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

Optimal. Leaf size=122 \[ -\frac{2 a (3 A+5 B) \cos (e+f x)}{3 f \sqrt{c-c \sin (e+f x)}}+\frac{2 \sqrt{2} a (A+B) \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{\sqrt{c} f}+\frac{2 a B \cos (e+f x) \sqrt{c-c \sin (e+f x)}}{3 c f} \]

[Out]

(2*Sqrt[2]*a*(A + B)*ArcTanh[(Sqrt[c]*Cos[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sin[e + f*x]])])/(Sqrt[c]*f) - (2*a*(3
*A + 5*B)*Cos[e + f*x])/(3*f*Sqrt[c - c*Sin[e + f*x]]) + (2*a*B*Cos[e + f*x]*Sqrt[c - c*Sin[e + f*x]])/(3*c*f)

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Rubi [A]  time = 0.335919, antiderivative size = 122, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.139, Rules used = {2967, 2858, 2751, 2649, 206} \[ -\frac{2 a (3 A+5 B) \cos (e+f x)}{3 f \sqrt{c-c \sin (e+f x)}}+\frac{2 \sqrt{2} a (A+B) \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{\sqrt{c} f}+\frac{2 a B \cos (e+f x) \sqrt{c-c \sin (e+f x)}}{3 c f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[2]*a*(A + B)*ArcTanh[(Sqrt[c]*Cos[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sin[e + f*x]])])/(Sqrt[c]*f) - (2*a*(3
*A + 5*B)*Cos[e + f*x])/(3*f*Sqrt[c - c*Sin[e + f*x]]) + (2*a*B*Cos[e + f*x]*Sqrt[c - c*Sin[e + f*x]])/(3*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 2858

Int[cos[(e_.) + (f_.)*(x_)]^2*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_
)]), x_Symbol] :> Simp[(d*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 2))/(b^2*f*(m + 3)), x] - Dist[1/(b^2*(m + 3)
), Int[(a + b*Sin[e + f*x])^(m + 1)*(b*d*(m + 2) - a*c*(m + 3) + (b*c*(m + 3) - a*d*(m + 4))*Sin[e + f*x]), x]
, x] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[a^2 - b^2, 0] && GeQ[m, -3/2] && LtQ[m, 0]

Rule 2751

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[(d
*Cos[e + f*x]*(a + b*Sin[e + f*x])^m)/(f*(m + 1)), x] + Dist[(a*d*m + b*c*(m + 1))/(b*(m + 1)), Int[(a + b*Sin
[e + f*x])^m, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] &&  !LtQ[m,
-2^(-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+a \sin (e+f x)) (A+B \sin (e+f x))}{\sqrt{c-c \sin (e+f x)}} \, dx &=(a c) \int \frac{\cos ^2(e+f x) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{3/2}} \, dx\\ &=\frac{2 a B \cos (e+f x) \sqrt{c-c \sin (e+f x)}}{3 c f}-\frac{(2 a) \int \frac{-\frac{3 A c}{2}-\frac{B c}{2}+\left (-\frac{3 A c}{2}-\frac{5 B c}{2}\right ) \sin (e+f x)}{\sqrt{c-c \sin (e+f x)}} \, dx}{3 c}\\ &=-\frac{2 a (3 A+5 B) \cos (e+f x)}{3 f \sqrt{c-c \sin (e+f x)}}+\frac{2 a B \cos (e+f x) \sqrt{c-c \sin (e+f x)}}{3 c f}+(2 a (A+B)) \int \frac{1}{\sqrt{c-c \sin (e+f x)}} \, dx\\ &=-\frac{2 a (3 A+5 B) \cos (e+f x)}{3 f \sqrt{c-c \sin (e+f x)}}+\frac{2 a B \cos (e+f x) \sqrt{c-c \sin (e+f x)}}{3 c f}-\frac{(4 a (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 )}{f}\\ &=\frac{2 \sqrt{2} a (A+B) \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 (3 A+5 B) \cos (e+f x)}{3 f \sqrt{c-c \sin (e+f x)}}+\frac{2 a B \cos (e+f x) \sqrt{c-c \sin (e+f x)}}{3 c f}\\ \end{align*}

Mathematica [A]  time = 1.26508, size = 166, normalized size = 1.36 \[ -\frac{a \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (6 \sqrt{2} (A+B) \sqrt{-c (\sin (e+f x)+1)} \tan ^{-1}\left (\frac{\sqrt{-c (\sin (e+f x)+1)}}{\sqrt{2} \sqrt{c}}\right )+\sqrt{c} (2 (3 A+5 B) \sin (e+f x)+6 A-B \cos (2 (e+f x))+9 B)\right )}{3 \sqrt{c} f \sqrt{c-c \sin (e+f x)} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(6*Sqrt[2]*(A + B)*ArcTan[Sqrt[-(c*(1 + Sin[e + f*x]))]/(Sqrt[2]*Sqr
t[c])]*Sqrt[-(c*(1 + Sin[e + f*x]))] + Sqrt[c]*(6*A + 9*B - B*Cos[2*(e + f*x)] + 2*(3*A + 5*B)*Sin[e + f*x])))
/(3*Sqrt[c]*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*Sqrt[c - c*Sin[e + f*x]])

