3.268 \(\int \frac {\sec (e+f x)}{\sqrt {4-5 \sec (e+f x)} \sqrt {2+3 \sec (e+f x)}} \, dx\)

Optimal. Leaf size=125 \[ \frac {2 i \cot (e+f x) \sqrt {\frac {1-\sec (e+f x)}{3 \sec (e+f x)+2}} \sqrt {\frac {\sec (e+f x)+1}{3 \sec (e+f x)+2}} (3 \sec (e+f x)+2) F\left (i \sinh ^{-1}\left (\frac {\sqrt {5} \sqrt {4-5 \sec (e+f x)}}{\sqrt {3 \sec (e+f x)+2}}\right )|\frac {1}{45}\right )}{3 \sqrt {5} f} \]

[Out]

2/15*I*cot(f*x+e)*EllipticF(I*5^(1/2)*(4-5*sec(f*x+e))^(1/2)/(2+3*sec(f*x+e))^(1/2),1/15*5^(1/2))*(2+3*sec(f*x
+e))*((1-sec(f*x+e))/(2+3*sec(f*x+e)))^(1/2)*((1+sec(f*x+e))/(2+3*sec(f*x+e)))^(1/2)/f*5^(1/2)

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Rubi [A]  time = 0.12, antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.029, Rules used = {3984} \[ \frac {2 i \cot (e+f x) \sqrt {\frac {1-\sec (e+f x)}{3 \sec (e+f x)+2}} \sqrt {\frac {\sec (e+f x)+1}{3 \sec (e+f x)+2}} (3 \sec (e+f x)+2) F\left (i \sinh ^{-1}\left (\frac {\sqrt {5} \sqrt {4-5 \sec (e+f x)}}{\sqrt {3 \sec (e+f x)+2}}\right )|\frac {1}{45}\right )}{3 \sqrt {5} f} \]

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]/(Sqrt[4 - 5*Sec[e + f*x]]*Sqrt[2 + 3*Sec[e + f*x]]),x]

[Out]

(((2*I)/3)*Cot[e + f*x]*EllipticF[I*ArcSinh[(Sqrt[5]*Sqrt[4 - 5*Sec[e + f*x]])/Sqrt[2 + 3*Sec[e + f*x]]], 1/45
]*Sqrt[(1 - Sec[e + f*x])/(2 + 3*Sec[e + f*x])]*Sqrt[(1 + Sec[e + f*x])/(2 + 3*Sec[e + f*x])]*(2 + 3*Sec[e + f
*x]))/(Sqrt[5]*f)

Rule 3984

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

Rubi steps

\begin {align*} \int \frac {\sec (e+f x)}{\sqrt {4-5 \sec (e+f x)} \sqrt {2+3 \sec (e+f x)}} \, dx &=\frac {2 i \cot (e+f x) F\left (i \sinh ^{-1}\left (\frac {\sqrt {5} \sqrt {4-5 \sec (e+f x)}}{\sqrt {2+3 \sec (e+f x)}}\right )|\frac {1}{45}\right ) \sqrt {\frac {1-\sec (e+f x)}{2+3 \sec (e+f x)}} \sqrt {\frac {1+\sec (e+f x)}{2+3 \sec (e+f x)}} (2+3 \sec (e+f x))}{3 \sqrt {5} f}\\ \end {align*}

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Mathematica [A]  time = 0.47, size = 176, normalized size = 1.41 \[ -\frac {4 \sin ^4\left (\frac {1}{2} (e+f x)\right ) \sqrt {-\cot ^2\left (\frac {1}{2} (e+f x)\right )} \csc (e+f x) \sec (e+f x) \sqrt {-\left ((2 \cos (e+f x)+3) \csc ^2\left (\frac {1}{2} (e+f x)\right )\right )} \sqrt {-\left ((4 \cos (e+f x)-5) \csc ^2\left (\frac {1}{2} (e+f x)\right )\right )} F\left (\sin ^{-1}\left (\sqrt {\frac {5}{22}} \sqrt {\frac {4 \cos (e+f x)-5}{\cos (e+f x)-1}}\right )|\frac {44}{45}\right )}{3 \sqrt {5} f \sqrt {4-5 \sec (e+f x)} \sqrt {3 \sec (e+f x)+2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[e + f*x]/(Sqrt[4 - 5*Sec[e + f*x]]*Sqrt[2 + 3*Sec[e + f*x]]),x]

[Out]

