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

Optimal. Leaf size=161 \[ -\frac{4 a (c+d) \left (15 c^2+10 c d+7 d^2\right ) \cos (e+f x)}{35 f \sqrt{a \sin (e+f x)+a}}-\frac{12 d^2 (c+d) \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{35 a f}-\frac{2 a \cos (e+f x) (c+d \sin (e+f x))^3}{7 f \sqrt{a \sin (e+f x)+a}}-\frac{8 d (5 c-d) (c+d) \cos (e+f x) \sqrt{a \sin (e+f x)+a}}{35 f} \]

[Out]

(-4*a*(c + d)*(15*c^2 + 10*c*d + 7*d^2)*Cos[e + f*x])/(35*f*Sqrt[a + a*Sin[e + f*x]]) - (8*(5*c - d)*d*(c + d)
*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]])/(35*f) - (12*d^2*(c + d)*Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2))/(35*
a*f) - (2*a*Cos[e + f*x]*(c + d*Sin[e + f*x])^3)/(7*f*Sqrt[a + a*Sin[e + f*x]])

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Rubi [A]  time = 0.278762, antiderivative size = 161, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.148, Rules used = {2770, 2761, 2751, 2646} \[ -\frac{4 a (c+d) \left (15 c^2+10 c d+7 d^2\right ) \cos (e+f x)}{35 f \sqrt{a \sin (e+f x)+a}}-\frac{12 d^2 (c+d) \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{35 a f}-\frac{2 a \cos (e+f x) (c+d \sin (e+f x))^3}{7 f \sqrt{a \sin (e+f x)+a}}-\frac{8 d (5 c-d) (c+d) \cos (e+f x) \sqrt{a \sin (e+f x)+a}}{35 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*a*(c + d)*(15*c^2 + 10*c*d + 7*d^2)*Cos[e + f*x])/(35*f*Sqrt[a + a*Sin[e + f*x]]) - (8*(5*c - d)*d*(c + d)
*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]])/(35*f) - (12*d^2*(c + d)*Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2))/(35*
a*f) - (2*a*Cos[e + f*x]*(c + d*Sin[e + f*x])^3)/(7*f*Sqrt[a + a*Sin[e + f*x]])

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)} (c+d \sin (e+f x))^3 \, dx &=-\frac{2 a \cos (e+f x) (c+d \sin (e+f x))^3}{7 f \sqrt{a+a \sin (e+f x)}}+\frac{1}{7} (6 (c+d)) \int \sqrt{a+a \sin (e+f x)} (c+d \sin (e+f x))^2 \, dx\\ &=-\frac{12 d^2 (c+d) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{35 a f}-\frac{2 a \cos (e+f x) (c+d \sin (e+f x))^3}{7 f \sqrt{a+a \sin (e+f x)}}+\frac{(12 (c+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}{35 a}\\ &=-\frac{8 (5 c-d) d (c+d) \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{35 f}-\frac{12 d^2 (c+d) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{35 a f}-\frac{2 a \cos (e+f x) (c+d \sin (e+f x))^3}{7 f \sqrt{a+a \sin (e+f x)}}+\frac{1}{35} \left (2 (c+d) \left (15 c^2+10 c d+7 d^2\right )\right ) \int \sqrt{a+a \sin (e+f x)} \, dx\\ &=-\frac{4 a (c+d) \left (15 c^2+10 c d+7 d^2\right ) \cos (e+f x)}{35 f \sqrt{a+a \sin (e+f x)}}-\frac{8 (5 c-d) d (c+d) \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{35 f}-\frac{12 d^2 (c+d) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{35 a f}-\frac{2 a \cos (e+f x) (c+d \sin (e+f x))^3}{7 f \sqrt{a+a \sin (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 0.520631, size = 146, normalized size = 0.91 \[ -\frac{\sqrt{a (\sin (e+f x)+1)} \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (d \left (140 c^2+112 c d+47 d^2\right ) \sin (e+f x)+280 c^2 d+140 c^3-6 d^2 (7 c+2 d) \cos (2 (e+f x))+266 c d^2-5 d^3 \sin (3 (e+f x))+76 d^3\right )}{70 f \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*Sqrt[a*(1 + Sin[e + f*x])]*(140*c^3 + 280*c^2*d + 266*c*d^2 + 76*d^3 -
6*d^2*(7*c + 2*d)*Cos[2*(e + f*x)] + d*(140*c^2 + 112*c*d + 47*d^2)*Sin[e + f*x] - 5*d^3*Sin[3*(e + f*x)]))/(
70*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]))

