3.109 \(\int \frac{(A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2}}{a+a \sin (e+f x)} \, dx\)

Optimal. Leaf size=159 \[ -\frac{8 c^2 (5 A-7 B) \cos (e+f x) \sqrt{c-c \sin (e+f x)}}{15 a f}-\frac{32 c^3 (5 A-7 B) \cos (e+f x)}{15 a f \sqrt{c-c \sin (e+f x)}}-\frac{c (5 A-7 B) \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{5 a f}-\frac{(A-B) \sec (e+f x) (c-c \sin (e+f x))^{7/2}}{a c f} \]

[Out]

(-32*(5*A - 7*B)*c^3*Cos[e + f*x])/(15*a*f*Sqrt[c - c*Sin[e + f*x]]) - (8*(5*A - 7*B)*c^2*Cos[e + f*x]*Sqrt[c
- c*Sin[e + f*x]])/(15*a*f) - ((5*A - 7*B)*c*Cos[e + f*x]*(c - c*Sin[e + f*x])^(3/2))/(5*a*f) - ((A - B)*Sec[e
+ f*x]*(c - c*Sin[e + f*x])^(7/2))/(a*c*f)

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Rubi [A]  time = 0.350131, antiderivative size = 159, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {2967, 2855, 2647, 2646} \[ -\frac{8 c^2 (5 A-7 B) \cos (e+f x) \sqrt{c-c \sin (e+f x)}}{15 a f}-\frac{32 c^3 (5 A-7 B) \cos (e+f x)}{15 a f \sqrt{c-c \sin (e+f x)}}-\frac{c (5 A-7 B) \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{5 a f}-\frac{(A-B) \sec (e+f x) (c-c \sin (e+f x))^{7/2}}{a c f} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^(5/2))/(a + a*Sin[e + f*x]),x]

[Out]

(-32*(5*A - 7*B)*c^3*Cos[e + f*x])/(15*a*f*Sqrt[c - c*Sin[e + f*x]]) - (8*(5*A - 7*B)*c^2*Cos[e + f*x]*Sqrt[c
- c*Sin[e + f*x]])/(15*a*f) - ((5*A - 7*B)*c*Cos[e + f*x]*(c - c*Sin[e + f*x])^(3/2))/(5*a*f) - ((A - B)*Sec[e
+ f*x]*(c - c*Sin[e + f*x])^(7/2))/(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 2647

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(a + b*Sin[c + d*x])^(n -
1))/(d*n), x] + Dist[(a*(2*n - 1))/n, Int[(a + b*Sin[c + d*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && Eq
Q[a^2 - b^2, 0] && IGtQ[n - 1/2, 0]

Rule 2646

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(-2*b*Cos[c + d*x])/(d*Sqrt[a + b*Sin[c + d*
x]]), x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rubi steps

\begin{align*} \int \frac{(A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2}}{a+a \sin (e+f x)} \, dx &=\frac{\int \sec ^2(e+f x) (A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2} \, dx}{a c}\\ &=-\frac{(A-B) \sec (e+f x) (c-c \sin (e+f x))^{7/2}}{a c f}-\frac{(5 A-7 B) \int (c-c \sin (e+f x))^{5/2} \, dx}{2 a}\\ &=-\frac{(5 A-7 B) c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{5 a f}-\frac{(A-B) \sec (e+f x) (c-c \sin (e+f x))^{7/2}}{a c f}-\frac{(4 (5 A-7 B) c) \int (c-c \sin (e+f x))^{3/2} \, dx}{5 a}\\ &=-\frac{8 (5 A-7 B) c^2 \cos (e+f x) \sqrt{c-c \sin (e+f x)}}{15 a f}-\frac{(5 A-7 B) c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{5 a f}-\frac{(A-B) \sec (e+f x) (c-c \sin (e+f x))^{7/2}}{a c f}-\frac{\left (16 (5 A-7 B) c^2\right ) \int \sqrt{c-c \sin (e+f x)} \, dx}{15 a}\\ &=-\frac{32 (5 A-7 B) c^3 \cos (e+f x)}{15 a f \sqrt{c-c \sin (e+f x)}}-\frac{8 (5 A-7 B) c^2 \cos (e+f x) \sqrt{c-c \sin (e+f x)}}{15 a f}-\frac{(5 A-7 B) c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{5 a f}-\frac{(A-B) \sec (e+f x) (c-c \sin (e+f x))^{7/2}}{a c f}\\ \end{align*}

