3.153 \(\int \frac {\cos ^2(c+d x)}{a \cos (c+d x)+i a \sin (c+d x)} \, dx\)

Optimal. Leaf size=52 \[ -\frac {\sin ^3(c+d x)}{3 a d}+\frac {\sin (c+d x)}{a d}+\frac {i \cos ^3(c+d x)}{3 a d} \]

[Out]

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

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Rubi [A]  time = 0.12, antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {3092, 3090, 2633, 2565, 30} \[ -\frac {\sin ^3(c+d x)}{3 a d}+\frac {\sin (c+d x)}{a d}+\frac {i \cos ^3(c+d x)}{3 a d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^2/(a*Cos[c + d*x] + I*a*Sin[c + d*x]),x]

[Out]

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2565

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[(a*f)^(-1), Subst[
Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2]
 &&  !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, n])

Rule 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rule 3090

Int[cos[(c_.) + (d_.)*(x_)]^(m_.)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_.), x_Sym
bol] :> Int[ExpandTrig[cos[c + d*x]^m*(a*cos[c + d*x] + b*sin[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d}, x] &&
 IntegerQ[m] && IGtQ[n, 0]

Rule 3092

Int[cos[(c_.) + (d_.)*(x_)]^(m_.)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symb
ol] :> Dist[a^n*b^n, Int[Cos[c + d*x]^m/(b*Cos[c + d*x] + a*Sin[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d, m},
x] && EqQ[a^2 + b^2, 0] && ILtQ[n, 0]

Rubi steps

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

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Mathematica [A]  time = 0.07, size = 73, normalized size = 1.40 \[ \frac {3 \sin (c+d x)}{4 a d}+\frac {\sin (3 (c+d x))}{12 a d}+\frac {i \cos (c+d x)}{4 a d}+\frac {i \cos (3 (c+d x))}{12 a d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^2/(a*Cos[c + d*x] + I*a*Sin[c + d*x]),x]

[Out]

((I/4)*Cos[c + d*x])/(a*d) + ((I/12)*Cos[3*(c + d*x)])/(a*d) + (3*Sin[c + d*x])/(4*a*d) + Sin[3*(c + d*x)]/(12
*a*d)

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fricas [A]  time = 0.49, size = 41, normalized size = 0.79 \[ \frac {{\left (-3 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 6 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i\right )} e^{\left (-3 i \, d x - 3 i \, c\right )}}{12 \, a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2/(a*cos(d*x+c)+I*a*sin(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 = 1.80, size = 67, normalized size = 1.29 \[ \frac {\frac {3}{a {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + i\right )}} + \frac {9 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 12 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 7}{a {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - i\right )}^{3}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2/(a*cos(d*x+c)+I*a*sin(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 = 0.16, size = 75, normalized size = 1.44 \[ \frac {\frac {2}{4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+4 i}-\frac {2}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{3}}+\frac {i}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{2}}+\frac {3}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )}}{a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2/(a*cos(d*x+c)+I*a*sin(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)^2/(a*cos(d*x+c)+I*a*sin(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 = 0.77, size = 78, normalized size = 1.50 \[ \frac {\left (-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,3{}\mathrm {i}+\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1{}\mathrm {i}\right )\,2{}\mathrm {i}}{3\,a\,d\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1{}\mathrm {i}\right )\,{\left (1+\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^2/(a*cos(c + d*x) + a*sin(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 = 0.48, size = 129, normalized size = 2.48 \[ \begin {cases} - \frac {\left (24 i a^{2} d^{2} e^{5 i c} e^{i d x} - 48 i a^{2} d^{2} e^{3 i c} e^{- i d x} - 8 i a^{2} d^{2} e^{i c} e^{- 3 i d x}\right ) e^{- 4 i c}}{96 a^{3} d^{3}} & \text {for}\: 96 a^{3} d^{3} e^{4 i c} \neq 0 \\\frac {x \left (e^{4 i c} + 2 e^{2 i c} + 1\right ) e^{- 3 i c}}{4 a} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**2/(a*cos(d*x+c)+I*a*sin(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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