3.286 \(\int \sqrt{a+a \sin (e+f x)} (A+B \sin (e+f x)) (c+d \sin (e+f x))^3 \, dx\)

Optimal. Leaf size=256 \[ \frac{4 a (c+d) \left (15 c^2+10 c d+7 d^2\right ) (-9 A d+B c-8 B d) \cos (e+f x)}{315 d f \sqrt{a \sin (e+f x)+a}}+\frac{2 a (-9 A d+B c-8 B d) \cos (e+f x) (c+d \sin (e+f x))^3}{63 d f \sqrt{a \sin (e+f x)+a}}+\frac{4 d (c+d) (-9 A d+B c-8 B d) \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{105 a f}+\frac{8 (5 c-d) (c+d) (-9 A d+B c-8 B d) \cos (e+f x) \sqrt{a \sin (e+f x)+a}}{315 f}-\frac{2 a B \cos (e+f x) (c+d \sin (e+f x))^4}{9 d f \sqrt{a \sin (e+f x)+a}} \]

[Out]

(4*a*(c + d)*(B*c - 9*A*d - 8*B*d)*(15*c^2 + 10*c*d + 7*d^2)*Cos[e + f*x])/(315*d*f*Sqrt[a + a*Sin[e + f*x]])
+ (8*(5*c - d)*(c + d)*(B*c - 9*A*d - 8*B*d)*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]])/(315*f) + (4*d*(c + d)*(B*
c - 9*A*d - 8*B*d)*Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2))/(105*a*f) + (2*a*(B*c - 9*A*d - 8*B*d)*Cos[e + f*x
]*(c + d*Sin[e + f*x])^3)/(63*d*f*Sqrt[a + a*Sin[e + f*x]]) - (2*a*B*Cos[e + f*x]*(c + d*Sin[e + f*x])^4)/(9*d
*f*Sqrt[a + a*Sin[e + f*x]])

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Rubi [A]  time = 0.459978, antiderivative size = 256, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 37, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.135, Rules used = {2981, 2770, 2761, 2751, 2646} \[ \frac{4 a (c+d) \left (15 c^2+10 c d+7 d^2\right ) (-9 A d+B c-8 B d) \cos (e+f x)}{315 d f \sqrt{a \sin (e+f x)+a}}+\frac{2 a (-9 A d+B c-8 B d) \cos (e+f x) (c+d \sin (e+f x))^3}{63 d f \sqrt{a \sin (e+f x)+a}}+\frac{4 d (c+d) (-9 A d+B c-8 B d) \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{105 a f}+\frac{8 (5 c-d) (c+d) (-9 A d+B c-8 B d) \cos (e+f x) \sqrt{a \sin (e+f x)+a}}{315 f}-\frac{2 a B \cos (e+f x) (c+d \sin (e+f x))^4}{9 d f \sqrt{a \sin (e+f x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + a*Sin[e + f*x]]*(A + B*Sin[e + f*x])*(c + d*Sin[e + f*x])^3,x]

[Out]

(4*a*(c + d)*(B*c - 9*A*d - 8*B*d)*(15*c^2 + 10*c*d + 7*d^2)*Cos[e + f*x])/(315*d*f*Sqrt[a + a*Sin[e + f*x]])
+ (8*(5*c - d)*(c + d)*(B*c - 9*A*d - 8*B*d)*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]])/(315*f) + (4*d*(c + d)*(B*
c - 9*A*d - 8*B*d)*Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2))/(105*a*f) + (2*a*(B*c - 9*A*d - 8*B*d)*Cos[e + f*x
]*(c + d*Sin[e + f*x])^3)/(63*d*f*Sqrt[a + a*Sin[e + f*x]]) - (2*a*B*Cos[e + f*x]*(c + d*Sin[e + f*x])^4)/(9*d
*f*Sqrt[a + a*Sin[e + f*x]])

Rule 2981

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-2*b*B*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(2*n + 3)*Sqr
t[a + b*Sin[e + f*x]]), x] + Dist[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(b*d*(2*n + 3)), Int[Sqrt[a + b*
Sin[e + f*x]]*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] &&
EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &&  !LtQ[n, -1]

Rule 2770

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp
[(-2*b*Cos[e + f*x]*(c + d*Sin[e + f*x])^n)/(f*(2*n + 1)*Sqrt[a + b*Sin[e + f*x]]), x] + Dist[(2*n*(b*c + a*d)
)/(b*(2*n + 1)), Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d, e, f}
, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[n, 0] && IntegerQ[2*n]

