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

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

[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]))

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Rubi [A]  time = 0.19137, antiderivative size = 154, normalized size of antiderivative = 1., 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 = {3515, 3500, 3769, 3771, 2639} \[ \frac{4 i \cos ^2(c+d x) (e \cos (c+d x))^{5/2}}{13 d \left (a^2+i a^2 \tan (c+d x)\right )}+\frac{42 E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) (e \cos (c+d x))^{5/2}}{65 a^2 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 \sin (c+d x) \cos (c+d x) (e \cos (c+d x))^{5/2}}{13 a^2 d}+\frac{14 \tan (c+d x) (e \cos (c+d x))^{5/2}}{65 a^2 d} \]

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 3515

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 3500

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 3769

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 3771

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]

Rule 2639

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

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*}

Mathematica [C]  time = 3.12367, size = 471, normalized size = 3.06 \[ \frac{(\cos (d x)+i \sin (d x))^2 (e \cos (c+d x))^{5/2} \left (\frac{14 \sqrt{2} \csc (c) e^{-i d x} (\cos (2 c)+i \sin (2 c)) \left (e^{2 i d x} \sqrt{1+e^{2 i (c+d x)}} \, _2F_1\left (\frac{1}{2},\frac{3}{4};\frac{7}{4};-e^{2 i (c+d x)}\right )+3 e^{2 i (c+d x)}-3 \sqrt{1-i e^{i (c+d x)}} \sqrt{e^{i (c+d x)} \left (e^{i (c+d x)}-i\right )} F\left (\left .\sin ^{-1}\left (\sqrt{\sin (c+d x)-i \cos (c+d x)}\right )\right |-1\right )+3 \sqrt{1-i e^{i (c+d x)}} \sqrt{e^{i (c+d x)} \left (e^{i (c+d x)}-i\right )} E\left (\left .\sin ^{-1}\left (\sqrt{\sin (c+d x)-i \cos (c+d x)}\right )\right |-1\right )+3\right )}{65 \sqrt{e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )}}-\frac{1}{260} \csc (c) \sqrt{\cos (c+d x)} (\cos (2 d x)-i \sin (2 d x)) (208 i \sin (c+2 d x)+128 i \sin (3 c+2 d x)-4 i \sin (3 c+4 d x)+4 i \sin (5 c+4 d x)+178 \cos (c+2 d x)+158 \cos (3 c+2 d x)-9 \cos (3 c+4 d x)+9 \cos (5 c+4 d x)-88 i \sin (c))\right )}{2 d \cos ^{\frac{9}{2}}(c+d x) (a+i a \tan (c+d x))^2} \]

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]  time = 3.112, size = 351, normalized size = 2.3 \begin{align*} -{\frac{2\,{e}^{3}}{65\,{a}^{2}d} \left ( 1280\,i \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{15}-1280\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{14}\cos \left ( 1/2\,dx+c/2 \right ) -5600\,i \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{9}+3840\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{12}\cos \left ( 1/2\,dx+c/2 \right ) -10\,i\sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -4960\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{10}\cos \left ( 1/2\,dx+c/2 \right ) -4480\,i \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{13}+3520\,\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{8}+140\,i \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}-1496\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{6}\cos \left ( 1/2\,dx+c/2 \right ) +2800\,i \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{7}+376\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}\cos \left ( 1/2\,dx+c/2 \right ) -840\,i \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}-21\,{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) \sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}-44\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}\cos \left ( 1/2\,dx+c/2 \right ) +6720\,i \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{11} \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}{\frac{1}{\sqrt{-2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}e+e}}}} \end{align*}

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)

[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*(1280*I*sin(1/2*d*x+1/2*c)^15-1280*sin(1/
2*d*x+1/2*c)^14*cos(1/2*d*x+1/2*c)-5600*I*sin(1/2*d*x+1/2*c)^9+3840*sin(1/2*d*x+1/2*c)^12*cos(1/2*d*x+1/2*c)-1
0*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)-4480*I*sin(1/2*d*x+1/2*c)^13+3520*cos(1/2
*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^8+140*I*sin(1/2*d*x+1/2*c)^3-1496*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+2800*
I*sin(1/2*d*x+1/2*c)^7+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)^5-21*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)-44*sin(1/2*d*x+1/2*c)^2
*cos(1/2*d*x+1/2*c)+6720*I*sin(1/2*d*x+1/2*c)^11)/d

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

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\sqrt{\frac{1}{2}}{\left (-13 i \, e^{2} e^{\left (9 i \, d x + 9 i \, c\right )} + 13 i \, e^{2} e^{\left (8 i \, d x + 8 i \, c\right )} - 286 i \, e^{2} e^{\left (7 i \, d x + 7 i \, c\right )} - 386 i \, e^{2} e^{\left (6 i \, d x + 6 i \, c\right )} + 88 i \, e^{2} e^{\left (5 i \, d x + 5 i \, c\right )} - 88 i \, e^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + 30 i \, e^{2} e^{\left (3 i \, d x + 3 i \, c\right )} - 30 i \, e^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + 5 i \, e^{2} e^{\left (i \, d x + 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 )} + 520 \,{\left (a^{2} d e^{\left (7 i \, d x + 7 i \, c\right )} - a^{2} d e^{\left (6 i \, d x + 6 i \, c\right )}\right )}{\rm integral}\left (\frac{\sqrt{\frac{1}{2}}{\left (-42 i \, e^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - 84 i \, e^{2} e^{\left (i \, d x + i \, c\right )} - 42 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 )}}{65 \,{\left (a^{2} d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, a^{2} d e^{\left (3 i \, d x + 3 i \, c\right )} + 2 \, a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} - 2 \, a^{2} d e^{\left (i \, d x + i \, c\right )} + a^{2} d\right )}}, x\right )}{520 \,{\left (a^{2} d e^{\left (7 i \, d x + 7 i \, c\right )} - a^{2} d e^{\left (6 i \, d x + 6 i \, c\right )}\right )}} \end{align*}

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)*(-13*I*e^2*e^(9*I*d*x + 9*I*c) + 13*I*e^2*e^(8*I*d*x + 8*I*c) - 286*I*e^2*e^(7*I*d*x + 7*I*c)
- 386*I*e^2*e^(6*I*d*x + 6*I*c) + 88*I*e^2*e^(5*I*d*x + 5*I*c) - 88*I*e^2*e^(4*I*d*x + 4*I*c) + 30*I*e^2*e^(3
*I*d*x + 3*I*c) - 30*I*e^2*e^(2*I*d*x + 2*I*c) + 5*I*e^2*e^(I*d*x + 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) + 520*(a^2*d*e^(7*I*d*x + 7*I*c) - a^2*d*e^(6*I*d*x + 6*I*c))*integral(1/65*sqr
t(1/2)*(-42*I*e^2*e^(2*I*d*x + 2*I*c) - 84*I*e^2*e^(I*d*x + I*c) - 42*I*e^2)*sqrt(e*e^(2*I*d*x + 2*I*c) + e)*e
^(-1/2*I*d*x - 1/2*I*c)/(a^2*d*e^(4*I*d*x + 4*I*c) - 2*a^2*d*e^(3*I*d*x + 3*I*c) + 2*a^2*d*e^(2*I*d*x + 2*I*c)
- 2*a^2*d*e^(I*d*x + I*c) + a^2*d), x))/(a^2*d*e^(7*I*d*x + 7*I*c) - a^2*d*e^(6*I*d*x + 6*I*c))

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

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: AttributeError

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

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