\(\int \frac {1}{\sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^2} \, dx\) [241]

Optimal result

Integrand size = 28, antiderivative size = 116 \[ \int \frac {1}{\sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^2} \, dx=\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 a^2 d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {2 e \sin (c+d x)}{9 a^2 d (e \sec (c+d x))^{3/2}}+\frac {4 i e^2}{9 d (e \sec (c+d x))^{5/2} \left (a^2+i a^2 \tan (c+d x)\right )} \] Output:

2/3*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/a^2/d/cos(d*x+c)^(1/2)/(e*sec(d* 
x+c))^(1/2)+2/9*e*sin(d*x+c)/a^2/d/(e*sec(d*x+c))^(3/2)+4/9*I*e^2/d/(e*sec 
(d*x+c))^(5/2)/(a^2+I*a^2*tan(d*x+c))
 

Mathematica [C] (verified)

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

Time = 1.97 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.06 \[ \int \frac {1}{\sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^2} \, dx=\frac {\left (-\frac {8 e^{4 i (c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )}{\sqrt {1+e^{2 i (c+d x)}}}+2 (2+8 \cos (2 (c+d x))+7 i \sin (2 (c+d x)))\right ) (i \cos (2 (c+d x))+\sin (2 (c+d x)))}{18 a^2 d \sqrt {e \sec (c+d x)}} \] Input:

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

Output:

(((-8*E^((4*I)*(c + d*x))*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + 
d*x))])/Sqrt[1 + E^((2*I)*(c + d*x))] + 2*(2 + 8*Cos[2*(c + d*x)] + (7*I)* 
Sin[2*(c + d*x)]))*(I*Cos[2*(c + d*x)] + Sin[2*(c + d*x)]))/(18*a^2*d*Sqrt 
[e*Sec[c + d*x]])
 

Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.09, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 3981, 3042, 4256, 3042, 4258, 3042, 3119}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3981

\(\displaystyle \frac {5 e^2 \int \frac {1}{(e \sec (c+d x))^{5/2}}dx}{9 a^2}+\frac {4 i e^2}{9 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {5 e^2 \int \frac {1}{\left (e \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}}dx}{9 a^2}+\frac {4 i e^2}{9 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{5/2}}\)

\(\Big \downarrow \) 4256

\(\displaystyle \frac {5 e^2 \left (\frac {3 \int \frac {1}{\sqrt {e \sec (c+d x)}}dx}{5 e^2}+\frac {2 \sin (c+d x)}{5 d e (e \sec (c+d x))^{3/2}}\right )}{9 a^2}+\frac {4 i e^2}{9 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {5 e^2 \left (\frac {3 \int \frac {1}{\sqrt {e \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{5 e^2}+\frac {2 \sin (c+d x)}{5 d e (e \sec (c+d x))^{3/2}}\right )}{9 a^2}+\frac {4 i e^2}{9 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{5/2}}\)

\(\Big \downarrow \) 4258

\(\displaystyle \frac {5 e^2 \left (\frac {3 \int \sqrt {\cos (c+d x)}dx}{5 e^2 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {2 \sin (c+d x)}{5 d e (e \sec (c+d x))^{3/2}}\right )}{9 a^2}+\frac {4 i e^2}{9 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {5 e^2 \left (\frac {3 \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{5 e^2 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {2 \sin (c+d x)}{5 d e (e \sec (c+d x))^{3/2}}\right )}{9 a^2}+\frac {4 i e^2}{9 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{5/2}}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {5 e^2 \left (\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d e^2 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {2 \sin (c+d x)}{5 d e (e \sec (c+d x))^{3/2}}\right )}{9 a^2}+\frac {4 i e^2}{9 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{5/2}}\)

Input:

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

Output:

(5*e^2*((6*EllipticE[(c + d*x)/2, 2])/(5*d*e^2*Sqrt[Cos[c + d*x]]*Sqrt[e*S 
ec[c + d*x]]) + (2*Sin[c + d*x])/(5*d*e*(e*Sec[c + d*x])^(3/2))))/(9*a^2) 
+ (((4*I)/9)*e^2)/(d*(e*Sec[c + d*x])^(5/2)*(a^2 + I*a^2*Tan[c + d*x]))
 

Defintions of rubi rules used

rule 3042

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

rule 3119

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 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 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. 479 vs. \(2 (102 ) = 204\).