Rule 2761

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> -Simp[(
d^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*(m + 2)), Int[(a + b*Sin[e + f*x]
)^m*Simp[b*(d^2*(m + 1) + c^2*(m + 2)) - d*(a*d - 2*b*c*(m + 2))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d
, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, -1]

Rule 2751

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[(d
*Cos[e + f*x]*(a + b*Sin[e + f*x])^m)/(f*(m + 1)), x] + Dist[(a*d*m + b*c*(m + 1))/(b*(m + 1)), Int[(a + b*Sin
[e + f*x])^m, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] &&  !LtQ[m,
-2^(-1)]

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 \sqrt{a+a \sin (e+f x)} (A+B \sin (e+f x)) (c+d \sin (e+f x))^3 \, dx &=-\frac{2 a B \cos (e+f x) (c+d \sin (e+f x))^4}{9 d f \sqrt{a+a \sin (e+f x)}}+\frac{(9 a A d-B (a c-8 a d)) \int \sqrt{a+a \sin (e+f x)} (c+d \sin (e+f x))^3 \, dx}{9 a d}\\ &=\frac{2 a (B c-9 A d-8 B d) \cos (e+f x) (c+d \sin (e+f x))^3}{63 d f \sqrt{a+a \sin (e+f x)}}-\frac{2 a B \cos (e+f x) (c+d \sin (e+f x))^4}{9 d f \sqrt{a+a \sin (e+f x)}}+\frac{(2 (c+d) (9 a A d-B (a c-8 a d))) \int \sqrt{a+a \sin (e+f x)} (c+d \sin (e+f x))^2 \, dx}{21 a d}\\ &=\frac{4 d (c+d) (B c-9 A d-8 B d) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{105 a f}+\frac{2 a (B c-9 A d-8 B d) \cos (e+f x) (c+d \sin (e+f x))^3}{63 d f \sqrt{a+a \sin (e+f x)}}-\frac{2 a B \cos (e+f x) (c+d \sin (e+f x))^4}{9 d f \sqrt{a+a \sin (e+f x)}}+\frac{(4 (c+d) (9 a A d-B (a c-8 a d))) \int \sqrt{a+a \sin (e+f x)} \left (\frac{1}{2} a \left (5 c^2+3 d^2\right )+a (5 c-d) d \sin (e+f x)\right ) \, dx}{105 a^2 d}\\ &=\frac{8 (5 c-d) (c+d) (B c-9 A d-8 B d) \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{315 f}+\frac{4 d (c+d) (B c-9 A d-8 B d) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{105 a f}+\frac{2 a (B c-9 A d-8 B d) \cos (e+f x) (c+d \sin (e+f x))^3}{63 d f \sqrt{a+a \sin (e+f x)}}-\frac{2 a B \cos (e+f x) (c+d \sin (e+f x))^4}{9 d f \sqrt{a+a \sin (e+f x)}}+\frac{\left (2 (c+d) \left (15 c^2+10 c d+7 d^2\right ) (9 a A d-B (a c-8 a d))\right ) \int \sqrt{a+a \sin (e+f x)} \, dx}{315 a d}\\ &=\frac{4 a (c+d) (B c-9 A d-8 B d) \left (15 c^2+10 c d+7 d^2\right ) \cos (e+f x)}{315 d f \sqrt{a+a \sin (e+f x)}}+\frac{8 (5 c-d) (c+d) (B c-9 A d-8 B d) \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{315 f}+\frac{4 d (c+d) (B c-9 A d-8 B d) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{105 a f}+\frac{2 a (B c-9 A d-8 B d) \cos (e+f x) (c+d \sin (e+f x))^3}{63 d f \sqrt{a+a \sin (e+f x)}}-\frac{2 a B \cos (e+f x) (c+d \sin (e+f x))^4}{9 d f \sqrt{a+a \sin (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 1.27302, size = 305, normalized size = 1.19 \[ -\frac{\sqrt{a (\sin (e+f x)+1)} \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (-4 d \left (27 A d (7 c+2 d)+B \left (189 c^2+162 c d+83 d^2\right )\right ) \cos (2 (e+f x))+2520 A c^2 d \sin (e+f x)+5040 A c^2 d+2520 A c^3+2016 A c d^2 \sin (e+f x)+4788 A c d^2+846 A d^3 \sin (e+f x)-90 A d^3 \sin (3 (e+f x))+1368 A d^3+2016 B c^2 d \sin (e+f x)+4788 B c^2 d+840 B c^3 \sin (e+f x)+1680 B c^3+2538 B c d^2 \sin (e+f x)-270 B c d^2 \sin (3 (e+f x))+4104 B c d^2+752 B d^3 \sin (e+f x)-80 B d^3 \sin (3 (e+f x))+35 B d^3 \cos (4 (e+f x))+1321 B d^3\right )}{1260 f \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + a*Sin[e + f*x]]*(A + B*Sin[e + f*x])*(c + d*Sin[e + f*x])^3,x]