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Maple [A]  time = 1.253, size = 159, normalized size = 1.3 \begin{align*} -{\frac{ \left ( -2+2\,\sin \left ( fx+e \right ) \right ) a}{3\,{c}^{2}\cos \left ( fx+e \right ) f}\sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) } \left ( 3\,{c}^{3/2}\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) }\sqrt{2}}{\sqrt{c}}} \right ) A+3\,{c}^{3/2}\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) }\sqrt{2}}{\sqrt{c}}} \right ) B-B \left ( c \left ( 1+\sin \left ( fx+e \right ) \right ) \right ) ^{{\frac{3}{2}}}-3\,Ac\sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) }-3\,Bc\sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) } \right ){\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+a*sin(f*x+e))*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(1/2),x)

[Out]

-2/3*(-1+sin(f*x+e))*(c*(1+sin(f*x+e)))^(1/2)*a*(3*c^(3/2)*2^(1/2)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2
)/c^(1/2))*A+3*c^(3/2)*2^(1/2)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2))*B-B*(c*(1+sin(f*x+e)))^(3
/2)-3*A*c*(c*(1+sin(f*x+e)))^(1/2)-3*B*c*(c*(1+sin(f*x+e)))^(1/2))/c^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{{\left (B \sin \left (f x + e\right ) + A\right )}{\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+a*sin(f*x+e))*(A+B*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 = 1.77234, size = 690, normalized size = 5.66 \begin{align*} \frac{\frac{3 \, \sqrt{2}{\left ({\left (A + B\right )} a c \cos \left (f x + e\right ) -{\left (A + B\right )} a c \sin \left (f x + e\right ) +{\left (A + B\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 (B a \cos \left (f x + e\right )^{2} -{\left (3 \, A + 4 \, B\right )} a \cos \left (f x + e\right ) -{\left (3 \, A + 5 \, B\right )} a -{\left (B a \cos \left (f x + e\right ) +{\left (3 \, A + 5 \, B\right )} a\right )} \sin \left (f x + e\right )\right )} \sqrt{-c \sin \left (f x + e\right ) + c}}{3 \,{\left (c f \cos \left (f x + e\right ) - c f \sin \left (f x + e\right ) + c f\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(3*sqrt(2)*((A + B)*a*c*cos(f*x + e) - (A + B)*a*c*sin(f*x + e) + (A + B)*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*(B*a*co
s(f*x + e)^2 - (3*A + 4*B)*a*cos(f*x + e) - (3*A + 5*B)*a - (B*a*cos(f*x + e) + (3*A + 5*B)*a)*sin(f*x + e))*s
qrt(-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{A}{\sqrt{- c \sin{\left (e + f x \right )} + c}}\, dx + \int \frac{A \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}}\, dx + \int \frac{B \sin ^{2}{\left (e + f x \right )}}{\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))*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))**(1/2),x)

[Out]

a*(Integral(A/sqrt(-c*sin(e + f*x) + c), x) + Integral(A*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), x) + Integral(B*sin(e + f*x)**2/sqrt(-c*sin(e + f*x) + c), x))

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Giac [B]  time = 2.41465, size = 541, normalized size = 4.43 \begin{align*} \frac{\frac{12 \, \sqrt{2}{\left (A a + B a\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 )}{\sqrt{-c} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1\right )} + \frac{{\left ({\left (\frac{{\left (3 \, A a c \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1\right ) + 4 \, B a c \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1\right )\right )} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}{c^{6}} + \frac{3 \,{\left (A a c \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1\right ) + 2 \, B a c \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1\right )\right )}}{c^{6}}\right )} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + \frac{3 \,{\left (A a c \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1\right ) + 2 \, B a c \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1\right )\right )}}{c^{6}}\right )} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + \frac{3 \, A a c \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1\right ) + 4 \, B a c \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1\right )}{c^{6}}}{{\left (c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + c\right )}^{\frac{3}{2}}} - \frac{{\left (12 \, \sqrt{2} A a c^{7} \arctan \left (\frac{\sqrt{c}}{\sqrt{-c}}\right ) + 12 \, \sqrt{2} B a c^{7} \arctan \left (\frac{\sqrt{c}}{\sqrt{-c}}\right ) + 3 \, \sqrt{2} A a \sqrt{-c} \sqrt{c} + 5 \, \sqrt{2} B 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^{7}}}{3 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(12*sqrt(2)*(A*a + B*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)) + ((((3*A*a*c*sgn(tan(1/2*f*x + 1/2*e) - 1)
+ 4*B*a*c*sgn(tan(1/2*f*x + 1/2*e) - 1))*tan(1/2*f*x + 1/2*e)/c^6 + 3*(A*a*c*sgn(tan(1/2*f*x + 1/2*e) - 1) +
2*B*a*c*sgn(tan(1/2*f*x + 1/2*e) - 1))/c^6)*tan(1/2*f*x + 1/2*e) + 3*(A*a*c*sgn(tan(1/2*f*x + 1/2*e) - 1) + 2*
B*a*c*sgn(tan(1/2*f*x + 1/2*e) - 1))/c^6)*tan(1/2*f*x + 1/2*e) + (3*A*a*c*sgn(tan(1/2*f*x + 1/2*e) - 1) + 4*B*
a*c*sgn(tan(1/2*f*x + 1/2*e) - 1))/c^6)/(c*tan(1/2*f*x + 1/2*e)^2 + c)^(3/2) - (12*sqrt(2)*A*a*c^7*arctan(sqrt
(c)/sqrt(-c)) + 12*sqrt(2)*B*a*c^7*arctan(sqrt(c)/sqrt(-c)) + 3*sqrt(2)*A*a*sqrt(-c)*sqrt(c) + 5*sqrt(2)*B*a*s
qrt(-c)*sqrt(c))*sgn(tan(1/2*f*x + 1/2*e) - 1)/(sqrt(-c)*c^7))/f