3.110 \(\int \frac{(A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2}}{a+a \sin (e+f x)} \, dx\)

Optimal. Leaf size=118 \[ -\frac{4 c^2 (3 A-5 B) \cos (e+f x)}{3 a f \sqrt{c-c \sin (e+f x)}}-\frac{c (3 A-5 B) \cos (e+f x) \sqrt{c-c \sin (e+f x)}}{3 a f}-\frac{(A-B) \sec (e+f x) (c-c \sin (e+f x))^{5/2}}{a c f} \]

[Out]

(-4*(3*A - 5*B)*c^2*Cos[e + f*x])/(3*a*f*Sqrt[c - c*Sin[e + f*x]]) - ((3*A - 5*B)*c*Cos[e + f*x]*Sqrt[c - c*Si
n[e + f*x]])/(3*a*f) - ((A - B)*Sec[e + f*x]*(c - c*Sin[e + f*x])^(5/2))/(a*c*f)

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Rubi [A]  time = 0.316923, antiderivative size = 118, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {2967, 2855, 2647, 2646} \[ -\frac{4 c^2 (3 A-5 B) \cos (e+f x)}{3 a f \sqrt{c-c \sin (e+f x)}}-\frac{c (3 A-5 B) \cos (e+f x) \sqrt{c-c \sin (e+f x)}}{3 a f}-\frac{(A-B) \sec (e+f x) (c-c \sin (e+f x))^{5/2}}{a c f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*(3*A - 5*B)*c^2*Cos[e + f*x])/(3*a*f*Sqrt[c - c*Sin[e + f*x]]) - ((3*A - 5*B)*c*Cos[e + f*x]*Sqrt[c - c*Si
n[e + f*x]])/(3*a*f) - ((A - B)*Sec[e + f*x]*(c - c*Sin[e + f*x])^(5/2))/(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))^{3/2}}{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))^{5/2} \, dx}{a c}\\ &=-\frac{(A-B) \sec (e+f x) (c-c \sin (e+f x))^{5/2}}{a c f}-\frac{(3 A-5 B) \int (c-c \sin (e+f x))^{3/2} \, dx}{2 a}\\ &=-\frac{(3 A-5 B) c \cos (e+f x) \sqrt{c-c \sin (e+f x)}}{3 a f}-\frac{(A-B) \sec (e+f x) (c-c \sin (e+f x))^{5/2}}{a c f}-\frac{(2 (3 A-5 B) c) \int \sqrt{c-c \sin (e+f x)} \, dx}{3 a}\\ &=-\frac{4 (3 A-5 B) c^2 \cos (e+f x)}{3 a f \sqrt{c-c \sin (e+f x)}}-\frac{(3 A-5 B) c \cos (e+f x) \sqrt{c-c \sin (e+f x)}}{3 a f}-\frac{(A-B) \sec (e+f x) (c-c \sin (e+f x))^{5/2}}{a c f}\\ \end{align*}

Mathematica [A]  time = 0.636489, size = 113, normalized size = 0.96 \[ \frac{c \sqrt{c-c \sin (e+f x)} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) ((14 B-6 A) \sin (e+f x)-18 A+B \cos (2 (e+f x))+27 B)}{3 a f (\sin (e+f x)+1) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(-18*A + 27*B + B*Cos[2*(e + f*x)] + (-6*A + 14*B)*Sin[e + f*x])*Sqrt
[c - c*Sin[e + f*x]])/(3*a*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(1 + Sin[e + f*x]))

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Maple [A]  time = 0.811, size = 73, normalized size = 0.6 \begin{align*}{\frac{2\,{c}^{2} \left ( -1+\sin \left ( fx+e \right ) \right ) \left ( \sin \left ( fx+e \right ) \left ( 3\,A-7\,B \right ) -B \left ( \cos \left ( fx+e \right ) \right ) ^{2}+9\,A-13\,B \right ) }{3\,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))^(3/2)/(a+a*sin(f*x+e)),x)

[Out]

2/3*c^2/a*(-1+sin(f*x+e))*(sin(f*x+e)*(3*A-7*B)-B*cos(f*x+e)^2+9*A-13*B)/cos(f*x+e)/(c-c*sin(f*x+e))^(1/2)/f

