\(\int \frac {1}{\sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}} \, dx\) [114]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [B] (verification not implemented)
   Sympy [F]
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [F(-1)]

Optimal result

Integrand size = 30, antiderivative size = 46 \[ \int \frac {1}{\sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}} \, dx=\frac {\log (\sin (e+f x)) \tan (e+f x)}{f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}} \]

[Out]

ln(sin(f*x+e))*tan(f*x+e)/f/(a+a*sec(f*x+e))^(1/2)/(c-c*sec(f*x+e))^(1/2)

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {3990, 3556} \[ \int \frac {1}{\sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}} \, dx=\frac {\tan (e+f x) \log (\sin (e+f x))}{f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}} \]

[In]

Int[1/(Sqrt[a + a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]]),x]

[Out]

(Log[Sin[e + f*x]]*Tan[e + f*x])/(f*Sqrt[a + a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]])

Rule 3556

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3990

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(m_), x_Symbol] :> Dist
[((-a)*c)^(m + 1/2)*(Cot[e + f*x]/(Sqrt[a + b*Csc[e + f*x]]*Sqrt[c + d*Csc[e + f*x]])), Int[Cot[e + f*x]^(2*m)
, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m + 1/2]

Rubi steps \begin{align*} \text {integral}& = \frac {\tan (e+f x) \int \cot (e+f x) \, dx}{\sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}} \\ & = \frac {\log (\sin (e+f x)) \tan (e+f x)}{f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.32 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.17 \[ \int \frac {1}{\sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}} \, dx=\frac {(\log (\cos (e+f x))+\log (\tan (e+f x))) \tan (e+f x)}{f \sqrt {a (1+\sec (e+f x))} \sqrt {c-c \sec (e+f x)}} \]

[In]

Integrate[1/(Sqrt[a + a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]]),x]

[Out]

((Log[Cos[e + f*x]] + Log[Tan[e + f*x]])*Tan[e + f*x])/(f*Sqrt[a*(1 + Sec[e + f*x])]*Sqrt[c - c*Sec[e + f*x]])

Maple [A] (verified)

Time = 1.98 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.76

method result size
default \(-\frac {\sqrt {a \left (\sec \left (f x +e \right )+1\right )}\, \left (\ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )-\ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )\right ) \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )\right )}{f a \sqrt {-c \left (\sec \left (f x +e \right )-1\right )}}\) \(81\)
risch \(\frac {\left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) x}{\sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{1+{\mathrm e}^{2 i \left (f x +e \right )}}}\, \left (1+{\mathrm e}^{2 i \left (f x +e \right )}\right ) \sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{1+{\mathrm e}^{2 i \left (f x +e \right )}}}}-\frac {2 \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) \left (f x +e \right )}{\sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{1+{\mathrm e}^{2 i \left (f x +e \right )}}}\, \left (1+{\mathrm e}^{2 i \left (f x +e \right )}\right ) \sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{1+{\mathrm e}^{2 i \left (f x +e \right )}}}\, f}-\frac {i \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )}{\sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{1+{\mathrm e}^{2 i \left (f x +e \right )}}}\, \left (1+{\mathrm e}^{2 i \left (f x +e \right )}\right ) \sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{1+{\mathrm e}^{2 i \left (f x +e \right )}}}\, f}\) \(317\)

[In]

int(1/(c-c*sec(f*x+e))^(1/2)/(a+a*sec(f*x+e))^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/f/a*(a*(sec(f*x+e)+1))^(1/2)*(ln(-cot(f*x+e)+csc(f*x+e))-ln(2/(cos(f*x+e)+1)))/(-c*(sec(f*x+e)-1))^(1/2)*(c
ot(f*x+e)-csc(f*x+e))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 113 vs. \(2 (42) = 84\).

