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

Optimal. Leaf size=165 \[ -\frac{8 a^2 (35 A c+21 A d+21 B c+19 B d) \cos (e+f x)}{105 f \sqrt{a \sin (e+f x)+a}}-\frac{2 (7 A d+7 B c-2 B d) \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{35 f}-\frac{2 a (35 A c+21 A d+21 B c+19 B d) \cos (e+f x) \sqrt{a \sin (e+f x)+a}}{105 f}-\frac{2 B d \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{7 a f} \]

[Out]

(-8*a^2*(35*A*c + 21*B*c + 21*A*d + 19*B*d)*Cos[e + f*x])/(105*f*Sqrt[a + a*Sin[e + f*x]]) - (2*a*(35*A*c + 21
*B*c + 21*A*d + 19*B*d)*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]])/(105*f) - (2*(7*B*c + 7*A*d - 2*B*d)*Cos[e + f*
x]*(a + a*Sin[e + f*x])^(3/2))/(35*f) - (2*B*d*Cos[e + f*x]*(a + a*Sin[e + f*x])^(5/2))/(7*a*f)

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Rubi [A]  time = 0.315633, antiderivative size = 165, normalized size of antiderivative = 1., number of steps used = 5, 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} \[ -\frac{8 a^2 (35 A c+21 A d+21 B c+19 B d) \cos (e+f x)}{105 f \sqrt{a \sin (e+f x)+a}}-\frac{2 (7 A d+7 B c-2 B d) \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{35 f}-\frac{2 a (35 A c+21 A d+21 B c+19 B d) \cos (e+f x) \sqrt{a \sin (e+f x)+a}}{105 f}-\frac{2 B d \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{7 a f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-8*a^2*(35*A*c + 21*B*c + 21*A*d + 19*B*d)*Cos[e + f*x])/(105*f*Sqrt[a + a*Sin[e + f*x]]) - (2*a*(35*A*c + 21
*B*c + 21*A*d + 19*B*d)*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]])/(105*f) - (2*(7*B*c + 7*A*d - 2*B*d)*Cos[e + f*
x]*(a + a*Sin[e + f*x])^(3/2))/(35*f) - (2*B*d*Cos[e + f*x]*(a + a*Sin[e + f*x])^(5/2))/(7*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))^{3/2} (A+B \sin (e+f x)) (c+d \sin (e+f x)) \, dx &=\int (a+a \sin (e+f x))^{3/2} \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))^{5/2}}{7 a f}+\frac{2 \int (a+a \sin (e+f x))^{3/2} \left (\frac{1}{2} a (7 A c+5 B d)+\frac{1}{2} a (7 B c+7 A d-2 B d) \sin (e+f x)\right ) \, dx}{7 a}\\ &=-\frac{2 (7 B c+7 A d-2 B d) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{35 f}-\frac{2 B d \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{7 a f}+\frac{1}{35} (35 A c+21 B c+21 A d+19 B d) \int (a+a \sin (e+f x))^{3/2} \, dx\\ &=-\frac{2 a (35 A c+21 B c+21 A d+19 B d) \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{105 f}-\frac{2 (7 B c+7 A d-2 B d) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{35 f}-\frac{2 B d \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{7 a f}+\frac{1}{105} (4 a (35 A c+21 B c+21 A d+19 B d)) \int \sqrt{a+a \sin (e+f x)} \, dx\\ &=-\frac{8 a^2 (35 A c+21 B c+21 A d+19 B d) \cos (e+f x)}{105 f \sqrt{a+a \sin (e+f x)}}-\frac{2 a (35 A c+21 B c+21 A d+19 B d) \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{105 f}-\frac{2 (7 B c+7 A d-2 B d) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{35 f}-\frac{2 B d \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{7 a f}\\ \end{align*}

Mathematica [A]  time = 1.07593, size = 144, normalized size = 0.87 \[ -\frac{a \sqrt{a (\sin (e+f x)+1)} \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) ((140 A c+252 A d+252 B c+253 B d) \sin (e+f x)-6 (7 A d+7 B c+13 B d) \cos (2 (e+f x))+700 A c+546 A d+546 B c-15 B d \sin (3 (e+f x))+494 B d)}{210 f \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*Sqrt[a*(1 + Sin[e + f*x])]*(700*A*c + 546*B*c + 546*A*d + 494*B*d -
6*(7*B*c + 7*A*d + 13*B*d)*Cos[2*(e + f*x)] + (140*A*c + 252*B*c + 252*A*d + 253*B*d)*Sin[e + f*x] - 15*B*d*Si
n[3*(e + f*x)]))/(210*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]))

