\(\int (e \cos (c+d x))^{5/2} \sqrt {a+i a \tan (c+d x)} \, dx\) [674]

Optimal result

Integrand size = 30, antiderivative size = 132 \[ \int (e \cos (c+d x))^{5/2} \sqrt {a+i a \tan (c+d x)} \, dx=\frac {8 i a (e \cos (c+d x))^{5/2} \sec ^2(c+d x)}{15 d \sqrt {a+i a \tan (c+d x)}}-\frac {2 i (e \cos (c+d x))^{5/2} \sqrt {a+i a \tan (c+d x)}}{5 d}-\frac {16 i (e \cos (c+d x))^{5/2} \sec ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{15 d} \]

[Out]

8/15*I*a*(e*cos(d*x+c))^(5/2)*sec(d*x+c)^2/d/(a+I*a*tan(d*x+c))^(1/2)-2/5*I*(e*cos(d*x+c))^(5/2)*(a+I*a*tan(d*
x+c))^(1/2)/d-16/15*I*(e*cos(d*x+c))^(5/2)*sec(d*x+c)^2*(a+I*a*tan(d*x+c))^(1/2)/d

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {3596, 3578, 3583, 3569} \[ \int (e \cos (c+d x))^{5/2} \sqrt {a+i a \tan (c+d x)} \, dx=-\frac {2 i \sqrt {a+i a \tan (c+d x)} (e \cos (c+d x))^{5/2}}{5 d}-\frac {16 i \sec ^2(c+d x) \sqrt {a+i a \tan (c+d x)} (e \cos (c+d x))^{5/2}}{15 d}+\frac {8 i a \sec ^2(c+d x) (e \cos (c+d x))^{5/2}}{15 d \sqrt {a+i a \tan (c+d x)}} \]

[In]

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

[Out]

(((8*I)/15)*a*(e*Cos[c + d*x])^(5/2)*Sec[c + d*x]^2)/(d*Sqrt[a + I*a*Tan[c + d*x]]) - (((2*I)/5)*(e*Cos[c + d*
x])^(5/2)*Sqrt[a + I*a*Tan[c + d*x]])/d - (((16*I)/15)*(e*Cos[c + d*x])^(5/2)*Sec[c + d*x]^2*Sqrt[a + I*a*Tan[
c + d*x]])/d

Rule 3569

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

Rule 3578

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

Rule 3583

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

Rule 3596

Int[(cos[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[(d*Co
s[e + f*x])^m*(d*Sec[e + f*x])^m, Int[(a + b*Tan[e + f*x])^n/(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e,
f, m, n}, x] &&  !IntegerQ[m]

Rubi steps \begin{align*} \text {integral}& = \left ((e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}\right ) \int \frac {\sqrt {a+i a \tan (c+d x)}}{(e \sec (c+d x))^{5/2}} \, dx \\ & = -\frac {2 i (e \cos (c+d x))^{5/2} \sqrt {a+i a \tan (c+d x)}}{5 d}+\frac {\left (4 a (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}\right ) \int \frac {1}{\sqrt {e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)}} \, dx}{5 e^2} \\ & = \frac {8 i a (e \cos (c+d x))^{5/2} \sec ^2(c+d x)}{15 d \sqrt {a+i a \tan (c+d x)}}-\frac {2 i (e \cos (c+d x))^{5/2} \sqrt {a+i a \tan (c+d x)}}{5 d}+\frac {\left (8 (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}\right ) \int \frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}} \, dx}{15 e^2} \\ & = \frac {8 i a (e \cos (c+d x))^{5/2} \sec ^2(c+d x)}{15 d \sqrt {a+i a \tan (c+d x)}}-\frac {2 i (e \cos (c+d x))^{5/2} \sqrt {a+i a \tan (c+d x)}}{5 d}-\frac {16 i (e \cos (c+d x))^{5/2} \sec ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{15 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.71 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.48 \[ \int (e \cos (c+d x))^{5/2} \sqrt {a+i a \tan (c+d x)} \, dx=\frac {i e^2 \sqrt {e \cos (c+d x)} (-15+\cos (2 (c+d x))-4 i \sin (2 (c+d x))) \sqrt {a+i a \tan (c+d x)}}{15 d} \]

[In]

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

[Out]

((I/15)*e^2*Sqrt[e*Cos[c + d*x]]*(-15 + Cos[2*(c + d*x)] - (4*I)*Sin[2*(c + d*x)])*Sqrt[a + I*a*Tan[c + d*x]])
/d

