\(\int \cot (c+d x) (a+i a \tan (c+d x))^4 \, dx\) [38]

Optimal result

Integrand size = 22, antiderivative size = 86 \[ \int \cot (c+d x) (a+i a \tan (c+d x))^4 \, dx=8 i a^4 x+\frac {7 a^4 \log (\cos (c+d x))}{d}+\frac {a^4 \log (\sin (c+d x))}{d}-\frac {\left (a^2+i a^2 \tan (c+d x)\right )^2}{2 d}-\frac {3 \left (a^4+i a^4 \tan (c+d x)\right )}{d} \]

[Out]

8*I*a^4*x+7*a^4*ln(cos(d*x+c))/d+a^4*ln(sin(d*x+c))/d-1/2*(a^2+I*a^2*tan(d*x+c))^2/d-3*(a^4+I*a^4*tan(d*x+c))/
d

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {3637, 3675, 3670, 3556, 3612} \[ \int \cot (c+d x) (a+i a \tan (c+d x))^4 \, dx=-\frac {3 \left (a^4+i a^4 \tan (c+d x)\right )}{d}+\frac {a^4 \log (\sin (c+d x))}{d}+\frac {7 a^4 \log (\cos (c+d x))}{d}+8 i a^4 x-\frac {\left (a^2+i a^2 \tan (c+d x)\right )^2}{2 d} \]

[In]

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

[Out]

(8*I)*a^4*x + (7*a^4*Log[Cos[c + d*x]])/d + (a^4*Log[Sin[c + d*x]])/d - (a^2 + I*a^2*Tan[c + d*x])^2/(2*d) - (
3*(a^4 + I*a^4*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 3637

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[b^2*(a + b*Tan[e + f*x])^(m - 2)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(m + n - 1))), x] + Dist[a/(d*(m + n - 1
)), Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^n*Simp[b*c*(m - 2) + a*d*(m + 2*n) + (a*c*(m - 2) +
b*d*(3*m + 2*n - 4))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a
^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && IntegerQ[2*m] && GtQ[m, 1] && NeQ[m + n - 1, 0] && (IntegerQ[m] || Intege
rsQ[2*m, 2*n])

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 3675

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

Rubi steps \begin{align*} \text {integral}& = -\frac {\left (a^2+i a^2 \tan (c+d x)\right )^2}{2 d}+\frac {1}{2} a \int \cot (c+d x) (a+i a \tan (c+d x))^2 (2 a+6 i a \tan (c+d x)) \, dx \\ & = -\frac {\left (a^2+i a^2 \tan (c+d x)\right )^2}{2 d}-\frac {3 \left (a^4+i a^4 \tan (c+d x)\right )}{d}+\frac {1}{2} a \int \cot (c+d x) (a+i a \tan (c+d x)) \left (2 a^2+14 i a^2 \tan (c+d x)\right ) \, dx \\ & = -\frac {\left (a^2+i a^2 \tan (c+d x)\right )^2}{2 d}-\frac {3 \left (a^4+i a^4 \tan (c+d x)\right )}{d}+\frac {1}{2} a \int \cot (c+d x) \left (2 a^3+16 i a^3 \tan (c+d x)\right ) \, dx-\left (7 a^4\right ) \int \tan (c+d x) \, dx \\ & = 8 i a^4 x+\frac {7 a^4 \log (\cos (c+d x))}{d}-\frac {\left (a^2+i a^2 \tan (c+d x)\right )^2}{2 d}-\frac {3 \left (a^4+i a^4 \tan (c+d x)\right )}{d}+a^4 \int \cot (c+d x) \, dx \\ & = 8 i a^4 x+\frac {7 a^4 \log (\cos (c+d x))}{d}+\frac {a^4 \log (\sin (c+d x))}{d}-\frac {\left (a^2+i a^2 \tan (c+d x)\right )^2}{2 d}-\frac {3 \left (a^4+i a^4 \tan (c+d x)\right )}{d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.36 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.79 \[ \int \cot (c+d x) (a+i a \tan (c+d x))^4 \, dx=\frac {a^4 \log (\tan (c+d x))}{d}-\frac {8 a^4 \log (i+\tan (c+d x))}{d}-\frac {4 i a^4 \tan (c+d x)}{d}+\frac {a^4 \tan ^2(c+d x)}{2 d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.45 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.60

