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

Optimal. Leaf size=83 \[ -12 a^5 x+\frac {12 i a^5 \log (\cos (c+d x))}{d}+\frac {5 a^5 \tan (c+d x)}{d}+\frac {i a^5 \tan ^2(c+d x)}{2 d}-\frac {8 i a^6}{d (a-i a \tan (c+d x))} \]

[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]
time = 0.05, antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {3568, 45} \begin {gather*} -\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 \end {gather*}

Antiderivative was successfully verified.

[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*} \int \cos ^2(c+d x) (a+i a \tan (c+d x))^5 \, dx &=-\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 [B] Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(649\) vs. \(2(83)=166\).
time = 6.31, size = 649, normalized size = 7.82 \begin {gather*} -\frac {12 x \cos (5 c) \cos ^5(c+d x) (a+i a \tan (c+d x))^5}{(\cos (d x)+i \sin (d x))^5}+\frac {6 i \cos (5 c) \cos ^5(c+d x) \log \left (\cos ^2(c+d x)\right ) (a+i a \tan (c+d x))^5}{d (\cos (d x)+i \sin (d x))^5}+\frac {\cos (2 d x) \cos ^5(c+d x) (-4 i \cos (3 c)-4 \sin (3 c)) (a+i a \tan (c+d x))^5}{d (\cos (d x)+i \sin (d x))^5}+\frac {\cos ^3(c+d x) \left (\frac {1}{2} i \cos (5 c)+\frac {1}{2} \sin (5 c)\right ) (a+i a \tan (c+d x))^5}{d (\cos (d x)+i \sin (d x))^5}+\frac {12 i x \cos ^5(c+d x) \sin (5 c) (a+i a \tan (c+d x))^5}{(\cos (d x)+i \sin (d x))^5}+\frac {6 \cos ^5(c+d x) \log \left (\cos ^2(c+d x)\right ) \sin (5 c) (a+i a \tan (c+d x))^5}{d (\cos (d x)+i \sin (d x))^5}+\frac {\cos ^4(c+d x) (5 \cos (5 c)-5 i \sin (5 c)) \sin (d x) (a+i a \tan (c+d x))^5}{d \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) (\cos (d x)+i \sin (d x))^5}+\frac {\cos ^5(c+d x) (4 \cos (3 c)-4 i \sin (3 c)) \sin (2 d x) (a+i a \tan (c+d x))^5}{d (\cos (d x)+i \sin (d x))^5}+\frac {x \cos ^5(c+d x) \left (6 \cos ^3(c)-6 \cos ^5(c)-24 i \cos ^2(c) \sin (c)+36 i \cos ^4(c) \sin (c)-36 \cos (c) \sin ^2(c)+90 \cos ^3(c) \sin ^2(c)+24 i \sin ^3(c)-120 i \cos ^2(c) \sin ^3(c)-90 \cos (c) \sin ^4(c)+36 i \sin ^5(c)+6 \sin ^3(c) \tan (c)+6 \sin ^5(c) \tan (c)-i (12 \cos (5 c)-12 i \sin (5 c)) \tan (c)\right ) (a+i a \tan (c+d x))^5}{(\cos (d x)+i \sin (d x))^5} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-12*x*Cos[5*c]*Cos[c + d*x]^5*(a + I*a*Tan[c + d*x])^5)/(Cos[d*x] + I*Sin[d*x])^5 + ((6*I)*Cos[5*c]*Cos[c + d
*x]^5*Log[Cos[c + d*x]^2]*(a + I*a*Tan[c + d*x])^5)/(d*(Cos[d*x] + I*Sin[d*x])^5) + (Cos[2*d*x]*Cos[c + d*x]^5
*((-4*I)*Cos[3*c] - 4*Sin[3*c])*(a + I*a*Tan[c + d*x])^5)/(d*(Cos[d*x] + I*Sin[d*x])^5) + (Cos[c + d*x]^3*((I/
2)*Cos[5*c] + Sin[5*c]/2)*(a + I*a*Tan[c + d*x])^5)/(d*(Cos[d*x] + I*Sin[d*x])^5) + ((12*I)*x*Cos[c + d*x]^5*S
in[5*c]*(a + I*a*Tan[c + d*x])^5)/(Cos[d*x] + I*Sin[d*x])^5 + (6*Cos[c + d*x]^5*Log[Cos[c + d*x]^2]*Sin[5*c]*(
a + I*a*Tan[c + d*x])^5)/(d*(Cos[d*x] + I*Sin[d*x])^5) + (Cos[c + d*x]^4*(5*Cos[5*c] - (5*I)*Sin[5*c])*Sin[d*x
]*(a + I*a*Tan[c + d*x])^5)/(d*(Cos[c/2] - Sin[c/2])*(Cos[c/2] + Sin[c/2])*(Cos[d*x] + I*Sin[d*x])^5) + (Cos[c
+ d*x]^5*(4*Cos[3*c] - (4*I)*Sin[3*c])*Sin[2*d*x]*(a + I*a*Tan[c + d*x])^5)/(d*(Cos[d*x] + I*Sin[d*x])^5) + (
x*Cos[c + d*x]^5*(6*Cos[c]^3 - 6*Cos[c]^5 - (24*I)*Cos[c]^2*Sin[c] + (36*I)*Cos[c]^4*Sin[c] - 36*Cos[c]*Sin[c]
^2 + 90*Cos[c]^3*Sin[c]^2 + (24*I)*Sin[c]^3 - (120*I)*Cos[c]^2*Sin[c]^3 - 90*Cos[c]*Sin[c]^4 + (36*I)*Sin[c]^5
+ 6*Sin[c]^3*Tan[c] + 6*Sin[c]^5*Tan[c] - I*(12*Cos[5*c] - (12*I)*Sin[5*c])*Tan[c])*(a + I*a*Tan[c + d*x])^5)
/(Cos[d*x] + I*Sin[d*x])^5

