3.82 \(\int \cos ^2(c+d x) (a+i a \tan (c+d x))^8 \, dx\)

Optimal. Leaf size=133 \[ -\frac {64 i a^9}{d (a-i a \tan (c+d x))}+\frac {a^8 \tan ^5(c+d x)}{5 d}-\frac {2 i a^8 \tan ^4(c+d x)}{d}-\frac {10 a^8 \tan ^3(c+d x)}{d}+\frac {36 i a^8 \tan ^2(c+d x)}{d}+\frac {129 a^8 \tan (c+d x)}{d}+\frac {192 i a^8 \log (\cos (c+d x))}{d}-192 a^8 x \]

[Out]

-192*a^8*x+192*I*a^8*ln(cos(d*x+c))/d+129*a^8*tan(d*x+c)/d+36*I*a^8*tan(d*x+c)^2/d-10*a^8*tan(d*x+c)^3/d-2*I*a
^8*tan(d*x+c)^4/d+1/5*a^8*tan(d*x+c)^5/d-64*I*a^9/d/(a-I*a*tan(d*x+c))

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Rubi [A]  time = 0.08, antiderivative size = 133, 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 = {3487, 43} \[ \frac {a^8 \tan ^5(c+d x)}{5 d}-\frac {2 i a^8 \tan ^4(c+d x)}{d}-\frac {10 a^8 \tan ^3(c+d x)}{d}+\frac {36 i a^8 \tan ^2(c+d x)}{d}-\frac {64 i a^9}{d (a-i a \tan (c+d x))}+\frac {129 a^8 \tan (c+d x)}{d}+\frac {192 i a^8 \log (\cos (c+d x))}{d}-192 a^8 x \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^2*(a + I*a*Tan[c + d*x])^8,x]

[Out]

-192*a^8*x + ((192*I)*a^8*Log[Cos[c + d*x]])/d + (129*a^8*Tan[c + d*x])/d + ((36*I)*a^8*Tan[c + d*x]^2)/d - (1
0*a^8*Tan[c + d*x]^3)/d - ((2*I)*a^8*Tan[c + d*x]^4)/d + (a^8*Tan[c + d*x]^5)/(5*d) - ((64*I)*a^9)/(d*(a - I*a
*Tan[c + d*x]))

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])

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]

Rubi steps

\begin {align*} \int \cos ^2(c+d x) (a+i a \tan (c+d x))^8 \, dx &=-\frac {\left (i a^3\right ) \operatorname {Subst}\left (\int \frac {(a+x)^6}{(a-x)^2} \, dx,x,i a \tan (c+d x)\right )}{d}\\ &=-\frac {\left (i a^3\right ) \operatorname {Subst}\left (\int \left (129 a^4+\frac {64 a^6}{(a-x)^2}-\frac {192 a^5}{a-x}+72 a^3 x+30 a^2 x^2+8 a x^3+x^4\right ) \, dx,x,i a \tan (c+d x)\right )}{d}\\ &=-192 a^8 x+\frac {192 i a^8 \log (\cos (c+d x))}{d}+\frac {129 a^8 \tan (c+d x)}{d}+\frac {36 i a^8 \tan ^2(c+d x)}{d}-\frac {10 a^8 \tan ^3(c+d x)}{d}-\frac {2 i a^8 \tan ^4(c+d x)}{d}+\frac {a^8 \tan ^5(c+d x)}{5 d}-\frac {64 i a^9}{d (a-i a \tan (c+d x))}\\ \end {align*}

