\(\int \cot ^5(c+d x) (a+i a \tan (c+d x))^4 \, dx\) [42]

Optimal result

Integrand size = 24, antiderivative size = 134 \[ \int \cot ^5(c+d x) (a+i a \tan (c+d x))^4 \, dx=8 i a^4 x+\frac {4 i a^4 \cot (c+d x)}{d}+\frac {8 a^4 \log (\sin (c+d x))}{d}-\frac {i a \cot ^3(c+d x) (a+i a \tan (c+d x))^3}{3 d}-\frac {\cot ^4(c+d x) (a+i a \tan (c+d x))^4}{4 d}+\frac {\cot ^2(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {3629, 3626, 3623, 3612, 3556} \[ \int \cot ^5(c+d x) (a+i a \tan (c+d x))^4 \, dx=\frac {4 i a^4 \cot (c+d x)}{d}+\frac {8 a^4 \log (\sin (c+d x))}{d}+8 i a^4 x+\frac {\cot ^2(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{d}-\frac {\cot ^4(c+d x) (a+i a \tan (c+d x))^4}{4 d}-\frac {i a \cot ^3(c+d x) (a+i a \tan (c+d x))^3}{3 d} \]

[In]

Int[Cot[c + d*x]^5*(a + I*a*Tan[c + d*x])^4,x]

[Out]

(8*I)*a^4*x + ((4*I)*a^4*Cot[c + d*x])/d + (8*a^4*Log[Sin[c + d*x]])/d - ((I/3)*a*Cot[c + d*x]^3*(a + I*a*Tan[
c + d*x])^3)/d - (Cot[c + d*x]^4*(a + I*a*Tan[c + d*x])^4)/(4*d) + (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 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]

Rule 3626

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

Rule 3629

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

Rubi steps \begin{align*} \text {integral}& = -\frac {\cot ^4(c+d x) (a+i a \tan (c+d x))^4}{4 d}+i \int \cot ^4(c+d x) (a+i a \tan (c+d x))^4 \, dx \\ & = -\frac {i a \cot ^3(c+d x) (a+i a \tan (c+d x))^3}{3 d}-\frac {\cot ^4(c+d x) (a+i a \tan (c+d x))^4}{4 d}-(2 a) \int \cot ^3(c+d x) (a+i a \tan (c+d x))^3 \, dx \\ & = -\frac {i a \cot ^3(c+d x) (a+i a \tan (c+d x))^3}{3 d}-\frac {\cot ^4(c+d x) (a+i a \tan (c+d x))^4}{4 d}+\frac {\cot ^2(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{d}-\left (4 i a^2\right ) \int \cot ^2(c+d x) (a+i a \tan (c+d x))^2 \, dx \\ & = \frac {4 i a^4 \cot (c+d x)}{d}-\frac {i a \cot ^3(c+d x) (a+i a \tan (c+d x))^3}{3 d}-\frac {\cot ^4(c+d x) (a+i a \tan (c+d x))^4}{4 d}+\frac {\cot ^2(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{d}-\left (4 i a^2\right ) \int \cot (c+d x) \left (2 i a^2-2 a^2 \tan (c+d x)\right ) \, dx \\ & = 8 i a^4 x+\frac {4 i a^4 \cot (c+d x)}{d}-\frac {i a \cot ^3(c+d x) (a+i a \tan (c+d x))^3}{3 d}-\frac {\cot ^4(c+d x) (a+i a \tan (c+d x))^4}{4 d}+\frac {\cot ^2(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{d}+\left (8 a^4\right ) \int \cot (c+d x) \, dx \\ & = 8 i a^4 x+\frac {4 i a^4 \cot (c+d x)}{d}+\frac {8 a^4 \log (\sin (c+d x))}{d}-\frac {i a \cot ^3(c+d x) (a+i a \tan (c+d x))^3}{3 d}-\frac {\cot ^4(c+d x) (a+i a \tan (c+d x))^4}{4 d}+\frac {\cot ^2(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.40 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.69 \[ \int \cot ^5(c+d x) (a+i a \tan (c+d x))^4 \, dx=a^4 \left (\frac {8 i \cot (c+d x)}{d}+\frac {7 \cot ^2(c+d x)}{2 d}-\frac {4 i \cot ^3(c+d x)}{3 d}-\frac {\cot ^4(c+d x)}{4 d}+\frac {8 \log (\tan (c+d x))}{d}-\frac {8 \log (i+\tan (c+d x))}{d}\right ) \]

[In]

Integrate[Cot[c + d*x]^5*(a + I*a*Tan[c + d*x])^4,x]

[Out]

a^4*(((8*I)*Cot[c + d*x])/d + (7*Cot[c + d*x]^2)/(2*d) - (((4*I)/3)*Cot[c + d*x]^3)/d - Cot[c + d*x]^4/(4*d) +
(8*Log[Tan[c + d*x]])/d - (8*Log[I + Tan[c + d*x]])/d)

