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

Optimal. Leaf size=90 \[ -\frac{2 i a (e \cos (c+d x))^{5/2}}{5 d}+\frac{6 a E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) (e \cos (c+d x))^{5/2}}{5 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 a \tan (c+d x) (e \cos (c+d x))^{5/2}}{5 d} \]

[Out]

(((-2*I)/5)*a*(e*Cos[c + d*x])^(5/2))/d + (6*a*(e*Cos[c + d*x])^(5/2)*EllipticE[(c + d*x)/2, 2])/(5*d*Cos[c +
d*x]^(5/2)) + (2*a*(e*Cos[c + d*x])^(5/2)*Tan[c + d*x])/(5*d)

________________________________________________________________________________________

Rubi [A]  time = 0.111353, antiderivative size = 90, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.192, Rules used = {3515, 3486, 3769, 3771, 2639} \[ -\frac{2 i a (e \cos (c+d x))^{5/2}}{5 d}+\frac{6 a E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) (e \cos (c+d x))^{5/2}}{5 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 a \tan (c+d x) (e \cos (c+d x))^{5/2}}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((-2*I)/5)*a*(e*Cos[c + d*x])^(5/2))/d + (6*a*(e*Cos[c + d*x])^(5/2)*EllipticE[(c + d*x)/2, 2])/(5*d*Cos[c +
d*x]^(5/2)) + (2*a*(e*Cos[c + d*x])^(5/2)*Tan[c + d*x])/(5*d)

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 3486

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*(d*Sec[
e + f*x])^m)/(f*m), x] + Dist[a, Int[(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m}, x] && (IntegerQ[2
*m] || NeQ[a^2 + b^2, 0])

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 (e \cos (c+d x))^{5/2} (a+i a \tan (c+d x)) \, dx &=\left ((e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}\right ) \int \frac{a+i a \tan (c+d x)}{(e \sec (c+d x))^{5/2}} \, dx\\ &=-\frac{2 i a (e \cos (c+d x))^{5/2}}{5 d}+\left (a (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\\ &=-\frac{2 i a (e \cos (c+d x))^{5/2}}{5 d}+\frac{2 a (e \cos (c+d x))^{5/2} \tan (c+d x)}{5 d}+\frac{\left (3 a (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}{5 e^2}\\ &=-\frac{2 i a (e \cos (c+d x))^{5/2}}{5 d}+\frac{2 a (e \cos (c+d x))^{5/2} \tan (c+d x)}{5 d}+\frac{\left (3 a (e \cos (c+d x))^{5/2}\right ) \int \sqrt{\cos (c+d x)} \, dx}{5 \cos ^{\frac{5}{2}}(c+d x)}\\ &=-\frac{2 i a (e \cos (c+d x))^{5/2}}{5 d}+\frac{6 a (e \cos (c+d x))^{5/2} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 a (e \cos (c+d x))^{5/2} \tan (c+d x)}{5 d}\\ \end{align*}

Mathematica [C]  time = 12.4758, size = 387, normalized size = 4.3 \[ \frac{(\cos (d x)-i \sin (d x)) (a+i a \tan (c+d x)) (e \cos (c+d x))^{5/2} \left (\frac{2 \sqrt{2} (\cot (c)-i) e^{-i d x} \left (e^{2 i d x} \sqrt{1+e^{2 i (c+d x)}} \, _2F_1\left (\frac{1}{2},\frac{3}{4};\frac{7}{4};-e^{2 i (c+d x)}\right )+3 e^{2 i (c+d x)}-3 \sqrt{1-i e^{i (c+d x)}} \sqrt{e^{i (c+d x)} \left (e^{i (c+d x)}-i\right )} F\left (\left .\sin ^{-1}\left (\sqrt{\sin (c+d x)-i \cos (c+d x)}\right )\right |-1\right )+3 \sqrt{1-i e^{i (c+d x)}} \sqrt{e^{i (c+d x)} \left (e^{i (c+d x)}-i\right )} E\left (\left .\sin ^{-1}\left (\sqrt{\sin (c+d x)-i \cos (c+d x)}\right )\right |-1\right )+3\right )}{5 \sqrt{e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )}}+\frac{2}{5} \sin (c) \sqrt{\cos (c+d x)} \left ((1-i \cot (c)) \cos (2 d x)+\cot (c) (\sin (2 d x)+5 i)-6 \cot ^2(c)+i \sin (2 d x)-1\right )\right )}{2 d \cos ^{\frac{3}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((e*Cos[c + d*x])^(5/2)*(Cos[d*x] - I*Sin[d*x])*((2*Sqrt[2]*(-I + Cot[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[Arc
Sin[Sqrt[(-I)*Cos[c + d*x] + Sin[c + d*x]]], -1] + E^((2*I)*d*x)*Sqrt[1 + E^((2*I)*(c + d*x))]*Hypergeometric2
F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))]))/(5*E^(I*d*x)*Sqrt[(1 + E^((2*I)*(c + d*x)))/E^(I*(c + d*x))]) + (2*S
qrt[Cos[c + d*x]]*Sin[c]*(-1 + Cos[2*d*x]*(1 - I*Cot[c]) - 6*Cot[c]^2 + I*Sin[2*d*x] + Cot[c]*(5*I + Sin[2*d*x
])))/5)*(a + I*a*Tan[c + d*x]))/(2*d*Cos[c + d*x]^(3/2))

