3.18 \(\int \frac{1}{\sqrt{g \sin (e+f x)} \sqrt{a+a \sin (e+f x)} (c-c \sin (e+f x))} \, dx\)

Optimal. Leaf size=118 \[ \frac{\sec (e+f x) \sqrt{a \sin (e+f x)+a} \sqrt{g \sin (e+f x)}}{a c f g}-\frac{\tan ^{-1}\left (\frac{\sqrt{a} \sqrt{g} \cos (e+f x)}{\sqrt{2} \sqrt{a \sin (e+f x)+a} \sqrt{g \sin (e+f x)}}\right )}{\sqrt{2} \sqrt{a} c f \sqrt{g}} \]

[Out]

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

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Rubi [A]  time = 0.495929, antiderivative size = 118, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {2938, 2782, 205, 2930, 12, 30} \[ \frac{\sec (e+f x) \sqrt{a \sin (e+f x)+a} \sqrt{g \sin (e+f x)}}{a c f g}-\frac{\tan ^{-1}\left (\frac{\sqrt{a} \sqrt{g} \cos (e+f x)}{\sqrt{2} \sqrt{a \sin (e+f x)+a} \sqrt{g \sin (e+f x)}}\right )}{\sqrt{2} \sqrt{a} c f \sqrt{g}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2938

Int[1/(Sqrt[(g_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_) + (d_.)*sin[(e_.)
+ (f_.)*(x_)])), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(Sqrt[g*Sin[e + f*x]]*Sqrt[a + b*Sin[e + f*x]]), x], x
] - Dist[d/(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] /; Fr
eeQ[{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])

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 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 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 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]

Rubi steps

\begin{align*} \int \frac{1}{\sqrt{g \sin (e+f x)} \sqrt{a+a \sin (e+f x)} (c-c \sin (e+f x))} \, dx &=\frac{\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{\int \frac{1}{\sqrt{g \sin (e+f x)} \sqrt{a+a \sin (e+f x)}} \, dx}{2 c}\\ &=-\frac{\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 \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{\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 \sqrt{g}}-\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 g}\\ &=-\frac{\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 \sqrt{g}}+\frac{\sec (e+f x) \sqrt{g \sin (e+f x)} \sqrt{a+a \sin (e+f x)}}{a c f g}\\ \end{align*}

Mathematica [A]  time = 0.288262, size = 132, normalized size = 1.12 \[ \frac{\sin ^{\frac{3}{2}}(e+f x) \csc (2 (e+f x)) \sqrt{a (\sin (e+f x)+1)} \left (2 \sqrt{c} \sqrt{\sin (e+f x)}+\sqrt{2} \sqrt{c-c \sin (e+f x)} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{\sin (e+f x)}}{\sqrt{c-c \sin (e+f x)}}\right )\right )}{a c^{3/2} f \sqrt{g \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(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)]*Sin[e + f*x]^(3/2)*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*Sqrt[g
*Sin[e + f*x]])

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Maple [A]  time = 0.331, size = 117, normalized size = 1. \begin{align*}{\frac{\sin \left ( fx+e \right ) }{cf \left ( -1+\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) \right ) } \left ( 2\,\cos \left ( fx+e \right ) \sqrt{-{\frac{-1+\cos \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }}}\arctan \left ( \sqrt{-{\frac{-1+\cos \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }}} \right ) -\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) +1 \right ){\frac{1}{\sqrt{g\sin \left ( fx+e \right ) }}}{\frac{1}{\sqrt{a \left ( 1+\sin \left ( fx+e \right ) \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\int \frac{1}{\sqrt{a \sin \left (f x + e\right ) + a}{\left (c \sin \left (f x + e\right ) - c\right )} \sqrt{g \sin \left (f x + e\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-integrate(1/(sqrt(a*sin(f*x + e) + a)*(c*sin(f*x + e) - c)*sqrt(g*sin(f*x + e))), x)

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Fricas [A]  time = 3.73188, size = 1017, normalized size = 8.62 \begin{align*} \left [\frac{\sqrt{2} a g \sqrt{-\frac{1}{a g}} \cos \left (f x + e\right ) \log \left (-\frac{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{1}{a g}} - 17 \, \cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )^{2} -{\left (17 \, \cos \left (f x + e\right )^{2} + 14 \, \cos \left (f x + e\right ) - 4\right )} \sin \left (f x + e\right ) + 18 \, \cos \left (f x + e\right ) + 4}{\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 g \cos \left (f x + e\right )}, \frac{\sqrt{2} a g \sqrt{\frac{1}{a g}} \arctan \left (\frac{\sqrt{2} \sqrt{a \sin \left (f x + e\right ) + a} \sqrt{g \sin \left (f x + e\right )} \sqrt{\frac{1}{a g}}{\left (3 \, \sin \left (f x + e\right ) - 1\right )}}{4 \, \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 g \cos \left (f x + e\right )}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*(sqrt(2)*a*g*sqrt(-1/(a*g))*cos(f*x + e)*log(-(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(-1/(a*g)) - 17*cos(f*x + e)^3 - 3
*cos(f*x + e)^2 - (17*cos(f*x + e)^2 + 14*cos(f*x + e) - 4)*sin(f*x + e) + 18*cos(f*x + e) + 4)/(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*si
n(f*x + e) + a)*sqrt(g*sin(f*x + e)))/(a*c*f*g*cos(f*x + e)), 1/4*(sqrt(2)*a*g*sqrt(1/(a*g))*arctan(1/4*sqrt(2
)*sqrt(a*sin(f*x + e) + a)*sqrt(g*sin(f*x + e))*sqrt(1/(a*g))*(3*sin(f*x + e) - 1)/(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*g*cos(f*x + e))]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \frac{\int \frac{1}{\sqrt{g \sin{\left (e + f x \right )}} \sqrt{a \sin{\left (e + f x \right )} + a} \sin{\left (e + f x \right )} - \sqrt{g \sin{\left (e + f x \right )}} \sqrt{a \sin{\left (e + f x \right )} + a}}\, dx}{c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c-c*sin(f*x+e))/(g*sin(f*x+e))**(1/2)/(a+a*sin(f*x+e))**(1/2),x)

[Out]

-Integral(1/(sqrt(g*sin(e + f*x))*sqrt(a*sin(e + f*x) + a)*sin(e + f*x) - sqrt(g*sin(e + f*x))*sqrt(a*sin(e +
f*x) + a)), x)/c

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int -\frac{1}{\sqrt{a \sin \left (f x + e\right ) + a}{\left (c \sin \left (f x + e\right ) - c\right )} \sqrt{g \sin \left (f x + e\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(-1/(sqrt(a*sin(f*x + e) + a)*(c*sin(f*x + e) - c)*sqrt(g*sin(f*x + e))), x)