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

Optimal. Leaf size=212 \[ -\frac{16 a^2 (21 A c+15 A d+15 B c+13 B d) \cos (e+f x) \sqrt{a \sin (e+f x)+a}}{315 f}-\frac{64 a^3 (21 A c+15 A d+15 B c+13 B d) \cos (e+f x)}{315 f \sqrt{a \sin (e+f x)+a}}-\frac{2 (9 A d+9 B c-2 B d) \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{63 f}-\frac{2 a (21 A c+15 A d+15 B c+13 B d) \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{105 f}-\frac{2 B d \cos (e+f x) (a \sin (e+f x)+a)^{7/2}}{9 a f} \]

[Out]

(-64*a^3*(21*A*c + 15*B*c + 15*A*d + 13*B*d)*Cos[e + f*x])/(315*f*Sqrt[a + a*Sin[e + f*x]]) - (16*a^2*(21*A*c
+ 15*B*c + 15*A*d + 13*B*d)*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]])/(315*f) - (2*a*(21*A*c + 15*B*c + 15*A*d +
13*B*d)*Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2))/(105*f) - (2*(9*B*c + 9*A*d - 2*B*d)*Cos[e + f*x]*(a + a*Sin[
e + f*x])^(5/2))/(63*f) - (2*B*d*Cos[e + f*x]*(a + a*Sin[e + f*x])^(7/2))/(9*a*f)

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Rubi [A]  time = 0.367651, antiderivative size = 212, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {2968, 3023, 2751, 2647, 2646} \[ -\frac{16 a^2 (21 A c+15 A d+15 B c+13 B d) \cos (e+f x) \sqrt{a \sin (e+f x)+a}}{315 f}-\frac{64 a^3 (21 A c+15 A d+15 B c+13 B d) \cos (e+f x)}{315 f \sqrt{a \sin (e+f x)+a}}-\frac{2 (9 A d+9 B c-2 B d) \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{63 f}-\frac{2 a (21 A c+15 A d+15 B c+13 B d) \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{105 f}-\frac{2 B d \cos (e+f x) (a \sin (e+f x)+a)^{7/2}}{9 a f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-64*a^3*(21*A*c + 15*B*c + 15*A*d + 13*B*d)*Cos[e + f*x])/(315*f*Sqrt[a + a*Sin[e + f*x]]) - (16*a^2*(21*A*c
+ 15*B*c + 15*A*d + 13*B*d)*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]])/(315*f) - (2*a*(21*A*c + 15*B*c + 15*A*d +
13*B*d)*Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2))/(105*f) - (2*(9*B*c + 9*A*d - 2*B*d)*Cos[e + f*x]*(a + a*Sin[
e + f*x])^(5/2))/(63*f) - (2*B*d*Cos[e + f*x]*(a + a*Sin[e + f*x])^(7/2))/(9*a*f)

Rule 2968

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x
]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]

