∫ ∘ ________ g sin(e+fx) _______∘a+asin (e+fx)(c−csin (e+fx)) dx

3.17 \(\int \frac {\sqrt {g \sin (e+f x)}}{\sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))} \, dx\)

Optimal. Leaf size=114 \[ \frac {\sec (e+f x) \sqrt {a \sin (e+f x)+a} \sqrt {g \sin (e+f x)}}{a c f}+\frac {\sqrt {g} \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {g} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a} \sqrt {g \sin (e+f x)}}\right )}{\sqrt {2} \sqrt {a} c f} \]

[Out]

1/2*arctan(1/2*cos(f*x+e)*a^(1/2)*g^(1/2)*2^(1/2)/(g*sin(f*x+e))^(1/2)/(a+a*sin(f*x+e))^(1/2))*g^(1/2)/c/f*2^(

1/2)/a^(1/2)+sec(f*x+e)*(g*sin(f*x+e))^(1/2)*(a+a*sin(f*x+e))^(1/2)/a/c/f

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Rubi [A] time = 0.49, antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {2936, 2782, 205, 2930, 12, 30} \[ \frac {\sec (e+f x) \sqrt {a \sin (e+f x)+a} \sqrt {g \sin (e+f x)}}{a c f}+\frac {\sqrt {g} \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {g} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a} \sqrt {g \sin (e+f x)}}\right )}{\sqrt {2} \sqrt {a} c f} \]

Antiderivative was successfully veriﬁed.

[In]

Int[Sqrt[g*Sin[e + f*x]]/(Sqrt[a + a*Sin[e + f*x]]*(c - c*Sin[e + f*x])),x]

[Out]

(Sqrt[g]*ArcTan[(Sqrt[a]*Sqrt[g]*Cos[e + f*x])/(Sqrt[2]*Sqrt[g*Sin[e + f*x]]*Sqrt[a + a*Sin[e + f*x]])])/(Sqrt

[2]*Sqrt[a]*c*f) + (Sec[e + f*x]*Sqrt[g*Sin[e + f*x]]*Sqrt[a + a*Sin[e + f*x]])/(a*c*f)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] && !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]

&& PosQ[a/b]

Rule 2782

Int[1/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> D

ist[(-2*a)/f, Subst[Int[1/(2*b^2 - (a*c - b*d)*x^2), x], x, (b*Cos[e + f*x])/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c

+ d*Sin[e + f*x]])], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 -

d^2, 0]

Rule 2930

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/(Sqrt[(g_.)*sin[(e_.) + (f_.)*(x_)]]*((c_) + (d_.)*sin[(e_.) +

(f_.)*(x_)])), x_Symbol] :> Dist[(-2*b)/f, Subst[Int[1/(b*c + a*d + c*g*x^2), x], x, (b*Cos[e + f*x])/(Sqrt[g*

Sin[e + f*x]]*Sqrt[a + b*Sin[e + f*x]])], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b*c - a*d, 0] && EqQ[a^

2 - b^2, 0]

Rule 2936

Int[Sqrt[(g_.)*sin[(e_.) + (f_.)*(x_)]]/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_) + (d_.)*sin[(e_.) +

(f_.)*(x_)])), x_Symbol] :> -Dist[(a*g)/(b*c - a*d), Int[1/(Sqrt[g*Sin[e + f*x]]*Sqrt[a + b*Sin[e + f*x]]), x]

, x] + Dist[(c*g)/(b*c - a*d), Int[Sqrt[a + b*Sin[e + f*x]]/(Sqrt[g*Sin[e + f*x]]*(c + d*Sin[e + f*x])), x], x

] /; FreeQ[{a, b, c, d, e, f, g}, 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 {\sqrt {g \sin (e+f x)}}{\sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))} \, dx &=\frac {g \int \frac {\sqrt {a+a \sin (e+f x)}}{\sqrt {g \sin (e+f x)} (c-c \sin (e+f x))} \, dx}{2 a}-\frac {g \int \frac {1}{\sqrt {g \sin (e+f x)} \sqrt {a+a \sin (e+f x)}} \, dx}{2 c}\\ &=-\frac {g \operatorname {Subst}\left (\int \frac {1}{c g x^2} \, dx,x,\frac {a \cos (e+f x)}{\sqrt {g \sin (e+f x)} \sqrt {a+a \sin (e+f x)}}\right )}{f}+\frac {(a g) \operatorname {Subst}\left (\int \frac {1}{2 a^2+a g x^2} \, dx,x,\frac {a \cos (e+f x)}{\sqrt {g \sin (e+f x)} \sqrt {a+a \sin (e+f x)}}\right )}{c f}\\ &=\frac {\sqrt {g} \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {g} \cos (e+f x)}{\sqrt {2} \sqrt {g \sin (e+f x)} \sqrt {a+a \sin (e+f x)}}\right )}{\sqrt {2} \sqrt {a} c f}-\frac {\operatorname {Subst}\left (\int \frac {1}{x^2} \, dx,x,\frac {a \cos (e+f x)}{\sqrt {g \sin (e+f x)} \sqrt {a+a \sin (e+f x)}}\right )}{c f}\\ &=\frac {\sqrt {g} \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {g} \cos (e+f x)}{\sqrt {2} \sqrt {g \sin (e+f x)} \sqrt {a+a \sin (e+f x)}}\right )}{\sqrt {2} \sqrt {a} c f}+\frac {\sec (e+f x) \sqrt {g \sin (e+f x)} \sqrt {a+a \sin (e+f x)}}{a c f}\\ \end {align*}