________________________________________________________________________________________

Maple [B]  time = 2.346, size = 205, normalized size = 2.3 \begin{align*}{\frac{2\,a{e}^{3}}{5\,d} \left ( 8\,i \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{7}+8\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{6}\cos \left ( 1/2\,dx+c/2 \right ) -12\,i \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}-8\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}\cos \left ( 1/2\,dx+c/2 \right ) +6\,i \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}+3\,{\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}}+2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}\cos \left ( 1/2\,dx+c/2 \right ) -i\sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \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)),x)

[Out]

2/5/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)*a*e^3*(8*I*sin(1/2*d*x+1/2*c)^7+8*sin(1/2*d*x+1/2*c
)^6*cos(1/2*d*x+1/2*c)-12*I*sin(1/2*d*x+1/2*c)^5-8*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+6*I*sin(1/2*d*x+1/2
*c)^3+3*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)+2*
sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-I*sin(1/2*d*x+1/2*c))/d

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (e \cos \left (d x + c\right )\right )^{\frac{5}{2}}{\left (i \, a \tan \left (d x + c\right ) + a\right )}\,{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)),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\sqrt{\frac{1}{2}}{\left (-i \, a e^{2} e^{\left (3 i \, d x + 3 i \, c\right )} + i \, a e^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - 7 i \, a e^{2} e^{\left (i \, d x + i \, c\right )} - 5 i \, a 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 )} + 5 \,{\left (d e^{\left (i \, d x + i \, c\right )} - d\right )}{\rm integral}\left (\frac{\sqrt{\frac{1}{2}}{\left (-6 i \, a e^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - 12 i \, a e^{2} e^{\left (i \, d x + i \, c\right )} - 6 i \, a 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 )}}{5 \,{\left (d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (3 i \, d x + 3 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} - 2 \, d e^{\left (i \, d x + i \, c\right )} + d\right )}}, x\right )}{5 \,{\left (d e^{\left (i \, d x + i \, c\right )} - d\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)),x, algorithm="fricas")

[Out]

1/5*(sqrt(1/2)*(-I*a*e^2*e^(3*I*d*x + 3*I*c) + I*a*e^2*e^(2*I*d*x + 2*I*c) - 7*I*a*e^2*e^(I*d*x + I*c) - 5*I*a
*e^2)*sqrt(e*e^(2*I*d*x + 2*I*c) + e)*e^(-1/2*I*d*x - 1/2*I*c) + 5*(d*e^(I*d*x + I*c) - d)*integral(1/5*sqrt(1
/2)*(-6*I*a*e^2*e^(2*I*d*x + 2*I*c) - 12*I*a*e^2*e^(I*d*x + I*c) - 6*I*a*e^2)*sqrt(e*e^(2*I*d*x + 2*I*c) + e)*
e^(-1/2*I*d*x - 1/2*I*c)/(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(3*I*d*x + 3*I*c) + 2*d*e^(2*I*d*x + 2*I*c) - 2*d*e^(I
*d*x + I*c) + d), x))/(d*e^(I*d*x + I*c) - d)

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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)),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (e \cos \left (d x + c\right )\right )^{\frac{5}{2}}{\left (i \, a \tan \left (d x + c\right ) + a\right )}\,{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)),x, algorithm="giac")

[Out]

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