Rule 3023

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*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[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !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 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 (a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x)) (c+d \sin (e+f x)) \, dx &=\int (a+a \sin (e+f x))^{5/2} \left (A c+(B c+A d) \sin (e+f x)+B d \sin ^2(e+f x)\right ) \, dx\\ &=-\frac{2 B d \cos (e+f x) (a+a \sin (e+f x))^{7/2}}{9 a f}+\frac{2 \int (a+a \sin (e+f x))^{5/2} \left (\frac{1}{2} a (9 A c+7 B d)+\frac{1}{2} a (9 B c+9 A d-2 B d) \sin (e+f x)\right ) \, dx}{9 a}\\ &=-\frac{2 (9 B c+9 A d-2 B d) \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{63 f}-\frac{2 B d \cos (e+f x) (a+a \sin (e+f x))^{7/2}}{9 a f}+\frac{1}{21} (21 A c+15 B c+15 A d+13 B d) \int (a+a \sin (e+f x))^{5/2} \, dx\\ &=-\frac{2 a (21 A c+15 B c+15 A d+13 B d) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{105 f}-\frac{2 (9 B c+9 A d-2 B d) \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{63 f}-\frac{2 B d \cos (e+f x) (a+a \sin (e+f x))^{7/2}}{9 a f}+\frac{1}{105} (8 a (21 A c+15 B c+15 A d+13 B d)) \int (a+a \sin (e+f x))^{3/2} \, dx\\ &=-\frac{16 a^2 (21 A c+15 B c+15 A d+13 B d) \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{315 f}-\frac{2 a (21 A c+15 B c+15 A d+13 B d) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{105 f}-\frac{2 (9 B c+9 A d-2 B d) \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{63 f}-\frac{2 B d \cos (e+f x) (a+a \sin (e+f x))^{7/2}}{9 a f}+\frac{1}{315} \left (32 a^2 (21 A c+15 B c+15 A d+13 B d)\right ) \int \sqrt{a+a \sin (e+f x)} \, dx\\ &=-\frac{64 a^3 (21 A c+15 B c+15 A d+13 B d) \cos (e+f x)}{315 f \sqrt{a+a \sin (e+f x)}}-\frac{16 a^2 (21 A c+15 B c+15 A d+13 B d) \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{315 f}-\frac{2 a (21 A c+15 B c+15 A d+13 B d) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{105 f}-\frac{2 (9 B c+9 A d-2 B d) \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{63 f}-\frac{2 B d \cos (e+f x) (a+a \sin (e+f x))^{7/2}}{9 a f}\\ \end{align*}

Mathematica [A]  time = 4.20475, size = 202, normalized size = 0.95 \[ -\frac{a^2 \sqrt{a (\sin (e+f x)+1)} \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) (-4 (63 A c+180 A d+180 B c+254 B d) \cos (2 (e+f x))+2352 A c \sin (e+f x)+7476 A c+3030 A d \sin (e+f x)-90 A d \sin (3 (e+f x))+6240 A d+3030 B c \sin (e+f x)-90 B c \sin (3 (e+f x))+6240 B c+3116 B d \sin (e+f x)-260 B d \sin (3 (e+f x))+35 B d \cos (4 (e+f x))+5653 B d)}{1260 f \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^2*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*Sqrt[a*(1 + Sin[e + f*x])]*(7476*A*c + 6240*B*c + 6240*A*d + 5653*
B*d - 4*(63*A*c + 180*B*c + 180*A*d + 254*B*d)*Cos[2*(e + f*x)] + 35*B*d*Cos[4*(e + f*x)] + 2352*A*c*Sin[e + f
*x] + 3030*B*c*Sin[e + f*x] + 3030*A*d*Sin[e + f*x] + 3116*B*d*Sin[e + f*x] - 90*B*c*Sin[3*(e + f*x)] - 90*A*d
*Sin[3*(e + f*x)] - 260*B*d*Sin[3*(e + f*x)]))/(1260*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]))

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Maple [A]  time = 1.167, size = 152, normalized size = 0.7 \begin{align*}{\frac{ \left ( 2+2\,\sin \left ( fx+e \right ) \right ){a}^{3} \left ( -1+\sin \left ( fx+e \right ) \right ) \left ( \left ( -45\,Ad-45\,Bc-130\,Bd \right ) \sin \left ( fx+e \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}+ \left ( 294\,Ac+390\,Ad+390\,Bc+422\,Bd \right ) \sin \left ( fx+e \right ) +35\,Bd \left ( \cos \left ( fx+e \right ) \right ) ^{4}+ \left ( -63\,Ac-180\,Ad-180\,Bc-289\,Bd \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}+966\,Ac+870\,Ad+870\,Bc+838\,Bd \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+a*sin(f*x+e))^(5/2)*(A+B*sin(f*x+e))*(c+d*sin(f*x+e)),x)

[Out]

