3.1.33 \(\int \cos ^5(c+d x) (a+i a \tan (c+d x))^2 \, dx\) [33]

Optimal. Leaf size=69 \[ \frac {3 a^2 \sin (c+d x)}{5 d}-\frac {a^2 \sin ^3(c+d x)}{5 d}-\frac {2 i \cos ^5(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{5 d} \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {3577, 2713} \begin {gather*} -\frac {a^2 \sin ^3(c+d x)}{5 d}+\frac {3 a^2 \sin (c+d x)}{5 d}-\frac {2 i \cos ^5(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{5 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^5*(a + I*a*Tan[c + d*x])^2,x]

[Out]

(3*a^2*Sin[c + d*x])/(5*d) - (a^2*Sin[c + d*x]^3)/(5*d) - (((2*I)/5)*Cos[c + d*x]^5*(a^2 + I*a^2*Tan[c + d*x])
)/d

Rule 2713

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 3577

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[2*b*(d
*Sec[e + f*x])^m*((a + b*Tan[e + f*x])^(n - 1)/(f*m)), x] - Dist[b^2*((m + 2*n - 2)/(d^2*m)), Int[(d*Sec[e + f
*x])^(m + 2)*(a + b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 + b^2, 0] && GtQ[n,
1] && ((IGtQ[n/2, 0] && ILtQ[m - 1/2, 0]) || (EqQ[n, 2] && LtQ[m, 0]) || (LeQ[m, -1] && GtQ[m + n, 0]) || (ILt
Q[m, 0] && LtQ[m/2 + n - 1, 0] && IntegerQ[n]) || (EqQ[n, 3/2] && EqQ[m, -2^(-1)])) && IntegerQ[2*m]

Rubi steps

\begin {align*} \int \cos ^5(c+d x) (a+i a \tan (c+d x))^2 \, dx &=-\frac {2 i \cos ^5(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{5 d}+\frac {1}{5} \left (3 a^2\right ) \int \cos ^3(c+d x) \, dx\\ &=-\frac {2 i \cos ^5(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{5 d}-\frac {\left (3 a^2\right ) \text {Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{5 d}\\ &=\frac {3 a^2 \sin (c+d x)}{5 d}-\frac {a^2 \sin ^3(c+d x)}{5 d}-\frac {2 i \cos ^5(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{5 d}\\ \end {align*}

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Mathematica [A]
time = 0.37, size = 72, normalized size = 1.04 \begin {gather*} \frac {a^2 (-i \cos (2 (c+d x))+\sin (2 (c+d x))) (10 \cos (c+d x)-2 \cos (3 (c+d x))-5 i \sin (c+d x)+3 i \sin (3 (c+d x)))}{20 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^5*(a + I*a*Tan[c + d*x])^2,x]

[Out]

(a^2*((-I)*Cos[2*(c + d*x)] + Sin[2*(c + d*x)])*(10*Cos[c + d*x] - 2*Cos[3*(c + d*x)] - (5*I)*Sin[c + d*x] + (
3*I)*Sin[3*(c + d*x)]))/(20*d)

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Maple [A]
time = 0.23, size = 91, normalized size = 1.32

method	result	size
risch	\(-\frac {i a^{2} {\mathrm e}^{5 i \left (d x +c \right )}}{40 d}-\frac {i a^{2} {\mathrm e}^{3 i \left (d x +c \right )}}{8 d}-\frac {i a^{2} \cos \left (d x +c \right )}{4 d}+\frac {a^{2} \sin \left (d x +c \right )}{2 d}\)	\(67\)
derivativedivides	\(\frac {-a^{2} \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{4}\left (d x +c \right )\right )}{5}+\frac {\left (\cos ^{2}\left (d x +c \right )+2\right ) \sin \left (d x +c \right )}{15}\right )-\frac {2 i a^{2} \left (\cos ^{5}\left (d x +c \right )\right )}{5}+\frac {a^{2} \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}}{d}\)	\(91\)
default	\(\frac {-a^{2} \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{4}\left (d x +c \right )\right )}{5}+\frac {\left (\cos ^{2}\left (d x +c \right )+2\right ) \sin \left (d x +c \right )}{15}\right )-\frac {2 i a^{2} \left (\cos ^{5}\left (d x +c \right )\right )}{5}+\frac {a^{2} \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}}{d}\)	\(91\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^5*(a+I*a*tan(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

1/d*(-a^2*(-1/5*sin(d*x+c)*cos(d*x+c)^4+1/15*(cos(d*x+c)^2+2)*sin(d*x+c))-2/5*I*a^2*cos(d*x+c)^5+1/5*a^2*(8/3+
cos(d*x+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c))

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Maxima [A]
time = 0.28, size = 79, normalized size = 1.14 \begin {gather*} -\frac {6 i \, a^{2} \cos \left (d x + c\right )^{5} - {\left (3 \, \sin \left (d x + c\right )^{5} - 5 \, \sin \left (d x + c\right )^{3}\right )} a^{2} - {\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^{2}}{15 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a+I*a*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/15*(6*I*a^2*cos(d*x + c)^5 - (3*sin(d*x + c)^5 - 5*sin(d*x + c)^3)*a^2 - (3*sin(d*x + c)^5 - 10*sin(d*x + c
)^3 + 15*sin(d*x + c))*a^2)/d