Mathematica [A]  time = 1.77777, size = 134, normalized size = 0.84 \[ -\frac{c^2 \sqrt{c-c \sin (e+f x)} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) (25 (8 A-13 B) \sin (e+f x)+2 (5 A-16 B) \cos (2 (e+f x))+450 A+3 B \sin (3 (e+f x))-600 B)}{30 a f (\sin (e+f x)+1) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^(5/2))/(a + a*Sin[e + f*x]),x]

[Out]

-(c^2*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*Sqrt[c - c*Sin[e + f*x]]*(450*A - 600*B + 2*(5*A - 16*B)*Cos[2*(e
+ f*x)] + 25*(8*A - 13*B)*Sin[e + f*x] + 3*B*Sin[3*(e + f*x)]))/(30*a*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*
(1 + Sin[e + f*x]))

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Maple [A]  time = 0.828, size = 95, normalized size = 0.6 \begin{align*} -{\frac{2\,{c}^{3} \left ( -1+\sin \left ( fx+e \right ) \right ) \left ( -3\,B \left ( \cos \left ( fx+e \right ) \right ) ^{2}\sin \left ( fx+e \right ) + \left ( -50\,A+82\,B \right ) \sin \left ( fx+e \right ) + \left ( -5\,A+16\,B \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}-110\,A+142\,B \right ) }{15\,af\cos \left ( fx+e \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+B*sin(f*x+e))*(c-c*sin(f*x+e))^(5/2)/(a+a*sin(f*x+e)),x)

[Out]

-2/15*c^3/a*(-1+sin(f*x+e))*(-3*B*cos(f*x+e)^2*sin(f*x+e)+(-50*A+82*B)*sin(f*x+e)+(-5*A+16*B)*cos(f*x+e)^2-110
*A+142*B)/cos(f*x+e)/(c-c*sin(f*x+e))^(1/2)/f

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Maxima [B]  time = 1.5615, size = 521, normalized size = 3.28 \begin{align*} \frac{2 \,{\left (\frac{5 \,{\left (23 \, c^{\frac{5}{2}} + \frac{20 \, c^{\frac{5}{2}} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{65 \, c^{\frac{5}{2}} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac{40 \, c^{\frac{5}{2}} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac{65 \, c^{\frac{5}{2}} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac{20 \, c^{\frac{5}{2}} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} + \frac{23 \, c^{\frac{5}{2}} \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}}\right )} A}{{\left (a + \frac{a \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{\left (\frac{\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )}^{\frac{5}{2}}} - \frac{2 \,{\left (79 \, c^{\frac{5}{2}} + \frac{79 \, c^{\frac{5}{2}} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{205 \, c^{\frac{5}{2}} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac{170 \, c^{\frac{5}{2}} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac{205 \, c^{\frac{5}{2}} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac{79 \, c^{\frac{5}{2}} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} + \frac{79 \, c^{\frac{5}{2}} \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}}\right )} B}{{\left (a + \frac{a \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{\left (\frac{\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )}^{\frac{5}{2}}}\right )}}{15 \, f} \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))^(5/2)/(a+a*sin(f*x+e)),x, algorithm="maxima")

[Out]

2/15*(5*(23*c^(5/2) + 20*c^(5/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 65*c^(5/2)*sin(f*x + e)^2/(cos(f*x + e) + 1
)^2 + 40*c^(5/2)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 65*c^(5/2)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 20*c^(
5/2)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 23*c^(5/2)*sin(f*x + e)^6/(cos(f*x + e) + 1)^6)*A/((a + a*sin(f*x +
e)/(cos(f*x + e) + 1))*(sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1)^(5/2)) - 2*(79*c^(5/2) + 79*c^(5/2)*sin(f*x
+ e)/(cos(f*x + e) + 1) + 205*c^(5/2)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 170*c^(5/2)*sin(f*x + e)^3/(cos(f*
x + e) + 1)^3 + 205*c^(5/2)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 79*c^(5/2)*sin(f*x + e)^5/(cos(f*x + e) + 1)
^5 + 79*c^(5/2)*sin(f*x + e)^6/(cos(f*x + e) + 1)^6)*B/((a + a*sin(f*x + e)/(cos(f*x + e) + 1))*(sin(f*x + e)^
2/(cos(f*x + e) + 1)^2 + 1)^(5/2)))/f