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Maple [A]  time = 0.996, size = 150, normalized size = 0.9 \begin{align*}{\frac{ \left ( 2+2\,\sin \left ( fx+e \right ) \right ){a}^{2} \left ( -1+\sin \left ( fx+e \right ) \right ) \left ( 15\,Bd \left ( \sin \left ( fx+e \right ) \right ) ^{3}+21\,Ad \left ( \sin \left ( fx+e \right ) \right ) ^{2}+21\,Bc \left ( \sin \left ( fx+e \right ) \right ) ^{2}+39\,B \left ( \sin \left ( fx+e \right ) \right ) ^{2}d+35\,A\sin \left ( fx+e \right ) c+63\,A\sin \left ( fx+e \right ) d+63\,B\sin \left ( fx+e \right ) c+52\,B\sin \left ( fx+e \right ) d+175\,Ac+126\,Ad+126\,Bc+104\,Bd \right ) }{105\,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))^(3/2)*(A+B*sin(f*x+e))*(c+d*sin(f*x+e)),x)

[Out]

2/105*(1+sin(f*x+e))*a^2*(-1+sin(f*x+e))*(15*B*d*sin(f*x+e)^3+21*A*d*sin(f*x+e)^2+21*B*c*sin(f*x+e)^2+39*B*sin
(f*x+e)^2*d+35*A*sin(f*x+e)*c+63*A*sin(f*x+e)*d+63*B*sin(f*x+e)*c+52*B*sin(f*x+e)*d+175*A*c+126*A*d+126*B*c+10
4*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{3}{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))^(3/2)*(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)^(3/2)*(d*sin(f*x + e) + c), x)

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Fricas [A]  time = 1.7824, size = 660, normalized size = 4. \begin{align*} \frac{2 \,{\left (15 \, B a d \cos \left (f x + e\right )^{4} + 3 \,{\left (7 \, B a c +{\left (7 \, A + 13 \, B\right )} a d\right )} \cos \left (f x + e\right )^{3} - 28 \,{\left (5 \, A + 3 \, B\right )} a c - 4 \,{\left (21 \, A + 19 \, B\right )} a d -{\left (7 \,{\left (5 \, A + 6 \, B\right )} a c +{\left (42 \, A + 43 \, B\right )} a d\right )} \cos \left (f x + e\right )^{2} -{\left (7 \,{\left (25 \, A + 21 \, B\right )} a c +{\left (147 \, A + 143 \, B\right )} a d\right )} \cos \left (f x + e\right ) +{\left (15 \, B a d \cos \left (f x + e\right )^{3} + 28 \,{\left (5 \, A + 3 \, B\right )} a c + 4 \,{\left (21 \, A + 19 \, B\right )} a d - 3 \,{\left (7 \, B a c +{\left (7 \, A + 8 \, B\right )} a d\right )} \cos \left (f x + e\right )^{2} -{\left (7 \,{\left (5 \, A + 9 \, B\right )} a c +{\left (63 \, A + 67 \, B\right )} a 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}}{105 \,{\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))^(3/2)*(A+B*sin(f*x+e))*(c+d*sin(f*x+e)),x, algorithm="fricas")

[Out]

2/105*(15*B*a*d*cos(f*x + e)^4 + 3*(7*B*a*c + (7*A + 13*B)*a*d)*cos(f*x + e)^3 - 28*(5*A + 3*B)*a*c - 4*(21*A
+ 19*B)*a*d - (7*(5*A + 6*B)*a*c + (42*A + 43*B)*a*d)*cos(f*x + e)^2 - (7*(25*A + 21*B)*a*c + (147*A + 143*B)*
a*d)*cos(f*x + e) + (15*B*a*d*cos(f*x + e)^3 + 28*(5*A + 3*B)*a*c + 4*(21*A + 19*B)*a*d - 3*(7*B*a*c + (7*A +
8*B)*a*d)*cos(f*x + e)^2 - (7*(5*A + 9*B)*a*c + (63*A + 67*B)*a*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))**(3/2)*(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))^(3/2)*(A+B*sin(f*x+e))*(c+d*sin(f*x+e)),x, algorithm="giac")

[Out]

Timed out