3.188 \(\int \sqrt{e \sec (c+d x)} (a+i a \tan (c+d x)) \, dx\)

Optimal. Leaf size=60 \[ \frac{2 a \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \sec (c+d x)}}{d}+\frac{2 i a \sqrt{e \sec (c+d x)}}{d} \]

[Out]

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

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Rubi [A]  time = 0.0434359, antiderivative size = 60, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {3486, 3771, 2641} \[ \frac{2 a \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \sec (c+d x)}}{d}+\frac{2 i a \sqrt{e \sec (c+d x)}}{d} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[e*Sec[c + d*x]]*(a + I*a*Tan[c + d*x]),x]

[Out]

((2*I)*a*Sqrt[e*Sec[c + d*x]])/d + (2*a*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[e*Sec[c + d*x]])/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 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 \sqrt{e \sec (c+d x)} (a+i a \tan (c+d x)) \, dx &=\frac{2 i a \sqrt{e \sec (c+d x)}}{d}+a \int \sqrt{e \sec (c+d x)} \, dx\\ &=\frac{2 i a \sqrt{e \sec (c+d x)}}{d}+\left (a \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{2 i a \sqrt{e \sec (c+d x)}}{d}+\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)}}{d}\\ \end{align*}

Mathematica [A]  time = 0.236105, size = 44, normalized size = 0.73 \[ \frac{2 a \sqrt{e \sec (c+d x)} \left (\sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )+i\right )}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[e*Sec[c + d*x]]*(a + I*a*Tan[c + d*x]),x]

[Out]

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

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Maple [B]  time = 0.238, size = 164, normalized size = 2.7 \begin{align*}{\frac{2\,ia \left ( \cos \left ( dx+c \right ) +1 \right ) ^{2} \left ( \cos \left ( dx+c \right ) -1 \right ) ^{2}}{d \left ( \sin \left ( dx+c \right ) \right ) ^{4}}\sqrt{{\frac{e}{\cos \left ( dx+c \right ) }}} \left ( \cos \left ( dx+c \right ){\it EllipticF} \left ({\frac{i \left ( \cos \left ( dx+c \right ) -1 \right ) }{\sin \left ( dx+c \right ) }},i \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 ) \sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}+1 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*sec(d*x+c))^(1/2)*(a+I*a*tan(d*x+c)),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{e \sec \left (d x + c\right )}{\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))^(1/2)*(a+I*a*tan(d*x+c)),x, algorithm="maxima")

[Out]

integrate(sqrt(e*sec(d*x + c))*(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{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 )} + d{\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 )}}{d}, x\right )}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))^(1/2)*(a+I*a*tan(d*x+c)),x, algorithm="fricas")

[Out]

(2*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*integral(-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, x))/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))**(1/2)*(a+I*a*tan(d*x+c)),x)

[Out]

a*(Integral(sqrt(e*sec(c + d*x)), x) + Integral(I*sqrt(e*sec(c + d*x))*tan(c + d*x), x))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{e \sec \left (d x + c\right )}{\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))^(1/2)*(a+I*a*tan(d*x+c)),x, algorithm="giac")

[Out]

integrate(sqrt(e*sec(d*x + c))*(I*a*tan(d*x + c) + a), x)