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

Optimal. Leaf size=67 \[ -\frac{i a \cos ^4(c+d x)}{4 d}+\frac{a \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac{3 a \sin (c+d x) \cos (c+d x)}{8 d}+\frac{3 a x}{8} \]

[Out]

(3*a*x)/8 - ((I/4)*a*Cos[c + d*x]^4)/d + (3*a*Cos[c + d*x]*Sin[c + d*x])/(8*d) + (a*Cos[c + d*x]^3*Sin[c + d*x
])/(4*d)

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Rubi [A]  time = 0.03995, antiderivative size = 67, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {3486, 2635, 8} \[ -\frac{i a \cos ^4(c+d x)}{4 d}+\frac{a \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac{3 a \sin (c+d x) \cos (c+d x)}{8 d}+\frac{3 a x}{8} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*a*x)/8 - ((I/4)*a*Cos[c + d*x]^4)/d + (3*a*Cos[c + d*x]*Sin[c + d*x])/(8*d) + (a*Cos[c + d*x]^3*Sin[c + d*x
])/(4*d)

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 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int \cos ^4(c+d x) (a+i a \tan (c+d x)) \, dx &=-\frac{i a \cos ^4(c+d x)}{4 d}+a \int \cos ^4(c+d x) \, dx\\ &=-\frac{i a \cos ^4(c+d x)}{4 d}+\frac{a \cos ^3(c+d x) \sin (c+d x)}{4 d}+\frac{1}{4} (3 a) \int \cos ^2(c+d x) \, dx\\ &=-\frac{i a \cos ^4(c+d x)}{4 d}+\frac{3 a \cos (c+d x) \sin (c+d x)}{8 d}+\frac{a \cos ^3(c+d x) \sin (c+d x)}{4 d}+\frac{1}{8} (3 a) \int 1 \, dx\\ &=\frac{3 a x}{8}-\frac{i a \cos ^4(c+d x)}{4 d}+\frac{3 a \cos (c+d x) \sin (c+d x)}{8 d}+\frac{a \cos ^3(c+d x) \sin (c+d x)}{4 d}\\ \end{align*}

Mathematica [A]  time = 0.0422158, size = 46, normalized size = 0.69 \[ \frac{a \left (8 \sin (2 (c+d x))+\sin (4 (c+d x))-8 i \cos ^4(c+d x)+12 c+12 d x\right )}{32 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(12*c + 12*d*x - (8*I)*Cos[c + d*x]^4 + 8*Sin[2*(c + d*x)] + Sin[4*(c + d*x)]))/(32*d)

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Maple [A]  time = 0.084, size = 53, normalized size = 0.8 \begin{align*}{\frac{1}{d} \left ( -{\frac{i}{4}}a \left ( \cos \left ( dx+c \right ) \right ) ^{4}+a \left ({\frac{\sin \left ( dx+c \right ) }{4} \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{3}+{\frac{3\,\cos \left ( dx+c \right ) }{2}} \right ) }+{\frac{3\,dx}{8}}+{\frac{3\,c}{8}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.6623, size = 82, normalized size = 1.22 \begin{align*} \frac{3 \,{\left (d x + c\right )} a + \frac{3 \, a \tan \left (d x + c\right )^{3} + 5 \, a \tan \left (d x + c\right ) - 2 i \, a}{\tan \left (d x + c\right )^{4} + 2 \, \tan \left (d x + c\right )^{2} + 1}}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(3*(d*x + c)*a + (3*a*tan(d*x + c)^3 + 5*a*tan(d*x + c) - 2*I*a)/(tan(d*x + c)^4 + 2*tan(d*x + c)^2 + 1))/
d

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Fricas [A]  time = 1.13684, size = 165, normalized size = 2.46 \begin{align*} \frac{{\left (12 \, a d x e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a e^{\left (6 i \, d x + 6 i \, c\right )} - 6 i \, a e^{\left (4 i \, d x + 4 i \, c\right )} + 2 i \, a\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{32 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*(12*a*d*x*e^(2*I*d*x + 2*I*c) - I*a*e^(6*I*d*x + 6*I*c) - 6*I*a*e^(4*I*d*x + 4*I*c) + 2*I*a)*e^(-2*I*d*x
- 2*I*c)/d

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Sympy [A]  time = 0.4609, size = 138, normalized size = 2.06 \begin{align*} \frac{3 a x}{8} + \begin{cases} \frac{\left (- 256 i a d^{2} e^{6 i c} e^{4 i d x} - 1536 i a d^{2} e^{4 i c} e^{2 i d x} + 512 i a d^{2} e^{- 2 i d x}\right ) e^{- 2 i c}}{8192 d^{3}} & \text{for}\: 8192 d^{3} e^{2 i c} \neq 0 \\x \left (- \frac{3 a}{8} + \frac{\left (a e^{6 i c} + 3 a e^{4 i c} + 3 a e^{2 i c} + a\right ) e^{- 2 i c}}{8}\right ) & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*a*x/8 + Piecewise(((-256*I*a*d**2*exp(6*I*c)*exp(4*I*d*x) - 1536*I*a*d**2*exp(4*I*c)*exp(2*I*d*x) + 512*I*a*
d**2*exp(-2*I*d*x))*exp(-2*I*c)/(8192*d**3), Ne(8192*d**3*exp(2*I*c), 0)), (x*(-3*a/8 + (a*exp(6*I*c) + 3*a*ex
p(4*I*c) + 3*a*exp(2*I*c) + a)*exp(-2*I*c)/8), True))

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Giac [A]  time = 1.14142, size = 139, normalized size = 2.07 \begin{align*} \frac{{\left (12 \, a d x e^{\left (2 i \, d x + 2 i \, c\right )} + i \, a e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - i \, a e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (e^{\left (2 i \, d x\right )} + e^{\left (-2 i \, c\right )}\right ) - i \, a e^{\left (6 i \, d x + 6 i \, c\right )} - 6 i \, a e^{\left (4 i \, d x + 4 i \, c\right )} + 2 i \, a\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{32 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*(12*a*d*x*e^(2*I*d*x + 2*I*c) + I*a*e^(2*I*d*x + 2*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) - I*a*e^(2*I*d*x + 2
*I*c)*log(e^(2*I*d*x) + e^(-2*I*c)) - I*a*e^(6*I*d*x + 6*I*c) - 6*I*a*e^(4*I*d*x + 4*I*c) + 2*I*a)*e^(-2*I*d*x
 - 2*I*c)/d