3.16 \(\int \cos ^3(c+d x) (a+i a \tan (c+d x)) \, dx\)

Optimal. Leaf size=46 \[ -\frac {a \sin ^3(c+d x)}{3 d}+\frac {a \sin (c+d x)}{d}-\frac {i a \cos ^3(c+d x)}{3 d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.03, antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {3486, 2633} \[ -\frac {a \sin ^3(c+d x)}{3 d}+\frac {a \sin (c+d x)}{d}-\frac {i a \cos ^3(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^3*(a + I*a*Tan[c + d*x]),x]

[Out]

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

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 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])

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.01, size = 46, normalized size = 1.00 \[ -\frac {a \sin ^3(c+d x)}{3 d}+\frac {a \sin (c+d x)}{d}-\frac {i a \cos ^3(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^3*(a + I*a*Tan[c + d*x]),x]

[Out]

((-1/3*I)*a*Cos[c + d*x]^3)/d + (a*Sin[c + d*x])/d - (a*Sin[c + d*x]^3)/(3*d)

________________________________________________________________________________________

fricas [A]  time = 0.41, size = 42, normalized size = 0.91 \[ \frac {{\left (-i \, a e^{\left (4 i \, d x + 4 i \, c\right )} - 6 i \, a e^{\left (2 i \, d x + 2 i \, c\right )} + 3 i \, a\right )} e^{\left (-i \, d x - i \, c\right )}}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(a+I*a*tan(d*x+c)),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

giac [B]  time = 1.18, size = 196, normalized size = 4.26 \[ -\frac {{\left (9 \, a e^{\left (i \, d x + i \, c\right )} \log \left (i \, e^{\left (i \, d x + i \, c\right )} + 1\right ) + 6 \, a e^{\left (i \, d x + i \, c\right )} \log \left (i \, e^{\left (i \, d x + i \, c\right )} - 1\right ) - 9 \, a e^{\left (i \, d x + i \, c\right )} \log \left (-i \, e^{\left (i \, d x + i \, c\right )} + 1\right ) - 6 \, a e^{\left (i \, d x + i \, c\right )} \log \left (-i \, e^{\left (i \, d x + i \, c\right )} - 1\right ) - 3 \, a e^{\left (i \, d x + i \, c\right )} \log \left (i \, e^{\left (i \, d x\right )} + e^{\left (-i \, c\right )}\right ) + 3 \, a e^{\left (i \, d x + i \, c\right )} \log \left (-i \, e^{\left (i \, d x\right )} + e^{\left (-i \, c\right )}\right ) + 4 i \, a e^{\left (4 i \, d x + 4 i \, c\right )} + 24 i \, a e^{\left (2 i \, d x + 2 i \, c\right )} - 12 i \, a\right )} e^{\left (-i \, d x - i \, c\right )}}{48 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(a+I*a*tan(d*x+c)),x, algorithm="giac")

[Out]

-1/48*(9*a*e^(I*d*x + I*c)*log(I*e^(I*d*x + I*c) + 1) + 6*a*e^(I*d*x + I*c)*log(I*e^(I*d*x + I*c) - 1) - 9*a*e
^(I*d*x + I*c)*log(-I*e^(I*d*x + I*c) + 1) - 6*a*e^(I*d*x + I*c)*log(-I*e^(I*d*x + I*c) - 1) - 3*a*e^(I*d*x +
I*c)*log(I*e^(I*d*x) + e^(-I*c)) + 3*a*e^(I*d*x + I*c)*log(-I*e^(I*d*x) + e^(-I*c)) + 4*I*a*e^(4*I*d*x + 4*I*c
) + 24*I*a*e^(2*I*d*x + 2*I*c) - 12*I*a)*e^(-I*d*x - I*c)/d

________________________________________________________________________________________

maple [A]  time = 0.41, size = 37, normalized size = 0.80 \[ \frac {-\frac {i a \left (\cos ^{3}\left (d x +c \right )\right )}{3}+\frac {a \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^3*(a+I*a*tan(d*x+c)),x)

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.32, size = 36, normalized size = 0.78 \[ -\frac {i \, a \cos \left (d x + c\right )^{3} + {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} a}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(a+I*a*tan(d*x+c)),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 3.41, size = 54, normalized size = 1.17 \[ \frac {2\,a\,\left (-\frac {{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,3{}\mathrm {i}}{4}-\frac {{\cos \left (\frac {3\,c}{2}+\frac {3\,d\,x}{2}\right )}^2\,1{}\mathrm {i}}{4}+\frac {9\,\sin \left (c+d\,x\right )}{8}+\frac {\sin \left (3\,c+3\,d\,x\right )}{8}\right )}{3\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^3*(a + a*tan(c + d*x)*1i),x)

[Out]

(2*a*((9*sin(c + d*x))/8 + sin(3*c + 3*d*x)/8 - (cos(c/2 + (d*x)/2)^2*3i)/4 - (cos((3*c)/2 + (3*d*x)/2)^2*1i)/
4))/(3*d)

________________________________________________________________________________________

sympy [A]  time = 0.30, size = 109, normalized size = 2.37 \[ \begin {cases} - \frac {\left (8 i a d^{2} e^{4 i c} e^{3 i d x} + 48 i a d^{2} e^{2 i c} e^{i d x} - 24 i a d^{2} e^{- i d x}\right ) e^{- i c}}{96 d^{3}} & \text {for}\: 96 d^{3} e^{i c} \neq 0 \\\frac {x \left (a e^{4 i c} + 2 a e^{2 i c} + a\right ) e^{- i c}}{4} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**3*(a+I*a*tan(d*x+c)),x)

[Out]

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

________________________________________________________________________________________