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

Optimal. Leaf size=118 \[ -\frac{2 (5 A d+5 B c-2 B d) \cos (e+f x) \sqrt{a \sin (e+f x)+a}}{15 f}-\frac{2 a (15 A c+5 A d+5 B c+7 B d) \cos (e+f x)}{15 f \sqrt{a \sin (e+f x)+a}}-\frac{2 B d \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{5 a f} \]

[Out]

(-2*a*(15*A*c + 5*B*c + 5*A*d + 7*B*d)*Cos[e + f*x])/(15*f*Sqrt[a + a*Sin[e + f*x]]) - (2*(5*B*c + 5*A*d - 2*B
*d)*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]])/(15*f) - (2*B*d*Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2))/(5*a*f)

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Rubi [A]  time = 0.249048, antiderivative size = 118, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.114, Rules used = {2968, 3023, 2751, 2646} \[ -\frac{2 (5 A d+5 B c-2 B d) \cos (e+f x) \sqrt{a \sin (e+f x)+a}}{15 f}-\frac{2 a (15 A c+5 A d+5 B c+7 B d) \cos (e+f x)}{15 f \sqrt{a \sin (e+f x)+a}}-\frac{2 B d \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{5 a f} \]

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]),x]

[Out]

(-2*a*(15*A*c + 5*B*c + 5*A*d + 7*B*d)*Cos[e + f*x])/(15*f*Sqrt[a + a*Sin[e + f*x]]) - (2*(5*B*c + 5*A*d - 2*B
*d)*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]])/(15*f) - (2*B*d*Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2))/(5*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 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)) \, dx &=\int \sqrt{a+a \sin (e+f x)} \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))^{3/2}}{5 a f}+\frac{2 \int \sqrt{a+a \sin (e+f x)} \left (\frac{1}{2} a (5 A c+3 B d)+\frac{1}{2} a (5 B c+5 A d-2 B d) \sin (e+f x)\right ) \, dx}{5 a}\\ &=-\frac{2 (5 B c+5 A d-2 B d) \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{15 f}-\frac{2 B d \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{5 a f}+\frac{1}{15} (15 A c+5 B c+5 A d+7 B d) \int \sqrt{a+a \sin (e+f x)} \, dx\\ &=-\frac{2 a (15 A c+5 B c+5 A d+7 B d) \cos (e+f x)}{15 f \sqrt{a+a \sin (e+f x)}}-\frac{2 (5 B c+5 A d-2 B d) \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{15 f}-\frac{2 B d \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{5 a f}\\ \end{align*}

Mathematica [A]  time = 0.36511, size = 117, normalized size = 0.99 \[ -\frac{\sqrt{a (\sin (e+f x)+1)} \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) (2 (5 A d+5 B c+4 B d) \sin (e+f x)+30 A c+20 A d+20 B c-3 B d \cos (2 (e+f x))+19 B d)}{15 f \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )} \]

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]),x]

[Out]

-((Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*Sqrt[a*(1 + Sin[e + f*x])]*(30*A*c + 20*B*c + 20*A*d + 19*B*d - 3*B*d*
Cos[2*(e + f*x)] + 2*(5*B*c + 5*A*d + 4*B*d)*Sin[e + f*x]))/(15*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]))

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Maple [A]  time = 0.954, size = 102, normalized size = 0.9 \begin{align*}{\frac{ \left ( 2+2\,\sin \left ( fx+e \right ) \right ) a \left ( -1+\sin \left ( fx+e \right ) \right ) \left ( 3\,B \left ( \sin \left ( fx+e \right ) \right ) ^{2}d+5\,A\sin \left ( fx+e \right ) d+5\,B\sin \left ( fx+e \right ) c+4\,B\sin \left ( fx+e \right ) d+15\,Ac+10\,Ad+10\,Bc+8\,Bd \right ) }{15\,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))*(a+a*sin(f*x+e))^(1/2),x)

[Out]

2/15*(1+sin(f*x+e))*a*(-1+sin(f*x+e))*(3*B*sin(f*x+e)^2*d+5*A*sin(f*x+e)*d+5*B*sin(f*x+e)*c+4*B*sin(f*x+e)*d+1
5*A*c+10*A*d+10*B*c+8*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 )} \sqrt{a \sin \left (f x + e\right ) + a}{\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+B*sin(f*x+e))*(c+d*sin(f*x+e))*(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), x)

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Fricas [A]  time = 1.99923, size = 436, normalized size = 3.69 \begin{align*} \frac{2 \,{\left (3 \, B d \cos \left (f x + e\right )^{3} -{\left (5 \, B c +{\left (5 \, A + B\right )} d\right )} \cos \left (f x + e\right )^{2} - 5 \,{\left (3 \, A + B\right )} c -{\left (5 \, A + 7 \, B\right )} d -{\left (5 \,{\left (3 \, A + 2 \, B\right )} c +{\left (10 \, A + 11 \, B\right )} d\right )} \cos \left (f x + e\right ) -{\left (3 \, B d \cos \left (f x + e\right )^{2} - 5 \,{\left (3 \, A + B\right )} c -{\left (5 \, A + 7 \, B\right )} d +{\left (5 \, B c +{\left (5 \, A + 4 \, B\right )} 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}}{15 \,{\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))*(a+a*sin(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

2/15*(3*B*d*cos(f*x + e)^3 - (5*B*c + (5*A + B)*d)*cos(f*x + e)^2 - 5*(3*A + B)*c - (5*A + 7*B)*d - (5*(3*A +
2*B)*c + (10*A + 11*B)*d)*cos(f*x + e) - (3*B*d*cos(f*x + e)^2 - 5*(3*A + B)*c - (5*A + 7*B)*d + (5*B*c + (5*A
+ 4*B)*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]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{a \left (\sin{\left (e + f x \right )} + 1\right )} \left (A + B \sin{\left (e + f x \right )}\right ) \left (c + d \sin{\left (e + f x \right )}\right )\, dx \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))*(a+a*sin(f*x+e))**(1/2),x)

[Out]

Integral(sqrt(a*(sin(e + f*x) + 1))*(A + B*sin(e + f*x))*(c + d*sin(e + f*x)), x)

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

[Out]

Timed out