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

Optimal. Leaf size=171 \[ -\frac {i a^6}{32 d (a-i a \tan (c+d x))^4}-\frac {i a^5}{16 d (a-i a \tan (c+d x))^3}-\frac {3 i a^4}{32 d (a-i a \tan (c+d x))^2}+\frac {i a^4}{64 d (a+i a \tan (c+d x))^2}-\frac {5 i a^3}{32 d (a-i a \tan (c+d x))}+\frac {5 i a^3}{64 d (a+i a \tan (c+d x))}+\frac {15 a^2 x}{64} \]

[Out]

15/64*a^2*x-1/32*I*a^6/d/(a-I*a*tan(d*x+c))^4-1/16*I*a^5/d/(a-I*a*tan(d*x+c))^3-3/32*I*a^4/d/(a-I*a*tan(d*x+c)
)^2-5/32*I*a^3/d/(a-I*a*tan(d*x+c))+1/64*I*a^4/d/(a+I*a*tan(d*x+c))^2+5/64*I*a^3/d/(a+I*a*tan(d*x+c))

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Rubi [A]  time = 0.11, antiderivative size = 171, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {3487, 44, 206} \[ -\frac {i a^6}{32 d (a-i a \tan (c+d x))^4}-\frac {i a^5}{16 d (a-i a \tan (c+d x))^3}-\frac {3 i a^4}{32 d (a-i a \tan (c+d x))^2}+\frac {i a^4}{64 d (a+i a \tan (c+d x))^2}-\frac {5 i a^3}{32 d (a-i a \tan (c+d x))}+\frac {5 i a^3}{64 d (a+i a \tan (c+d x))}+\frac {15 a^2 x}{64} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(15*a^2*x)/64 - ((I/32)*a^6)/(d*(a - I*a*Tan[c + d*x])^4) - ((I/16)*a^5)/(d*(a - I*a*Tan[c + d*x])^3) - (((3*I
)/32)*a^4)/(d*(a - I*a*Tan[c + d*x])^2) - (((5*I)/32)*a^3)/(d*(a - I*a*Tan[c + d*x])) + ((I/64)*a^4)/(d*(a + I
*a*Tan[c + d*x])^2) + (((5*I)/64)*a^3)/(d*(a + I*a*Tan[c + d*x]))

Rule 44

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 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 ^8(c+d x) (a+i a \tan (c+d x))^2 \, dx &=-\frac {\left (i a^9\right ) \operatorname {Subst}\left (\int \frac {1}{(a-x)^5 (a+x)^3} \, dx,x,i a \tan (c+d x)\right )}{d}\\ &=-\frac {\left (i a^9\right ) \operatorname {Subst}\left (\int \left (\frac {1}{8 a^3 (a-x)^5}+\frac {3}{16 a^4 (a-x)^4}+\frac {3}{16 a^5 (a-x)^3}+\frac {5}{32 a^6 (a-x)^2}+\frac {1}{32 a^5 (a+x)^3}+\frac {5}{64 a^6 (a+x)^2}+\frac {15}{64 a^6 \left (a^2-x^2\right )}\right ) \, dx,x,i a \tan (c+d x)\right )}{d}\\ &=-\frac {i a^6}{32 d (a-i a \tan (c+d x))^4}-\frac {i a^5}{16 d (a-i a \tan (c+d x))^3}-\frac {3 i a^4}{32 d (a-i a \tan (c+d x))^2}-\frac {5 i a^3}{32 d (a-i a \tan (c+d x))}+\frac {i a^4}{64 d (a+i a \tan (c+d x))^2}+\frac {5 i a^3}{64 d (a+i a \tan (c+d x))}-\frac {\left (15 i a^3\right ) \operatorname {Subst}\left (\int \frac {1}{a^2-x^2} \, dx,x,i a \tan (c+d x)\right )}{64 d}\\ &=\frac {15 a^2 x}{64}-\frac {i a^6}{32 d (a-i a \tan (c+d x))^4}-\frac {i a^5}{16 d (a-i a \tan (c+d x))^3}-\frac {3 i a^4}{32 d (a-i a \tan (c+d x))^2}-\frac {5 i a^3}{32 d (a-i a \tan (c+d x))}+\frac {i a^4}{64 d (a+i a \tan (c+d x))^2}+\frac {5 i a^3}{64 d (a+i a \tan (c+d x))}\\ \end {align*}

