\(\int \frac {\cos ^4(c+d x)}{(a+i a \tan (c+d x))^2} \, dx\) [121]

Optimal result

Integrand size = 24, antiderivative size = 165 \[ \int \frac {\cos ^4(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\frac {15 x}{64 a^2}-\frac {i}{64 d (a-i a \tan (c+d x))^2}+\frac {i a^2}{32 d (a+i a \tan (c+d x))^4}+\frac {i a}{16 d (a+i a \tan (c+d x))^3}+\frac {3 i}{32 d (a+i a \tan (c+d x))^2}-\frac {5 i}{64 d \left (a^2-i a^2 \tan (c+d x)\right )}+\frac {5 i}{32 d \left (a^2+i a^2 \tan (c+d x)\right )} \]

[Out]

15/64*x/a^2-1/64*I/d/(a-I*a*tan(d*x+c))^2+1/32*I*a^2/d/(a+I*a*tan(d*x+c))^4+1/16*I*a/d/(a+I*a*tan(d*x+c))^3+3/
32*I/d/(a+I*a*tan(d*x+c))^2-5/64*I/d/(a^2-I*a^2*tan(d*x+c))+5/32*I/d/(a^2+I*a^2*tan(d*x+c))

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {3568, 46, 212} \[ \int \frac {\cos ^4(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\frac {i a^2}{32 d (a+i a \tan (c+d x))^4}-\frac {5 i}{64 d \left (a^2-i a^2 \tan (c+d x)\right )}+\frac {5 i}{32 d \left (a^2+i a^2 \tan (c+d x)\right )}+\frac {15 x}{64 a^2}+\frac {i a}{16 d (a+i a \tan (c+d x))^3}-\frac {i}{64 d (a-i a \tan (c+d x))^2}+\frac {3 i}{32 d (a+i a \tan (c+d x))^2} \]

[In]

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

[Out]

(15*x)/(64*a^2) - (I/64)/(d*(a - I*a*Tan[c + d*x])^2) + ((I/32)*a^2)/(d*(a + I*a*Tan[c + d*x])^4) + ((I/16)*a)
/(d*(a + I*a*Tan[c + d*x])^3) + ((3*I)/32)/(d*(a + I*a*Tan[c + d*x])^2) - ((5*I)/64)/(d*(a^2 - I*a^2*Tan[c + d
*x])) + ((5*I)/32)/(d*(a^2 + I*a^2*Tan[c + d*x]))

Rule 46

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x
)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[m + n + 2, 0])

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 3568

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[1/(a^(m - 2)*b
*f), Subst[Int[(a - x)^(m/2 - 1)*(a + x)^(n + m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x
] && EqQ[a^2 + b^2, 0] && IntegerQ[m/2]

Rubi steps \begin{align*} \text {integral}& = -\frac {\left (i a^5\right ) \text {Subst}\left (\int \frac {1}{(a-x)^3 (a+x)^5} \, dx,x,i a \tan (c+d x)\right )}{d} \\ & = -\frac {\left (i a^5\right ) \text {Subst}\left (\int \left (\frac {1}{32 a^5 (a-x)^3}+\frac {5}{64 a^6 (a-x)^2}+\frac {1}{8 a^3 (a+x)^5}+\frac {3}{16 a^4 (a+x)^4}+\frac {3}{16 a^5 (a+x)^3}+\frac {5}{32 a^6 (a+x)^2}+\frac {15}{64 a^6 \left (a^2-x^2\right )}\right ) \, dx,x,i a \tan (c+d x)\right )}{d} \\ & = -\frac {i}{64 d (a-i a \tan (c+d x))^2}+\frac {i a^2}{32 d (a+i a \tan (c+d x))^4}+\frac {i a}{16 d (a+i a \tan (c+d x))^3}+\frac {3 i}{32 d (a+i a \tan (c+d x))^2}-\frac {5 i}{64 d \left (a^2-i a^2 \tan (c+d x)\right )}+\frac {5 i}{32 d \left (a^2+i a^2 \tan (c+d x)\right )}-\frac {(15 i) \text {Subst}\left (\int \frac {1}{a^2-x^2} \, dx,x,i a \tan (c+d x)\right )}{64 a d} \\ & = \frac {15 x}{64 a^2}-\frac {i}{64 d (a-i a \tan (c+d x))^2}+\frac {i a^2}{32 d (a+i a \tan (c+d x))^4}+\frac {i a}{16 d (a+i a \tan (c+d x))^3}+\frac {3 i}{32 d (a+i a \tan (c+d x))^2}-\frac {5 i}{64 d \left (a^2-i a^2 \tan (c+d x)\right )}+\frac {5 i}{32 d \left (a^2+i a^2 \tan (c+d x)\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.31 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.86 \[ \int \frac {\cos ^4(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\frac {i \sec ^6(c+d x) (-80-65 \cos (2 (c+d x))+16 \cos (4 (c+d x))+\cos (6 (c+d x))+120 i \arctan (\tan (c+d x)) (\cos (2 (c+d x))+i \sin (2 (c+d x)))-5 i \sin (2 (c+d x))+32 i \sin (4 (c+d x))+3 i \sin (6 (c+d x)))}{512 a^2 d (-i+\tan (c+d x))^4 (i+\tan (c+d x))^2} \]

