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*(e*sec(d*x+c))^(1/2)/d+2*a*(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)/d

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Rubi [A]  time = 0.04, antiderivative size = 60, normalized size of antiderivative = 1.00, 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 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]

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]

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*}

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Mathematica [A]  time = 0.26, 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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fricas [F]  time = 0.86, size = 0, normalized size = 0.00 \[ \frac {2 i \, \sqrt {2} a \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )} + d {\rm integral}\left (-\frac {i \, \sqrt {2} a \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (-\frac {1}{2} i \, d x - \frac {1}{2} i \, c\right )}}{d}, x\right )}{d} \]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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)

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maple [B]  time = 0.89, size = 164, normalized size = 2.73 \[ \frac {2 i a \sqrt {\frac {e}{\cos \left (d x +c \right )}}\, \left (1+\cos \left (d x +c \right )\right )^{2} \left (-1+\cos \left (d x +c \right )\right )^{2} \left (\sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \cos \left (d x +c \right ) \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )+\sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )+1\right )}{d \sin \left (d x +c \right )^{4}} \]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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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mupad [B]  time = 3.78, size = 40, normalized size = 0.67 \[ \frac {2\,a\,\left (\sqrt {\cos \left (c+d\,x\right )}\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )+1{}\mathrm {i}\right )\,\sqrt {\frac {e}{\cos \left (c+d\,x\right )}}}{d} \]

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]

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

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

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

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