\(\int \cos ^4(c+d x) (a+i a \tan (c+d x))^5 \, dx\) [65]

Optimal result

Integrand size = 24, antiderivative size = 73 \[ \int \cos ^4(c+d x) (a+i a \tan (c+d x))^5 \, dx=a^5 x-\frac {i a^5 \log (\cos (c+d x))}{d}-\frac {2 i a^7}{d (a-i a \tan (c+d x))^2}+\frac {4 i a^6}{d (a-i a \tan (c+d x))} \]

[Out]

a^5*x-I*a^5*ln(cos(d*x+c))/d-2*I*a^7/d/(a-I*a*tan(d*x+c))^2+4*I*a^6/d/(a-I*a*tan(d*x+c))

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {3568, 45} \[ \int \cos ^4(c+d x) (a+i a \tan (c+d x))^5 \, dx=-\frac {2 i a^7}{d (a-i a \tan (c+d x))^2}+\frac {4 i a^6}{d (a-i a \tan (c+d x))}-\frac {i a^5 \log (\cos (c+d x))}{d}+a^5 x \]

[In]

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

[Out]

a^5*x - (I*a^5*Log[Cos[c + d*x]])/d - ((2*I)*a^7)/(d*(a - I*a*Tan[c + d*x])^2) + ((4*I)*a^6)/(d*(a - I*a*Tan[c
+ d*x]))

Rule 45

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 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 {(a+x)^2}{(a-x)^3} \, dx,x,i a \tan (c+d x)\right )}{d} \\ & = -\frac {\left (i a^5\right ) \text {Subst}\left (\int \left (\frac {4 a^2}{(a-x)^3}-\frac {4 a}{(a-x)^2}+\frac {1}{a-x}\right ) \, dx,x,i a \tan (c+d x)\right )}{d} \\ & = a^5 x-\frac {i a^5 \log (\cos (c+d x))}{d}-\frac {2 i a^7}{d (a-i a \tan (c+d x))^2}+\frac {4 i a^6}{d (a-i a \tan (c+d x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.27 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.67 \[ \int \cos ^4(c+d x) (a+i a \tan (c+d x))^5 \, dx=-\frac {i a^5 \left (-\log (i+\tan (c+d x))+\frac {2-4 i \tan (c+d x)}{(i+\tan (c+d x))^2}\right )}{d} \]

[In]

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

[Out]

((-I)*a^5*(-Log[I + Tan[c + d*x]] + (2 - (4*I)*Tan[c + d*x])/(I + Tan[c + d*x])^2))/d

Maple [A] (verified)

Time = 48.94 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.93

method	result	size
risch	\(-\frac {i a^{5} {\mathrm e}^{4 i \left (d x +c \right )}}{2 d}+\frac {i a^{5} {\mathrm e}^{2 i \left (d x +c \right )}}{d}-\frac {2 a^{5} c}{d}-\frac {i a^{5} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}\)	\(68\)
derivativedivides	\(\frac {i a^{5} \left (-\frac {\left (\sin ^{4}\left (d x +c \right )\right )}{4}-\frac {\left (\sin ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )+5 a^{5} \left (-\frac {\left (\sin ^{3}\left (d x +c \right )+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )-\frac {5 i a^{5} \left (\sin ^{4}\left (d x +c \right )\right )}{2}-10 a^{5} \left (-\frac {\left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{4}+\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{8}+\frac {d x}{8}+\frac {c}{8}\right )-\frac {5 i a^{5} \left (\cos ^{4}\left (d x +c \right )\right )}{4}+a^{5} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}\)	\(188\)
default	\(\frac {i a^{5} \left (-\frac {\left (\sin ^{4}\left (d x +c \right )\right )}{4}-\frac {\left (\sin ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )+5 a^{5} \left (-\frac {\left (\sin ^{3}\left (d x +c \right )+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )-\frac {5 i a^{5} \left (\sin ^{4}\left (d x +c \right )\right )}{2}-10 a^{5} \left (-\frac {\left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{4}+\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{8}+\frac {d x}{8}+\frac {c}{8}\right )-\frac {5 i a^{5} \left (\cos ^{4}\left (d x +c \right )\right )}{4}+a^{5} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}\)	\(188\)