[In]

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

[Out]

((I/512)*Sec[c + d*x]^6*(-80 - 65*Cos[2*(c + d*x)] + 16*Cos[4*(c + d*x)] + Cos[6*(c + d*x)] + (120*I)*ArcTan[T
an[c + d*x]]*(Cos[2*(c + d*x)] + I*Sin[2*(c + d*x)]) - (5*I)*Sin[2*(c + d*x)] + (32*I)*Sin[4*(c + d*x)] + (3*I
)*Sin[6*(c + d*x)]))/(a^2*d*(-I + Tan[c + d*x])^4*(I + Tan[c + d*x])^2)

Maple [A] (verified)

Time = 0.75 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.69

method	result	size
risch	\(\frac {15 x}{64 a^{2}}+\frac {i {\mathrm e}^{-6 i \left (d x +c \right )}}{64 a^{2} d}+\frac {i {\mathrm e}^{-8 i \left (d x +c \right )}}{512 a^{2} d}+\frac {7 i \cos \left (4 d x +4 c \right )}{128 a^{2} d}+\frac {\sin \left (4 d x +4 c \right )}{16 a^{2} d}+\frac {7 i \cos \left (2 d x +2 c \right )}{64 a^{2} d}+\frac {13 \sin \left (2 d x +2 c \right )}{64 a^{2} d}\)	\(114\)
derivativedivides	\(\frac {-\frac {15 i \ln \left (\tan \left (d x +c \right )-i\right )}{128}+\frac {i}{32 \left (\tan \left (d x +c \right )-i\right )^{4}}-\frac {3 i}{32 \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {1}{16 \left (\tan \left (d x +c \right )-i\right )^{3}}+\frac {5}{32 \left (\tan \left (d x +c \right )-i\right )}+\frac {i}{64 \left (\tan \left (d x +c \right )+i\right )^{2}}+\frac {15 i \ln \left (\tan \left (d x +c \right )+i\right )}{128}+\frac {5}{64 \left (\tan \left (d x +c \right )+i\right )}}{d \,a^{2}}\)	\(116\)
default	\(\frac {-\frac {15 i \ln \left (\tan \left (d x +c \right )-i\right )}{128}+\frac {i}{32 \left (\tan \left (d x +c \right )-i\right )^{4}}-\frac {3 i}{32 \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {1}{16 \left (\tan \left (d x +c \right )-i\right )^{3}}+\frac {5}{32 \left (\tan \left (d x +c \right )-i\right )}+\frac {i}{64 \left (\tan \left (d x +c \right )+i\right )^{2}}+\frac {15 i \ln \left (\tan \left (d x +c \right )+i\right )}{128}+\frac {5}{64 \left (\tan \left (d x +c \right )+i\right )}}{d \,a^{2}}\)	\(116\)

[In]

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

[Out]

15/64*x/a^2+1/64*I/a^2/d*exp(-6*I*(d*x+c))+1/512*I/a^2/d*exp(-8*I*(d*x+c))+7/128*I/a^2/d*cos(4*d*x+4*c)+1/16/a
^2/d*sin(4*d*x+4*c)+7/64*I/a^2/d*cos(2*d*x+2*c)+13/64/a^2/d*sin(2*d*x+2*c)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.53 \[ \int \frac {\cos ^4(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\frac {{\left (120 \, d x e^{\left (8 i \, d x + 8 i \, c\right )} - 2 i \, e^{\left (12 i \, d x + 12 i \, c\right )} - 24 i \, e^{\left (10 i \, d x + 10 i \, c\right )} + 80 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 30 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 8 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i\right )} e^{\left (-8 i \, d x - 8 i \, c\right )}}{512 \, a^{2} d} \]

[In]

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

[Out]

1/512*(120*d*x*e^(8*I*d*x + 8*I*c) - 2*I*e^(12*I*d*x + 12*I*c) - 24*I*e^(10*I*d*x + 10*I*c) + 80*I*e^(6*I*d*x
+ 6*I*c) + 30*I*e^(4*I*d*x + 4*I*c) + 8*I*e^(2*I*d*x + 2*I*c) + I)*e^(-8*I*d*x - 8*I*c)/(a^2*d)

Sympy [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.56 \[ \int \frac {\cos ^4(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\begin {cases} \frac {\left (- 17179869184 i a^{10} d^{5} e^{24 i c} e^{4 i d x} - 206158430208 i a^{10} d^{5} e^{22 i c} e^{2 i d x} + 687194767360 i a^{10} d^{5} e^{18 i c} e^{- 2 i d x} + 257698037760 i a^{10} d^{5} e^{16 i c} e^{- 4 i d x} + 68719476736 i a^{10} d^{5} e^{14 i c} e^{- 6 i d x} + 8589934592 i a^{10} d^{5} e^{12 i c} e^{- 8 i d x}\right ) e^{- 20 i c}}{4398046511104 a^{12} d^{6}} & \text {for}\: a^{12} d^{6} e^{20 i c} \neq 0 \\x \left (\frac {\left (e^{12 i c} + 6 e^{10 i c} + 15 e^{8 i c} + 20 e^{6 i c} + 15 e^{4 i c} + 6 e^{2 i c} + 1\right ) e^{- 8 i c}}{64 a^{2}} - \frac {15}{64 a^{2}}\right ) & \text {otherwise} \end {cases} + \frac {15 x}{64 a^{2}} \]

