\(\int \frac {(e \cos (c+d x))^{5/2}}{(a+i a \tan (c+d x))^2} \, dx\) [673]

Optimal result
Mathematica [C] (warning: unable to verify)
Rubi [A] (verified)
Maple [B] (verified)
Fricas [A] (verification not implemented)
Sympy [F(-1)]
Maxima [F(-2)]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 28, antiderivative size = 154 \[ \int \frac {(e \cos (c+d x))^{5/2}}{(a+i a \tan (c+d x))^2} \, dx=\frac {42 (e \cos (c+d x))^{5/2} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{65 a^2 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 \cos (c+d x) (e \cos (c+d x))^{5/2} \sin (c+d x)}{13 a^2 d}+\frac {14 (e \cos (c+d x))^{5/2} \tan (c+d x)}{65 a^2 d}+\frac {4 i \cos ^2(c+d x) (e \cos (c+d x))^{5/2}}{13 d \left (a^2+i a^2 \tan (c+d x)\right )} \] Output:

42/65*(e*cos(d*x+c))^(5/2)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/a^2/d/cos 
(d*x+c)^(5/2)+2/13*cos(d*x+c)*(e*cos(d*x+c))^(5/2)*sin(d*x+c)/a^2/d+14/65* 
(e*cos(d*x+c))^(5/2)*tan(d*x+c)/a^2/d+4/13*I*cos(d*x+c)^2*(e*cos(d*x+c))^( 
5/2)/d/(a^2+I*a^2*tan(d*x+c))
 

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 3.99 (sec) , antiderivative size = 464, normalized size of antiderivative = 3.01 \[ \int \frac {(e \cos (c+d x))^{5/2}}{(a+i a \tan (c+d x))^2} \, dx=\frac {(e \cos (c+d x))^{5/2} \sec ^5(c+d x) (\cos (d x)+i \sin (d x))^2 \left (-21 \cos (c) \, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2(d x+\arctan (\tan (c)))\right ) \sin (d x+\arctan (\tan (c)))-42 i \, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2(d x+\arctan (\tan (c)))\right ) \sin (c) \sin (d x+\arctan (\tan (c)))+21 i (3 \cos (c-d x-\arctan (\tan (c)))+\cos (c+d x+\arctan (\tan (c)))) \sqrt {\sin ^2(d x+\arctan (\tan (c)))}+\frac {21}{2} (3 \cos (c-d x-\arctan (\tan (c)))+\cos (c+d x+\arctan (\tan (c)))) \cot (c) \sqrt {\sin ^2(d x+\arctan (\tan (c)))}-\frac {1}{8} \cos (c+d x) \csc (c) \sqrt {\sec ^2(c)} (\cos (2 d x)-i \sin (2 d x)) (178 \cos (c+2 d x)+158 \cos (3 c+2 d x)-9 \cos (3 c+4 d x)+9 \cos (5 c+4 d x)-88 i \sin (c)+208 i \sin (c+2 d x)+128 i \sin (3 c+2 d x)-4 i \sin (3 c+4 d x)+4 i \sin (5 c+4 d x)) \sqrt {\sin ^2(d x+\arctan (\tan (c)))}+21 \, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2(d x+\arctan (\tan (c)))\right ) \sin (c) \sin (d x+\arctan (\tan (c))) \tan (c)-\frac {21}{2} (3 \cos (c-d x-\arctan (\tan (c)))+\cos (c+d x+\arctan (\tan (c)))) \sqrt {\sin ^2(d x+\arctan (\tan (c)))} \tan (c)\right )}{65 d \sqrt {\sec ^2(c)} \sqrt {\sin ^2(d x+\arctan (\tan (c)))} (a+i a \tan (c+d x))^2} \] Input:

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

Output:

