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

Optimal. Leaf size=96 \[ \frac{2 a \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \sec (c+d x)}}{3 d e^2}-\frac{2 i a}{3 d (e \sec (c+d x))^{3/2}}+\frac{2 a \sin (c+d x)}{3 d e \sqrt{e \sec (c+d x)}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0680948, antiderivative size = 96, 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, 3769, 3771, 2641} \[ \frac{2 a \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \sec (c+d x)}}{3 d e^2}-\frac{2 i a}{3 d (e \sec (c+d x))^{3/2}}+\frac{2 a \sin (c+d x)}{3 d e \sqrt{e \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.394509, size = 62, normalized size = 0.65 \[ \frac{2 a \left (\sin (c+d x)-i \cos (c+d x)+\frac{F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{\sqrt{\cos (c+d x)}}\right )}{3 d e \sqrt{e \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.199, size = 170, normalized size = 1.8 \begin{align*}{\frac{2\,a}{3\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}} \left ( i\cos \left ( dx+c \right ) \sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}{\it EllipticF} \left ({\frac{i \left ( \cos \left ( dx+c \right ) -1 \right ) }{\sin \left ( dx+c \right ) }},i \right ) +i\sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}{\it EllipticF} \left ({\frac{i \left ( \cos \left ( dx+c \right ) -1 \right ) }{\sin \left ( dx+c \right ) }},i \right ) -i \left ( \cos \left ( dx+c \right ) \right ) ^{2}+\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) \right ) \left ({\frac{e}{\cos \left ( dx+c \right ) }} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{i \, a \tan \left (d x + c\right ) + a}{\left (e \sec \left (d x + c\right )\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a \left (\int \frac{1}{\left (e \sec{\left (c + d x \right )}\right )^{\frac{3}{2}}}\, dx + \int \frac{i \tan{\left (c + d x \right )}}{\left (e \sec{\left (c + d x \right )}\right )^{\frac{3}{2}}}\, dx\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*(Integral((e*sec(c + d*x))**(-3/2), x) + Integral(I*tan(c + d*x)/(e*sec(c + d*x))**(3/2), x))

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{i \, a \tan \left (d x + c\right ) + a}{\left (e \sec \left (d x + c\right )\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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