3.44 \(\int \cos ^8(c+d x) (a+i a \tan (c+d x))^3 \, dx\)

Optimal. Leaf size=144 \[ -\frac{i a^7}{16 d (a-i a \tan (c+d x))^4}-\frac{i a^6}{12 d (a-i a \tan (c+d x))^3}-\frac{3 i a^5}{32 d (a-i a \tan (c+d x))^2}-\frac{i a^4}{8 d (a-i a \tan (c+d x))}+\frac{i a^4}{32 d (a+i a \tan (c+d x))}+\frac{5 a^3 x}{32} \]

[Out]

(5*a^3*x)/32 - ((I/16)*a^7)/(d*(a - I*a*Tan[c + d*x])^4) - ((I/12)*a^6)/(d*(a - I*a*Tan[c + d*x])^3) - (((3*I)
/32)*a^5)/(d*(a - I*a*Tan[c + d*x])^2) - ((I/8)*a^4)/(d*(a - I*a*Tan[c + d*x])) + ((I/32)*a^4)/(d*(a + I*a*Tan
[c + d*x]))

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Rubi [A]  time = 0.091226, antiderivative size = 144, normalized size of antiderivative = 1., 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} \[ -\frac{i a^7}{16 d (a-i a \tan (c+d x))^4}-\frac{i a^6}{12 d (a-i a \tan (c+d x))^3}-\frac{3 i a^5}{32 d (a-i a \tan (c+d x))^2}-\frac{i a^4}{8 d (a-i a \tan (c+d x))}+\frac{i a^4}{32 d (a+i a \tan (c+d x))}+\frac{5 a^3 x}{32} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^8*(a + I*a*Tan[c + d*x])^3,x]

[Out]

(5*a^3*x)/32 - ((I/16)*a^7)/(d*(a - I*a*Tan[c + d*x])^4) - ((I/12)*a^6)/(d*(a - I*a*Tan[c + d*x])^3) - (((3*I)
/32)*a^5)/(d*(a - I*a*Tan[c + d*x])^2) - ((I/8)*a^4)/(d*(a - I*a*Tan[c + d*x])) + ((I/32)*a^4)/(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 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])

Rubi steps

\begin{align*} \int \cos ^8(c+d x) (a+i a \tan (c+d x))^3 \, dx &=-\frac{\left (i a^9\right ) \operatorname{Subst}\left (\int \frac{1}{(a-x)^5 (a+x)^2} \, dx,x,i a \tan (c+d x)\right )}{d}\\ &=-\frac{\left (i a^9\right ) \operatorname{Subst}\left (\int \left (\frac{1}{4 a^2 (a-x)^5}+\frac{1}{4 a^3 (a-x)^4}+\frac{3}{16 a^4 (a-x)^3}+\frac{1}{8 a^5 (a-x)^2}+\frac{1}{32 a^5 (a+x)^2}+\frac{5}{32 a^5 \left (a^2-x^2\right )}\right ) \, dx,x,i a \tan (c+d x)\right )}{d}\\ &=-\frac{i a^7}{16 d (a-i a \tan (c+d x))^4}-\frac{i a^6}{12 d (a-i a \tan (c+d x))^3}-\frac{3 i a^5}{32 d (a-i a \tan (c+d x))^2}-\frac{i a^4}{8 d (a-i a \tan (c+d x))}+\frac{i a^4}{32 d (a+i a \tan (c+d x))}-\frac{\left (5 i a^4\right ) \operatorname{Subst}\left (\int \frac{1}{a^2-x^2} \, dx,x,i a \tan (c+d x)\right )}{32 d}\\ &=\frac{5 a^3 x}{32}-\frac{i a^7}{16 d (a-i a \tan (c+d x))^4}-\frac{i a^6}{12 d (a-i a \tan (c+d x))^3}-\frac{3 i a^5}{32 d (a-i a \tan (c+d x))^2}-\frac{i a^4}{8 d (a-i a \tan (c+d x))}+\frac{i a^4}{32 d (a+i a \tan (c+d x))}\\ \end{align*}

Mathematica [A]  time = 0.581832, size = 131, normalized size = 0.91 \[ \frac{a^3 (-60 \sin (c+d x)-120 i d x \sin (3 (c+d x))+20 \sin (3 (c+d x))+15 \sin (5 (c+d x))-180 i \cos (c+d x)+20 (6 d x-i) \cos (3 (c+d x))+9 i \cos (5 (c+d x))) (\cos (3 (c+2 d x))+i \sin (3 (c+2 d x)))}{768 d (\cos (d x)+i \sin (d x))^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^8*(a + I*a*Tan[c + d*x])^3,x]

