3.229 \(\int \frac {1}{\sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))} \, dx\)

Optimal. Leaf size=80 \[ \frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 a d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {2 i}{5 d (a+i a \tan (c+d x)) \sqrt {e \sec (c+d x)}} \]

[Out]

6/5*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/a/d/cos(d*x+c)^(1/2)
/(e*sec(d*x+c))^(1/2)+2/5*I/d/(e*sec(d*x+c))^(1/2)/(a+I*a*tan(d*x+c))

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Rubi [A]  time = 0.07, antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {3502, 3771, 2639} \[ \frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 a d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {2 i}{5 d (a+i a \tan (c+d x)) \sqrt {e \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[e*Sec[c + d*x]]*(a + I*a*Tan[c + d*x])),x]

[Out]

(6*EllipticE[(c + d*x)/2, 2])/(5*a*d*Sqrt[Cos[c + d*x]]*Sqrt[e*Sec[c + d*x]]) + ((2*I)/5)/(d*Sqrt[e*Sec[c + d*
x]]*(a + I*a*Tan[c + d*x]))

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]

Rule 3502

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(a*(d
*Sec[e + f*x])^m*(a + b*Tan[e + f*x])^n)/(b*f*(m + 2*n)), x] + Dist[Simplify[m + n]/(a*(m + 2*n)), Int[(d*Sec[
e + f*x])^m*(a + b*Tan[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, m}, x] && EqQ[a^2 + b^2, 0] && LtQ[n
, 0] && NeQ[m + 2*n, 0] && IntegersQ[2*m, 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]

Rubi steps

\begin {align*} \int \frac {1}{\sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))} \, dx &=\frac {2 i}{5 d \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))}+\frac {3 \int \frac {1}{\sqrt {e \sec (c+d x)}} \, dx}{5 a}\\ &=\frac {2 i}{5 d \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))}+\frac {3 \int \sqrt {\cos (c+d x)} \, dx}{5 a \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}\\ &=\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 a d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {2 i}{5 d \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))}\\ \end {align*}

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Mathematica [C]  time = 0.88, size = 109, normalized size = 1.36 \[ \frac {(\tan (c+d x)+i) \left (-2 e^{2 i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-e^{2 i (c+d x)}\right )+3 i \sin (2 (c+d x))+4 \cos (2 (c+d x))+4\right )}{5 a d \sqrt {e \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[e*Sec[c + d*x]]*(a + I*a*Tan[c + d*x])),x]

[Out]

((4 + 4*Cos[2*(c + d*x)] - 2*E^((2*I)*(c + d*x))*Sqrt[1 + E^((2*I)*(c + d*x))]*Hypergeometric2F1[1/2, 3/4, 7/4
, -E^((2*I)*(c + d*x))] + (3*I)*Sin[2*(c + d*x)])*(I + Tan[c + d*x]))/(5*a*d*Sqrt[e*Sec[c + d*x]])

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fricas [F]  time = 0.53, size = 0, normalized size = 0.00 \[ \frac {\sqrt {2} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-5 i \, e^{\left (5 i \, d x + 5 i \, c\right )} - 7 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 4 i \, e^{\left (3 i \, d x + 3 i \, c\right )} - 8 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i \, e^{\left (i \, d x + i \, c\right )} - i\right )} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )} + 10 \, {\left (a d e e^{\left (4 i \, d x + 4 i \, c\right )} - a d e e^{\left (3 i \, d x + 3 i \, c\right )}\right )} {\rm integral}\left (\frac {\sqrt {2} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-3 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - 6 i \, e^{\left (i \, d x + i \, c\right )} - 3 i\right )} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )}}{5 \, {\left (a d e e^{\left (3 i \, d x + 3 i \, c\right )} - 2 \, a d e e^{\left (2 i \, d x + 2 i \, c\right )} + a d e e^{\left (i \, d x + i \, c\right )}\right )}}, x\right )}{10 \, {\left (a d e e^{\left (4 i \, d x + 4 i \, c\right )} - a d e e^{\left (3 i \, d x + 3 i \, c\right )}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/10*(sqrt(2)*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*(-5*I*e^(5*I*d*x + 5*I*c) - 7*I*e^(4*I*d*x + 4*I*c) - 4*I*e^(3
*I*d*x + 3*I*c) - 8*I*e^(2*I*d*x + 2*I*c) + I*e^(I*d*x + I*c) - I)*e^(1/2*I*d*x + 1/2*I*c) + 10*(a*d*e*e^(4*I*
d*x + 4*I*c) - a*d*e*e^(3*I*d*x + 3*I*c))*integral(1/5*sqrt(2)*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*(-3*I*e^(2*I*
d*x + 2*I*c) - 6*I*e^(I*d*x + I*c) - 3*I)*e^(1/2*I*d*x + 1/2*I*c)/(a*d*e*e^(3*I*d*x + 3*I*c) - 2*a*d*e*e^(2*I*
d*x + 2*I*c) + a*d*e*e^(I*d*x + I*c)), x))/(a*d*e*e^(4*I*d*x + 4*I*c) - a*d*e*e^(3*I*d*x + 3*I*c))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {e \sec \left (d x + c\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 1.63, size = 358, normalized size = 4.48 \[ \frac {2 \left (3 i \cos \left (d x +c \right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \sin \left (d x +c \right )-3 i \cos \left (d x +c \right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticE \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \sin \left (d x +c \right )+3 i \sin \left (d x +c \right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )-3 i \sin \left (d x +c \right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticE \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )+i \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )-\left (\cos ^{4}\left (d x +c \right )\right )-2 \left (\cos ^{2}\left (d x +c \right )\right )+3 \cos \left (d x +c \right )\right ) \left (1+\cos \left (d x +c \right )\right )^{2} \left (-1+\cos \left (d x +c \right )\right )^{2} \sqrt {\frac {e}{\cos \left (d x +c \right )}}}{5 a d \sin \left (d x +c \right )^{5} e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is undefined.

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{\sqrt {\frac {e}{\cos \left (c+d\,x\right )}}\,\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \frac {i \int \frac {1}{\sqrt {e \sec {\left (c + d x \right )}} \tan {\left (c + d x \right )} - i \sqrt {e \sec {\left (c + d x \right )}}}\, dx}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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