Maple [A] (verified)

Time = 0.44 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.56

method	result	size
parallelrisch	\(\frac {a^{4} \left (-3 \left (\cot ^{4}\left (d x +c \right )\right )-16 i \left (\cot ^{3}\left (d x +c \right )\right )+96 i d x +42 \left (\cot ^{2}\left (d x +c \right )\right )+96 i \cot \left (d x +c \right )+96 \ln \left (\tan \left (d x +c \right )\right )-48 \ln \left (\sec ^{2}\left (d x +c \right )\right )\right )}{12 d}\)	\(75\)
derivativedivides	\(\frac {a^{4} \left (8 i \cot \left (d x +c \right )-\frac {\left (\cot ^{4}\left (d x +c \right )\right )}{4}-\frac {4 i \left (\cot ^{3}\left (d x +c \right )\right )}{3}+\frac {7 \left (\cot ^{2}\left (d x +c \right )\right )}{2}-4 \ln \left (\cot ^{2}\left (d x +c \right )+1\right )-8 i \left (\frac {\pi }{2}-\operatorname {arccot}\left (\cot \left (d x +c \right )\right )\right )\right )}{d}\)	\(78\)
default	\(\frac {a^{4} \left (8 i \cot \left (d x +c \right )-\frac {\left (\cot ^{4}\left (d x +c \right )\right )}{4}-\frac {4 i \left (\cot ^{3}\left (d x +c \right )\right )}{3}+\frac {7 \left (\cot ^{2}\left (d x +c \right )\right )}{2}-4 \ln \left (\cot ^{2}\left (d x +c \right )+1\right )-8 i \left (\frac {\pi }{2}-\operatorname {arccot}\left (\cot \left (d x +c \right )\right )\right )\right )}{d}\)	\(78\)
risch	\(-\frac {16 i a^{4} c}{d}-\frac {4 a^{4} \left (30 \,{\mathrm e}^{6 i \left (d x +c \right )}-63 \,{\mathrm e}^{4 i \left (d x +c \right )}+50 \,{\mathrm e}^{2 i \left (d x +c \right )}-14\right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{4}}+\frac {8 a^{4} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\)	\(88\)
norman	\(\frac {-\frac {a^{4}}{4 d}+\frac {7 a^{4} \left (\tan ^{2}\left (d x +c \right )\right )}{2 d}-\frac {4 i a^{4} \tan \left (d x +c \right )}{3 d}+\frac {8 i a^{4} \left (\tan ^{3}\left (d x +c \right )\right )}{d}+8 i a^{4} x \left (\tan ^{4}\left (d x +c \right )\right )}{\tan \left (d x +c \right )^{4}}+\frac {8 a^{4} \ln \left (\tan \left (d x +c \right )\right )}{d}-\frac {4 a^{4} \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{d}\)	\(117\)

[In]

int(cot(d*x+c)^5*(a+I*a*tan(d*x+c))^4,x,method=_RETURNVERBOSE)

[Out]

1/12*a^4*(-3*cot(d*x+c)^4-16*I*cot(d*x+c)^3+96*I*d*x+42*cot(d*x+c)^2+96*I*cot(d*x+c)+96*ln(tan(d*x+c))-48*ln(s
ec(d*x+c)^2))/d

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.30 \[ \int \cot ^5(c+d x) (a+i a \tan (c+d x))^4 \, dx=-\frac {4 \, {\left (30 \, a^{4} e^{\left (6 i \, d x + 6 i \, c\right )} - 63 \, a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} + 50 \, a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} - 14 \, a^{4} - 6 \, {\left (a^{4} e^{\left (8 i \, d x + 8 i \, c\right )} - 4 \, a^{4} e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} - 4 \, a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{4}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right )\right )}}{3 \, {\left (d e^{\left (8 i \, d x + 8 i \, c\right )} - 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} - 4 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

[In]

integrate(cot(d*x+c)^5*(a+I*a*tan(d*x+c))^4,x, algorithm="fricas")

[Out]

-4/3*(30*a^4*e^(6*I*d*x + 6*I*c) - 63*a^4*e^(4*I*d*x + 4*I*c) + 50*a^4*e^(2*I*d*x + 2*I*c) - 14*a^4 - 6*(a^4*e
^(8*I*d*x + 8*I*c) - 4*a^4*e^(6*I*d*x + 6*I*c) + 6*a^4*e^(4*I*d*x + 4*I*c) - 4*a^4*e^(2*I*d*x + 2*I*c) + a^4)*
log(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 = 1.27 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.25 \[ \int \cot ^5(c+d x) (a+i a \tan (c+d x))^4 \, dx=\frac {8 a^{4} \log {\left (e^{2 i d x} - e^{- 2 i c} \right )}}{d} + \frac {- 120 a^{4} e^{6 i c} e^{6 i d x} + 252 a^{4} e^{4 i c} e^{4 i d x} - 200 a^{4} e^{2 i c} e^{2 i d x} + 56 a^{4}}{3 d e^{8 i c} e^{8 i d x} - 12 d e^{6 i c} e^{6 i d x} + 18 d e^{4 i c} e^{4 i d x} - 12 d e^{2 i c} e^{2 i d x} + 3 d} \]