((e*Cos[c + d*x])^(5/2)*Sec[c + d*x]^5*(Cos[d*x] + I*Sin[d*x])^2*(-21*Cos[ 
c]*HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sin 
[d*x + ArcTan[Tan[c]]] - (42*I)*HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cos 
[d*x + ArcTan[Tan[c]]]^2]*Sin[c]*Sin[d*x + ArcTan[Tan[c]]] + (21*I)*(3*Cos 
[c - d*x - ArcTan[Tan[c]]] + Cos[c + d*x + ArcTan[Tan[c]]])*Sqrt[Sin[d*x + 
 ArcTan[Tan[c]]]^2] + (21*(3*Cos[c - d*x - ArcTan[Tan[c]]] + Cos[c + d*x + 
 ArcTan[Tan[c]]])*Cot[c]*Sqrt[Sin[d*x + ArcTan[Tan[c]]]^2])/2 - (Cos[c + d 
*x]*Csc[c]*Sqrt[Sec[c]^2]*(Cos[2*d*x] - I*Sin[2*d*x])*(178*Cos[c + 2*d*x] 
+ 158*Cos[3*c + 2*d*x] - 9*Cos[3*c + 4*d*x] + 9*Cos[5*c + 4*d*x] - (88*I)* 
Sin[c] + (208*I)*Sin[c + 2*d*x] + (128*I)*Sin[3*c + 2*d*x] - (4*I)*Sin[3*c 
 + 4*d*x] + (4*I)*Sin[5*c + 4*d*x])*Sqrt[Sin[d*x + ArcTan[Tan[c]]]^2])/8 + 
 21*HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Si 
n[c]*Sin[d*x + ArcTan[Tan[c]]]*Tan[c] - (21*(3*Cos[c - d*x - ArcTan[Tan[c] 
]] + Cos[c + d*x + ArcTan[Tan[c]]])*Sqrt[Sin[d*x + ArcTan[Tan[c]]]^2]*Tan[ 
c])/2))/(65*d*Sqrt[Sec[c]^2]*Sqrt[Sin[d*x + ArcTan[Tan[c]]]^2]*(a + I*a*Ta 
n[c + d*x])^2)
 

Rubi [A] (verified)

Time = 0.86 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.21, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3042, 3998, 3042, 3981, 3042, 4256, 3042, 4256, 3042, 4258, 3042, 3119}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(e \cos (c+d x))^{5/2}}{(a+i a \tan (c+d x))^2}dx\)

\(\Big \downarrow \) 3998

\(\displaystyle (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2} \int \frac {1}{(e \sec (c+d x))^{5/2} (i \tan (c+d x) a+a)^2}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2} \int \frac {1}{(e \sec (c+d x))^{5/2} (i \tan (c+d x) a+a)^2}dx\)

\(\Big \downarrow \) 3981

\(\displaystyle (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2} \left (\frac {9 e^2 \int \frac {1}{(e \sec (c+d x))^{9/2}}dx}{13 a^2}+\frac {4 i e^2}{13 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{9/2}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2} \left (\frac {9 e^2 \int \frac {1}{\left (e \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{9/2}}dx}{13 a^2}+\frac {4 i e^2}{13 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{9/2}}\right )\)

\(\Big \downarrow \) 4256

\(\displaystyle (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2} \left (\frac {9 e^2 \left (\frac {7 \int \frac {1}{(e \sec (c+d x))^{5/2}}dx}{9 e^2}+\frac {2 \sin (c+d x)}{9 d e (e \sec (c+d x))^{7/2}}\right )}{13 a^2}+\frac {4 i e^2}{13 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{9/2}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2} \left (\frac {9 e^2 \left (\frac {7 \int \frac {1}{\left (e \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}}dx}{9 e^2}+\frac {2 \sin (c+d x)}{9 d e (e \sec (c+d x))^{7/2}}\right )}{13 a^2}+\frac {4 i e^2}{13 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{9/2}}\right )\)

