\(\int \cot ^4(c+d x) (a+i a \tan (c+d x))^2 \, dx\) [21]

Optimal result

Integrand size = 24, antiderivative size = 74 \[ \int \cot ^4(c+d x) (a+i a \tan (c+d x))^2 \, dx=2 a^2 x+\frac {2 a^2 \cot (c+d x)}{d}-\frac {i a^2 \cot ^2(c+d x)}{d}-\frac {a^2 \cot ^3(c+d x)}{3 d}-\frac {2 i a^2 \log (\sin (c+d x))}{d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3623, 3610, 3612, 3556} \[ \int \cot ^4(c+d x) (a+i a \tan (c+d x))^2 \, dx=-\frac {a^2 \cot ^3(c+d x)}{3 d}-\frac {i a^2 \cot ^2(c+d x)}{d}+\frac {2 a^2 \cot (c+d x)}{d}-\frac {2 i a^2 \log (\sin (c+d x))}{d}+2 a^2 x \]

[In]

Int[Cot[c + d*x]^4*(a + I*a*Tan[c + d*x])^2,x]

[Out]

2*a^2*x + (2*a^2*Cot[c + d*x])/d - (I*a^2*Cot[c + d*x]^2)/d - (a^2*Cot[c + d*x]^3)/(3*d) - ((2*I)*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 ^3(c+d x)}{3 d}+\int \cot ^3(c+d x) \left (2 i a^2-2 a^2 \tan (c+d x)\right ) \, dx \\ & = -\frac {i a^2 \cot ^2(c+d x)}{d}-\frac {a^2 \cot ^3(c+d x)}{3 d}+\int \cot ^2(c+d x) \left (-2 a^2-2 i a^2 \tan (c+d x)\right ) \, dx \\ & = \frac {2 a^2 \cot (c+d x)}{d}-\frac {i a^2 \cot ^2(c+d x)}{d}-\frac {a^2 \cot ^3(c+d x)}{3 d}+\int \cot (c+d x) \left (-2 i a^2+2 a^2 \tan (c+d x)\right ) \, dx \\ & = 2 a^2 x+\frac {2 a^2 \cot (c+d x)}{d}-\frac {i a^2 \cot ^2(c+d x)}{d}-\frac {a^2 \cot ^3(c+d x)}{3 d}-\left (2 i a^2\right ) \int \cot (c+d x) \, dx \\ & = 2 a^2 x+\frac {2 a^2 \cot (c+d x)}{d}-\frac {i a^2 \cot ^2(c+d x)}{d}-\frac {a^2 \cot ^3(c+d x)}{3 d}-\frac {2 i 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 = 0.50 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.42 \[ \int \cot ^4(c+d x) (a+i a \tan (c+d x))^2 \, dx=-\frac {a^2 \cot ^3(c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},-\tan ^2(c+d x)\right )}{3 d}+\frac {a^2 \cot (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},-\tan ^2(c+d x)\right )}{d}-\frac {i a^2 \left (\cot ^2(c+d x)+2 \log (\cos (c+d x))+2 \log (\tan (c+d x))\right )}{d} \]

[In]

Integrate[Cot[c + d*x]^4*(a + I*a*Tan[c + d*x])^2,x]

[Out]

-1/3*(a^2*Cot[c + d*x]^3*Hypergeometric2F1[-3/2, 1, -1/2, -Tan[c + d*x]^2])/d + (a^2*Cot[c + d*x]*Hypergeometr
ic2F1[-1/2, 1, 1/2, -Tan[c + d*x]^2])/d - (I*a^2*(Cot[c + d*x]^2 + 2*Log[Cos[c + d*x]] + 2*Log[Tan[c + d*x]]))
/d

Maple [A] (verified)

Time = 0.51 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.85

method	result	size
parallelrisch	\(-\frac {a^{2} \left (\cot ^{3}\left (d x +c \right )+3 i \left (\cot ^{2}\left (d x +c \right )\right )+6 i \ln \left (\tan \left (d x +c \right )\right )-3 i \ln \left (\sec ^{2}\left (d x +c \right )\right )-6 d x -6 \cot \left (d x +c \right )\right )}{3 d}\)	\(63\)
derivativedivides	\(\frac {-a^{2} \left (-\cot \left (d x +c \right )-d x -c \right )+2 i a^{2} \left (-\frac {\left (\cot ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\sin \left (d x +c \right )\right )\right )+a^{2} \left (-\frac {\left (\cot ^{3}\left (d x +c \right )\right )}{3}+\cot \left (d x +c \right )+d x +c \right )}{d}\)	\(78\)
default	\(\frac {-a^{2} \left (-\cot \left (d x +c \right )-d x -c \right )+2 i a^{2} \left (-\frac {\left (\cot ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\sin \left (d x +c \right )\right )\right )+a^{2} \left (-\frac {\left (\cot ^{3}\left (d x +c \right )\right )}{3}+\cot \left (d x +c \right )+d x +c \right )}{d}\)	\(78\)
risch	\(-\frac {4 a^{2} c}{d}+\frac {2 i a^{2} \left (15 \,{\mathrm e}^{4 i \left (d x +c \right )}-18 \,{\mathrm e}^{2 i \left (d x +c \right )}+7\right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{3}}-\frac {2 i a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\)	\(78\)
norman	\(\frac {-\frac {a^{2}}{3 d}+2 a^{2} x \left (\tan ^{3}\left (d x +c \right )\right )+\frac {2 a^{2} \left (\tan ^{2}\left (d x +c \right )\right )}{d}-\frac {i a^{2} \tan \left (d x +c \right )}{d}}{\tan \left (d x +c \right )^{3}}+\frac {i a^{2} \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{d}-\frac {2 i a^{2} \ln \left (\tan \left (d x +c \right )\right )}{d}\)	\(101\)

[In]

int(cot(d*x+c)^4*(a+I*a*tan(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

-1/3*a^2*(cot(d*x+c)^3+3*I*cot(d*x+c)^2+6*I*ln(tan(d*x+c))-3*I*ln(sec(d*x+c)^2)-6*d*x-6*cot(d*x+c))/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. 139 vs. \(2 (68) = 136\).

