3.34 \(\int \cos ^7(c+d x) (a+i a \tan (c+d x))^2 \, dx\)

Optimal. Leaf size=87 \[ \frac{a^2 \sin ^5(c+d x)}{7 d}-\frac{10 a^2 \sin ^3(c+d x)}{21 d}+\frac{5 a^2 \sin (c+d x)}{7 d}-\frac{2 i \cos ^7(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{7 d} \]

[Out]

(5*a^2*Sin[c + d*x])/(7*d) - (10*a^2*Sin[c + d*x]^3)/(21*d) + (a^2*Sin[c + d*x]^5)/(7*d) - (((2*I)/7)*Cos[c +
d*x]^7*(a^2 + I*a^2*Tan[c + d*x]))/d

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Rubi [A]  time = 0.0523546, antiderivative size = 87, 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 = {3496, 2633} \[ \frac{a^2 \sin ^5(c+d x)}{7 d}-\frac{10 a^2 \sin ^3(c+d x)}{21 d}+\frac{5 a^2 \sin (c+d x)}{7 d}-\frac{2 i \cos ^7(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{7 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^7*(a + I*a*Tan[c + d*x])^2,x]

[Out]

(5*a^2*Sin[c + d*x])/(7*d) - (10*a^2*Sin[c + d*x]^3)/(21*d) + (a^2*Sin[c + d*x]^5)/(7*d) - (((2*I)/7)*Cos[c +
d*x]^7*(a^2 + I*a^2*Tan[c + d*x]))/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 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 ^7(c+d x) (a+i a \tan (c+d x))^2 \, dx &=-\frac{2 i \cos ^7(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{7 d}+\frac{1}{7} \left (5 a^2\right ) \int \cos ^5(c+d x) \, dx\\ &=-\frac{2 i \cos ^7(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{7 d}-\frac{\left (5 a^2\right ) \operatorname{Subst}\left (\int \left (1-2 x^2+x^4\right ) \, dx,x,-\sin (c+d x)\right )}{7 d}\\ &=\frac{5 a^2 \sin (c+d x)}{7 d}-\frac{10 a^2 \sin ^3(c+d x)}{21 d}+\frac{a^2 \sin ^5(c+d x)}{7 d}-\frac{2 i \cos ^7(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{7 d}\\ \end{align*}

Mathematica [A]  time = 0.414968, size = 111, normalized size = 1.28 \[ \frac{a^2 (-70 \sin (c+d x)+63 \sin (3 (c+d x))+5 \sin (5 (c+d x))-140 i \cos (c+d x)+42 i \cos (3 (c+d x))+2 i \cos (5 (c+d x))) (\cos (2 (c+2 d x))+i \sin (2 (c+2 d x)))}{336 d (\cos (d x)+i \sin (d x))^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^7*(a + I*a*Tan[c + d*x])^2,x]

[Out]

(a^2*((-140*I)*Cos[c + d*x] + (42*I)*Cos[3*(c + d*x)] + (2*I)*Cos[5*(c + d*x)] - 70*Sin[c + d*x] + 63*Sin[3*(c
+ d*x)] + 5*Sin[5*(c + d*x)])*(Cos[2*(c + 2*d*x)] + I*Sin[2*(c + 2*d*x)]))/(336*d*(Cos[d*x] + I*Sin[d*x])^2)

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Maple [A]  time = 0.055, size = 111, normalized size = 1.3 \begin{align*}{\frac{1}{d} \left ( -{a}^{2} \left ( -{\frac{ \left ( \cos \left ( dx+c \right ) \right ) ^{6}\sin \left ( dx+c \right ) }{7}}+{\frac{\sin \left ( dx+c \right ) }{35} \left ({\frac{8}{3}}+ \left ( \cos \left ( dx+c \right ) \right ) ^{4}+{\frac{4\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{3}} \right ) } \right ) -{\frac{2\,i}{7}}{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{7}+{\frac{{a}^{2}\sin \left ( dx+c \right ) }{7} \left ({\frac{16}{5}}+ \left ( \cos \left ( dx+c \right ) \right ) ^{6}+{\frac{6\, \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{5}}+{\frac{8\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{5}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^7*(a+I*a*tan(d*x+c))^2,x)

