3.3.75 \(\int \frac {\sec (e+f x)}{\sqrt {a+b \sec (e+f x)} (c+c \sec (e+f x))} \, dx\) [275]

Optimal. Leaf size=209 \[ -\frac {2 \sqrt {a+b} \cot (e+f x) F\left (\text {ArcSin}\left (\frac {\sqrt {a+b \sec (e+f x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (e+f x))}{a+b}} \sqrt {-\frac {b (1+\sec (e+f x))}{a-b}}}{(a-b) c f}+\frac {E\left (\text {ArcSin}\left (\frac {\tan (e+f x)}{1+\sec (e+f x)}\right )|\frac {a-b}{a+b}\right ) \sqrt {\frac {1}{1+\sec (e+f x)}} \sqrt {a+b \sec (e+f x)}}{(a-b) c f \sqrt {\frac {a+b \sec (e+f x)}{(a+b) (1+\sec (e+f x))}}} \]

[Out]

-2*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)/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 = 0.20, antiderivative size = 209, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {4057, 3917, 4053} \begin {gather*} \frac {\sqrt {\frac {1}{\sec (e+f x)+1}} \sqrt {a+b \sec (e+f x)} E\left (\text {ArcSin}\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)}}}-\frac {2 \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 (\text {ArcSin}\left (\frac {\sqrt {a+b \sec (e+f x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{c f (a-b)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*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)*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 3917

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

Rule 4053

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])]/(d*f*Sqrt[c*d*((a + b*Csc[e + f
*x])/((b*c + a*d)*(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)], 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 4057

Int[csc[(e_.) + (f_.)*(x_)]/(Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))
), x_Symbol] :> Dist[b/(b*c - a*d), Int[Csc[e + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] - Dist[d/(b*c - a*d), In
t[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])

Rubi steps

\begin {align*} \int \frac {\sec (e+f x)}{\sqrt {a+b \sec (e+f x)} (c+c \sec (e+f x))} \, dx &=-\frac {b \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 \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) 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 = 13.98, size = 375, normalized size = 1.79 \begin {gather*} \frac {\cos ^2\left (\frac {e}{2}+\frac {f x}{2}\right ) (b+a \cos (e+f x)) \sec ^2(e+f x) \left (\frac {2 \sin (e+f x)}{-a+b}-\frac {2 \tan \left (\frac {1}{2} (e+f x)\right )}{-a+b}\right )}{f \sqrt {a+b \sec (e+f x)} (c+c \sec (e+f x))}-\frac {2 \cos ^2\left (\frac {e}{2}+\frac {f x}{2}\right ) \sec ^{\frac {3}{2}}(e+f x) \sqrt {\cos ^2\left (\frac {1}{2} (e+f x)\right ) \sec (e+f x)} \left ((a-b) E\left (\text {ArcSin}\left (\sqrt {\frac {a-b}{a+b}} \tan \left (\frac {1}{2} (e+f x)\right )\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {(b+a \cos (e+f x)) \sec ^2\left (\frac {1}{2} (e+f x)\right )}{a+b}}+\sqrt {2} \sqrt {\frac {a-b}{a+b}} \sqrt {\frac {\cos (e+f x)}{1+\cos (e+f x)}} (b+a \cos (e+f x)) \tan \left (\frac {1}{2} (e+f x)\right )\right ) \left (-1+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )}{\left (\frac {a-b}{a+b}\right )^{3/2} (a+b) f \sqrt {\cos (e+f x) \sec ^4\left (\frac {1}{2} (e+f x)\right )} \sqrt {a+b \sec (e+f x)} (c+c \sec (e+f x))} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Cos[e/2 + (f*x)/2]^2*(b + a*Cos[e + f*x])*Sec[e + f*x]^2*((2*Sin[e + f*x])/(-a + b) - (2*Tan[(e + f*x)/2])/(-
a + b)))/(f*Sqrt[a + b*Sec[e + f*x]]*(c + c*Sec[e + f*x])) - (2*Cos[e/2 + (f*x)/2]^2*Sec[e + f*x]^(3/2)*Sqrt[C
os[(e + f*x)/2]^2*Sec[e + f*x]]*((a - b)*EllipticE[ArcSin[Sqrt[(a - b)/(a + b)]*Tan[(e + f*x)/2]], (a + b)/(a
- b)]*Sqrt[((b + a*Cos[e + f*x])*Sec[(e + f*x)/2]^2)/(a + b)] + Sqrt[2]*Sqrt[(a - b)/(a + b)]*Sqrt[Cos[e + f*x
]/(1 + Cos[e + f*x])]*(b + a*Cos[e + f*x])*Tan[(e + f*x)/2])*(-1 + Tan[(e + f*x)/2]^2))/(((a - b)/(a + b))^(3/
2)*(a + b)*f*Sqrt[Cos[e + f*x]*Sec[(e + f*x)/2]^4]*Sqrt[a + b*Sec[e + f*x]]*(c + c*Sec[e + f*x]))

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Maple [A]
time = 4.81, size = 225, normalized size = 1.08

method	result	size
default	\(-\frac {\sqrt {\frac {a \cos \left (f x +e \right )+b}{\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {a \cos \left (f x +e \right )+b}{\left (\cos \left (f x +e \right )+1\right ) \left (a +b \right )}}\, \left (\cos \left (f x +e \right )+1\right )^{2} \left (-1+\cos \left (f x +e \right )\right ) \left (2 \EllipticF \left (\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}, \sqrt {\frac {a -b}{a +b}}\right ) b -a \EllipticE \left (\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}, \sqrt {\frac {a -b}{a +b}}\right )-b \EllipticE \left (\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}, \sqrt {\frac {a -b}{a +b}}\right )\right )}{c f \left (a \cos \left (f x +e \right )+b \right ) \sin \left (f x +e \right )^{2} \left (a -b \right )}\)	\(225\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {\sec {\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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sec(f*x + e)/(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 \begin {gather*} \int \frac {1}{\cos \left (e+f\,x\right )\,\sqrt {a+\frac {b}{\cos \left (e+f\,x\right )}}\,\left (c+\frac {c}{\cos \left (e+f\,x\right )}\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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