\(\int \frac {1}{\sqrt {e \cos (c+d x)} (a+i a \tan (c+d x))^2} \, dx\) [667]

Optimal result

Integrand size = 28, antiderivative size = 120 \[ \int \frac {1}{\sqrt {e \cos (c+d x)} (a+i a \tan (c+d x))^2} \, dx=\frac {2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{7 a^2 d \sqrt {e \cos (c+d x)}}+\frac {2 i}{7 d \sqrt {e \cos (c+d x)} (a+i a \tan (c+d x))^2}+\frac {2 i}{7 d \sqrt {e \cos (c+d x)} \left (a^2+i a^2 \tan (c+d x)\right )} \]

[Out]

2/7*(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)/a^2
/d/(e*cos(d*x+c))^(1/2)+2/7*I/d/(e*cos(d*x+c))^(1/2)/(a+I*a*tan(d*x+c))^2+2/7*I/d/(e*cos(d*x+c))^(1/2)/(a^2+I*
a^2*tan(d*x+c))

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.05, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {3596, 3581, 3854, 3856, 2720} \[ \int \frac {1}{\sqrt {e \cos (c+d x)} (a+i a \tan (c+d x))^2} \, dx=\frac {4 i \cos ^2(c+d x)}{7 d \left (a^2+i a^2 \tan (c+d x)\right ) \sqrt {e \cos (c+d x)}}+\frac {2 \sin (c+d x) \cos (c+d x)}{7 a^2 d \sqrt {e \cos (c+d x)}}+\frac {2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{7 a^2 d \sqrt {e \cos (c+d x)}} \]

[In]

Int[1/(Sqrt[e*Cos[c + d*x]]*(a + I*a*Tan[c + d*x])^2),x]

[Out]

(2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(7*a^2*d*Sqrt[e*Cos[c + d*x]]) + (2*Cos[c + d*x]*Sin[c + d*x]
)/(7*a^2*d*Sqrt[e*Cos[c + d*x]]) + (((4*I)/7)*Cos[c + d*x]^2)/(d*Sqrt[e*Cos[c + d*x]]*(a^2 + I*a^2*Tan[c + d*x
]))

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 3581

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

Rule 3596

Int[(cos[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[(d*Co
s[e + f*x])^m*(d*Sec[e + f*x])^m, Int[(a + b*Tan[e + f*x])^n/(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e,
f, m, n}, x] &&  !IntegerQ[m]

Rule 3854

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[Cos[c + d*x]*((b*Csc[c + d*x])^(n + 1)/(b*d*n)), x
] + Dist[(n + 1)/(b^2*n), Int[(b*Csc[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && Integer
Q[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 {\int \frac {\sqrt {e \sec (c+d x)}}{(a+i a \tan (c+d x))^2} \, dx}{\sqrt {e \cos (c+d x)} \sqrt {e \sec (c+d x)}} \\ & = \frac {4 i \cos ^2(c+d x)}{7 d \sqrt {e \cos (c+d x)} \left (a^2+i a^2 \tan (c+d x)\right )}+\frac {\left (3 e^2\right ) \int \frac {1}{(e \sec (c+d x))^{3/2}} \, dx}{7 a^2 \sqrt {e \cos (c+d x)} \sqrt {e \sec (c+d x)}} \\ & = \frac {2 \cos (c+d x) \sin (c+d x)}{7 a^2 d \sqrt {e \cos (c+d x)}}+\frac {4 i \cos ^2(c+d x)}{7 d \sqrt {e \cos (c+d x)} \left (a^2+i a^2 \tan (c+d x)\right )}+\frac {\int \sqrt {e \sec (c+d x)} \, dx}{7 a^2 \sqrt {e \cos (c+d x)} \sqrt {e \sec (c+d x)}} \\ & = \frac {2 \cos (c+d x) \sin (c+d x)}{7 a^2 d \sqrt {e \cos (c+d x)}}+\frac {4 i \cos ^2(c+d x)}{7 d \sqrt {e \cos (c+d x)} \left (a^2+i a^2 \tan (c+d x)\right )}+\frac {\sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{7 a^2 \sqrt {e \cos (c+d x)}} \\ & = \frac {2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{7 a^2 d \sqrt {e \cos (c+d x)}}+\frac {2 \cos (c+d x) \sin (c+d x)}{7 a^2 d \sqrt {e \cos (c+d x)}}+\frac {4 i \cos ^2(c+d x)}{7 d \sqrt {e \cos (c+d x)} \left (a^2+i a^2 \tan (c+d x)\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.90 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.32 \[ \int \frac {1}{\sqrt {e \cos (c+d x)} (a+i a \tan (c+d x))^2} \, dx=\frac {\left (-i \cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (\sqrt {\cos (c+d x)} \left (3 \cos \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {3}{2} (c+d x)\right )+4 i \sin ^3\left (\frac {1}{2} (c+d x)\right )\right )+2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \left (-i \cos \left (\frac {3}{2} (c+d x)\right )+\sin \left (\frac {3}{2} (c+d x)\right )\right )\right )}{7 a^2 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {e \cos (c+d x)} (-i+\tan (c+d x))^2} \]