[Out]

1/d*(-a^2*(-1/7*cos(d*x+c)^6*sin(d*x+c)+1/35*(8/3+cos(d*x+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c))-2/7*I*a^2*cos(d*x
+c)^7+1/7*a^2*(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))

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Maxima [A]  time = 1.14223, size = 132, normalized size = 1.52 \begin{align*} -\frac{30 i \, a^{2} \cos \left (d x + c\right )^{7} +{\left (15 \, \sin \left (d x + c\right )^{7} - 42 \, \sin \left (d x + c\right )^{5} + 35 \, \sin \left (d x + c\right )^{3}\right )} a^{2} + 3 \,{\left (5 \, \sin \left (d x + c\right )^{7} - 21 \, \sin \left (d x + c\right )^{5} + 35 \, \sin \left (d x + c\right )^{3} - 35 \, \sin \left (d x + c\right )\right )} a^{2}}{105 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^7*(a+I*a*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/105*(30*I*a^2*cos(d*x + c)^7 + (15*sin(d*x + c)^7 - 42*sin(d*x + c)^5 + 35*sin(d*x + c)^3)*a^2 + 3*(5*sin(d
*x + c)^7 - 21*sin(d*x + c)^5 + 35*sin(d*x + c)^3 - 35*sin(d*x + c))*a^2)/d

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Fricas [A]  time = 1.01903, size = 269, normalized size = 3.09 \begin{align*} \frac{{\left (-3 i \, a^{2} e^{\left (10 i \, d x + 10 i \, c\right )} - 21 i \, a^{2} e^{\left (8 i \, d x + 8 i \, c\right )} - 70 i \, a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} - 210 i \, a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + 105 i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + 7 i \, a^{2}\right )} e^{\left (-3 i \, d x - 3 i \, c\right )}}{672 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^7*(a+I*a*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

1/672*(-3*I*a^2*e^(10*I*d*x + 10*I*c) - 21*I*a^2*e^(8*I*d*x + 8*I*c) - 70*I*a^2*e^(6*I*d*x + 6*I*c) - 210*I*a^
2*e^(4*I*d*x + 4*I*c) + 105*I*a^2*e^(2*I*d*x + 2*I*c) + 7*I*a^2)*e^(-3*I*d*x - 3*I*c)/d

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Sympy [A]  time = 0.976075, size = 240, normalized size = 2.76 \begin{align*} \begin{cases} \frac{\left (- 75497472 i a^{2} d^{5} e^{11 i c} e^{7 i d x} - 528482304 i a^{2} d^{5} e^{9 i c} e^{5 i d x} - 1761607680 i a^{2} d^{5} e^{7 i c} e^{3 i d x} - 5284823040 i a^{2} d^{5} e^{5 i c} e^{i d x} + 2642411520 i a^{2} d^{5} e^{3 i c} e^{- i d x} + 176160768 i a^{2} d^{5} e^{i c} e^{- 3 i d x}\right ) e^{- 4 i c}}{16911433728 d^{6}} & \text{for}\: 16911433728 d^{6} e^{4 i c} \neq 0 \\\frac{x \left (a^{2} e^{10 i c} + 5 a^{2} e^{8 i c} + 10 a^{2} e^{6 i c} + 10 a^{2} e^{4 i c} + 5 a^{2} e^{2 i c} + a^{2}\right ) e^{- 3 i c}}{32} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**7*(a+I*a*tan(d*x+c))**2,x)

[Out]

