\(\int \frac {(3+3 \sin (e+f x))^{3/2}}{(c-c \sin (e+f x))^{9/2}} \, dx\) [356]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [F(-1)]
   Maxima [F]
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 30, antiderivative size = 88 \[ \int \frac {(3+3 \sin (e+f x))^{3/2}}{(c-c \sin (e+f x))^{9/2}} \, dx=\frac {3 \cos (e+f x) \sqrt {3+3 \sin (e+f x)}}{4 f (c-c \sin (e+f x))^{9/2}}-\frac {3 \cos (e+f x)}{4 c f \sqrt {3+3 \sin (e+f x)} (c-c \sin (e+f x))^{7/2}} \]

[Out]

-1/12*a^2*cos(f*x+e)/c/f/(c-c*sin(f*x+e))^(7/2)/(a+a*sin(f*x+e))^(1/2)+1/4*a*cos(f*x+e)*(a+a*sin(f*x+e))^(1/2)
/f/(c-c*sin(f*x+e))^(9/2)

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.05, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {2818, 2817} \[ \int \frac {(3+3 \sin (e+f x))^{3/2}}{(c-c \sin (e+f x))^{9/2}} \, dx=\frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{4 f (c-c \sin (e+f x))^{9/2}}-\frac {a^2 \cos (e+f x)}{12 c f \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{7/2}} \]

[In]

Int[(a + a*Sin[e + f*x])^(3/2)/(c - c*Sin[e + f*x])^(9/2),x]

[Out]

(a*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]])/(4*f*(c - c*Sin[e + f*x])^(9/2)) - (a^2*Cos[e + f*x])/(12*c*f*Sqrt[a
 + a*Sin[e + f*x]]*(c - c*Sin[e + f*x])^(7/2))

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)]

Rule 2818

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp
[-2*b*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[e + f*x])^n/(f*(2*n + 1))), x] - Dist[b*((2*m - 1)
/(d*(2*n + 1))), Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e
, f}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IGtQ[m - 1/2, 0] && LtQ[n, -1] &&  !(ILtQ[m + n, 0] && G
tQ[2*m + n + 1, 0])

Rubi steps \begin{align*} \text {integral}& = \frac {a \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{4 f (c-c \sin (e+f x))^{9/2}}-\frac {a \int \frac {\sqrt {a+a \sin (e+f x)}}{(c-c \sin (e+f x))^{7/2}} \, dx}{4 c} \\ & = \frac {a \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{4 f (c-c \sin (e+f x))^{9/2}}-\frac {a^2 \cos (e+f x)}{12 c f \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{7/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.12 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.23 \[ \int \frac {(3+3 \sin (e+f x))^{3/2}}{(c-c \sin (e+f x))^{9/2}} \, dx=\frac {\sqrt {3} \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) (1+\sin (e+f x))^{3/2} (1+2 \sin (e+f x))}{2 c^4 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3 (-1+\sin (e+f x))^4 \sqrt {c-c \sin (e+f x)}} \]

[In]

Integrate[(3 + 3*Sin[e + f*x])^(3/2)/(c - c*Sin[e + f*x])^(9/2),x]

[Out]

(Sqrt[3]*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(1 + Sin[e + f*x])^(3/2)*(1 + 2*Sin[e + f*x]))/(2*c^4*f*(Cos[(e
 + f*x)/2] + Sin[(e + f*x)/2])^3*(-1 + Sin[e + f*x])^4*Sqrt[c - c*Sin[e + f*x]])

Maple [A] (verified)

Time = 3.12 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.35

method result size
default \(-\frac {\sqrt {a \left (\sin \left (f x +e \right )+1\right )}\, a \left (\cos ^{3}\left (f x +e \right )+4 \sin \left (f x +e \right ) \cos \left (f x +e \right )-8 \cos \left (f x +e \right )-10 \tan \left (f x +e \right )+7 \sec \left (f x +e \right )\right )}{6 f \left (\sin \left (f x +e \right ) \left (\cos ^{2}\left (f x +e \right )\right )-3 \left (\cos ^{2}\left (f x +e \right )\right )-4 \sin \left (f x +e \right )+4\right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, c^{4}}\) \(119\)

[In]

int((a+a*sin(f*x+e))^(3/2)/(c-c*sin(f*x+e))^(9/2),x,method=_RETURNVERBOSE)

[Out]

-1/6/f*(a*(sin(f*x+e)+1))^(1/2)*a/(sin(f*x+e)*cos(f*x+e)^2-3*cos(f*x+e)^2-4*sin(f*x+e)+4)/(-c*(sin(f*x+e)-1))^
(1/2)/c^4*(cos(f*x+e)^3+4*sin(f*x+e)*cos(f*x+e)-8*cos(f*x+e)-10*tan(f*x+e)+7*sec(f*x+e))

