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

Optimal. Leaf size=78 \[ -\frac {(A-B) \sec ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{3 a^2 c^2 f}-\frac {(A+5 B) \sec (e+f x) \sqrt {c-c \sin (e+f x)}}{3 a^2 f} \]

[Out]

-1/3*(A-B)*sec(f*x+e)^3*(c-c*sin(f*x+e))^(5/2)/a^2/c^2/f-1/3*(A+5*B)*sec(f*x+e)*(c-c*sin(f*x+e))^(1/2)/a^2/f

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Rubi [A]  time = 0.31, antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.079, Rules used = {2967, 2855, 2673} \[ -\frac {(A-B) \sec ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{3 a^2 c^2 f}-\frac {(A+5 B) \sec (e+f x) \sqrt {c-c \sin (e+f x)}}{3 a^2 f} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*Sin[e + f*x])*Sqrt[c - c*Sin[e + f*x]])/(a + a*Sin[e + f*x])^2,x]

[Out]

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

Rule 2673

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 - 1))/(f*g*(m - 1)), x] /; FreeQ[{a, b, e, f, g, m, p}, x] && Eq
Q[a^2 - b^2, 0] && EqQ[2*m + p - 1, 0] && NeQ[m, 1]

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 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]))

Rubi steps

\begin {align*} \int \frac {(A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)}}{(a+a \sin (e+f x))^2} \, dx &=\frac {\int \sec ^4(e+f x) (A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2} \, dx}{a^2 c^2}\\ &=-\frac {(A-B) \sec ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{3 a^2 c^2 f}+\frac {(A+5 B) \int \sec ^2(e+f x) (c-c \sin (e+f x))^{3/2} \, dx}{6 a^2 c}\\ &=-\frac {(A+5 B) \sec (e+f x) \sqrt {c-c \sin (e+f x)}}{3 a^2 f}-\frac {(A-B) \sec ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{3 a^2 c^2 f}\\ \end {align*}

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Mathematica [A]  time = 0.28, size = 87, normalized size = 1.12 \[ -\frac {2 \sqrt {c-c \sin (e+f x)} (A+3 B \sin (e+f x)+2 B)}{3 a^2 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.43, size = 60, normalized size = 0.77 \[ -\frac {2 \, {\left (3 \, B \sin \left (f x + e\right ) + A + 2 \, B\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{3 \, {\left (a^{2} f \cos \left (f x + e\right ) \sin \left (f x + e\right ) + a^{2} f \cos \left (f x + e\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*(3*B*sin(f*x + e) + A + 2*B)*sqrt(-c*sin(f*x + e) + c)/(a^2*f*cos(f*x + e)*sin(f*x + e) + a^2*f*cos(f*x +
e))

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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maple [A]  time = 1.31, size = 63, normalized size = 0.81 \[ \frac {2 c \left (\sin \left (f x +e \right )-1\right ) \left (3 B \sin \left (f x +e \right )+A +2 B \right )}{3 a^{2} \left (1+\sin \left (f x +e \right )\right ) \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.46, size = 343, normalized size = 4.40 \[ \frac {2 \, {\left (\frac {2 \, B {\left (\sqrt {c} + \frac {3 \, \sqrt {c} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {2 \, \sqrt {c} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {3 \, \sqrt {c} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {\sqrt {c} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}}\right )}}{{\left (a^{2} + \frac {3 \, a^{2} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {3 \, a^{2} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {a^{2} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}\right )} \sqrt {\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1}} + \frac {A {\left (\sqrt {c} + \frac {2 \, \sqrt {c} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {\sqrt {c} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}}\right )}}{{\left (a^{2} + \frac {3 \, a^{2} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {3 \, a^{2} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {a^{2} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}\right )} \sqrt {\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1}}\right )}}{3 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 17.51, size = 137, normalized size = 1.76 \[ \frac {4\,{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,\sqrt {c-c\,\left (\frac {{\mathrm {e}}^{-e\,1{}\mathrm {i}-f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}\right )}\,\left (B\,3{}\mathrm {i}+2\,A\,{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}+4\,B\,{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}-B\,{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,3{}\mathrm {i}\right )}{3\,a^2\,f\,{\left ({\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}+1{}\mathrm {i}\right )}^3\,\left (1+{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(4*exp(e*1i + f*x*1i)*(c - c*((exp(- e*1i - f*x*1i)*1i)/2 - (exp(e*1i + f*x*1i)*1i)/2))^(1/2)*(B*3i + 2*A*exp(
e*1i + f*x*1i) + 4*B*exp(e*1i + f*x*1i) - B*exp(e*2i + f*x*2i)*3i))/(3*a^2*f*(exp(e*1i + f*x*1i) + 1i)^3*(exp(
e*1i + f*x*1i)*1i + 1))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {A \sqrt {- c \sin {\left (e + f x \right )} + c}}{\sin ^{2}{\left (e + f x \right )} + 2 \sin {\left (e + f x \right )} + 1}\, dx + \int \frac {B \sqrt {- c \sin {\left (e + f x \right )} + c} \sin {\left (e + f x \right )}}{\sin ^{2}{\left (e + f x \right )} + 2 \sin {\left (e + f x \right )} + 1}\, dx}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(A*sqrt(-c*sin(e + f*x) + c)/(sin(e + f*x)**2 + 2*sin(e + f*x) + 1), x) + Integral(B*sqrt(-c*sin(e +
f*x) + c)*sin(e + f*x)/(sin(e + f*x)**2 + 2*sin(e + f*x) + 1), x))/a**2

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