3.111 \(\int \frac {\cos (c+d x)}{a+i a \tan (c+d x)} \, dx\)

Optimal. Leaf size=47 \[ \frac {2 \sin (c+d x)}{3 a d}+\frac {i \cos (c+d x)}{3 d (a+i a \tan (c+d x))} \]

[Out]

2/3*sin(d*x+c)/a/d+1/3*I*cos(d*x+c)/d/(a+I*a*tan(d*x+c))

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Rubi [A]  time = 0.04, antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {3502, 2637} \[ \frac {2 \sin (c+d x)}{3 a d}+\frac {i \cos (c+d x)}{3 d (a+i a \tan (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sin[c + d*x])/(3*a*d) + ((I/3)*Cos[c + d*x])/(d*(a + I*a*Tan[c + d*x]))

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3502

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(a*(d
*Sec[e + f*x])^m*(a + b*Tan[e + f*x])^n)/(b*f*(m + 2*n)), x] + Dist[Simplify[m + n]/(a*(m + 2*n)), Int[(d*Sec[
e + f*x])^m*(a + b*Tan[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, m}, x] && EqQ[a^2 + b^2, 0] && LtQ[n
, 0] && NeQ[m + 2*n, 0] && IntegersQ[2*m, 2*n]

Rubi steps

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

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Mathematica [A]  time = 0.13, size = 50, normalized size = 1.06 \[ -\frac {\sec (c+d x) (2 i \sin (2 (c+d x))+\cos (2 (c+d x))-3)}{6 a d (\tan (c+d x)-i)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/6*(Sec[c + d*x]*(-3 + Cos[2*(c + d*x)] + (2*I)*Sin[2*(c + d*x)]))/(a*d*(-I + Tan[c + d*x]))

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fricas [A]  time = 0.52, size = 41, normalized size = 0.87 \[ \frac {{\left (-3 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 6 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i\right )} e^{\left (-3 i \, d x - 3 i \, c\right )}}{12 \, a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(-3*I*e^(4*I*d*x + 4*I*c) + 6*I*e^(2*I*d*x + 2*I*c) + I)*e^(-3*I*d*x - 3*I*c)/(a*d)

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giac [A]  time = 0.65, size = 67, normalized size = 1.43 \[ \frac {\frac {3}{a {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + i\right )}} + \frac {9 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 12 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 7}{a {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - i\right )}^{3}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(3/(a*(tan(1/2*d*x + 1/2*c) + I)) + (9*tan(1/2*d*x + 1/2*c)^2 - 12*I*tan(1/2*d*x + 1/2*c) - 7)/(a*(tan(1/2
*d*x + 1/2*c) - I)^3))/d

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maple [A]  time = 0.44, size = 75, normalized size = 1.60 \[ \frac {\frac {2}{4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+4 i}-\frac {2}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{3}}+\frac {i}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{2}}+\frac {3}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )}}{d a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: expt: undefined: 0 to a negative e
xponent.

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mupad [B]  time = 3.55, size = 78, normalized size = 1.66 \[ \frac {\left (-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,3{}\mathrm {i}+\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1{}\mathrm {i}\right )\,2{}\mathrm {i}}{3\,a\,d\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1{}\mathrm {i}\right )\,{\left (1+\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.30, size = 129, normalized size = 2.74 \[ \begin {cases} - \frac {\left (24 i a^{2} d^{2} e^{5 i c} e^{i d x} - 48 i a^{2} d^{2} e^{3 i c} e^{- i d x} - 8 i a^{2} d^{2} e^{i c} e^{- 3 i d x}\right ) e^{- 4 i c}}{96 a^{3} d^{3}} & \text {for}\: 96 a^{3} d^{3} e^{4 i c} \neq 0 \\\frac {x \left (e^{4 i c} + 2 e^{2 i c} + 1\right ) e^{- 3 i c}}{4 a} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-(24*I*a**2*d**2*exp(5*I*c)*exp(I*d*x) - 48*I*a**2*d**2*exp(3*I*c)*exp(-I*d*x) - 8*I*a**2*d**2*exp(
I*c)*exp(-3*I*d*x))*exp(-4*I*c)/(96*a**3*d**3), Ne(96*a**3*d**3*exp(4*I*c), 0)), (x*(exp(4*I*c) + 2*exp(2*I*c)
+ 1)*exp(-3*I*c)/(4*a), True))

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