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Maple [A]  time = 0.647, size = 141, 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 ( 5\,{d}^{3} \left ( \sin \left ( fx+e \right ) \right ) ^{3}+21\,c{d}^{2} \left ( \sin \left ( fx+e \right ) \right ) ^{2}+6\,{d}^{3} \left ( \sin \left ( fx+e \right ) \right ) ^{2}+35\,{c}^{2}d\sin \left ( fx+e \right ) +28\,\sin \left ( fx+e \right ){d}^{2}c+8\,{d}^{3}\sin \left ( fx+e \right ) +35\,{c}^{3}+70\,{c}^{2}d+56\,c{d}^{2}+16\,{d}^{3} \right ) }{35\,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))^3,x)

[Out]

2/35*(1+sin(f*x+e))*a*(-1+sin(f*x+e))*(5*d^3*sin(f*x+e)^3+21*c*d^2*sin(f*x+e)^2+6*d^3*sin(f*x+e)^2+35*c^2*d*si
n(f*x+e)+28*sin(f*x+e)*d^2*c+8*d^3*sin(f*x+e)+35*c^3+70*c^2*d+56*c*d^2+16*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 \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+a*sin(f*x+e))^(1/2)*(c+d*sin(f*x+e))^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.96251, size = 576, normalized size = 3.58 \begin{align*} \frac{2 \,{\left (5 \, d^{3} \cos \left (f x + e\right )^{4} + 3 \,{\left (7 \, c d^{2} + 2 \, d^{3}\right )} \cos \left (f x + e\right )^{3} - 35 \, c^{3} - 35 \, c^{2} d - 49 \, c d^{2} - 9 \, d^{3} -{\left (35 \, c^{2} d + 7 \, c d^{2} + 12 \, d^{3}\right )} \cos \left (f x + e\right )^{2} -{\left (35 \, c^{3} + 70 \, c^{2} d + 77 \, c d^{2} + 22 \, d^{3}\right )} \cos \left (f x + e\right ) +{\left (5 \, d^{3} \cos \left (f x + e\right )^{3} + 35 \, c^{3} + 35 \, c^{2} d + 49 \, c d^{2} + 9 \, d^{3} -{\left (21 \, c d^{2} + d^{3}\right )} \cos \left (f x + e\right )^{2} -{\left (35 \, c^{2} d + 28 \, c d^{2} + 13 \, 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}}{35 \,{\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))^3,x, algorithm="fricas")

[Out]

2/35*(5*d^3*cos(f*x + e)^4 + 3*(7*c*d^2 + 2*d^3)*cos(f*x + e)^3 - 35*c^3 - 35*c^2*d - 49*c*d^2 - 9*d^3 - (35*c
^2*d + 7*c*d^2 + 12*d^3)*cos(f*x + e)^2 - (35*c^3 + 70*c^2*d + 77*c*d^2 + 22*d^3)*cos(f*x + e) + (5*d^3*cos(f*
x + e)^3 + 35*c^3 + 35*c^2*d + 49*c*d^2 + 9*d^3 - (21*c*d^2 + d^3)*cos(f*x + e)^2 - (35*c^2*d + 28*c*d^2 + 13*
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]  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 )^{3}\, 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))**3,x)

[Out]

Integral(sqrt(a*(sin(e + f*x) + 1))*(c + d*sin(e + f*x))**3, 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))^3,x, algorithm="giac")

[Out]

Timed out