Time = 4.46 (sec) , antiderivative size = 480, normalized size of antiderivative = 4.14

method	result	size
default	\(\frac {\frac {2 i \sin \left (d x +c \right ) \left (6 \cos \left (d x +c \right )^{2}+12 \cos \left (d x +c \right )+6\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}}\, \operatorname {EllipticE}\left (i \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ), i\right )}{9}+\frac {2 \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}}\, \operatorname {EllipticE}\left (i \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ), i\right ) \left (6 \cos \left (d x +c \right )^{3}+12 \cos \left (d x +c \right )^{2}+3 \cos \left (d x +c \right )-6-3 \sec \left (d x +c \right )\right )}{9}+\frac {2 i \sin \left (d x +c \right ) \left (-6 \cos \left (d x +c \right )^{2}-12 \cos \left (d x +c \right )-6\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}}\, \operatorname {EllipticF}\left (i \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ), i\right )}{9}+\frac {2 \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}}\, \operatorname {EllipticF}\left (i \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ), i\right ) \left (-6 \cos \left (d x +c \right )^{3}-12 \cos \left (d x +c \right )^{2}-3 \cos \left (d x +c \right )+6+3 \sec \left (d x +c \right )\right )}{9}+\frac {2 i \sin \left (d x +c \right ) \left (5 \cos \left (d x +c \right )^{2}-\cos \left (d x +c \right )-3\right )}{9}+\frac {8 \cos \left (d x +c \right )^{3}}{9}-\frac {4 \cos \left (d x +c \right )^{2}}{9}-\frac {4 \cos \left (d x +c \right )}{3}}{a^{2} d \left (\sin \left (d x +c \right ) \cos \left (d x +c \right ) \left (-2 \cos \left (d x +c \right )-2\right )+i \left (2 \cos \left (d x +c \right )^{3}+2 \cos \left (d x +c \right )^{2}-\cos \left (d x +c \right )-1\right )\right ) \sqrt {e \sec \left (d x +c \right )}}\)	\(480\)

Input:

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

Output:

2/9/a^2/d/(sin(d*x+c)*cos(d*x+c)*(-2*cos(d*x+c)-2)+I*(2*cos(d*x+c)^3+2*cos 
(d*x+c)^2-cos(d*x+c)-1))/(e*sec(d*x+c))^(1/2)*(I*sin(d*x+c)*(6*cos(d*x+c)^ 
2+12*cos(d*x+c)+6)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x+c)+1))^(1 
/2)*EllipticE(I*(cot(d*x+c)-csc(d*x+c)),I)+(cos(d*x+c)/(cos(d*x+c)+1))^(1/ 
2)*(1/(cos(d*x+c)+1))^(1/2)*EllipticE(I*(cot(d*x+c)-csc(d*x+c)),I)*(6*cos( 
d*x+c)^3+12*cos(d*x+c)^2+3*cos(d*x+c)-6-3*sec(d*x+c))+I*sin(d*x+c)*(-6*cos 
(d*x+c)^2-12*cos(d*x+c)-6)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x+c 
)+1))^(1/2)*EllipticF(I*(cot(d*x+c)-csc(d*x+c)),I)+(cos(d*x+c)/(cos(d*x+c) 
+1))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*EllipticF(I*(cot(d*x+c)-csc(d*x+c)),I) 
*(-6*cos(d*x+c)^3-12*cos(d*x+c)^2-3*cos(d*x+c)+6+3*sec(d*x+c))+I*sin(d*x+c 
)*(5*cos(d*x+c)^2-cos(d*x+c)-3)+4*cos(d*x+c)^3-2*cos(d*x+c)^2-6*cos(d*x+c) 
)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.02 \[ \int \frac {1}{\sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^2} \, dx=\frac {{\left (\sqrt {2} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (15 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 19 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 5 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i\right )} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )} + 24 i \, \sqrt {2} \sqrt {e} e^{\left (5 i \, d x + 5 i \, c\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right )\right )} e^{\left (-5 i \, d x - 5 i \, c\right )}}{36 \, a^{2} d e} \] Input:

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

Output:

1/36*(sqrt(2)*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*(15*I*e^(6*I*d*x + 6*I*c) 
+ 19*I*e^(4*I*d*x + 4*I*c) + 5*I*e^(2*I*d*x + 2*I*c) + I)*e^(1/2*I*d*x + 1 
/2*I*c) + 24*I*sqrt(2)*sqrt(e)*e^(5*I*d*x + 5*I*c)*weierstrassZeta(-4, 0, 
weierstrassPInverse(-4, 0, e^(I*d*x + I*c))))*e^(-5*I*d*x - 5*I*c)/(a^2*d* 
e)
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

integrate(1/(e*sec(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]

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

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

Output:

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

Mupad [F(-1)]

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

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

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

Output:

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