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Mathematica [A]  time = 0.59, size = 138, normalized size = 0.81 \[ \frac {a^2 (-120 i d x \sin (2 (c+d x))+30 \sin (2 (c+d x))+32 \sin (4 (c+d x))+3 \sin (6 (c+d x))+30 (4 d x-i) \cos (2 (c+d x))+16 i \cos (4 (c+d x))+i \cos (6 (c+d x))-80 i) (\cos (2 (c+2 d x))+i \sin (2 (c+2 d x)))}{512 d (\cos (d x)+i \sin (d x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(-80*I + 30*(-I + 4*d*x)*Cos[2*(c + d*x)] + (16*I)*Cos[4*(c + d*x)] + I*Cos[6*(c + d*x)] + 30*Sin[2*(c +
d*x)] - (120*I)*d*x*Sin[2*(c + d*x)] + 32*Sin[4*(c + d*x)] + 3*Sin[6*(c + d*x)])*(Cos[2*(c + 2*d*x)] + I*Sin[2
*(c + 2*d*x)]))/(512*d*(Cos[d*x] + I*Sin[d*x])^2)

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fricas [A]  time = 0.56, size = 106, normalized size = 0.62 \[ \frac {{\left (120 \, a^{2} d x e^{\left (4 i \, d x + 4 i \, c\right )} - i \, a^{2} e^{\left (12 i \, d x + 12 i \, c\right )} - 8 i \, a^{2} e^{\left (10 i \, d x + 10 i \, c\right )} - 30 i \, a^{2} e^{\left (8 i \, d x + 8 i \, c\right )} - 80 i \, a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} + 24 i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + 2 i \, a^{2}\right )} e^{\left (-4 i \, d x - 4 i \, c\right )}}{512 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/512*(120*a^2*d*x*e^(4*I*d*x + 4*I*c) - I*a^2*e^(12*I*d*x + 12*I*c) - 8*I*a^2*e^(10*I*d*x + 10*I*c) - 30*I*a^
2*e^(8*I*d*x + 8*I*c) - 80*I*a^2*e^(6*I*d*x + 6*I*c) + 24*I*a^2*e^(2*I*d*x + 2*I*c) + 2*I*a^2)*e^(-4*I*d*x - 4
*I*c)/d

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giac [B]  time = 2.03, size = 342, normalized size = 2.00 \[ \frac {120 \, a^{2} d x e^{\left (8 i \, d x + 4 i \, c\right )} + 240 \, a^{2} d x e^{\left (6 i \, d x + 2 i \, c\right )} + 120 \, a^{2} d x e^{\left (4 i \, d x\right )} + 8 i \, a^{2} e^{\left (8 i \, d x + 4 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 16 i \, a^{2} e^{\left (6 i \, d x + 2 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 8 i \, a^{2} e^{\left (4 i \, d x\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 8 i \, a^{2} e^{\left (8 i \, d x + 4 i \, c\right )} \log \left (e^{\left (2 i \, d x\right )} + e^{\left (-2 i \, c\right )}\right ) - 16 i \, a^{2} e^{\left (6 i \, d x + 2 i \, c\right )} \log \left (e^{\left (2 i \, d x\right )} + e^{\left (-2 i \, c\right )}\right ) - 8 i \, a^{2} e^{\left (4 i \, d x\right )} \log \left (e^{\left (2 i \, d x\right )} + e^{\left (-2 i \, c\right )}\right ) - i \, a^{2} e^{\left (16 i \, d x + 12 i \, c\right )} - 10 i \, a^{2} e^{\left (14 i \, d x + 10 i \, c\right )} - 47 i \, a^{2} e^{\left (12 i \, d x + 8 i \, c\right )} - 148 i \, a^{2} e^{\left (10 i \, d x + 6 i \, c\right )} - 190 i \, a^{2} e^{\left (8 i \, d x + 4 i \, c\right )} - 56 i \, a^{2} e^{\left (6 i \, d x + 2 i \, c\right )} + 28 i \, a^{2} e^{\left (2 i \, d x - 2 i \, c\right )} + 50 i \, a^{2} e^{\left (4 i \, d x\right )} + 2 i \, a^{2} e^{\left (-4 i \, c\right )}}{512 \, {\left (d e^{\left (8 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (6 i \, d x + 2 i \, c\right )} + d e^{\left (4 i \, d x\right )}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/512*(120*a^2*d*x*e^(8*I*d*x + 4*I*c) + 240*a^2*d*x*e^(6*I*d*x + 2*I*c) + 120*a^2*d*x*e^(4*I*d*x) + 8*I*a^2*e
^(8*I*d*x + 4*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 16*I*a^2*e^(6*I*d*x + 2*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) +
8*I*a^2*e^(4*I*d*x)*log(e^(2*I*d*x + 2*I*c) + 1) - 8*I*a^2*e^(8*I*d*x + 4*I*c)*log(e^(2*I*d*x) + e^(-2*I*c)) -
16*I*a^2*e^(6*I*d*x + 2*I*c)*log(e^(2*I*d*x) + e^(-2*I*c)) - 8*I*a^2*e^(4*I*d*x)*log(e^(2*I*d*x) + e^(-2*I*c)
) - I*a^2*e^(16*I*d*x + 12*I*c) - 10*I*a^2*e^(14*I*d*x + 10*I*c) - 47*I*a^2*e^(12*I*d*x + 8*I*c) - 148*I*a^2*e
^(10*I*d*x + 6*I*c) - 190*I*a^2*e^(8*I*d*x + 4*I*c) - 56*I*a^2*e^(6*I*d*x + 2*I*c) + 28*I*a^2*e^(2*I*d*x - 2*I
*c) + 50*I*a^2*e^(4*I*d*x) + 2*I*a^2*e^(-4*I*c))/(d*e^(8*I*d*x + 4*I*c) + 2*d*e^(6*I*d*x + 2*I*c) + d*e^(4*I*d
*x))