________________________________________________________________________________________

Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 204 vs. \(2 (77 ) = 154\).
time = 0.26, size = 205, normalized size = 2.47

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	\(\frac {i a^{5} \left (\frac {\sin ^{6}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}+\frac {\left (\sin ^{4}\left (d x +c \right )\right )}{2}+\sin ^{2}\left (d x +c \right )+2 \ln \left (\cos \left (d x +c \right )\right )\right )+5 a^{5} \left (\frac {\sin ^{5}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\sin ^{3}\left (d x +c \right )+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )-\frac {3 d x}{2}-\frac {3 c}{2}\right )-10 i a^{5} \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )-10 a^{5} \left (-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )-\frac {5 i a^{5} \left (\cos ^{2}\left (d x +c \right )\right )}{2}+a^{5} \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(205\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(I*a^5*(1/2*sin(d*x+c)^6/cos(d*x+c)^2+1/2*sin(d*x+c)^4+sin(d*x+c)^2+2*ln(cos(d*x+c)))+5*a^5*(sin(d*x+c)^5/
cos(d*x+c)+(sin(d*x+c)^3+3/2*sin(d*x+c))*cos(d*x+c)-3/2*d*x-3/2*c)-10*I*a^5*(-1/2*sin(d*x+c)^2-ln(cos(d*x+c)))
-10*a^5*(-1/2*sin(d*x+c)*cos(d*x+c)+1/2*d*x+1/2*c)-5/2*I*a^5*cos(d*x+c)^2+a^5*(1/2*sin(d*x+c)*cos(d*x+c)+1/2*d
*x+1/2*c))

________________________________________________________________________________________

Maxima [A]
time = 0.49, size = 86, normalized size = 1.04 \begin {gather*} -\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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

________________________________________________________________________________________

Fricas [A]
time = 0.38, size = 125, normalized size = 1.51 \begin {gather*} -\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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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]
time = 0.26, size = 131, normalized size = 1.58 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

________________________________________________________________________________________

Giac [A]
time = 0.96, size = 146, normalized size = 1.76 \begin {gather*} -\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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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]
time = 3.30, size = 70, normalized size = 0.84 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________