3.73 \(\int \cos ^5(c+d x) (a+i a \tan (c+d x))^5 \, dx\)

Optimal. Leaf size=32 \[ -\frac {i \cos ^5(c+d x) (a+i a \tan (c+d x))^5}{5 d} \]

[Out]

-1/5*I*cos(d*x+c)^5*(a+I*a*tan(d*x+c))^5/d

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Rubi [A]  time = 0.04, antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.042, Rules used = {3488} \[ -\frac {i \cos ^5(c+d x) (a+i a \tan (c+d x))^5}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^5*(a + I*a*Tan[c + d*x])^5,x]

[Out]

((-I/5)*Cos[c + d*x]^5*(a + I*a*Tan[c + d*x])^5)/d

Rule 3488

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

Rubi steps

\begin {align*} \int \cos ^5(c+d x) (a+i a \tan (c+d x))^5 \, dx &=-\frac {i \cos ^5(c+d x) (a+i a \tan (c+d x))^5}{5 d}\\ \end {align*}

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Mathematica [A]  time = 0.16, size = 31, normalized size = 0.97 \[ -\frac {i a^5 (\cos (c+d x)+i \sin (c+d x))^5}{5 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^5*(a + I*a*Tan[c + d*x])^5,x]

[Out]

((-1/5*I)*a^5*(Cos[c + d*x] + I*Sin[c + d*x])^5)/d

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fricas [A]  time = 0.42, size = 17, normalized size = 0.53 \[ -\frac {i \, a^{5} e^{\left (5 i \, d x + 5 i \, c\right )}}{5 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a+I*a*tan(d*x+c))^5,x, algorithm="fricas")

[Out]

-1/5*I*a^5*e^(5*I*d*x + 5*I*c)/d

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giac [B]  time = 6.63, size = 1669, normalized size = 52.16 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a+I*a*tan(d*x+c))^5,x, algorithm="giac")

[Out]

