\(\int \cot ^3(c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx\) [14]

Optimal result

Integrand size = 34, antiderivative size = 94 \[ \int \cot ^3(c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=-2 a^2 (i A+B) x-\frac {a^2 (3 i A+2 B) \cot (c+d x)}{2 d}-\frac {2 a^2 (A-i B) \log (\sin (c+d x))}{d}-\frac {A \cot ^2(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{2 d} \]

[Out]

-2*a^2*(I*A+B)*x-1/2*a^2*(3*I*A+2*B)*cot(d*x+c)/d-2*a^2*(A-I*B)*ln(sin(d*x+c))/d-1/2*A*cot(d*x+c)^2*(a^2+I*a^2
*tan(d*x+c))/d

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {3674, 3672, 3612, 3556} \[ \int \cot ^3(c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=-\frac {a^2 (2 B+3 i A) \cot (c+d x)}{2 d}-\frac {2 a^2 (A-i B) \log (\sin (c+d x))}{d}-2 a^2 x (B+i A)-\frac {A \cot ^2(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{2 d} \]

[In]

Int[Cot[c + d*x]^3*(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*((3*I)*A + 2*B)*Cot[c + d*x])/(2*d) - (2*a^2*(A - I*B)*Log[Sin[c + d*x]])/d - (A*Cot
[c + d*x]^2*(a^2 + I*a^2*Tan[c + d*x]))/(2*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 3672

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e
_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*c - a*d)*(A*b - a*B)*((a + b*Tan[e + f*x])^(m + 1)/(b*f*(m + 1)*(a^2
+ b^2))), x] + Dist[1/(a^2 + b^2), Int[(a + b*Tan[e + f*x])^(m + 1)*Simp[a*A*c + b*B*c + A*b*d - a*B*d - (A*b*
c - a*B*c - a*A*d - b*B*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0]
&& LtQ[m, -1] && NeQ[a^2 + b^2, 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 ^2(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{2 d}+\frac {1}{2} \int \cot ^2(c+d x) (a+i a \tan (c+d x)) (a (3 i A+2 B)-a (A-2 i B) \tan (c+d x)) \, dx \\ & = -\frac {a^2 (3 i A+2 B) \cot (c+d x)}{2 d}-\frac {A \cot ^2(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{2 d}+\frac {1}{2} \int \cot (c+d x) \left (-4 a^2 (A-i B)-4 a^2 (i A+B) \tan (c+d x)\right ) \, dx \\ & = -2 a^2 (i A+B) x-\frac {a^2 (3 i A+2 B) \cot (c+d x)}{2 d}-\frac {A \cot ^2(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{2 d}-\left (2 a^2 (A-i B)\right ) \int \cot (c+d x) \, dx \\ & = -2 a^2 (i A+B) x-\frac {a^2 (3 i A+2 B) \cot (c+d x)}{2 d}-\frac {2 a^2 (A-i B) \log (\sin (c+d x))}{d}-\frac {A \cot ^2(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.50 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.18 \[ \int \cot ^3(c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=a^2 \left (-\frac {2 i A \cot (c+d x)}{d}-\frac {B \cot (c+d x)}{d}-\frac {A \cot ^2(c+d x)}{2 d}-\frac {2 A \log (\tan (c+d x))}{d}+\frac {2 i B \log (\tan (c+d x))}{d}+\frac {2 A \log (i+\tan (c+d x))}{d}-\frac {2 i B \log (i+\tan (c+d x))}{d}\right ) \]

[In]

Integrate[Cot[c + d*x]^3*(a + I*a*Tan[c + d*x])^2*(A + B*Tan[c + d*x]),x]

[Out]

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

Maple [A] (verified)

Time = 0.24 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.82

method	result	size
parallelrisch	\(-\frac {2 a^{2} \left (\left (-\frac {A}{2}+\frac {i B}{2}\right ) \ln \left (\sec ^{2}\left (d x +c \right )\right )+\left (-i B +A \right ) \ln \left (\tan \left (d x +c \right )\right )+\frac {A \left (\cot ^{2}\left (d x +c \right )\right )}{4}+\cot \left (d x +c \right ) \left (i A +\frac {B}{2}\right )+\left (i A +B \right ) x d \right )}{d}\)	\(77\)
derivativedivides	\(\frac {-A \,a^{2} \ln \left (\sin \left (d x +c \right )\right )-B \,a^{2} \left (d x +c \right )+2 i A \,a^{2} \left (-\cot \left (d x +c \right )-d x -c \right )+2 i B \,a^{2} \ln \left (\sin \left (d x +c \right )\right )+A \,a^{2} \left (-\frac {\left (\cot ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\sin \left (d x +c \right )\right )\right )+B \,a^{2} \left (-\cot \left (d x +c \right )-d x -c \right )}{d}\)	\(113\)
default	\(\frac {-A \,a^{2} \ln \left (\sin \left (d x +c \right )\right )-B \,a^{2} \left (d x +c \right )+2 i A \,a^{2} \left (-\cot \left (d x +c \right )-d x -c \right )+2 i B \,a^{2} \ln \left (\sin \left (d x +c \right )\right )+A \,a^{2} \left (-\frac {\left (\cot ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\sin \left (d x +c \right )\right )\right )+B \,a^{2} \left (-\cot \left (d x +c \right )-d x -c \right )}{d}\)	\(113\)
risch	\(\frac {4 a^{2} B c}{d}+\frac {4 i a^{2} A c}{d}-\frac {2 i a^{2} \left (3 i A \,{\mathrm e}^{2 i \left (d x +c \right )}+B \,{\mathrm e}^{2 i \left (d x +c \right )}-2 i A -B \right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}+\frac {2 i a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) B}{d}-\frac {2 A \,a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\)	\(120\)
norman	\(\frac {\left (-2 i A \,a^{2}-2 B \,a^{2}\right ) x \left (\tan ^{2}\left (d x +c \right )\right )-\frac {A \,a^{2}}{2 d}-\frac {\left (2 i A \,a^{2}+B \,a^{2}\right ) \tan \left (d x +c \right )}{d}}{\tan \left (d x +c \right )^{2}}+\frac {\left (-i B \,a^{2}+A \,a^{2}\right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{d}-\frac {2 \left (-i B \,a^{2}+A \,a^{2}\right ) \ln \left (\tan \left (d x +c \right )\right )}{d}\)	\(122\)

