3.51 \(\int \cos ^9(c+d x) (a+i a \tan (c+d x))^3 \, dx\)

Optimal. Leaf size=124 \[ -\frac{5 a^3 \sin ^7(c+d x)}{63 d}+\frac{a^3 \sin ^5(c+d x)}{3 d}-\frac{5 a^3 \sin ^3(c+d x)}{9 d}+\frac{5 a^3 \sin (c+d x)}{9 d}-\frac{5 i a^3 \cos ^7(c+d x)}{63 d}-\frac{2 i a \cos ^9(c+d x) (a+i a \tan (c+d x))^2}{9 d} \]

[Out]

(((-5*I)/63)*a^3*Cos[c + d*x]^7)/d + (5*a^3*Sin[c + d*x])/(9*d) - (5*a^3*Sin[c + d*x]^3)/(9*d) + (a^3*Sin[c +
d*x]^5)/(3*d) - (5*a^3*Sin[c + d*x]^7)/(63*d) - (((2*I)/9)*a*Cos[c + d*x]^9*(a + I*a*Tan[c + d*x])^2)/d

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Rubi [A]  time = 0.083663, antiderivative size = 124, 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 = {3496, 3486, 2633} \[ -\frac{5 a^3 \sin ^7(c+d x)}{63 d}+\frac{a^3 \sin ^5(c+d x)}{3 d}-\frac{5 a^3 \sin ^3(c+d x)}{9 d}+\frac{5 a^3 \sin (c+d x)}{9 d}-\frac{5 i a^3 \cos ^7(c+d x)}{63 d}-\frac{2 i a \cos ^9(c+d x) (a+i a \tan (c+d x))^2}{9 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^9*(a + I*a*Tan[c + d*x])^3,x]

[Out]

(((-5*I)/63)*a^3*Cos[c + d*x]^7)/d + (5*a^3*Sin[c + d*x])/(9*d) - (5*a^3*Sin[c + d*x]^3)/(9*d) + (a^3*Sin[c +
d*x]^5)/(3*d) - (5*a^3*Sin[c + d*x]^7)/(63*d) - (((2*I)/9)*a*Cos[c + d*x]^9*(a + I*a*Tan[c + d*x])^2)/d

Rule 3496

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(2*b*(
d*Sec[e + f*x])^m*(a + b*Tan[e + f*x])^(n - 1))/(f*m), x] - Dist[(b^2*(m + 2*n - 2))/(d^2*m), Int[(d*Sec[e + f
*x])^(m + 2)*(a + b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 + b^2, 0] && GtQ[n,
1] && ((IGtQ[n/2, 0] && ILtQ[m - 1/2, 0]) || (EqQ[n, 2] && LtQ[m, 0]) || (LeQ[m, -1] && GtQ[m + n, 0]) || (ILt
Q[m, 0] && LtQ[m/2 + n - 1, 0] && IntegerQ[n]) || (EqQ[n, 3/2] && EqQ[m, -2^(-1)])) && IntegerQ[2*m]

Rule 3486

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*(d*Sec[
e + f*x])^m)/(f*m), x] + Dist[a, Int[(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m}, x] && (IntegerQ[2
*m] || NeQ[a^2 + b^2, 0])

Rule 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rubi steps

\begin{align*} \int \cos ^9(c+d x) (a+i a \tan (c+d x))^3 \, dx &=-\frac{2 i a \cos ^9(c+d x) (a+i a \tan (c+d x))^2}{9 d}+\frac{1}{9} \left (5 a^2\right ) \int \cos ^7(c+d x) (a+i a \tan (c+d x)) \, dx\\ &=-\frac{5 i a^3 \cos ^7(c+d x)}{63 d}-\frac{2 i a \cos ^9(c+d x) (a+i a \tan (c+d x))^2}{9 d}+\frac{1}{9} \left (5 a^3\right ) \int \cos ^7(c+d x) \, dx\\ &=-\frac{5 i a^3 \cos ^7(c+d x)}{63 d}-\frac{2 i a \cos ^9(c+d x) (a+i a \tan (c+d x))^2}{9 d}-\frac{\left (5 a^3\right ) \operatorname{Subst}\left (\int \left (1-3 x^2+3 x^4-x^6\right ) \, dx,x,-\sin (c+d x)\right )}{9 d}\\ &=-\frac{5 i a^3 \cos ^7(c+d x)}{63 d}+\frac{5 a^3 \sin (c+d x)}{9 d}-\frac{5 a^3 \sin ^3(c+d x)}{9 d}+\frac{a^3 \sin ^5(c+d x)}{3 d}-\frac{5 a^3 \sin ^7(c+d x)}{63 d}-\frac{2 i a \cos ^9(c+d x) (a+i a \tan (c+d x))^2}{9 d}\\ \end{align*}

