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

Optimal. Leaf size=83 \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{\sqrt{2} a \sqrt{c} f}-\frac{\sec (e+f x) \sqrt{c-c \sin (e+f x)}}{a c f} \]

[Out]

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

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Rubi [A]  time = 0.158731, antiderivative size = 83, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {2736, 2675, 2649, 206} \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{\sqrt{2} a \sqrt{c} f}-\frac{\sec (e+f x) \sqrt{c-c \sin (e+f x)}}{a c f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2736

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Di
st[a^m*c^m, Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&
EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] &&  !(IntegerQ[n] && ((LtQ[m, 0] && GtQ[n, 0]) || LtQ[0,
n, m] || LtQ[m, n, 0]))

Rule 2675

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> -Simp[(b*(g
*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m)/(a*f*g*(p + 1)), x] + Dist[(a*(m + p + 1))/(g^2*(p + 1)), Int[(
g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^(m - 1), x], x] /; FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 - b^2, 0]
&& GtQ[m, 0] && LeQ[p, -2*m] && IntegersQ[m + 1/2, 2*p]

Rule 2649

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[-2/d, Subst[Int[1/(2*a - x^2), x], x, (b*C
os[c + d*x])/Sqrt[a + b*Sin[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{1}{(a+a \sin (e+f x)) \sqrt{c-c \sin (e+f x)}} \, dx &=\frac{\int \sec ^2(e+f x) \sqrt{c-c \sin (e+f x)} \, dx}{a c}\\ &=-\frac{\sec (e+f x) \sqrt{c-c \sin (e+f x)}}{a c f}+\frac{\int \frac{1}{\sqrt{c-c \sin (e+f x)}} \, dx}{2 a}\\ &=-\frac{\sec (e+f x) \sqrt{c-c \sin (e+f x)}}{a c f}-\frac{\operatorname{Subst}\left (\int \frac{1}{2 c-x^2} \, dx,x,-\frac{c \cos (e+f x)}{\sqrt{c-c \sin (e+f x)}}\right )}{a f}\\ &=\frac{\tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{\sqrt{2} a \sqrt{c} f}-\frac{\sec (e+f x) \sqrt{c-c \sin (e+f x)}}{a c f}\\ \end{align*}

Mathematica [C]  time = 0.31445, size = 97, normalized size = 1.17 \[ -\frac{\cos (e+f x) \left (1+(1+i) \sqrt [4]{-1} \tan ^{-1}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) \sqrt [4]{-1} \left (\tan \left (\frac{1}{4} (e+f x)\right )+1\right )\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )\right )}{a f (\sin (e+f x)+1) \sqrt{c-c \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Cos[e + f*x]*(1 + (1 + I)*(-1)^(1/4)*ArcTan[(1/2 + I/2)*(-1)^(1/4)*(1 + Tan[(e + f*x)/4])]*(Cos[(e + f*x)/2
] + Sin[(e + f*x)/2])))/(a*f*(1 + Sin[e + f*x])*Sqrt[c - c*Sin[e + f*x]]))

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Maple [A]  time = 0.625, size = 85, normalized size = 1. \begin{align*} -{\frac{-1+\sin \left ( fx+e \right ) }{2\,af\cos \left ( fx+e \right ) } \left ( \sqrt{2}{\it Artanh} \left ({\frac{\sqrt{2}}{2}\sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) }{\frac{1}{\sqrt{c}}}} \right ) c\sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) }-2\,{c}^{3/2} \right ){c}^{-{\frac{3}{2}}}{\frac{1}{\sqrt{c-c\sin \left ( fx+e \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/a*(-1+sin(f*x+e))*(2^(1/2)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2))*c*(c*(1+sin(f*x+e)))^(1/
2)-2*c^(3/2))/c^(3/2)/cos(f*x+e)/(c-c*sin(f*x+e))^(1/2)/f

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.09838, size = 425, normalized size = 5.12 \begin{align*} \frac{\sqrt{2} \sqrt{c} \cos \left (f x + e\right ) \log \left (-\frac{\cos \left (f x + e\right )^{2} +{\left (\cos \left (f x + e\right ) - 2\right )} \sin \left (f x + e\right ) + \frac{2 \, \sqrt{2} \sqrt{-c \sin \left (f x + e\right ) + c}{\left (\cos \left (f x + e\right ) + \sin \left (f x + e\right ) + 1\right )}}{\sqrt{c}} + 3 \, \cos \left (f x + e\right ) + 2}{\cos \left (f x + e\right )^{2} +{\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right ) - 4 \, \sqrt{-c \sin \left (f x + e\right ) + c}}{4 \, a c f \cos \left (f x + e\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(sqrt(2)*sqrt(c)*cos(f*x + e)*log(-(cos(f*x + e)^2 + (cos(f*x + e) - 2)*sin(f*x + e) + 2*sqrt(2)*sqrt(-c*s
in(f*x + e) + c)*(cos(f*x + e) + sin(f*x + e) + 1)/sqrt(c) + 3*cos(f*x + e) + 2)/(cos(f*x + e)^2 + (cos(f*x +
e) + 2)*sin(f*x + e) - cos(f*x + e) - 2)) - 4*sqrt(-c*sin(f*x + e) + c))/(a*c*f*cos(f*x + e))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.6202, size = 408, normalized size = 4.92 \begin{align*} \frac{\frac{{\left (2 \, \sqrt{2} c \arctan \left (\frac{\sqrt{c}}{\sqrt{-c}}\right ) - 2 \, c \arctan \left (\frac{\sqrt{c}}{\sqrt{-c}}\right ) + \sqrt{-c} \sqrt{c}\right )} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1\right )}{\sqrt{2} a \sqrt{-c} c - 2 \, a \sqrt{-c} c} + \frac{\sqrt{2} \arctan \left (-\frac{\sqrt{2}{\left (\sqrt{c} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - \sqrt{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + c} - \sqrt{c}\right )}}{2 \, \sqrt{-c}}\right )}{a \sqrt{-c} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1\right )} + \frac{2 \,{\left (\sqrt{c} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - \sqrt{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + c} - \sqrt{c}\right )}}{{\left ({\left (\sqrt{c} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - \sqrt{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + c}\right )}^{2} + 2 \,{\left (\sqrt{c} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - \sqrt{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + c}\right )} \sqrt{c} - c\right )} a \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1\right )}}{f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((2*sqrt(2)*c*arctan(sqrt(c)/sqrt(-c)) - 2*c*arctan(sqrt(c)/sqrt(-c)) + sqrt(-c)*sqrt(c))*sgn(tan(1/2*f*x + 1/
2*e) - 1)/(sqrt(2)*a*sqrt(-c)*c - 2*a*sqrt(-c)*c) + sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(c)*tan(1/2*f*x + 1/2*e)
- sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c) - sqrt(c))/sqrt(-c))/(a*sqrt(-c)*sgn(tan(1/2*f*x + 1/2*e) - 1)) + 2*(sqrt
(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c) - sqrt(c))/(((sqrt(c)*tan(1/2*f*x + 1/2*e) - sqr
t(c*tan(1/2*f*x + 1/2*e)^2 + c))^2 + 2*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))*sqr
t(c) - c)*a*sgn(tan(1/2*f*x + 1/2*e) - 1)))/f