\(\int \frac {(a+i a \tan (c+d x))^2}{\sqrt {e \sec (c+d x)}} \, dx\) [195]

Optimal result

Integrand size = 28, antiderivative size = 107 \[ \int \frac {(a+i a \tan (c+d x))^2}{\sqrt {e \sec (c+d x)}} \, dx=\frac {6 a^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}-\frac {6 a^2 \sqrt {e \sec (c+d x)} \sin (c+d x)}{d e}-\frac {4 i \left (a^2+i a^2 \tan (c+d x)\right )}{d \sqrt {e \sec (c+d x)}} \]

[Out]

6*a^2*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/d/cos(d*x+c)^(1/2)
/(e*sec(d*x+c))^(1/2)-6*a^2*sin(d*x+c)*(e*sec(d*x+c))^(1/2)/d/e-4*I*(a^2+I*a^2*tan(d*x+c))/d/(e*sec(d*x+c))^(1
/2)

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 107, 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 = {3577, 3853, 3856, 2719} \[ \int \frac {(a+i a \tan (c+d x))^2}{\sqrt {e \sec (c+d x)}} \, dx=-\frac {6 a^2 \sin (c+d x) \sqrt {e \sec (c+d x)}}{d e}-\frac {4 i \left (a^2+i a^2 \tan (c+d x)\right )}{d \sqrt {e \sec (c+d x)}}+\frac {6 a^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}} \]

[In]

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

[Out]

(6*a^2*EllipticE[(c + d*x)/2, 2])/(d*Sqrt[Cos[c + d*x]]*Sqrt[e*Sec[c + d*x]]) - (6*a^2*Sqrt[e*Sec[c + d*x]]*Si
n[c + d*x])/(d*e) - ((4*I)*(a^2 + I*a^2*Tan[c + d*x]))/(d*Sqrt[e*Sec[c + d*x]])

Rule 2719

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

Rule 3577

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[2*b*(d
*Sec[e + f*x])^m*((a + b*Tan[e + f*x])^(n - 1)/(f*m)), x] - Dist[b^2*((m + 2*n - 2)/(d^2*m)), Int[(d*Sec[e + f
*x])^(m + 2)*(a + b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 + b^2, 0] && GtQ[n,
1] && ((IGtQ[n/2, 0] && ILtQ[m - 1/2, 0]) || (EqQ[n, 2] && LtQ[m, 0]) || (LeQ[m, -1] && GtQ[m + n, 0]) || (ILt
Q[m, 0] && LtQ[m/2 + n - 1, 0] && IntegerQ[n]) || (EqQ[n, 3/2] && EqQ[m, -2^(-1)])) && IntegerQ[2*m]

Rule 3853

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Csc[c + d*x])^(n - 1)/(d*(n
- 1))), x] + Dist[b^2*((n - 2)/(n - 1)), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n,
1] && IntegerQ[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 {4 i \left (a^2+i a^2 \tan (c+d x)\right )}{d \sqrt {e \sec (c+d x)}}-\frac {\left (3 a^2\right ) \int (e \sec (c+d x))^{3/2} \, dx}{e^2} \\ & = -\frac {6 a^2 \sqrt {e \sec (c+d x)} \sin (c+d x)}{d e}-\frac {4 i \left (a^2+i a^2 \tan (c+d x)\right )}{d \sqrt {e \sec (c+d x)}}+\left (3 a^2\right ) \int \frac {1}{\sqrt {e \sec (c+d x)}} \, dx \\ & = -\frac {6 a^2 \sqrt {e \sec (c+d x)} \sin (c+d x)}{d e}-\frac {4 i \left (a^2+i a^2 \tan (c+d x)\right )}{d \sqrt {e \sec (c+d x)}}+\frac {\left (3 a^2\right ) \int \sqrt {\cos (c+d x)} \, dx}{\sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}} \\ & = \frac {6 a^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}-\frac {6 a^2 \sqrt {e \sec (c+d x)} \sin (c+d x)}{d e}-\frac {4 i \left (a^2+i a^2 \tan (c+d x)\right )}{d \sqrt {e \sec (c+d x)}} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 2.18 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.23 \[ \int \frac {(a+i a \tan (c+d x))^2}{\sqrt {e \sec (c+d x)}} \, dx=-\frac {2 i \sqrt {2} a^2 e^{2 i (c+d x)} \left (-\sqrt {1+e^{2 i (c+d x)}}+\left (1+e^{2 i (c+d x)}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )\right )}{d \sqrt {\frac {e e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \left (1+e^{2 i (c+d x)}\right )^{3/2}} \]

