3.219 \(\int \frac {1}{\sqrt {a+b \sec (e+f x)} \sqrt {c+d \sec (e+f x)}} \, dx\)

Optimal. Leaf size=398 \[ -\frac {2 b \sqrt {a+b} \cot (e+f x) (c+d \sec (e+f x)) \sqrt {\frac {(b c-a d) (1-\sec (e+f x))}{(a+b) (c+d \sec (e+f x))}} \sqrt {-\frac {(b c-a d) (\sec (e+f x)+1)}{(a-b) (c+d \sec (e+f x))}} F\left (\sin ^{-1}\left (\frac {\sqrt {c+d} \sqrt {a+b \sec (e+f x)}}{\sqrt {a+b} \sqrt {c+d \sec (e+f x)}}\right )|\frac {(a+b) (c-d)}{(a-b) (c+d)}\right )}{a f \sqrt {c+d} (b c-a d)}-\frac {2 \sqrt {c+d} \cot (e+f x) (a+b \sec (e+f x)) \sqrt {-\frac {(b c-a d) (1-\sec (e+f x))}{(c+d) (a+b \sec (e+f x))}} \sqrt {\frac {(b c-a d) (\sec (e+f x)+1)}{(c-d) (a+b \sec (e+f x))}} \Pi \left (\frac {a (c+d)}{(a+b) c};\sin ^{-1}\left (\frac {\sqrt {a+b} \sqrt {c+d \sec (e+f x)}}{\sqrt {c+d} \sqrt {a+b \sec (e+f x)}}\right )|\frac {(a-b) (c+d)}{(a+b) (c-d)}\right )}{a c f \sqrt {a+b}} \]

[Out]

-2*cot(f*x+e)*EllipticPi((a+b)^(1/2)*(c+d*sec(f*x+e))^(1/2)/(c+d)^(1/2)/(a+b*sec(f*x+e))^(1/2),a*(c+d)/(a+b)/c
,((a-b)*(c+d)/(a+b)/(c-d))^(1/2))*(a+b*sec(f*x+e))*(c+d)^(1/2)*(-(-a*d+b*c)*(1-sec(f*x+e))/(c+d)/(a+b*sec(f*x+
e)))^(1/2)*((-a*d+b*c)*(1+sec(f*x+e))/(c-d)/(a+b*sec(f*x+e)))^(1/2)/a/c/f/(a+b)^(1/2)-2*b*cot(f*x+e)*EllipticF
((c+d)^(1/2)*(a+b*sec(f*x+e))^(1/2)/(a+b)^(1/2)/(c+d*sec(f*x+e))^(1/2),((a+b)*(c-d)/(a-b)/(c+d))^(1/2))*(c+d*s
ec(f*x+e))*(a+b)^(1/2)*((-a*d+b*c)*(1-sec(f*x+e))/(a+b)/(c+d*sec(f*x+e)))^(1/2)*(-(-a*d+b*c)*(1+sec(f*x+e))/(a
-b)/(c+d*sec(f*x+e)))^(1/2)/a/(-a*d+b*c)/f/(c+d)^(1/2)

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Rubi [A]  time = 0.44, antiderivative size = 398, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {3938, 3936, 3984} \[ -\frac {2 b \sqrt {a+b} \cot (e+f x) (c+d \sec (e+f x)) \sqrt {\frac {(b c-a d) (1-\sec (e+f x))}{(a+b) (c+d \sec (e+f x))}} \sqrt {-\frac {(b c-a d) (\sec (e+f x)+1)}{(a-b) (c+d \sec (e+f x))}} F\left (\sin ^{-1}\left (\frac {\sqrt {c+d} \sqrt {a+b \sec (e+f x)}}{\sqrt {a+b} \sqrt {c+d \sec (e+f x)}}\right )|\frac {(a+b) (c-d)}{(a-b) (c+d)}\right )}{a f \sqrt {c+d} (b c-a d)}-\frac {2 \sqrt {c+d} \cot (e+f x) (a+b \sec (e+f x)) \sqrt {-\frac {(b c-a d) (1-\sec (e+f x))}{(c+d) (a+b \sec (e+f x))}} \sqrt {\frac {(b c-a d) (\sec (e+f x)+1)}{(c-d) (a+b \sec (e+f x))}} \Pi \left (\frac {a (c+d)}{(a+b) c};\sin ^{-1}\left (\frac {\sqrt {a+b} \sqrt {c+d \sec (e+f x)}}{\sqrt {c+d} \sqrt {a+b \sec (e+f x)}}\right )|\frac {(a-b) (c+d)}{(a+b) (c-d)}\right )}{a c f \sqrt {a+b}} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[a + b*Sec[e + f*x]]*Sqrt[c + d*Sec[e + f*x]]),x]

[Out]

