3.3.28 \(\int \frac {\sqrt {e \sec (c+d x)}}{a+i a \tan (c+d x)} \, dx\) [228]

Optimal. Leaf size=80 \[ \frac {2 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \sec (c+d x)}}{3 a d}+\frac {2 i \sqrt {e \sec (c+d x)}}{3 d (a+i a \tan (c+d x))} \]

[Out]

2/3*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*(e*
sec(d*x+c))^(1/2)/a/d+2/3*I*(e*sec(d*x+c))^(1/2)/d/(a+I*a*tan(d*x+c))

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Rubi [A]
time = 0.06, 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 = {3583, 3856, 2720} \begin {gather*} \frac {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 a d}+\frac {2 i \sqrt {e \sec (c+d x)}}{3 d (a+i a \tan (c+d x))} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ
[{c, d}, x]

Rule 3583

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 3856

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

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Mathematica [A]
time = 0.31, size = 83, normalized size = 1.04 \begin {gather*} \frac {2 (e \sec (c+d x))^{3/2} \left (\cos (c+d x)+\sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) (-i \cos (c+d x)+\sin (c+d x))\right )}{3 a d e (-i+\tan (c+d x))} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.58, size = 164, normalized size = 2.05

method	result	size
default	\(\frac {2 \sqrt {\frac {e}{\cos \left (d x +c \right )}}\, \left (i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \cos \left (d x +c \right ) \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )+i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )+i \left (\cos ^{2}\left (d x +c \right )\right )+\sin \left (d x +c \right ) \cos \left (d x +c \right )\right )}{3 a d}\)	\(164\)

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,method=_RETURNVERBOSE)

[Out]

2/3/a/d*(e/cos(d*x+c))^(1/2)*(I*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)*Elliptic
F(I*(-1+cos(d*x+c))/sin(d*x+c),I)+I*(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)+I*cos(d*x+c)^2+sin(d*x+c)*cos(d*x+c))

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

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]

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

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.09, size = 88, normalized size = 1.10 \begin {gather*} \frac {{\left (-2 i \, \sqrt {2} e^{\left (2 i \, d x + 2 i \, c + \frac {1}{2}\right )} {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right ) + \frac {\sqrt {2} {\left (i \, e^{\frac {1}{2}} + i \, e^{\left (2 i \, d x + 2 i \, c + \frac {1}{2}\right )}\right )} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )}}{\sqrt {e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{3 \, a d} \end {gather*}

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]

1/3*(-2*I*sqrt(2)*e^(2*I*d*x + 2*I*c + 1/2)*weierstrassPInverse(-4, 0, e^(I*d*x + I*c)) + sqrt(2)*(I*e^(1/2) +
I*e^(2*I*d*x + 2*I*c + 1/2))*e^(1/2*I*d*x + 1/2*I*c)/sqrt(e^(2*I*d*x + 2*I*c) + 1))*e^(-2*I*d*x - 2*I*c)/(a*d
)

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

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]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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