Maple [A] (verified)

Time = 8.99 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.47

method	result	size
default	\(\frac {2 \left (i \left (\cos ^{2}\left (d x +c \right )\right )+4 \sin \left (d x +c \right ) \cos \left (d x +c \right )-8 i\right ) \sqrt {e \cos \left (d x +c \right )}\, \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, e^{2}}{15 d}\)	\(62\)
risch	\(-\frac {i e^{2} \sqrt {2}\, \sqrt {e \cos \left (d x +c \right )}\, \sqrt {\frac {a \,{\mathrm e}^{2 i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, \left (30-2 \cos \left (2 d x +2 c \right )+8 i \sin \left (2 d x +2 c \right )\right )}{30 d}\)	\(74\)

[In]

int((e*cos(d*x+c))^(5/2)*(a+I*a*tan(d*x+c))^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.65 \[ \int (e \cos (c+d x))^{5/2} \sqrt {a+i a \tan (c+d x)} \, dx=\frac {\sqrt {2} \sqrt {\frac {1}{2}} {\left (-3 i \, e^{2} e^{\left (4 i \, d x + 4 i \, c\right )} - 30 i \, e^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + 5 i \, e^{2}\right )} \sqrt {e e^{\left (2 i \, d x + 2 i \, c\right )} + e} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (-\frac {3}{2} i \, d x - \frac {3}{2} i \, c\right )}}{30 \, d} \]

[In]

integrate((e*cos(d*x+c))^(5/2)*(a+I*a*tan(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

1/30*sqrt(2)*sqrt(1/2)*(-3*I*e^2*e^(4*I*d*x + 4*I*c) - 30*I*e^2*e^(2*I*d*x + 2*I*c) + 5*I*e^2)*sqrt(e*e^(2*I*d
*x + 2*I*c) + e)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(-3/2*I*d*x - 3/2*I*c)/d

Sympy [F(-1)]

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

[In]

integrate((e*cos(d*x+c))**(5/2)*(a+I*a*tan(d*x+c))**(1/2),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.39 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.12 \[ \int (e \cos (c+d x))^{5/2} \sqrt {a+i a \tan (c+d x)} \, dx=\frac {{\left (5 i \, e^{2} \cos \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) - 3 i \, e^{2} \cos \left (\frac {5}{3} \, \arctan \left (\sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ), \cos \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right )\right )\right ) - 30 i \, e^{2} \cos \left (\frac {1}{3} \, \arctan \left (\sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ), \cos \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right )\right )\right ) + 5 \, e^{2} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 3 \, e^{2} \sin \left (\frac {5}{3} \, \arctan \left (\sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ), \cos \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right )\right )\right ) + 30 \, e^{2} \sin \left (\frac {1}{3} \, \arctan \left (\sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ), \cos \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right )\right )\right )\right )} \sqrt {a} \sqrt {e}}{30 \, d} \]

[In]

integrate((e*cos(d*x+c))^(5/2)*(a+I*a*tan(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

1/30*(5*I*e^2*cos(3/2*d*x + 3/2*c) - 3*I*e^2*cos(5/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) - 30
*I*e^2*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 5*e^2*sin(3/2*d*x + 3/2*c) + 3*e^2*sin(5
/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 30*e^2*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2
*d*x + 3/2*c))))*sqrt(a)*sqrt(e)/d

Giac [F]

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

[In]

integrate((e*cos(d*x+c))^(5/2)*(a+I*a*tan(d*x+c))^(1/2),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.84 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.64 \[ \int (e \cos (c+d x))^{5/2} \sqrt {a+i a \tan (c+d x)} \, dx=\frac {e^2\,\sqrt {e\,\cos \left (c+d\,x\right )}\,\sqrt {\frac {a\,\left (\cos \left (2\,c+2\,d\,x\right )+1+\sin \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,c+2\,d\,x\right )+1}}\,\left (\cos \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}+4\,\sin \left (2\,c+2\,d\,x\right )-15{}\mathrm {i}\right )}{15\,d} \]

[In]

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

[Out]

(e^2*(e*cos(c + d*x))^(1/2)*((a*(cos(2*c + 2*d*x) + sin(2*c + 2*d*x)*1i + 1))/(cos(2*c + 2*d*x) + 1))^(1/2)*(c
os(2*c + 2*d*x)*1i + 4*sin(2*c + 2*d*x) - 15i))/(15*d)