3.84 \(\int \cos ^6(c+d x) (a+i a \tan (c+d x))^8 \, dx\)

Optimal. Leaf size=114 \[ -\frac{16 i a^{11}}{3 d (a-i a \tan (c+d x))^3}+\frac{16 i a^{10}}{d (a-i a \tan (c+d x))^2}-\frac{24 i a^9}{d (a-i a \tan (c+d x))}+\frac{a^8 \tan (c+d x)}{d}+\frac{8 i a^8 \log (\cos (c+d x))}{d}-8 a^8 x \]

[Out]

-8*a^8*x + ((8*I)*a^8*Log[Cos[c + d*x]])/d + (a^8*Tan[c + d*x])/d - (((16*I)/3)*a^11)/(d*(a - I*a*Tan[c + d*x]
)^3) + ((16*I)*a^10)/(d*(a - I*a*Tan[c + d*x])^2) - ((24*I)*a^9)/(d*(a - I*a*Tan[c + d*x]))

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Rubi [A]  time = 0.0709375, antiderivative size = 114, normalized size of antiderivative = 1., 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 = {3487, 43} \[ -\frac{16 i a^{11}}{3 d (a-i a \tan (c+d x))^3}+\frac{16 i a^{10}}{d (a-i a \tan (c+d x))^2}-\frac{24 i a^9}{d (a-i a \tan (c+d x))}+\frac{a^8 \tan (c+d x)}{d}+\frac{8 i a^8 \log (\cos (c+d x))}{d}-8 a^8 x \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^6*(a + I*a*Tan[c + d*x])^8,x]

[Out]

-8*a^8*x + ((8*I)*a^8*Log[Cos[c + d*x]])/d + (a^8*Tan[c + d*x])/d - (((16*I)/3)*a^11)/(d*(a - I*a*Tan[c + d*x]
)^3) + ((16*I)*a^10)/(d*(a - I*a*Tan[c + d*x])^2) - ((24*I)*a^9)/(d*(a - I*a*Tan[c + d*x]))

Rule 3487

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]

Rule 43

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])

Rubi steps

\begin{align*} \int \cos ^6(c+d x) (a+i a \tan (c+d x))^8 \, dx &=-\frac{\left (i a^7\right ) \operatorname{Subst}\left (\int \frac{(a+x)^4}{(a-x)^4} \, dx,x,i a \tan (c+d x)\right )}{d}\\ &=-\frac{\left (i a^7\right ) \operatorname{Subst}\left (\int \left (1+\frac{16 a^4}{(a-x)^4}-\frac{32 a^3}{(a-x)^3}+\frac{24 a^2}{(a-x)^2}-\frac{8 a}{a-x}\right ) \, dx,x,i a \tan (c+d x)\right )}{d}\\ &=-8 a^8 x+\frac{8 i a^8 \log (\cos (c+d x))}{d}+\frac{a^8 \tan (c+d x)}{d}-\frac{16 i a^{11}}{3 d (a-i a \tan (c+d x))^3}+\frac{16 i a^{10}}{d (a-i a \tan (c+d x))^2}-\frac{24 i a^9}{d (a-i a \tan (c+d x))}\\ \end{align*}

Mathematica [B]  time = 2.09391, size = 414, normalized size = 3.63 \[ -\frac{a^8 \sec (c) \sec (c+d x) (\cos (3 c+11 d x)+i \sin (3 c+11 d x)) \left (-12 i d x \sin (c+2 d x)+11 \sin (c+2 d x)-12 i d x \sin (3 c+2 d x)+14 \sin (3 c+2 d x)-12 i d x \sin (3 c+4 d x)-4 \sin (3 c+4 d x)-12 i d x \sin (5 c+4 d x)-\sin (5 c+4 d x)+12 d x \cos (3 c+2 d x)+10 i \cos (3 c+2 d x)+12 d x \cos (3 c+4 d x)-2 i \cos (3 c+4 d x)+12 d x \cos (5 c+4 d x)+i \cos (5 c+4 d x)+\cos (c+2 d x) \left (-6 i \log \left (\cos ^2(c+d x)\right )+12 d x+7 i\right )-6 i \cos (3 c+2 d x) \log \left (\cos ^2(c+d x)\right )-6 i \cos (3 c+4 d x) \log \left (\cos ^2(c+d x)\right )-6 i \cos (5 c+4 d x) \log \left (\cos ^2(c+d x)\right )-6 \sin (c+2 d x) \log \left (\cos ^2(c+d x)\right )-6 \sin (3 c+2 d x) \log \left (\cos ^2(c+d x)\right )-6 \sin (3 c+4 d x) \log \left (\cos ^2(c+d x)\right )-6 \sin (5 c+4 d x) \log \left (\cos ^2(c+d x)\right )+12 i \cos (c)\right )}{6 d (\cos (d x)+i \sin (d x))^8} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^6*(a + I*a*Tan[c + d*x])^8,x]

