\(\int \cot ^5(c+d x) (a+i a \tan (c+d x))^2 \, dx\) [22]

Optimal result

Integrand size = 24, antiderivative size = 93 \[ \int \cot ^5(c+d x) (a+i a \tan (c+d x))^2 \, dx=2 i a^2 x+\frac {2 i a^2 \cot (c+d x)}{d}+\frac {a^2 \cot ^2(c+d x)}{d}-\frac {2 i a^2 \cot ^3(c+d x)}{3 d}-\frac {a^2 \cot ^4(c+d x)}{4 d}+\frac {2 a^2 \log (\sin (c+d x))}{d} \]

[Out]

2*I*a^2*x+2*I*a^2*cot(d*x+c)/d+a^2*cot(d*x+c)^2/d-2/3*I*a^2*cot(d*x+c)^3/d-1/4*a^2*cot(d*x+c)^4/d+2*a^2*ln(sin
(d*x+c))/d

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3623, 3610, 3612, 3556} \[ \int \cot ^5(c+d x) (a+i a \tan (c+d x))^2 \, dx=-\frac {a^2 \cot ^4(c+d x)}{4 d}-\frac {2 i a^2 \cot ^3(c+d x)}{3 d}+\frac {a^2 \cot ^2(c+d x)}{d}+\frac {2 i a^2 \cot (c+d x)}{d}+\frac {2 a^2 \log (\sin (c+d x))}{d}+2 i a^2 x \]

[In]

Int[Cot[c + d*x]^5*(a + I*a*Tan[c + d*x])^2,x]

[Out]

(2*I)*a^2*x + ((2*I)*a^2*Cot[c + d*x])/d + (a^2*Cot[c + d*x]^2)/d - (((2*I)/3)*a^2*Cot[c + d*x]^3)/d - (a^2*Co
t[c + d*x]^4)/(4*d) + (2*a^2*Log[Sin[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 3610

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

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 3623

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

Rubi steps \begin{align*} \text {integral}& = -\frac {a^2 \cot ^4(c+d x)}{4 d}+\int \cot ^4(c+d x) \left (2 i a^2-2 a^2 \tan (c+d x)\right ) \, dx \\ & = -\frac {2 i a^2 \cot ^3(c+d x)}{3 d}-\frac {a^2 \cot ^4(c+d x)}{4 d}+\int \cot ^3(c+d x) \left (-2 a^2-2 i a^2 \tan (c+d x)\right ) \, dx \\ & = \frac {a^2 \cot ^2(c+d x)}{d}-\frac {2 i a^2 \cot ^3(c+d x)}{3 d}-\frac {a^2 \cot ^4(c+d x)}{4 d}+\int \cot ^2(c+d x) \left (-2 i a^2+2 a^2 \tan (c+d x)\right ) \, dx \\ & = \frac {2 i a^2 \cot (c+d x)}{d}+\frac {a^2 \cot ^2(c+d x)}{d}-\frac {2 i a^2 \cot ^3(c+d x)}{3 d}-\frac {a^2 \cot ^4(c+d x)}{4 d}+\int \cot (c+d x) \left (2 a^2+2 i a^2 \tan (c+d x)\right ) \, dx \\ & = 2 i a^2 x+\frac {2 i a^2 \cot (c+d x)}{d}+\frac {a^2 \cot ^2(c+d x)}{d}-\frac {2 i a^2 \cot ^3(c+d x)}{3 d}-\frac {a^2 \cot ^4(c+d x)}{4 d}+\left (2 a^2\right ) \int \cot (c+d x) \, dx \\ & = 2 i a^2 x+\frac {2 i a^2 \cot (c+d x)}{d}+\frac {a^2 \cot ^2(c+d x)}{d}-\frac {2 i a^2 \cot ^3(c+d x)}{3 d}-\frac {a^2 \cot ^4(c+d x)}{4 d}+\frac {2 a^2 \log (\sin (c+d x))}{d} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.70 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.84 \[ \int \cot ^5(c+d x) (a+i a \tan (c+d x))^2 \, dx=\frac {a^2 \left (12 \cot ^2(c+d x)-3 \cot ^4(c+d x)-8 i \cot ^3(c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},-\tan ^2(c+d x)\right )+24 (\log (\cos (c+d x))+\log (\tan (c+d x)))\right )}{12 d} \]

[In]

Integrate[Cot[c + d*x]^5*(a + I*a*Tan[c + d*x])^2,x]

[Out]

(a^2*(12*Cot[c + d*x]^2 - 3*Cot[c + d*x]^4 - (8*I)*Cot[c + d*x]^3*Hypergeometric2F1[-3/2, 1, -1/2, -Tan[c + d*
x]^2] + 24*(Log[Cos[c + d*x]] + Log[Tan[c + d*x]])))/(12*d)

Maple [A] (verified)

