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

Optimal. Leaf size=46 \[ \frac{\tan (e+f x) \log (\sin (e+f x))}{f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}} \]

[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]])

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Rubi [A]  time = 0.0896478, antiderivative size = 46, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.067, Rules used = {3905, 3475} \[ \frac{\tan (e+f x) \log (\sin (e+f x))}{f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}} \]

Antiderivative was successfully verified.

[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 3905

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]

Rule 3475

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

Rubi steps

\begin{align*} \int \frac{1}{\sqrt{a+a \sec (e+f x)} \sqrt{c-c \sec (e+f x)}} \, dx &=\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 [C]  time = 1.06858, size = 104, normalized size = 2.26 \[ -\frac{2 \left (-1+e^{i (e+f x)}\right ) \left (f x+i \log \left (1-e^{2 i (e+f x)}\right )\right ) \cos ^2\left (\frac{1}{2} (e+f x)\right ) \sec (e+f x)}{f \left (1+e^{i (e+f x)}\right ) \sqrt{a (\sec (e+f x)+1)} \sqrt{c-c \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(-1 + E^(I*(e + f*x)))*Cos[(e + f*x)/2]^2*(f*x + I*Log[1 - E^((2*I)*(e + f*x))])*Sec[e + f*x])/((1 + E^(I*
(e + f*x)))*f*Sqrt[a*(1 + Sec[e + f*x])]*Sqrt[c - c*Sec[e + f*x]])

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Maple [B]  time = 0.26, size = 100, normalized size = 2.2 \begin{align*}{\frac{\cos \left ( fx+e \right ) }{af\sin \left ( fx+e \right ) c}\sqrt{{\frac{a \left ( 1+\cos \left ( fx+e \right ) \right ) }{\cos \left ( fx+e \right ) }}} \left ( \ln \left ( 2\, \left ( 1+\cos \left ( fx+e \right ) \right ) ^{-1} \right ) -\ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) \right ) \sqrt{{\frac{c \left ( -1+\cos \left ( fx+e \right ) \right ) }{\cos \left ( fx+e \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.77173, size = 53, normalized size = 1.15 \begin{align*} -\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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [B]  time = 1.8759, size = 682, normalized size = 14.83 \begin{align*} \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 ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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 \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out