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

Optimal. Leaf size=65 \[ \frac{2 i a \cos (c+d x) (a+i a \tan (c+d x))^{3/2}}{d}-\frac{8 i a^2 \cos (c+d x) \sqrt{a+i a \tan (c+d x)}}{d} \]

[Out]

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

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Rubi [A]  time = 0.0966806, antiderivative size = 65, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.083, Rules used = {3494, 3493} \[ \frac{2 i a \cos (c+d x) (a+i a \tan (c+d x))^{3/2}}{d}-\frac{8 i a^2 \cos (c+d x) \sqrt{a+i a \tan (c+d x)}}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-8*I)*a^2*Cos[c + d*x]*Sqrt[a + I*a*Tan[c + d*x]])/d + ((2*I)*a*Cos[c + d*x]*(a + I*a*Tan[c + d*x])^(3/2))/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 (c+d x) (a+i a \tan (c+d x))^{5/2} \, dx &=\frac{2 i a \cos (c+d x) (a+i a \tan (c+d x))^{3/2}}{d}+(4 a) \int \cos (c+d x) (a+i a \tan (c+d x))^{3/2} \, dx\\ &=-\frac{8 i a^2 \cos (c+d x) \sqrt{a+i a \tan (c+d x)}}{d}+\frac{2 i a \cos (c+d x) (a+i a \tan (c+d x))^{3/2}}{d}\\ \end{align*}

Mathematica [A]  time = 0.253252, size = 46, normalized size = 0.71 \[ -\frac{2 i a^2 \sqrt{a+i a \tan (c+d x)} (3 \cos (c+d x)-i \sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.279, size = 53, normalized size = 0.8 \begin{align*} -2\,{\frac{{a}^{2} \left ( 3\,i\cos \left ( dx+c \right ) +\sin \left ( dx+c \right ) \right ) }{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)*(a+I*a*tan(d*x+c))^(5/2),x)

[Out]

-2/d*a^2*(3*I*cos(d*x+c)+sin(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 = 1.93972, size = 447, normalized size = 6.88 \begin{align*} \frac{2 \,{\left (-3 i \, a^{\frac{5}{2}} - \frac{2 \, a^{\frac{5}{2}} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{9 i \, a^{\frac{5}{2}} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{4 \, a^{\frac{5}{2}} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac{9 i \, a^{\frac{5}{2}} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac{2 \, a^{\frac{5}{2}} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac{3 i \, a^{\frac{5}{2}} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}\right )}{\left (-\frac{2 i \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - 1\right )}^{\frac{5}{2}}}{d{\left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac{5}{2}}{\left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}^{\frac{5}{2}}{\left (\frac{4 i \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{5 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac{5 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac{4 i \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac{\sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(-3*I*a^(5/2) - 2*a^(5/2)*sin(d*x + c)/(cos(d*x + c) + 1) + 9*I*a^(5/2)*sin(d*x + c)^2/(cos(d*x + c) + 1)^2
+ 4*a^(5/2)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 9*I*a^(5/2)*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 2*a^(5/2)*
sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 3*I*a^(5/2)*sin(d*x + c)^6/(cos(d*x + c) + 1)^6)*(-2*I*sin(d*x + c)/(cos
(d*x + c) + 1) + sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 1)^(5/2)/(d*(sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(5/2)
*(sin(d*x + c)/(cos(d*x + c) + 1) - 1)^(5/2)*(4*I*sin(d*x + c)/(cos(d*x + c) + 1) - 5*sin(d*x + c)^2/(cos(d*x
+ c) + 1)^2 - 5*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 4*I*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + sin(d*x + c)^6
/(cos(d*x + c) + 1)^6 + 1))

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Fricas [A]  time = 1.97035, size = 116, normalized size = 1.78 \begin{align*} \frac{\sqrt{2}{\left (-2 i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - 4 i \, a^{2}\right )} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(2)*(-2*I*a^2*e^(2*I*d*x + 2*I*c) - 4*I*a^2)*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)*(a+I*a*tan(d*x+c))**(5/2),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{5}{2}} \cos \left (d x + c\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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