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

Optimal. Leaf size=225 \[ -\frac {2 \cot (e+f x) \sqrt {-\frac {d (1-\sec (e+f x))}{c+d \sec (e+f x)}} \sqrt {\frac {d (\sec (e+f x)+1)}{c+d \sec (e+f x)}} (c+d \sec (e+f x)) \Pi \left (\frac {c}{c+d};\sin ^{-1}\left (\frac {\sqrt {c+d}}{\sqrt {c+d \sec (e+f x)}}\right )|\frac {c-d}{c+d}\right )}{a f \sqrt {c+d}}-\frac {\sqrt {\frac {1}{\sec (e+f x)+1}} \sqrt {c+d \sec (e+f x)} E\left (\sin ^{-1}\left (\frac {\tan (e+f x)}{\sec (e+f x)+1}\right )|\frac {c-d}{c+d}\right )}{a f \sqrt {\frac {c+d \sec (e+f x)}{(c+d) (\sec (e+f x)+1)}}} \]

[Out]

-2*cot(f*x+e)*EllipticPi((c+d)^(1/2)/(c+d*sec(f*x+e))^(1/2),c/(c+d),((c-d)/(c+d))^(1/2))*(c+d*sec(f*x+e))*(-d*
(1-sec(f*x+e))/(c+d*sec(f*x+e)))^(1/2)*(d*(1+sec(f*x+e))/(c+d*sec(f*x+e)))^(1/2)/a/f/(c+d)^(1/2)-EllipticE(tan
(f*x+e)/(1+sec(f*x+e)),((c-d)/(c+d))^(1/2))*(1/(1+sec(f*x+e)))^(1/2)*(c+d*sec(f*x+e))^(1/2)/a/f/((c+d*sec(f*x+
e))/(c+d)/(1+sec(f*x+e)))^(1/2)

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Rubi [A]  time = 0.21, antiderivative size = 225, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {3925, 3780, 3968} \[ -\frac {2 \cot (e+f x) \sqrt {-\frac {d (1-\sec (e+f x))}{c+d \sec (e+f x)}} \sqrt {\frac {d (\sec (e+f x)+1)}{c+d \sec (e+f x)}} (c+d \sec (e+f x)) \Pi \left (\frac {c}{c+d};\sin ^{-1}\left (\frac {\sqrt {c+d}}{\sqrt {c+d \sec (e+f x)}}\right )|\frac {c-d}{c+d}\right )}{a f \sqrt {c+d}}-\frac {\sqrt {\frac {1}{\sec (e+f x)+1}} \sqrt {c+d \sec (e+f x)} E\left (\sin ^{-1}\left (\frac {\tan (e+f x)}{\sec (e+f x)+1}\right )|\frac {c-d}{c+d}\right )}{a f \sqrt {\frac {c+d \sec (e+f x)}{(c+d) (\sec (e+f x)+1)}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3780

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

Rule 3925

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

Rule 3968

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

Rubi steps

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

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Mathematica [A]  time = 8.75, size = 178, normalized size = 0.79 \[ -\frac {4 \cos ^4\left (\frac {1}{2} (e+f x)\right ) \sqrt {\frac {1}{\sec (e+f x)+1}} \sqrt {c+d \sec (e+f x)} \left (2 (c-d) F\left (\sin ^{-1}\left (\tan \left (\frac {1}{2} (e+f x)\right )\right )|\frac {c-d}{c+d}\right )+(c+d) E\left (\sin ^{-1}\left (\tan \left (\frac {1}{2} (e+f x)\right )\right )|\frac {c-d}{c+d}\right )-4 c \Pi \left (-1;\sin ^{-1}\left (\tan \left (\frac {1}{2} (e+f x)\right )\right )|\frac {c-d}{c+d}\right )\right )}{a f (c+d) (\cos (e+f x)+1)^2 \sqrt {\frac {c \cos (e+f x)+d}{(c+d) (\cos (e+f x)+1)}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*Cos[(e + f*x)/2]^4*((c + d)*EllipticE[ArcSin[Tan[(e + f*x)/2]], (c - d)/(c + d)] + 2*(c - d)*EllipticF[Arc
Sin[Tan[(e + f*x)/2]], (c - d)/(c + d)] - 4*c*EllipticPi[-1, ArcSin[Tan[(e + f*x)/2]], (c - d)/(c + d)])*Sqrt[
(1 + Sec[e + f*x])^(-1)]*Sqrt[c + d*Sec[e + f*x]])/(a*(c + d)*f*(1 + Cos[e + f*x])^2*Sqrt[(d + c*Cos[e + f*x])
/((c + d)*(1 + Cos[e + f*x]))])

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fricas [F]  time = 22.53, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {d \sec \left (f x + e\right ) + c}}{a \sec \left (f x + e\right ) + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 1.98, size = 285, normalized size = 1.27 \[ -\frac {\sqrt {\frac {d +c \cos \left (f x +e \right )}{\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {d +c \cos \left (f x +e \right )}{\left (1+\cos \left (f x +e \right )\right ) \left (c +d \right )}}\, \left (1+\cos \left (f x +e \right )\right )^{2} \left (2 \EllipticF \left (\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}, \sqrt {\frac {c -d}{c +d}}\right ) c -2 \EllipticF \left (\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}, \sqrt {\frac {c -d}{c +d}}\right ) d +c \EllipticE \left (\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}, \sqrt {\frac {c -d}{c +d}}\right )+d \EllipticE \left (\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}, \sqrt {\frac {c -d}{c +d}}\right )-4 c \EllipticPi \left (\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}, -1, \sqrt {\frac {c -d}{c +d}}\right )\right ) \left (-1+\cos \left (f x +e \right )\right )}{a f \left (d +c \cos \left (f x +e \right )\right ) \sin \left (f x +e \right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/a/f*((d+c*cos(f*x+e))/cos(f*x+e))^(1/2)*(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*((d+c*cos(f*x+e))/(1+cos(f*x+e))/
(c+d))^(1/2)*(1+cos(f*x+e))^2*(2*EllipticF((-1+cos(f*x+e))/sin(f*x+e),((c-d)/(c+d))^(1/2))*c-2*EllipticF((-1+c
os(f*x+e))/sin(f*x+e),((c-d)/(c+d))^(1/2))*d+c*EllipticE((-1+cos(f*x+e))/sin(f*x+e),((c-d)/(c+d))^(1/2))+d*Ell
ipticE((-1+cos(f*x+e))/sin(f*x+e),((c-d)/(c+d))^(1/2))-4*c*EllipticPi((-1+cos(f*x+e))/sin(f*x+e),-1,((c-d)/(c+
d))^(1/2)))*(-1+cos(f*x+e))/(d+c*cos(f*x+e))/sin(f*x+e)^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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