Time = 0.62 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.81

method	result	size
parallelrisch	\(-\frac {a^{2} \left (3 \left (\cot ^{4}\left (d x +c \right )\right )+8 i \left (\cot ^{3}\left (d x +c \right )\right )-24 i d x -12 \left (\cot ^{2}\left (d x +c \right )\right )-24 i \cot \left (d x +c \right )-24 \ln \left (\tan \left (d x +c \right )\right )+12 \ln \left (\sec ^{2}\left (d x +c \right )\right )\right )}{12 d}\)	\(75\)
risch	\(-\frac {4 i a^{2} c}{d}-\frac {2 a^{2} \left (21 \,{\mathrm e}^{6 i \left (d x +c \right )}-36 \,{\mathrm e}^{4 i \left (d x +c \right )}+29 \,{\mathrm e}^{2 i \left (d x +c \right )}-8\right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{4}}+\frac {2 a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\)	\(88\)
derivativedivides	\(\frac {-a^{2} \left (-\frac {\left (\cot ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\sin \left (d x +c \right )\right )\right )+2 i a^{2} \left (-\frac {\left (\cot ^{3}\left (d x +c \right )\right )}{3}+\cot \left (d x +c \right )+d x +c \right )+a^{2} \left (-\frac {\left (\cot ^{4}\left (d x +c \right )\right )}{4}+\frac {\left (\cot ^{2}\left (d x +c \right )\right )}{2}+\ln \left (\sin \left (d x +c \right )\right )\right )}{d}\)	\(90\)
default	\(\frac {-a^{2} \left (-\frac {\left (\cot ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\sin \left (d x +c \right )\right )\right )+2 i a^{2} \left (-\frac {\left (\cot ^{3}\left (d x +c \right )\right )}{3}+\cot \left (d x +c \right )+d x +c \right )+a^{2} \left (-\frac {\left (\cot ^{4}\left (d x +c \right )\right )}{4}+\frac {\left (\cot ^{2}\left (d x +c \right )\right )}{2}+\ln \left (\sin \left (d x +c \right )\right )\right )}{d}\)	\(90\)
norman	\(\frac {\frac {a^{2} \left (\tan ^{2}\left (d x +c \right )\right )}{d}-\frac {a^{2}}{4 d}+2 i a^{2} x \left (\tan ^{4}\left (d x +c \right )\right )-\frac {2 i a^{2} \tan \left (d x +c \right )}{3 d}+\frac {2 i a^{2} \left (\tan ^{3}\left (d x +c \right )\right )}{d}}{\tan \left (d x +c \right )^{4}}+\frac {2 a^{2} \ln \left (\tan \left (d x +c \right )\right )}{d}-\frac {a^{2} \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{d}\)	\(116\)

[In]

int(cot(d*x+c)^5*(a+I*a*tan(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

-1/12*a^2*(3*cot(d*x+c)^4+8*I*cot(d*x+c)^3-24*I*d*x-12*cot(d*x+c)^2-24*I*cot(d*x+c)-24*ln(tan(d*x+c))+12*ln(se
c(d*x+c)^2))/d

Fricas [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. \(2 (83) = 166\).

Time = 0.25 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.87 \[ \int \cot ^5(c+d x) (a+i a \tan (c+d x))^2 \, dx=-\frac {2 \, {\left (21 \, a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} - 36 \, a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + 29 \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - 8 \, a^{2} - 3 \, {\left (a^{2} e^{\left (8 i \, d x + 8 i \, c\right )} - 4 \, a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} - 4 \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right )\right )}}{3 \, {\left (d e^{\left (8 i \, d x + 8 i \, c\right )} - 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} - 4 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

[In]

integrate(cot(d*x+c)^5*(a+I*a*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

-2/3*(21*a^2*e^(6*I*d*x + 6*I*c) - 36*a^2*e^(4*I*d*x + 4*I*c) + 29*a^2*e^(2*I*d*x + 2*I*c) - 8*a^2 - 3*(a^2*e^
(8*I*d*x + 8*I*c) - 4*a^2*e^(6*I*d*x + 6*I*c) + 6*a^2*e^(4*I*d*x + 4*I*c) - 4*a^2*e^(2*I*d*x + 2*I*c) + a^2)*l
og(e^(2*I*d*x + 2*I*c) - 1))/(d*e^(8*I*d*x + 8*I*c) - 4*d*e^(6*I*d*x + 6*I*c) + 6*d*e^(4*I*d*x + 4*I*c) - 4*d*
e^(2*I*d*x + 2*I*c) + d)

Sympy [A] (verification not implemented)

