-2 i B) \log (\cos (c+d x))}{d}+\frac {a^2 A \log (\sin (c+d x))}{d}+\frac {i B \left (a^2+i a^2 \tan (c+d x)\right )}{d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.50 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.61 \[ \int \cot (c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=\frac {a^2 (A \log (\tan (c+d x))-2 (A-i B) \log (i+\tan (c+d x))-B \tan (c+d x))}{d} \]

[In]

Integrate[Cot[c + d*x]*(a + I*a*Tan[c + d*x])^2*(A + B*Tan[c + d*x]),x]

[Out]

(a^2*(A*Log[Tan[c + d*x]] - 2*(A - I*B)*Log[I + Tan[c + d*x]] - B*Tan[c + d*x]))/d

Maple [A] (verified)

Time = 0.17 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.84

method	result	size
parallelrisch	\(\frac {a^{2} \left (2 i A x d +i B \ln \left (\sec ^{2}\left (d x +c \right )\right )+2 B d x -A \ln \left (\sec ^{2}\left (d x +c \right )\right )+A \ln \left (\tan \left (d x +c \right )\right )-B \tan \left (d x +c \right )\right )}{d}\)	\(63\)
norman	\(\left (2 i A \,a^{2}+2 B \,a^{2}\right ) x -\frac {B \,a^{2} \tan \left (d x +c \right )}{d}+\frac {A \,a^{2} \ln \left (\tan \left (d x +c \right )\right )}{d}-\frac {\left (-i B \,a^{2}+A \,a^{2}\right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{d}\)	\(77\)
derivativedivides	\(-\frac {a^{2} \left (\frac {\left (-2 i B +2 A \right ) \ln \left (\cot ^{2}\left (d x +c \right )+1\right )}{2}+\left (2 i A +2 B \right ) \left (\frac {\pi }{2}-\operatorname {arccot}\left (\cot \left (d x +c \right )\right )\right )+\left (2 i B -A \right ) \ln \left (\cot \left (d x +c \right )\right )+\frac {B}{\cot \left (d x +c \right )}\right )}{d}\)	\(79\)
default	\(-\frac {a^{2} \left (\frac {\left (-2 i B +2 A \right ) \ln \left (\cot ^{2}\left (d x +c \right )+1\right )}{2}+\left (2 i A +2 B \right ) \left (\frac {\pi }{2}-\operatorname {arccot}\left (\cot \left (d x +c \right )\right )\right )+\left (2 i B -A \right ) \ln \left (\cot \left (d x +c \right )\right )+\frac {B}{\cot \left (d x +c \right )}\right )}{d}\)	\(79\)
risch	\(-\frac {4 i a^{2} A c}{d}-\frac {4 a^{2} B c}{d}-\frac {2 i B \,a^{2}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}+\frac {A \,a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}-\frac {2 i a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) B}{d}+\frac {a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) A}{d}\)	\(108\)

[In]

int(cot(d*x+c)*(a+I*a*tan(d*x+c))^2*(A+B*tan(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

1/d*a^2*(2*I*A*x*d+I*B*ln(sec(d*x+c)^2)+2*B*d*x-A*ln(sec(d*x+c)^2)+A*ln(tan(d*x+c))-B*tan(d*x+c))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.29 \[ \int \cot (c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=\frac {-2 i \, B a^{2} + {\left ({\left (A - 2 i \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (A - 2 i \, B\right )} a^{2}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + {\left (A a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + A a^{2}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right )}{d e^{\left (2 i \, d x + 2 i \, c\right )} + d} \]

[In]

integrate(cot(d*x+c)*(a+I*a*tan(d*x+c))^2*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

(-2*I*B*a^2 + ((A - 2*I*B)*a^2*e^(2*I*d*x + 2*I*c) + (A - 2*I*B)*a^2)*log(e^(2*I*d*x + 2*I*c) + 1) + (A*a^2*e^
(2*I*d*x + 2*I*c) + A*a^2)*log(e^(2*I*d*x + 2*I*c) - 1))/(d*e^(2*I*d*x + 2*I*c) + d)

