\(\int \frac {(e \cos (c+d x))^{5/2}}{(a+i a \tan (c+d x))^2} \, dx\) [664]

Optimal result

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

[Out]

42/65*(e*cos(d*x+c))^(5/2)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*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))

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {3596, 3581, 3854, 3856, 2719} \[ \int \frac {(e \cos (c+d x))^{5/2}}{(a+i a \tan (c+d x))^2} \, dx=\frac {42 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) (e \cos (c+d x))^{5/2}}{65 a^2 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {4 i \cos ^2(c+d x) (e \cos (c+d x))^{5/2}}{13 d \left (a^2+i a^2 \tan (c+d x)\right )}+\frac {2 \sin (c+d x) \cos (c+d x) (e \cos (c+d x))^{5/2}}{13 a^2 d}+\frac {14 \tan (c+d x) (e \cos (c+d x))^{5/2}}{65 a^2 d} \]

[In]

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

[Out]

(42*(e*Cos[c + d*x])^(5/2)*EllipticE[(c + d*x)/2, 2])/(65*a^2*d*Cos[c + d*x]^(5/2)) + (2*Cos[c + d*x]*(e*Cos[c
+ d*x])^(5/2)*Sin[c + d*x])/(13*a^2*d) + (14*(e*Cos[c + d*x])^(5/2)*Tan[c + d*x])/(65*a^2*d) + (((4*I)/13)*Co
s[c + d*x]^2*(e*Cos[c + d*x])^(5/2))/(d*(a^2 + I*a^2*Tan[c + d*x]))

Rule 2719

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 3581

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] - Dist[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] || (Integers
Q[n, m + 1/2] && GtQ[2*m + n + 1, 0])) && IntegerQ[2*m]

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]

Rule 3854

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
] + Dist[(n + 1)/(b^2*n), Int[(b*Csc[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && Integer
Q[2*n]

Rule 3856

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(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]

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} (a+i a \tan (c+d x))^2} \, dx \\ & = \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 )}+\frac {\left (9 e^2 (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}\right ) \int \frac {1}{(e \sec (c+d x))^{9/2}} \, dx}{13 a^2} \\ & = \frac {2 \cos (c+d x) (e \cos (c+d x))^{5/2} \sin (c+d x)}{13 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 )}+\frac {\left (7 (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}} \, dx}{13 a^2} \\ & = \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 )}+\frac {\left (21 (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}\right ) \int \frac {1}{\sqrt {e \sec (c+d x)}} \, dx}{65 a^2 e^2} \\ & = \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 )}+\frac {\left (21 (e \cos (c+d x))^{5/2}\right ) \int \sqrt {\cos (c+d x)} \, dx}{65 a^2 \cos ^{\frac {5}{2}}(c+d x)} \\ & = \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 )} \\ \end{align*}

Mathematica [C] (warning: unable to verify)

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

Time = 6.00 (sec) , antiderivative size = 464, normalized size of antiderivative = 3.01 \[ \int \frac {(e \cos (c+d x))^{5/2}}{(a+i a \tan (c+d x))^2} \, dx=\frac {(e \cos (c+d x))^{5/2} \sec ^5(c+d x) (\cos (d x)+i \sin (d x))^2 \left (-21 \cos (c) \, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2(d x+\arctan (\tan (c)))\right ) \sin (d x+\arctan (\tan (c)))-42 i \, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2(d x+\arctan (\tan (c)))\right ) \sin (c) \sin (d x+\arctan (\tan (c)))+21 i (3 \cos (c-d x-\arctan (\tan (c)))+\cos (c+d x+\arctan (\tan (c)))) \sqrt {\sin ^2(d x+\arctan (\tan (c)))}+\frac {21}{2} (3 \cos (c-d x-\arctan (\tan (c)))+\cos (c+d x+\arctan (\tan (c)))) \cot (c) \sqrt {\sin ^2(d x+\arctan (\tan (c)))}-\frac {1}{8} \cos (c+d x) \csc (c) \sqrt {\sec ^2(c)} (\cos (2 d x)-i \sin (2 d x)) (178 \cos (c+2 d x)+158 \cos (3 c+2 d x)-9 \cos (3 c+4 d x)+9 \cos (5 c+4 d x)-88 i \sin (c)+208 i \sin (c+2 d x)+128 i \sin (3 c+2 d x)-4 i \sin (3 c+4 d x)+4 i \sin (5 c+4 d x)) \sqrt {\sin ^2(d x+\arctan (\tan (c)))}+21 \, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2(d x+\arctan (\tan (c)))\right ) \sin (c) \sin (d x+\arctan (\tan (c))) \tan (c)-\frac {21}{2} (3 \cos (c-d x-\arctan (\tan (c)))+\cos (c+d x+\arctan (\tan (c)))) \sqrt {\sin ^2(d x+\arctan (\tan (c)))} \tan (c)\right )}{65 d \sqrt {\sec ^2(c)} \sqrt {\sin ^2(d x+\arctan (\tan (c)))} (a+i a \tan (c+d x))^2} \]

[In]

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

[Out]

((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}, C
os[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]*Sin[c]*Sin[d*x + ArcTan[Tan[c]]]*Tan[c] - (21*(3*Cos[c - d*x - ArcTan[Tan[c]]] + Cos[c + d*x + ArcTan[Ta
n[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*Tan[c + 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. 350 vs. \(2 (160 ) = 320\).

Time = 8.42 (sec) , antiderivative size = 351, normalized size of antiderivative = 2.28

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

[In]

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

[Out]

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*(840*I*sin(1/2*d*x+1/2*c)^5+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)-6720*
I*sin(1/2*d*x+1/2*c)^11+4960*sin(1/2*d*x+1/2*c)^10*cos(1/2*d*x+1/2*c)-1280*I*sin(1/2*d*x+1/2*c)^15-3520*cos(1/
2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^8-2800*I*sin(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)+10*
I*sin(1/2*d*x+1/2*c)-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*cos(1/2*d*x+1/
2*c)*sin(1/2*d*x+1/2*c)^2+21*EllipticE(cos(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)+4480*I*sin(1/2*d*x+1/2*c)^13)/d

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.09 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.91 \[ \int \frac {(e \cos (c+d x))^{5/2}}{(a+i a \tan (c+d x))^2} \, dx=\frac {{\left (336 i \, \sqrt {2} e^{\frac {5}{2}} e^{\left (6 i \, d x + 6 i \, c\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right ) + \sqrt {\frac {1}{2}} {\left (-13 i \, e^{2} e^{\left (8 i \, d x + 8 i \, c\right )} + 386 i \, e^{2} e^{\left (6 i \, d x + 6 i \, c\right )} + 88 i \, e^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + 30 i \, e^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + 5 i \, e^{2}\right )} \sqrt {e e^{\left (2 i \, d x + 2 i \, c\right )} + e} e^{\left (-\frac {1}{2} i \, d x - \frac {1}{2} i \, c\right )}\right )} e^{\left (-6 i \, d x - 6 i \, c\right )}}{520 \, a^{2} d} \]

[In]

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

[Out]

1/520*(336*I*sqrt(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. \[ \int \frac {(e \cos (c+d x))^{5/2}}{(a+i a \tan (c+d x))^2} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F(-2)]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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