[Out]

-(a^8*Sec[c]*Sec[c + d*x]*((12*I)*Cos[c] + (10*I)*Cos[3*c + 2*d*x] + 12*d*x*Cos[3*c + 2*d*x] - (2*I)*Cos[3*c +
4*d*x] + 12*d*x*Cos[3*c + 4*d*x] + I*Cos[5*c + 4*d*x] + 12*d*x*Cos[5*c + 4*d*x] + Cos[c + 2*d*x]*(7*I + 12*d*
x - (6*I)*Log[Cos[c + d*x]^2]) - (6*I)*Cos[3*c + 2*d*x]*Log[Cos[c + d*x]^2] - (6*I)*Cos[3*c + 4*d*x]*Log[Cos[c
+ d*x]^2] - (6*I)*Cos[5*c + 4*d*x]*Log[Cos[c + d*x]^2] + 11*Sin[c + 2*d*x] - (12*I)*d*x*Sin[c + 2*d*x] - 6*Lo
g[Cos[c + d*x]^2]*Sin[c + 2*d*x] + 14*Sin[3*c + 2*d*x] - (12*I)*d*x*Sin[3*c + 2*d*x] - 6*Log[Cos[c + d*x]^2]*S
in[3*c + 2*d*x] - 4*Sin[3*c + 4*d*x] - (12*I)*d*x*Sin[3*c + 4*d*x] - 6*Log[Cos[c + d*x]^2]*Sin[3*c + 4*d*x] -
Sin[5*c + 4*d*x] - (12*I)*d*x*Sin[5*c + 4*d*x] - 6*Log[Cos[c + d*x]^2]*Sin[5*c + 4*d*x])*(Cos[3*c + 11*d*x] +
I*Sin[3*c + 11*d*x]))/(6*d*(Cos[d*x] + I*Sin[d*x])^8)

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Maple [B]  time = 0.07, size = 319, normalized size = 2.8 \begin{align*}{\frac{{a}^{8}\cos \left ( dx+c \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{7}}{d}}+{\frac{8\,i{a}^{8}\ln \left ( \cos \left ( dx+c \right ) \right ) }{d}}-{\frac{35\,{a}^{8} \left ( \sin \left ( dx+c \right ) \right ) ^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{3\,d}}+{\frac{{\frac{14\,i}{3}}{a}^{8} \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{d}}+{\frac{{\frac{28\,i}{3}}{a}^{8} \left ( \sin \left ( dx+c \right ) \right ) ^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{d}}+{\frac{2\,i{a}^{8} \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{d}}+{\frac{{\frac{32\,i}{3}}{a}^{8} \left ( \sin \left ( dx+c \right ) \right ) ^{6}}{d}}-8\,{a}^{8}x+{\frac{{a}^{8} \left ( \sin \left ( dx+c \right ) \right ) ^{9}}{d\cos \left ( dx+c \right ) }}+{\frac{35\,{a}^{8}\cos \left ( dx+c \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{6\,d}}+{\frac{175\,{a}^{8}\cos \left ( dx+c \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{24\,d}}+{\frac{4\,i{a}^{8} \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{d}}+{\frac{29\,{a}^{8} \left ( \cos \left ( dx+c \right ) \right ) ^{5}\sin \left ( dx+c \right ) }{6\,d}}-{\frac{233\,{a}^{8} \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) }{24\,d}}+{\frac{111\,{a}^{8}\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) }{8\,d}}-8\,{\frac{{a}^{8}c}{d}}-{\frac{{\frac{4\,i}{3}}{a}^{8} \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^6*(a+I*a*tan(d*x+c))^8,x)

[Out]

