3.120 \(\int \frac{A+B \sin (e+f x)}{(a+a \sin (e+f x))^2 \sqrt{c-c \sin (e+f x)}} \, dx\)

Optimal. Leaf size=135 \[ -\frac{(A-B) \sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{3 a^2 c^2 f}-\frac{(A+B) \sec (e+f x) \sqrt{c-c \sin (e+f x)}}{2 a^2 c f}+\frac{(A+B) \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{2 \sqrt{2} a^2 \sqrt{c} f} \]

[Out]

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

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Rubi [A]  time = 0.353657, antiderivative size = 135, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.132, Rules used = {2967, 2855, 2675, 2649, 206} \[ -\frac{(A-B) \sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{3 a^2 c^2 f}-\frac{(A+B) \sec (e+f x) \sqrt{c-c \sin (e+f x)}}{2 a^2 c f}+\frac{(A+B) \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{2 \sqrt{2} a^2 \sqrt{c} f} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Sin[e + f*x])/((a + a*Sin[e + f*x])^2*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]])])/(2*Sqrt[2]*a^2*Sqrt[c]*f) - ((A +
B)*Sec[e + f*x]*Sqrt[c - c*Sin[e + f*x]])/(2*a^2*c*f) - ((A - B)*Sec[e + f*x]^3*(c - c*Sin[e + f*x])^(3/2))/(
3*a^2*c^2*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 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{A+B \sin (e+f x)}{(a+a \sin (e+f x))^2 \sqrt{c-c \sin (e+f x)}} \, dx &=\frac{\int \sec ^4(e+f x) (A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2} \, dx}{a^2 c^2}\\ &=-\frac{(A-B) \sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{3 a^2 c^2 f}+\frac{(A+B) \int \sec ^2(e+f x) \sqrt{c-c \sin (e+f x)} \, dx}{2 a^2 c}\\ &=-\frac{(A+B) \sec (e+f x) \sqrt{c-c \sin (e+f x)}}{2 a^2 c f}-\frac{(A-B) \sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{3 a^2 c^2 f}+\frac{(A+B) \int \frac{1}{\sqrt{c-c \sin (e+f x)}} \, dx}{4 a^2}\\ &=-\frac{(A+B) \sec (e+f x) \sqrt{c-c \sin (e+f x)}}{2 a^2 c f}-\frac{(A-B) \sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{3 a^2 c^2 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 )}{2 a^2 f}\\ &=\frac{(A+B) \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{2 \sqrt{2} a^2 \sqrt{c} f}-\frac{(A+B) \sec (e+f x) \sqrt{c-c \sin (e+f x)}}{2 a^2 c f}-\frac{(A-B) \sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{3 a^2 c^2 f}\\ \end{align*}

Mathematica [C]  time = 0.533808, size = 176, normalized size = 1.3 \[ \frac{\left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) \left (-3 (A+B) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2+(-3-3 i) \sqrt [4]{-1} (A+B) \tan ^{-1}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) \sqrt [4]{-1} \left (\tan \left (\frac{1}{4} (e+f x)\right )+1\right )\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^3+2 (B-A)\right )}{6 a^2 f (\sin (e+f x)+1)^2 \sqrt{c-c \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Sin[e + f*x])/((a + a*Sin[e + f*x])^2*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])*(2*(-A + B) - 3*(A + B)*(Cos[(e +
f*x)/2] + Sin[(e + f*x)/2])^2 - (3 + 3*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])^3))/(6*a^2*f*(1 + Sin[e + f*x])^2*Sqrt[c - c*Sin[e + f*x]])

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Maple [A]  time = 1.105, size = 168, normalized size = 1.2 \begin{align*} -{\frac{-1+\sin \left ( fx+e \right ) }{12\,{a}^{2} \left ( 1+\sin \left ( fx+e \right ) \right ) \cos \left ( fx+e \right ) f} \left ( 3\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) }\sqrt{2}}{\sqrt{c}}} \right ) \left ( c \left ( 1+\sin \left ( fx+e \right ) \right ) \right ) ^{3/2}cA-6\,A{c}^{5/2}\sin \left ( fx+e \right ) +3\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) }\sqrt{2}}{\sqrt{c}}} \right ) \left ( c \left ( 1+\sin \left ( fx+e \right ) \right ) \right ) ^{3/2}cB-6\,B{c}^{5/2}\sin \left ( fx+e \right ) -10\,A{c}^{5/2}-2\,B{c}^{5/2} \right ){c}^{-{\frac{5}{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((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^2/(c-c*sin(f*x+e))^(1/2),x)

[Out]