[Out]

-((Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*Sqrt[a*(1 + Sin[e + f*x])]*(2520*A*c^3 + 1680*B*c^3 + 5040*A*c^2*d + 4
788*B*c^2*d + 4788*A*c*d^2 + 4104*B*c*d^2 + 1368*A*d^3 + 1321*B*d^3 - 4*d*(27*A*d*(7*c + 2*d) + B*(189*c^2 + 1
62*c*d + 83*d^2))*Cos[2*(e + f*x)] + 35*B*d^3*Cos[4*(e + f*x)] + 840*B*c^3*Sin[e + f*x] + 2520*A*c^2*d*Sin[e +
f*x] + 2016*B*c^2*d*Sin[e + f*x] + 2016*A*c*d^2*Sin[e + f*x] + 2538*B*c*d^2*Sin[e + f*x] + 846*A*d^3*Sin[e +
f*x] + 752*B*d^3*Sin[e + f*x] - 270*B*c*d^2*Sin[3*(e + f*x)] - 90*A*d^3*Sin[3*(e + f*x)] - 80*B*d^3*Sin[3*(e +
f*x)]))/(1260*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]))

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Maple [A]  time = 1.207, size = 242, normalized size = 1. \begin{align*}{\frac{ \left ( 2+2\,\sin \left ( fx+e \right ) \right ) a \left ( -1+\sin \left ( fx+e \right ) \right ) \left ( \left ( -45\,A{d}^{3}-135\,Bc{d}^{2}-40\,B{d}^{3} \right ) \sin \left ( fx+e \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}+ \left ( 315\,A{c}^{2}d+252\,Ac{d}^{2}+117\,A{d}^{3}+105\,B{c}^{3}+252\,B{c}^{2}d+351\,Bc{d}^{2}+104\,B{d}^{3} \right ) \sin \left ( fx+e \right ) +35\,B \left ( \cos \left ( fx+e \right ) \right ) ^{4}{d}^{3}+ \left ( -189\,Ac{d}^{2}-54\,A{d}^{3}-189\,B{c}^{2}d-162\,Bc{d}^{2}-118\,B{d}^{3} \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}+315\,A{c}^{3}+630\,A{c}^{2}d+693\,Ac{d}^{2}+198\,A{d}^{3}+210\,B{c}^{3}+693\,B{c}^{2}d+594\,Bc{d}^{2}+211\,B{d}^{3} \right ) }{315\,f\cos \left ( fx+e \right ) }{\frac{1}{\sqrt{a+a\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+d*sin(f*x+e))^3*(a+a*sin(f*x+e))^(1/2),x)

[Out]

2/315*(1+sin(f*x+e))*a*(-1+sin(f*x+e))*((-45*A*d^3-135*B*c*d^2-40*B*d^3)*sin(f*x+e)*cos(f*x+e)^2+(315*A*c^2*d+
252*A*c*d^2+117*A*d^3+105*B*c^3+252*B*c^2*d+351*B*c*d^2+104*B*d^3)*sin(f*x+e)+35*B*cos(f*x+e)^4*d^3+(-189*A*c*
d^2-54*A*d^3-189*B*c^2*d-162*B*c*d^2-118*B*d^3)*cos(f*x+e)^2+315*A*c^3+630*A*c^2*d+693*A*c*d^2+198*A*d^3+210*B
*c^3+693*B*c^2*d+594*B*c*d^2+211*B*d^3)/cos(f*x+e)/(a+a*sin(f*x+e))^(1/2)/f

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (B \sin \left (f x + e\right ) + A\right )} \sqrt{a \sin \left (f x + e\right ) + a}{\left (d \sin \left (f x + e\right ) + c\right )}^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sin(f*x+e))*(c+d*sin(f*x+e))^3*(a+a*sin(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

integrate((B*sin(f*x + e) + A)*sqrt(a*sin(f*x + e) + a)*(d*sin(f*x + e) + c)^3, x)

