3.328 \(\int \cos ^3(c+d x) (a+i a \tan (c+d x))^{7/2} \, dx\)

Optimal. Leaf size=71 \[ \frac{8 i a^2 \cos ^3(c+d x) (a+i a \tan (c+d x))^{3/2}}{3 d}-\frac{2 i a \cos ^3(c+d x) (a+i a \tan (c+d x))^{5/2}}{d} \]

[Out]

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

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Rubi [A]  time = 0.121975, antiderivative size = 71, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.077, Rules used = {3494, 3493} \[ \frac{8 i a^2 \cos ^3(c+d x) (a+i a \tan (c+d x))^{3/2}}{3 d}-\frac{2 i a \cos ^3(c+d x) (a+i a \tan (c+d x))^{5/2}}{d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^3*(a + I*a*Tan[c + d*x])^(7/2),x]

[Out]

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

Rule 3494

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

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 ^3(c+d x) (a+i a \tan (c+d x))^{7/2} \, dx &=-\frac{2 i a \cos ^3(c+d x) (a+i a \tan (c+d x))^{5/2}}{d}-(4 a) \int \cos ^3(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx\\ &=\frac{8 i a^2 \cos ^3(c+d x) (a+i a \tan (c+d x))^{3/2}}{3 d}-\frac{2 i a \cos ^3(c+d x) (a+i a \tan (c+d x))^{5/2}}{d}\\ \end{align*}

Mathematica [A]  time = 0.346773, size = 86, normalized size = 1.21 \[ \frac{2 a^3 \cos (c+d x) \sqrt{a+i a \tan (c+d x)} (3 \sin (c+d x)+i \cos (c+d x)) (\cos (c+4 d x)+i \sin (c+4 d x))}{3 d (\cos (d x)+i \sin (d x))^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^3*(a + I*a*Tan[c + d*x])^(7/2),x]

[Out]

(2*a^3*Cos[c + d*x]*(I*Cos[c + d*x] + 3*Sin[c + d*x])*(Cos[c + 4*d*x] + I*Sin[c + 4*d*x])*Sqrt[a + I*a*Tan[c +
d*x]])/(3*d*(Cos[d*x] + I*Sin[d*x])^3)

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Maple [A]  time = 0.278, size = 71, normalized size = 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 ) -3\,i \right ) \cos \left ( dx+c \right ) }{3\,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)^3*(a+I*a*tan(d*x+c))^(7/2),x)

[Out]

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

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Maxima [B]  time = 2.17934, size = 680, normalized size = 9.58 \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)^3*(a+I*a*tan(d*x+c))^(7/2),x, algorithm="maxima")

[Out]

-2*(-I*a^(7/2) - 6*a^(7/2)*sin(d*x + c)/(cos(d*x + c) + 1) + 5*I*a^(7/2)*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 +
24*a^(7/2)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 10*I*a^(7/2)*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 36*a^(7/2
)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 10*I*a^(7/2)*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 24*a^(7/2)*sin(d*x
+ c)^7/(cos(d*x + c) + 1)^7 - 5*I*a^(7/2)*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 - 6*a^(7/2)*sin(d*x + c)^9/(cos(
d*x + c) + 1)^9 + I*a^(7/2)*sin(d*x + c)^10/(cos(d*x + c) + 1)^10)*(-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)*(12*I*sin(d*x + c)/(cos(d*x + c) + 1) - 9*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 24*I*
sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 42*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 42*sin(d*x + c)^6/(cos(d*x + c)
+ 1)^6 - 24*I*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 - 9*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 - 12*I*sin(d*x + c)
^9/(cos(d*x + c) + 1)^9 + 3*sin(d*x + c)^10/(cos(d*x + c) + 1)^10 + 3))

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Fricas [A]  time = 1.9417, size = 157, normalized size = 2.21 \begin{align*} \frac{\sqrt{2}{\left (-i \, a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + 2 i \, a^{3}\right )} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(a+I*a*tan(d*x+c))^(7/2),x, algorithm="fricas")

[Out]

1/3*sqrt(2)*(-I*a^3*e^(4*I*d*x + 4*I*c) + I*a^3*e^(2*I*d*x + 2*I*c) + 2*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)**3*(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 )^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(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)^3, x)