Mathematica [A]  time = 0.651795, size = 116, normalized size = 0.94 \[ \frac{a^3 (-378 i \sin (2 (c+d x))+216 i \sin (4 (c+d x))+14 i \sin (6 (c+d x))+567 \cos (2 (c+d x))-162 \cos (4 (c+d x))-7 \cos (6 (c+d x))+210) (\sin (3 (c+2 d x))-i \cos (3 (c+2 d x)))}{2016 d (\cos (d x)+i \sin (d x))^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^9*(a + I*a*Tan[c + d*x])^3,x]

[Out]

(a^3*(210 + 567*Cos[2*(c + d*x)] - 162*Cos[4*(c + d*x)] - 7*Cos[6*(c + d*x)] - (378*I)*Sin[2*(c + d*x)] + (216
*I)*Sin[4*(c + d*x)] + (14*I)*Sin[6*(c + d*x)])*((-I)*Cos[3*(c + 2*d*x)] + Sin[3*(c + 2*d*x)]))/(2016*d*(Cos[d
*x] + I*Sin[d*x])^3)

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Maple [A]  time = 0.063, size = 166, normalized size = 1.3 \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 ) ^{7}}{9}}-{\frac{2\, \left ( \cos \left ( dx+c \right ) \right ) ^{7}}{63}} \right ) -3\,{a}^{3} \left ( -1/9\,\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{8}+{\frac{\sin \left ( dx+c \right ) }{63} \left ({\frac{16}{5}}+ \left ( \cos \left ( dx+c \right ) \right ) ^{6}+6/5\, \left ( \cos \left ( dx+c \right ) \right ) ^{4}+8/5\, \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) } \right ) -{\frac{i}{3}}{a}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{9}+{\frac{{a}^{3}\sin \left ( dx+c \right ) }{9} \left ({\frac{128}{35}}+ \left ( \cos \left ( dx+c \right ) \right ) ^{8}+{\frac{8\, \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{7}}+{\frac{48\, \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{35}}+{\frac{64\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{35}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^9*(a+I*a*tan(d*x+c))^3,x)

[Out]

1/d*(-I*a^3*(-1/9*sin(d*x+c)^2*cos(d*x+c)^7-2/63*cos(d*x+c)^7)-3*a^3*(-1/9*sin(d*x+c)*cos(d*x+c)^8+1/63*(16/5+
cos(d*x+c)^6+6/5*cos(d*x+c)^4+8/5*cos(d*x+c)^2)*sin(d*x+c))-1/3*I*a^3*cos(d*x+c)^9+1/9*a^3*(128/35+cos(d*x+c)^
8+8/7*cos(d*x+c)^6+48/35*cos(d*x+c)^4+64/35*cos(d*x+c)^2)*sin(d*x+c))

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Maxima [A]  time = 1.20519, size = 196, normalized size = 1.58 \begin{align*} -\frac{105 i \, a^{3} \cos \left (d x + c\right )^{9} + 5 i \,{\left (7 \, \cos \left (d x + c\right )^{9} - 9 \, \cos \left (d x + c\right )^{7}\right )} a^{3} - 3 \,{\left (35 \, \sin \left (d x + c\right )^{9} - 135 \, \sin \left (d x + c\right )^{7} + 189 \, \sin \left (d x + c\right )^{5} - 105 \, \sin \left (d x + c\right )^{3}\right )} a^{3} -{\left (35 \, \sin \left (d x + c\right )^{9} - 180 \, \sin \left (d x + c\right )^{7} + 378 \, \sin \left (d x + c\right )^{5} - 420 \, \sin \left (d x + c\right )^{3} + 315 \, \sin \left (d x + c\right )\right )} a^{3}}{315 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^9*(a+I*a*tan(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/315*(105*I*a^3*cos(d*x + c)^9 + 5*I*(7*cos(d*x + c)^9 - 9*cos(d*x + c)^7)*a^3 - 3*(35*sin(d*x + c)^9 - 135*
sin(d*x + c)^7 + 189*sin(d*x + c)^5 - 105*sin(d*x + c)^3)*a^3 - (35*sin(d*x + c)^9 - 180*sin(d*x + c)^7 + 378*
sin(d*x + c)^5 - 420*sin(d*x + c)^3 + 315*sin(d*x + c))*a^3)/d

