3.127 \(\int \frac{(A+B \sin (e+f x)) \sqrt{c-c \sin (e+f x)}}{(a+a \sin (e+f x))^3} \, dx\)
Optimal. Leaf size=85 \[ -\frac{(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{7/2}}{5 a^3 c^3 f}-\frac{(3 A+7 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{15 a^3 c f} \]
[Out]
-((3*A + 7*B)*Sec[e + f*x]^3*(c - c*Sin[e + f*x])^(3/2))/(15*a^3*c*f) - ((A - B)*Sec[e + f*x]^5*(c - c*Sin[e +
f*x])^(7/2))/(5*a^3*c^3*f)
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Rubi [A] time = 0.315198, antiderivative size = 85, normalized size of antiderivative = 1.,
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} \[ -\frac{(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{7/2}}{5 a^3 c^3 f}-\frac{(3 A+7 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{15 a^3 c f} \]
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])^3,x]
[Out]
-((3*A + 7*B)*Sec[e + f*x]^3*(c - c*Sin[e + f*x])^(3/2))/(15*a^3*c*f) - ((A - B)*Sec[e + f*x]^5*(c - c*Sin[e +
f*x])^(7/2))/(5*a^3*c^3*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 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]
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))^3} \, dx &=\frac{\int \sec ^6(e+f x) (A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2} \, dx}{a^3 c^3}\\ &=-\frac{(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{7/2}}{5 a^3 c^3 f}+\frac{(3 A+7 B) \int \sec ^4(e+f x) (c-c \sin (e+f x))^{5/2} \, dx}{10 a^3 c^2}\\ &=-\frac{(3 A+7 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{15 a^3 c f}-\frac{(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{7/2}}{5 a^3 c^3 f}\\ \end{align*}
Mathematica [A] time = 0.306657, size = 89, normalized size = 1.05 \[ -\frac{2 \sqrt{c-c \sin (e+f x)} (3 A+5 B \sin (e+f x)+2 B)}{15 a^3 f \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^5} \]
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])^3,x]
[Out]
(-2*(3*A + 2*B + 5*B*Sin[e + f*x])*Sqrt[c - c*Sin[e + f*x]])/(15*a^3*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(
Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^5)
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Maple [A] time = 1.001, size = 65, normalized size = 0.8 \begin{align*}{\frac{2\,c \left ( -1+\sin \left ( fx+e \right ) \right ) \left ( 5\,B\sin \left ( fx+e \right ) +3\,A+2\,B \right ) }{15\,{a}^{3} \left ( 1+\sin \left ( fx+e \right ) \right ) ^{2}\cos \left ( fx+e \right ) f}{\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))*(c-c*sin(f*x+e))^(1/2)/(a+a*sin(f*x+e))^3,x)
[Out]
2/15*c/a^3*(-1+sin(f*x+e))/(1+sin(f*x+e))^2*(5*B*sin(f*x+e)+3*A+2*B)/cos(f*x+e)/(c-c*sin(f*x+e))^(1/2)/f
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Maxima [B] time = 1.56314, size = 682, normalized size = 8.02 \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))*(c-c*sin(f*x+e))^(1/2)/(a+a*sin(f*x+e))^3,x, algorithm="maxima")
[Out]
2/15*(2*B*(sqrt(c) + 5*sqrt(c)*sin(f*x + e)/(cos(f*x + e) + 1) + 3*sqrt(c)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2
+ 10*sqrt(c)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 3*sqrt(c)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 5*sqrt(c)*
sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + sqrt(c)*sin(f*x + e)^6/(cos(f*x + e) + 1)^6)/((a^3 + 5*a^3*sin(f*x + e)/
(cos(f*x + e) + 1) + 10*a^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 10*a^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 +
5*a^3*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + a^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5)*sqrt(sin(f*x + e)^2/(cos
(f*x + e) + 1)^2 + 1)) + 3*A*(sqrt(c) + 3*sqrt(c)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 3*sqrt(c)*sin(f*x + e)