2/315*(1+sin(f*x+e))*a^3*(-1+sin(f*x+e))*((-45*A*d-45*B*c-130*B*d)*sin(f*x+e)*cos(f*x+e)^2+(294*A*c+390*A*d+39
0*B*c+422*B*d)*sin(f*x+e)+35*B*d*cos(f*x+e)^4+(-63*A*c-180*A*d-180*B*c-289*B*d)*cos(f*x+e)^2+966*A*c+870*A*d+8
70*B*c+838*B*d)/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 )}{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac{5}{2}}{\left (d \sin \left (f x + e\right ) + c\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.76944, size = 900, normalized size = 4.25 \begin{align*} -\frac{2 \,{\left (35 \, B a^{2} d \cos \left (f x + e\right )^{5} - 5 \,{\left (9 \, B a^{2} c +{\left (9 \, A + 19 \, B\right )} a^{2} d\right )} \cos \left (f x + e\right )^{4} + 96 \,{\left (7 \, A + 5 \, B\right )} a^{2} c + 32 \,{\left (15 \, A + 13 \, B\right )} a^{2} d -{\left (9 \,{\left (7 \, A + 20 \, B\right )} a^{2} c +{\left (180 \, A + 289 \, B\right )} a^{2} d\right )} \cos \left (f x + e\right )^{3} +{\left (3 \,{\left (77 \, A + 85 \, B\right )} a^{2} c +{\left (255 \, A + 263 \, B\right )} a^{2} d\right )} \cos \left (f x + e\right )^{2} + 2 \,{\left (3 \,{\left (161 \, A + 145 \, B\right )} a^{2} c +{\left (435 \, A + 419 \, B\right )} a^{2} d\right )} \cos \left (f x + e\right ) -{\left (35 \, B a^{2} d \cos \left (f x + e\right )^{4} + 96 \,{\left (7 \, A + 5 \, B\right )} a^{2} c + 32 \,{\left (15 \, A + 13 \, B\right )} a^{2} d + 5 \,{\left (9 \, B a^{2} c +{\left (9 \, A + 26 \, B\right )} a^{2} d\right )} \cos \left (f x + e\right )^{3} - 3 \,{\left (3 \,{\left (7 \, A + 15 \, B\right )} a^{2} c +{\left (45 \, A + 53 \, B\right )} a^{2} d\right )} \cos \left (f x + e\right )^{2} - 2 \,{\left (3 \,{\left (49 \, A + 65 \, B\right )} a^{2} c +{\left (195 \, A + 211 \, B\right )} a^{2} d\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+a*sin(f*x+e))^(5/2)*(A+B*sin(f*x+e))*(c+d*sin(f*x+e)),x, algorithm="fricas")

[Out]

-2/315*(35*B*a^2*d*cos(f*x + e)^5 - 5*(9*B*a^2*c + (9*A + 19*B)*a^2*d)*cos(f*x + e)^4 + 96*(7*A + 5*B)*a^2*c +
32*(15*A + 13*B)*a^2*d - (9*(7*A + 20*B)*a^2*c + (180*A + 289*B)*a^2*d)*cos(f*x + e)^3 + (3*(77*A + 85*B)*a^2
*c + (255*A + 263*B)*a^2*d)*cos(f*x + e)^2 + 2*(3*(161*A + 145*B)*a^2*c + (435*A + 419*B)*a^2*d)*cos(f*x + e)
- (35*B*a^2*d*cos(f*x + e)^4 + 96*(7*A + 5*B)*a^2*c + 32*(15*A + 13*B)*a^2*d + 5*(9*B*a^2*c + (9*A + 26*B)*a^2
*d)*cos(f*x + e)^3 - 3*(3*(7*A + 15*B)*a^2*c + (45*A + 53*B)*a^2*d)*cos(f*x + e)^2 - 2*(3*(49*A + 65*B)*a^2*c
+ (195*A + 211*B)*a^2*d)*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+a*sin(f*x+e))**(5/2)*(A+B*sin(f*x+e))*(c+d*sin(f*x+e)),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+a*sin(f*x+e))^(5/2)*(A+B*sin(f*x+e))*(c+d*sin(f*x+e)),x, algorithm="giac")

[Out]

Timed out