3.329 \(\int \cos ^5(c+d x) (a+i a \tan (c+d x))^{7/2} \, dx\)

Optimal. Leaf size=35 \[ -\frac{2 i a \cos ^5(c+d x) (a+i a \tan (c+d x))^{5/2}}{5 d} \]

[Out]

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

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Rubi [A]  time = 0.0586437, antiderivative size = 35, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.038, Rules used = {3493} \[ -\frac{2 i a \cos ^5(c+d x) (a+i a \tan (c+d x))^{5/2}}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^5*(a + I*a*Tan[c + d*x])^(7/2),x]

[Out]

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

Rule 3493

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] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 + b^2
, 0] && EqQ[Simplify[m/2 + n - 1], 0]

Rubi steps

\begin{align*} \int \cos ^5(c+d x) (a+i a \tan (c+d x))^{7/2} \, dx &=-\frac{2 i a \cos ^5(c+d x) (a+i a \tan (c+d x))^{5/2}}{5 d}\\ \end{align*}

Mathematica [B]  time = 0.514704, size = 73, normalized size = 2.09 \[ \frac{2 a^3 \cos ^3(c+d x) \sqrt{a+i a \tan (c+d x)} (\sin (2 c+5 d x)-i \cos (2 c+5 d x))}{5 d (\cos (d x)+i \sin (d x))^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^5*(a + I*a*Tan[c + d*x])^(7/2),x]

[Out]

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

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Maple [B]  time = 0.349, size = 73, normalized size = 2.1 \begin{align*} -{\frac{2\,{a}^{3} \left ( 2\,i \left ( \cos \left ( dx+c \right ) \right ) ^{2}-2\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) -i \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{5\,d}\sqrt{{\frac{a \left ( i\sin \left ( dx+c \right ) +\cos \left ( dx+c \right ) \right ) }{\cos \left ( dx+c \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^5*(a+I*a*tan(d*x+c))^(7/2),x)

[Out]

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

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Maxima [B]  time = 2.14331, size = 613, normalized size = 17.51 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a+I*a*tan(d*x+c))^(7/2),x, algorithm="maxima")

[Out]

2*(I*a^(7/2) - 6*I*a^(7/2)*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 15*I*a^(7/2)*sin(d*x + c)^4/(cos(d*x + c) + 1
)^4 - 20*I*a^(7/2)*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 15*I*a^(7/2)*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 - 6*
I*a^(7/2)*sin(d*x + c)^10/(cos(d*x + c) + 1)^10 + I*a^(7/2)*sin(d*x + c)^12/(cos(d*x + c) + 1)^12)*(-2*I*sin(d
*x + c)/(cos(d*x + c) + 1) + sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 1)^(7/2)/(d*(sin(d*x + c)/(cos(d*x + c) + 1
) + 1)^(7/2)*(sin(d*x + c)/(cos(d*x + c) + 1) - 1)^(7/2)*(-10*I*sin(d*x + c)/(cos(d*x + c) + 1) - 20*sin(d*x +
c)^2/(cos(d*x + c) + 1)^2 - 50*I*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 25*sin(d*x + c)^4/(cos(d*x + c) + 1)^4
- 100*I*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 100*I*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 25*sin(d*x + c)^8/(
cos(d*x + c) + 1)^8 - 50*I*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 + 20*sin(d*x + c)^10/(cos(d*x + c) + 1)^10 - 10
*I*sin(d*x + c)^11/(cos(d*x + c) + 1)^11 + 5*sin(d*x + c)^12/(cos(d*x + c) + 1)^12 - 5))

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Fricas [B]  time = 2.22783, size = 198, normalized size = 5.66 \begin{align*} \frac{\sqrt{2}{\left (-i \, a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} - 3 i \, a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} - 3 i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a^{3}\right )} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{20 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a+I*a*tan(d*x+c))^(7/2),x, algorithm="fricas")

[Out]

1/20*sqrt(2)*(-I*a^3*e^(6*I*d*x + 6*I*c) - 3*I*a^3*e^(4*I*d*x + 4*I*c) - 3*I*a^3*e^(2*I*d*x + 2*I*c) - I*a^3)*
sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))/d

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**5*(a+I*a*tan(d*x+c))**(7/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{7}{2}} \cos \left (d x + c\right )^{5}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a+I*a*tan(d*x+c))^(7/2),x, algorithm="giac")

[Out]

integrate((I*a*tan(d*x + c) + a)^(7/2)*cos(d*x + c)^5, x)