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

Optimal. Leaf size=157 \[ \frac{64 a c^4 (3 A-B) \cos ^3(e+f x)}{315 f (c-c \sin (e+f x))^{3/2}}+\frac{16 a c^3 (3 A-B) \cos ^3(e+f x)}{105 f \sqrt{c-c \sin (e+f x)}}+\frac{2 a c^2 (3 A-B) \cos ^3(e+f x) \sqrt{c-c \sin (e+f x)}}{21 f}-\frac{2 a B c \cos ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{9 f} \]

[Out]

(64*a*(3*A - B)*c^4*Cos[e + f*x]^3)/(315*f*(c - c*Sin[e + f*x])^(3/2)) + (16*a*(3*A - B)*c^3*Cos[e + f*x]^3)/(
105*f*Sqrt[c - c*Sin[e + f*x]]) + (2*a*(3*A - B)*c^2*Cos[e + f*x]^3*Sqrt[c - c*Sin[e + f*x]])/(21*f) - (2*a*B*
c*Cos[e + f*x]^3*(c - c*Sin[e + f*x])^(3/2))/(9*f)

________________________________________________________________________________________

Rubi [A]  time = 0.412332, antiderivative size = 157, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {2967, 2856, 2674, 2673} \[ \frac{64 a c^4 (3 A-B) \cos ^3(e+f x)}{315 f (c-c \sin (e+f x))^{3/2}}+\frac{16 a c^3 (3 A-B) \cos ^3(e+f x)}{105 f \sqrt{c-c \sin (e+f x)}}+\frac{2 a c^2 (3 A-B) \cos ^3(e+f x) \sqrt{c-c \sin (e+f x)}}{21 f}-\frac{2 a B c \cos ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{9 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(64*a*(3*A - B)*c^4*Cos[e + f*x]^3)/(315*f*(c - c*Sin[e + f*x])^(3/2)) + (16*a*(3*A - B)*c^3*Cos[e + f*x]^3)/(
105*f*Sqrt[c - c*Sin[e + f*x]]) + (2*a*(3*A - B)*c^2*Cos[e + f*x]^3*Sqrt[c - c*Sin[e + f*x]])/(21*f) - (2*a*B*
c*Cos[e + f*x]^3*(c - c*Sin[e + f*x])^(3/2))/(9*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 2856

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> -Simp[(d*(g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m)/(f*g*(m + p + 1)), x]
+ Dist[(a*d*m + b*c*(m + p + 1))/(b*(m + p + 1)), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^m, x], x] /; Fre
eQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && IGtQ[Simplify[(2*m + p + 1)/2], 0] && NeQ[m + p + 1
, 0]

Rule 2674

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 + p)), x] + Dist[(a*(2*m + p - 1))/(m + p), Int[(
g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m - 1), x], x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0]
&& IGtQ[Simplify[(2*m + p - 1)/2], 0] && NeQ[m + p, 0]

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 (a+a \sin (e+f x)) (A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2} \, dx &=(a c) \int \cos ^2(e+f x) (A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2} \, dx\\ &=-\frac{2 a B c \cos ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{9 f}+\frac{1}{3} (a (3 A-B) c) \int \cos ^2(e+f x) (c-c \sin (e+f x))^{3/2} \, dx\\ &=\frac{2 a (3 A-B) c^2 \cos ^3(e+f x) \sqrt{c-c \sin (e+f x)}}{21 f}-\frac{2 a B c \cos ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{9 f}+\frac{1}{21} \left (8 a (3 A-B) c^2\right ) \int \cos ^2(e+f x) \sqrt{c-c \sin (e+f x)} \, dx\\ &=\frac{16 a (3 A-B) c^3 \cos ^3(e+f x)}{105 f \sqrt{c-c \sin (e+f x)}}+\frac{2 a (3 A-B) c^2 \cos ^3(e+f x) \sqrt{c-c \sin (e+f x)}}{21 f}-\frac{2 a B c \cos ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{9 f}+\frac{1}{105} \left (32 a (3 A-B) c^3\right ) \int \frac{\cos ^2(e+f x)}{\sqrt{c-c \sin (e+f x)}} \, dx\\ &=\frac{64 a (3 A-B) c^4 \cos ^3(e+f x)}{315 f (c-c \sin (e+f x))^{3/2}}+\frac{16 a (3 A-B) c^3 \cos ^3(e+f x)}{105 f \sqrt{c-c \sin (e+f x)}}+\frac{2 a (3 A-B) c^2 \cos ^3(e+f x) \sqrt{c-c \sin (e+f x)}}{21 f}-\frac{2 a B c \cos ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{9 f}\\ \end{align*}

