\(\int \frac {1}{\sqrt {e \cos (c+d x)} (a+i a \tan (c+d x))^2} \, dx\) [676]

Optimal result

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

2/7*cos(d*x+c)^(1/2)*InverseJacobiAM(1/2*d*x+1/2*c,2^(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))
 

Mathematica [A] (verified)

Time = 1.10 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.32 \[ \int \frac {1}{\sqrt {e \cos (c+d x)} (a+i a \tan (c+d x))^2} \, dx=\frac {\left (-i \cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (\sqrt {\cos (c+d x)} \left (3 \cos \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {3}{2} (c+d x)\right )+4 i \sin ^3\left (\frac {1}{2} (c+d x)\right )\right )+2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \left (-i \cos \left (\frac {3}{2} (c+d x)\right )+\sin \left (\frac {3}{2} (c+d x)\right )\right )\right )}{7 a^2 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {e \cos (c+d x)} (-i+\tan (c+d x))^2} \] Input:

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

Output:

(((-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)
 

Rubi [A] (verified)

Time = 0.70 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.26, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {3042, 3998, 3042, 3981, 3042, 4256, 3042, 4258, 3042, 3120}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3998

\(\displaystyle \frac {\int \frac {\sqrt {e \sec (c+d x)}}{(i \tan (c+d x) a+a)^2}dx}{\sqrt {e \cos (c+d x)} \sqrt {e \sec (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {\sqrt {e \sec (c+d x)}}{(i \tan (c+d x) a+a)^2}dx}{\sqrt {e \cos (c+d x)} \sqrt {e \sec (c+d x)}}\)

\(\Big \downarrow \) 3981

\(\displaystyle \frac {\frac {3 e^2 \int \frac {1}{(e \sec (c+d x))^{3/2}}dx}{7 a^2}+\frac {4 i e^2}{7 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{3/2}}}{\sqrt {e \cos (c+d x)} \sqrt {e \sec (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {3 e^2 \int \frac {1}{\left (e \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx}{7 a^2}+\frac {4 i e^2}{7 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{3/2}}}{\sqrt {e \cos (c+d x)} \sqrt {e \sec (c+d x)}}\)

\(\Big \downarrow \) 4256

\(\displaystyle \frac {\frac {3 e^2 \left (\frac {\int \sqrt {e \sec (c+d x)}dx}{3 e^2}+\frac {2 \sin (c+d x)}{3 d e \sqrt {e \sec (c+d x)}}\right )}{7 a^2}+\frac {4 i e^2}{7 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{3/2}}}{\sqrt {e \cos (c+d x)} \sqrt {e \sec (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {3 e^2 \left (\frac {\int \sqrt {e \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{3 e^2}+\frac {2 \sin (c+d x)}{3 d e \sqrt {e \sec (c+d x)}}\right )}{7 a^2}+\frac {4 i e^2}{7 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{3/2}}}{\sqrt {e \cos (c+d x)} \sqrt {e \sec (c+d x)}}\)

\(\Big \downarrow \) 4258

\(\displaystyle \frac {\frac {3 e^2 \left (\frac {\sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{3 e^2}+\frac {2 \sin (c+d x)}{3 d e \sqrt {e \sec (c+d x)}}\right )}{7 a^2}+\frac {4 i e^2}{7 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{3/2}}}{\sqrt {e \cos (c+d x)} \sqrt {e \sec (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {3 e^2 \left (\frac {\sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 e^2}+\frac {2 \sin (c+d x)}{3 d e \sqrt {e \sec (c+d x)}}\right )}{7 a^2}+\frac {4 i e^2}{7 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{3/2}}}{\sqrt {e \cos (c+d x)} \sqrt {e \sec (c+d x)}}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {\frac {3 e^2 \left (\frac {2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {e \sec (c+d x)}}{3 d e^2}+\frac {2 \sin (c+d x)}{3 d e \sqrt {e \sec (c+d x)}}\right )}{7 a^2}+\frac {4 i e^2}{7 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{3/2}}}{\sqrt {e \cos (c+d x)} \sqrt {e \sec (c+d x)}}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3120

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 3981

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] - Simp[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] || (IntegersQ[n, m + 
1/2] && GtQ[2*m + n + 1, 0])) && IntegerQ[2*m]
 

rule 3998

Int[(cos[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x 
_)])^(n_.), x_Symbol] :> Simp[(d*Cos[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 4256

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] + Simp[(n + 1)/(b^2*n)   Int[(b*Csc[c 
+ d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && IntegerQ[2* 
n]
 

rule 4258

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(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]
 

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 238 vs. \(2 (103 ) = 206\).

Time = 6.01 (sec) , antiderivative size = 239, normalized size of antiderivative = 1.99

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

Input:

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

Output:

-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*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+(sin(1/2*d*x+1/2*c)^2)^(1/2)* 
(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))-2*I 
*sin(1/2*d*x+1/2*c))/d
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.76 \[ \int \frac {1}{\sqrt {e \cos (c+d x)} (a+i a \tan (c+d x))^2} \, dx=\frac {{\left (\sqrt {\frac {1}{2}} \sqrt {e e^{\left (2 i \, d x + 2 i \, c\right )} + e} {\left (3 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i\right )} e^{\left (-\frac {1}{2} i \, d x - \frac {1}{2} i \, c\right )} - 4 i \, \sqrt {\frac {1}{2}} \sqrt {e} e^{\left (3 i \, d x + 3 i \, c\right )} {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right )} e^{\left (-3 i \, d x - 3 i \, c\right )}}{7 \, a^{2} d e} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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