\(\int \cot ^2(c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx\) [13]

Optimal result

Integrand size = 34, antiderivative size = 79 \[ \int \cot ^2(c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=-2 a^2 (A-i B) x+\frac {a^2 B \log (\cos (c+d x))}{d}+\frac {a^2 (2 i A+B) \log (\sin (c+d x))}{d}-\frac {A \cot (c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{d} \]

[Out]

-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*cot(d*x+c)*(a^2+I*a^2*tan(d*x+c))/d

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {3674, 3670, 3556, 3612} \[ \int \cot ^2(c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=\frac {a^2 (B+2 i A) \log (\sin (c+d x))}{d}-2 a^2 x (A-i B)-\frac {A \cot (c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{d}+\frac {a^2 B \log (\cos (c+d x))}{d} \]

[In]

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

[Out]

-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

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 3674

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-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] - Dist[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] && E
qQ[a^2 + b^2, 0] && GtQ[m, 1] && LtQ[n, -1]

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

Mathematica [A] (verified)

Time = 0.75 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.68 \[ \int \cot ^2(c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=\frac {a^2 (-A \cot (c+d x)+(2 i A+B) \log (\tan (c+d x))-2 i (A-i B) \log (i+\tan (c+d x)))}{d} \]

[In]

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

[Out]

(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

Maple [A] (verified)

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	\(-\frac {4 i a^{2} B c}{d}+\frac {4 a^{2} A c}{d}-\frac {2 i A \,a^{2}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}+\frac {a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) B}{d}+\frac {2 i a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) A}{d}+\frac {a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) B}{d}\)	\(108\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.29 \[ \int \cot ^2(c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=\frac {-2 i \, A a^{2} + {\left (B a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - B a^{2}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + {\left ({\left (2 i \, A + B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (-2 i \, A - B\right )} a^{2}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right )}{d e^{\left (2 i \, d x + 2 i \, c\right )} - d} \]

[In]

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

[Out]

(-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)

Sympy [A] (verification not implemented)

Time = 1.32 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.38 \[ \int \cot ^2(c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=- \frac {2 i A a^{2}}{d e^{2 i c} e^{2 i d x} - d} + \frac {B a^{2} \log {\left (e^{2 i d x} + e^{- 2 i c} \right )}}{d} + \frac {i a^{2} \cdot \left (2 A - i B\right ) \log {\left (e^{2 i d x} + \frac {\left (A a^{2} - i B a^{2} - a^{2} \cdot \left (2 A - i B\right )\right ) e^{- 2 i c}}{A a^{2}} \right )}}{d} \]

[In]

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

[Out]

-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) + exp(-2*I*c))/d + I*a**2*(2*A - I*B)*lo
g(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

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.94 \[ \int \cot ^2(c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=-\frac {2 \, {\left (d x + c\right )} {\left (A - i \, B\right )} a^{2} - {\left (-i \, A - B\right )} a^{2} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - {\left (2 i \, A + B\right )} a^{2} \log \left (\tan \left (d x + c\right )\right ) + \frac {A a^{2}}{\tan \left (d x + c\right )}}{d} \]

[In]

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

[Out]

-(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

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. 155 vs. \(2 (73) = 146\).

Time = 0.86 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.96 \[ \int \cot ^2(c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=\frac {2 \, B a^{2} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right ) + 2 \, B a^{2} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right ) + A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 8 \, {\left (i \, A a^{2} + B a^{2}\right )} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + i\right ) + 2 \, {\left (2 i \, A a^{2} + B a^{2}\right )} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + \frac {-4 i \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - A a^{2}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}}{2 \, d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 7.69 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.10 \[ \int \cot ^2(c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=\frac {B\,a^2\,\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )}{d}-\frac {2\,B\,a^2\,\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )}{d}-\frac {A\,a^2\,\mathrm {cot}\left (c+d\,x\right )}{d}+\frac {A\,a^2\,\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )\,2{}\mathrm {i}}{d}-\frac {A\,a^2\,\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,2{}\mathrm {i}}{d} \]

[In]

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

[Out]

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