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Maxima [B]  time = 1.52605, size = 397, normalized size = 3.36 \begin{align*} \frac{2 \,{\left (\frac{3 \,{\left (3 \, c^{\frac{3}{2}} + \frac{2 \, c^{\frac{3}{2}} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{6 \, c^{\frac{3}{2}} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac{2 \, c^{\frac{3}{2}} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac{3 \, c^{\frac{3}{2}} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}}\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{3}{2}}} - \frac{2 \,{\left (7 \, c^{\frac{3}{2}} + \frac{7 \, c^{\frac{3}{2}} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{12 \, c^{\frac{3}{2}} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac{7 \, c^{\frac{3}{2}} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac{7 \, c^{\frac{3}{2}} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}}\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{3}{2}}}\right )}}{3 \, 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))^(3/2)/(a+a*sin(f*x+e)),x, algorithm="maxima")

[Out]

2/3*(3*(3*c^(3/2) + 2*c^(3/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 6*c^(3/2)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2
+ 2*c^(3/2)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 3*c^(3/2)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4)*A/((a + a*sin
(f*x + e)/(cos(f*x + e) + 1))*(sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1)^(3/2)) - 2*(7*c^(3/2) + 7*c^(3/2)*sin(
f*x + e)/(cos(f*x + e) + 1) + 12*c^(3/2)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 7*c^(3/2)*sin(f*x + e)^3/(cos(f
*x + e) + 1)^3 + 7*c^(3/2)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4)*B/((a + a*sin(f*x + e)/(cos(f*x + e) + 1))*(si
n(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1)^(3/2)))/f

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Fricas [A]  time = 1.39912, size = 158, normalized size = 1.34 \begin{align*} \frac{2 \,{\left (B c \cos \left (f x + e\right )^{2} -{\left (3 \, A - 7 \, B\right )} c \sin \left (f x + e\right ) -{\left (9 \, A - 13 \, B\right )} c\right )} \sqrt{-c \sin \left (f x + e\right ) + c}}{3 \, 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))^(3/2)/(a+a*sin(f*x+e)),x, algorithm="fricas")

[Out]

2/3*(B*c*cos(f*x + e)^2 - (3*A - 7*B)*c*sin(f*x + e) - (9*A - 13*B)*c)*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))**(3/2)/(a+a*sin(f*x+e)),x)

[Out]

Timed out

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Giac [B]  time = 1.59174, size = 786, normalized size = 6.66 \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))^(3/2)/(a+a*sin(f*x+e)),x, algorithm="giac")

[Out]

-1/3*((6*sqrt(2)*A*c^(13/2) - 6*sqrt(2)*B*c^(13/2) - 3*sqrt(2)*A*a^4*sqrt(c) + 7*sqrt(2)*B*a^4*sqrt(c) + 6*A*a
^4*sqrt(c) - 14*B*a^4*sqrt(c))*sgn(tan(1/2*f*x + 1/2*e) - 1)/(sqrt(2)*a*c^5 - a*c^5) - ((((3*A*a^3*c^3*sgn(tan
(1/2*f*x + 1/2*e) - 1) - 8*B*a^3*c^3*sgn(tan(1/2*f*x + 1/2*e) - 1))*tan(1/2*f*x + 1/2*e)/c^6 + 3*(A*a^3*c^3*sg
n(tan(1/2*f*x + 1/2*e) - 1) - 2*B*a^3*c^3*sgn(tan(1/2*f*x + 1/2*e) - 1))/c^6)*tan(1/2*f*x + 1/2*e) + 3*(A*a^3*
c^3*sgn(tan(1/2*f*x + 1/2*e) - 1) - 2*B*a^3*c^3*sgn(tan(1/2*f*x + 1/2*e) - 1))/c^6)*tan(1/2*f*x + 1/2*e) + (3*
A*a^3*c^3*sgn(tan(1/2*f*x + 1/2*e) - 1) - 8*B*a^3*c^3*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) - 24*((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*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^2*sgn(tan(1/2*f*x +
1/2*e) - 1) - A*c^(5/2)*sgn(tan(1/2*f*x + 1/2*e) - 1) + B*c^(5/2)*sgn(tan(1/2*f*x + 1/2*e) - 1))/(((sqrt(c)*t
an(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)*a))/f