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Mathematica [B]  time = 7.01, size = 321, normalized size = 2.41 \[ \frac {\cos ^3(c+d x) (a+i a \tan (c+d x))^8 \left (-960 d x \cos (8 c) \cos ^5(c+d x)-160 i (\cos (6 c)-i \sin (6 c)) \cos (2 d x) \cos ^5(c+d x)+960 i d x \sin (8 c) \cos ^5(c+d x)+160 (\cos (6 c)-i \sin (6 c)) \sin (2 d x) \cos ^5(c+d x)+480 i \cos (8 c) \cos ^5(c+d x) \log \left (\cos ^2(c+d x)\right )+696 \sec (c) (\cos (8 c)-i \sin (8 c)) \sin (d x) \cos ^4(c+d x)-4 (13 \tan (c)-50 i) (\cos (8 c)-i \sin (8 c)) \cos ^3(c+d x)-52 \sec (c) (\cos (8 c)-i \sin (8 c)) \sin (d x) \cos ^2(c+d x)+480 \sin (8 c) \cos ^5(c+d x) \log \left (\cos ^2(c+d x)\right )+(\tan (c)-10 i) (\cos (8 c)-i \sin (8 c)) \cos (c+d x)+\sec (c) (\cos (8 c)-i \sin (8 c)) \sin (d x)\right )}{5 d (\cos (d x)+i \sin (d x))^8} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^2*(a + I*a*Tan[c + d*x])^8,x]

[Out]

(Cos[c + d*x]^3*(-960*d*x*Cos[8*c]*Cos[c + d*x]^5 + (480*I)*Cos[8*c]*Cos[c + d*x]^5*Log[Cos[c + d*x]^2] - (160
*I)*Cos[2*d*x]*Cos[c + d*x]^5*(Cos[6*c] - I*Sin[6*c]) + (960*I)*d*x*Cos[c + d*x]^5*Sin[8*c] + 480*Cos[c + d*x]
^5*Log[Cos[c + d*x]^2]*Sin[8*c] + Sec[c]*(Cos[8*c] - I*Sin[8*c])*Sin[d*x] - 52*Cos[c + d*x]^2*Sec[c]*(Cos[8*c]
- I*Sin[8*c])*Sin[d*x] + 696*Cos[c + d*x]^4*Sec[c]*(Cos[8*c] - I*Sin[8*c])*Sin[d*x] + 160*Cos[c + d*x]^5*(Cos
[6*c] - I*Sin[6*c])*Sin[2*d*x] + Cos[c + d*x]*(Cos[8*c] - I*Sin[8*c])*(-10*I + Tan[c]) - 4*Cos[c + d*x]^3*(Cos
[8*c] - I*Sin[8*c])*(-50*I + 13*Tan[c]))*(a + I*a*Tan[c + d*x])^8)/(5*d*(Cos[d*x] + I*Sin[d*x])^8)

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fricas [B]  time = 0.58, size = 244, normalized size = 1.83 \[ \frac {-160 i \, a^{8} e^{\left (12 i \, d x + 12 i \, c\right )} - 800 i \, a^{8} e^{\left (10 i \, d x + 10 i \, c\right )} + 800 i \, a^{8} e^{\left (8 i \, d x + 8 i \, c\right )} + 6400 i \, a^{8} e^{\left (6 i \, d x + 6 i \, c\right )} + 9600 i \, a^{8} e^{\left (4 i \, d x + 4 i \, c\right )} + 6000 i \, a^{8} e^{\left (2 i \, d x + 2 i \, c\right )} + 1392 i \, a^{8} + {\left (960 i \, a^{8} e^{\left (10 i \, d x + 10 i \, c\right )} + 4800 i \, a^{8} e^{\left (8 i \, d x + 8 i \, c\right )} + 9600 i \, a^{8} e^{\left (6 i \, d x + 6 i \, c\right )} + 9600 i \, a^{8} e^{\left (4 i \, d x + 4 i \, c\right )} + 4800 i \, a^{8} e^{\left (2 i \, d x + 2 i \, c\right )} + 960 i \, a^{8}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )}{5 \, {\left (d e^{\left (10 i \, d x + 10 i \, c\right )} + 5 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 10 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(a+I*a*tan(d*x+c))^8,x, algorithm="fricas")

[Out]