-1/122880*(34125*a^5*e^(16*I*d*x + 8*I*c)*log(I*e^(I*d*x + I*c) + 1) + 273000*a^5*e^(14*I*d*x + 6*I*c)*log(I*e
^(I*d*x + I*c) + 1) + 955500*a^5*e^(12*I*d*x + 4*I*c)*log(I*e^(I*d*x + I*c) + 1) + 1911000*a^5*e^(10*I*d*x + 2
*I*c)*log(I*e^(I*d*x + I*c) + 1) + 1911000*a^5*e^(6*I*d*x - 2*I*c)*log(I*e^(I*d*x + I*c) + 1) + 955500*a^5*e^(
4*I*d*x - 4*I*c)*log(I*e^(I*d*x + I*c) + 1) + 273000*a^5*e^(2*I*d*x - 6*I*c)*log(I*e^(I*d*x + I*c) + 1) + 2388
750*a^5*e^(8*I*d*x)*log(I*e^(I*d*x + I*c) + 1) + 34125*a^5*e^(-8*I*c)*log(I*e^(I*d*x + I*c) + 1) + 34770*a^5*e
^(16*I*d*x + 8*I*c)*log(I*e^(I*d*x + I*c) - 1) + 278160*a^5*e^(14*I*d*x + 6*I*c)*log(I*e^(I*d*x + I*c) - 1) +
973560*a^5*e^(12*I*d*x + 4*I*c)*log(I*e^(I*d*x + I*c) - 1) + 1947120*a^5*e^(10*I*d*x + 2*I*c)*log(I*e^(I*d*x +
I*c) - 1) + 1947120*a^5*e^(6*I*d*x - 2*I*c)*log(I*e^(I*d*x + I*c) - 1) + 973560*a^5*e^(4*I*d*x - 4*I*c)*log(I
*e^(I*d*x + I*c) - 1) + 278160*a^5*e^(2*I*d*x - 6*I*c)*log(I*e^(I*d*x + I*c) - 1) + 2433900*a^5*e^(8*I*d*x)*lo
g(I*e^(I*d*x + I*c) - 1) + 34770*a^5*e^(-8*I*c)*log(I*e^(I*d*x + I*c) - 1) - 34125*a^5*e^(16*I*d*x + 8*I*c)*lo
g(-I*e^(I*d*x + I*c) + 1) - 273000*a^5*e^(14*I*d*x + 6*I*c)*log(-I*e^(I*d*x + I*c) + 1) - 955500*a^5*e^(12*I*d
*x + 4*I*c)*log(-I*e^(I*d*x + I*c) + 1) - 1911000*a^5*e^(10*I*d*x + 2*I*c)*log(-I*e^(I*d*x + I*c) + 1) - 19110
00*a^5*e^(6*I*d*x - 2*I*c)*log(-I*e^(I*d*x + I*c) + 1) - 955500*a^5*e^(4*I*d*x - 4*I*c)*log(-I*e^(I*d*x + I*c)
+ 1) - 273000*a^5*e^(2*I*d*x - 6*I*c)*log(-I*e^(I*d*x + I*c) + 1) - 2388750*a^5*e^(8*I*d*x)*log(-I*e^(I*d*x +
I*c) + 1) - 34125*a^5*e^(-8*I*c)*log(-I*e^(I*d*x + I*c) + 1) - 34770*a^5*e^(16*I*d*x + 8*I*c)*log(-I*e^(I*d*x
+ I*c) - 1) - 278160*a^5*e^(14*I*d*x + 6*I*c)*log(-I*e^(I*d*x + I*c) - 1) - 973560*a^5*e^(12*I*d*x + 4*I*c)*l
og(-I*e^(I*d*x + I*c) - 1) - 1947120*a^5*e^(10*I*d*x + 2*I*c)*log(-I*e^(I*d*x + I*c) - 1) - 1947120*a^5*e^(6*I
*d*x - 2*I*c)*log(-I*e^(I*d*x + I*c) - 1) - 973560*a^5*e^(4*I*d*x - 4*I*c)*log(-I*e^(I*d*x + I*c) - 1) - 27816
0*a^5*e^(2*I*d*x - 6*I*c)*log(-I*e^(I*d*x + I*c) - 1) - 2433900*a^5*e^(8*I*d*x)*log(-I*e^(I*d*x + I*c) - 1) -
34770*a^5*e^(-8*I*c)*log(-I*e^(I*d*x + I*c) - 1) + 645*a^5*e^(16*I*d*x + 8*I*c)*log(I*e^(I*d*x) + e^(-I*c)) +
5160*a^5*e^(14*I*d*x + 6*I*c)*log(I*e^(I*d*x) + e^(-I*c)) + 18060*a^5*e^(12*I*d*x + 4*I*c)*log(I*e^(I*d*x) + e
^(-I*c)) + 36120*a^5*e^(10*I*d*x + 2*I*c)*log(I*e^(I*d*x) + e^(-I*c)) + 36120*a^5*e^(6*I*d*x - 2*I*c)*log(I*e^
(I*d*x) + e^(-I*c)) + 18060*a^5*e^(4*I*d*x - 4*I*c)*log(I*e^(I*d*x) + e^(-I*c)) + 5160*a^5*e^(2*I*d*x - 6*I*c)
*log(I*e^(I*d*x) + e^(-I*c)) + 45150*a^5*e^(8*I*d*x)*log(I*e^(I*d*x) + e^(-I*c)) + 645*a^5*e^(-8*I*c)*log(I*e^
(I*d*x) + e^(-I*c)) - 645*a^5*e^(16*I*d*x + 8*I*c)*log(-I*e^(I*d*x) + e^(-I*c)) - 5160*a^5*e^(14*I*d*x + 6*I*c
)*log(-I*e^(I*d*x) + e^(-I*c)) - 18060*a^5*e^(12*I*d*x + 4*I*c)*log(-I*e^(I*d*x) + e^(-I*c)) - 36120*a^5*e^(10
*I*d*x + 2*I*c)*log(-I*e^(I*d*x) + e^(-I*c)) - 36120*a^5*e^(6*I*d*x - 2*I*c)*log(-I*e^(I*d*x) + e^(-I*c)) - 18
060*a^5*e^(4*I*d*x - 4*I*c)*log(-I*e^(I*d*x) + e^(-I*c)) - 5160*a^5*e^(2*I*d*x - 6*I*c)*log(-I*e^(I*d*x) + e^(
-I*c)) - 45150*a^5*e^(8*I*d*x)*log(-I*e^(I*d*x) + e^(-I*c)) - 645*a^5*e^(-8*I*c)*log(-I*e^(I*d*x) + e^(-I*c))
+ 24576*I*a^5*e^(21*I*d*x + 13*I*c) + 196608*I*a^5*e^(19*I*d*x + 11*I*c) + 688128*I*a^5*e^(17*I*d*x + 9*I*c) +
1376256*I*a^5*e^(15*I*d*x + 7*I*c) + 1720320*I*a^5*e^(13*I*d*x + 5*I*c) + 1376256*I*a^5*e^(11*I*d*x + 3*I*c)
+ 688128*I*a^5*e^(9*I*d*x + I*c) + 196608*I*a^5*e^(7*I*d*x - I*c) + 24576*I*a^5*e^(5*I*d*x - 3*I*c))/(d*e^(16*
I*d*x + 8*I*c) + 8*d*e^(14*I*d*x + 6*I*c) + 28*d*e^(12*I*d*x + 4*I*c) + 56*d*e^(10*I*d*x + 2*I*c) + 56*d*e^(6*
I*d*x - 2*I*c) + 28*d*e^(4*I*d*x - 4*I*c) + 8*d*e^(2*I*d*x - 6*I*c) + 70*d*e^(8*I*d*x) + d*e^(-8*I*c))