[In]

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

[Out]

-1/2*I/d*a^5*exp(4*I*(d*x+c))+I/d*a^5*exp(2*I*(d*x+c))-2/d*a^5*c-I/d*a^5*ln(exp(2*I*(d*x+c))+1)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.70 \[ \int \cos ^4(c+d x) (a+i a \tan (c+d x))^5 \, dx=\frac {-i \, a^{5} e^{\left (4 i \, d x + 4 i \, c\right )} + 2 i \, a^{5} e^{\left (2 i \, d x + 2 i \, c\right )} - 2 i \, a^{5} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )}{2 \, d} \]

[In]

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

[Out]

1/2*(-I*a^5*e^(4*I*d*x + 4*I*c) + 2*I*a^5*e^(2*I*d*x + 2*I*c) - 2*I*a^5*log(e^(2*I*d*x + 2*I*c) + 1))/d

Sympy [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.40 \[ \int \cos ^4(c+d x) (a+i a \tan (c+d x))^5 \, dx=- \frac {i a^{5} \log {\left (e^{2 i d x} + e^{- 2 i c} \right )}}{d} + \begin {cases} \frac {- i a^{5} d e^{4 i c} e^{4 i d x} + 2 i a^{5} d e^{2 i c} e^{2 i d x}}{2 d^{2}} & \text {for}\: d^{2} \neq 0 \\x \left (2 a^{5} e^{4 i c} - 2 a^{5} e^{2 i c}\right ) & \text {otherwise} \end {cases} \]

[In]

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

[Out]

-I*a**5*log(exp(2*I*d*x) + exp(-2*I*c))/d + Piecewise(((-I*a**5*d*exp(4*I*c)*exp(4*I*d*x) + 2*I*a**5*d*exp(2*I
*c)*exp(2*I*d*x))/(2*d**2), Ne(d**2, 0)), (x*(2*a**5*exp(4*I*c) - 2*a**5*exp(2*I*c)), True))

Maxima [A] (verification not implemented)

none

Time = 0.39 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.21 \[ \int \cos ^4(c+d x) (a+i a \tan (c+d x))^5 \, dx=\frac {2 \, {\left (d x + c\right )} a^{5} + i \, a^{5} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - \frac {4 \, {\left (2 \, a^{5} \tan \left (d x + c\right )^{3} - 3 i \, a^{5} \tan \left (d x + c\right )^{2} - i \, a^{5}\right )}}{\tan \left (d x + c\right )^{4} + 2 \, \tan \left (d x + c\right )^{2} + 1}}{2 \, d} \]

[In]

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

[Out]

1/2*(2*(d*x + c)*a^5 + I*a^5*log(tan(d*x + c)^2 + 1) - 4*(2*a^5*tan(d*x + c)^3 - 3*I*a^5*tan(d*x + c)^2 - I*a^
5)/(tan(d*x + c)^4 + 2*tan(d*x + c)^2 + 1))/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. 450 vs. \(2 (63) = 126\).