Time = 0.43 (sec) , antiderivative size = 272, normalized size of antiderivative = 5.91 \[ \int \frac {1}{\sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}} \, dx=\left [-\frac {\sqrt {-a c} \log \left (-\frac {8 \, {\left ({\left (256 \, \cos \left (f x + e\right )^{5} - 512 \, \cos \left (f x + e\right )^{3} + 175 \, \cos \left (f x + e\right )\right )} \sqrt {-a c} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}} - {\left (256 \, a c \cos \left (f x + e\right )^{4} - 512 \, a c \cos \left (f x + e\right )^{2} + 337 \, a c\right )} \sin \left (f x + e\right )\right )}}{{\left (\cos \left (f x + e\right )^{2} - 1\right )} \sin \left (f x + e\right )}\right )}{2 \, a c f}, -\frac {\sqrt {a c} \arctan \left (\frac {{\left (16 \, \cos \left (f x + e\right )^{3} - 7 \, \cos \left (f x + e\right )\right )} \sqrt {a c} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{{\left (16 \, a c \cos \left (f x + e\right )^{2} - 25 \, a c\right )} \sin \left (f x + e\right )}\right )}{a c f}\right ] \]

[In]

integrate(1/(c-c*sec(f*x+e))^(1/2)/(a+a*sec(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

[-1/2*sqrt(-a*c)*log(-8*((256*cos(f*x + e)^5 - 512*cos(f*x + e)^3 + 175*cos(f*x + e))*sqrt(-a*c)*sqrt((a*cos(f
*x + e) + a)/cos(f*x + e))*sqrt((c*cos(f*x + e) - c)/cos(f*x + e)) - (256*a*c*cos(f*x + e)^4 - 512*a*c*cos(f*x
 + e)^2 + 337*a*c)*sin(f*x + e))/((cos(f*x + e)^2 - 1)*sin(f*x + e)))/(a*c*f), -sqrt(a*c)*arctan((16*cos(f*x +
 e)^3 - 7*cos(f*x + e))*sqrt(a*c)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sqrt((c*cos(f*x + e) - c)/cos(f*x +
e))/((16*a*c*cos(f*x + e)^2 - 25*a*c)*sin(f*x + e)))/(a*c*f)]

Sympy [F]

\[ \int \frac {1}{\sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}} \, dx=\int \frac {1}{\sqrt {a \left (\sec {\left (e + f x \right )} + 1\right )} \sqrt {- c \left (\sec {\left (e + f x \right )} - 1\right )}}\, dx \]

[In]

integrate(1/(c-c*sec(f*x+e))**(1/2)/(a+a*sec(f*x+e))**(1/2),x)

[Out]

Integral(1/(sqrt(a*(sec(e + f*x) + 1))*sqrt(-c*(sec(e + f*x) - 1))), x)

Maxima [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.85 \[ \int \frac {1}{\sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}} \, dx=-\frac {f x + e - \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right ) - 1\right )}{\sqrt {a} \sqrt {c} f} \]

[In]

integrate(1/(c-c*sec(f*x+e))^(1/2)/(a+a*sec(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

-(f*x + e - arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) - 1))/(sqrt(a)*sqrt(c)*f)

Giac [A] (verification not implemented)

none

Time = 1.20 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.46 \[ \int \frac {1}{\sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}} \, dx=\frac {\frac {\sqrt {-a c} \log \left ({\left | c \right |} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2}\right )}{a {\left | c \right |}} - \frac {2 \, \sqrt {-a c} \log \left ({\left | c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + c \right |}\right )}{a {\left | c \right |}}}{2 \, f} \]

[In]

integrate(1/(c-c*sec(f*x+e))^(1/2)/(a+a*sec(f*x+e))^(1/2),x, algorithm="giac")

[Out]

1/2*(sqrt(-a*c)*log(abs(c)*tan(1/2*f*x + 1/2*e)^2)/(a*abs(c)) - 2*sqrt(-a*c)*log(abs(c*tan(1/2*f*x + 1/2*e)^2
+ c))/(a*abs(c)))/f

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}} \, dx=\int \frac {1}{\sqrt {a+\frac {a}{\cos \left (e+f\,x\right )}}\,\sqrt {c-\frac {c}{\cos \left (e+f\,x\right )}}} \,d x \]

[In]

int(1/((a + a/cos(e + f*x))^(1/2)*(c - c/cos(e + f*x))^(1/2)),x)

[Out]

int(1/((a + a/cos(e + f*x))^(1/2)*(c - c/cos(e + f*x))^(1/2)), x)