Mathematica [A]  time = 1.44892, size = 123, normalized size = 0.78 \[ -\frac{a c^2 \sqrt{c-c \sin (e+f x)} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^3 ((648 A-741 B) \sin (e+f x)+30 (3 A-8 B) \cos (2 (e+f x))-942 A+35 B \sin (3 (e+f x))+664 B)}{630 f \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*c^2*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3*Sqrt[c - c*Sin[e + f*x]]*(-942*A + 664*B + 30*(3*A - 8*B)*Cos[
2*(e + f*x)] + (648*A - 741*B)*Sin[e + f*x] + 35*B*Sin[3*(e + f*x)]))/(630*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)
/2]))

________________________________________________________________________________________

Maple [A]  time = 1.018, size = 103, normalized size = 0.7 \begin{align*} -{\frac{ \left ( -2+2\,\sin \left ( fx+e \right ) \right ){c}^{3} \left ( 1+\sin \left ( fx+e \right ) \right ) ^{2}a \left ( -35\,B \left ( \cos \left ( fx+e \right ) \right ) ^{2}\sin \left ( fx+e \right ) + \left ( -162\,A+194\,B \right ) \sin \left ( fx+e \right ) + \left ( -45\,A+120\,B \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}+258\,A-226\,B \right ) }{315\,f\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+a*sin(f*x+e))*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(5/2),x)

[Out]

-2/315*(-1+sin(f*x+e))*c^3*(1+sin(f*x+e))^2*a*(-35*B*cos(f*x+e)^2*sin(f*x+e)+(-162*A+194*B)*sin(f*x+e)+(-45*A+
120*B)*cos(f*x+e)^2+258*A-226*B)/cos(f*x+e)/(c-c*sin(f*x+e))^(1/2)/f

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (B \sin \left (f x + e\right ) + A\right )}{\left (a \sin \left (f x + e\right ) + a\right )}{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac{5}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

integrate((B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)*(-c*sin(f*x + e) + c)^(5/2), x)

________________________________________________________________________________________

Fricas [A]  time = 1.75947, size = 591, normalized size = 3.76 \begin{align*} \frac{2 \,{\left (35 \, B a c^{2} \cos \left (f x + e\right )^{5} + 5 \,{\left (9 \, A - 10 \, B\right )} a c^{2} \cos \left (f x + e\right )^{4} +{\left (117 \, A - 109 \, B\right )} a c^{2} \cos \left (f x + e\right )^{3} - 8 \,{\left (3 \, A - B\right )} a c^{2} \cos \left (f x + e\right )^{2} + 32 \,{\left (3 \, A - B\right )} a c^{2} \cos \left (f x + e\right ) + 64 \,{\left (3 \, A - B\right )} a c^{2} +{\left (35 \, B a c^{2} \cos \left (f x + e\right )^{4} - 5 \,{\left (9 \, A - 17 \, B\right )} a c^{2} \cos \left (f x + e\right )^{3} + 24 \,{\left (3 \, A - B\right )} a c^{2} \cos \left (f x + e\right )^{2} + 32 \,{\left (3 \, A - B\right )} a c^{2} \cos \left (f x + e\right ) + 64 \,{\left (3 \, A - B\right )} a c^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt{-c \sin \left (f x + e\right ) + c}}{315 \,{\left (f \cos \left (f x + e\right ) - f \sin \left (f x + e\right ) + f\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

2/315*(35*B*a*c^2*cos(f*x + e)^5 + 5*(9*A - 10*B)*a*c^2*cos(f*x + e)^4 + (117*A - 109*B)*a*c^2*cos(f*x + e)^3
- 8*(3*A - B)*a*c^2*cos(f*x + e)^2 + 32*(3*A - B)*a*c^2*cos(f*x + e) + 64*(3*A - B)*a*c^2 + (35*B*a*c^2*cos(f*
x + e)^4 - 5*(9*A - 17*B)*a*c^2*cos(f*x + e)^3 + 24*(3*A - B)*a*c^2*cos(f*x + e)^2 + 32*(3*A - B)*a*c^2*cos(f*
x + e) + 64*(3*A - B)*a*c^2)*sin(f*x + e))*sqrt(-c*sin(f*x + e) + c)/(f*cos(f*x + e) - f*sin(f*x + e) + f)

________________________________________________________________________________________

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+a*sin(f*x+e))*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))**(5/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (B \sin \left (f x + e\right ) + A\right )}{\left (a \sin \left (f x + e\right ) + a\right )}{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac{5}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(5/2),x, algorithm="giac")

[Out]

integrate((B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)*(-c*sin(f*x + e) + c)^(5/2), x)