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

Optimal. Leaf size=91 \[ \frac{(A+B) \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{\sqrt{2} a \sqrt{c} f}-\frac{(A-B) \sec (e+f x) \sqrt{c-c \sin (e+f x)}}{a c f} \]

[Out]

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

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Rubi [A]  time = 0.283368, antiderivative size = 91, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {2967, 2855, 2649, 206} \[ \frac{(A+B) \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{\sqrt{2} a \sqrt{c} f}-\frac{(A-B) \sec (e+f x) \sqrt{c-c \sin (e+f x)}}{a c f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2967

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

Rule 2855

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

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

Mathematica [C]  time = 0.460433, size = 140, normalized size = 1.54 \[ \frac{\left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) \left (-(1+i) \sqrt [4]{-1} (A+B) \tan ^{-1}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) \sqrt [4]{-1} \left (\tan \left (\frac{1}{4} (e+f x)\right )+1\right )\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )-A+B\right )}{a f (\sin (e+f x)+1) \sqrt{c-c \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(-A + B - (1 + I)*(-1)^(1/4)*(A +
B)*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 + Si
n[e + f*x])*Sqrt[c - c*Sin[e + f*x]])

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Maple [A]  time = 1.197, size = 130, normalized size = 1.4 \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 ) \sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) }A+\sqrt{2}{\it Artanh} \left ({\frac{\sqrt{2}}{2}\sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) }{\frac{1}{\sqrt{c}}}} \right ) \sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) }B-2\,\sqrt{c}A+2\,\sqrt{c}B \right ){\frac{1}{\sqrt{c}}}{\frac{1}{\sqrt{c-c\sin \left ( fx+e \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.39279, size = 447, normalized size = 4.91 \begin{align*} \frac{\sqrt{2}{\left (A + B\right )} \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}{\left (A - B\right )}}{4 \, a c f \cos \left (f x + e\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(sqrt(2)*(A + B)*sqrt(c)*cos(f*x + e)*log(-(cos(f*x + e)^2 + (cos(f*x + e) - 2)*sin(f*x + e) + 2*sqrt(2)*s
qrt(-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 + (co
s(f*x + e) + 2)*sin(f*x + e) - cos(f*x + e) - 2)) - 4*sqrt(-c*sin(f*x + e) + c)*(A - B))/(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{A}{\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 + \int \frac{B \sin{\left (e + f x \right )}}{\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((A+B*sin(f*x+e))/(a+a*sin(f*x+e))/(c-c*sin(f*x+e))**(1/2),x)

[Out]

(Integral(A/(sqrt(-c*sin(e + f*x) + c)*sin(e + f*x) + sqrt(-c*sin(e + f*x) + c)), x) + Integral(B*sin(e + f*x)
/(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 = 1.75312, size = 536, normalized size = 5.89 \begin{align*} \frac{\frac{{\left (2 \, \sqrt{2} A c \arctan \left (\frac{\sqrt{c}}{\sqrt{-c}}\right ) + 2 \, \sqrt{2} B c \arctan \left (\frac{\sqrt{c}}{\sqrt{-c}}\right ) - 2 \, A c \arctan \left (\frac{\sqrt{c}}{\sqrt{-c}}\right ) - 2 \, B c \arctan \left (\frac{\sqrt{c}}{\sqrt{-c}}\right ) + A \sqrt{-c} \sqrt{c} - B \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}{\left (A + B\right )} \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 ({\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 )} A -{\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 )} B - A \sqrt{c} + B \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((A+B*sin(f*x+e))/(a+a*sin(f*x+e))/(c-c*sin(f*x+e))^(1/2),x, algorithm="giac")

[Out]

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