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Mathematica [A] time = 0.32, size = 133, normalized size = 1.17 \[ \frac {\sqrt {\sin (e+f x)} \csc (2 (e+f x)) \sqrt {a (\sin (e+f x)+1)} \sqrt {g \sin (e+f x)} \left (2 \sqrt {c} \sqrt {\sin (e+f x)}-\sqrt {2} \sqrt {c-c \sin (e+f x)} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {\sin (e+f x)}}{\sqrt {c-c \sin (e+f x)}}\right )\right )}{a c^{3/2} f} \]

Antiderivative was successfully veriﬁed.

[In]

Integrate[Sqrt[g*Sin[e + f*x]]/(Sqrt[a + a*Sin[e + f*x]]*(c - c*Sin[e + f*x])),x]

[Out]

(Csc[2*(e + f*x)]*Sqrt[Sin[e + f*x]]*Sqrt[g*Sin[e + f*x]]*Sqrt[a*(1 + Sin[e + f*x])]*(2*Sqrt[c]*Sqrt[Sin[e + f

*x]] - Sqrt[2]*ArcTan[(Sqrt[2]*Sqrt[c]*Sqrt[Sin[e + f*x]])/Sqrt[c - c*Sin[e + f*x]]]*Sqrt[c - c*Sin[e + f*x]])

)/(a*c^(3/2)*f)

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fricas [A] time = 0.66, size = 385, normalized size = 3.38 \[ \left [\frac {\sqrt {2} a \sqrt {-\frac {g}{a}} \cos \left (f x + e\right ) \log \left (\frac {17 \, g \cos \left (f x + e\right )^{3} + 4 \, \sqrt {2} {\left (3 \, \cos \left (f x + e\right )^{2} + {\left (3 \, \cos \left (f x + e\right ) + 4\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 4\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {g \sin \left (f x + e\right )} \sqrt {-\frac {g}{a}} + 3 \, g \cos \left (f x + e\right )^{2} - 18 \, g \cos \left (f x + e\right ) + {\left (17 \, g \cos \left (f x + e\right )^{2} + 14 \, g \cos \left (f x + e\right ) - 4 \, g\right )} \sin \left (f x + e\right ) - 4 \, g}{\cos \left (f x + e\right )^{3} + 3 \, \cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right )^{2} - 2 \, \cos \left (f x + e\right ) - 4\right )} \sin \left (f x + e\right ) - 2 \, \cos \left (f x + e\right ) - 4}\right ) + 8 \, \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {g \sin \left (f x + e\right )}}{8 \, a c f \cos \left (f x + e\right )}, -\frac {\sqrt {2} a \sqrt {\frac {g}{a}} \arctan \left (\frac {\sqrt {2} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {g \sin \left (f x + e\right )} \sqrt {\frac {g}{a}} {\left (3 \, \sin \left (f x + e\right ) - 1\right )}}{4 \, g \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right ) \cos \left (f x + e\right ) - 4 \, \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {g \sin \left (f x + e\right )}}{4 \, a c f \cos \left (f x + e\right )}\right ] \]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*(sqrt(2)*a*sqrt(-g/a)*cos(f*x + e)*log((17*g*cos(f*x + e)^3 + 4*sqrt(2)*(3*cos(f*x + e)^2 + (3*cos(f*x +

e) + 4)*sin(f*x + e) - cos(f*x + e) - 4)*sqrt(a*sin(f*x + e) + a)*sqrt(g*sin(f*x + e))*sqrt(-g/a) + 3*g*cos(f*

x + e)^2 - 18*g*cos(f*x + e) + (17*g*cos(f*x + e)^2 + 14*g*cos(f*x + e) - 4*g)*sin(f*x + e) - 4*g)/(cos(f*x +

e)^3 + 3*cos(f*x + e)^2 + (cos(f*x + e)^2 - 2*cos(f*x + e) - 4)*sin(f*x + e) - 2*cos(f*x + e) - 4)) + 8*sqrt(a

*sin(f*x + e) + a)*sqrt(g*sin(f*x + e)))/(a*c*f*cos(f*x + e)), -1/4*(sqrt(2)*a*sqrt(g/a)*arctan(1/4*sqrt(2)*sq

rt(a*sin(f*x + e) + a)*sqrt(g*sin(f*x + e))*sqrt(g/a)*(3*sin(f*x + e) - 1)/(g*cos(f*x + e)*sin(f*x + e)))*cos(

f*x + e) - 4*sqrt(a*sin(f*x + e) + a)*sqrt(g*sin(f*x + e)))/(a*c*f*cos(f*x + e))]

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giac [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Unab

le to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*p

i/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unab

le to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*p

i/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unab

le to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/

2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Warning, integration of abs or sign assumes consta

nt sign by intervals (correct if the argument is real):Check [abs(cos((f*t_nostep+exp(1))/2-pi/4))]Unable to c

heck sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Disco

ntinuities at zeroes of cos((f*t_nostep+exp(1))/2-pi/4) were not checkedUnable to check sign: (4*pi/t_nostep/2

)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_no

step/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*p

i/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign