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

Optimal. Leaf size=37 \[ \frac {2 \sqrt {a} \sinh ^{-1}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a \sec (e+f x)+a}}\right )}{f} \]

[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 = 37, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {3801, 215} \[ \frac {2 \sqrt {a} \sinh ^{-1}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a \sec (e+f x)+a}}\right )}{f} \]

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 [A]  time = 0.15, size = 54, normalized size = 1.46 \[ -\frac {2 \tan \left (\frac {1}{2} (e+f x)\right ) \sqrt {a (\sec (e+f x)+1)} \sin ^{-1}\left (\sqrt {\sec (e+f x)}\right )}{f \sqrt {1-\sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.77, size = 189, normalized size = 5.11 \[ \left [\frac {\sqrt {a} \log \left (\frac {a \cos \left (f x + e\right )^{3} - 7 \, a \cos \left (f x + e\right )^{2} - \frac {4 \, {\left (\cos \left (f x + e\right )^{2} - 2 \, \cos \left (f x + e\right )\right )} \sqrt {a} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sin \left (f x + e\right )}{\sqrt {\cos \left (f x + e\right )}} + 8 \, a}{\cos \left (f x + e\right )^{3} + \cos \left (f x + e\right )^{2}}\right )}{2 \, f}, \frac {\sqrt {-a} \arctan \left (\frac {2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt {\cos \left (f x + e\right )} \sin \left (f x + e\right )}{a \cos \left (f x + e\right )^{2} - a \cos \left (f x + e\right ) - 2 \, a}\right )}{f}\right ] \]

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((a*cos(f*x + e)^3 - 7*a*cos(f*x + e)^2 - 4*(cos(f*x + e)^2 - 2*cos(f*x + e))*sqrt(a)*sqrt((a*
cos(f*x + e) + a)/cos(f*x + e))*sin(f*x + e)/sqrt(cos(f*x + e)) + 8*a)/(cos(f*x + e)^3 + cos(f*x + e)^2))/f, s
qrt(-a)*arctan(2*sqrt(-a)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sqrt(cos(f*x + e))*sin(f*x + e)/(a*cos(f*x +
e)^2 - a*cos(f*x + e) - 2*a))/f]

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

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]

integrate(sqrt(a*sec(f*x + e) + a)*sqrt(sec(f*x + e)), x)

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maple [B]  time = 1.61, size = 150, normalized size = 4.05 \[ \frac {\sqrt {\frac {a \left (1+\cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )}}\, \sqrt {\frac {1}{\cos \left (f x +e \right )}}\, \cos \left (f x +e \right ) \left (\arctan \left (\frac {\sqrt {-\frac {2}{1+\cos \left (f x +e \right )}}\, \left (\cos \left (f x +e \right )+1+\sin \left (f x +e \right )\right ) \sqrt {2}}{4}\right )-\arctan \left (\frac {\sqrt {-\frac {2}{1+\cos \left (f x +e \right )}}\, \left (\cos \left (f x +e \right )+1-\sin \left (f x +e \right )\right ) \sqrt {2}}{4}\right )\right ) \sqrt {-\frac {2}{1+\cos \left (f x +e \right )}}\, \left (-1+\cos ^{2}\left (f x +e \right )\right ) \sqrt {2}}{2 f \sin \left (f x +e \right )^{2}} \]

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]

1/2/f*(a*(1+cos(f*x+e))/cos(f*x+e))^(1/2)*(1/cos(f*x+e))^(1/2)*cos(f*x+e)*(arctan(1/4*(-2/(1+cos(f*x+e)))^(1/2
)*(cos(f*x+e)+1+sin(f*x+e))*2^(1/2))-arctan(1/4*(-2/(1+cos(f*x+e)))^(1/2)*(cos(f*x+e)+1-sin(f*x+e))*2^(1/2)))*
(-2/(1+cos(f*x+e)))^(1/2)/sin(f*x+e)^2*(-1+cos(f*x+e)^2)*2^(1/2)

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maxima [B]  time = 0.62, size = 241, normalized size = 6.51 \[ \frac {\sqrt {a} {\left (\log \left (2 \, \cos \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2 \, \sin \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2 \, \sqrt {2} \cos \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 2 \, \sqrt {2} \sin \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 2\right ) - \log \left (2 \, \cos \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2 \, \sin \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2 \, \sqrt {2} \cos \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 2 \, \sqrt {2} \sin \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 2\right ) + \log \left (2 \, \cos \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2 \, \sin \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 2 \, \sqrt {2} \cos \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 2 \, \sqrt {2} \sin \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 2\right ) - \log \left (2 \, \cos \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2 \, \sin \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 2 \, \sqrt {2} \cos \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 2 \, \sqrt {2} \sin \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 2\right )\right )}}{2 \, f} \]

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*f*x + 1/2*e)^2 + 2*sin(1/2*f*x + 1/2*e)^2 + 2*sqrt(2)*cos(1/2*f*x + 1/2*e) + 2*sqrt
(2)*sin(1/2*f*x + 1/2*e) + 2) - log(2*cos(1/2*f*x + 1/2*e)^2 + 2*sin(1/2*f*x + 1/2*e)^2 + 2*sqrt(2)*cos(1/2*f*
x + 1/2*e) - 2*sqrt(2)*sin(1/2*f*x + 1/2*e) + 2) + log(2*cos(1/2*f*x + 1/2*e)^2 + 2*sin(1/2*f*x + 1/2*e)^2 - 2
*sqrt(2)*cos(1/2*f*x + 1/2*e) + 2*sqrt(2)*sin(1/2*f*x + 1/2*e) + 2) - log(2*cos(1/2*f*x + 1/2*e)^2 + 2*sin(1/2
*f*x + 1/2*e)^2 - 2*sqrt(2)*cos(1/2*f*x + 1/2*e) - 2*sqrt(2)*sin(1/2*f*x + 1/2*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 \[ \int \sqrt {a \left (\sec {\left (e + f x \right )} + 1\right )} \sqrt {\sec {\left (e + f x \right )}}\, dx \]

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

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