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.30 \[ \int \frac {(3+3 \sin (e+f x))^{3/2}}{(c-c \sin (e+f x))^{9/2}} \, dx=\frac {{\left (2 \, a \sin \left (f x + e\right ) + a\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{6 \, {\left (c^{5} f \cos \left (f x + e\right )^{5} - 8 \, c^{5} f \cos \left (f x + e\right )^{3} + 8 \, c^{5} f \cos \left (f x + e\right ) + 4 \, {\left (c^{5} f \cos \left (f x + e\right )^{3} - 2 \, c^{5} f \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )}} \]

[In]

integrate((a+a*sin(f*x+e))^(3/2)/(c-c*sin(f*x+e))^(9/2),x, algorithm="fricas")

[Out]

1/6*(2*a*sin(f*x + e) + a)*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c)/(c^5*f*cos(f*x + e)^5 - 8*c^5*f*
cos(f*x + e)^3 + 8*c^5*f*cos(f*x + e) + 4*(c^5*f*cos(f*x + e)^3 - 2*c^5*f*cos(f*x + e))*sin(f*x + e))

Sympy [F(-1)]

Timed out. \[ \int \frac {(3+3 \sin (e+f x))^{3/2}}{(c-c \sin (e+f x))^{9/2}} \, dx=\text {Timed out} \]

[In]

integrate((a+a*sin(f*x+e))**(3/2)/(c-c*sin(f*x+e))**(9/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {(3+3 \sin (e+f x))^{3/2}}{(c-c \sin (e+f x))^{9/2}} \, dx=\int { \frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {3}{2}}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {9}{2}}} \,d x } \]

[In]

integrate((a+a*sin(f*x+e))^(3/2)/(c-c*sin(f*x+e))^(9/2),x, algorithm="maxima")

[Out]

integrate((a*sin(f*x + e) + a)^(3/2)/(-c*sin(f*x + e) + c)^(9/2), x)

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.99 \[ \int \frac {(3+3 \sin (e+f x))^{3/2}}{(c-c \sin (e+f x))^{9/2}} \, dx=\frac {{\left (4 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 3 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \sqrt {a}}{96 \, c^{\frac {9}{2}} f \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{8}} \]

[In]

integrate((a+a*sin(f*x+e))^(3/2)/(c-c*sin(f*x+e))^(9/2),x, algorithm="giac")

[Out]

1/96*(4*a*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e)^2 - 3*a*sgn(cos(-1/4*pi + 1/2*f*x
 + 1/2*e)))*sqrt(a)/(c^(9/2)*f*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e)^8)

Mupad [B] (verification not implemented)

Time = 11.91 (sec) , antiderivative size = 195, normalized size of antiderivative = 2.22 \[ \int \frac {(3+3 \sin (e+f x))^{3/2}}{(c-c \sin (e+f x))^{9/2}} \, dx=\frac {\left (\frac {16\,a\,{\mathrm {e}}^{e\,5{}\mathrm {i}+f\,x\,5{}\mathrm {i}}\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{3\,c^5\,f}+\frac {32\,a\,{\mathrm {e}}^{e\,5{}\mathrm {i}+f\,x\,5{}\mathrm {i}}\,\sin \left (e+f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{3\,c^5\,f}\right )\,\sqrt {c-c\,\sin \left (e+f\,x\right )}}{84\,\cos \left (e+f\,x\right )\,{\mathrm {e}}^{e\,5{}\mathrm {i}+f\,x\,5{}\mathrm {i}}-54\,{\mathrm {e}}^{e\,5{}\mathrm {i}+f\,x\,5{}\mathrm {i}}\,\cos \left (3\,e+3\,f\,x\right )+2\,{\mathrm {e}}^{e\,5{}\mathrm {i}+f\,x\,5{}\mathrm {i}}\,\cos \left (5\,e+5\,f\,x\right )-96\,{\mathrm {e}}^{e\,5{}\mathrm {i}+f\,x\,5{}\mathrm {i}}\,\sin \left (2\,e+2\,f\,x\right )+16\,{\mathrm {e}}^{e\,5{}\mathrm {i}+f\,x\,5{}\mathrm {i}}\,\sin \left (4\,e+4\,f\,x\right )} \]

[In]

int((a + a*sin(e + f*x))^(3/2)/(c - c*sin(e + f*x))^(9/2),x)

[Out]

(((16*a*exp(e*5i + f*x*5i)*(a + a*sin(e + f*x))^(1/2))/(3*c^5*f) + (32*a*exp(e*5i + f*x*5i)*sin(e + f*x)*(a +
a*sin(e + f*x))^(1/2))/(3*c^5*f))*(c - c*sin(e + f*x))^(1/2))/(84*cos(e + f*x)*exp(e*5i + f*x*5i) - 54*exp(e*5
i + f*x*5i)*cos(3*e + 3*f*x) + 2*exp(e*5i + f*x*5i)*cos(5*e + 5*f*x) - 96*exp(e*5i + f*x*5i)*sin(2*e + 2*f*x)
+ 16*exp(e*5i + f*x*5i)*sin(4*e + 4*f*x))