Piecewise(((-75497472*I*a**2*d**5*exp(11*I*c)*exp(7*I*d*x) - 528482304*I*a**2*d**5*exp(9*I*c)*exp(5*I*d*x) - 1
761607680*I*a**2*d**5*exp(7*I*c)*exp(3*I*d*x) - 5284823040*I*a**2*d**5*exp(5*I*c)*exp(I*d*x) + 2642411520*I*a*
*2*d**5*exp(3*I*c)*exp(-I*d*x) + 176160768*I*a**2*d**5*exp(I*c)*exp(-3*I*d*x))*exp(-4*I*c)/(16911433728*d**6),
Ne(16911433728*d**6*exp(4*I*c), 0)), (x*(a**2*exp(10*I*c) + 5*a**2*exp(8*I*c) + 10*a**2*exp(6*I*c) + 10*a**2*
exp(4*I*c) + 5*a**2*exp(2*I*c) + a**2)*exp(-3*I*c)/32, True))

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Giac [B]  time = 1.34624, size = 865, normalized size = 9.94 \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)^7*(a+I*a*tan(d*x+c))^2,x, algorithm="giac")

[Out]

-1/10752*(2583*a^2*e^(7*I*d*x + 3*I*c)*log(I*e^(I*d*x + I*c) + 1) + 5166*a^2*e^(5*I*d*x + I*c)*log(I*e^(I*d*x
+ I*c) + 1) + 2583*a^2*e^(3*I*d*x - I*c)*log(I*e^(I*d*x + I*c) + 1) + 2121*a^2*e^(7*I*d*x + 3*I*c)*log(I*e^(I*
d*x + I*c) - 1) + 4242*a^2*e^(5*I*d*x + I*c)*log(I*e^(I*d*x + I*c) - 1) + 2121*a^2*e^(3*I*d*x - I*c)*log(I*e^(
I*d*x + I*c) - 1) - 2583*a^2*e^(7*I*d*x + 3*I*c)*log(-I*e^(I*d*x + I*c) + 1) - 5166*a^2*e^(5*I*d*x + I*c)*log(
-I*e^(I*d*x + I*c) + 1) - 2583*a^2*e^(3*I*d*x - I*c)*log(-I*e^(I*d*x + I*c) + 1) - 2121*a^2*e^(7*I*d*x + 3*I*c
)*log(-I*e^(I*d*x + I*c) - 1) - 4242*a^2*e^(5*I*d*x + I*c)*log(-I*e^(I*d*x + I*c) - 1) - 2121*a^2*e^(3*I*d*x -
I*c)*log(-I*e^(I*d*x + I*c) - 1) - 462*a^2*e^(7*I*d*x + 3*I*c)*log(I*e^(I*d*x) + e^(-I*c)) - 924*a^2*e^(5*I*d
*x + I*c)*log(I*e^(I*d*x) + e^(-I*c)) - 462*a^2*e^(3*I*d*x - I*c)*log(I*e^(I*d*x) + e^(-I*c)) + 462*a^2*e^(7*I
*d*x + 3*I*c)*log(-I*e^(I*d*x) + e^(-I*c)) + 924*a^2*e^(5*I*d*x + I*c)*log(-I*e^(I*d*x) + e^(-I*c)) + 462*a^2*
e^(3*I*d*x - I*c)*log(-I*e^(I*d*x) + e^(-I*c)) + 48*I*a^2*e^(14*I*d*x + 10*I*c) + 432*I*a^2*e^(12*I*d*x + 8*I*
c) + 1840*I*a^2*e^(10*I*d*x + 6*I*c) + 5936*I*a^2*e^(8*I*d*x + 4*I*c) + 6160*I*a^2*e^(6*I*d*x + 2*I*c) - 1904*
I*a^2*e^(2*I*d*x - 2*I*c) - 112*I*a^2*e^(4*I*d*x) - 112*I*a^2*e^(-4*I*c))/(d*e^(7*I*d*x + 3*I*c) + 2*d*e^(5*I*
d*x + I*c) + d*e^(3*I*d*x - I*c))