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

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

[Out]

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

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Rubi [A]  time = 0.0467685, 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, 2639} \[ \frac{2 a E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}}-\frac{2 i a}{d \sqrt{e \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [C]  time = 0.343589, size = 73, normalized size = 1.22 \[ -\frac{4 i a e^{2 i (c+d x)} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},-e^{2 i (c+d x)}\right )}{3 d \sqrt{1+e^{2 i (c+d x)}} \sqrt{e \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((-4*I)/3)*a*E^((2*I)*(c + d*x))*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))])/(d*Sqrt[1 + E^((2*I)
*(c + d*x))]*Sqrt[e*Sec[c + d*x]])

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Maple [B]  time = 0.243, size = 910, normalized size = 15.2 \begin{align*} \text{result too large to display} \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))^(1/2),x)

[Out]

-1/2*a/d*(cos(d*x+c)-1)*(4*I*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*sin(d*x+c)*cos(d*x+c)^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)-4*I*(-cos(d*x+c)/(cos(d*x+c)+1
)^2)^(1/2)*sin(d*x+c)*cos(d*x+c)^2*EllipticE(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)+8*I*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*sin(d*x+c)*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)-8*I*(-cos(d*x+c)/(cos(d*x+c)+1)^
2)^(1/2)*sin(d*x+c)*cos(d*x+c)*EllipticE(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)+4*I*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(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)*sin(d*x+c)-4*I*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d
*x+c)+1))^(1/2)*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*EllipticE(I*(cos(d*x+c)-1)/sin(d*x+c),I)*sin(d*x+c)-4*I*s
in(d*x+c)*cos(d*x+c)^2*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)-4*I*sin(d*x+c)*cos(d*x+c)*(-cos(d*x+c)/(cos(d*x+c)
+1)^2)^(1/2)+I*ln(-2*(2*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*cos(d*x+c)^2-cos(d*x+c)^2-2*(-cos(d*x+c)/(cos(d*x
+c)+1)^2)^(1/2)+2*cos(d*x+c)-1)/sin(d*x+c)^2)*cos(d*x+c)*sin(d*x+c)-I*cos(d*x+c)*ln(-(2*(-cos(d*x+c)/(cos(d*x+
c)+1)^2)^(1/2)*cos(d*x+c)^2-cos(d*x+c)^2-2*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)+2*cos(d*x+c)-1)/sin(d*x+c)^2)*
sin(d*x+c)-4*cos(d*x+c)^3*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)+4*cos(d*x+c)*(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/
2))/(-cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)/sin(d*x+c)^3/cos(d*x+c)/(e/cos(d*x+c))^(1/2)

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

[Out]

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

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

[Out]

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

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

[Out]

a*(Integral(1/sqrt(e*sec(c + d*x)), x) + Integral(I*tan(c + d*x)/sqrt(e*sec(c + d*x)), x))

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

[Out]

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