3.657 \(\int (e \cos (c+d x))^{3/2} (a+i a \tan (c+d x)) \, dx\)

Optimal. Leaf size=90 \[ -\frac{2 i a (e \cos (c+d x))^{3/2}}{3 d}+\frac{2 a F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) (e \cos (c+d x))^{3/2}}{3 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 a \tan (c+d x) (e \cos (c+d x))^{3/2}}{3 d} \]

[Out]

(((-2*I)/3)*a*(e*Cos[c + d*x])^(3/2))/d + (2*a*(e*Cos[c + d*x])^(3/2)*EllipticF[(c + d*x)/2, 2])/(3*d*Cos[c +
d*x]^(3/2)) + (2*a*(e*Cos[c + d*x])^(3/2)*Tan[c + d*x])/(3*d)

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Rubi [A]  time = 0.109314, 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, 2641} \[ -\frac{2 i a (e \cos (c+d x))^{3/2}}{3 d}+\frac{2 a F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) (e \cos (c+d x))^{3/2}}{3 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 a \tan (c+d x) (e \cos (c+d x))^{3/2}}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[(e*Cos[c + d*x])^(3/2)*(a + I*a*Tan[c + d*x]),x]

[Out]

(((-2*I)/3)*a*(e*Cos[c + d*x])^(3/2))/d + (2*a*(e*Cos[c + d*x])^(3/2)*EllipticF[(c + d*x)/2, 2])/(3*d*Cos[c +
d*x]^(3/2)) + (2*a*(e*Cos[c + d*x])^(3/2)*Tan[c + d*x])/(3*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 2641

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

Rubi steps

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

Mathematica [A]  time = 0.342929, size = 100, normalized size = 1.11 \[ \frac{2 a e \sqrt{\cos (c+d x)} (\tan (c+d x)-i) (\cos (d x)-i \sin (d x)) \sqrt{e \cos (c+d x)} \left (\sqrt{\cos (c+d x)} (\cos (d x)+i \sin (d x))+(\sin (c)+i \cos (c)) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )\right )}{3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(e*Cos[c + d*x])^(3/2)*(a + I*a*Tan[c + d*x]),x]

[Out]

(2*a*e*Sqrt[Cos[c + d*x]]*Sqrt[e*Cos[c + d*x]]*(EllipticF[(c + d*x)/2, 2]*(I*Cos[c] + Sin[c]) + Sqrt[Cos[c + d
*x]]*(Cos[d*x] + I*Sin[d*x]))*(Cos[d*x] - I*Sin[d*x])*(-I + Tan[c + d*x]))/(3*d)

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Maple [A]  time = 2.457, size = 168, normalized size = 1.9 \begin{align*} -{\frac{2\,a{e}^{2}}{3\,d} \left ( 4\,i \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}+4\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}\cos \left ( 1/2\,dx+c/2 \right ) -4\,i \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}+\sqrt{ \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}{\it EllipticF} \left ( \cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) ,\sqrt{2} \right ) -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))^(3/2)*(a+I*a*tan(d*x+c)),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (e \cos \left (d x + c\right )\right )^{\frac{3}{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))^(3/2)*(a+I*a*tan(d*x+c)),x, algorithm="maxima")

[Out]

integrate((e*cos(d*x + c))^(3/2)*(I*a*tan(d*x + c) + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(3/2)*(a+I*a*tan(d*x+c)),x, algorithm="fricas")

[Out]

1/3*(-2*I*sqrt(1/2)*sqrt(e*e^(2*I*d*x + 2*I*c) + e)*a*e*e^(1/2*I*d*x + 1/2*I*c) + 3*d*integral(-2/3*I*sqrt(1/2
)*sqrt(e*e^(2*I*d*x + 2*I*c) + e)*a*e*e^(-1/2*I*d*x - 1/2*I*c)/(d*e^(2*I*d*x + 2*I*c) + d), x))/d

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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))**(3/2)*(a+I*a*tan(d*x+c)),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (e \cos \left (d x + c\right )\right )^{\frac{3}{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))^(3/2)*(a+I*a*tan(d*x+c)),x, algorithm="giac")

[Out]

integrate((e*cos(d*x + c))^(3/2)*(I*a*tan(d*x + c) + a), x)