3.615 \(\int \frac {1}{\sqrt {c \cos (e+f x)} \sqrt {a+b \sin (e+f x)}} \, dx\)

Optimal. Leaf size=183 \[ \frac {2 \sqrt {2} \sqrt [4]{b-a} \sqrt {c \cos (e+f x)} \sqrt {\frac {a+b \sin (e+f x)}{(a-b) (1-\sin (e+f x))}} F\left (\left .\sin ^{-1}\left (\frac {\sqrt [4]{a+b} \sqrt {\frac {\cos (e+f x)+\sin (e+f x)+1}{\cos (e+f x)-\sin (e+f x)+1}}}{\sqrt [4]{b-a}}\right )\right |-1\right )}{c f \sqrt [4]{a+b} \sqrt {\frac {\sin (e+f x)+\cos (e+f x)+1}{-\sin (e+f x)+\cos (e+f x)+1}} \sqrt {a+b \sin (e+f x)}} \]

[Out]

2*(-a+b)^(1/4)*EllipticF((a+b)^(1/4)*((1+cos(f*x+e)+sin(f*x+e))/(1+cos(f*x+e)-sin(f*x+e)))^(1/2)/(-a+b)^(1/4),
I)*2^(1/2)*(c*cos(f*x+e))^(1/2)*((a+b*sin(f*x+e))/(a-b)/(1-sin(f*x+e)))^(1/2)/(a+b)^(1/4)/c/f/((1+cos(f*x+e)+s
in(f*x+e))/(1+cos(f*x+e)-sin(f*x+e)))^(1/2)/(a+b*sin(f*x+e))^(1/2)

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Rubi [B]  time = 0.43, antiderivative size = 374, normalized size of antiderivative = 2.04, number of steps used = 2, number of rules used = 2, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {2697, 220} \[ \frac {\sqrt {2} \sqrt [4]{a-b} \sqrt {c \cos (e+f x)} \sqrt {\frac {a+b \sin (e+f x)}{(a-b) (1-\sin (e+f x))}} \sqrt {\frac {a+b \sin (e+f x)}{(a-b) (\sin (e) (-\cos (f x))-\cos (e) \sin (f x)+1) \left (\frac {\sqrt {a+b} (\sin (e+f x)+\cos (e+f x)+1)}{\sqrt {a-b} (-\sin (e+f x)+\cos (e+f x)+1)}+1\right )^2}} \left (\frac {\sqrt {a+b} (\sin (e+f x)+\cos (e+f x)+1)}{\sqrt {a-b} (-\sin (e+f x)+\cos (e+f x)+1)}+1\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{a+b} \sqrt {\frac {\cos (e+f x)+\sin (e+f x)+1}{\cos (e+f x)-\sin (e+f x)+1}}}{\sqrt [4]{a-b}}\right )|\frac {1}{2}\right )}{c f \sqrt [4]{a+b} \sqrt {\frac {\sin (e+f x)+\cos (e+f x)+1}{-\sin (e+f x)+\cos (e+f x)+1}} \sqrt {a+b \sin (e+f x)} \sqrt {\frac {a+b \sin (e+f x)}{(a-b) (\sin (e) (-\cos (f x))-\cos (e) \sin (f x)+1)}}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Sqrt[2]*(a - b)^(1/4)*Sqrt[c*Cos[e + f*x]]*EllipticF[2*ArcTan[((a + b)^(1/4)*Sqrt[(1 + Cos[e + f*x] + Sin[e +
f*x])/(1 + Cos[e + f*x] - Sin[e + f*x])])/(a - b)^(1/4)], 1/2]*Sqrt[(a + b*Sin[e + f*x])/((a - b)*(1 - Sin[e
+ f*x]))]*Sqrt[(a + b*Sin[e + f*x])/((a - b)*(1 - Cos[f*x]*Sin[e] - Cos[e]*Sin[f*x])*(1 + (Sqrt[a + b]*(1 + Co
s[e + f*x] + Sin[e + f*x]))/(Sqrt[a - b]*(1 + Cos[e + f*x] - Sin[e + f*x])))^2)]*(1 + (Sqrt[a + b]*(1 + Cos[e
+ f*x] + Sin[e + f*x]))/(Sqrt[a - b]*(1 + Cos[e + f*x] - Sin[e + f*x]))))/((a + b)^(1/4)*c*f*Sqrt[(1 + Cos[e +
f*x] + Sin[e + f*x])/(1 + Cos[e + f*x] - Sin[e + f*x])]*Sqrt[a + b*Sin[e + f*x]]*Sqrt[(a + b*Sin[e + f*x])/((
a - b)*(1 - Cos[f*x]*Sin[e] - Cos[e]*Sin[f*x]))])

