3.523 \(\int \sqrt{a+a \sin (e+f x)} (c+d \sin (e+f x))^2 \, dx\)

Optimal. Leaf size=112 \[ -\frac{2 a \left (15 c^2+10 c d+7 d^2\right ) \cos (e+f x)}{15 f \sqrt{a \sin (e+f x)+a}}-\frac{4 d (5 c-d) \cos (e+f x) \sqrt{a \sin (e+f x)+a}}{15 f}-\frac{2 d^2 \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{5 a f} \]

[Out]

(-2*a*(15*c^2 + 10*c*d + 7*d^2)*Cos[e + f*x])/(15*f*Sqrt[a + a*Sin[e + f*x]]) - (4*(5*c - d)*d*Cos[e + f*x]*Sq
rt[a + a*Sin[e + f*x]])/(15*f) - (2*d^2*Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2))/(5*a*f)

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Rubi [A]  time = 0.168584, antiderivative size = 112, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {2761, 2751, 2646} \[ -\frac{2 a \left (15 c^2+10 c d+7 d^2\right ) \cos (e+f x)}{15 f \sqrt{a \sin (e+f x)+a}}-\frac{4 d (5 c-d) \cos (e+f x) \sqrt{a \sin (e+f x)+a}}{15 f}-\frac{2 d^2 \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{5 a f} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + a*Sin[e + f*x]]*(c + d*Sin[e + f*x])^2,x]

[Out]

(-2*a*(15*c^2 + 10*c*d + 7*d^2)*Cos[e + f*x])/(15*f*Sqrt[a + a*Sin[e + f*x]]) - (4*(5*c - d)*d*Cos[e + f*x]*Sq
rt[a + a*Sin[e + f*x]])/(15*f) - (2*d^2*Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2))/(5*a*f)

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

Mathematica [A]  time = 0.286979, size = 111, normalized size = 0.99 \[ -\frac{\sqrt{a (\sin (e+f x)+1)} \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (30 c^2+4 d (5 c+2 d) \sin (e+f x)+40 c d-3 d^2 \cos (2 (e+f x))+19 d^2\right )}{15 f \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + a*Sin[e + f*x]]*(c + d*Sin[e + f*x])^2,x]

[Out]

-((Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*Sqrt[a*(1 + Sin[e + f*x])]*(30*c^2 + 40*c*d + 19*d^2 - 3*d^2*Cos[2*(e
+ f*x)] + 4*d*(5*c + 2*d)*Sin[e + f*x]))/(15*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]))

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Maple [A]  time = 0.648, size = 92, normalized size = 0.8 \begin{align*}{\frac{ \left ( 2+2\,\sin \left ( fx+e \right ) \right ) a \left ( -1+\sin \left ( fx+e \right ) \right ) \left ( 3\,{d}^{2} \left ( \sin \left ( fx+e \right ) \right ) ^{2}+10\,\sin \left ( fx+e \right ) cd+4\,\sin \left ( fx+e \right ){d}^{2}+15\,{c}^{2}+20\,cd+8\,{d}^{2} \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+a*sin(f*x+e))^(1/2)*(c+d*sin(f*x+e))^2,x)

[Out]

2/15*(1+sin(f*x+e))*a*(-1+sin(f*x+e))*(3*d^2*sin(f*x+e)^2+10*sin(f*x+e)*c*d+4*sin(f*x+e)*d^2+15*c^2+20*c*d+8*d
^2)/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 \sqrt{a \sin \left (f x + e\right ) + a}{\left (d \sin \left (f x + e\right ) + c\right )}^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(a*sin(f*x + e) + a)*(d*sin(f*x + e) + c)^2, x)

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Fricas [A]  time = 1.87626, size = 390, normalized size = 3.48 \begin{align*} \frac{2 \,{\left (3 \, d^{2} \cos \left (f x + e\right )^{3} -{\left (10 \, c d + d^{2}\right )} \cos \left (f x + e\right )^{2} - 15 \, c^{2} - 10 \, c d - 7 \, d^{2} -{\left (15 \, c^{2} + 20 \, c d + 11 \, d^{2}\right )} \cos \left (f x + e\right ) -{\left (3 \, d^{2} \cos \left (f x + e\right )^{2} - 15 \, c^{2} - 10 \, c d - 7 \, d^{2} + 2 \,{\left (5 \, c d + 2 \, d^{2}\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+a*sin(f*x+e))^(1/2)*(c+d*sin(f*x+e))^2,x, algorithm="fricas")

[Out]

2/15*(3*d^2*cos(f*x + e)^3 - (10*c*d + d^2)*cos(f*x + e)^2 - 15*c^2 - 10*c*d - 7*d^2 - (15*c^2 + 20*c*d + 11*d
^2)*cos(f*x + e) - (3*d^2*cos(f*x + e)^2 - 15*c^2 - 10*c*d - 7*d^2 + 2*(5*c*d + 2*d^2)*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 (c + d \sin{\left (e + f x \right )}\right )^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out