-1/12*(-1+sin(f*x+e))*(3*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)))^(3/2
)*c*A-6*A*c^(5/2)*sin(f*x+e)+3*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))
)^(3/2)*c*B-6*B*c^(5/2)*sin(f*x+e)-10*A*c^(5/2)-2*B*c^(5/2))/a^2/c^(5/2)/(1+sin(f*x+e))/cos(f*x+e)/(c-c*sin(f*
x+e))^(1/2)/f

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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))^2/(c-c*sin(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 1.77789, size = 603, normalized size = 4.47 \begin{align*} \frac{3 \, \sqrt{2}{\left ({\left (A + B\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) +{\left (A + B\right )} \cos \left (f x + e\right )\right )} \sqrt{c} \log \left (-\frac{c \cos \left (f x + e\right )^{2} + 2 \, \sqrt{2} \sqrt{-c \sin \left (f x + e\right ) + c} \sqrt{c}{\left (\cos \left (f x + e\right ) + \sin \left (f x + e\right ) + 1\right )} + 3 \, c \cos \left (f x + e\right ) +{\left (c \cos \left (f x + e\right ) - 2 \, c\right )} \sin \left (f x + e\right ) + 2 \, c}{\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 \,{\left (3 \,{\left (A + B\right )} \sin \left (f x + e\right ) + 5 \, A + B\right )} \sqrt{-c \sin \left (f x + e\right ) + c}}{24 \,{\left (a^{2} c f \cos \left (f x + e\right ) \sin \left (f x + e\right ) + a^{2} c f \cos \left (f x + e\right )\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))^2/(c-c*sin(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

1/24*(3*sqrt(2)*((A + B)*cos(f*x + e)*sin(f*x + e) + (A + B)*cos(f*x + e))*sqrt(c)*log(-(c*cos(f*x + e)^2 + 2*
sqrt(2)*sqrt(-c*sin(f*x + e) + c)*sqrt(c)*(cos(f*x + e) + sin(f*x + e) + 1) + 3*c*cos(f*x + e) + (c*cos(f*x +
e) - 2*c)*sin(f*x + e) + 2*c)/(cos(f*x + e)^2 + (cos(f*x + e) + 2)*sin(f*x + e) - cos(f*x + e) - 2)) - 4*(3*(A
+ B)*sin(f*x + e) + 5*A + B)*sqrt(-c*sin(f*x + e) + c))/(a^2*c*f*cos(f*x + e)*sin(f*x + e) + a^2*c*f*cos(f*x
+ e))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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))**2/(c-c*sin(f*x+e))**(1/2),x)

[Out]

Timed out

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Giac [B]  time = 1.87945, size = 1046, normalized size = 7.75 \begin{align*} \text{result too large to display} \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))^2/(c-c*sin(f*x+e))^(1/2),x, algorithm="giac")

[Out]

1/6*((30*sqrt(2)*A*c*arctan(sqrt(c)/sqrt(-c)) + 30*sqrt(2)*B*c*arctan(sqrt(c)/sqrt(-c)) - 42*A*c*arctan(sqrt(c
)/sqrt(-c)) - 42*B*c*arctan(sqrt(c)/sqrt(-c)) - 15*sqrt(2)*A*sqrt(-c)*sqrt(c) + 3*sqrt(2)*B*sqrt(-c)*sqrt(c) +
22*A*sqrt(-c)*sqrt(c) - 4*B*sqrt(-c)*sqrt(c))*sgn(tan(1/2*f*x + 1/2*e) - 1)/(7*sqrt(2)*a^2*sqrt(-c)*c - 10*a^
2*sqrt(-c)*c) + 3*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^2*sqrt(-c)*sgn(tan(1/2*f*x + 1/2*e) - 1)) + 2*(9*(sqrt(c)*tan(1/2*f*x + 1/2
*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^5*A - 3*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^
2 + c))^5*B + 15*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^4*A*sqrt(c) + 3*(sqrt(c)*
tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^4*B*sqrt(c) - 10*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sq
rt(c*tan(1/2*f*x + 1/2*e)^2 + c))^3*A*c - 2*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c)
)^3*B*c - 30*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^2*A*c^(3/2) - 6*(sqrt(c)*tan(
1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^2*B*c^(3/2) + 21*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c
*tan(1/2*f*x + 1/2*e)^2 + c))*A*c^2 + 9*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))*B*
c^2 - 5*A*c^(5/2) - B*c^(5/2))/(((sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^2 + 2*(sq
rt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))*sqrt(c) - c)^3*a^2*sgn(tan(1/2*f*x + 1/2*e) -
1)))/f