\(\int \sqrt {3+3 \sin (e+f x)} (c+d \sin (e+f x))^2 \, dx\) [523]

Optimal result

Integrand size = 27, antiderivative size = 108 \[ \int \sqrt {3+3 \sin (e+f x)} (c+d \sin (e+f x))^2 \, dx=-\frac {2 \left (15 c^2+10 c d+7 d^2\right ) \cos (e+f x)}{5 f \sqrt {3+3 \sin (e+f x)}}-\frac {4 (5 c-d) d \cos (e+f x) \sqrt {3+3 \sin (e+f x)}}{15 f}-\frac {2 d^2 \cos (e+f x) (3+3 \sin (e+f x))^{3/2}}{15 f} \]

[Out]

-2/5*d^2*cos(f*x+e)*(a+a*sin(f*x+e))^(3/2)/a/f-2/15*a*(15*c^2+10*c*d+7*d^2)*cos(f*x+e)/f/(a+a*sin(f*x+e))^(1/2
)-4/15*(5*c-d)*d*cos(f*x+e)*(a+a*sin(f*x+e))^(1/2)/f

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.04, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2840, 2830, 2725} \[ \int \sqrt {3+3 \sin (e+f x)} (c+d \sin (e+f x))^2 \, dx=-\frac {2 a \left (15 c^2+10 c d+7 d^2\right ) \cos (e+f x)}{15 f \sqrt {a \sin (e+f x)+a}}-\frac {4 d (5 c-d) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{15 f}-\frac {2 d^2 \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{5 a f} \]

[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 2725

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]

Rule 2830

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*S
in[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 2840

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]

Rubi steps \begin{align*} \text {integral}& = -\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] (verified)

Time = 0.41 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.06 \[ \int \sqrt {3+3 \sin (e+f x)} (c+d \sin (e+f x))^2 \, dx=-\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {1+\sin (e+f x)} \left (30 c^2+40 c d+19 d^2-3 d^2 \cos (2 (e+f x))+4 d (5 c+2 d) \sin (e+f x)\right )}{5 \sqrt {3} f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )} \]

[In]

Integrate[Sqrt[3 + 3*Sin[e + f*x]]*(c + d*Sin[e + f*x])^2,x]

[Out]

-1/5*((Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*Sqrt[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]))/(Sqrt[3]*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]))

Maple [A] (verified)

Time = 1.83 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.85

method	result	size
default	\(\frac {2 \left (\sin \left (f x +e \right )+1\right ) a \left (\sin \left (f x +e \right )-1\right ) \left (3 \left (\sin ^{2}\left (f x +e \right )\right ) d^{2}+10 c d \sin \left (f x +e \right )+4 \sin \left (f x +e \right ) d^{2}+15 c^{2}+20 c d +8 d^{2}\right )}{15 \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}\)	\(92\)
parts	\(\frac {2 c^{2} \left (\sin \left (f x +e \right )+1\right ) \left (\sin \left (f x +e \right )-1\right ) a}{\cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}+\frac {2 d^{2} \left (\sin \left (f x +e \right )+1\right ) a \left (\sin \left (f x +e \right )-1\right ) \left (3 \left (\sin ^{2}\left (f x +e \right )\right )+4 \sin \left (f x +e \right )+8\right )}{15 \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}+\frac {4 c d \left (\sin \left (f x +e \right )+1\right ) a \left (\sin \left (f x +e \right )-1\right ) \left (\sin \left (f x +e \right )+2\right )}{3 \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}\)	\(164\)

[In]

int((a+a*sin(f*x+e))^(1/2)*(c+d*sin(f*x+e))^2,x,method=_RETURNVERBOSE)

[Out]

2/15*(sin(f*x+e)+1)*a*(sin(f*x+e)-1)*(3*sin(f*x+e)^2*d^2+10*c*d*sin(f*x+e)+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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.45 \[ \int \sqrt {3+3 \sin (e+f x)} (c+d \sin (e+f x))^2 \, dx=\frac {2 \, {\left (3 \, d^{2} \cos \left (f x + e\right )^{3} - {\left (10 \, c d + d^{2}\right )} \cos \left (f x + e\right )^{2} - 15 \, c^{2} - 10 \, c d - 7 \, d^{2} - {\left (15 \, c^{2} + 20 \, c d + 11 \, d^{2}\right )} \cos \left (f x + e\right ) - {\left (3 \, d^{2} \cos \left (f x + e\right )^{2} - 15 \, c^{2} - 10 \, c d - 7 \, d^{2} + 2 \, {\left (5 \, c d + 2 \, d^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a}}{15 \, {\left (f \cos \left (f x + e\right ) + f \sin \left (f x + e\right ) + f\right )}} \]

[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)

Sympy [F]

\[ \int \sqrt {3+3 \sin (e+f x)} (c+d \sin (e+f x))^2 \, dx=\int \sqrt {a \left (\sin {\left (e + f x \right )} + 1\right )} \left (c + d \sin {\left (e + f x \right )}\right )^{2}\, dx \]

[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)

Maxima [F]

\[ \int \sqrt {3+3 \sin (e+f x)} (c+d \sin (e+f x))^2 \, dx=\int { \sqrt {a \sin \left (f x + e\right ) + a} {\left (d \sin \left (f x + e\right ) + c\right )}^{2} \,d x } \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.49 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.46 \[ \int \sqrt {3+3 \sin (e+f x)} (c+d \sin (e+f x))^2 \, dx=\frac {\sqrt {2} {\left (3 \, d^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {5}{4} \, \pi + \frac {5}{2} \, f x + \frac {5}{2} \, e\right ) + 30 \, {\left (2 \, c^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 2 \, c d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + d^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 5 \, {\left (4 \, c d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + d^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \sin \left (-\frac {3}{4} \, \pi + \frac {3}{2} \, f x + \frac {3}{2} \, e\right )\right )} \sqrt {a}}{30 \, f} \]

[In]

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

[Out]

1/30*sqrt(2)*(3*d^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-5/4*pi + 5/2*f*x + 5/2*e) + 30*(2*c^2*sgn(cos(-1/
4*pi + 1/2*f*x + 1/2*e)) + 2*c*d*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + d^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))
)*sin(-1/4*pi + 1/2*f*x + 1/2*e) + 5*(4*c*d*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + d^2*sgn(cos(-1/4*pi + 1/2*f*
x + 1/2*e)))*sin(-3/4*pi + 3/2*f*x + 3/2*e))*sqrt(a)/f

Mupad [F(-1)]

Timed out. \[ \int \sqrt {3+3 \sin (e+f x)} (c+d \sin (e+f x))^2 \, dx=\int \sqrt {a+a\,\sin \left (e+f\,x\right )}\,{\left (c+d\,\sin \left (e+f\,x\right )\right )}^2 \,d x \]

[In]

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

[Out]

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