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Fricas [A]
time = 0.36, size = 62, normalized size = 0.90 \begin {gather*} \frac {{\left (-i \, a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} - 5 i \, a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} - 15 i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + 5 i \, a^{2}\right )} e^{\left (-i \, d x - i \, c\right )}}{40 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a+I*a*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

1/40*(-I*a^2*e^(6*I*d*x + 6*I*c) - 5*I*a^2*e^(4*I*d*x + 4*I*c) - 15*I*a^2*e^(2*I*d*x + 2*I*c) + 5*I*a^2)*e^(-I
*d*x - I*c)/d

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Sympy [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 153 vs. \(2 (60) = 120\).
time = 0.22, size = 153, normalized size = 2.22 \begin {gather*} \begin {cases} \frac {\left (- 512 i a^{2} d^{3} e^{6 i c} e^{5 i d x} - 2560 i a^{2} d^{3} e^{4 i c} e^{3 i d x} - 7680 i a^{2} d^{3} e^{2 i c} e^{i d x} + 2560 i a^{2} d^{3} e^{- i d x}\right ) e^{- i c}}{20480 d^{4}} & \text {for}\: d^{4} e^{i c} \neq 0 \\\frac {x \left (a^{2} e^{6 i c} + 3 a^{2} e^{4 i c} + 3 a^{2} e^{2 i c} + a^{2}\right ) e^{- i c}}{8} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**5*(a+I*a*tan(d*x+c))**2,x)

[Out]

Piecewise(((-512*I*a**2*d**3*exp(6*I*c)*exp(5*I*d*x) - 2560*I*a**2*d**3*exp(4*I*c)*exp(3*I*d*x) - 7680*I*a**2*
d**3*exp(2*I*c)*exp(I*d*x) + 2560*I*a**2*d**3*exp(-I*d*x))*exp(-I*c)/(20480*d**4), Ne(d**4*exp(I*c), 0)), (x*(
a**2*exp(6*I*c) + 3*a**2*exp(4*I*c) + 3*a**2*exp(2*I*c) + a**2)*exp(-I*c)/8, True))

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Giac [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 613 vs. \(2 (59) = 118\).
time = 0.72, size = 613, normalized size = 8.88 \begin {gather*} -\frac {45 \, a^{2} e^{\left (5 i \, d x + 3 i \, c\right )} \log \left (i \, e^{\left (i \, d x + i \, c\right )} + 1\right ) + 90 \, a^{2} e^{\left (3 i \, d x + i \, c\right )} \log \left (i \, e^{\left (i \, d x + i \, c\right )} + 1\right ) + 45 \, a^{2} e^{\left (i \, d x - i \, c\right )} \log \left (i \, e^{\left (i \, d x + i \, c\right )} + 1\right ) + 40 \, a^{2} e^{\left (5 i \, d x + 3 i \, c\right )} \log \left (i \, e^{\left (i \, d x + i \, c\right )} - 1\right ) + 80 \, a^{2} e^{\left (3 i \, d x + i \, c\right )} \log \left (i \, e^{\left (i \, d x + i \, c\right )} - 1\right ) + 40 \, a^{2} e^{\left (i \, d x - i \, c\right )} \log \left (i \, e^{\left (i \, d x + i \, c\right )} - 1\right ) - 45 \, a^{2} e^{\left (5 i \, d x + 3 i \, c\right )} \log \left (-i \, e^{\left (i \, d x + i \, c\right )} + 1\right ) - 90 \, a^{2} e^{\left (3 i \, d x + i \, c\right )} \log \left (-i \, e^{\left (i \, d x + i \, c\right )} + 1\right ) - 45 \, a^{2} e^{\left (i \, d x - i \, c\right )} \log \left (-i \, e^{\left (i \, d x + i \, c\right )} + 1\right ) - 40 \, a^{2} e^{\left (5 i \, d x + 3 i \, c\right )} \log \left (-i \, e^{\left (i \, d x + i \, c\right )} - 1\right ) - 80 \, a^{2} e^{\left (3 i \, d x + i \, c\right )} \log \left (-i \, e^{\left (i \, d x + i \, c\right )} - 1\right ) - 40 \, a^{2} e^{\left (i \, d x - i \, c\right )} \log \left (-i \, e^{\left (i \, d x + i \, c\right )} - 1\right ) - 5 \, a^{2} e^{\left (5 i \, d x + 3 i \, c\right )} \log \left (i \, e^{\left (i \, d x\right )} + e^{\left (-i \, c\right )}\right ) - 10 \, a^{2} e^{\left (3 i \, d x + i \, c\right )} \log \left (i \, e^{\left (i \, d x\right )} + e^{\left (-i \, c\right )}\right ) - 5 \, a^{2} e^{\left (i \, d x - i \, c\right )} \log \left (i \, e^{\left (i \, d x\right )} + e^{\left (-i \, c\right )}\right ) + 5 \, a^{2} e^{\left (5 i \, d x + 3 i \, c\right )} \log \left (-i \, e^{\left (i \, d x\right )} + e^{\left (-i \, c\right )}\right ) + 10 \, a^{2} e^{\left (3 i \, d x + i \, c\right )} \log \left (-i \, e^{\left (i \, d x\right )} + e^{\left (-i \, c\right )}\right ) + 5 \, a^{2} 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^{2} e^{\left (10 i \, d x + 8 i \, c\right )} + 28 i \, a^{2} e^{\left (8 i \, d x + 6 i \, c\right )} + 104 i \, a^{2} e^{\left (6 i \, d x + 4 i \, c\right )} + 120 i \, a^{2} e^{\left (4 i \, d x + 2 i \, c\right )} + 20 i \, a^{2} e^{\left (2 i \, d x\right )} - 20 i \, a^{2} e^{\left (-2 i \, c\right )}}{160 \, {\left (d e^{\left (5 i \, d x + 3 i \, c\right )} + 2 \, d e^{\left (3 i \, d x + i \, c\right )} + d e^{\left (i \, d x - i \, c\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a+I*a*tan(d*x+c))^2,x, algorithm="giac")