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Fricas [A]  time = 1.49436, size = 230, normalized size = 1.45 \begin{align*} -\frac{2 \,{\left ({\left (5 \, A - 16 \, B\right )} c^{2} \cos \left (f x + e\right )^{2} + 2 \,{\left (55 \, A - 71 \, B\right )} c^{2} +{\left (3 \, B c^{2} \cos \left (f x + e\right )^{2} + 2 \,{\left (25 \, A - 41 \, B\right )} c^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt{-c \sin \left (f x + e\right ) + c}}{15 \, a 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))*(c-c*sin(f*x+e))^(5/2)/(a+a*sin(f*x+e)),x, algorithm="fricas")

[Out]

-2/15*((5*A - 16*B)*c^2*cos(f*x + e)^2 + 2*(55*A - 71*B)*c^2 + (3*B*c^2*cos(f*x + e)^2 + 2*(25*A - 41*B)*c^2)*
sin(f*x + e))*sqrt(-c*sin(f*x + e) + c)/(a*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))**(5/2)/(a+a*sin(f*x+e)),x)

[Out]

Timed out

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Giac [B]  time = 1.92908, size = 957, normalized size = 6.02 \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))^(5/2)/(a+a*sin(f*x+e)),x, algorithm="giac")

[Out]

-1/60*(2*(120*sqrt(2)*A*c^(19/2) - 120*sqrt(2)*B*c^(19/2) - 25*sqrt(2)*A*a^6*sqrt(c) + 41*sqrt(2)*B*a^6*sqrt(c
) + 50*A*a^6*sqrt(c) - 82*B*a^6*sqrt(c))*sgn(tan(1/2*f*x + 1/2*e) - 1)/(sqrt(2)*a*c^7 - a*c^7) - 960*((sqrt(c)
*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))*A*c^3*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))*B*c^3*sgn(tan(1/2*f*x + 1/2*e) - 1) - A*c^(7/2)*sgn(ta
n(1/2*f*x + 1/2*e) - 1) + B*c^(7/2)*sgn(tan(1/2*f*x + 1/2*e) - 1))/(((sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*ta
n(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) - ((((((55*A*a^5*c^5*sgn(tan(1/2*f*x + 1/2*e) - 1) - 98*B*a^5*c^5*sgn(tan(1/2*f*x + 1/2*e) - 1))*tan(1/
2*f*x + 1/2*e)/c^9 + 15*(3*A*a^5*c^5*sgn(tan(1/2*f*x + 1/2*e) - 1) - 4*B*a^5*c^5*sgn(tan(1/2*f*x + 1/2*e) - 1)
)/c^9)*tan(1/2*f*x + 1/2*e) + 10*(10*A*a^5*c^5*sgn(tan(1/2*f*x + 1/2*e) - 1) - 17*B*a^5*c^5*sgn(tan(1/2*f*x +
1/2*e) - 1))/c^9)*tan(1/2*f*x + 1/2*e) + 10*(10*A*a^5*c^5*sgn(tan(1/2*f*x + 1/2*e) - 1) - 17*B*a^5*c^5*sgn(tan
(1/2*f*x + 1/2*e) - 1))/c^9)*tan(1/2*f*x + 1/2*e) + 15*(3*A*a^5*c^5*sgn(tan(1/2*f*x + 1/2*e) - 1) - 4*B*a^5*c^
5*sgn(tan(1/2*f*x + 1/2*e) - 1))/c^9)*tan(1/2*f*x + 1/2*e) + (55*A*a^5*c^5*sgn(tan(1/2*f*x + 1/2*e) - 1) - 98*
B*a^5*c^5*sgn(tan(1/2*f*x + 1/2*e) - 1))/c^9)/(c*tan(1/2*f*x + 1/2*e)^2 + c)^(5/2))/f