3.56 \(\int \cos ^5(c+d x) (a+i a \tan (c+d x))^4 \, dx\)

Optimal. Leaf size=66 \[ -\frac {i \cos ^5(c+d x) (a+i a \tan (c+d x))^4}{5 d}-\frac {i a \cos ^3(c+d x) (a+i a \tan (c+d x))^3}{15 d} \]

[Out]

-1/15*I*a*cos(d*x+c)^3*(a+I*a*tan(d*x+c))^3/d-1/5*I*cos(d*x+c)^5*(a+I*a*tan(d*x+c))^4/d

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Rubi [A]  time = 0.07, antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {3497, 3488} \[ -\frac {i \cos ^5(c+d x) (a+i a \tan (c+d x))^4}{5 d}-\frac {i a \cos ^3(c+d x) (a+i a \tan (c+d x))^3}{15 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^5*(a + I*a*Tan[c + d*x])^4,x]

[Out]

((-I/15)*a*Cos[c + d*x]^3*(a + I*a*Tan[c + d*x])^3)/d - ((I/5)*Cos[c + d*x]^5*(a + I*a*Tan[c + d*x])^4)/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]

Rule 3497

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] + Dist[(a*(m + n))/(m*d^2), Int[(d*Sec[e + f*x])^(m + 2)*(
a + b*Tan[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 + b^2, 0] && GtQ[n, 0] && LtQ[m, -
1] && IntegersQ[2*m, 2*n]

Rubi steps

\begin {align*} \int \cos ^5(c+d x) (a+i a \tan (c+d x))^4 \, dx &=-\frac {i \cos ^5(c+d x) (a+i a \tan (c+d x))^4}{5 d}+\frac {1}{5} a \int \cos ^3(c+d x) (a+i a \tan (c+d x))^3 \, dx\\ &=-\frac {i a \cos ^3(c+d x) (a+i a \tan (c+d x))^3}{15 d}-\frac {i \cos ^5(c+d x) (a+i a \tan (c+d x))^4}{5 d}\\ \end {align*}

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Mathematica [A]  time = 0.41, size = 50, normalized size = 0.76 \[ \frac {a^4 (4 \cos (c+d x)-i \sin (c+d x)) (\sin (4 (c+d x))-i \cos (4 (c+d x)))}{15 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^5*(a + I*a*Tan[c + d*x])^4,x]

[Out]

(a^4*(4*Cos[c + d*x] - I*Sin[c + d*x])*((-I)*Cos[4*(c + d*x)] + Sin[4*(c + d*x)]))/(15*d)

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fricas [A]  time = 0.62, size = 34, normalized size = 0.52 \[ \frac {-3 i \, a^{4} e^{\left (5 i \, d x + 5 i \, c\right )} - 5 i \, a^{4} e^{\left (3 i \, d x + 3 i \, c\right )}}{30 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a+I*a*tan(d*x+c))^4,x, algorithm="fricas")

[Out]

1/30*(-3*I*a^4*e^(5*I*d*x + 5*I*c) - 5*I*a^4*e^(3*I*d*x + 3*I*c))/d

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giac [B]  time = 4.03, size = 915, normalized size = 13.86 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a+I*a*tan(d*x+c))^4,x, algorithm="giac")

[Out]

1/7680*(9075*a^4*e^(8*I*d*x + 4*I*c)*log(I*e^(I*d*x + I*c) + 1) + 36300*a^4*e^(6*I*d*x + 2*I*c)*log(I*e^(I*d*x
+ I*c) + 1) + 36300*a^4*e^(2*I*d*x - 2*I*c)*log(I*e^(I*d*x + I*c) + 1) + 54450*a^4*e^(4*I*d*x)*log(I*e^(I*d*x
+ I*c) + 1) + 9075*a^4*e^(-4*I*c)*log(I*e^(I*d*x + I*c) + 1) + 9000*a^4*e^(8*I*d*x + 4*I*c)*log(I*e^(I*d*x +
I*c) - 1) + 36000*a^4*e^(6*I*d*x + 2*I*c)*log(I*e^(I*d*x + I*c) - 1) + 36000*a^4*e^(2*I*d*x - 2*I*c)*log(I*e^(
I*d*x + I*c) - 1) + 54000*a^4*e^(4*I*d*x)*log(I*e^(I*d*x + I*c) - 1) + 9000*a^4*e^(-4*I*c)*log(I*e^(I*d*x + I*
c) - 1) - 9075*a^4*e^(8*I*d*x + 4*I*c)*log(-I*e^(I*d*x + I*c) + 1) - 36300*a^4*e^(6*I*d*x + 2*I*c)*log(-I*e^(I
*d*x + I*c) + 1) - 36300*a^4*e^(2*I*d*x - 2*I*c)*log(-I*e^(I*d*x + I*c) + 1) - 54450*a^4*e^(4*I*d*x)*log(-I*e^
(I*d*x + I*c) + 1) - 9075*a^4*e^(-4*I*c)*log(-I*e^(I*d*x + I*c) + 1) - 9000*a^4*e^(8*I*d*x + 4*I*c)*log(-I*e^(
I*d*x + I*c) - 1) - 36000*a^4*e^(6*I*d*x + 2*I*c)*log(-I*e^(I*d*x + I*c) - 1) - 36000*a^4*e^(2*I*d*x - 2*I*c)*
log(-I*e^(I*d*x + I*c) - 1) - 54000*a^4*e^(4*I*d*x)*log(-I*e^(I*d*x + I*c) - 1) - 9000*a^4*e^(-4*I*c)*log(-I*e
^(I*d*x + I*c) - 1) - 75*a^4*e^(8*I*d*x + 4*I*c)*log(I*e^(I*d*x) + e^(-I*c)) - 300*a^4*e^(6*I*d*x + 2*I*c)*log
(I*e^(I*d*x) + e^(-I*c)) - 300*a^4*e^(2*I*d*x - 2*I*c)*log(I*e^(I*d*x) + e^(-I*c)) - 450*a^4*e^(4*I*d*x)*log(I
*e^(I*d*x) + e^(-I*c)) - 75*a^4*e^(-4*I*c)*log(I*e^(I*d*x) + e^(-I*c)) + 75*a^4*e^(8*I*d*x + 4*I*c)*log(-I*e^(
I*d*x) + e^(-I*c)) + 300*a^4*e^(6*I*d*x + 2*I*c)*log(-I*e^(I*d*x) + e^(-I*c)) + 300*a^4*e^(2*I*d*x - 2*I*c)*lo
g(-I*e^(I*d*x) + e^(-I*c)) + 450*a^4*e^(4*I*d*x)*log(-I*e^(I*d*x) + e^(-I*c)) + 75*a^4*e^(-4*I*c)*log(-I*e^(I*
d*x) + e^(-I*c)) - 768*I*a^4*e^(13*I*d*x + 9*I*c) - 4352*I*a^4*e^(11*I*d*x + 7*I*c) - 9728*I*a^4*e^(9*I*d*x +
5*I*c) - 10752*I*a^4*e^(7*I*d*x + 3*I*c) - 5888*I*a^4*e^(5*I*d*x + I*c) - 1280*I*a^4*e^(3*I*d*x - I*c))/(d*e^(
8*I*d*x + 4*I*c) + 4*d*e^(6*I*d*x + 2*I*c) + 4*d*e^(2*I*d*x - 2*I*c) + 6*d*e^(4*I*d*x) + d*e^(-4*I*c))

