\(\int \frac {\sqrt {3+3 \sin (e+f x)}}{(c-c \sin (e+f x))^{5/2}} \, dx\) [346]

Optimal result

Integrand size = 30, antiderivative size = 42 \[ \int \frac {\sqrt {3+3 \sin (e+f x)}}{(c-c \sin (e+f x))^{5/2}} \, dx=\frac {3 \cos (e+f x)}{2 f \sqrt {3+3 \sin (e+f x)} (c-c \sin (e+f x))^{5/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.02, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.033, Rules used = {2817} \[ \int \frac {\sqrt {3+3 \sin (e+f x)}}{(c-c \sin (e+f x))^{5/2}} \, dx=\frac {a \cos (e+f x)}{2 f \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{5/2}} \]

[In]

Int[Sqrt[a + a*Sin[e + f*x]]/(c - c*Sin[e + f*x])^(5/2),x]

[Out]

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

Rule 2817

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] /; FreeQ[{a, b, c, d, e,
f, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[n, -2^(-1)]

Rubi steps \begin{align*} \text {integral}& = \frac {a \cos (e+f x)}{2 f \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{5/2}} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(90\) vs. \(2(42)=84\).

Time = 0.80 (sec) , antiderivative size = 90, normalized size of antiderivative = 2.14 \[ \int \frac {\sqrt {3+3 \sin (e+f x)}}{(c-c \sin (e+f x))^{5/2}} \, dx=\frac {\sqrt {3} \sqrt {1+\sin (e+f x)} \sqrt {c-c \sin (e+f x)}}{2 c^3 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^5 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )} \]

[In]

Integrate[Sqrt[3 + 3*Sin[e + f*x]]/(c - c*Sin[e + f*x])^(5/2),x]

[Out]

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

Maple [A] (verified)

Time = 2.47 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.60

method	result	size
default	\(-\frac {\sqrt {a \left (\sin \left (f x +e \right )+1\right )}\, \left (\cos \left (f x +e \right )+2 \tan \left (f x +e \right )-\sec \left (f x +e \right )\right )}{2 f \left (\sin \left (f x +e \right )-1\right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, c^{2}}\)	\(67\)

[In]

int((a+a*sin(f*x+e))^(1/2)/(c-c*sin(f*x+e))^(5/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.74 \[ \int \frac {\sqrt {3+3 \sin (e+f x)}}{(c-c \sin (e+f x))^{5/2}} \, dx=-\frac {\sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{2 \, {\left (c^{3} f \cos \left (f x + e\right )^{3} + 2 \, c^{3} f \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, c^{3} f \cos \left (f x + e\right )\right )}} \]

[In]

integrate((a+a*sin(f*x+e))^(1/2)/(c-c*sin(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {\sqrt {3+3 \sin (e+f x)}}{(c-c \sin (e+f x))^{5/2}} \, dx=\int \frac {\sqrt {a \left (\sin {\left (e + f x \right )} + 1\right )}}{\left (- c \left (\sin {\left (e + f x \right )} - 1\right )\right )^{\frac {5}{2}}}\, dx \]

[In]

integrate((a+a*sin(f*x+e))**(1/2)/(c-c*sin(f*x+e))**(5/2),x)

[Out]

Integral(sqrt(a*(sin(e + f*x) + 1))/(-c*(sin(e + f*x) - 1))**(5/2), x)

Maxima [F]

\[ \int \frac {\sqrt {3+3 \sin (e+f x)}}{(c-c \sin (e+f x))^{5/2}} \, dx=\int { \frac {\sqrt {a \sin \left (f x + e\right ) + a}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}}} \,d x } \]

[In]

integrate((a+a*sin(f*x+e))^(1/2)/(c-c*sin(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.26 \[ \int \frac {\sqrt {3+3 \sin (e+f x)}}{(c-c \sin (e+f x))^{5/2}} \, dx=-\frac {\sqrt {a} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{8 \, c^{\frac {5}{2}} f \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4}} \]

[In]

integrate((a+a*sin(f*x+e))^(1/2)/(c-c*sin(f*x+e))^(5/2),x, algorithm="giac")

[Out]

-1/8*sqrt(a)*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))/(c^(5/2)*f*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi +
1/2*f*x + 1/2*e)^4)

Mupad [B] (verification not implemented)

Time = 8.97 (sec) , antiderivative size = 142, normalized size of antiderivative = 3.38 \[ \int \frac {\sqrt {3+3 \sin (e+f x)}}{(c-c \sin (e+f x))^{5/2}} \, dx=\frac {4\,\sqrt {a\,\left (\sin \left (e+f\,x\right )+1\right )}\,\sqrt {-c\,\left (\sin \left (e+f\,x\right )-1\right )}\,\left (10\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-2\,{\sin \left (\frac {3\,e}{2}+\frac {3\,f\,x}{2}\right )}^2+4\,\sin \left (2\,e+2\,f\,x\right )-4\right )}{c^3\,f\,\left (30\,{\sin \left (e+f\,x\right )}^2+48\,\sin \left (e+f\,x\right )-52\,{\sin \left (2\,e+2\,f\,x\right )}^2+2\,{\sin \left (3\,e+3\,f\,x\right )}^2+40\,\sin \left (3\,e+3\,f\,x\right )-8\,\sin \left (5\,e+5\,f\,x\right )-32\right )} \]

[In]

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

[Out]

(4*(a*(sin(e + f*x) + 1))^(1/2)*(-c*(sin(e + f*x) - 1))^(1/2)*(4*sin(2*e + 2*f*x) + 10*sin(e/2 + (f*x)/2)^2 -
2*sin((3*e)/2 + (3*f*x)/2)^2 - 4))/(c^3*f*(48*sin(e + f*x) + 40*sin(3*e + 3*f*x) - 8*sin(5*e + 5*f*x) - 52*sin
(2*e + 2*f*x)^2 + 2*sin(3*e + 3*f*x)^2 + 30*sin(e + f*x)^2 - 32))