[In]

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

[Out]

Piecewise(((-17179869184*I*a**10*d**5*exp(24*I*c)*exp(4*I*d*x) - 206158430208*I*a**10*d**5*exp(22*I*c)*exp(2*I
*d*x) + 687194767360*I*a**10*d**5*exp(18*I*c)*exp(-2*I*d*x) + 257698037760*I*a**10*d**5*exp(16*I*c)*exp(-4*I*d
*x) + 68719476736*I*a**10*d**5*exp(14*I*c)*exp(-6*I*d*x) + 8589934592*I*a**10*d**5*exp(12*I*c)*exp(-8*I*d*x))*
exp(-20*I*c)/(4398046511104*a**12*d**6), Ne(a**12*d**6*exp(20*I*c), 0)), (x*((exp(12*I*c) + 6*exp(10*I*c) + 15
*exp(8*I*c) + 20*exp(6*I*c) + 15*exp(4*I*c) + 6*exp(2*I*c) + 1)*exp(-8*I*c)/(64*a**2) - 15/(64*a**2)), True))
+ 15*x/(64*a**2)

Maxima [F(-2)]

Exception generated. \[ \int \frac {\cos ^4(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\text {Exception raised: RuntimeError} \]

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: expt: undefined: 0 to a negative exponent.

Giac [A] (verification not implemented)

none

Time = 0.51 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.75 \[ \int \frac {\cos ^4(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=-\frac {-\frac {60 i \, \log \left (\tan \left (d x + c\right ) + i\right )}{a^{2}} + \frac {60 i \, \log \left (\tan \left (d x + c\right ) - i\right )}{a^{2}} + \frac {2 \, {\left (45 i \, \tan \left (d x + c\right )^{2} - 110 \, \tan \left (d x + c\right ) - 69 i\right )}}{a^{2} {\left (\tan \left (d x + c\right ) + i\right )}^{2}} + \frac {-125 i \, \tan \left (d x + c\right )^{4} - 580 \, \tan \left (d x + c\right )^{3} + 1038 i \, \tan \left (d x + c\right )^{2} + 868 \, \tan \left (d x + c\right ) - 301 i}{a^{2} {\left (\tan \left (d x + c\right ) - i\right )}^{4}}}{512 \, d} \]

[In]

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

[Out]

-1/512*(-60*I*log(tan(d*x + c) + I)/a^2 + 60*I*log(tan(d*x + c) - I)/a^2 + 2*(45*I*tan(d*x + c)^2 - 110*tan(d*
x + c) - 69*I)/(a^2*(tan(d*x + c) + I)^2) + (-125*I*tan(d*x + c)^4 - 580*tan(d*x + c)^3 + 1038*I*tan(d*x + c)^
2 + 868*tan(d*x + c) - 301*I)/(a^2*(tan(d*x + c) - I)^4))/d

Mupad [B] (verification not implemented)

Time = 5.25 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.90 \[ \int \frac {\cos ^4(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\frac {15\,x}{64\,a^2}+\frac {\frac {1}{4\,a^2}-\frac {\mathrm {tan}\left (c+d\,x\right )\,17{}\mathrm {i}}{64\,a^2}+\frac {25\,{\mathrm {tan}\left (c+d\,x\right )}^2}{32\,a^2}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^3\,5{}\mathrm {i}}{32\,a^2}+\frac {15\,{\mathrm {tan}\left (c+d\,x\right )}^4}{32\,a^2}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^5\,15{}\mathrm {i}}{64\,a^2}}{d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^6\,1{}\mathrm {i}+2\,{\mathrm {tan}\left (c+d\,x\right )}^5+{\mathrm {tan}\left (c+d\,x\right )}^4\,1{}\mathrm {i}+4\,{\mathrm {tan}\left (c+d\,x\right )}^3-{\mathrm {tan}\left (c+d\,x\right )}^2\,1{}\mathrm {i}+2\,\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )} \]

[In]

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

[Out]

(15*x)/(64*a^2) + (1/(4*a^2) - (tan(c + d*x)*17i)/(64*a^2) + (25*tan(c + d*x)^2)/(32*a^2) + (tan(c + d*x)^3*5i
)/(32*a^2) + (15*tan(c + d*x)^4)/(32*a^2) + (tan(c + d*x)^5*15i)/(64*a^2))/(d*(2*tan(c + d*x) - tan(c + d*x)^2
*1i + 4*tan(c + d*x)^3 + tan(c + d*x)^4*1i + 2*tan(c + d*x)^5 + tan(c + d*x)^6*1i - 1i))