1/5*(-160*I*a^8*e^(12*I*d*x + 12*I*c) - 800*I*a^8*e^(10*I*d*x + 10*I*c) + 800*I*a^8*e^(8*I*d*x + 8*I*c) + 6400
*I*a^8*e^(6*I*d*x + 6*I*c) + 9600*I*a^8*e^(4*I*d*x + 4*I*c) + 6000*I*a^8*e^(2*I*d*x + 2*I*c) + 1392*I*a^8 + (9
60*I*a^8*e^(10*I*d*x + 10*I*c) + 4800*I*a^8*e^(8*I*d*x + 8*I*c) + 9600*I*a^8*e^(6*I*d*x + 6*I*c) + 9600*I*a^8*
e^(4*I*d*x + 4*I*c) + 4800*I*a^8*e^(2*I*d*x + 2*I*c) + 960*I*a^8)*log(e^(2*I*d*x + 2*I*c) + 1))/(d*e^(10*I*d*x
+ 10*I*c) + 5*d*e^(8*I*d*x + 8*I*c) + 10*d*e^(6*I*d*x + 6*I*c) + 10*d*e^(4*I*d*x + 4*I*c) + 5*d*e^(2*I*d*x +
2*I*c) + d)

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giac [B]  time = 5.53, size = 302, normalized size = 2.27 \[ \frac {960 i \, a^{8} e^{\left (10 i \, d x + 10 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 4800 i \, a^{8} e^{\left (8 i \, d x + 8 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 9600 i \, a^{8} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 9600 i \, a^{8} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 4800 i \, a^{8} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 160 i \, a^{8} e^{\left (12 i \, d x + 12 i \, c\right )} - 800 i \, a^{8} e^{\left (10 i \, d x + 10 i \, c\right )} + 800 i \, a^{8} e^{\left (8 i \, d x + 8 i \, c\right )} + 6400 i \, a^{8} e^{\left (6 i \, d x + 6 i \, c\right )} + 9600 i \, a^{8} e^{\left (4 i \, d x + 4 i \, c\right )} + 6000 i \, a^{8} e^{\left (2 i \, d x + 2 i \, c\right )} + 960 i \, a^{8} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 1392 i \, a^{8}}{5 \, {\left (d e^{\left (10 i \, d x + 10 i \, c\right )} + 5 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 10 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(a+I*a*tan(d*x+c))^8,x, algorithm="giac")

[Out]

1/5*(960*I*a^8*e^(10*I*d*x + 10*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 4800*I*a^8*e^(8*I*d*x + 8*I*c)*log(e^(2*I*
d*x + 2*I*c) + 1) + 9600*I*a^8*e^(6*I*d*x + 6*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 9600*I*a^8*e^(4*I*d*x + 4*I*
c)*log(e^(2*I*d*x + 2*I*c) + 1) + 4800*I*a^8*e^(2*I*d*x + 2*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) - 160*I*a^8*e^(1
2*I*d*x + 12*I*c) - 800*I*a^8*e^(10*I*d*x + 10*I*c) + 800*I*a^8*e^(8*I*d*x + 8*I*c) + 6400*I*a^8*e^(6*I*d*x +
6*I*c) + 9600*I*a^8*e^(4*I*d*x + 4*I*c) + 6000*I*a^8*e^(2*I*d*x + 2*I*c) + 960*I*a^8*log(e^(2*I*d*x + 2*I*c) +
1) + 1392*I*a^8)/(d*e^(10*I*d*x + 10*I*c) + 5*d*e^(8*I*d*x + 8*I*c) + 10*d*e^(6*I*d*x + 6*I*c) + 10*d*e^(4*I*
d*x + 4*I*c) + 5*d*e^(2*I*d*x + 2*I*c) + d)

