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

Optimal. Leaf size=77 \[ \frac{2 \sqrt{2} a \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{\sqrt{c} f}-\frac{2 a \cos (e+f x)}{f \sqrt{c-c \sin (e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.138476, antiderivative size = 77, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {2736, 2679, 2649, 206} \[ \frac{2 \sqrt{2} a \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{\sqrt{c} f}-\frac{2 a \cos (e+f x)}{f \sqrt{c-c \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 2679

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(g*(g*
Cos[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + p)), x] + Dist[(g^2*(p - 1))/(a*(m + p)), 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]
&& LtQ[m, -1] && GtQ[p, 1] && (GtQ[m, -2] || EqQ[2*m + p + 1, 0] || (EqQ[m, -2] && IntegerQ[p])) && NeQ[m + p,
0] && IntegersQ[2*m, 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{a+a \sin (e+f x)}{\sqrt{c-c \sin (e+f x)}} \, dx &=(a c) \int \frac{\cos ^2(e+f x)}{(c-c \sin (e+f x))^{3/2}} \, dx\\ &=-\frac{2 a \cos (e+f x)}{f \sqrt{c-c \sin (e+f x)}}+(2 a) \int \frac{1}{\sqrt{c-c \sin (e+f x)}} \, dx\\ &=-\frac{2 a \cos (e+f x)}{f \sqrt{c-c \sin (e+f x)}}-\frac{(4 a) \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 )}{f}\\ &=\frac{2 \sqrt{2} a \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{\sqrt{c} f}-\frac{2 a \cos (e+f x)}{f \sqrt{c-c \sin (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 0.585694, size = 135, normalized size = 1.75 \[ -\frac{2 a \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (\sqrt{c} (\sin (e+f x)+1)+\sqrt{2} \sqrt{-c (\sin (e+f x)+1)} \tan ^{-1}\left (\frac{\sqrt{-c (\sin (e+f x)+1)}}{\sqrt{2} \sqrt{c}}\right )\right )}{\sqrt{c} f \sqrt{c-c \sin (e+f x)} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(Sqrt[c]*(1 + Sin[e + f*x]) + Sqrt[2]*ArcTan[Sqrt[-(c*(1 + Sin[e +
f*x]))]/(Sqrt[2]*Sqrt[c])]*Sqrt[-(c*(1 + Sin[e + f*x]))]))/(Sqrt[c]*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*S
qrt[c - c*Sin[e + f*x]])

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Maple [A]  time = 0.576, size = 94, normalized size = 1.2 \begin{align*} -2\,{\frac{ \left ( -1+\sin \left ( fx+e \right ) \right ) \sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) }a}{c\cos \left ( fx+e \right ) \sqrt{c-c\sin \left ( fx+e \right ) }f} \left ( \sqrt{c}\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) }\sqrt{2}}{\sqrt{c}}} \right ) -\sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(-1+sin(f*x+e))*(c*(1+sin(f*x+e)))^(1/2)*a*(c^(1/2)*2^(1/2)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^
(1/2))-(c*(1+sin(f*x+e)))^(1/2))/c/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{a \sin \left (f x + e\right ) + a}{\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((a+a*sin(f*x+e))/(c-c*sin(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.0731, size = 544, normalized size = 7.06 \begin{align*} \frac{\frac{\sqrt{2}{\left (a c \cos \left (f x + e\right ) - a c \sin \left (f x + e\right ) + a c\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 )}{\sqrt{c}} - 2 \,{\left (a \cos \left (f x + e\right ) + a \sin \left (f x + e\right ) + a\right )} \sqrt{-c \sin \left (f x + e\right ) + c}}{c f \cos \left (f x + e\right ) - c f \sin \left (f x + e\right ) + c f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(sqrt(2)*(a*c*cos(f*x + e) - a*c*sin(f*x + e) + a*c)*log(-(cos(f*x + e)^2 + (cos(f*x + e) - 2)*sin(f*x + e) +
2*sqrt(2)*sqrt(-c*sin(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))/sqrt(c) - 2*(a*cos(f*x + e) + a*sin(f*x + e) + a)
*sqrt(-c*sin(f*x + e) + c))/(c*f*cos(f*x + e) - c*f*sin(f*x + e) + c*f)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 2.39374, size = 267, normalized size = 3.47 \begin{align*} \frac{2 \,{\left (\frac{2 \, \sqrt{2} a \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 )}{\sqrt{-c} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1\right )} - \frac{\sqrt{2}{\left (2 \, a c \arctan \left (\frac{\sqrt{c}}{\sqrt{-c}}\right ) + a \sqrt{-c} \sqrt{c}\right )} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1\right )}{\sqrt{-c} c} + \frac{\frac{a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}{\mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1\right )} + \frac{a}{\mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1\right )}}{\sqrt{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + c}}\right )}}{f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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