\(\Big \downarrow \) 4256

\(\displaystyle (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2} \left (\frac {9 e^2 \left (\frac {7 \left (\frac {3 \int \frac {1}{\sqrt {e \sec (c+d x)}}dx}{5 e^2}+\frac {2 \sin (c+d x)}{5 d e (e \sec (c+d x))^{3/2}}\right )}{9 e^2}+\frac {2 \sin (c+d x)}{9 d e (e \sec (c+d x))^{7/2}}\right )}{13 a^2}+\frac {4 i e^2}{13 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{9/2}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2} \left (\frac {9 e^2 \left (\frac {7 \left (\frac {3 \int \frac {1}{\sqrt {e \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{5 e^2}+\frac {2 \sin (c+d x)}{5 d e (e \sec (c+d x))^{3/2}}\right )}{9 e^2}+\frac {2 \sin (c+d x)}{9 d e (e \sec (c+d x))^{7/2}}\right )}{13 a^2}+\frac {4 i e^2}{13 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{9/2}}\right )\)

\(\Big \downarrow \) 4258

\(\displaystyle (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2} \left (\frac {9 e^2 \left (\frac {7 \left (\frac {3 \int \sqrt {\cos (c+d x)}dx}{5 e^2 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {2 \sin (c+d x)}{5 d e (e \sec (c+d x))^{3/2}}\right )}{9 e^2}+\frac {2 \sin (c+d x)}{9 d e (e \sec (c+d x))^{7/2}}\right )}{13 a^2}+\frac {4 i e^2}{13 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{9/2}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2} \left (\frac {9 e^2 \left (\frac {7 \left (\frac {3 \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{5 e^2 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {2 \sin (c+d x)}{5 d e (e \sec (c+d x))^{3/2}}\right )}{9 e^2}+\frac {2 \sin (c+d x)}{9 d e (e \sec (c+d x))^{7/2}}\right )}{13 a^2}+\frac {4 i e^2}{13 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{9/2}}\right )\)

\(\Big \downarrow \) 3119

\(\displaystyle (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2} \left (\frac {9 e^2 \left (\frac {7 \left (\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d e^2 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {2 \sin (c+d x)}{5 d e (e \sec (c+d x))^{3/2}}\right )}{9 e^2}+\frac {2 \sin (c+d x)}{9 d e (e \sec (c+d x))^{7/2}}\right )}{13 a^2}+\frac {4 i e^2}{13 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{9/2}}\right )\)

Input:

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

Output:

(e*Cos[c + d*x])^(5/2)*(e*Sec[c + d*x])^(5/2)*((9*e^2*((2*Sin[c + d*x])/(9 
*d*e*(e*Sec[c + d*x])^(7/2)) + (7*((6*EllipticE[(c + d*x)/2, 2])/(5*d*e^2* 
Sqrt[Cos[c + d*x]]*Sqrt[e*Sec[c + d*x]]) + (2*Sin[c + d*x])/(5*d*e*(e*Sec[ 
c + d*x])^(3/2))))/(9*e^2)))/(13*a^2) + (((4*I)/13)*e^2)/(d*(e*Sec[c + d*x 
])^(9/2)*(a^2 + I*a^2*Tan[c + d*x])))
 

Defintions of rubi rules used

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3119
Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* 
(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

rule 3981
Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x 
_)])^(n_), x_Symbol] :> Simp[2*d^2*(d*Sec[e + f*x])^(m - 2)*((a + b*Tan[e + 
 f*x])^(n + 1)/(b*f*(m + 2*n))), x] - Simp[d^2*((m - 2)/(b^2*(m + 2*n))) 
Int[(d*Sec[e + f*x])^(m - 2)*(a + b*Tan[e + f*x])^(n + 2), x], x] /; FreeQ[ 
{a, b, d, e, f, m}, x] && EqQ[a^2 + b^2, 0] && LtQ[n, -1] && ((ILtQ[n/2, 0] 
 && IGtQ[m - 1/2, 0]) || EqQ[n, -2] || IGtQ[m + n, 0] || (IntegersQ[n, m + 
1/2] && GtQ[2*m + n + 1, 0])) && IntegerQ[2*m]
 

rule 3998
Int[(cos[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x 
_)])^(n_.), x_Symbol] :> Simp[(d*Cos[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]
 

rule 4256
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( 
b*Csc[c + d*x])^(n + 1)/(b*d*n)), x] + Simp[(n + 1)/(b^2*n)   Int[(b*Csc[c 
+ d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && IntegerQ[2* 
n]
 

rule 4258
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x] 
)^n*Sin[c + d*x]^n   Int[1/Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && 
 EqQ[n^2, 1/4]
 
Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 350 vs. \(2 (136 ) = 272\).