(-2*Sqrt[c + d]*Cot[e + f*x]*EllipticPi[(a*(c + d))/((a + b)*c), ArcSin[(Sqrt[a + b]*Sqrt[c + d*Sec[e + f*x]])
/(Sqrt[c + d]*Sqrt[a + b*Sec[e + f*x]])], ((a - b)*(c + d))/((a + b)*(c - d))]*Sqrt[-(((b*c - a*d)*(1 - Sec[e
+ f*x]))/((c + d)*(a + b*Sec[e + f*x])))]*Sqrt[((b*c - a*d)*(1 + Sec[e + f*x]))/((c - d)*(a + b*Sec[e + f*x]))
]*(a + b*Sec[e + f*x]))/(a*Sqrt[a + b]*c*f) - (2*b*Sqrt[a + b]*Cot[e + f*x]*EllipticF[ArcSin[(Sqrt[c + d]*Sqrt
[a + b*Sec[e + f*x]])/(Sqrt[a + b]*Sqrt[c + d*Sec[e + f*x]])], ((a + b)*(c - d))/((a - b)*(c + d))]*Sqrt[((b*c
- a*d)*(1 - Sec[e + f*x]))/((a + b)*(c + d*Sec[e + f*x]))]*Sqrt[-(((b*c - a*d)*(1 + Sec[e + f*x]))/((a - b)*(
c + d*Sec[e + f*x])))]*(c + d*Sec[e + f*x]))/(a*Sqrt[c + d]*(b*c - a*d)*f)

Rule 3936

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)], x_Symbol] :> Simp[(
2*(a + b*Csc[e + f*x])*Sqrt[((b*c - a*d)*(1 + Csc[e + f*x]))/((c - d)*(a + b*Csc[e + f*x]))]*Sqrt[-(((b*c - a*
d)*(1 - Csc[e + f*x]))/((c + d)*(a + b*Csc[e + f*x])))]*EllipticPi[(a*(c + d))/(c*(a + b)), ArcSin[(Rt[(a + b)
/(c + d), 2]*Sqrt[c + d*Csc[e + f*x]])/Sqrt[a + b*Csc[e + f*x]]], ((a - b)*(c + d))/((a + b)*(c - d))])/(c*f*R
t[(a + b)/(c + d), 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]

Rule 3938

Int[1/(Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)]), x_Symbol] :> Di
st[1/a, Int[Sqrt[a + b*Csc[e + f*x]]/Sqrt[c + d*Csc[e + f*x]], x], x] - Dist[b/a, Int[Csc[e + f*x]/(Sqrt[a + b
*Csc[e + f*x]]*Sqrt[c + d*Csc[e + f*x]]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]

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 {1}{\sqrt {a+b \sec (e+f x)} \sqrt {c+d \sec (e+f x)}} \, dx &=\frac {\int \frac {\sqrt {a+b \sec (e+f x)}}{\sqrt {c+d \sec (e+f x)}} \, dx}{a}-\frac {b \int \frac {\sec (e+f x)}{\sqrt {a+b \sec (e+f x)} \sqrt {c+d \sec (e+f x)}} \, dx}{a}\\ &=-\frac {2 \sqrt {c+d} \cot (e+f x) \Pi \left (\frac {a (c+d)}{(a+b) c};\sin ^{-1}\left (\frac {\sqrt {a+b} \sqrt {c+d \sec (e+f x)}}{\sqrt {c+d} \sqrt {a+b \sec (e+f x)}}\right )|\frac {(a-b) (c+d)}{(a+b) (c-d)}\right ) \sqrt {-\frac {(b c-a d) (1-\sec (e+f x))}{(c+d) (a+b \sec (e+f x))}} \sqrt {\frac {(b c-a d) (1+\sec (e+f x))}{(c-d) (a+b \sec (e+f x))}} (a+b \sec (e+f x))}{a \sqrt {a+b} c f}-\frac {2 b \sqrt {a+b} \cot (e+f x) F\left (\sin ^{-1}\left (\frac {\sqrt {c+d} \sqrt {a+b \sec (e+f x)}}{\sqrt {a+b} \sqrt {c+d \sec (e+f x)}}\right )|\frac {(a+b) (c-d)}{(a-b) (c+d)}\right ) \sqrt {\frac {(b c-a d) (1-\sec (e+f x))}{(a+b) (c+d \sec (e+f x))}} \sqrt {-\frac {(b c-a d) (1+\sec (e+f x))}{(a-b) (c+d \sec (e+f x))}} (c+d \sec (e+f x))}{a \sqrt {c+d} (b c-a d) f}\\ \end {align*}

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Mathematica [C]  time = 2.37, size = 249, normalized size = 0.63 \[ \frac {4 i \cos ^2\left (\frac {1}{2} (e+f x)\right ) \sec (e+f x) \sqrt {\frac {a \cos (e+f x)+b}{(a+b) (\cos (e+f x)+1)}} \sqrt {\frac {c \cos (e+f x)+d}{(c+d) (\cos (e+f x)+1)}} \left (F\left (i \sinh ^{-1}\left (\sqrt {\frac {b-a}{a+b}} \tan \left (\frac {1}{2} (e+f x)\right )\right )|\frac {(a+b) (c-d)}{(a-b) (c+d)}\right )-2 \Pi \left (-\frac {a+b}{a-b};i \sinh ^{-1}\left (\sqrt {\frac {b-a}{a+b}} \tan \left (\frac {1}{2} (e+f x)\right )\right )|\frac {(a+b) (c-d)}{(a-b) (c+d)}\right )\right )}{f \sqrt {\frac {b-a}{a+b}} \sqrt {a+b \sec (e+f x)} \sqrt {c+d \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[a + b*Sec[e + f*x]]*Sqrt[c + d*Sec[e + f*x]]),x]