[Out]

-1/160*(45*a^2*e^(5*I*d*x + 3*I*c)*log(I*e^(I*d*x + I*c) + 1) + 90*a^2*e^(3*I*d*x + I*c)*log(I*e^(I*d*x + I*c)
+ 1) + 45*a^2*e^(I*d*x - I*c)*log(I*e^(I*d*x + I*c) + 1) + 40*a^2*e^(5*I*d*x + 3*I*c)*log(I*e^(I*d*x + I*c) -
1) + 80*a^2*e^(3*I*d*x + I*c)*log(I*e^(I*d*x + I*c) - 1) + 40*a^2*e^(I*d*x - I*c)*log(I*e^(I*d*x + I*c) - 1)
- 45*a^2*e^(5*I*d*x + 3*I*c)*log(-I*e^(I*d*x + I*c) + 1) - 90*a^2*e^(3*I*d*x + I*c)*log(-I*e^(I*d*x + I*c) + 1
) - 45*a^2*e^(I*d*x - I*c)*log(-I*e^(I*d*x + I*c) + 1) - 40*a^2*e^(5*I*d*x + 3*I*c)*log(-I*e^(I*d*x + I*c) - 1
) - 80*a^2*e^(3*I*d*x + I*c)*log(-I*e^(I*d*x + I*c) - 1) - 40*a^2*e^(I*d*x - I*c)*log(-I*e^(I*d*x + I*c) - 1)
- 5*a^2*e^(5*I*d*x + 3*I*c)*log(I*e^(I*d*x) + e^(-I*c)) - 10*a^2*e^(3*I*d*x + I*c)*log(I*e^(I*d*x) + e^(-I*c))
- 5*a^2*e^(I*d*x - I*c)*log(I*e^(I*d*x) + e^(-I*c)) + 5*a^2*e^(5*I*d*x + 3*I*c)*log(-I*e^(I*d*x) + e^(-I*c))
+ 10*a^2*e^(3*I*d*x + I*c)*log(-I*e^(I*d*x) + e^(-I*c)) + 5*a^2*e^(I*d*x - I*c)*log(-I*e^(I*d*x) + e^(-I*c)) +
4*I*a^2*e^(10*I*d*x + 8*I*c) + 28*I*a^2*e^(8*I*d*x + 6*I*c) + 104*I*a^2*e^(6*I*d*x + 4*I*c) + 120*I*a^2*e^(4*
I*d*x + 2*I*c) + 20*I*a^2*e^(2*I*d*x) - 20*I*a^2*e^(-2*I*c))/(d*e^(5*I*d*x + 3*I*c) + 2*d*e^(3*I*d*x + I*c) +
d*e^(I*d*x - I*c))

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Mupad [B]
time = 4.17, size = 71, normalized size = 1.03 \begin {gather*} \frac {2\,a^2\,\left (\frac {5\,\sin \left (3\,c+3\,d\,x\right )}{16}-\frac {\cos \left (5\,c+5\,d\,x\right )\,1{}\mathrm {i}}{16}-\frac {\cos \left (3\,c+3\,d\,x\right )\,5{}\mathrm {i}}{16}+\frac {\sin \left (5\,c+5\,d\,x\right )}{16}+\frac {5\,\sqrt {3}\,\sin \left (c+d\,x-\frac {\ln \left (3\right )\,1{}\mathrm {i}}{2}\right )}{8}\right )}{5\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^5*(a + a*tan(c + d*x)*1i)^2,x)

[Out]

(2*a^2*((5*sin(3*c + 3*d*x))/16 - (cos(5*c + 5*d*x)*1i)/16 - (cos(3*c + 3*d*x)*5i)/16 + sin(5*c + 5*d*x)/16 +
(5*3^(1/2)*sin(c - (log(3)*1i)/2 + d*x))/8))/(5*d)

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