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

Optimal. Leaf size=38 \[ \frac {2 \sqrt {a} \sinh ^{-1}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a-a \sec (e+f x)}}\right )}{f} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {3801, 215} \[ \frac {2 \sqrt {a} \sinh ^{-1}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a-a \sec (e+f x)}}\right )}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 215

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[(Rt[b, 2]*x)/Sqrt[a]]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

Rule 3801

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[(-2*a*Sq
rt[(a*d)/b])/(b*f), Subst[Int[1/Sqrt[1 + x^2/a], x], x, (b*Cot[e + f*x])/Sqrt[a + b*Csc[e + f*x]]], x] /; Free
Q[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && GtQ[(a*d)/b, 0]

Rubi steps

\begin {align*} \int \sqrt {-\sec (e+f x)} \sqrt {a-a \sec (e+f x)} \, dx &=\frac {2 \operatorname {Subst}\left (\int \frac {1}{\sqrt {1+\frac {x^2}{a}}} \, dx,x,\frac {a \tan (e+f x)}{\sqrt {a-a \sec (e+f x)}}\right )}{f}\\ &=\frac {2 \sqrt {a} \sinh ^{-1}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a-a \sec (e+f x)}}\right )}{f}\\ \end {align*}

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Mathematica [C]  time = 2.06, size = 299, normalized size = 7.87 \[ \frac {\csc \left (\frac {1}{2} (e+f x)\right ) \sqrt {a-a \sec (e+f x)} \left (\log \left (-\sqrt {2} \sin \left (\frac {1}{2} (e+f x)\right )-\sqrt {2} \cos \left (\frac {1}{2} (e+f x)\right )+2\right )-\log \left (-\sqrt {2} \sin \left (\frac {1}{2} (e+f x)\right )+\sqrt {2} \cos \left (\frac {1}{2} (e+f x)\right )+2\right )-2 i \tan ^{-1}\left (\frac {\cos \left (\frac {1}{4} (e+f x)\right )-\left (\sqrt {2}-1\right ) \sin \left (\frac {1}{4} (e+f x)\right )}{\left (1+\sqrt {2}\right ) \cos \left (\frac {1}{4} (e+f x)\right )-\sin \left (\frac {1}{4} (e+f x)\right )}\right )+2 i \tan ^{-1}\left (\frac {\cos \left (\frac {1}{4} (e+f x)\right )-\left (1+\sqrt {2}\right ) \sin \left (\frac {1}{4} (e+f x)\right )}{\left (\sqrt {2}-1\right ) \cos \left (\frac {1}{4} (e+f x)\right )-\sin \left (\frac {1}{4} (e+f x)\right )}\right )-4 \tanh ^{-1}\left (\sqrt {2} \sec \left (\frac {f x}{4}\right ) \cos \left (\frac {1}{4} (2 e+f x)\right )+\tan \left (\frac {f x}{4}\right )\right )\right )}{2 \sqrt {2} f \sqrt {-\sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Csc[(e + f*x)/2]*((-2*I)*ArcTan[(Cos[(e + f*x)/4] - (-1 + Sqrt[2])*Sin[(e + f*x)/4])/((1 + Sqrt[2])*Cos[(e +
f*x)/4] - Sin[(e + f*x)/4])] + (2*I)*ArcTan[(Cos[(e + f*x)/4] - (1 + Sqrt[2])*Sin[(e + f*x)/4])/((-1 + Sqrt[2]
)*Cos[(e + f*x)/4] - Sin[(e + f*x)/4])] - 4*ArcTanh[Sqrt[2]*Cos[(2*e + f*x)/4]*Sec[(f*x)/4] + Tan[(f*x)/4]] +
Log[2 - Sqrt[2]*Cos[(e + f*x)/2] - Sqrt[2]*Sin[(e + f*x)/2]] - Log[2 + Sqrt[2]*Cos[(e + f*x)/2] - Sqrt[2]*Sin[
(e + f*x)/2]])*Sqrt[a - a*Sec[e + f*x]])/(2*Sqrt[2]*f*Sqrt[-Sec[e + f*x]])