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maple [B]  time = 0.59, size = 406, normalized size = 3.05 \[ \frac {4 i a^{8} \left (\sin ^{6}\left (d x +c \right )\right )}{d}+\frac {8 a^{8} \cos \left (d x +c \right ) \left (\sin ^{7}\left (d x +c \right )\right )}{5 d}-\frac {192 a^{8} c}{d}+\frac {34 i a^{8} \left (\sin ^{4}\left (d x +c \right )\right )}{d}+\frac {28 i a^{8} \left (\sin ^{6}\left (d x +c \right )\right )}{d \cos \left (d x +c \right )^{2}}-\frac {2 i a^{8} \left (\sin ^{8}\left (d x +c \right )\right )}{d \cos \left (d x +c \right )^{4}}+\frac {4 i a^{8} \left (\sin ^{8}\left (d x +c \right )\right )}{d \cos \left (d x +c \right )^{2}}+\frac {70 a^{8} \left (\sin ^{5}\left (d x +c \right )\right )}{d \cos \left (d x +c \right )}+\frac {192 i a^{8} \ln \left (\cos \left (d x +c \right )\right )}{d}-\frac {4 i a^{8} \left (\cos ^{2}\left (d x +c \right )\right )}{d}+\frac {96 i a^{8} \left (\sin ^{2}\left (d x +c \right )\right )}{d}+\frac {193 a^{8} \sin \left (d x +c \right ) \cos \left (d x +c \right )}{d}-\frac {28 a^{8} \left (\sin ^{7}\left (d x +c \right )\right )}{3 d \cos \left (d x +c \right )^{3}}+\frac {112 a^{8} \left (\sin ^{7}\left (d x +c \right )\right )}{3 d \cos \left (d x +c \right )}-192 a^{8} x +\frac {a^{8} \left (\sin ^{9}\left (d x +c \right )\right )}{5 d \cos \left (d x +c \right )^{5}}-\frac {4 a^{8} \left (\sin ^{9}\left (d x +c \right )\right )}{15 d \cos \left (d x +c \right )^{3}}+\frac {8 a^{8} \left (\sin ^{9}\left (d x +c \right )\right )}{5 d \cos \left (d x +c \right )}+\frac {196 a^{8} \cos \left (d x +c \right ) \left (\sin ^{5}\left (d x +c \right )\right )}{5 d}+\frac {119 a^{8} \cos \left (d x +c \right ) \left (\sin ^{3}\left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2*(a+I*a*tan(d*x+c))^8,x)

[Out]

4*I/d*a^8*sin(d*x+c)^6+8/5/d*a^8*cos(d*x+c)*sin(d*x+c)^7-192/d*a^8*c+34*I/d*a^8*sin(d*x+c)^4+28*I/d*a^8*sin(d*
x+c)^6/cos(d*x+c)^2-2*I/d*a^8*sin(d*x+c)^8/cos(d*x+c)^4+4*I/d*a^8*sin(d*x+c)^8/cos(d*x+c)^2+70/d*a^8*sin(d*x+c
)^5/cos(d*x+c)+192*I*a^8*ln(cos(d*x+c))/d-4*I/d*a^8*cos(d*x+c)^2+96*I/d*a^8*sin(d*x+c)^2+193/d*a^8*sin(d*x+c)*
cos(d*x+c)-28/3/d*a^8*sin(d*x+c)^7/cos(d*x+c)^3+112/3/d*a^8*sin(d*x+c)^7/cos(d*x+c)-192*a^8*x+1/5/d*a^8*sin(d*
x+c)^9/cos(d*x+c)^5-4/15/d*a^8*sin(d*x+c)^9/cos(d*x+c)^3+8/5/d*a^8*sin(d*x+c)^9/cos(d*x+c)+196/5/d*a^8*cos(d*x
+c)*sin(d*x+c)^5+119/d*a^8*cos(d*x+c)*sin(d*x+c)^3

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maxima [A]  time = 0.71, size = 124, normalized size = 0.93 \[ \frac {a^{8} \tan \left (d x + c\right )^{5} - 10 i \, a^{8} \tan \left (d x + c\right )^{4} - 50 \, a^{8} \tan \left (d x + c\right )^{3} + 180 i \, a^{8} \tan \left (d x + c\right )^{2} - 960 \, {\left (d x + c\right )} a^{8} - 480 i \, a^{8} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 645 \, a^{8} \tan \left (d x + c\right ) + \frac {320 \, {\left (a^{8} \tan \left (d x + c\right ) - i \, a^{8}\right )}}{\tan \left (d x + c\right )^{2} + 1}}{5 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(a+I*a*tan(d*x+c))^8,x, algorithm="maxima")

