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

Optimal. Leaf size=49 \[ \frac{c \tan (e+f x) \log (\cos (e+f x)+1)}{f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}} \]

[Out]

(c*Log[1 + Cos[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 = 0.084185, antiderivative size = 49, 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 = {3911, 31} \[ \frac{c \tan (e+f x) \log (\cos (e+f x)+1)}{f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3911

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

\begin{align*} \int \frac{\sqrt{c-c \sec (e+f x)}}{\sqrt{a+a \sec (e+f x)}} \, dx &=\frac{(a c \tan (e+f x)) \operatorname{Subst}\left (\int \frac{1}{a+a x} \, dx,x,\cos (e+f x)\right )}{f \sqrt{a+a \sec (e+f x)} \sqrt{c-c \sec (e+f x)}}\\ &=\frac{c \log (1+\cos (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 = 0.910697, size = 127, normalized size = 2.59 \[ -\frac{\left (1+e^{i (e+f x)}\right ) \sqrt{\frac{c \left (-1+e^{i (e+f x)}\right )^2}{1+e^{2 i (e+f x)}}} \left (f x+2 i \log \left (1+e^{i (e+f x)}\right )\right )}{f \left (-1+e^{i (e+f x)}\right ) \sqrt{\frac{a \left (1+e^{i (e+f x)}\right )^2}{1+e^{2 i (e+f x)}}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.302, size = 75, normalized size = 1.5 \begin{align*}{\frac{\cos \left ( fx+e \right ) }{af\sin \left ( fx+e \right ) }\sqrt{{\frac{c \left ( -1+\cos \left ( fx+e \right ) \right ) }{\cos \left ( fx+e \right ) }}}\sqrt{{\frac{a \left ( 1+\cos \left ( fx+e \right ) \right ) }{\cos \left ( fx+e \right ) }}}\ln \left ( 2\, \left ( 1+\cos \left ( fx+e \right ) \right ) ^{-1} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.57561, size = 46, normalized size = 0.94 \begin{align*} \frac{\sqrt{c} \log \left (\frac{\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )}{\sqrt{-a} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{-c \sec \left (f x + e\right ) + c}}{\sqrt{a \sec \left (f x + e\right ) + a}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{- c \left (\sec{\left (e + f x \right )} - 1\right )}}{\sqrt{a \left (\sec{\left (e + f x \right )} + 1\right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.79559, size = 92, normalized size = 1.88 \begin{align*} \frac{\sqrt{-a c} c \log \left ({\left | c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + c \right |}\right ) \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )\right ) \mathrm{sgn}\left (\cos \left (f x + e\right )\right )}{a f{\left | c \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(-a*c)*c*log(abs(c*tan(1/2*f*x + 1/2*e)^2 + c))*sgn(tan(1/2*f*x + 1/2*e)^3 + tan(1/2*f*x + 1/2*e))*sgn(cos
(f*x + e))/(a*f*abs(c))