[Out]

((4*I)*Cos[(e + f*x)/2]^2*Sqrt[(b + a*Cos[e + f*x])/((a + b)*(1 + Cos[e + f*x]))]*Sqrt[(d + c*Cos[e + f*x])/((
c + d)*(1 + Cos[e + f*x]))]*(EllipticF[I*ArcSinh[Sqrt[(-a + b)/(a + b)]*Tan[(e + f*x)/2]], ((a + b)*(c - d))/(
(a - b)*(c + d))] - 2*EllipticPi[-((a + b)/(a - b)), I*ArcSinh[Sqrt[(-a + b)/(a + b)]*Tan[(e + f*x)/2]], ((a +
b)*(c - d))/((a - b)*(c + d))])*Sec[e + f*x])/(Sqrt[(-a + b)/(a + b)]*f*Sqrt[a + b*Sec[e + f*x]]*Sqrt[c + d*S
ec[e + f*x]])

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fricas [F]  time = 5.84, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {b \sec \left (f x + e\right ) + a} \sqrt {d \sec \left (f x + e\right ) + c}}{b d \sec \left (f x + e\right )^{2} + a c + {\left (b c + a d\right )} \sec \left (f x + e\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(b*sec(f*x + e) + a)*sqrt(d*sec(f*x + e) + c)/(b*d*sec(f*x + e)^2 + a*c + (b*c + a*d)*sec(f*x + e
)), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {b \sec \left (f x + e\right ) + a} \sqrt {d \sec \left (f x + e\right ) + c}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/(sqrt(b*sec(f*x + e) + a)*sqrt(d*sec(f*x + e) + c)), x)

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maple [A]  time = 2.15, size = 292, normalized size = 0.73 \[ -\frac {2 \left (\EllipticF \left (\frac {\left (-1+\cos \left (f x +e \right )\right ) \sqrt {\frac {a -b}{a +b}}}{\sin \left (f x +e \right )}, \sqrt {\frac {\left (a +b \right ) \left (c -d \right )}{\left (a -b \right ) \left (c +d \right )}}\right )-2 \EllipticPi \left (\frac {\left (-1+\cos \left (f x +e \right )\right ) \sqrt {\frac {a -b}{a +b}}}{\sin \left (f x +e \right )}, -\frac {a +b}{a -b}, \frac {\sqrt {\frac {c -d}{c +d}}}{\sqrt {\frac {a -b}{a +b}}}\right )\right ) \cos \left (f x +e \right ) \left (\sin ^{2}\left (f x +e \right )\right ) \sqrt {\frac {d +c \cos \left (f x +e \right )}{\cos \left (f x +e \right )}}\, \sqrt {\frac {b +a \cos \left (f x +e \right )}{\cos \left (f x +e \right )}}\, \sqrt {\frac {b +a \cos \left (f x +e \right )}{\left (1+\cos \left (f x +e \right )\right ) \left (a +b \right )}}\, \sqrt {\frac {d +c \cos \left (f x +e \right )}{\left (1+\cos \left (f x +e \right )\right ) \left (c +d \right )}}}{f \left (-1+\cos \left (f x +e \right )\right ) \left (d +c \cos \left (f x +e \right )\right ) \left (b +a \cos \left (f x +e \right )\right ) \sqrt {\frac {a -b}{a +b}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(c+d*sec(f*x+e))^(1/2)/(a+b*sec(f*x+e))^(1/2),x)

[Out]

-2/f*(EllipticF((-1+cos(f*x+e))*((a-b)/(a+b))^(1/2)/sin(f*x+e),((a+b)*(c-d)/(a-b)/(c+d))^(1/2))-2*EllipticPi((
-1+cos(f*x+e))*((a-b)/(a+b))^(1/2)/sin(f*x+e),-(a+b)/(a-b),((c-d)/(c+d))^(1/2)/((a-b)/(a+b))^(1/2)))*cos(f*x+e
)*sin(f*x+e)^2*((d+c*cos(f*x+e))/cos(f*x+e))^(1/2)*((b+a*cos(f*x+e))/cos(f*x+e))^(1/2)*((b+a*cos(f*x+e))/(1+co
s(f*x+e))/(a+b))^(1/2)*((d+c*cos(f*x+e))/(1+cos(f*x+e))/(c+d))^(1/2)/(-1+cos(f*x+e))/(d+c*cos(f*x+e))/(b+a*cos
(f*x+e))/((a-b)/(a+b))^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {b \sec \left (f x + e\right ) + a} \sqrt {d \sec \left (f x + e\right ) + c}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/(sqrt(b*sec(f*x + e) + a)*sqrt(d*sec(f*x + e) + c)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/(sqrt(a + b*sec(e + f*x))*sqrt(c + d*sec(e + f*x))), x)

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