Rule 220

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[((1 + q^2*x^2)*Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 2697

Int[1/(Sqrt[cos[(e_.) + (f_.)*(x_)]*(g_.)]*Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist[(2*S
qrt[2]*Sqrt[g*Cos[e + f*x]]*Sqrt[(a + b*Sin[e + f*x])/((a - b)*(1 - Sin[e + f*x]))])/(f*g*Sqrt[a + b*Sin[e + f
*x]]*Sqrt[(1 + Cos[e + f*x] + Sin[e + f*x])/(1 + Cos[e + f*x] - Sin[e + f*x])]), Subst[Int[1/Sqrt[1 + ((a + b)
*x^4)/(a - b)], x], x, Sqrt[(1 + Cos[e + f*x] + Sin[e + f*x])/(1 + Cos[e + f*x] - Sin[e + f*x])]], x] /; FreeQ
[{a, b, e, f, g}, x] && NeQ[a^2 - b^2, 0]

Rubi steps

\begin {align*} \int \frac {1}{\sqrt {c \cos (e+f x)} \sqrt {a+b \sin (e+f x)}} \, dx &=\frac {\left (2 \sqrt {2} \sqrt {c \cos (e+f x)} \sqrt {\frac {a+b \sin (e+f x)}{(a-b) (1-\sin (e+f x))}}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1+\frac {(a+b) x^4}{a-b}}} \, dx,x,\sqrt {\frac {1+\cos (e+f x)+\sin (e+f x)}{1+\cos (e+f x)-\sin (e+f x)}}\right )}{c f \sqrt {\frac {1+\cos (e+f x)+\sin (e+f x)}{1+\cos (e+f x)-\sin (e+f x)}} \sqrt {a+b \sin (e+f x)}}\\ &=\frac {\sqrt {2} \sqrt [4]{a-b} \sqrt {c \cos (e+f x)} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{a+b} \sqrt {\frac {1+\cos (e+f x)+\sin (e+f x)}{1+\cos (e+f x)-\sin (e+f x)}}}{\sqrt [4]{a-b}}\right )|\frac {1}{2}\right ) \sqrt {\frac {a+b \sin (e+f x)}{(a-b) (1-\sin (e+f x))}} \sqrt {\frac {a+b \sin (e+f x)}{(a-b) (1-\cos (f x) \sin (e)-\cos (e) \sin (f x)) \left (1+\frac {\sqrt {a+b} (1+\cos (e+f x)+\sin (e+f x))}{\sqrt {a-b} (1+\cos (e+f x)-\sin (e+f x))}\right )^2}} \left (1+\frac {\sqrt {a+b} (1+\cos (e+f x)+\sin (e+f x))}{\sqrt {a-b} (1+\cos (e+f x)-\sin (e+f x))}\right )}{\sqrt [4]{a+b} c f \sqrt {\frac {1+\cos (e+f x)+\sin (e+f x)}{1+\cos (e+f x)-\sin (e+f x)}} \sqrt {a+b \sin (e+f x)} \sqrt {\frac {a+b \sin (e+f x)}{(a-b) (1-\cos (f x) \sin (e)-\cos (e) \sin (f x))}}}\\ \end {align*}

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Mathematica [C]  time = 0.32, size = 117, normalized size = 0.64 \[ -\frac {2 c (\sin (e+f x)-1) \left (\frac {(a+b) (\sin (e+f x)+1)}{(a-b) (\sin (e+f x)-1)}\right )^{3/4} \sqrt {a+b \sin (e+f x)} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {3}{2};-\frac {2 (a+b \sin (e+f x))}{(a-b) (\sin (e+f x)-1)}\right )}{f (a+b) (c \cos (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*c*Hypergeometric2F1[1/2, 3/4, 3/2, (-2*(a + b*Sin[e + f*x]))/((a - b)*(-1 + Sin[e + f*x]))]*(-1 + Sin[e +
f*x])*(((a + b)*(1 + Sin[e + f*x]))/((a - b)*(-1 + Sin[e + f*x])))^(3/4)*Sqrt[a + b*Sin[e + f*x]])/((a + b)*f*
(c*Cos[e + f*x])^(3/2))

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fricas [F]  time = 1.07, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {c \cos \left (f x + e\right )} \sqrt {b \sin \left (f x + e\right ) + a}}{b c \cos \left (f x + e\right ) \sin \left (f x + e\right ) + a c \cos \left (f x + e\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(c*cos(f*x + e))*sqrt(b*sin(f*x + e) + a)/(b*c*cos(f*x + e)*sin(f*x + e) + a*c*cos(f*x + e)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/(sqrt(c*cos(f*x + e))*sqrt(b*sin(f*x + e) + a)), x)