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fricas [B]  time = 0.70, size = 215, normalized size = 5.66 \[ \left [\frac {\sqrt {a} \log \left (\frac {4 \, {\left (\cos \left (f x + e\right )^{3} + 3 \, \cos \left (f x + e\right )^{2} + 2 \, \cos \left (f x + e\right )\right )} \sqrt {a} \sqrt {\frac {a \cos \left (f x + e\right ) - a}{\cos \left (f x + e\right )}} \sqrt {-\frac {1}{\cos \left (f x + e\right )}} + {\left (a \cos \left (f x + e\right )^{2} + 8 \, a \cos \left (f x + e\right ) + 8 \, a\right )} \sin \left (f x + e\right )}{\cos \left (f x + e\right )^{2} \sin \left (f x + e\right )}\right )}{2 \, f}, -\frac {\sqrt {-a} \arctan \left (\frac {2 \, {\left (\cos \left (f x + e\right )^{2} + \cos \left (f x + e\right )\right )} \sqrt {-a} \sqrt {\frac {a \cos \left (f x + e\right ) - a}{\cos \left (f x + e\right )}} \sqrt {-\frac {1}{\cos \left (f x + e\right )}}}{{\left (a \cos \left (f x + e\right ) + 2 \, a\right )} \sin \left (f x + e\right )}\right )}{f}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*sqrt(a)*log((4*(cos(f*x + e)^3 + 3*cos(f*x + e)^2 + 2*cos(f*x + e))*sqrt(a)*sqrt((a*cos(f*x + e) - a)/cos
(f*x + e))*sqrt(-1/cos(f*x + e)) + (a*cos(f*x + e)^2 + 8*a*cos(f*x + e) + 8*a)*sin(f*x + e))/(cos(f*x + e)^2*s
in(f*x + e)))/f, -sqrt(-a)*arctan(2*(cos(f*x + e)^2 + cos(f*x + e))*sqrt(-a)*sqrt((a*cos(f*x + e) - a)/cos(f*x
+ e))*sqrt(-1/cos(f*x + e))/((a*cos(f*x + e) + 2*a)*sin(f*x + e)))/f]

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)2*sqrt(2)*(-a^2*atan(sqrt(a)/sqrt(-a))/sqrt(2)/sqrt(-a)+a^2*at
an(sqrt(a*tan(1/2*(f*x+exp(1)))^2+a)/sqrt(2)/sqrt(-a))/sqrt(2)/sqrt(-a))*abs(a)*sign(tan(1/2*(f*x+exp(1)))^3+t
an(1/2*(f*x+exp(1))))/a^2/f

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maple [B]  time = 1.70, size = 126, normalized size = 3.32 \[ \frac {\left (\arctanh \left (\frac {\sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \left (\cos \left (f x +e \right )+1+\sin \left (f x +e \right )\right )}{2}\right )-\arctanh \left (\frac {\sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \left (-\cos \left (f x +e \right )-1+\sin \left (f x +e \right )\right )}{2}\right )\right ) \cos \left (f x +e \right ) \sqrt {-\frac {1}{\cos \left (f x +e \right )}}\, \sqrt {\frac {a \left (-1+\cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )}}}{f \sin \left (f x +e \right ) \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/f*(arctanh(1/2*(1/(1+cos(f*x+e)))^(1/2)*(cos(f*x+e)+1+sin(f*x+e)))-arctanh(1/2*(1/(1+cos(f*x+e)))^(1/2)*(-co
s(f*x+e)-1+sin(f*x+e))))*cos(f*x+e)*(-1/cos(f*x+e))^(1/2)*(a*(-1+cos(f*x+e))/cos(f*x+e))^(1/2)/sin(f*x+e)/(1/(
1+cos(f*x+e)))^(1/2)