[Out]

(a^3*((-180*I)*Cos[c + d*x] + 20*(-I + 6*d*x)*Cos[3*(c + d*x)] + (9*I)*Cos[5*(c + d*x)] - 60*Sin[c + d*x] + 20
*Sin[3*(c + d*x)] - (120*I)*d*x*Sin[3*(c + d*x)] + 15*Sin[5*(c + d*x)])*(Cos[3*(c + 2*d*x)] + I*Sin[3*(c + 2*d
*x)]))/(768*d*(Cos[d*x] + I*Sin[d*x])^3)

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Maple [A]  time = 0.062, size = 176, normalized size = 1.2 \begin{align*}{\frac{1}{d} \left ( -i{a}^{3} \left ( -{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{8}}-{\frac{ \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{24}} \right ) -3\,{a}^{3} \left ( -1/8\,\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{7}+1/48\, \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{5}+5/4\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}+{\frac{15\,\cos \left ( dx+c \right ) }{8}} \right ) \sin \left ( dx+c \right ) +{\frac{5\,dx}{128}}+{\frac{5\,c}{128}} \right ) -{\frac{3\,i}{8}}{a}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{8}+{a}^{3} \left ({\frac{\sin \left ( dx+c \right ) }{8} \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{7}+{\frac{7\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{6}}+{\frac{35\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{24}}+{\frac{35\,\cos \left ( dx+c \right ) }{16}} \right ) }+{\frac{35\,dx}{128}}+{\frac{35\,c}{128}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^8*(a+I*a*tan(d*x+c))^3,x)

[Out]

1/d*(-I*a^3*(-1/8*sin(d*x+c)^2*cos(d*x+c)^6-1/24*cos(d*x+c)^6)-3*a^3*(-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)-3/8*I*a^3*cos(d*x+c)^8+a^3*(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 = 1.63956, size = 173, normalized size = 1.2 \begin{align*} \frac{60 \,{\left (d x + c\right )} a^{3} + \frac{60 \, a^{3} \tan \left (d x + c\right )^{7} + 220 \, a^{3} \tan \left (d x + c\right )^{5} + 292 \, a^{3} \tan \left (d x + c\right )^{3} + 64 i \, a^{3} \tan \left (d x + c\right )^{2} + 324 \, a^{3} \tan \left (d x + c\right ) - 128 i \, a^{3}}{\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}}{384 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^8*(a+I*a*tan(d*x+c))^3,x, algorithm="maxima")

[Out]

1/384*(60*(d*x + c)*a^3 + (60*a^3*tan(d*x + c)^7 + 220*a^3*tan(d*x + c)^5 + 292*a^3*tan(d*x + c)^3 + 64*I*a^3*
tan(d*x + c)^2 + 324*a^3*tan(d*x + c) - 128*I*a^3)/(tan(d*x + c)^8 + 4*tan(d*x + c)^6 + 6*tan(d*x + c)^4 + 4*t
an(d*x + c)^2 + 1))/d

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Fricas [A]  time = 1.25267, size = 271, normalized size = 1.88 \begin{align*} \frac{{\left (120 \, a^{3} d x e^{\left (2 i \, d x + 2 i \, c\right )} - 3 i \, a^{3} e^{\left (10 i \, d x + 10 i \, c\right )} - 20 i \, a^{3} e^{\left (8 i \, d x + 8 i \, c\right )} - 60 i \, a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} - 120 i \, a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + 12 i \, a^{3}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{768 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^8*(a+I*a*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

1/768*(120*a^3*d*x*e^(2*I*d*x + 2*I*c) - 3*I*a^3*e^(10*I*d*x + 10*I*c) - 20*I*a^3*e^(8*I*d*x + 8*I*c) - 60*I*a
^3*e^(6*I*d*x + 6*I*c) - 120*I*a^3*e^(4*I*d*x + 4*I*c) + 12*I*a^3)*e^(-2*I*d*x - 2*I*c)/d