method	result	size
parallelrisch	\(\frac {a^{4} \left (16 i d x -8 i \tan \left (d x +c \right )+\tan ^{2}\left (d x +c \right )+2 \ln \left (\tan \left (d x +c \right )\right )-8 \ln \left (\sec ^{2}\left (d x +c \right )\right )\right )}{2 d}\)	\(52\)
derivativedivides	\(\frac {a^{4} \left (-4 \ln \left (\cot ^{2}\left (d x +c \right )+1\right )-8 i \left (\frac {\pi }{2}-\operatorname {arccot}\left (\cot \left (d x +c \right )\right )\right )-\frac {4 i}{\cot \left (d x +c \right )}+\frac {1}{2 \cot \left (d x +c \right )^{2}}+7 \ln \left (\cot \left (d x +c \right )\right )\right )}{d}\)	\(68\)
default	\(\frac {a^{4} \left (-4 \ln \left (\cot ^{2}\left (d x +c \right )+1\right )-8 i \left (\frac {\pi }{2}-\operatorname {arccot}\left (\cot \left (d x +c \right )\right )\right )-\frac {4 i}{\cot \left (d x +c \right )}+\frac {1}{2 \cot \left (d x +c \right )^{2}}+7 \ln \left (\cot \left (d x +c \right )\right )\right )}{d}\)	\(68\)
norman	\(\frac {a^{4} \left (\tan ^{2}\left (d x +c \right )\right )}{2 d}+8 i a^{4} x -\frac {4 i a^{4} \tan \left (d x +c \right )}{d}+\frac {a^{4} \ln \left (\tan \left (d x +c \right )\right )}{d}-\frac {4 a^{4} \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{d}\)	\(73\)
risch	\(-\frac {16 i a^{4} c}{d}+\frac {2 a^{4} \left (5 \,{\mathrm e}^{2 i \left (d x +c \right )}+4\right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}+\frac {7 a^{4} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}+\frac {a^{4} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\)	\(85\)

[In]

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

[Out]

1/2*a^4*(16*I*d*x-8*I*tan(d*x+c)+tan(d*x+c)^2+2*ln(tan(d*x+c))-8*ln(sec(d*x+c)^2))/d

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.59 \[ \int \cot (c+d x) (a+i a \tan (c+d x))^4 \, dx=\frac {10 \, a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} + 8 \, a^{4} + 7 \, {\left (a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{4}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + {\left (a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{4}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right )}{d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d} \]

[In]

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

[Out]

(10*a^4*e^(2*I*d*x + 2*I*c) + 8*a^4 + 7*(a^4*e^(4*I*d*x + 4*I*c) + 2*a^4*e^(2*I*d*x + 2*I*c) + a^4)*log(e^(2*I
*d*x + 2*I*c) + 1) + (a^4*e^(4*I*d*x + 4*I*c) + 2*a^4*e^(2*I*d*x + 2*I*c) + a^4)*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 = 0.26 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.22 \[ \int \cot (c+d x) (a+i a \tan (c+d x))^4 \, dx=\frac {a^{4} \left (\log {\left (e^{2 i d x} - e^{- 2 i c} \right )} + 7 \log {\left (e^{2 i d x} + e^{- 2 i c} \right )}\right )}{d} + \frac {10 a^{4} e^{2 i c} e^{2 i d x} + 8 a^{4}}{d e^{4 i c} e^{4 i d x} + 2 d e^{2 i c} e^{2 i d x} + d} \]

[In]

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

[Out]

a**4*(log(exp(2*I*d*x) - exp(-2*I*c)) + 7*log(exp(2*I*d*x) + exp(-2*I*c)))/d + (10*a**4*exp(2*I*c)*exp(2*I*d*x
) + 8*a**4)/(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 = 0.44 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.78 \[ \int \cot (c+d x) (a+i a \tan (c+d x))^4 \, dx=\frac {a^{4} \tan \left (d x + c\right )^{2} + 16 i \, {\left (d x + c\right )} a^{4} - 8 \, a^{4} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 2 \, a^{4} \log \left (\tan \left (d x + c\right )\right ) - 8 i \, a^{4} \tan \left (d x + c\right )}{2 \, d} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.86 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.83 \[ \int \cot (c+d x) (a+i a \tan (c+d x))^4 \, dx=\frac {14 \, a^{4} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right ) - 32 \, a^{4} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + i\right ) + 14 \, a^{4} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right ) + 2 \, a^{4} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) - \frac {21 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 16 i \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 46 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 16 i \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 21 \, a^{4}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}}}{2 \, d} \]

[In]

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

[Out]

1/2*(14*a^4*log(tan(1/2*d*x + 1/2*c) + 1) - 32*a^4*log(tan(1/2*d*x + 1/2*c) + I) + 14*a^4*log(tan(1/2*d*x + 1/
2*c) - 1) + 2*a^4*log(tan(1/2*d*x + 1/2*c)) - (21*a^4*tan(1/2*d*x + 1/2*c)^4 - 16*I*a^4*tan(1/2*d*x + 1/2*c)^3
- 46*a^4*tan(1/2*d*x + 1/2*c)^2 + 16*I*a^4*tan(1/2*d*x + 1/2*c) + 21*a^4)/(tan(1/2*d*x + 1/2*c)^2 - 1)^2)/d

Mupad [B] (verification not implemented)

Time = 4.73 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.74 \[ \int \cot (c+d x) (a+i a \tan (c+d x))^4 \, dx=\frac {a^4\,{\mathrm {tan}\left (c+d\,x\right )}^2}{2\,d}-\frac {8\,a^4\,\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )}{d}+\frac {a^4\,\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )}{d}-\frac {a^4\,\mathrm {tan}\left (c+d\,x\right )\,4{}\mathrm {i}}{d} \]

[In]

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

[Out]

(a^4*tan(c + d*x)^2)/(2*d) - (a^4*tan(c + d*x)*4i)/d - (8*a^4*log(tan(c + d*x) + 1i))/d + (a^4*log(tan(c + d*x
)))/d