[Out]

1/5*(a^8*tan(d*x + c)^5 - 10*I*a^8*tan(d*x + c)^4 - 50*a^8*tan(d*x + c)^3 + 180*I*a^8*tan(d*x + c)^2 - 960*(d*
x + c)*a^8 - 480*I*a^8*log(tan(d*x + c)^2 + 1) + 645*a^8*tan(d*x + c) + 320*(a^8*tan(d*x + c) - I*a^8)/(tan(d*
x + c)^2 + 1))/d

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mupad [B]  time = 3.35, size = 102, normalized size = 0.77 \[ \frac {\frac {64\,a^8}{\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}}+129\,a^8\,\mathrm {tan}\left (c+d\,x\right )-10\,a^8\,{\mathrm {tan}\left (c+d\,x\right )}^3+\frac {a^8\,{\mathrm {tan}\left (c+d\,x\right )}^5}{5}-a^8\,\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,192{}\mathrm {i}+a^8\,{\mathrm {tan}\left (c+d\,x\right )}^2\,36{}\mathrm {i}-a^8\,{\mathrm {tan}\left (c+d\,x\right )}^4\,2{}\mathrm {i}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^2*(a + a*tan(c + d*x)*1i)^8,x)

[Out]

((64*a^8)/(tan(c + d*x) + 1i) - a^8*log(tan(c + d*x) + 1i)*192i + 129*a^8*tan(c + d*x) + a^8*tan(c + d*x)^2*36
i - 10*a^8*tan(c + d*x)^3 - a^8*tan(c + d*x)^4*2i + (a^8*tan(c + d*x)^5)/5)/d

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sympy [A]  time = 0.78, size = 257, normalized size = 1.93 \[ \frac {192 i a^{8} \log {\left (e^{2 i d x} + e^{- 2 i c} \right )}}{d} + \frac {2400 i a^{8} e^{8 i c} e^{8 i d x} + 8000 i a^{8} e^{6 i c} e^{6 i d x} + 10400 i a^{8} e^{4 i c} e^{4 i d x} + 6160 i a^{8} e^{2 i c} e^{2 i d x} + 1392 i a^{8}}{5 d e^{10 i c} e^{10 i d x} + 25 d e^{8 i c} e^{8 i d x} + 50 d e^{6 i c} e^{6 i d x} + 50 d e^{4 i c} e^{4 i d x} + 25 d e^{2 i c} e^{2 i d x} + 5 d} + \begin {cases} - \frac {32 i a^{8} e^{2 i c} e^{2 i d x}}{d} & \text {for}\: d \neq 0 \\64 a^{8} x e^{2 i c} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**2*(a+I*a*tan(d*x+c))**8,x)

[Out]

192*I*a**8*log(exp(2*I*d*x) + exp(-2*I*c))/d + (2400*I*a**8*exp(8*I*c)*exp(8*I*d*x) + 8000*I*a**8*exp(6*I*c)*e
xp(6*I*d*x) + 10400*I*a**8*exp(4*I*c)*exp(4*I*d*x) + 6160*I*a**8*exp(2*I*c)*exp(2*I*d*x) + 1392*I*a**8)/(5*d*e
xp(10*I*c)*exp(10*I*d*x) + 25*d*exp(8*I*c)*exp(8*I*d*x) + 50*d*exp(6*I*c)*exp(6*I*d*x) + 50*d*exp(4*I*c)*exp(4
*I*d*x) + 25*d*exp(2*I*c)*exp(2*I*d*x) + 5*d) + Piecewise((-32*I*a**8*exp(2*I*c)*exp(2*I*d*x)/d, Ne(d, 0)), (6
4*a**8*x*exp(2*I*c), True))

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