3.2.9 \(\int \sec (e+f x) \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)} \, dx\) [109]

Optimal. Leaf size=41 \[ -\frac {c \sqrt {a+a \sec (e+f x)} \tan (e+f x)}{f \sqrt {c-c \sec (e+f x)}} \]

[Out]

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

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Rubi [A]
time = 0.08, antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.028, Rules used = {4038} \begin {gather*} -\frac {c \tan (e+f x) \sqrt {a \sec (e+f x)+a}}{f \sqrt {c-c \sec (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4038

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

Rubi steps

\begin {align*} \int \sec (e+f x) \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)} \, dx &=-\frac {c \sqrt {a+a \sec (e+f x)} \tan (e+f x)}{f \sqrt {c-c \sec (e+f x)}}\\ \end {align*}

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Mathematica [A]
time = 0.18, size = 56, normalized size = 1.37 \begin {gather*} \frac {\csc \left (\frac {1}{2} (e+f x)\right ) \sec \left (\frac {1}{2} (e+f x)\right ) \sqrt {a (1+\sec (e+f x))} \sqrt {c-c \sec (e+f x)}}{2 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Csc[(e + f*x)/2]*Sec[(e + f*x)/2]*Sqrt[a*(1 + Sec[e + f*x])]*Sqrt[c - c*Sec[e + f*x]])/(2*f)

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Maple [A]
time = 2.58, size = 62, normalized size = 1.51

method	result	size
default	\(-\frac {\sqrt {\frac {a \left (\cos \left (f x +e \right )+1\right )}{\cos \left (f x +e \right )}}\, \sqrt {\frac {c \left (-1+\cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )}}\, \sin \left (f x +e \right )}{f \left (-1+\cos \left (f x +e \right )\right )}\)	\(62\)
risch	\(\frac {2 i \sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, {\mathrm e}^{i \left (f x +e \right )}}{\left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) f}\)	\(102\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.51, size = 59, normalized size = 1.44 \begin {gather*} \frac {2 \, \sqrt {-a} \sqrt {c}}{f {\left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )} {\left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 2.10, size = 61, normalized size = 1.49 \begin {gather*} \frac {\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 )}}}{f \sin \left (f x + e\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sqrt((c*cos(f*x + e) - c)/cos(f*x + e))/(f*sin(f*x + e))

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt {a \left (\sec {\left (e + f x \right )} + 1\right )} \sqrt {- c \left (\sec {\left (e + f x \right )} - 1\right )} \sec {\left (e + f x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 1.40, size = 54, normalized size = 1.32 \begin {gather*} \frac {2 \, \sqrt {-a c} {\left | c \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 )}{{\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c\right )} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 1.94, size = 47, normalized size = 1.15 \begin {gather*} \frac {\sqrt {c-\frac {c}{\cos \left (e+f\,x\right )}}\,\sqrt {\frac {a\,\left (\cos \left (e+f\,x\right )+1\right )}{\cos \left (e+f\,x\right )}}}{f\,\sin \left (e+f\,x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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