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maple [B]  time = 0.54, size = 170, normalized size = 5.31 \[ \frac {-\frac {i a^{5} \left (\frac {8}{3}+\sin ^{4}\left (d x +c \right )+\frac {4 \left (\sin ^{2}\left (d x +c \right )\right )}{3}\right ) \cos \left (d x +c \right )}{5}+a^{5} \left (\sin ^{5}\left (d x +c \right )\right )-10 i a^{5} \left (-\frac {\left (\cos ^{3}\left (d x +c \right )\right ) \left (\sin ^{2}\left (d x +c \right )\right )}{5}-\frac {2 \left (\cos ^{3}\left (d x +c \right )\right )}{15}\right )-10 a^{5} \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{4}\left (d x +c \right )\right )}{5}+\frac {\left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{15}\right )-i a^{5} \left (\cos ^{5}\left (d x +c \right )\right )+\frac {a^{5} \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^5*(a+I*a*tan(d*x+c))^5,x)

[Out]

1/d*(-1/5*I*a^5*(8/3+sin(d*x+c)^4+4/3*sin(d*x+c)^2)*cos(d*x+c)+a^5*sin(d*x+c)^5-10*I*a^5*(-1/5*cos(d*x+c)^3*si
n(d*x+c)^2-2/15*cos(d*x+c)^3)-10*a^5*(-1/5*sin(d*x+c)*cos(d*x+c)^4+1/15*(2+cos(d*x+c)^2)*sin(d*x+c))-I*a^5*cos
(d*x+c)^5+1/5*a^5*(8/3+cos(d*x+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c))

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maxima [B]  time = 0.40, size = 152, normalized size = 4.75 \[ -\frac {15 i \, a^{5} \cos \left (d x + c\right )^{5} - 15 \, a^{5} \sin \left (d x + c\right )^{5} + 10 i \, {\left (3 \, \cos \left (d x + c\right )^{5} - 5 \, \cos \left (d x + c\right )^{3}\right )} a^{5} + i \, {\left (3 \, \cos \left (d x + c\right )^{5} - 10 \, \cos \left (d x + c\right )^{3} + 15 \, \cos \left (d x + c\right )\right )} a^{5} - 10 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 5 \, \sin \left (d x + c\right )^{3}\right )} a^{5} - {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} a^{5}}{15 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a+I*a*tan(d*x+c))^5,x, algorithm="maxima")

[Out]

-1/15*(15*I*a^5*cos(d*x + c)^5 - 15*a^5*sin(d*x + c)^5 + 10*I*(3*cos(d*x + c)^5 - 5*cos(d*x + c)^3)*a^5 + I*(3
*cos(d*x + c)^5 - 10*cos(d*x + c)^3 + 15*cos(d*x + c))*a^5 - 10*(3*sin(d*x + c)^5 - 5*sin(d*x + c)^3)*a^5 - (3
*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*a^5)/d

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mupad [B]  time = 3.45, size = 104, normalized size = 3.25 \[ \frac {2\,a^5\,\left (5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{5\,d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,5{}\mathrm {i}-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,10{}\mathrm {i}+5\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1{}\mathrm {i}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^5*(a + a*tan(c + d*x)*1i)^5,x)

[Out]

(2*a^5*(5*tan(c/2 + (d*x)/2)^4 - 10*tan(c/2 + (d*x)/2)^2 + 1))/(5*d*(5*tan(c/2 + (d*x)/2) - tan(c/2 + (d*x)/2)
^2*10i - 10*tan(c/2 + (d*x)/2)^3 + tan(c/2 + (d*x)/2)^4*5i + tan(c/2 + (d*x)/2)^5 + 1i))

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sympy [A]  time = 0.39, size = 37, normalized size = 1.16 \[ \begin {cases} - \frac {i a^{5} e^{5 i c} e^{5 i d x}}{5 d} & \text {for}\: 5 d \neq 0 \\a^{5} x e^{5 i c} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**5*(a+I*a*tan(d*x+c))**5,x)

[Out]

Piecewise((-I*a**5*exp(5*I*c)*exp(5*I*d*x)/(5*d), Ne(5*d, 0)), (a**5*x*exp(5*I*c), True))

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