3.194 \(\int \sqrt{e \sec (c+d x)} (a+i a \tan (c+d x))^2 \, dx\)

Optimal. Leaf size=106 \[ \frac{10 i a^2 \sqrt{e \sec (c+d x)}}{3 d}+\frac{2 i \left (a^2+i a^2 \tan (c+d x)\right ) \sqrt{e \sec (c+d x)}}{3 d}+\frac{10 a^2 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \sec (c+d x)}}{3 d} \]

[Out]

(((10*I)/3)*a^2*Sqrt[e*Sec[c + d*x]])/d + (10*a^2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[e*Sec[c +
d*x]])/(3*d) + (((2*I)/3)*Sqrt[e*Sec[c + d*x]]*(a^2 + I*a^2*Tan[c + d*x]))/d

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Rubi [A]  time = 0.0894341, antiderivative size = 106, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3498, 3486, 3771, 2641} \[ \frac{10 i a^2 \sqrt{e \sec (c+d x)}}{3 d}+\frac{2 i \left (a^2+i a^2 \tan (c+d x)\right ) \sqrt{e \sec (c+d x)}}{3 d}+\frac{10 a^2 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \sec (c+d x)}}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[e*Sec[c + d*x]]*(a + I*a*Tan[c + d*x])^2,x]

[Out]

(((10*I)/3)*a^2*Sqrt[e*Sec[c + d*x]])/d + (10*a^2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[e*Sec[c +
d*x]])/(3*d) + (((2*I)/3)*Sqrt[e*Sec[c + d*x]]*(a^2 + I*a^2*Tan[c + d*x]))/d

Rule 3498

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

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))^2 \, dx &=\frac{2 i \sqrt{e \sec (c+d x)} \left (a^2+i a^2 \tan (c+d x)\right )}{3 d}+\frac{1}{3} (5 a) \int \sqrt{e \sec (c+d x)} (a+i a \tan (c+d x)) \, dx\\ &=\frac{10 i a^2 \sqrt{e \sec (c+d x)}}{3 d}+\frac{2 i \sqrt{e \sec (c+d x)} \left (a^2+i a^2 \tan (c+d x)\right )}{3 d}+\frac{1}{3} \left (5 a^2\right ) \int \sqrt{e \sec (c+d x)} \, dx\\ &=\frac{10 i a^2 \sqrt{e \sec (c+d x)}}{3 d}+\frac{2 i \sqrt{e \sec (c+d x)} \left (a^2+i a^2 \tan (c+d x)\right )}{3 d}+\frac{1}{3} \left (5 a^2 \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{10 i a^2 \sqrt{e \sec (c+d x)}}{3 d}+\frac{10 a^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 d}+\frac{2 i \sqrt{e \sec (c+d x)} \left (a^2+i a^2 \tan (c+d x)\right )}{3 d}\\ \end{align*}

Mathematica [A]  time = 0.614256, size = 67, normalized size = 0.63 \[ \frac{2 a^2 (e \sec (c+d x))^{3/2} \left (-\sin (c+d x)+6 i \cos (c+d x)+5 \cos ^{\frac{3}{2}}(c+d x) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )\right )}{3 d e} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[e*Sec[c + d*x]]*(a + I*a*Tan[c + d*x])^2,x]

[Out]

(2*a^2*(e*Sec[c + d*x])^(3/2)*((6*I)*Cos[c + d*x] + 5*Cos[c + d*x]^(3/2)*EllipticF[(c + d*x)/2, 2] - Sin[c + d
*x]))/(3*d*e)

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Maple [A]  time = 0.237, size = 201, normalized size = 1.9 \begin{align*}{\frac{2\,{a}^{2} \left ( \cos \left ( dx+c \right ) -1 \right ) ^{2} \left ( \cos \left ( dx+c \right ) +1 \right ) ^{2}}{3\,d\cos \left ( dx+c \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{4}}\sqrt{{\frac{e}{\cos \left ( dx+c \right ) }}} \left ( 5\,i\sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}} \left ( \cos \left ( dx+c \right ) \right ) ^{2}{\it EllipticF} \left ({\frac{i \left ( \cos \left ( dx+c \right ) -1 \right ) }{\sin \left ( dx+c \right ) }},i \right ) +5\,i\cos \left ( dx+c \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 ) +6\,i\cos \left ( dx+c \right ) -\sin \left ( dx+c \right ) \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))^2,x)

[Out]

2/3*a^2/d*(e/cos(d*x+c))^(1/2)*(cos(d*x+c)-1)^2*(5*I*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2
)*cos(d*x+c)^2*EllipticF(I*(cos(d*x+c)-1)/sin(d*x+c),I)+5*I*cos(d*x+c)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(c
os(d*x+c)+1))^(1/2)*EllipticF(I*(cos(d*x+c)-1)/sin(d*x+c),I)+6*I*cos(d*x+c)-sin(d*x+c))*(cos(d*x+c)+1)^2/cos(d
*x+c)/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 )}^{2}\,{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))^2,x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\sqrt{2}{\left (14 i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + 10 i \, a^{2}\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 )} + 3 \,{\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}{\rm integral}\left (-\frac{5 i \, \sqrt{2} a^{2} \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 )}}{3 \, d}, x\right )}{3 \,{\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\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))^2,x, algorithm="fricas")

[Out]

1/3*(sqrt(2)*(14*I*a^2*e^(2*I*d*x + 2*I*c) + 10*I*a^2)*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(1/2*I*d*x + 1/2*I*
c) + 3*(d*e^(2*I*d*x + 2*I*c) + d)*integral(-5/3*I*sqrt(2)*a^2*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(-1/2*I*d*x
- 1/2*I*c)/d, x))/(d*e^(2*I*d*x + 2*I*c) + d)

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

[Out]

a**2*(Integral(sqrt(e*sec(c + d*x)), x) + Integral(-sqrt(e*sec(c + d*x))*tan(c + d*x)**2, x) + Integral(2*I*sq
rt(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 )}^{2}\,{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))^2,x, algorithm="giac")

[Out]

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