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Fricas [A]  time = 2.16706, size = 1156, normalized size = 4.52 \begin{align*} -\frac{2 \,{\left (35 \, B d^{3} \cos \left (f x + e\right )^{5} - 5 \,{\left (27 \, B c d^{2} +{\left (9 \, A + B\right )} d^{3}\right )} \cos \left (f x + e\right )^{4} + 105 \,{\left (3 \, A + B\right )} c^{3} + 63 \,{\left (5 \, A + 7 \, B\right )} c^{2} d + 9 \,{\left (49 \, A + 27 \, B\right )} c d^{2} +{\left (81 \, A + 107 \, B\right )} d^{3} -{\left (189 \, B c^{2} d + 27 \,{\left (7 \, A + 6 \, B\right )} c d^{2} + 2 \,{\left (27 \, A + 59 \, B\right )} d^{3}\right )} \cos \left (f x + e\right )^{3} +{\left (105 \, B c^{3} + 63 \,{\left (5 \, A + B\right )} c^{2} d + 9 \,{\left (7 \, A + 36 \, B\right )} c d^{2} + 2 \,{\left (54 \, A + 13 \, B\right )} d^{3}\right )} \cos \left (f x + e\right )^{2} +{\left (105 \,{\left (3 \, A + 2 \, B\right )} c^{3} + 63 \,{\left (10 \, A + 11 \, B\right )} c^{2} d + 99 \,{\left (7 \, A + 6 \, B\right )} c d^{2} +{\left (198 \, A + 211 \, B\right )} d^{3}\right )} \cos \left (f x + e\right ) -{\left (35 \, B d^{3} \cos \left (f x + e\right )^{4} + 105 \,{\left (3 \, A + B\right )} c^{3} + 63 \,{\left (5 \, A + 7 \, B\right )} c^{2} d + 9 \,{\left (49 \, A + 27 \, B\right )} c d^{2} +{\left (81 \, A + 107 \, B\right )} d^{3} + 5 \,{\left (27 \, B c d^{2} +{\left (9 \, A + 8 \, B\right )} d^{3}\right )} \cos \left (f x + e\right )^{3} - 3 \,{\left (63 \, B c^{2} d + 9 \,{\left (7 \, A + B\right )} c d^{2} +{\left (3 \, A + 26 \, B\right )} d^{3}\right )} \cos \left (f x + e\right )^{2} -{\left (105 \, B c^{3} + 63 \,{\left (5 \, A + 4 \, B\right )} c^{2} d + 9 \,{\left (28 \, A + 39 \, B\right )} c d^{2} + 13 \,{\left (9 \, A + 8 \, B\right )} d^{3}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )} \sqrt{a \sin \left (f x + e\right ) + a}}{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+B*sin(f*x+e))*(c+d*sin(f*x+e))^3*(a+a*sin(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

-2/315*(35*B*d^3*cos(f*x + e)^5 - 5*(27*B*c*d^2 + (9*A + B)*d^3)*cos(f*x + e)^4 + 105*(3*A + B)*c^3 + 63*(5*A
+ 7*B)*c^2*d + 9*(49*A + 27*B)*c*d^2 + (81*A + 107*B)*d^3 - (189*B*c^2*d + 27*(7*A + 6*B)*c*d^2 + 2*(27*A + 59
*B)*d^3)*cos(f*x + e)^3 + (105*B*c^3 + 63*(5*A + B)*c^2*d + 9*(7*A + 36*B)*c*d^2 + 2*(54*A + 13*B)*d^3)*cos(f*
x + e)^2 + (105*(3*A + 2*B)*c^3 + 63*(10*A + 11*B)*c^2*d + 99*(7*A + 6*B)*c*d^2 + (198*A + 211*B)*d^3)*cos(f*x
+ e) - (35*B*d^3*cos(f*x + e)^4 + 105*(3*A + B)*c^3 + 63*(5*A + 7*B)*c^2*d + 9*(49*A + 27*B)*c*d^2 + (81*A +
107*B)*d^3 + 5*(27*B*c*d^2 + (9*A + 8*B)*d^3)*cos(f*x + e)^3 - 3*(63*B*c^2*d + 9*(7*A + B)*c*d^2 + (3*A + 26*B
)*d^3)*cos(f*x + e)^2 - (105*B*c^3 + 63*(5*A + 4*B)*c^2*d + 9*(28*A + 39*B)*c*d^2 + 13*(9*A + 8*B)*d^3)*cos(f*
x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)/(f*cos(f*x + e) + f*sin(f*x + e) + f)

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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+d*sin(f*x+e))**3*(a+a*sin(f*x+e))**(1/2),x)

[Out]

Timed out

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Giac [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+d*sin(f*x+e))^3*(a+a*sin(f*x+e))^(1/2),x, algorithm="giac")

[Out]

Timed out