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maple [B]  time = 0.51, size = 139, normalized size = 2.11 \[ \frac {\frac {a^{4} \left (\sin ^{5}\left (d x +c \right )\right )}{5}-4 i a^{4} \left (-\frac {\left (\cos ^{3}\left (d x +c \right )\right ) \left (\sin ^{2}\left (d x +c \right )\right )}{5}-\frac {2 \left (\cos ^{3}\left (d x +c \right )\right )}{15}\right )-6 a^{4} \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{4}\left (d x +c \right )\right )}{5}+\frac {\left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{15}\right )-\frac {4 i a^{4} \left (\cos ^{5}\left (d x +c \right )\right )}{5}+\frac {a^{4} \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^5*(a+I*a*tan(d*x+c))^4,x)

[Out]

1/d*(1/5*a^4*sin(d*x+c)^5-4*I*a^4*(-1/5*cos(d*x+c)^3*sin(d*x+c)^2-2/15*cos(d*x+c)^3)-6*a^4*(-1/5*sin(d*x+c)*co
s(d*x+c)^4+1/15*(2+cos(d*x+c)^2)*sin(d*x+c))-4/5*I*a^4*cos(d*x+c)^5+1/5*a^4*(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 = 118, normalized size = 1.79 \[ -\frac {12 i \, a^{4} \cos \left (d x + c\right )^{5} - 3 \, a^{4} \sin \left (d x + c\right )^{5} + 4 i \, {\left (3 \, \cos \left (d x + c\right )^{5} - 5 \, \cos \left (d x + c\right )^{3}\right )} a^{4} - 6 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 5 \, \sin \left (d x + c\right )^{3}\right )} a^{4} - {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} a^{4}}{15 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a+I*a*tan(d*x+c))^4,x, algorithm="maxima")

[Out]

-1/15*(12*I*a^4*cos(d*x + c)^5 - 3*a^4*sin(d*x + c)^5 + 4*I*(3*cos(d*x + c)^5 - 5*cos(d*x + c)^3)*a^4 - 6*(3*s
in(d*x + c)^5 - 5*sin(d*x + c)^3)*a^4 - (3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*a^4)/d

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mupad [B]  time = 3.55, size = 130, normalized size = 1.97 \[ \frac {2\,a^4\,\left (15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,15{}\mathrm {i}-25\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,5{}\mathrm {i}+4\right )}{15\,d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,5{}\mathrm {i}-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,10{}\mathrm {i}+5\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1{}\mathrm {i}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^5*(a + a*tan(c + d*x)*1i)^4,x)

[Out]

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

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sympy [A]  time = 0.42, size = 82, normalized size = 1.24 \[ \begin {cases} \frac {- 6 i a^{4} d e^{5 i c} e^{5 i d x} - 10 i a^{4} d e^{3 i c} e^{3 i d x}}{60 d^{2}} & \text {for}\: 60 d^{2} \neq 0 \\x \left (\frac {a^{4} e^{5 i c}}{2} + \frac {a^{4} e^{3 i c}}{2}\right ) & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**5*(a+I*a*tan(d*x+c))**4,x)

[Out]

Piecewise(((-6*I*a**4*d*exp(5*I*c)*exp(5*I*d*x) - 10*I*a**4*d*exp(3*I*c)*exp(3*I*d*x))/(60*d**2), Ne(60*d**2,
0)), (x*(a**4*exp(5*I*c)/2 + a**4*exp(3*I*c)/2), True))

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