\(\int (a+a \sin (e+f x)) \sqrt {c-c \sin (e+f x)} \, dx\) [293]

Optimal result

Integrand size = 26, antiderivative size = 34 \[ \int (a+a \sin (e+f x)) \sqrt {c-c \sin (e+f x)} \, dx=\frac {2 a c^2 \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^{3/2}} \]

[Out]

2/3*a*c^2*cos(f*x+e)^3/f/(c-c*sin(f*x+e))^(3/2)

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {2815, 2752} \[ \int (a+a \sin (e+f x)) \sqrt {c-c \sin (e+f x)} \, dx=\frac {2 a c^2 \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^{3/2}} \]

[In]

Int[(a + a*Sin[e + f*x])*Sqrt[c - c*Sin[e + f*x]],x]

[Out]

(2*a*c^2*Cos[e + f*x]^3)/(3*f*(c - c*Sin[e + f*x])^(3/2))

Rule 2752

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[b*(g*C
os[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^(m - 1)/(f*g*(m - 1))), x] /; FreeQ[{a, b, e, f, g, m, p}, x] && Eq
Q[a^2 - b^2, 0] && EqQ[2*m + p - 1, 0] && NeQ[m, 1]

Rule 2815

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Di
st[a^m*c^m, Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&
EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] &&  !(IntegerQ[n] && ((LtQ[m, 0] && GtQ[n, 0]) || LtQ[0,
n, m] || LtQ[m, n, 0]))

Rubi steps \begin{align*} \text {integral}& = (a c) \int \frac {\cos ^2(e+f x)}{\sqrt {c-c \sin (e+f x)}} \, dx \\ & = \frac {2 a c^2 \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^{3/2}} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(71\) vs. \(2(34)=68\).

Time = 0.10 (sec) , antiderivative size = 71, normalized size of antiderivative = 2.09 \[ \int (a+a \sin (e+f x)) \sqrt {c-c \sin (e+f x)} \, dx=\frac {2 a \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3 \sqrt {c-c \sin (e+f x)}}{3 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )} \]

[In]

Integrate[(a + a*Sin[e + f*x])*Sqrt[c - c*Sin[e + f*x]],x]

[Out]

(2*a*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3*Sqrt[c - c*Sin[e + f*x]])/(3*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/
2]))

Maple [A] (verified)

Time = 1.30 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.38

method	result	size
default	\(-\frac {2 \left (\sin \left (f x +e \right )-1\right ) c \left (\sin \left (f x +e \right )+1\right )^{2} a}{3 \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\)	\(47\)
parts	\(-\frac {2 a \left (\sin \left (f x +e \right )-1\right ) \left (\sin \left (f x +e \right )+1\right ) c}{\cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}-\frac {2 a \left (\sin \left (f x +e \right )-1\right ) c \left (\sin \left (f x +e \right )+1\right ) \left (\sin \left (f x +e \right )-2\right )}{3 \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\)	\(98\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 79 vs. \(2 (30) = 60\).

Time = 0.26 (sec) , antiderivative size = 79, normalized size of antiderivative = 2.32 \[ \int (a+a \sin (e+f x)) \sqrt {c-c \sin (e+f x)} \, dx=-\frac {2 \, {\left (a \cos \left (f x + e\right )^{2} - a \cos \left (f x + e\right ) - {\left (a \cos \left (f x + e\right ) + 2 \, a\right )} \sin \left (f x + e\right ) - 2 \, a\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{3 \, {\left (f \cos \left (f x + e\right ) - f \sin \left (f x + e\right ) + f\right )}} \]

[In]

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

[Out]

-2/3*(a*cos(f*x + e)^2 - a*cos(f*x + e) - (a*cos(f*x + e) + 2*a)*sin(f*x + e) - 2*a)*sqrt(-c*sin(f*x + e) + c)
/(f*cos(f*x + e) - f*sin(f*x + e) + f)

Sympy [F]

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

[In]

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

[Out]

a*(Integral(sqrt(-c*sin(e + f*x) + c)*sin(e + f*x), x) + Integral(sqrt(-c*sin(e + f*x) + c), x))

Maxima [F]

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

[In]

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

[Out]

integrate((a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c), x)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 67 vs. \(2 (30) = 60\).

Time = 0.35 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.97 \[ \int (a+a \sin (e+f x)) \sqrt {c-c \sin (e+f x)} \, dx=-\frac {\sqrt {2} {\left (3 \, a \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + a \cos \left (-\frac {3}{4} \, \pi + \frac {3}{2} \, f x + \frac {3}{2} \, e\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \sqrt {c}}{3 \, f} \]

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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