Time = 0.70 (sec) , antiderivative size = 450, normalized size of antiderivative = 6.16 \[ \int \cos ^4(c+d x) (a+i a \tan (c+d x))^5 \, dx=-\frac {2 i \, a^{5} e^{\left (16 i \, d x + 8 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 16 i \, a^{5} e^{\left (14 i \, d x + 6 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 56 i \, a^{5} e^{\left (12 i \, d x + 4 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 112 i \, a^{5} e^{\left (10 i \, d x + 2 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 112 i \, a^{5} e^{\left (6 i \, d x - 2 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 56 i \, a^{5} e^{\left (4 i \, d x - 4 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 16 i \, a^{5} e^{\left (2 i \, d x - 6 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 140 i \, a^{5} e^{\left (8 i \, d x\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 2 i \, a^{5} e^{\left (-8 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + i \, a^{5} e^{\left (20 i \, d x + 12 i \, c\right )} + 6 i \, a^{5} e^{\left (18 i \, d x + 10 i \, c\right )} + 12 i \, a^{5} e^{\left (16 i \, d x + 8 i \, c\right )} - 42 i \, a^{5} e^{\left (12 i \, d x + 4 i \, c\right )} - 84 i \, a^{5} e^{\left (10 i \, d x + 2 i \, c\right )} - 48 i \, a^{5} e^{\left (6 i \, d x - 2 i \, c\right )} - 15 i \, a^{5} e^{\left (4 i \, d x - 4 i \, c\right )} - 2 i \, a^{5} e^{\left (2 i \, d x - 6 i \, c\right )} - 84 i \, a^{5} e^{\left (8 i \, d x\right )}}{2 \, {\left (d e^{\left (16 i \, d x + 8 i \, c\right )} + 8 \, d e^{\left (14 i \, d x + 6 i \, c\right )} + 28 \, d e^{\left (12 i \, d x + 4 i \, c\right )} + 56 \, d e^{\left (10 i \, d x + 2 i \, c\right )} + 56 \, d e^{\left (6 i \, d x - 2 i \, c\right )} + 28 \, d e^{\left (4 i \, d x - 4 i \, c\right )} + 8 \, d e^{\left (2 i \, d x - 6 i \, c\right )} + 70 \, d e^{\left (8 i \, d x\right )} + d e^{\left (-8 i \, c\right )}\right )}} \]

[In]

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

[Out]

-1/2*(2*I*a^5*e^(16*I*d*x + 8*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 16*I*a^5*e^(14*I*d*x + 6*I*c)*log(e^(2*I*d*x
+ 2*I*c) + 1) + 56*I*a^5*e^(12*I*d*x + 4*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 112*I*a^5*e^(10*I*d*x + 2*I*c)*l
og(e^(2*I*d*x + 2*I*c) + 1) + 112*I*a^5*e^(6*I*d*x - 2*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 56*I*a^5*e^(4*I*d*x
- 4*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 16*I*a^5*e^(2*I*d*x - 6*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 140*I*a^5
*e^(8*I*d*x)*log(e^(2*I*d*x + 2*I*c) + 1) + 2*I*a^5*e^(-8*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + I*a^5*e^(20*I*d*
x + 12*I*c) + 6*I*a^5*e^(18*I*d*x + 10*I*c) + 12*I*a^5*e^(16*I*d*x + 8*I*c) - 42*I*a^5*e^(12*I*d*x + 4*I*c) -
84*I*a^5*e^(10*I*d*x + 2*I*c) - 48*I*a^5*e^(6*I*d*x - 2*I*c) - 15*I*a^5*e^(4*I*d*x - 4*I*c) - 2*I*a^5*e^(2*I*d
*x - 6*I*c) - 84*I*a^5*e^(8*I*d*x))/(d*e^(16*I*d*x + 8*I*c) + 8*d*e^(14*I*d*x + 6*I*c) + 28*d*e^(12*I*d*x + 4*
I*c) + 56*d*e^(10*I*d*x + 2*I*c) + 56*d*e^(6*I*d*x - 2*I*c) + 28*d*e^(4*I*d*x - 4*I*c) + 8*d*e^(2*I*d*x - 6*I*
c) + 70*d*e^(8*I*d*x) + d*e^(-8*I*c))

Mupad [B] (verification not implemented)

Time = 4.01 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.88 \[ \int \cos ^4(c+d x) (a+i a \tan (c+d x))^5 \, dx=\frac {a^5\,\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,1{}\mathrm {i}}{d}-\frac {4\,a^5\,\mathrm {tan}\left (c+d\,x\right )+a^5\,2{}\mathrm {i}}{d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^2+\mathrm {tan}\left (c+d\,x\right )\,2{}\mathrm {i}-1\right )} \]

[In]

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

[Out]

(a^5*log(tan(c + d*x) + 1i)*1i)/d - (4*a^5*tan(c + d*x) + a^5*2i)/(d*(tan(c + d*x)*2i + tan(c + d*x)^2 - 1))