\(\int \cos ^2(c+d x) (a+i a \tan (c+d x))^5 \, dx\) [64]

Optimal result

Integrand size = 24, antiderivative size = 83 \[ \int \cos ^2(c+d x) (a+i a \tan (c+d x))^5 \, dx=-12 a^5 x+\frac {12 i a^5 \log (\cos (c+d x))}{d}+\frac {5 a^5 \tan (c+d x)}{d}+\frac {i a^5 \tan ^2(c+d x)}{2 d}-\frac {8 i a^6}{d (a-i a \tan (c+d x))} \]

[Out]

-12*a^5*x+12*I*a^5*ln(cos(d*x+c))/d+5*a^5*tan(d*x+c)/d+1/2*I*a^5*tan(d*x+c)^2/d-8*I*a^6/d/(a-I*a*tan(d*x+c))

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 83, 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} \[ \int \cos ^2(c+d x) (a+i a \tan (c+d x))^5 \, dx=-\frac {8 i a^6}{d (a-i a \tan (c+d x))}+\frac {i a^5 \tan ^2(c+d x)}{2 d}+\frac {5 a^5 \tan (c+d x)}{d}+\frac {12 i a^5 \log (\cos (c+d x))}{d}-12 a^5 x \]

[In]

Int[Cos[c + d*x]^2*(a + I*a*Tan[c + d*x])^5,x]

[Out]

-12*a^5*x + ((12*I)*a^5*Log[Cos[c + d*x]])/d + (5*a^5*Tan[c + d*x])/d + ((I/2)*a^5*Tan[c + d*x]^2)/d - ((8*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^3\right ) \text {Subst}\left (\int \frac {(a+x)^3}{(a-x)^2} \, dx,x,i a \tan (c+d x)\right )}{d} \\ & = -\frac {\left (i a^3\right ) \text {Subst}\left (\int \left (5 a+\frac {8 a^3}{(a-x)^2}-\frac {12 a^2}{a-x}+x\right ) \, dx,x,i a \tan (c+d x)\right )}{d} \\ & = -12 a^5 x+\frac {12 i a^5 \log (\cos (c+d x))}{d}+\frac {5 a^5 \tan (c+d x)}{d}+\frac {i a^5 \tan ^2(c+d x)}{2 d}-\frac {8 i a^6}{d (a-i a \tan (c+d x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.59 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.75 \[ \int \cos ^2(c+d x) (a+i a \tan (c+d x))^5 \, dx=-\frac {i a^5 \left (24 \log (i+\tan (c+d x))+10 i \tan (c+d x)-\tan ^2(c+d x)+\frac {16 i}{i+\tan (c+d x)}\right )}{2 d} \]

[In]

Integrate[Cos[c + d*x]^2*(a + I*a*Tan[c + d*x])^5,x]

[Out]

((-1/2*I)*a^5*(24*Log[I + Tan[c + d*x]] + (10*I)*Tan[c + d*x] - Tan[c + d*x]^2 + (16*I)/(I + Tan[c + d*x])))/d

Maple [A] (verified)

Time = 14.13 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.02

method	result	size
risch	\(-\frac {4 i a^{5} {\mathrm e}^{2 i \left (d x +c \right )}}{d}+\frac {24 a^{5} c}{d}+\frac {2 i a^{5} \left (6 \,{\mathrm e}^{2 i \left (d x +c \right )}+5\right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}+\frac {12 i a^{5} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}\)	\(85\)
derivativedivides	\(\frac {i a^{5} \left (\frac {\sin ^{6}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}+\frac {\left (\sin ^{4}\left (d x +c \right )\right )}{2}+\sin ^{2}\left (d x +c \right )+2 \ln \left (\cos \left (d x +c \right )\right )\right )+5 a^{5} \left (\frac {\sin ^{5}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\sin ^{3}\left (d x +c \right )+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )-\frac {3 d x}{2}-\frac {3 c}{2}\right )-10 i a^{5} \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )-10 a^{5} \left (-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )-\frac {5 i a^{5} \left (\cos ^{2}\left (d x +c \right )\right )}{2}+a^{5} \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(205\)
default	\(\frac {i a^{5} \left (\frac {\sin ^{6}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}+\frac {\left (\sin ^{4}\left (d x +c \right )\right )}{2}+\sin ^{2}\left (d x +c \right )+2 \ln \left (\cos \left (d x +c \right )\right )\right )+5 a^{5} \left (\frac {\sin ^{5}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\sin ^{3}\left (d x +c \right )+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )-\frac {3 d x}{2}-\frac {3 c}{2}\right )-10 i a^{5} \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )-10 a^{5} \left (-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )-\frac {5 i a^{5} \left (\cos ^{2}\left (d x +c \right )\right )}{2}+a^{5} \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(205\)

[In]

int(cos(d*x+c)^2*(a+I*a*tan(d*x+c))^5,x,method=_RETURNVERBOSE)

[Out]

-4*I/d*a^5*exp(2*I*(d*x+c))+24/d*a^5*c+2*I*a^5*(6*exp(2*I*(d*x+c))+5)/d/(exp(2*I*(d*x+c))+1)^2+12*I/d*a^5*ln(e
xp(2*I*(d*x+c))+1)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.51 \[ \int \cos ^2(c+d x) (a+i a \tan (c+d x))^5 \, dx=-\frac {2 \, {\left (2 i \, a^{5} e^{\left (6 i \, d x + 6 i \, c\right )} + 4 i \, a^{5} e^{\left (4 i \, d x + 4 i \, c\right )} - 4 i \, a^{5} e^{\left (2 i \, d x + 2 i \, c\right )} - 5 i \, a^{5} + 6 \, {\left (-i \, a^{5} e^{\left (4 i \, d x + 4 i \, c\right )} - 2 i \, a^{5} e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a^{5}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )\right )}}{d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d} \]

