\(\int \frac {\sqrt {c-c \sin (e+f x)}}{(3+3 \sin (e+f x))^3} \, dx\) [336]

   Optimal result
   Rubi [A] (verified)
   Mathematica [B] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [F]
   Maxima [B] (verification not implemented)
   Giac [B] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 28, antiderivative size = 33 \[ \int \frac {\sqrt {c-c \sin (e+f x)}}{(3+3 \sin (e+f x))^3} \, dx=-\frac {2 \sec ^5(e+f x) (c-c \sin (e+f x))^{5/2}}{135 c^2 f} \]

[Out]

-2/5*sec(f*x+e)^5*(c-c*sin(f*x+e))^(5/2)/a^3/c^2/f

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.09, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {2815, 2752} \[ \int \frac {\sqrt {c-c \sin (e+f x)}}{(3+3 \sin (e+f x))^3} \, dx=-\frac {2 \sec ^5(e+f x) (c-c \sin (e+f x))^{5/2}}{5 a^3 c^2 f} \]

[In]

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

[Out]

(-2*Sec[e + f*x]^5*(c - c*Sin[e + f*x])^(5/2))/(5*a^3*c^2*f)

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}& = \frac {\int \sec ^6(e+f x) (c-c \sin (e+f x))^{7/2} \, dx}{a^3 c^3} \\ & = -\frac {2 \sec ^5(e+f x) (c-c \sin (e+f x))^{5/2}}{5 a^3 c^2 f} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(70\) vs. \(2(33)=66\).

Time = 0.66 (sec) , antiderivative size = 70, normalized size of antiderivative = 2.12 \[ \int \frac {\sqrt {c-c \sin (e+f x)}}{(3+3 \sin (e+f x))^3} \, dx=-\frac {2 \sqrt {c-c \sin (e+f x)}}{135 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^5} \]

[In]

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

[Out]

(-2*Sqrt[c - c*Sin[e + f*x]])/(135*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2
])^5)

Maple [A] (verified)

Time = 0.63 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.48

method result size
default \(\frac {2 c \left (\sin \left (f x +e \right )-1\right )}{5 a^{3} \left (\sin \left (f x +e \right )+1\right )^{2} \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\) \(49\)

[In]

int((c-c*sin(f*x+e))^(1/2)/(a+a*sin(f*x+e))^3,x,method=_RETURNVERBOSE)

[Out]

2/5*c/a^3*(sin(f*x+e)-1)/(sin(f*x+e)+1)^2/cos(f*x+e)/(c-c*sin(f*x+e))^(1/2)/f

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.85 \[ \int \frac {\sqrt {c-c \sin (e+f x)}}{(3+3 \sin (e+f x))^3} \, dx=\frac {2 \, \sqrt {-c \sin \left (f x + e\right ) + c}}{5 \, {\left (a^{3} f \cos \left (f x + e\right )^{3} - 2 \, a^{3} f \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, a^{3} f \cos \left (f x + e\right )\right )}} \]

[In]

integrate((c-c*sin(f*x+e))^(1/2)/(a+a*sin(f*x+e))^3,x, algorithm="fricas")

[Out]

2/5*sqrt(-c*sin(f*x + e) + c)/(a^3*f*cos(f*x + e)^3 - 2*a^3*f*cos(f*x + e)*sin(f*x + e) - 2*a^3*f*cos(f*x + e)
)

Sympy [F]

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

[In]

integrate((c-c*sin(f*x+e))**(1/2)/(a+a*sin(f*x+e))**3,x)

[Out]

Integral(sqrt(-c*sin(e + f*x) + c)/(sin(e + f*x)**3 + 3*sin(e + f*x)**2 + 3*sin(e + f*x) + 1), x)/a**3

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 218 vs. \(2 (32) = 64\).

Time = 0.29 (sec) , antiderivative size = 218, normalized size of antiderivative = 6.61 \[ \int \frac {\sqrt {c-c \sin (e+f x)}}{(3+3 \sin (e+f x))^3} \, dx=\frac {2 \, {\left (\sqrt {c} + \frac {3 \, \sqrt {c} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {3 \, \sqrt {c} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac {\sqrt {c} \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}}\right )}}{5 \, {\left (a^{3} + \frac {5 \, a^{3} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {10 \, a^{3} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {10 \, a^{3} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {5 \, a^{3} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac {a^{3} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}\right )} f \sqrt {\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1}} \]

[In]

integrate((c-c*sin(f*x+e))^(1/2)/(a+a*sin(f*x+e))^3,x, algorithm="maxima")

[Out]

2/5*(sqrt(c) + 3*sqrt(c)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 3*sqrt(c)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 +
 sqrt(c)*sin(f*x + e)^6/(cos(f*x + e) + 1)^6)/((a^3 + 5*a^3*sin(f*x + e)/(cos(f*x + e) + 1) + 10*a^3*sin(f*x +
 e)^2/(cos(f*x + e) + 1)^2 + 10*a^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*a^3*sin(f*x + e)^4/(cos(f*x + e) +
 1)^4 + a^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5)*f*sqrt(sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1))

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 157 vs. \(2 (32) = 64\).

Time = 0.32 (sec) , antiderivative size = 157, normalized size of antiderivative = 4.76 \[ \int \frac {\sqrt {c-c \sin (e+f x)}}{(3+3 \sin (e+f x))^3} \, dx=\frac {\sqrt {2} \sqrt {c} {\left (\frac {10 \, {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{2}} + \frac {5 \, {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{4} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{4}} + \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )}}{10 \, a^{3} f {\left (\frac {\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1} + 1\right )}^{5}} \]

[In]

integrate((c-c*sin(f*x+e))^(1/2)/(a+a*sin(f*x+e))^3,x, algorithm="giac")

[Out]

1/10*sqrt(2)*sqrt(c)*(10*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)^2*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))/(cos(-1/4*
pi + 1/2*f*x + 1/2*e) + 1)^2 + 5*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)^4*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))/(c
os(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^4 + sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)))/(a^3*f*((cos(-1/4*pi + 1/2*f*x + 1
/2*e) - 1)/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1) + 1)^5)

Mupad [B] (verification not implemented)

Time = 9.66 (sec) , antiderivative size = 90, normalized size of antiderivative = 2.73 \[ \int \frac {\sqrt {c-c \sin (e+f x)}}{(3+3 \sin (e+f x))^3} \, dx=\frac {{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}\,\sqrt {c-c\,\left (\frac {{\mathrm {e}}^{-e\,1{}\mathrm {i}-f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}\right )}\,16{}\mathrm {i}}{5\,a^3\,f\,{\left ({\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}+1{}\mathrm {i}\right )}^5\,\left (1+{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}\right )} \]

[In]

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

[Out]

(exp(e*3i + f*x*3i)*(c - c*((exp(- e*1i - f*x*1i)*1i)/2 - (exp(e*1i + f*x*1i)*1i)/2))^(1/2)*16i)/(5*a^3*f*(exp
(e*1i + f*x*1i) + 1i)^5*(exp(e*1i + f*x*1i)*1i + 1))