[In]

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

[Out]

((-2*I)*Sqrt[2]*a^2*E^((2*I)*(c + d*x))*(-Sqrt[1 + E^((2*I)*(c + d*x))] + (1 + E^((2*I)*(c + d*x)))*Hypergeome
tric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))]))/(d*Sqrt[(e*E^(I*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]*(1 + E^(
(2*I)*(c + d*x)))^(3/2))

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 806 vs. \(2 (123 ) = 246\).

Time = 10.34 (sec) , antiderivative size = 807, normalized size of antiderivative = 7.54

method	result	size
parts	\(\frac {2 a^{2} \left (i \cos \left (d x +c \right ) E\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}}-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 E\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}}+i \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}}\, E\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right )-i \sec \left (d x +c \right ) \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 ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}+\sin \left (d x +c \right )\right )}{d \left (\cos \left (d x +c \right )+1\right ) \sqrt {e \sec \left (d x +c \right )}}-\frac {4 i a^{2}}{\sqrt {e \sec \left (d x +c \right )}\, d}+\frac {2 a^{2} \left (2 i \cos \left (d x +c \right ) E\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 \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}}+4 i E\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}}-4 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}}+2 i \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}}\, E\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right )-2 i \sec \left (d x +c \right ) \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 ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}+\sin \left (d x +c \right )-\tan \left (d x +c \right )\right )}{d \left (\cos \left (d x +c \right )+1\right ) \sqrt {e \sec \left (d x +c \right )}}\)	\(807\)
default	\(\text {Expression too large to display}\)	\(902\)

[In]

int((a+I*a*tan(d*x+c))^2/(e*sec(d*x+c))^(1/2),x,method=_RETURNVERBOSE)

[Out]

2*a^2/d/(cos(d*x+c)+1)/(e*sec(d*x+c))^(1/2)*(I*EllipticE(I*(csc(d*x+c)-cot(d*x+c)),I)*(1/(cos(d*x+c)+1))^(1/2)
*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)-I*cos(d*x+c)*EllipticF(I*(csc(d*x+c)-cot(d*x+c)),I)*(1/(cos(d*x+
c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+2*I*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*
EllipticE(I*(csc(d*x+c)-cot(d*x+c)),I)-2*I*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticF(I*(csc(d*x+c)-cot(d*x+c
)),I)*(1/(cos(d*x+c)+1))^(1/2)+I*sec(d*x+c)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*Ellipti
cE(I*(csc(d*x+c)-cot(d*x+c)),I)-I*sec(d*x+c)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticF(I*(csc(d*x+c)-cot(d*x
+c)),I)*(1/(cos(d*x+c)+1))^(1/2)+sin(d*x+c))-4*I*a^2/(e*sec(d*x+c))^(1/2)/d+2*a^2/d/(cos(d*x+c)+1)/(e*sec(d*x+
c))^(1/2)*(2*I*EllipticE(I*(csc(d*x+c)-cot(d*x+c)),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*I*EllipticF(I*(csc(d*x+c)-cot(d*x+c)),I)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^
(1/2)*cos(d*x+c)+4*I*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticE(I*(csc(d*x+c)-cot(d*
x+c)),I)-4*I*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticF(I*(csc(d*x+c)-cot(d*x+c)),I)*(1/(cos(d*x+c)+1))^(1/2)
+2*I*sec(d*x+c)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticE(I*(csc(d*x+c)-cot(d*x+c))
,I)-2*I*sec(d*x+c)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticF(I*(csc(d*x+c)-cot(d*x+c)),I)*(1/(cos(d*x+c)+1))
^(1/2)+sin(d*x+c)-tan(d*x+c))

Fricas [C] (verification not implemented)

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

Time = 0.08 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.64 \[ \int \frac {(a+i a \tan (c+d x))^2}{\sqrt {e \sec (c+d x)}} \, dx=-\frac {2 \, {\left (-i \, \sqrt {2} a^{2} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {3}{2} i \, d x + \frac {3}{2} i \, c\right )} - 3 i \, \sqrt {2} a^{2} \sqrt {e} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right )\right )}}{d e} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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