(-4*Sqrt[-Cot[(e + f*x)/2]^2]*Sqrt[-((3 + 2*Cos[e + f*x])*Csc[(e + f*x)/2]^2)]*Sqrt[-((-5 + 4*Cos[e + f*x])*Cs
c[(e + f*x)/2]^2)]*Csc[e + f*x]*EllipticF[ArcSin[Sqrt[5/22]*Sqrt[(-5 + 4*Cos[e + f*x])/(-1 + Cos[e + f*x])]],
44/45]*Sec[e + f*x]*Sin[(e + f*x)/2]^4)/(3*Sqrt[5]*f*Sqrt[4 - 5*Sec[e + f*x]]*Sqrt[2 + 3*Sec[e + f*x]])

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fricas [F]  time = 0.92, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {3 \, \sec \left (f x + e\right ) + 2} \sqrt {-5 \, \sec \left (f x + e\right ) + 4} \sec \left (f x + e\right )}{15 \, \sec \left (f x + e\right )^{2} - 2 \, \sec \left (f x + e\right ) - 8}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)/(4-5*sec(f*x+e))^(1/2)/(2+3*sec(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

integral(-sqrt(3*sec(f*x + e) + 2)*sqrt(-5*sec(f*x + e) + 4)*sec(f*x + e)/(15*sec(f*x + e)^2 - 2*sec(f*x + e)
- 8), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec \left (f x + e\right )}{\sqrt {3 \, \sec \left (f x + e\right ) + 2} \sqrt {-5 \, \sec \left (f x + e\right ) + 4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)/(4-5*sec(f*x+e))^(1/2)/(2+3*sec(f*x+e))^(1/2),x, algorithm="giac")

[Out]

integrate(sec(f*x + e)/(sqrt(3*sec(f*x + e) + 2)*sqrt(-5*sec(f*x + e) + 4)), x)

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maple [A]  time = 2.07, size = 170, normalized size = 1.36 \[ -\frac {i \sqrt {-\frac {2 \left (4 \cos \left (f x +e \right )-5\right )}{1+\cos \left (f x +e \right )}}\, \sqrt {10}\, \sqrt {\frac {2 \cos \left (f x +e \right )+3}{1+\cos \left (f x +e \right )}}\, \left (\sin ^{2}\left (f x +e \right )\right ) \EllipticF \left (\frac {3 i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, \frac {\sqrt {5}}{15}\right ) \cos \left (f x +e \right ) \sqrt {\frac {4 \cos \left (f x +e \right )-5}{\cos \left (f x +e \right )}}\, \sqrt {\frac {2 \cos \left (f x +e \right )+3}{\cos \left (f x +e \right )}}}{15 f \left (8 \left (\cos ^{3}\left (f x +e \right )\right )-6 \left (\cos ^{2}\left (f x +e \right )\right )-17 \cos \left (f x +e \right )+15\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)/(4-5*sec(f*x+e))^(1/2)/(2+3*sec(f*x+e))^(1/2),x)

[Out]

-1/15*I/f*(-2*(4*cos(f*x+e)-5)/(1+cos(f*x+e)))^(1/2)*10^(1/2)*((2*cos(f*x+e)+3)/(1+cos(f*x+e)))^(1/2)*sin(f*x+
e)^2*EllipticF(3*I*(-1+cos(f*x+e))/sin(f*x+e),1/15*5^(1/2))*cos(f*x+e)*((4*cos(f*x+e)-5)/cos(f*x+e))^(1/2)*((2
*cos(f*x+e)+3)/cos(f*x+e))^(1/2)/(8*cos(f*x+e)^3-6*cos(f*x+e)^2-17*cos(f*x+e)+15)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec \left (f x + e\right )}{\sqrt {3 \, \sec \left (f x + e\right ) + 2} \sqrt {-5 \, \sec \left (f x + e\right ) + 4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)/(4-5*sec(f*x+e))^(1/2)/(2+3*sec(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

integrate(sec(f*x + e)/(sqrt(3*sec(f*x + e) + 2)*sqrt(-5*sec(f*x + e) + 4)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{\cos \left (e+f\,x\right )\,\sqrt {\frac {3}{\cos \left (e+f\,x\right )}+2}\,\sqrt {4-\frac {5}{\cos \left (e+f\,x\right )}}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(e + f*x)*(3/cos(e + f*x) + 2)^(1/2)*(4 - 5/cos(e + f*x))^(1/2)),x)

[Out]

int(1/(cos(e + f*x)*(3/cos(e + f*x) + 2)^(1/2)*(4 - 5/cos(e + f*x))^(1/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec {\left (e + f x \right )}}{\sqrt {4 - 5 \sec {\left (e + f x \right )}} \sqrt {3 \sec {\left (e + f x \right )} + 2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)/(4-5*sec(f*x+e))**(1/2)/(2+3*sec(f*x+e))**(1/2),x)

[Out]

Integral(sec(e + f*x)/(sqrt(4 - 5*sec(e + f*x))*sqrt(3*sec(e + f*x) + 2)), x)

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