^4/(cos(f*x + e) + 1)^4 + sqrt(c)*sin(f*x + e)^6/(cos(f*x + e) + 1)^6)/((a^3 + 5*a^3*sin(f*x + e)/(cos(f*x + e
) + 1) + 10*a^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 10*a^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*a^3*sin(f
*x + e)^4/(cos(f*x + e) + 1)^4 + a^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5)*sqrt(sin(f*x + e)^2/(cos(f*x + e) +
1)^2 + 1)))/f
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Fricas [A] time = 1.62582, size = 196, normalized size = 2.31 \begin{align*} \frac{2 \,{\left (5 \, B \sin \left (f x + e\right ) + 3 \, A + 2 \, B\right )} \sqrt{-c \sin \left (f x + e\right ) + c}}{15 \,{\left (a^{3} f \cos \left (f x + e\right )^{3} - 2 \, a^{3} f \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, a^{3} 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))*(c-c*sin(f*x+e))^(1/2)/(a+a*sin(f*x+e))^3,x, algorithm="fricas")
[Out]
2/15*(5*B*sin(f*x + e) + 3*A + 2*B)*sqrt(-c*sin(f*x + e) + c)/(a^3*f*cos(f*x + e)^3 - 2*a^3*f*cos(f*x + e)*sin
(f*x + e) - 2*a^3*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))*(c-c*sin(f*x+e))**(1/2)/(a+a*sin(f*x+e))**3,x)
[Out]
Timed out
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Giac [B] time = 1.72661, size = 1539, normalized size = 18.11 \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))*(c-c*sin(f*x+e))^(1/2)/(a+a*sin(f*x+e))^3,x, algorithm="giac")
[Out]
-1/60*((573*sqrt(2)*A*sqrt(c) + 177*sqrt(2)*B*sqrt(c) - 810*A*sqrt(c) - 250*B*sqrt(c))*sgn(tan(1/2*f*x + 1/2*e
) - 1)/(29*sqrt(2)*a^3 - 41*a^3) - 16*(15*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^
9*A*c*sgn(tan(1/2*f*x + 1/2*e) - 1) + 45*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^8
*A*c^(3/2)*sgn(tan(1/2*f*x + 1/2*e) - 1) + 30*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 +
c))^8*B*c^(3/2)*sgn(tan(1/2*f*x + 1/2*e) - 1) + 60*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)
^2 + c))^7*A*c^2*sgn(tan(1/2*f*x + 1/2*e) - 1) + 20*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e
)^2 + c))^7*B*c^2*sgn(tan(1/2*f*x + 1/2*e) - 1) - 60*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*
e)^2 + c))^6*A*c^(5/2)*sgn(tan(1/2*f*x + 1/2*e) - 1) - 40*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x +
1/2*e)^2 + c))^6*B*c^(5/2)*sgn(tan(1/2*f*x + 1/2*e) - 1) - 102*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2
*f*x + 1/2*e)^2 + c))^5*A*c^3*sgn(tan(1/2*f*x + 1/2*e) - 1) - 68*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/
2*f*x + 1/2*e)^2 + c))^5*B*c^3*sgn(tan(1/2*f*x + 1/2*e) - 1) + 30*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1
/2*f*x + 1/2*e)^2 + c))^4*A*c^(7/2)*sgn(tan(1/2*f*x + 1/2*e) - 1) + 20*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*
tan(1/2*f*x + 1/2*e)^2 + c))^4*B*c^(7/2)*sgn(tan(1/2*f*x + 1/2*e) - 1) + 60*(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^4*sgn(tan(1/2*f*x + 1/2*e) - 1) + 60*(sqrt(c)*tan(1/2*f*x + 1/2*e) - s
qrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^3*B*c^4*sgn(tan(1/2*f*x + 1/2*e) - 1) - 60*(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^(9/2)*sgn(tan(1/2*f*x + 1/2*e) - 1) - 40*(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^(9/2)*sgn(tan(1/2*f*x + 1/2*e) - 1) + 15*(sqrt(c)*tan(1/2*f*x +
1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))*A*c^5*sgn(tan(1/2*f*x + 1/2*e) - 1) + 20*(sqrt(c)*tan(1/2*f*x +
1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))*B*c^5*sgn(tan(1/2*f*x + 1/2*e) - 1) - 3*A*c^(11/2)*sgn(tan(1/2*f*
x + 1/2*e) - 1) - 2*B*c^(11/2)*sgn(tan(1/2*f*x + 1/2*e) - 1))/(((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)^5
*a^3))/f