[In]

integrate(cos(d*x+c)^2*(a+I*a*tan(d*x+c))^5,x, algorithm="fricas")

[Out]

-2*(2*I*a^5*e^(6*I*d*x + 6*I*c) + 4*I*a^5*e^(4*I*d*x + 4*I*c) - 4*I*a^5*e^(2*I*d*x + 2*I*c) - 5*I*a^5 + 6*(-I*
a^5*e^(4*I*d*x + 4*I*c) - 2*I*a^5*e^(2*I*d*x + 2*I*c) - I*a^5)*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 = 0.24 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.58 \[ \int \cos ^2(c+d x) (a+i a \tan (c+d x))^5 \, dx=\frac {12 i a^{5} \log {\left (e^{2 i d x} + e^{- 2 i c} \right )}}{d} + \frac {12 i a^{5} e^{2 i c} e^{2 i d x} + 10 i a^{5}}{d e^{4 i c} e^{4 i d x} + 2 d e^{2 i c} e^{2 i d x} + d} + \begin {cases} - \frac {4 i a^{5} e^{2 i c} e^{2 i d x}}{d} & \text {for}\: d \neq 0 \\8 a^{5} x e^{2 i c} & \text {otherwise} \end {cases} \]

[In]

integrate(cos(d*x+c)**2*(a+I*a*tan(d*x+c))**5,x)

[Out]

12*I*a**5*log(exp(2*I*d*x) + exp(-2*I*c))/d + (12*I*a**5*exp(2*I*c)*exp(2*I*d*x) + 10*I*a**5)/(d*exp(4*I*c)*ex
p(4*I*d*x) + 2*d*exp(2*I*c)*exp(2*I*d*x) + d) + Piecewise((-4*I*a**5*exp(2*I*c)*exp(2*I*d*x)/d, Ne(d, 0)), (8*
a**5*x*exp(2*I*c), True))

Maxima [A] (verification not implemented)

none

Time = 0.45 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.04 \[ \int \cos ^2(c+d x) (a+i a \tan (c+d x))^5 \, dx=-\frac {-i \, a^{5} \tan \left (d x + c\right )^{2} + 24 \, {\left (d x + c\right )} a^{5} + 12 i \, a^{5} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 10 \, a^{5} \tan \left (d x + c\right ) - \frac {16 \, {\left (a^{5} \tan \left (d x + c\right ) - i \, a^{5}\right )}}{\tan \left (d x + c\right )^{2} + 1}}{2 \, d} \]

[In]

integrate(cos(d*x+c)^2*(a+I*a*tan(d*x+c))^5,x, algorithm="maxima")

[Out]

-1/2*(-I*a^5*tan(d*x + c)^2 + 24*(d*x + c)*a^5 + 12*I*a^5*log(tan(d*x + c)^2 + 1) - 10*a^5*tan(d*x + c) - 16*(
a^5*tan(d*x + c) - I*a^5)/(tan(d*x + c)^2 + 1))/d

Giac [A] (verification not implemented)

none

Time = 0.94 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.76 \[ \int \cos ^2(c+d x) (a+i a \tan (c+d x))^5 \, dx=-\frac {2 \, {\left (-6 i \, a^{5} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 12 i \, a^{5} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 2 i \, a^{5} e^{\left (6 i \, d x + 6 i \, c\right )} + 4 i \, a^{5} e^{\left (4 i \, d x + 4 i \, c\right )} - 4 i \, a^{5} e^{\left (2 i \, d x + 2 i \, c\right )} - 6 i \, a^{5} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 5 i \, a^{5}\right )}}{d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d} \]

[In]

integrate(cos(d*x+c)^2*(a+I*a*tan(d*x+c))^5,x, algorithm="giac")

[Out]

-2*(-6*I*a^5*e^(4*I*d*x + 4*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) - 12*I*a^5*e^(2*I*d*x + 2*I*c)*log(e^(2*I*d*x +
2*I*c) + 1) + 2*I*a^5*e^(6*I*d*x + 6*I*c) + 4*I*a^5*e^(4*I*d*x + 4*I*c) - 4*I*a^5*e^(2*I*d*x + 2*I*c) - 6*I*a^
5*log(e^(2*I*d*x + 2*I*c) + 1) - 5*I*a^5)/(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d)

Mupad [B] (verification not implemented)

Time = 3.72 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.84 \[ \int \cos ^2(c+d x) (a+i a \tan (c+d x))^5 \, dx=\frac {8\,a^5}{d\,\left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )}-\frac {a^5\,\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,12{}\mathrm {i}}{d}+\frac {5\,a^5\,\mathrm {tan}\left (c+d\,x\right )}{d}+\frac {a^5\,{\mathrm {tan}\left (c+d\,x\right )}^2\,1{}\mathrm {i}}{2\,d} \]

[In]

int(cos(c + d*x)^2*(a + a*tan(c + d*x)*1i)^5,x)

[Out]

(8*a^5)/(d*(tan(c + d*x) + 1i)) - (a^5*log(tan(c + d*x) + 1i)*12i)/d + (5*a^5*tan(c + d*x))/d + (a^5*tan(c + d
*x)^2*1i)/(2*d)