3.2.36 \(\int \frac {\sec (e+f x)}{\sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}} \, dx\) [136]

Optimal. Leaf size=47 \[ -\frac {\tanh ^{-1}(\cos (e+f x)) \tan (e+f x)}{f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}} \]

[Out]

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

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Rubi [A]
time = 0.09, antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {4044, 3855} \begin {gather*} -\frac {\tan (e+f x) \tanh ^{-1}(\cos (e+f x))}{f \sqrt {a \sec (e+f x)+a} \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]

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

Rule 3855

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

Rule 4044

Int[csc[(e_.) + (f_.)*(x_)]*(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[Csc[e + f*x]*Cot[e + f*x]^(2*m), 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]

Rubi steps

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

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Mathematica [C] Result contains complex when optimal does not.
time = 0.81, size = 94, normalized size = 2.00 \begin {gather*} \frac {4 i \left (-1+e^{i (e+f x)}\right ) \tanh ^{-1}\left (e^{i (e+f x)}\right ) \cos ^2\left (\frac {1}{2} (e+f x)\right ) \sec (e+f x)}{\left (1+e^{i (e+f x)}\right ) f \sqrt {a (1+\sec (e+f x))} \sqrt {c-c \sec (e+f x)}} \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]

((4*I)*(-1 + E^(I*(e + f*x)))*ArcTanh[E^(I*(e + f*x))]*Cos[(e + f*x)/2]^2*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 [A]
time = 3.11, size = 85, normalized size = 1.81

method	result	size
default	\(-\frac {\sqrt {\frac {a \left (\cos \left (f x +e \right )+1\right )}{\cos \left (f x +e \right )}}\, \ln \left (-\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}\right ) \sqrt {\frac {c \left (-1+\cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )}}\, \cos \left (f x +e \right )}{f \sin \left (f x +e \right ) c a}\)	\(85\)
risch	\(-\frac {i \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) \ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )}{\sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) \sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, f}+\frac {i \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) \ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )}{\sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) \sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, f}\)	\(228\)

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)*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/a

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Maxima [A]
time = 0.56, size = 48, normalized size = 1.02 \begin {gather*} -\frac {\arctan \left (\sin \left (f x + e\right ), \cos \left (f x + e\right ) + 1\right ) - \arctan \left (\sin \left (f x + e\right ), \cos \left (f x + e\right ) - 1\right )}{\sqrt {a} \sqrt {c} 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="maxima")

[Out]

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

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Fricas [A]
time = 3.17, size = 218, normalized size = 4.64 \begin {gather*} \left [-\frac {\sqrt {-a c} \log \left (-\frac {4 \, {\left (2 \, \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 )}} \cos \left (f x + e\right )^{2} + {\left (a c \cos \left (f x + e\right )^{2} + 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 {\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 )}}}{a c \sin \left (f x + e\right )}\right )}{a c f}\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]

[-1/2*sqrt(-a*c)*log(-4*(2*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))*cos(f*x + e)^2 + (a*c*cos(f*x + e)^2 + a*c)*sin(f*x + e))/((cos(f*x + e)^2 - 1)*sin(f*x + e)))/(a*c*f)
, sqrt(a*c)*arctan(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))/(
a*c*sin(f*x + e)))/(a*c*f)]

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

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Giac [A]
time = 1.68, size = 70, normalized size = 1.49 \begin {gather*} -\frac {c^{2} {\left (\frac {\log \left ({\left | c \right |} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2}\right )}{c} - \frac {\log \left ({\left | c \right |}\right )}{c}\right )}}{2 \, \sqrt {-a c} f {\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 )} \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]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {1}{\cos \left (e+f\,x\right )\,\sqrt {a+\frac {a}{\cos \left (e+f\,x\right )}}\,\sqrt {c-\frac {c}{\cos \left (e+f\,x\right )}}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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