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Sympy [A]  time = 0.825254, size = 228, normalized size = 1.58 \begin{align*} \frac{5 a^{3} x}{32} + \begin{cases} \frac{\left (- 25165824 i a^{3} d^{4} e^{10 i c} e^{8 i d x} - 167772160 i a^{3} d^{4} e^{8 i c} e^{6 i d x} - 503316480 i a^{3} d^{4} e^{6 i c} e^{4 i d x} - 1006632960 i a^{3} d^{4} e^{4 i c} e^{2 i d x} + 100663296 i a^{3} d^{4} e^{- 2 i d x}\right ) e^{- 2 i c}}{6442450944 d^{5}} & \text{for}\: 6442450944 d^{5} e^{2 i c} \neq 0 \\x \left (- \frac{5 a^{3}}{32} + \frac{\left (a^{3} e^{10 i c} + 5 a^{3} e^{8 i c} + 10 a^{3} e^{6 i c} + 10 a^{3} e^{4 i c} + 5 a^{3} e^{2 i c} + a^{3}\right ) e^{- 2 i c}}{32}\right ) & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**8*(a+I*a*tan(d*x+c))**3,x)

[Out]

5*a**3*x/32 + Piecewise(((-25165824*I*a**3*d**4*exp(10*I*c)*exp(8*I*d*x) - 167772160*I*a**3*d**4*exp(8*I*c)*ex
p(6*I*d*x) - 503316480*I*a**3*d**4*exp(6*I*c)*exp(4*I*d*x) - 1006632960*I*a**3*d**4*exp(4*I*c)*exp(2*I*d*x) +
100663296*I*a**3*d**4*exp(-2*I*d*x))*exp(-2*I*c)/(6442450944*d**5), Ne(6442450944*d**5*exp(2*I*c), 0)), (x*(-5
*a**3/32 + (a**3*exp(10*I*c) + 5*a**3*exp(8*I*c) + 10*a**3*exp(6*I*c) + 10*a**3*exp(4*I*c) + 5*a**3*exp(2*I*c)
+ a**3)*exp(-2*I*c)/32), True))

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Giac [B]  time = 1.3893, size = 694, normalized size = 4.82 \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)^8*(a+I*a*tan(d*x+c))^3,x, algorithm="giac")

[Out]

1/3072*(480*a^3*d*x*e^(10*I*d*x + 6*I*c) + 1920*a^3*d*x*e^(8*I*d*x + 4*I*c) + 2880*a^3*d*x*e^(6*I*d*x + 2*I*c)
+ 480*a^3*d*x*e^(2*I*d*x - 2*I*c) + 1920*a^3*d*x*e^(4*I*d*x) - 66*I*a^3*e^(10*I*d*x + 6*I*c)*log(e^(2*I*d*x +
2*I*c) + 1) - 264*I*a^3*e^(8*I*d*x + 4*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) - 396*I*a^3*e^(6*I*d*x + 2*I*c)*log(
e^(2*I*d*x + 2*I*c) + 1) - 66*I*a^3*e^(2*I*d*x - 2*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) - 264*I*a^3*e^(4*I*d*x)*l
og(e^(2*I*d*x + 2*I*c) + 1) + 66*I*a^3*e^(10*I*d*x + 6*I*c)*log(e^(2*I*d*x) + e^(-2*I*c)) + 264*I*a^3*e^(8*I*d
*x + 4*I*c)*log(e^(2*I*d*x) + e^(-2*I*c)) + 396*I*a^3*e^(6*I*d*x + 2*I*c)*log(e^(2*I*d*x) + e^(-2*I*c)) + 66*I
*a^3*e^(2*I*d*x - 2*I*c)*log(e^(2*I*d*x) + e^(-2*I*c)) + 264*I*a^3*e^(4*I*d*x)*log(e^(2*I*d*x) + e^(-2*I*c)) -
12*I*a^3*e^(18*I*d*x + 14*I*c) - 128*I*a^3*e^(16*I*d*x + 12*I*c) - 632*I*a^3*e^(14*I*d*x + 10*I*c) - 1968*I*a
^3*e^(12*I*d*x + 8*I*c) - 3692*I*a^3*e^(10*I*d*x + 6*I*c) - 3872*I*a^3*e^(8*I*d*x + 4*I*c) - 1968*I*a^3*e^(6*I
*d*x + 2*I*c) + 192*I*a^3*e^(2*I*d*x - 2*I*c) - 192*I*a^3*e^(4*I*d*x) + 48*I*a^3*e^(-4*I*c))/(d*e^(10*I*d*x +
6*I*c) + 4*d*e^(8*I*d*x + 4*I*c) + 6*d*e^(6*I*d*x + 2*I*c) + d*e^(2*I*d*x - 2*I*c) + 4*d*e^(4*I*d*x))