[In]

integrate(cot(d*x+c)**5*(a+I*a*tan(d*x+c))**4,x)

[Out]

8*a**4*log(exp(2*I*d*x) - exp(-2*I*c))/d + (-120*a**4*exp(6*I*c)*exp(6*I*d*x) + 252*a**4*exp(4*I*c)*exp(4*I*d*
x) - 200*a**4*exp(2*I*c)*exp(2*I*d*x) + 56*a**4)/(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 = 0.31 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.72 \[ \int \cot ^5(c+d x) (a+i a \tan (c+d x))^4 \, dx=-\frac {-96 i \, {\left (d x + c\right )} a^{4} + 48 \, a^{4} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 96 \, a^{4} \log \left (\tan \left (d x + c\right )\right ) + \frac {-96 i \, a^{4} \tan \left (d x + c\right )^{3} - 42 \, a^{4} \tan \left (d x + c\right )^{2} + 16 i \, a^{4} \tan \left (d x + c\right ) + 3 \, a^{4}}{\tan \left (d x + c\right )^{4}}}{12 \, d} \]

[In]

integrate(cot(d*x+c)^5*(a+I*a*tan(d*x+c))^4,x, algorithm="maxima")

[Out]

-1/12*(-96*I*(d*x + c)*a^4 + 48*a^4*log(tan(d*x + c)^2 + 1) - 96*a^4*log(tan(d*x + c)) + (-96*I*a^4*tan(d*x +
c)^3 - 42*a^4*tan(d*x + c)^2 + 16*I*a^4*tan(d*x + c) + 3*a^4)/tan(d*x + c)^4)/d

Giac [A] (verification not implemented)

none

Time = 1.34 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.34 \[ \int \cot ^5(c+d x) (a+i a \tan (c+d x))^4 \, dx=-\frac {3 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 32 i \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 180 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 3072 \, a^{4} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + i\right ) - 1536 \, a^{4} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 864 i \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {3200 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 864 i \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 180 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 32 i \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, a^{4}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4}}}{192 \, d} \]

[In]

integrate(cot(d*x+c)^5*(a+I*a*tan(d*x+c))^4,x, algorithm="giac")

[Out]

-1/192*(3*a^4*tan(1/2*d*x + 1/2*c)^4 - 32*I*a^4*tan(1/2*d*x + 1/2*c)^3 - 180*a^4*tan(1/2*d*x + 1/2*c)^2 + 3072
*a^4*log(tan(1/2*d*x + 1/2*c) + I) - 1536*a^4*log(tan(1/2*d*x + 1/2*c)) + 864*I*a^4*tan(1/2*d*x + 1/2*c) + (32
00*a^4*tan(1/2*d*x + 1/2*c)^4 - 864*I*a^4*tan(1/2*d*x + 1/2*c)^3 - 180*a^4*tan(1/2*d*x + 1/2*c)^2 + 32*I*a^4*t
an(1/2*d*x + 1/2*c) + 3*a^4)/tan(1/2*d*x + 1/2*c)^4)/d

Mupad [B] (verification not implemented)

Time = 4.57 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.60 \[ \int \cot ^5(c+d x) (a+i a \tan (c+d x))^4 \, dx=\frac {a^4\,\mathrm {atan}\left (2\,\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,16{}\mathrm {i}}{d}-\frac {-a^4\,{\mathrm {tan}\left (c+d\,x\right )}^3\,8{}\mathrm {i}-\frac {7\,a^4\,{\mathrm {tan}\left (c+d\,x\right )}^2}{2}+\frac {a^4\,\mathrm {tan}\left (c+d\,x\right )\,4{}\mathrm {i}}{3}+\frac {a^4}{4}}{d\,{\mathrm {tan}\left (c+d\,x\right )}^4} \]

[In]

int(cot(c + d*x)^5*(a + a*tan(c + d*x)*1i)^4,x)

[Out]

(a^4*atan(2*tan(c + d*x) + 1i)*16i)/d - ((a^4*tan(c + d*x)*4i)/3 + a^4/4 - (7*a^4*tan(c + d*x)^2)/2 - a^4*tan(
c + d*x)^3*8i)/(d*tan(c + d*x)^4)