Sympy [A] (verification not implemented)

Time = 1.22 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.45 \[ \int \cot (c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=\frac {A a^{2} \log {\left (e^{2 i d x} - e^{- 2 i c} \right )}}{d} - \frac {2 i B a^{2}}{d e^{2 i c} e^{2 i d x} + d} + \frac {a^{2} \left (A - 2 i B\right ) \log {\left (e^{2 i d x} + \frac {\left (- i A a^{2} - B a^{2} + i a^{2} \left (A - 2 i B\right )\right ) e^{- 2 i c}}{B a^{2}} \right )}}{d} \]

[In]

integrate(cot(d*x+c)*(a+I*a*tan(d*x+c))**2*(A+B*tan(d*x+c)),x)

[Out]

A*a**2*log(exp(2*I*d*x) - exp(-2*I*c))/d - 2*I*B*a**2/(d*exp(2*I*c)*exp(2*I*d*x) + d) + a**2*(A - 2*I*B)*log(e
xp(2*I*d*x) + (-I*A*a**2 - B*a**2 + I*a**2*(A - 2*I*B))*exp(-2*I*c)/(B*a**2))/d

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.89 \[ \int \cot (c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=-\frac {2 \, {\left (d x + c\right )} {\left (-i \, A - B\right )} a^{2} + {\left (A - i \, B\right )} a^{2} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - A a^{2} \log \left (\tan \left (d x + c\right )\right ) + B a^{2} \tan \left (d x + c\right )}{d} \]

[In]

integrate(cot(d*x+c)*(a+I*a*tan(d*x+c))^2*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

-(2*(d*x + c)*(-I*A - B)*a^2 + (A - I*B)*a^2*log(tan(d*x + c)^2 + 1) - A*a^2*log(tan(d*x + c)) + B*a^2*tan(d*x
+ c))/d

Giac [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 174 vs. \(2 (67) = 134\).

Time = 0.64 (sec) , antiderivative size = 174, normalized size of antiderivative = 2.32 \[ \int \cot (c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=\frac {A a^{2} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + {\left (A a^{2} - 2 i \, B a^{2}\right )} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right ) - 4 \, {\left (A a^{2} - i \, B a^{2}\right )} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + i\right ) + {\left (A a^{2} - 2 i \, B a^{2}\right )} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right ) - \frac {A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 2 i \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 2 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - A a^{2} + 2 i \, B a^{2}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1}}{d} \]

[In]

integrate(cot(d*x+c)*(a+I*a*tan(d*x+c))^2*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

(A*a^2*log(tan(1/2*d*x + 1/2*c)) + (A*a^2 - 2*I*B*a^2)*log(tan(1/2*d*x + 1/2*c) + 1) - 4*(A*a^2 - I*B*a^2)*log
(tan(1/2*d*x + 1/2*c) + I) + (A*a^2 - 2*I*B*a^2)*log(tan(1/2*d*x + 1/2*c) - 1) - (A*a^2*tan(1/2*d*x + 1/2*c)^2
- 2*I*B*a^2*tan(1/2*d*x + 1/2*c)^2 - 2*B*a^2*tan(1/2*d*x + 1/2*c) - A*a^2 + 2*I*B*a^2)/(tan(1/2*d*x + 1/2*c)^
2 - 1))/d

Mupad [B] (verification not implemented)

Time = 8.13 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.93 \[ \int \cot (c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=\frac {A\,a^2\,\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )}{d}-\frac {2\,A\,a^2\,\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )}{d}-\frac {B\,a^2\,\mathrm {tan}\left (c+d\,x\right )}{d}+\frac {B\,a^2\,\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,2{}\mathrm {i}}{d} \]

[In]

int(cot(c + d*x)*(A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^2,x)

[Out]

(A*a^2*log(tan(c + d*x)))/d - (2*A*a^2*log(tan(c + d*x) + 1i))/d + (B*a^2*log(tan(c + d*x) + 1i)*2i)/d - (B*a^
2*tan(c + d*x))/d