1/d*a^8*cos(d*x+c)*sin(d*x+c)^7+8*I*a^8*ln(cos(d*x+c))/d-35/3/d*a^8*sin(d*x+c)^3*cos(d*x+c)^3+14/3*I/d*a^8*cos
(d*x+c)^4+28/3*I/d*a^8*sin(d*x+c)^2*cos(d*x+c)^4+2*I/d*a^8*sin(d*x+c)^4+32/3*I/d*a^8*sin(d*x+c)^6-8*a^8*x+1/d*
a^8*sin(d*x+c)^9/cos(d*x+c)+35/6/d*a^8*cos(d*x+c)*sin(d*x+c)^5+175/24/d*a^8*cos(d*x+c)*sin(d*x+c)^3+4*I/d*a^8*
sin(d*x+c)^2+29/6/d*a^8*cos(d*x+c)^5*sin(d*x+c)-233/24/d*a^8*cos(d*x+c)^3*sin(d*x+c)+111/8/d*a^8*sin(d*x+c)*co
s(d*x+c)-8/d*a^8*c-4/3*I/d*a^8*cos(d*x+c)^6

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Maxima [A]  time = 1.63764, size = 197, normalized size = 1.73 \begin{align*} -\frac{384 \,{\left (d x + c\right )} a^{8} + 192 i \, a^{8} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 48 \, a^{8} \tan \left (d x + c\right ) - \frac{1152 \, a^{8} \tan \left (d x + c\right )^{5} - 1920 i \, a^{8} \tan \left (d x + c\right )^{4} + 512 \, a^{8} \tan \left (d x + c\right )^{3} - 1536 i \, a^{8} \tan \left (d x + c\right )^{2} + 384 \, a^{8} \tan \left (d x + c\right ) - 640 i \, a^{8}}{\tan \left (d x + c\right )^{6} + 3 \, \tan \left (d x + c\right )^{4} + 3 \, \tan \left (d x + c\right )^{2} + 1}}{48 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*(a+I*a*tan(d*x+c))^8,x, algorithm="maxima")

[Out]

-1/48*(384*(d*x + c)*a^8 + 192*I*a^8*log(tan(d*x + c)^2 + 1) - 48*a^8*tan(d*x + c) - (1152*a^8*tan(d*x + c)^5
- 1920*I*a^8*tan(d*x + c)^4 + 512*a^8*tan(d*x + c)^3 - 1536*I*a^8*tan(d*x + c)^2 + 384*a^8*tan(d*x + c) - 640*
I*a^8)/(tan(d*x + c)^6 + 3*tan(d*x + c)^4 + 3*tan(d*x + c)^2 + 1))/d

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Fricas [A]  time = 1.4936, size = 323, normalized size = 2.83 \begin{align*} \frac{-2 i \, a^{8} e^{\left (8 i \, d x + 8 i \, c\right )} + 4 i \, a^{8} e^{\left (6 i \, d x + 6 i \, c\right )} - 12 i \, a^{8} e^{\left (4 i \, d x + 4 i \, c\right )} - 18 i \, a^{8} e^{\left (2 i \, d x + 2 i \, c\right )} + 6 i \, a^{8} +{\left (24 i \, a^{8} e^{\left (2 i \, d x + 2 i \, c\right )} + 24 i \, a^{8}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )}{3 \,{\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*(a+I*a*tan(d*x+c))^8,x, algorithm="fricas")

[Out]

1/3*(-2*I*a^8*e^(8*I*d*x + 8*I*c) + 4*I*a^8*e^(6*I*d*x + 6*I*c) - 12*I*a^8*e^(4*I*d*x + 4*I*c) - 18*I*a^8*e^(2
*I*d*x + 2*I*c) + 6*I*a^8 + (24*I*a^8*e^(2*I*d*x + 2*I*c) + 24*I*a^8)*log(e^(2*I*d*x + 2*I*c) + 1))/(d*e^(2*I*
d*x + 2*I*c) + d)

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Sympy [A]  time = 1.62002, size = 144, normalized size = 1.26 \begin{align*} 12 a^{8} \left (\begin{cases} - \frac{i e^{2 i d x}}{2 d} & \text{for}\: d \neq 0 \\x & \text{otherwise} \end{cases}\right ) e^{2 i c} - 8 a^{8} \left (\begin{cases} - \frac{i e^{4 i d x}}{4 d} & \text{for}\: d \neq 0 \\x & \text{otherwise} \end{cases}\right ) e^{4 i c} + 4 a^{8} \left (\begin{cases} - \frac{i e^{6 i d x}}{6 d} & \text{for}\: d \neq 0 \\x & \text{otherwise} \end{cases}\right ) e^{6 i c} + \frac{8 i a^{8} \log{\left (e^{2 i d x} + e^{- 2 i c} \right )}}{d} + \frac{2 i a^{8} e^{- 2 i c}}{d \left (e^{2 i d x} + e^{- 2 i c}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**6*(a+I*a*tan(d*x+c))**8,x)