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Fricas [A]  time = 1.18284, size = 319, normalized size = 2.57 \begin{align*} \frac{{\left (-7 i \, a^{3} e^{\left (12 i \, d x + 12 i \, c\right )} - 54 i \, a^{3} e^{\left (10 i \, d x + 10 i \, c\right )} - 189 i \, a^{3} e^{\left (8 i \, d x + 8 i \, c\right )} - 420 i \, a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} - 945 i \, a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + 378 i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + 21 i \, a^{3}\right )} e^{\left (-3 i \, d x - 3 i \, c\right )}}{4032 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^9*(a+I*a*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

1/4032*(-7*I*a^3*e^(12*I*d*x + 12*I*c) - 54*I*a^3*e^(10*I*d*x + 10*I*c) - 189*I*a^3*e^(8*I*d*x + 8*I*c) - 420*
I*a^3*e^(6*I*d*x + 6*I*c) - 945*I*a^3*e^(4*I*d*x + 4*I*c) + 378*I*a^3*e^(2*I*d*x + 2*I*c) + 21*I*a^3)*e^(-3*I*
d*x - 3*I*c)/d

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Sympy [A]  time = 1.74558, size = 277, normalized size = 2.23 \begin{align*} \begin{cases} \frac{\left (- 270582939648 i a^{3} d^{6} e^{13 i c} e^{9 i d x} - 2087354105856 i a^{3} d^{6} e^{11 i c} e^{7 i d x} - 7305739370496 i a^{3} d^{6} e^{9 i c} e^{5 i d x} - 16234976378880 i a^{3} d^{6} e^{7 i c} e^{3 i d x} - 36528696852480 i a^{3} d^{6} e^{5 i c} e^{i d x} + 14611478740992 i a^{3} d^{6} e^{3 i c} e^{- i d x} + 811748818944 i a^{3} d^{6} e^{i c} e^{- 3 i d x}\right ) e^{- 4 i c}}{155855773237248 d^{7}} & \text{for}\: 155855773237248 d^{7} e^{4 i c} \neq 0 \\\frac{x \left (a^{3} e^{12 i c} + 6 a^{3} e^{10 i c} + 15 a^{3} e^{8 i c} + 20 a^{3} e^{6 i c} + 15 a^{3} e^{4 i c} + 6 a^{3} e^{2 i c} + a^{3}\right ) e^{- 3 i c}}{64} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**9*(a+I*a*tan(d*x+c))**3,x)

[Out]

Piecewise(((-270582939648*I*a**3*d**6*exp(13*I*c)*exp(9*I*d*x) - 2087354105856*I*a**3*d**6*exp(11*I*c)*exp(7*I
*d*x) - 7305739370496*I*a**3*d**6*exp(9*I*c)*exp(5*I*d*x) - 16234976378880*I*a**3*d**6*exp(7*I*c)*exp(3*I*d*x)
- 36528696852480*I*a**3*d**6*exp(5*I*c)*exp(I*d*x) + 14611478740992*I*a**3*d**6*exp(3*I*c)*exp(-I*d*x) + 8117
48818944*I*a**3*d**6*exp(I*c)*exp(-3*I*d*x))*exp(-4*I*c)/(155855773237248*d**7), Ne(155855773237248*d**7*exp(4
*I*c), 0)), (x*(a**3*exp(12*I*c) + 6*a**3*exp(10*I*c) + 15*a**3*exp(8*I*c) + 20*a**3*exp(6*I*c) + 15*a**3*exp(
4*I*c) + 6*a**3*exp(2*I*c) + a**3)*exp(-3*I*c)/64, True))

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Giac [B]  time = 1.62819, size = 1403, normalized size = 11.31 \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)^9*(a+I*a*tan(d*x+c))^3,x, algorithm="giac")

[Out]

