3.2.94 \(\int \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^2 \, dx\) [194]

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

[Out]

10/3*I*a^2*(e*sec(d*x+c))^(1/2)/d+10/3*a^2*(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+2/3*I*(e*sec(d*x+c))^(1/2)*(a^2+I*a^2*tan(d*x+c))/d

________________________________________________________________________________________

Rubi [A]
time = 0.07, antiderivative size = 106, normalized size of antiderivative = 1.00, 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 = {3579, 3567, 3856, 2720} \begin {gather*} \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} \end {gather*}

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 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 3567

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 3579

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 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 \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 = 0.56, size = 67, normalized size = 0.63 \begin {gather*} \frac {2 a^2 (e \sec (c+d x))^{3/2} \left (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 )-\sin (c+d x)\right )}{3 d e} \end {gather*}

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)

________________________________________________________________________________________

Maple [A]
time = 0.49, size = 201, normalized size = 1.90

method	result	size
default	\(\frac {2 a^{2} \sqrt {\frac {e}{\cos \left (d x +c \right )}}\, \left (-1+\cos \left (d x +c \right )\right )^{2} \left (5 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \left (\cos ^{2}\left (d x +c \right )\right )+5 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \cos \left (d x +c \right ) \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )+6 i \cos \left (d x +c \right )-\sin \left (d x +c \right )\right ) \left (1+\cos \left (d x +c \right )\right )^{2}}{3 d \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4}}\)	\(201\)

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,method=_RETURNVERBOSE)

[Out]

2/3*a^2/d*(e/cos(d*x+c))^(1/2)*(-1+cos(d*x+c))^2*(5*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)*cos(d*x+c)^2+5*I*(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)+6*I*cos(d*x+c)-sin(d*x+c))*(1+cos(d*x+c))^2/co
s(d*x+c)/sin(d*x+c)^4

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to 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))^2,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.11, size = 110, normalized size = 1.04 \begin {gather*} -\frac {2 \, {\left (\frac {\sqrt {2} {\left (-5 i \, a^{2} e^{\frac {1}{2}} - 7 i \, a^{2} 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}} + 5 \, {\left (i \, \sqrt {2} a^{2} e^{\frac {1}{2}} + i \, \sqrt {2} a^{2} e^{\left (2 i \, d x + 2 i \, c + \frac {1}{2}\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right )}}{3 \, {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \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))^2,x, algorithm="fricas")

[Out]

-2/3*(sqrt(2)*(-5*I*a^2*e^(1/2) - 7*I*a^2*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) + 5*(I*sqrt(2)*a^2*e^(1/2) + I*sqrt(2)*a^2*e^(2*I*d*x + 2*I*c + 1/2))*weierstrassPInverse(-4, 0,
e^(I*d*x + I*c)))/(d*e^(2*I*d*x + 2*I*c) + d)

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - a^{2} \left (\int \left (- \sqrt {e \sec {\left (c + d x \right )}}\right )\, dx + \int \sqrt {e \sec {\left (c + d x \right )}} \tan ^{2}{\left (c + d x \right )}\, dx + \int \left (- 2 i \sqrt {e \sec {\left (c + d x \right )}} \tan {\left (c + d x \right )}\right )\, dx\right ) \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))**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*
sqrt(e*sec(c + d*x))*tan(c + d*x), x))

________________________________________________________________________________________

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))^2,x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \sqrt {\frac {e}{\cos \left (c+d\,x\right )}}\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^2 \,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)^2,x)

[Out]

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

________________________________________________________________________________________