\(\int \frac {(e \cos (c+d x))^{5/2}}{\sqrt {a+i a \tan (c+d x)}} \, dx\) [681]

Optimal result

Integrand size = 30, antiderivative size = 175 \[ \int \frac {(e \cos (c+d x))^{5/2}}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {2 i (e \cos (c+d x))^{5/2}}{7 d \sqrt {a+i a \tan (c+d x)}}+\frac {16 i (e \cos (c+d x))^{5/2} \sec ^2(c+d x)}{35 d \sqrt {a+i a \tan (c+d x)}}-\frac {12 i (e \cos (c+d x))^{5/2} \sqrt {a+i a \tan (c+d x)}}{35 a d}-\frac {32 i (e \cos (c+d x))^{5/2} \sec ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{35 a d} \]

[Out]

2/7*I*(e*cos(d*x+c))^(5/2)/d/(a+I*a*tan(d*x+c))^(1/2)+16/35*I*(e*cos(d*x+c))^(5/2)*sec(d*x+c)^2/d/(a+I*a*tan(d
*x+c))^(1/2)-12/35*I*(e*cos(d*x+c))^(5/2)*(a+I*a*tan(d*x+c))^(1/2)/a/d-32/35*I*(e*cos(d*x+c))^(5/2)*sec(d*x+c)
^2*(a+I*a*tan(d*x+c))^(1/2)/a/d

Rubi [A] (verified)

Time = 0.59 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {3596, 3583, 3578, 3569} \[ \int \frac {(e \cos (c+d x))^{5/2}}{\sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {12 i \sqrt {a+i a \tan (c+d x)} (e \cos (c+d x))^{5/2}}{35 a d}+\frac {2 i (e \cos (c+d x))^{5/2}}{7 d \sqrt {a+i a \tan (c+d x)}}-\frac {32 i \sec ^2(c+d x) \sqrt {a+i a \tan (c+d x)} (e \cos (c+d x))^{5/2}}{35 a d}+\frac {16 i \sec ^2(c+d x) (e \cos (c+d x))^{5/2}}{35 d \sqrt {a+i a \tan (c+d x)}} \]

[In]

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

[Out]

(((2*I)/7)*(e*Cos[c + d*x])^(5/2))/(d*Sqrt[a + I*a*Tan[c + d*x]]) + (((16*I)/35)*(e*Cos[c + d*x])^(5/2)*Sec[c
+ d*x]^2)/(d*Sqrt[a + I*a*Tan[c + d*x]]) - (((12*I)/35)*(e*Cos[c + d*x])^(5/2)*Sqrt[a + I*a*Tan[c + d*x]])/(a*
d) - (((32*I)/35)*(e*Cos[c + d*x])^(5/2)*Sec[c + d*x]^2*Sqrt[a + I*a*Tan[c + d*x]])/(a*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 {1}{(e \sec (c+d x))^{5/2} \sqrt {a+i a \tan (c+d x)}} \, dx \\ & = \frac {2 i (e \cos (c+d x))^{5/2}}{7 d \sqrt {a+i a \tan (c+d x)}}+\frac {\left (6 (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}{7 a} \\ & = \frac {2 i (e \cos (c+d x))^{5/2}}{7 d \sqrt {a+i a \tan (c+d x)}}-\frac {12 i (e \cos (c+d x))^{5/2} \sqrt {a+i a \tan (c+d x)}}{35 a d}+\frac {\left (24 (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}{35 e^2} \\ & = \frac {2 i (e \cos (c+d x))^{5/2}}{7 d \sqrt {a+i a \tan (c+d x)}}+\frac {16 i (e \cos (c+d x))^{5/2} \sec ^2(c+d x)}{35 d \sqrt {a+i a \tan (c+d x)}}-\frac {12 i (e \cos (c+d x))^{5/2} \sqrt {a+i a \tan (c+d x)}}{35 a d}+\frac {\left (16 (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}{35 a e^2} \\ & = \frac {2 i (e \cos (c+d x))^{5/2}}{7 d \sqrt {a+i a \tan (c+d x)}}+\frac {16 i (e \cos (c+d x))^{5/2} \sec ^2(c+d x)}{35 d \sqrt {a+i a \tan (c+d x)}}-\frac {12 i (e \cos (c+d x))^{5/2} \sqrt {a+i a \tan (c+d x)}}{35 a d}-\frac {32 i (e \cos (c+d x))^{5/2} \sec ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{35 a d} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.06 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.46 \[ \int \frac {(e \cos (c+d x))^{5/2}}{\sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {i e^3 (35 \cos (c+d x)+\cos (3 (c+d x))+70 i \sin (c+d x)+6 i \sin (3 (c+d x)))}{70 d \sqrt {e \cos (c+d x)} \sqrt {a+i a \tan (c+d x)}} \]