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maple [A]  time = 0.54, size = 141, normalized size = 0.82 \[ \frac {-a^{2} \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{7}\left (d x +c \right )\right )}{8}+\frac {\left (\cos ^{5}\left (d x +c \right )+\frac {5 \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{48}+\frac {5 d x}{128}+\frac {5 c}{128}\right )-\frac {i a^{2} \left (\cos ^{8}\left (d x +c \right )\right )}{4}+a^{2} \left (\frac {\left (\cos ^{7}\left (d x +c \right )+\frac {7 \left (\cos ^{5}\left (d x +c \right )\right )}{6}+\frac {35 \left (\cos ^{3}\left (d x +c \right )\right )}{24}+\frac {35 \cos \left (d x +c \right )}{16}\right ) \sin \left (d x +c \right )}{8}+\frac {35 d x}{128}+\frac {35 c}{128}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(-a^2*(-1/8*sin(d*x+c)*cos(d*x+c)^7+1/48*(cos(d*x+c)^5+5/4*cos(d*x+c)^3+15/8*cos(d*x+c))*sin(d*x+c)+5/128*
d*x+5/128*c)-1/4*I*a^2*cos(d*x+c)^8+a^2*(1/8*(cos(d*x+c)^7+7/6*cos(d*x+c)^5+35/24*cos(d*x+c)^3+35/16*cos(d*x+c
))*sin(d*x+c)+35/128*d*x+35/128*c))

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maxima [A]  time = 0.49, size = 115, normalized size = 0.67 \[ \frac {15 \, {\left (d x + c\right )} a^{2} + \frac {15 \, a^{2} \tan \left (d x + c\right )^{7} + 55 \, a^{2} \tan \left (d x + c\right )^{5} + 73 \, a^{2} \tan \left (d x + c\right )^{3} + 49 \, a^{2} \tan \left (d x + c\right ) - 16 i \, a^{2}}{\tan \left (d x + c\right )^{8} + 4 \, \tan \left (d x + c\right )^{6} + 6 \, \tan \left (d x + c\right )^{4} + 4 \, \tan \left (d x + c\right )^{2} + 1}}{64 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64*(15*(d*x + c)*a^2 + (15*a^2*tan(d*x + c)^7 + 55*a^2*tan(d*x + c)^5 + 73*a^2*tan(d*x + c)^3 + 49*a^2*tan(d
*x + c) - 16*I*a^2)/(tan(d*x + c)^8 + 4*tan(d*x + c)^6 + 6*tan(d*x + c)^4 + 4*tan(d*x + c)^2 + 1))/d