1/516096*(119511*a^3*e^(11*I*d*x + 5*I*c)*log(I*e^(I*d*x + I*c) + 1) + 478044*a^3*e^(9*I*d*x + 3*I*c)*log(I*e^
(I*d*x + I*c) + 1) + 717066*a^3*e^(7*I*d*x + I*c)*log(I*e^(I*d*x + I*c) + 1) + 478044*a^3*e^(5*I*d*x - I*c)*lo
g(I*e^(I*d*x + I*c) + 1) + 119511*a^3*e^(3*I*d*x - 3*I*c)*log(I*e^(I*d*x + I*c) + 1) + 128898*a^3*e^(11*I*d*x
+ 5*I*c)*log(I*e^(I*d*x + I*c) - 1) + 515592*a^3*e^(9*I*d*x + 3*I*c)*log(I*e^(I*d*x + I*c) - 1) + 773388*a^3*e
^(7*I*d*x + I*c)*log(I*e^(I*d*x + I*c) - 1) + 515592*a^3*e^(5*I*d*x - I*c)*log(I*e^(I*d*x + I*c) - 1) + 128898
*a^3*e^(3*I*d*x - 3*I*c)*log(I*e^(I*d*x + I*c) - 1) - 119511*a^3*e^(11*I*d*x + 5*I*c)*log(-I*e^(I*d*x + I*c) +
1) - 478044*a^3*e^(9*I*d*x + 3*I*c)*log(-I*e^(I*d*x + I*c) + 1) - 717066*a^3*e^(7*I*d*x + I*c)*log(-I*e^(I*d*
x + I*c) + 1) - 478044*a^3*e^(5*I*d*x - I*c)*log(-I*e^(I*d*x + I*c) + 1) - 119511*a^3*e^(3*I*d*x - 3*I*c)*log(
-I*e^(I*d*x + I*c) + 1) - 128898*a^3*e^(11*I*d*x + 5*I*c)*log(-I*e^(I*d*x + I*c) - 1) - 515592*a^3*e^(9*I*d*x
+ 3*I*c)*log(-I*e^(I*d*x + I*c) - 1) - 773388*a^3*e^(7*I*d*x + I*c)*log(-I*e^(I*d*x + I*c) - 1) - 515592*a^3*e
^(5*I*d*x - I*c)*log(-I*e^(I*d*x + I*c) - 1) - 128898*a^3*e^(3*I*d*x - 3*I*c)*log(-I*e^(I*d*x + I*c) - 1) + 93
87*a^3*e^(11*I*d*x + 5*I*c)*log(I*e^(I*d*x) + e^(-I*c)) + 37548*a^3*e^(9*I*d*x + 3*I*c)*log(I*e^(I*d*x) + e^(-
I*c)) + 56322*a^3*e^(7*I*d*x + I*c)*log(I*e^(I*d*x) + e^(-I*c)) + 37548*a^3*e^(5*I*d*x - I*c)*log(I*e^(I*d*x)
+ e^(-I*c)) + 9387*a^3*e^(3*I*d*x - 3*I*c)*log(I*e^(I*d*x) + e^(-I*c)) - 9387*a^3*e^(11*I*d*x + 5*I*c)*log(-I*
e^(I*d*x) + e^(-I*c)) - 37548*a^3*e^(9*I*d*x + 3*I*c)*log(-I*e^(I*d*x) + e^(-I*c)) - 56322*a^3*e^(7*I*d*x + I*
c)*log(-I*e^(I*d*x) + e^(-I*c)) - 37548*a^3*e^(5*I*d*x - I*c)*log(-I*e^(I*d*x) + e^(-I*c)) - 9387*a^3*e^(3*I*d
*x - 3*I*c)*log(-I*e^(I*d*x) + e^(-I*c)) - 896*I*a^3*e^(20*I*d*x + 14*I*c) - 10496*I*a^3*e^(18*I*d*x + 12*I*c)
- 57216*I*a^3*e^(16*I*d*x + 10*I*c) - 195584*I*a^3*e^(14*I*d*x + 8*I*c) - 509696*I*a^3*e^(12*I*d*x + 6*I*c) -
861696*I*a^3*e^(10*I*d*x + 4*I*c) - 768768*I*a^3*e^(8*I*d*x + 2*I*c) + 88704*I*a^3*e^(4*I*d*x - 2*I*c) + 5913
6*I*a^3*e^(2*I*d*x - 4*I*c) - 236544*I*a^3*e^(6*I*d*x) + 2688*I*a^3*e^(-6*I*c))/(d*e^(11*I*d*x + 5*I*c) + 4*d*
e^(9*I*d*x + 3*I*c) + 6*d*e^(7*I*d*x + I*c) + 4*d*e^(5*I*d*x - I*c) + d*e^(3*I*d*x - 3*I*c))