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

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

[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))

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Rubi [A]
time = 0.15, antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {3596, 3581, 3854, 3856, 2719} \begin {gather*} \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} \end {gather*}

Antiderivative was successfully verified.

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

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 3.30, size = 471, normalized size = 3.06 \begin {gather*} \frac {(e \cos (c+d x))^{5/2} (\cos (d x)+i \sin (d x))^2 \left (\frac {14 \sqrt {2} e^{-i d x} \csc (c) \left (3+3 e^{2 i (c+d x)}+3 \sqrt {1-i e^{i (c+d x)}} \sqrt {e^{i (c+d x)} \left (-i+e^{i (c+d x)}\right )} E\left (\left .\text {ArcSin}\left (\sqrt {-i \cos (c+d x)+\sin (c+d x)}\right )\right |-1\right )-3 \sqrt {1-i e^{i (c+d x)}} \sqrt {e^{i (c+d x)} \left (-i+e^{i (c+d x)}\right )} F\left (\left .\text {ArcSin}\left (\sqrt {-i \cos (c+d x)+\sin (c+d x)}\right )\right |-1\right )+e^{2 i d x} \sqrt {1+e^{2 i (c+d x)}} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-e^{2 i (c+d x)}\right )\right ) (\cos (2 c)+i \sin (2 c))}{65 \sqrt {e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )}}-\frac {1}{260} \sqrt {\cos (c+d x)} \csc (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))\right )}{2 d \cos ^{\frac {9}{2}}(c+d x) (a+i a \tan (c+d x))^2} \end {gather*}

Antiderivative was successfully verified.

[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)*(Cos[d*x] + I*Sin[d*x])^2*((14*Sqrt[2]*Csc[c]*(3 + 3*E^((2*I)*(c + d*x)) + 3*Sqrt[1 -
I*E^(I*(c + d*x))]*Sqrt[E^(I*(c + d*x))*(-I + E^(I*(c + d*x)))]*EllipticE[ArcSin[Sqrt[(-I)*Cos[c + d*x] + Sin[
c + d*x]]], -1] - 3*Sqrt[1 - I*E^(I*(c + d*x))]*Sqrt[E^(I*(c + d*x))*(-I + E^(I*(c + d*x)))]*EllipticF[ArcSin[
Sqrt[(-I)*Cos[c + d*x] + Sin[c + d*x]]], -1] + E^((2*I)*d*x)*Sqrt[1 + E^((2*I)*(c + d*x))]*Hypergeometric2F1[1
/2, 3/4, 7/4, -E^((2*I)*(c + d*x))])*(Cos[2*c] + I*Sin[2*c]))/(65*E^(I*d*x)*Sqrt[(1 + E^((2*I)*(c + d*x)))/E^(
I*(c + d*x))]) - (Sqrt[Cos[c + d*x]]*Csc[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]))/260))/(2*d*Cos[c + d*x]^(9/2)*(a + I*a*Tan[c + d*x])^
2)

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Maple [B] 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 = 1.76, size = 351, normalized size = 2.28

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

Verification of antiderivative is not currently implemented for this CAS.

[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*(-2800*I*sin(1/2*d*x+1/2*c)^7+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)-128
0*I*sin(1/2*d*x+1/2*c)^15+4960*sin(1/2*d*x+1/2*c)^10*cos(1/2*d*x+1/2*c)+5600*I*sin(1/2*d*x+1/2*c)^9-3520*cos(1
/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^8-6720*I*sin(1/2*d*x+1/2*c)^11+1496*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6+1
0*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)+840*I*sin(1/2*d*x+1/2*c)^5+44*sin(1/2*d*x+1
/2*c)^2*cos(1/2*d*x+1/2*c)+21*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*
d*x+1/2*c),2^(1/2))+4480*I*sin(1/2*d*x+1/2*c)^13)/d

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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.

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.10, size = 127, normalized size = 0.82 \begin {gather*} \frac {{\left (\sqrt {\frac {1}{2}} {\left (5 i \, e^{\frac {5}{2}} - 13 i \, e^{\left (8 i \, d x + 8 i \, c + \frac {5}{2}\right )} + 386 i \, e^{\left (6 i \, d x + 6 i \, c + \frac {5}{2}\right )} + 88 i \, e^{\left (4 i \, d x + 4 i \, c + \frac {5}{2}\right )} + 30 i \, e^{\left (2 i \, d x + 2 i \, c + \frac {5}{2}\right )}\right )} \sqrt {e^{\left (2 i \, d x + 2 i \, c\right )} + 1} e^{\left (-\frac {1}{2} i \, d x - \frac {1}{2} i \, c\right )} + 336 i \, \sqrt {2} e^{\left (6 i \, d x + 6 i \, c + \frac {5}{2}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right )\right )} e^{\left (-6 i \, d x - 6 i \, c\right )}}{520 \, a^{2} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/520*(sqrt(1/2)*(5*I*e^(5/2) - 13*I*e^(8*I*d*x + 8*I*c + 5/2) + 386*I*e^(6*I*d*x + 6*I*c + 5/2) + 88*I*e^(4*I
*d*x + 4*I*c + 5/2) + 30*I*e^(2*I*d*x + 2*I*c + 5/2))*sqrt(e^(2*I*d*x + 2*I*c) + 1)*e^(-1/2*I*d*x - 1/2*I*c) +
336*I*sqrt(2)*e^(6*I*d*x + 6*I*c + 5/2)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, e^(I*d*x + I*c))))*
e^(-6*I*d*x - 6*I*c)/(a^2*d)

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: SystemError >> excessive stack use: stack is 4371 deep

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \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 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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