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]

((-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

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Rubi [A]  time = 0.0736642, antiderivative size = 66, normalized size of antiderivative = 1., 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 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]

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))^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*}

Mathematica [A]  time = 0.355633, 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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Maple [B]  time = 0.063, size = 139, normalized size = 2.1 \begin{align*}{\frac{1}{d} \left ({\frac{{a}^{4} \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{5}}-4\,i{a}^{4} \left ( -{\frac{ \left ( \cos \left ( dx+c \right ) \right ) ^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{5}}-{\frac{2\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{15}} \right ) -6\,{a}^{4} \left ( -1/5\,\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{4}+1/15\, \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) \right ) -{\frac{4\,i}{5}}{a}^{4} \left ( \cos \left ( dx+c \right ) \right ) ^{5}+{\frac{{a}^{4}\sin \left ( dx+c \right ) }{5} \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 ) } \end{align*}

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 = 1.14519, size = 159, normalized size = 2.41 \begin{align*} -\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} \end{align*}

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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Fricas [A]  time = 1.00192, size = 93, normalized size = 1.41 \begin{align*} \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} \end{align*}

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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Sympy [A]  time = 0.652151, size = 82, normalized size = 1.24 \begin{align*} \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} \end{align*}

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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Giac [B]  time = 1.6316, size = 1235, normalized size = 18.71 \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)^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))