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

Optimal. Leaf size=94 \[ \frac{2 a e^2 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \sec (c+d x)}}{3 d}+\frac{2 i a (e \sec (c+d x))^{5/2}}{5 d}+\frac{2 a e \sin (c+d x) (e \sec (c+d x))^{3/2}}{3 d} \]

[Out]

(2*a*e^2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[e*Sec[c + d*x]])/(3*d) + (((2*I)/5)*a*(e*Sec[c + d*
x])^(5/2))/d + (2*a*e*(e*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(3*d)

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Rubi [A]  time = 0.060312, antiderivative size = 94, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {3486, 3768, 3771, 2641} \[ \frac{2 a e^2 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \sec (c+d x)}}{3 d}+\frac{2 i a (e \sec (c+d x))^{5/2}}{5 d}+\frac{2 a e \sin (c+d x) (e \sec (c+d x))^{3/2}}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*e^2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[e*Sec[c + d*x]])/(3*d) + (((2*I)/5)*a*(e*Sec[c + d*
x])^(5/2))/d + (2*a*e*(e*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(3*d)

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 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[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 \sec (c+d x))^{5/2} (a+i a \tan (c+d x)) \, dx &=\frac{2 i a (e \sec (c+d x))^{5/2}}{5 d}+a \int (e \sec (c+d x))^{5/2} \, dx\\ &=\frac{2 i a (e \sec (c+d x))^{5/2}}{5 d}+\frac{2 a e (e \sec (c+d x))^{3/2} \sin (c+d x)}{3 d}+\frac{1}{3} \left (a e^2\right ) \int \sqrt{e \sec (c+d x)} \, dx\\ &=\frac{2 i a (e \sec (c+d x))^{5/2}}{5 d}+\frac{2 a e (e \sec (c+d x))^{3/2} \sin (c+d x)}{3 d}+\frac{1}{3} \left (a e^2 \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{2 a e^2 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \sec (c+d x)}}{3 d}+\frac{2 i a (e \sec (c+d x))^{5/2}}{5 d}+\frac{2 a e (e \sec (c+d x))^{3/2} \sin (c+d x)}{3 d}\\ \end{align*}

Mathematica [A]  time = 0.504491, size = 57, normalized size = 0.61 \[ \frac{a (e \sec (c+d x))^{5/2} \left (5 \sin (2 (c+d x))+10 \cos ^{\frac{5}{2}}(c+d x) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )+6 i\right )}{15 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(e*Sec[c + d*x])^(5/2)*(6*I + 10*Cos[c + d*x]^(5/2)*EllipticF[(c + d*x)/2, 2] + 5*Sin[2*(c + d*x)]))/(15*d)

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Maple [A]  time = 0.232, size = 192, normalized size = 2. \begin{align*}{\frac{2\,a \left ( \cos \left ( dx+c \right ) +1 \right ) ^{2} \left ( \cos \left ( dx+c \right ) -1 \right ) ^{2}}{15\,d \left ( \sin \left ( dx+c \right ) \right ) ^{4}} \left ( 5\,i\sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}} \left ( \cos \left ( dx+c \right ) \right ) ^{3}{\it EllipticF} \left ({\frac{i \left ( \cos \left ( dx+c \right ) -1 \right ) }{\sin \left ( dx+c \right ) }},i \right ) +5\,i\sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}} \left ( \cos \left ( dx+c \right ) \right ) ^{2}{\it EllipticF} \left ({\frac{i \left ( \cos \left ( dx+c \right ) -1 \right ) }{\sin \left ( dx+c \right ) }},i \right ) +5\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +3\,i \right ) \left ({\frac{e}{\cos \left ( dx+c \right ) }} \right ) ^{{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*a/d*(cos(d*x+c)+1)^2*(cos(d*x+c)-1)^2*(5*I*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*cos
(d*x+c)^3*EllipticF(I*(cos(d*x+c)-1)/sin(d*x+c),I)+5*I*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1
/2)*cos(d*x+c)^2*EllipticF(I*(cos(d*x+c)-1)/sin(d*x+c),I)+5*cos(d*x+c)*sin(d*x+c)+3*I)*(e/cos(d*x+c))^(5/2)/si
n(d*x+c)^4

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\sqrt{2}{\left (-10 i \, a e^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + 24 i \, a e^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + 10 i \, a e^{2}\right )} \sqrt{\frac{e}{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 )} + 15 \,{\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}{\rm integral}\left (-\frac{i \, \sqrt{2} a e^{2} \sqrt{\frac{e}{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 )}}{3 \, d}, x\right )}{15 \,{\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(sqrt(2)*(-10*I*a*e^2*e^(4*I*d*x + 4*I*c) + 24*I*a*e^2*e^(2*I*d*x + 2*I*c) + 10*I*a*e^2)*sqrt(e/(e^(2*I*d
*x + 2*I*c) + 1))*e^(1/2*I*d*x + 1/2*I*c) + 15*(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d)*integral(
-1/3*I*sqrt(2)*a*e^2*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(-1/2*I*d*x - 1/2*I*c)/d, x))/(d*e^(4*I*d*x + 4*I*c)
+ 2*d*e^(2*I*d*x + 2*I*c) + 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*sec(d*x+c))**(5/2)*(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 \sec \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*sec(d*x+c))^(5/2)*(a+I*a*tan(d*x+c)),x, algorithm="giac")

[Out]

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