\(\int \sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x)) (c+d \sin (e+f x)) \, dx\) [288]

Optimal result

Integrand size = 35, antiderivative size = 118 \[ \int \sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x)) (c+d \sin (e+f x)) \, dx=-\frac {2 a (15 A c+5 B c+5 A d+7 B d) \cos (e+f x)}{15 f \sqrt {a+a \sin (e+f x)}}-\frac {2 (5 B c+5 A d-2 B d) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{15 f}-\frac {2 B d \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{5 a f} \]

[Out]

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

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.114, Rules used = {3047, 3102, 2830, 2725} \[ \int \sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x)) (c+d \sin (e+f x)) \, dx=-\frac {2 (5 A d+5 B c-2 B d) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{15 f}-\frac {2 a (15 A c+5 A d+5 B c+7 B d) \cos (e+f x)}{15 f \sqrt {a \sin (e+f x)+a}}-\frac {2 B d \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{5 a f} \]

[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 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 3047

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 3102

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]

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

Time = 0.74 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.99 \[ \int \sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x)) (c+d \sin (e+f x)) \, dx=-\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {a (1+\sin (e+f x))} (30 A c+20 B c+20 A d+19 B d-3 B d \cos (2 (e+f x))+2 (5 B c+5 A d+4 B d) \sin (e+f x))}{15 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )} \]

[In]

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

[Out]

-1/15*((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]))/(f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]))

Maple [A] (verified)

Time = 1.54 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.86

method	result	size
default	\(\frac {2 \left (1+\sin \left (f x +e \right )\right ) a \left (\sin \left (f x +e \right )-1\right ) \left (3 B \left (\sin ^{2}\left (f x +e \right )\right ) d +5 A \sin \left (f x +e \right ) d +5 B \sin \left (f x +e \right ) c +4 B \sin \left (f x +e \right ) d +15 A c +10 d A +10 B c +8 d B \right )}{15 \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}\)	\(102\)
parts	\(\frac {2 A c \left (1+\sin \left (f x +e \right )\right ) \left (\sin \left (f x +e \right )-1\right ) a}{\cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}+\frac {2 \left (d A +B c \right ) \left (1+\sin \left (f x +e \right )\right ) a \left (\sin \left (f x +e \right )-1\right ) \left (\sin \left (f x +e \right )+2\right )}{3 \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}+\frac {2 d B \left (1+\sin \left (f x +e \right )\right ) a \left (\sin \left (f x +e \right )-1\right ) \left (3 \left (\sin ^{2}\left (f x +e \right )\right )+4 \sin \left (f x +e \right )+8\right )}{15 \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}\)	\(167\)

[In]

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

[Out]

2/15*(1+sin(f*x+e))*a*(sin(f*x+e)-1)*(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+15
*A*c+10*d*A+10*B*c+8*d*B)/cos(f*x+e)/(a+a*sin(f*x+e))^(1/2)/f

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.48 \[ \int \sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x)) (c+d \sin (e+f x)) \, dx=\frac {2 \, {\left (3 \, B d \cos \left (f x + e\right )^{3} - {\left (5 \, B c + {\left (5 \, A + B\right )} d\right )} \cos \left (f x + e\right )^{2} - 5 \, {\left (3 \, A + B\right )} c - {\left (5 \, A + 7 \, B\right )} d - {\left (5 \, {\left (3 \, A + 2 \, B\right )} c + {\left (10 \, A + 11 \, B\right )} d\right )} \cos \left (f x + e\right ) - {\left (3 \, B d \cos \left (f x + e\right )^{2} - 5 \, {\left (3 \, A + B\right )} c - {\left (5 \, A + 7 \, B\right )} d + {\left (5 \, B c + {\left (5 \, A + 4 \, B\right )} d\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a}}{15 \, {\left (f \cos \left (f x + e\right ) + f \sin \left (f x + e\right ) + f\right )}} \]

[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)

Sympy [F]

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

[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)

Maxima [F]

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

[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)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.58 \[ \int \sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x)) (c+d \sin (e+f x)) \, dx=\frac {\sqrt {2} {\left (3 \, B d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {5}{4} \, \pi + \frac {5}{2} \, f x + \frac {5}{2} \, e\right ) + 30 \, {\left (2 \, A c \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + B c \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + A d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + B d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 5 \, {\left (2 \, B c \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 2 \, A d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + B d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \sin \left (-\frac {3}{4} \, \pi + \frac {3}{2} \, f x + \frac {3}{2} \, e\right )\right )} \sqrt {a}}{30 \, f} \]

[In]

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

[Out]

1/30*sqrt(2)*(3*B*d*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*A*c*sgn(cos(-1/
4*pi + 1/2*f*x + 1/2*e)) + B*c*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + A*d*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) +
B*d*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)))*sin(-1/4*pi + 1/2*f*x + 1/2*e) + 5*(2*B*c*sgn(cos(-1/4*pi + 1/2*f*x
+ 1/2*e)) + 2*A*d*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + B*d*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. \[ \int \sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x)) (c+d \sin (e+f x)) \, dx=\int \left (A+B\,\sin \left (e+f\,x\right )\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}\,\left (c+d\,\sin \left (e+f\,x\right )\right ) \,d x \]

[In]

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

[Out]

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