\(\int \cos ^3(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx\) [316]

Optimal result

Integrand size = 26, antiderivative size = 35 \[ \int \cos ^3(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=-\frac {2 i a \cos ^3(c+d x) (a+i a \tan (c+d x))^{3/2}}{3 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {3574} \[ \int \cos ^3(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=-\frac {2 i a \cos ^3(c+d x) (a+i a \tan (c+d x))^{3/2}}{3 d} \]

[In]

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

[Out]

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

Rule 3574

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*} \text {integral}& = -\frac {2 i a \cos ^3(c+d x) (a+i a \tan (c+d x))^{3/2}}{3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.78 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.97 \[ \int \cos ^3(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=\frac {2 a^2 \cos ^2(c+d x) (-i \cos (c+3 d x)+\sin (c+3 d x)) \sqrt {a+i a \tan (c+d x)}}{3 d (\cos (d x)+i \sin (d x))^2} \]

[In]

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

[Out]

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

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 65 vs. \(2 (29 ) = 58\).

Time = 35.41 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.89

method	result	size
default	\(-\frac {2 i \left (-\tan \left (d x +c \right )+i\right )^{2} \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \left (i \sin \left (d x +c \right ) \left (\cos ^{4}\left (d x +c \right )\right )-\left (\cos ^{5}\left (d x +c \right )\right )\right ) a^{2}}{3 d}\)	\(66\)

[In]

int(cos(d*x+c)^3*(a+I*a*tan(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 59 vs. \(2 (27) = 54\).

Time = 0.24 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.69 \[ \int \cos ^3(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=\frac {\sqrt {2} {\left (-i \, a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} - 2 i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a^{2}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{6 \, d} \]

[In]

integrate(cos(d*x+c)^3*(a+I*a*tan(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int \cos ^3(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**3*(a+I*a*tan(d*x+c))**(5/2),x)

[Out]

Timed out

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 328 vs. \(2 (27) = 54\).

Time = 0.46 (sec) , antiderivative size = 328, normalized size of antiderivative = 9.37 \[ \int \cos ^3(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=\frac {2 \, {\left (i \, a^{\frac {5}{2}} - \frac {4 i \, a^{\frac {5}{2}} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {6 i \, a^{\frac {5}{2}} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {4 i \, a^{\frac {5}{2}} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {i \, a^{\frac {5}{2}} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}}\right )} {\left (-\frac {2 i \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - 1\right )}^{\frac {5}{2}}}{-3 \, d {\left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {5}{2}} {\left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}^{\frac {5}{2}} {\left (\frac {2 i \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {2 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {6 i \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {6 i \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {2 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {2 i \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac {\sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} + 1\right )}} \]

[In]

integrate(cos(d*x+c)^3*(a+I*a*tan(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

2*(I*a^(5/2) - 4*I*a^(5/2)*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 6*I*a^(5/2)*sin(d*x + c)^4/(cos(d*x + c) + 1)
^4 - 4*I*a^(5/2)*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + I*a^(5/2)*sin(d*x + c)^8/(cos(d*x + c) + 1)^8)*(-2*I*si
n(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)*(-6*I*sin(d*x + c)/(cos(d*x + c) + 1) - 6*sin(d*x
+ c)^2/(cos(d*x + c) + 1)^2 - 18*I*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 18*I*sin(d*x + c)^5/(cos(d*x + c) + 1
)^5 + 6*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 - 6*I*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 3*sin(d*x + c)^8/(cos(
d*x + c) + 1)^8 - 3))

Giac [F]

\[ \int \cos ^3(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=\int { {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \cos \left (d x + c\right )^{3} \,d x } \]

[In]

integrate(cos(d*x+c)^3*(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)^3, x)

Mupad [B] (verification not implemented)

Time = 0.97 (sec) , antiderivative size = 89, normalized size of antiderivative = 2.54 \[ \int \cos ^3(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=-\frac {a^2\,\sqrt {\frac {a\,\left (\cos \left (2\,c+2\,d\,x\right )+1+\sin \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,c+2\,d\,x\right )+1}}\,\left (-\sin \left (c+d\,x\right )-\sin \left (3\,c+3\,d\,x\right )+\cos \left (c+d\,x\right )\,3{}\mathrm {i}+\cos \left (3\,c+3\,d\,x\right )\,1{}\mathrm {i}\right )}{6\,d} \]

[In]

int(cos(c + d*x)^3*(a + a*tan(c + d*x)*1i)^(5/2),x)

[Out]

-(a^2*((a*(cos(2*c + 2*d*x) + sin(2*c + 2*d*x)*1i + 1))/(cos(2*c + 2*d*x) + 1))^(1/2)*(cos(c + d*x)*3i - sin(c
+ d*x) + cos(3*c + 3*d*x)*1i - sin(3*c + 3*d*x)))/(6*d)