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maxima [B]  time = 0.99, size = 353, normalized size = 9.29 \[ -\frac {\sqrt {a} {\left (\log \left (2 \, \cos \left (\frac {1}{2} \, \arctan \left (\sin \left (f x + e\right ), \cos \left (f x + e\right )\right )\right )^{2} + 2 \, \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (f x + e\right ), \cos \left (f x + e\right )\right )\right )^{2} + 2 \, \sqrt {2} \cos \left (\frac {1}{2} \, \arctan \left (\sin \left (f x + e\right ), \cos \left (f x + e\right )\right )\right ) + 2 \, \sqrt {2} \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (f x + e\right ), \cos \left (f x + e\right )\right )\right ) + 2\right ) + \log \left (2 \, \cos \left (\frac {1}{2} \, \arctan \left (\sin \left (f x + e\right ), \cos \left (f x + e\right )\right )\right )^{2} + 2 \, \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (f x + e\right ), \cos \left (f x + e\right )\right )\right )^{2} + 2 \, \sqrt {2} \cos \left (\frac {1}{2} \, \arctan \left (\sin \left (f x + e\right ), \cos \left (f x + e\right )\right )\right ) - 2 \, \sqrt {2} \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (f x + e\right ), \cos \left (f x + e\right )\right )\right ) + 2\right ) - \log \left (2 \, \cos \left (\frac {1}{2} \, \arctan \left (\sin \left (f x + e\right ), \cos \left (f x + e\right )\right )\right )^{2} + 2 \, \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (f x + e\right ), \cos \left (f x + e\right )\right )\right )^{2} - 2 \, \sqrt {2} \cos \left (\frac {1}{2} \, \arctan \left (\sin \left (f x + e\right ), \cos \left (f x + e\right )\right )\right ) + 2 \, \sqrt {2} \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (f x + e\right ), \cos \left (f x + e\right )\right )\right ) + 2\right ) - \log \left (2 \, \cos \left (\frac {1}{2} \, \arctan \left (\sin \left (f x + e\right ), \cos \left (f x + e\right )\right )\right )^{2} + 2 \, \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (f x + e\right ), \cos \left (f x + e\right )\right )\right )^{2} - 2 \, \sqrt {2} \cos \left (\frac {1}{2} \, \arctan \left (\sin \left (f x + e\right ), \cos \left (f x + e\right )\right )\right ) - 2 \, \sqrt {2} \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (f x + e\right ), \cos \left (f x + e\right )\right )\right ) + 2\right )\right )}}{2 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*sqrt(a)*(log(2*cos(1/2*arctan2(sin(f*x + e), cos(f*x + e)))^2 + 2*sin(1/2*arctan2(sin(f*x + e), cos(f*x +
e)))^2 + 2*sqrt(2)*cos(1/2*arctan2(sin(f*x + e), cos(f*x + e))) + 2*sqrt(2)*sin(1/2*arctan2(sin(f*x + e), cos
(f*x + e))) + 2) + log(2*cos(1/2*arctan2(sin(f*x + e), cos(f*x + e)))^2 + 2*sin(1/2*arctan2(sin(f*x + e), cos(
f*x + e)))^2 + 2*sqrt(2)*cos(1/2*arctan2(sin(f*x + e), cos(f*x + e))) - 2*sqrt(2)*sin(1/2*arctan2(sin(f*x + e)
, cos(f*x + e))) + 2) - log(2*cos(1/2*arctan2(sin(f*x + e), cos(f*x + e)))^2 + 2*sin(1/2*arctan2(sin(f*x + e),
cos(f*x + e)))^2 - 2*sqrt(2)*cos(1/2*arctan2(sin(f*x + e), cos(f*x + e))) + 2*sqrt(2)*sin(1/2*arctan2(sin(f*x
+ e), cos(f*x + e))) + 2) - log(2*cos(1/2*arctan2(sin(f*x + e), cos(f*x + e)))^2 + 2*sin(1/2*arctan2(sin(f*x
+ e), cos(f*x + e)))^2 - 2*sqrt(2)*cos(1/2*arctan2(sin(f*x + e), cos(f*x + e))) - 2*sqrt(2)*sin(1/2*arctan2(si
n(f*x + e), cos(f*x + e))) + 2))/f

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \int \sqrt {a-\frac {a}{\cos \left (e+f\,x\right )}}\,\sqrt {-\frac {1}{\cos \left (e+f\,x\right )}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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