3.276 \(\int \frac {\sec ^2(e+f x)}{\sqrt {a+b \sec (e+f x)} (c+c \sec (e+f x))} \, dx\)

Optimal. Leaf size=214 \[ \frac {2 a \sqrt {a+b} \cot (e+f x) \sqrt {\frac {b (1-\sec (e+f x))}{a+b}} \sqrt {-\frac {b (\sec (e+f x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (e+f x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b c f (a-b)}-\frac {\sqrt {\frac {1}{\sec (e+f x)+1}} \sqrt {a+b \sec (e+f x)} E\left (\sin ^{-1}\left (\frac {\tan (e+f x)}{\sec (e+f x)+1}\right )|\frac {a-b}{a+b}\right )}{c f (a-b) \sqrt {\frac {a+b \sec (e+f x)}{(a+b) (\sec (e+f x)+1)}}} \]

[Out]

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

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Rubi [A]  time = 0.36, antiderivative size = 214, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.086, Rules used = {3976, 3832, 3968} \[ \frac {2 a \sqrt {a+b} \cot (e+f x) \sqrt {\frac {b (1-\sec (e+f x))}{a+b}} \sqrt {-\frac {b (\sec (e+f x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (e+f x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b c f (a-b)}-\frac {\sqrt {\frac {1}{\sec (e+f x)+1}} \sqrt {a+b \sec (e+f x)} E\left (\sin ^{-1}\left (\frac {\tan (e+f x)}{\sec (e+f x)+1}\right )|\frac {a-b}{a+b}\right )}{c f (a-b) \sqrt {\frac {a+b \sec (e+f x)}{(a+b) (\sec (e+f x)+1)}}} \]

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]^2/(Sqrt[a + b*Sec[e + f*x]]*(c + c*Sec[e + f*x])),x]

[Out]

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

Rule 3832

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

Rule 3976

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

Rubi steps

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

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Mathematica [A]  time = 5.31, size = 156, normalized size = 0.73 \[ \frac {4 \cos ^4\left (\frac {1}{2} (e+f x)\right ) \sqrt {\frac {a \cos (e+f x)+b}{(a+b) (\cos (e+f x)+1)}} \left ((a+b) E\left (\sin ^{-1}\left (\tan \left (\frac {1}{2} (e+f x)\right )\right )|\frac {a-b}{a+b}\right )-2 a F\left (\sin ^{-1}\left (\tan \left (\frac {1}{2} (e+f x)\right )\right )|\frac {a-b}{a+b}\right )\right )}{c f (b-a) \sqrt {\frac {\cos (e+f x)}{\cos (e+f x)+1}} (\cos (e+f x)+1)^2 \sqrt {a+b \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[e + f*x]^2/(Sqrt[a + b*Sec[e + f*x]]*(c + c*Sec[e + f*x])),x]

[Out]

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

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fricas [F]  time = 0.88, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {b \sec \left (f x + e\right ) + a} \sec \left (f x + e\right )^{2}}{b c \sec \left (f x + e\right )^{2} + {\left (a + b\right )} c \sec \left (f x + e\right ) + a c}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 1.90, size = 224, normalized size = 1.05 \[ \frac {\sqrt {\frac {b +a \cos \left (f x +e \right )}{\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {b +a \cos \left (f x +e \right )}{\left (1+\cos \left (f x +e \right )\right ) \left (a +b \right )}}\, \left (1+\cos \left (f x +e \right )\right )^{2} \left (2 \EllipticF \left (\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}, \sqrt {\frac {a -b}{a +b}}\right ) a -a \EllipticE \left (\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}, \sqrt {\frac {a -b}{a +b}}\right )-b \EllipticE \left (\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}, \sqrt {\frac {a -b}{a +b}}\right )\right ) \left (-1+\cos \left (f x +e \right )\right )}{c f \left (b +a \cos \left (f x +e \right )\right ) \sin \left (f x +e \right )^{2} \left (a -b \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {1}{{\cos \left (e+f\,x\right )}^2\,\sqrt {a+\frac {b}{\cos \left (e+f\,x\right )}}\,\left (c+\frac {c}{\cos \left (e+f\,x\right )}\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\sec ^{2}{\left (e + f x \right )}}{\sqrt {a + b \sec {\left (e + f x \right )}} \sec {\left (e + f x \right )} + \sqrt {a + b \sec {\left (e + f x \right )}}}\, dx}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sec(e + f*x)**2/(sqrt(a + b*sec(e + f*x))*sec(e + f*x) + sqrt(a + b*sec(e + f*x))), x)/c

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