[Out]

12*a**8*Piecewise((-I*exp(2*I*d*x)/(2*d), Ne(d, 0)), (x, True))*exp(2*I*c) - 8*a**8*Piecewise((-I*exp(4*I*d*x)
/(4*d), Ne(d, 0)), (x, True))*exp(4*I*c) + 4*a**8*Piecewise((-I*exp(6*I*d*x)/(6*d), Ne(d, 0)), (x, True))*exp(
6*I*c) + 8*I*a**8*log(exp(2*I*d*x) + exp(-2*I*c))/d + 2*I*a**8*exp(-2*I*c)/(d*(exp(2*I*d*x) + exp(-2*I*c)))

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Giac [B]  time = 2.33369, size = 1079, normalized size = 9.46 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*(a+I*a*tan(d*x+c))^8,x, algorithm="giac")

[Out]

1/13440*(107520*I*a^8*e^(28*I*d*x + 14*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 1505280*I*a^8*e^(26*I*d*x + 12*I*c)
*log(e^(2*I*d*x + 2*I*c) + 1) + 9784320*I*a^8*e^(24*I*d*x + 10*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 39137280*I*
a^8*e^(22*I*d*x + 8*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 107627520*I*a^8*e^(20*I*d*x + 6*I*c)*log(e^(2*I*d*x +
2*I*c) + 1) + 215255040*I*a^8*e^(18*I*d*x + 4*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 322882560*I*a^8*e^(16*I*d*x
+ 2*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 322882560*I*a^8*e^(12*I*d*x - 2*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 21
5255040*I*a^8*e^(10*I*d*x - 4*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 107627520*I*a^8*e^(8*I*d*x - 6*I*c)*log(e^(2
*I*d*x + 2*I*c) + 1) + 39137280*I*a^8*e^(6*I*d*x - 8*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 9784320*I*a^8*e^(4*I*
d*x - 10*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 1505280*I*a^8*e^(2*I*d*x - 12*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) +
369008640*I*a^8*e^(14*I*d*x)*log(e^(2*I*d*x + 2*I*c) + 1) + 107520*I*a^8*e^(-14*I*c)*log(e^(2*I*d*x + 2*I*c)
+ 1) - 8960*I*a^8*e^(34*I*d*x + 20*I*c) - 98560*I*a^8*e^(32*I*d*x + 18*I*c) - 519680*I*a^8*e^(30*I*d*x + 16*I*
c) - 1944320*I*a^8*e^(28*I*d*x + 14*I*c) - 6496000*I*a^8*e^(26*I*d*x + 12*I*c) - 20034560*I*a^8*e^(24*I*d*x +
10*I*c) - 51717120*I*a^8*e^(22*I*d*x + 8*I*c) - 103783680*I*a^8*e^(20*I*d*x + 6*I*c) - 157597440*I*a^8*e^(18*I
*d*x + 4*I*c) - 179379200*I*a^8*e^(16*I*d*x + 2*I*c) - 91669760*I*a^8*e^(12*I*d*x - 2*I*c) - 37157120*I*a^8*e^
(10*I*d*x - 4*I*c) - 7813120*I*a^8*e^(8*I*d*x - 6*I*c) + 716800*I*a^8*e^(6*I*d*x - 8*I*c) + 994560*I*a^8*e^(4*
I*d*x - 10*I*c) + 268800*I*a^8*e^(2*I*d*x - 12*I*c) - 151191040*I*a^8*e^(14*I*d*x) + 26880*I*a^8*e^(-14*I*c))/
(d*e^(28*I*d*x + 14*I*c) + 14*d*e^(26*I*d*x + 12*I*c) + 91*d*e^(24*I*d*x + 10*I*c) + 364*d*e^(22*I*d*x + 8*I*c
) + 1001*d*e^(20*I*d*x + 6*I*c) + 2002*d*e^(18*I*d*x + 4*I*c) + 3003*d*e^(16*I*d*x + 2*I*c) + 3003*d*e^(12*I*d
*x - 2*I*c) + 2002*d*e^(10*I*d*x - 4*I*c) + 1001*d*e^(8*I*d*x - 6*I*c) + 364*d*e^(6*I*d*x - 8*I*c) + 91*d*e^(4
*I*d*x - 10*I*c) + 14*d*e^(2*I*d*x - 12*I*c) + 3432*d*e^(14*I*d*x) + d*e^(-14*I*c))