Time = 0.24 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.88 \[ \int \cot ^4(c+d x) (a+i a \tan (c+d x))^2 \, dx=-\frac {2 \, {\left (-15 i \, a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + 18 i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - 7 i \, a^{2} + 3 \, {\left (i \, a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} - 3 i \, a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + 3 i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a^{2}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right )\right )}}{3 \, {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} - 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )}} \]

[In]

integrate(cot(d*x+c)^4*(a+I*a*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

-2/3*(-15*I*a^2*e^(4*I*d*x + 4*I*c) + 18*I*a^2*e^(2*I*d*x + 2*I*c) - 7*I*a^2 + 3*(I*a^2*e^(6*I*d*x + 6*I*c) -
3*I*a^2*e^(4*I*d*x + 4*I*c) + 3*I*a^2*e^(2*I*d*x + 2*I*c) - I*a^2)*log(e^(2*I*d*x + 2*I*c) - 1))/(d*e^(6*I*d*x
+ 6*I*c) - 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*I*d*x + 2*I*c) - d)

Sympy [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 136 vs. \(2 (66) = 132\).

Time = 0.21 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.84 \[ \int \cot ^4(c+d x) (a+i a \tan (c+d x))^2 \, dx=- \frac {2 i a^{2} \log {\left (e^{2 i d x} - e^{- 2 i c} \right )}}{d} + \frac {30 i a^{2} e^{4 i c} e^{4 i d x} - 36 i a^{2} e^{2 i c} e^{2 i d x} + 14 i a^{2}}{3 d e^{6 i c} e^{6 i d x} - 9 d e^{4 i c} e^{4 i d x} + 9 d e^{2 i c} e^{2 i d x} - 3 d} \]

[In]

integrate(cot(d*x+c)**4*(a+I*a*tan(d*x+c))**2,x)

[Out]

-2*I*a**2*log(exp(2*I*d*x) - exp(-2*I*c))/d + (30*I*a**2*exp(4*I*c)*exp(4*I*d*x) - 36*I*a**2*exp(2*I*c)*exp(2*
I*d*x) + 14*I*a**2)/(3*d*exp(6*I*c)*exp(6*I*d*x) - 9*d*exp(4*I*c)*exp(4*I*d*x) + 9*d*exp(2*I*c)*exp(2*I*d*x) -
3*d)

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.12 \[ \int \cot ^4(c+d x) (a+i a \tan (c+d x))^2 \, dx=\frac {6 \, {\left (d x + c\right )} a^{2} + 3 i \, a^{2} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 6 i \, a^{2} \log \left (\tan \left (d x + c\right )\right ) + \frac {6 \, a^{2} \tan \left (d x + c\right )^{2} - 3 i \, a^{2} \tan \left (d x + c\right ) - a^{2}}{\tan \left (d x + c\right )^{3}}}{3 \, d} \]

[In]

integrate(cot(d*x+c)^4*(a+I*a*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

1/3*(6*(d*x + c)*a^2 + 3*I*a^2*log(tan(d*x + c)^2 + 1) - 6*I*a^2*log(tan(d*x + c)) + (6*a^2*tan(d*x + c)^2 - 3
*I*a^2*tan(d*x + c) - a^2)/tan(d*x + c)^3)/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. 146 vs. \(2 (68) = 136\).

Time = 0.81 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.97 \[ \int \cot ^4(c+d x) (a+i a \tan (c+d x))^2 \, dx=\frac {a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 6 i \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 96 i \, a^{2} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + i\right ) - 48 i \, a^{2} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) - 27 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \frac {-88 i \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 27 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 6 i \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{2}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}}}{24 \, d} \]

[In]

integrate(cot(d*x+c)^4*(a+I*a*tan(d*x+c))^2,x, algorithm="giac")

[Out]

1/24*(a^2*tan(1/2*d*x + 1/2*c)^3 - 6*I*a^2*tan(1/2*d*x + 1/2*c)^2 + 96*I*a^2*log(tan(1/2*d*x + 1/2*c) + I) - 4
8*I*a^2*log(tan(1/2*d*x + 1/2*c)) - 27*a^2*tan(1/2*d*x + 1/2*c) - (-88*I*a^2*tan(1/2*d*x + 1/2*c)^3 - 27*a^2*t
an(1/2*d*x + 1/2*c)^2 + 6*I*a^2*tan(1/2*d*x + 1/2*c) + a^2)/tan(1/2*d*x + 1/2*c)^3)/d

Mupad [B] (verification not implemented)

Time = 4.26 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.92 \[ \int \cot ^4(c+d x) (a+i a \tan (c+d x))^2 \, dx=\frac {2\,a^2\,\mathrm {cot}\left (c+d\,x\right )}{d}+\frac {4\,a^2\,\mathrm {atan}\left (2\,\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )}{d}-\frac {a^2\,{\mathrm {cot}\left (c+d\,x\right )}^3}{3\,d}-\frac {a^2\,{\mathrm {cot}\left (c+d\,x\right )}^2\,1{}\mathrm {i}}{d} \]

[In]

int(cot(c + d*x)^4*(a + a*tan(c + d*x)*1i)^2,x)

[Out]

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