[In]

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

[Out]

((-1/70*I)*e^3*(35*Cos[c + d*x] + Cos[3*(c + d*x)] + (70*I)*Sin[c + d*x] + (6*I)*Sin[3*(c + d*x)]))/(d*Sqrt[e*
Cos[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])

Maple [A] (verified)

Time = 7.94 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.40

method	result	size
default	\(-\frac {2 e^{2} \sqrt {e \cos \left (d x +c \right )}\, \left (i \left (\cos ^{2}\left (d x +c \right )\right )-6 \sin \left (d x +c \right ) \cos \left (d x +c \right )+8 i-16 \tan \left (d x +c \right )\right )}{35 d \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}}\)	\(70\)

[In]

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

[Out]

-2/35/d*e^2*(e*cos(d*x+c))^(1/2)/(a*(1+I*tan(d*x+c)))^(1/2)*(I*cos(d*x+c)^2-6*sin(d*x+c)*cos(d*x+c)+8*I-16*tan
(d*x+c))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.59 \[ \int \frac {(e \cos (c+d x))^{5/2}}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {\sqrt {2} \sqrt {\frac {1}{2}} {\left (-7 i \, e^{2} e^{\left (6 i \, d x + 6 i \, c\right )} - 105 i \, e^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + 35 i \, e^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + 5 i \, e^{2}\right )} \sqrt {e e^{\left (2 i \, d x + 2 i \, c\right )} + e} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (-\frac {7}{2} i \, d x - \frac {7}{2} i \, c\right )}}{140 \, a d} \]

[In]

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

[Out]

1/140*sqrt(2)*sqrt(1/2)*(-7*I*e^2*e^(6*I*d*x + 6*I*c) - 105*I*e^2*e^(4*I*d*x + 4*I*c) + 35*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^(-7/2*I*d*x - 7/2*I*c)/(
a*d)

Sympy [F(-1)]

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

[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 = 0.80 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.15 \[ \int \frac {(e \cos (c+d x))^{5/2}}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {{\left (5 i \, e^{2} \cos \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) - 7 i \, e^{2} \cos \left (\frac {5}{7} \, \arctan \left (\sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ), \cos \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right )\right )\right ) + 35 i \, e^{2} \cos \left (\frac {3}{7} \, \arctan \left (\sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ), \cos \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right )\right )\right ) - 105 i \, e^{2} \cos \left (\frac {1}{7} \, \arctan \left (\sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ), \cos \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right )\right )\right ) + 5 \, e^{2} \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 7 \, e^{2} \sin \left (\frac {5}{7} \, \arctan \left (\sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ), \cos \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right )\right )\right ) + 35 \, e^{2} \sin \left (\frac {3}{7} \, \arctan \left (\sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ), \cos \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right )\right )\right ) + 105 \, e^{2} \sin \left (\frac {1}{7} \, \arctan \left (\sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ), \cos \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right )\right )\right )\right )} \sqrt {e}}{140 \, \sqrt {a} d} \]

[In]

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

[Out]

1/140*(5*I*e^2*cos(7/2*d*x + 7/2*c) - 7*I*e^2*cos(5/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))) + 3
5*I*e^2*cos(3/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))) - 105*I*e^2*cos(1/7*arctan2(sin(7/2*d*x +
7/2*c), cos(7/2*d*x + 7/2*c))) + 5*e^2*sin(7/2*d*x + 7/2*c) + 7*e^2*sin(5/7*arctan2(sin(7/2*d*x + 7/2*c), cos
(7/2*d*x + 7/2*c))) + 35*e^2*sin(3/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))) + 105*e^2*sin(1/7*ar
ctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))))*sqrt(e)/(sqrt(a)*d)

Giac [F]

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

[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 = 6.14 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.63 \[ \int \frac {(e \cos (c+d x))^{5/2}}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {e^2\,\sqrt {e\,\cos \left (c+d\,x\right )}\,\sqrt {\frac {a\,\left (\cos \left (2\,c+2\,d\,x\right )+1+\sin \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,c+2\,d\,x\right )+1}}\,\left (\cos \left (2\,c+2\,d\,x\right )\,28{}\mathrm {i}+\cos \left (4\,c+4\,d\,x\right )\,5{}\mathrm {i}+42\,\sin \left (2\,c+2\,d\,x\right )+5\,\sin \left (4\,c+4\,d\,x\right )-105{}\mathrm {i}\right )}{140\,a\,d} \]

[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)*28i + cos(4*c + 4*d*x)*5i + 42*sin(2*c + 2*d*x) + 5*sin(4*c + 4*d*x) - 105i))/(140*a*d)