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maple [B]  time = 0.61, size = 442, normalized size = 2.42 \[ \frac {4 \EllipticF \left (\sqrt {\frac {\left (\sin \left (f x +e \right )-1\right ) \left (b +\sqrt {-a^{2}+b^{2}}-a \right )}{\cos \left (f x +e \right ) \left (b +\sqrt {-a^{2}+b^{2}}+a \right )}}, \sqrt {\frac {\left (a -b +\sqrt {-a^{2}+b^{2}}\right ) \left (b +\sqrt {-a^{2}+b^{2}}+a \right )}{\left (-b +\sqrt {-a^{2}+b^{2}}-a \right ) \left (b +\sqrt {-a^{2}+b^{2}}-a \right )}}\right ) \sqrt {\frac {\cos \left (f x +e \right ) \sqrt {-a^{2}+b^{2}}+a \sin \left (f x +e \right )+b \cos \left (f x +e \right )+\sqrt {-a^{2}+b^{2}}+b}{\left (1+\cos \left (f x +e \right )+\sin \left (f x +e \right )\right ) \left (b +\sqrt {-a^{2}+b^{2}}+a \right )}}\, \sqrt {\frac {\left (\sin \left (f x +e \right )-1\right ) \left (b +\sqrt {-a^{2}+b^{2}}-a \right )}{\cos \left (f x +e \right ) \left (b +\sqrt {-a^{2}+b^{2}}+a \right )}}\, \sqrt {-\frac {a \sin \left (f x +e \right )-\cos \left (f x +e \right ) \sqrt {-a^{2}+b^{2}}+b \cos \left (f x +e \right )-\sqrt {-a^{2}+b^{2}}+b}{\left (1+\cos \left (f x +e \right )+\sin \left (f x +e \right )\right ) \left (-b +\sqrt {-a^{2}+b^{2}}-a \right )}}\, \left (\cos \left (f x +e \right )+1\right )^{2} \left (-1+\cos \left (f x +e \right )\right )^{2} \left (b +\sqrt {-a^{2}+b^{2}}+a \right ) \left (1+\sin \left (f x +e \right )\right )}{f \sqrt {a +b \sin \left (f x +e \right )}\, \sin \left (f x +e \right )^{4} \sqrt {c \cos \left (f x +e \right )}\, \left (b +\sqrt {-a^{2}+b^{2}}-a \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/f*EllipticF(((sin(f*x+e)-1)/cos(f*x+e)*(b+(-a^2+b^2)^(1/2)-a)/(b+(-a^2+b^2)^(1/2)+a))^(1/2),((a-b+(-a^2+b^2)
^(1/2))*(b+(-a^2+b^2)^(1/2)+a)/(-b+(-a^2+b^2)^(1/2)-a)/(b+(-a^2+b^2)^(1/2)-a))^(1/2))*((cos(f*x+e)*(-a^2+b^2)^
(1/2)+a*sin(f*x+e)+b*cos(f*x+e)+(-a^2+b^2)^(1/2)+b)/(1+cos(f*x+e)+sin(f*x+e))/(b+(-a^2+b^2)^(1/2)+a))^(1/2)*((
sin(f*x+e)-1)/cos(f*x+e)*(b+(-a^2+b^2)^(1/2)-a)/(b+(-a^2+b^2)^(1/2)+a))^(1/2)*(-(a*sin(f*x+e)-cos(f*x+e)*(-a^2
+b^2)^(1/2)+b*cos(f*x+e)-(-a^2+b^2)^(1/2)+b)/(1+cos(f*x+e)+sin(f*x+e))/(-b+(-a^2+b^2)^(1/2)-a))^(1/2)*(cos(f*x
+e)+1)^2*(-1+cos(f*x+e))^2*(b+(-a^2+b^2)^(1/2)+a)*(1+sin(f*x+e))/(a+b*sin(f*x+e))^(1/2)/sin(f*x+e)^4/(c*cos(f*
x+e))^(1/2)/(b+(-a^2+b^2)^(1/2)-a)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/(sqrt(c*cos(f*x + e))*sqrt(b*sin(f*x + e) + a)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{\sqrt {c\,\cos \left (e+f\,x\right )}\,\sqrt {a+b\,\sin \left (e+f\,x\right )}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/(sqrt(c*cos(e + f*x))*sqrt(a + b*sin(e + f*x))), x)

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