Time = 12.76 (sec) , antiderivative size = 351, normalized size of antiderivative = 2.28

method result size
default \(\frac {2 e^{3} \left (4480 i \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}+1280 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{14} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-140 i \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}-3840 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{12} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+10 i \sin \left (\frac {d x}{2}+\frac {c}{2}\right )+4960 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{10} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-6720 i \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}-3520 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}-2800 i \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}+1496 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-1280 i \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{15}-376 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+5600 i \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}+44 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+21 \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}+840 i \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}\right )}{65 a^{2} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} e +e}\, d}\) \(351\)

Input:

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

Output:

2/65/a^2/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)*e^3*(4480* 
I*sin(1/2*d*x+1/2*c)^13+1280*sin(1/2*d*x+1/2*c)^14*cos(1/2*d*x+1/2*c)-140* 
I*sin(1/2*d*x+1/2*c)^3-3840*sin(1/2*d*x+1/2*c)^12*cos(1/2*d*x+1/2*c)+10*I* 
sin(1/2*d*x+1/2*c)+4960*sin(1/2*d*x+1/2*c)^10*cos(1/2*d*x+1/2*c)-6720*I*si 
n(1/2*d*x+1/2*c)^11-3520*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^8-2800*I*si 
n(1/2*d*x+1/2*c)^7+1496*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)-1280*I*sin 
(1/2*d*x+1/2*c)^15-376*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+5600*I*sin( 
1/2*d*x+1/2*c)^9+44*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+21*EllipticE(c 
os(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1 
/2*c)^2)^(1/2)+840*I*sin(1/2*d*x+1/2*c)^5)/d
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.91 \[ \int \frac {(e \cos (c+d x))^{5/2}}{(a+i a \tan (c+d x))^2} \, dx=\frac {{\left (672 i \, \sqrt {\frac {1}{2}} e^{\frac {5}{2}} e^{\left (6 i \, d x + 6 i \, c\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right ) + \sqrt {\frac {1}{2}} {\left (-13 i \, e^{2} e^{\left (8 i \, d x + 8 i \, c\right )} + 386 i \, e^{2} e^{\left (6 i \, d x + 6 i \, c\right )} + 88 i \, e^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + 30 i \, e^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + 5 i \, e^{2}\right )} \sqrt {e e^{\left (2 i \, d x + 2 i \, c\right )} + e} e^{\left (-\frac {1}{2} i \, d x - \frac {1}{2} i \, c\right )}\right )} e^{\left (-6 i \, d x - 6 i \, c\right )}}{520 \, a^{2} d} \] Input:

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

Output:

1/520*(672*I*sqrt(1/2)*e^(5/2)*e^(6*I*d*x + 6*I*c)*weierstrassZeta(-4, 0, 
weierstrassPInverse(-4, 0, e^(I*d*x + I*c))) + sqrt(1/2)*(-13*I*e^2*e^(8*I 
*d*x + 8*I*c) + 386*I*e^2*e^(6*I*d*x + 6*I*c) + 88*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)*e^(-1/2*I*d*x - 1/2*I*c))*e^(-6*I*d*x - 6*I*c)/(a^2*d)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {(e \cos (c+d x))^{5/2}}{(a+i a \tan (c+d x))^2} \, dx=\text {Exception raised: RuntimeError} \] Input:

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

Output:

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is un 
defined.
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {(e \cos (c+d x))^{5/2}}{(a+i a \tan (c+d x))^2} \, dx=\int \frac {{\left (e\,\cos \left (c+d\,x\right )\right )}^{5/2}}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^2} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {(e \cos (c+d x))^{5/2}}{(a+i a \tan (c+d x))^2} \, dx=-\frac {\sqrt {e}\, \left (\int \frac {\sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{2}}{\tan \left (d x +c \right )^{2}-2 \tan \left (d x +c \right ) i -1}d x \right ) e^{2}}{a^{2}} \] Input:

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

Output:

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