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

Optimal result

Integrand size = 28, antiderivative size = 106 \[ \int \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^2 \, dx=\frac {10 i a^2 \sqrt {e \sec (c+d x)}}{3 d}+\frac {10 a^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {e \sec (c+d x)}}{3 d}+\frac {2 i \sqrt {e \sec (c+d x)} \left (a^2+i a^2 \tan (c+d x)\right )}{3 d} \]

[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] (verified)

Time = 0.16 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3579, 3567, 3856, 2720} \[ \int \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^2 \, dx=\frac {10 i a^2 \sqrt {e \sec (c+d x)}}{3 d}+\frac {2 i \left (a^2+i a^2 \tan (c+d x)\right ) \sqrt {e \sec (c+d x)}}{3 d}+\frac {10 a^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {e \sec (c+d x)}}{3 d} \]

[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*} \text {integral}& = \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)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),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] (verified)

Time = 1.61 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.63 \[ \int \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^2 \, dx=\frac {2 a^2 (e \sec (c+d x))^{3/2} \left (6 i \cos (c+d x)+5 \cos ^{\frac {3}{2}}(c+d x) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-\sin (c+d x)\right )}{3 d e} \]

[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] (verified)

Time = 10.73 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.36

method	result	size
default	\(-\frac {2 a^{2} \sqrt {e \sec \left (d x +c \right )}\, \left (-5 i \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, F\left (i \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ), i\right ) \cos \left (d x +c \right )-5 i F\left (i \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-6 i+\tan \left (d x +c \right )\right )}{3 d}\)	\(144\)
parts	\(-\frac {2 i a^{2} \left (\cos \left (d x +c \right )+1\right ) \sqrt {e \sec \left (d x +c \right )}\, \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, F\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right )}{d}+\frac {4 i a^{2} \sqrt {e \sec \left (d x +c \right )}}{d}-\frac {2 a^{2} \sqrt {e \sec \left (d x +c \right )}\, \left (2 i \cos \left (d x +c \right ) F\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}+2 i F\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}+\tan \left (d x +c \right )\right )}{3 d}\)	\(241\)

[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*sec(d*x+c))^(1/2)*(-5*I*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*EllipticF(I*(
-csc(d*x+c)+cot(d*x+c)),I)*cos(d*x+c)-5*I*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*EllipticF
(I*(-csc(d*x+c)+cot(d*x+c)),I)-6*I+tan(d*x+c))

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.07 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.02 \[ \int \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^2 \, dx=-\frac {2 \, {\left (\sqrt {2} {\left (-7 i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - 5 i \, a^{2}\right )} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )} + 5 \, \sqrt {2} {\left (i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, a^{2}\right )} \sqrt {e} {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right )}}{3 \, {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

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

Sympy [F]

\[ \int \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^2 \, dx=- a^{2} \left (\int \left (- \sqrt {e \sec {\left (c + d x \right )}}\right )\, dx + \int \sqrt {e \sec {\left (c + d x \right )}} \tan ^{2}{\left (c + d x \right )}\, dx + \int \left (- 2 i \sqrt {e \sec {\left (c + d x \right )}} \tan {\left (c + d x \right )}\right )\, dx\right ) \]

[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))

Maxima [F]

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

[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)

Giac [F(-2)]

Exception generated. \[ \int \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^2 \, dx=\text {Exception raised: TypeError} \]

[In]

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

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Unable to divide, perhaps due to rounding error%%%{-1,[2,0]%%%}+%%%{%%{[-2,0]:[1,0,%%%{1,[1]%%%}]%%},[1,0]%
%%}+%%%{%%%

Mupad [F(-1)]

Timed out. \[ \int \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^2 \, dx=\int \sqrt {\frac {e}{\cos \left (c+d\,x\right )}}\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^2 \,d x \]

[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)