Integrand size = 30, antiderivative size = 88 \[ \int \frac {(3+3 \sin (e+f x))^{3/2}}{(c-c \sin (e+f x))^{11/2}} \, dx=\frac {3 \cos (e+f x) \sqrt {3+3 \sin (e+f x)}}{5 f (c-c \sin (e+f x))^{11/2}}-\frac {9 \cos (e+f x)}{20 c f \sqrt {3+3 \sin (e+f x)} (c-c \sin (e+f x))^{9/2}} \]
-1/20*a^2*cos(f*x+e)/c/f/(c-c*sin(f*x+e))^(9/2)/(a+a*sin(f*x+e))^(1/2)+1/5 *a*cos(f*x+e)*(a+a*sin(f*x+e))^(1/2)/f/(c-c*sin(f*x+e))^(11/2)
Time = 3.11 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.23 \[ \int \frac {(3+3 \sin (e+f x))^{3/2}}{(c-c \sin (e+f x))^{11/2}} \, dx=-\frac {3 \sqrt {3} \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) (1+\sin (e+f x))^{3/2} (3+5 \sin (e+f x))}{20 c^5 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3 (-1+\sin (e+f x))^5 \sqrt {c-c \sin (e+f x)}} \]
(-3*Sqrt[3]*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(1 + Sin[e + f*x])^(3/2) *(3 + 5*Sin[e + f*x]))/(20*c^5*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3*( -1 + Sin[e + f*x])^5*Sqrt[c - c*Sin[e + f*x]])
Time = 0.42 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.05, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {3042, 3218, 3042, 3217}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {(a \sin (e+f x)+a)^{3/2}}{(c-c \sin (e+f x))^{11/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a \sin (e+f x)+a)^{3/2}}{(c-c \sin (e+f x))^{11/2}}dx\) |
\(\Big \downarrow \) 3218 |
\(\displaystyle \frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{5 f (c-c \sin (e+f x))^{11/2}}-\frac {a \int \frac {\sqrt {\sin (e+f x) a+a}}{(c-c \sin (e+f x))^{9/2}}dx}{5 c}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{5 f (c-c \sin (e+f x))^{11/2}}-\frac {a \int \frac {\sqrt {\sin (e+f x) a+a}}{(c-c \sin (e+f x))^{9/2}}dx}{5 c}\) |
\(\Big \downarrow \) 3217 |
\(\displaystyle \frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{5 f (c-c \sin (e+f x))^{11/2}}-\frac {a^2 \cos (e+f x)}{20 c f \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{9/2}}\) |
(a*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]])/(5*f*(c - c*Sin[e + f*x])^(11/2) ) - (a^2*Cos[e + f*x])/(20*c*f*Sqrt[a + a*Sin[e + f*x]]*(c - c*Sin[e + f*x ])^(9/2))
3.4.57.3.1 Defintions of rubi rules used
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)]
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] - Simp[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] && GtQ[2*m + n + 1, 0])
Time = 3.26 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.66
method | result | size |
default | \(\frac {\sqrt {a \left (\sin \left (f x +e \right )+1\right )}\, a \left (3 \left (\cos ^{3}\left (f x +e \right )\right ) \sin \left (f x +e \right )-15 \left (\cos ^{3}\left (f x +e \right )\right )-36 \sin \left (f x +e \right ) \cos \left (f x +e \right )+60 \cos \left (f x +e \right )+53 \tan \left (f x +e \right )-45 \sec \left (f x +e \right )\right )}{20 f \left (\cos ^{4}\left (f x +e \right )+4 \sin \left (f x +e \right ) \left (\cos ^{2}\left (f x +e \right )\right )-8 \left (\cos ^{2}\left (f x +e \right )\right )-8 \sin \left (f x +e \right )+8\right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, c^{5}}\) | \(146\) |
1/20/f*(a*(sin(f*x+e)+1))^(1/2)*a/(cos(f*x+e)^4+4*sin(f*x+e)*cos(f*x+e)^2- 8*cos(f*x+e)^2-8*sin(f*x+e)+8)/(-c*(sin(f*x+e)-1))^(1/2)/c^5*(3*cos(f*x+e) ^3*sin(f*x+e)-15*cos(f*x+e)^3-36*sin(f*x+e)*cos(f*x+e)+60*cos(f*x+e)+53*ta n(f*x+e)-45*sec(f*x+e))