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mupad [B]  time = 4.03, size = 144, normalized size = 0.84 \[ \frac {15\,a^2\,x}{64}+\frac {\frac {15\,a^2\,{\mathrm {tan}\left (c+d\,x\right )}^5}{64}+\frac {a^2\,{\mathrm {tan}\left (c+d\,x\right )}^4\,15{}\mathrm {i}}{32}+\frac {5\,a^2\,{\mathrm {tan}\left (c+d\,x\right )}^3}{32}+\frac {a^2\,{\mathrm {tan}\left (c+d\,x\right )}^2\,25{}\mathrm {i}}{32}-\frac {17\,a^2\,\mathrm {tan}\left (c+d\,x\right )}{64}+\frac {a^2\,1{}\mathrm {i}}{4}}{d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^6+{\mathrm {tan}\left (c+d\,x\right )}^5\,2{}\mathrm {i}+{\mathrm {tan}\left (c+d\,x\right )}^4+{\mathrm {tan}\left (c+d\,x\right )}^3\,4{}\mathrm {i}-{\mathrm {tan}\left (c+d\,x\right )}^2+\mathrm {tan}\left (c+d\,x\right )\,2{}\mathrm {i}-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(15*a^2*x)/64 + ((a^2*1i)/4 - (17*a^2*tan(c + d*x))/64 + (a^2*tan(c + d*x)^2*25i)/32 + (5*a^2*tan(c + d*x)^3)/
32 + (a^2*tan(c + d*x)^4*15i)/32 + (15*a^2*tan(c + d*x)^5)/64)/(d*(tan(c + d*x)*2i - tan(c + d*x)^2 + tan(c +
d*x)^3*4i + tan(c + d*x)^4 + tan(c + d*x)^5*2i + tan(c + d*x)^6 - 1))

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sympy [A]  time = 0.53, size = 272, normalized size = 1.59 \[ \frac {15 a^{2} x}{64} + \begin {cases} \frac {\left (- 8589934592 i a^{2} d^{5} e^{14 i c} e^{8 i d x} - 68719476736 i a^{2} d^{5} e^{12 i c} e^{6 i d x} - 257698037760 i a^{2} d^{5} e^{10 i c} e^{4 i d x} - 687194767360 i a^{2} d^{5} e^{8 i c} e^{2 i d x} + 206158430208 i a^{2} d^{5} e^{4 i c} e^{- 2 i d x} + 17179869184 i a^{2} d^{5} e^{2 i c} e^{- 4 i d x}\right ) e^{- 6 i c}}{4398046511104 d^{6}} & \text {for}\: 4398046511104 d^{6} e^{6 i c} \neq 0 \\x \left (- \frac {15 a^{2}}{64} + \frac {\left (a^{2} e^{12 i c} + 6 a^{2} e^{10 i c} + 15 a^{2} e^{8 i c} + 20 a^{2} e^{6 i c} + 15 a^{2} e^{4 i c} + 6 a^{2} e^{2 i c} + a^{2}\right ) e^{- 4 i c}}{64}\right ) & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

15*a**2*x/64 + Piecewise(((-8589934592*I*a**2*d**5*exp(14*I*c)*exp(8*I*d*x) - 68719476736*I*a**2*d**5*exp(12*I
*c)*exp(6*I*d*x) - 257698037760*I*a**2*d**5*exp(10*I*c)*exp(4*I*d*x) - 687194767360*I*a**2*d**5*exp(8*I*c)*exp
(2*I*d*x) + 206158430208*I*a**2*d**5*exp(4*I*c)*exp(-2*I*d*x) + 17179869184*I*a**2*d**5*exp(2*I*c)*exp(-4*I*d*
x))*exp(-6*I*c)/(4398046511104*d**6), Ne(4398046511104*d**6*exp(6*I*c), 0)), (x*(-15*a**2/64 + (a**2*exp(12*I*
c) + 6*a**2*exp(10*I*c) + 15*a**2*exp(8*I*c) + 20*a**2*exp(6*I*c) + 15*a**2*exp(4*I*c) + 6*a**2*exp(2*I*c) + a
**2)*exp(-4*I*c)/64), True))

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