Time = 0.70 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.81 \[ \int \cot ^5(c+d x) (a+i a \tan (c+d x))^2 \, dx=\frac {2 a^{2} \log {\left (e^{2 i d x} - e^{- 2 i c} \right )}}{d} + \frac {- 42 a^{2} e^{6 i c} e^{6 i d x} + 72 a^{2} e^{4 i c} e^{4 i d x} - 58 a^{2} e^{2 i c} e^{2 i d x} + 16 a^{2}}{3 d e^{8 i c} e^{8 i d x} - 12 d e^{6 i c} e^{6 i d x} + 18 d e^{4 i c} e^{4 i d x} - 12 d e^{2 i c} e^{2 i d x} + 3 d} \]

[In]

integrate(cot(d*x+c)**5*(a+I*a*tan(d*x+c))**2,x)

[Out]

2*a**2*log(exp(2*I*d*x) - exp(-2*I*c))/d + (-42*a**2*exp(6*I*c)*exp(6*I*d*x) + 72*a**2*exp(4*I*c)*exp(4*I*d*x)
- 58*a**2*exp(2*I*c)*exp(2*I*d*x) + 16*a**2)/(3*d*exp(8*I*c)*exp(8*I*d*x) - 12*d*exp(6*I*c)*exp(6*I*d*x) + 18
*d*exp(4*I*c)*exp(4*I*d*x) - 12*d*exp(2*I*c)*exp(2*I*d*x) + 3*d)

Maxima [A] (verification not implemented)

none

Time = 0.45 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.04 \[ \int \cot ^5(c+d x) (a+i a \tan (c+d x))^2 \, dx=-\frac {-24 i \, {\left (d x + c\right )} a^{2} + 12 \, a^{2} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 24 \, a^{2} \log \left (\tan \left (d x + c\right )\right ) - \frac {24 i \, a^{2} \tan \left (d x + c\right )^{3} + 12 \, a^{2} \tan \left (d x + c\right )^{2} - 8 i \, a^{2} \tan \left (d x + c\right ) - 3 \, a^{2}}{\tan \left (d x + c\right )^{4}}}{12 \, d} \]

[In]

integrate(cot(d*x+c)^5*(a+I*a*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/12*(-24*I*(d*x + c)*a^2 + 12*a^2*log(tan(d*x + c)^2 + 1) - 24*a^2*log(tan(d*x + c)) - (24*I*a^2*tan(d*x + c
)^3 + 12*a^2*tan(d*x + c)^2 - 8*I*a^2*tan(d*x + c) - 3*a^2)/tan(d*x + c)^4)/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. 180 vs. \(2 (83) = 166\).

Time = 0.89 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.94 \[ \int \cot ^5(c+d x) (a+i a \tan (c+d x))^2 \, dx=-\frac {3 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 16 i \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 60 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 768 \, a^{2} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + i\right ) - 384 \, a^{2} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 240 i \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {800 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 240 i \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 60 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 16 i \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, a^{2}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4}}}{192 \, d} \]

[In]

integrate(cot(d*x+c)^5*(a+I*a*tan(d*x+c))^2,x, algorithm="giac")

[Out]

-1/192*(3*a^2*tan(1/2*d*x + 1/2*c)^4 - 16*I*a^2*tan(1/2*d*x + 1/2*c)^3 - 60*a^2*tan(1/2*d*x + 1/2*c)^2 + 768*a
^2*log(tan(1/2*d*x + 1/2*c) + I) - 384*a^2*log(tan(1/2*d*x + 1/2*c)) + 240*I*a^2*tan(1/2*d*x + 1/2*c) + (800*a
^2*tan(1/2*d*x + 1/2*c)^4 - 240*I*a^2*tan(1/2*d*x + 1/2*c)^3 - 60*a^2*tan(1/2*d*x + 1/2*c)^2 + 16*I*a^2*tan(1/
2*d*x + 1/2*c) + 3*a^2)/tan(1/2*d*x + 1/2*c)^4)/d

Mupad [B] (verification not implemented)

Time = 5.34 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.86 \[ \int \cot ^5(c+d x) (a+i a \tan (c+d x))^2 \, dx=\frac {a^2\,\mathrm {atan}\left (2\,\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,4{}\mathrm {i}}{d}-\frac {-a^2\,{\mathrm {tan}\left (c+d\,x\right )}^3\,2{}\mathrm {i}-a^2\,{\mathrm {tan}\left (c+d\,x\right )}^2+\frac {a^2\,\mathrm {tan}\left (c+d\,x\right )\,2{}\mathrm {i}}{3}+\frac {a^2}{4}}{d\,{\mathrm {tan}\left (c+d\,x\right )}^4} \]

[In]

int(cot(c + d*x)^5*(a + a*tan(c + d*x)*1i)^2,x)

[Out]

(a^2*atan(2*tan(c + d*x) + 1i)*4i)/d - ((a^2*tan(c + d*x)*2i)/3 + a^2/4 - a^2*tan(c + d*x)^2 - a^2*tan(c + d*x
)^3*2i)/(d*tan(c + d*x)^4)