[In]

Integrate[1/(Sqrt[e*Cos[c + d*x]]*(a + I*a*Tan[c + d*x])^2),x]

[Out]

(((-I)*Cos[(c + d*x)/2] + Sin[(c + d*x)/2])*(Sqrt[Cos[c + d*x]]*(3*Cos[(c + d*x)/2] + Cos[(3*(c + d*x))/2] + (
4*I)*Sin[(c + d*x)/2]^3) + 2*EllipticF[(c + d*x)/2, 2]*((-I)*Cos[(3*(c + d*x))/2] + Sin[(3*(c + d*x))/2])))/(7
*a^2*d*Cos[c + d*x]^(3/2)*Sqrt[e*Cos[c + d*x]]*(-I + Tan[c + d*x])^2)

Maple [A] (verified)

Time = 5.09 (sec) , antiderivative size = 239, normalized size of antiderivative = 1.99

method	result	size
default	\(-\frac {2 \left (-32 i \left (\sin ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+32 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+64 i \left (\sin ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-48 \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-48 i \left (\sin ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+28 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+16 i \left (\sin ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-6 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}-2 i \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{7 a^{2} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, d}\)	\(239\)

[In]

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

[Out]

-2/7/a^2/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)*(-32*I*sin(1/2*d*x+1/2*c)^9+32*cos(1/2*d*x+1/2
*c)*sin(1/2*d*x+1/2*c)^8+64*I*sin(1/2*d*x+1/2*c)^7-48*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)-48*I*sin(1/2*d*x
+1/2*c)^5+28*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+16*I*sin(1/2*d*x+1/2*c)^3-6*cos(1/2*d*x+1/2*c)*sin(1/2*d*
x+1/2*c)^2+(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)
-2*I*sin(1/2*d*x+1/2*c))/d

Fricas [C] (verification not implemented)

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

Time = 0.08 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.76 \[ \int \frac {1}{\sqrt {e \cos (c+d x)} (a+i a \tan (c+d x))^2} \, dx=\frac {{\left (\sqrt {\frac {1}{2}} \sqrt {e e^{\left (2 i \, d x + 2 i \, c\right )} + e} {\left (3 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i\right )} e^{\left (-\frac {1}{2} i \, d x - \frac {1}{2} i \, c\right )} - 2 i \, \sqrt {2} \sqrt {e} e^{\left (3 i \, d x + 3 i \, c\right )} {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right )} e^{\left (-3 i \, d x - 3 i \, c\right )}}{7 \, a^{2} d e} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{\sqrt {e \cos (c+d x)} (a+i a \tan (c+d x))^2} \, dx=\text {Exception raised: RuntimeError} \]

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is undefined.

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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