[In]

int(cot(d*x+c)^3*(a+I*a*tan(d*x+c))^2*(A+B*tan(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.31 \[ \int \cot ^3(c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=\frac {2 \, {\left ({\left (3 \, A - i \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - {\left (2 \, A - i \, B\right )} a^{2} - {\left ({\left (A - i \, B\right )} a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, {\left (A - i \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (A - i \, B\right )} a^{2}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right )\right )}}{d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d} \]

[In]

integrate(cot(d*x+c)^3*(a+I*a*tan(d*x+c))^2*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

2*((3*A - I*B)*a^2*e^(2*I*d*x + 2*I*c) - (2*A - I*B)*a^2 - ((A - I*B)*a^2*e^(4*I*d*x + 4*I*c) - 2*(A - I*B)*a^
2*e^(2*I*d*x + 2*I*c) + (A - I*B)*a^2)*log(e^(2*I*d*x + 2*I*c) - 1))/(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x +
2*I*c) + d)

Sympy [A] (verification not implemented)

Time = 0.33 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.27 \[ \int \cot ^3(c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=- \frac {2 a^{2} \left (A - i B\right ) \log {\left (e^{2 i d x} - e^{- 2 i c} \right )}}{d} + \frac {- 4 A a^{2} + 2 i B a^{2} + \left (6 A a^{2} e^{2 i c} - 2 i B a^{2} e^{2 i c}\right ) e^{2 i d x}}{d e^{4 i c} e^{4 i d x} - 2 d e^{2 i c} e^{2 i d x} + d} \]

[In]

integrate(cot(d*x+c)**3*(a+I*a*tan(d*x+c))**2*(A+B*tan(d*x+c)),x)

[Out]

-2*a**2*(A - I*B)*log(exp(2*I*d*x) - exp(-2*I*c))/d + (-4*A*a**2 + 2*I*B*a**2 + (6*A*a**2*exp(2*I*c) - 2*I*B*a
**2*exp(2*I*c))*exp(2*I*d*x))/(d*exp(4*I*c)*exp(4*I*d*x) - 2*d*exp(2*I*c)*exp(2*I*d*x) + d)

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.00 \[ \int \cot ^3(c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=-\frac {4 \, {\left (d x + c\right )} {\left (i \, A + B\right )} a^{2} - 2 \, {\left (A - i \, B\right )} a^{2} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 4 \, {\left (A - i \, B\right )} a^{2} \log \left (\tan \left (d x + c\right )\right ) - \frac {2 \, {\left (-2 i \, A - B\right )} a^{2} \tan \left (d x + c\right ) - A a^{2}}{\tan \left (d x + c\right )^{2}}}{2 \, d} \]

[In]

integrate(cot(d*x+c)^3*(a+I*a*tan(d*x+c))^2*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

-1/2*(4*(d*x + c)*(I*A + B)*a^2 - 2*(A - I*B)*a^2*log(tan(d*x + c)^2 + 1) + 4*(A - I*B)*a^2*log(tan(d*x + c))
- (2*(-2*I*A - B)*a^2*tan(d*x + c) - A*a^2)/tan(d*x + c)^2)/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. 186 vs. \(2 (82) = 164\).

Time = 1.08 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.98 \[ \int \cot ^3(c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=-\frac {A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 8 i \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 4 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 32 \, {\left (A a^{2} - i \, B a^{2}\right )} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + i\right ) + 16 \, {\left (A a^{2} - i \, B a^{2}\right )} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) - \frac {24 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 24 i \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 8 i \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 4 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - A a^{2}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}}{8 \, d} \]

[In]

integrate(cot(d*x+c)^3*(a+I*a*tan(d*x+c))^2*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

-1/8*(A*a^2*tan(1/2*d*x + 1/2*c)^2 - 8*I*A*a^2*tan(1/2*d*x + 1/2*c) - 4*B*a^2*tan(1/2*d*x + 1/2*c) - 32*(A*a^2
- I*B*a^2)*log(tan(1/2*d*x + 1/2*c) + I) + 16*(A*a^2 - I*B*a^2)*log(tan(1/2*d*x + 1/2*c)) - (24*A*a^2*tan(1/2
*d*x + 1/2*c)^2 - 24*I*B*a^2*tan(1/2*d*x + 1/2*c)^2 - 8*I*A*a^2*tan(1/2*d*x + 1/2*c) - 4*B*a^2*tan(1/2*d*x + 1
/2*c) - A*a^2)/tan(1/2*d*x + 1/2*c)^2)/d

Mupad [B] (verification not implemented)

Time = 7.40 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.71 \[ \int \cot ^3(c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=-\frac {\frac {A\,a^2}{2}+\mathrm {tan}\left (c+d\,x\right )\,\left (B\,a^2+A\,a^2\,2{}\mathrm {i}\right )}{d\,{\mathrm {tan}\left (c+d\,x\right )}^2}-\frac {4\,a^2\,\mathrm {atan}\left (2\,\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (B+A\,1{}\mathrm {i}\right )}{d} \]

[In]

int(cot(c + d*x)^3*(A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^2,x)

[Out]

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