Time = 0.28 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.49 \[ \int \frac {(3+3 \sin (e+f x))^{3/2}}{(c-c \sin (e+f x))^{11/2}} \, dx=\frac {{\left (5 \, a \sin \left (f x + e\right ) + 3 \, a\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{20 \, {\left (5 \, c^{6} f \cos \left (f x + e\right )^{5} - 20 \, c^{6} f \cos \left (f x + e\right )^{3} + 16 \, c^{6} f \cos \left (f x + e\right ) - {\left (c^{6} f \cos \left (f x + e\right )^{5} - 12 \, c^{6} f \cos \left (f x + e\right )^{3} + 16 \, c^{6} f \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )}} \]
1/20*(5*a*sin(f*x + e) + 3*a)*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e ) + c)/(5*c^6*f*cos(f*x + e)^5 - 20*c^6*f*cos(f*x + e)^3 + 16*c^6*f*cos(f* x + e) - (c^6*f*cos(f*x + e)^5 - 12*c^6*f*cos(f*x + e)^3 + 16*c^6*f*cos(f* x + e))*sin(f*x + e))
Timed out. \[ \int \frac {(3+3 \sin (e+f x))^{3/2}}{(c-c \sin (e+f x))^{11/2}} \, dx=\text {Timed out} \]
\[ \int \frac {(3+3 \sin (e+f x))^{3/2}}{(c-c \sin (e+f x))^{11/2}} \, dx=\int { \frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {3}{2}}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {11}{2}}} \,d x } \]
Time = 0.34 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.06 \[ \int \frac {(3+3 \sin (e+f x))^{3/2}}{(c-c \sin (e+f x))^{11/2}} \, dx=\frac {{\left (5 \, a \sqrt {c} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 4 \, a \sqrt {c} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \sqrt {a}}{320 \, c^{6} f \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{10}} \]
1/320*(5*a*sqrt(c)*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f *x + 1/2*e)^2 - 4*a*sqrt(c)*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)))*sqrt(a)/( c^6*f*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e)^1 0)
Time = 13.06 (sec) , antiderivative size = 225, normalized size of antiderivative = 2.56 \[ \int \frac {(3+3 \sin (e+f x))^{3/2}}{(c-c \sin (e+f x))^{11/2}} \, dx=\frac {\left (\frac {a\,{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\sqrt {a+a\,\sin \left (e+f\,x\right )}\,48{}\mathrm {i}}{5\,c^6\,f}+\frac {a\,{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\sin \left (e+f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}\,16{}\mathrm {i}}{c^6\,f}\right )\,\sqrt {c-c\,\sin \left (e+f\,x\right )}}{\cos \left (e+f\,x\right )\,{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,264{}\mathrm {i}-{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\cos \left (3\,e+3\,f\,x\right )\,220{}\mathrm {i}+{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\cos \left (5\,e+5\,f\,x\right )\,20{}\mathrm {i}-{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\sin \left (2\,e+2\,f\,x\right )\,330{}\mathrm {i}+{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\sin \left (4\,e+4\,f\,x\right )\,88{}\mathrm {i}-{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\sin \left (6\,e+6\,f\,x\right )\,2{}\mathrm {i}} \]
(((a*exp(e*6i + f*x*6i)*(a + a*sin(e + f*x))^(1/2)*48i)/(5*c^6*f) + (a*exp (e*6i + f*x*6i)*sin(e + f*x)*(a + a*sin(e + f*x))^(1/2)*16i)/(c^6*f))*(c - c*sin(e + f*x))^(1/2))/(cos(e + f*x)*exp(e*6i + f*x*6i)*264i - exp(e*6i + f*x*6i)*cos(3*e + 3*f*x)*220i + exp(e*6i + f*x*6i)*cos(5*e + 5*f*x)*20i - exp(e*6i + f*x*6i)*sin(2*e + 2*f*x)*330i + exp(e*6i + f*x*6i)*sin(4*e + 4 *f*x)*88i - exp(e*6i + f*x*6i)*sin(6*e + 6*f*x)*2i)