Integrand size = 30, antiderivative size = 224 \[ \int \frac {1}{(3+3 \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{5/2}} \, dx=-\frac {\cos (e+f x)}{4 f (3+3 \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{5/2}}-\frac {\cos (e+f x)}{6 f (3+3 \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2}}+\frac {\cos (e+f x)}{24 f \sqrt {3+3 \sin (e+f x)} (c-c \sin (e+f x))^{5/2}}+\frac {\cos (e+f x)}{24 c f \sqrt {3+3 \sin (e+f x)} (c-c \sin (e+f x))^{3/2}}+\frac {\text {arctanh}(\sin (e+f x)) \cos (e+f x)}{24 c^2 f \sqrt {3+3 \sin (e+f x)} \sqrt {c-c \sin (e+f x)}} \]
-1/4*cos(f*x+e)/f/(a+a*sin(f*x+e))^(5/2)/(c-c*sin(f*x+e))^(5/2)-1/2*cos(f* x+e)/a/f/(a+a*sin(f*x+e))^(3/2)/(c-c*sin(f*x+e))^(5/2)+3/8*cos(f*x+e)/a^2/ f/(c-c*sin(f*x+e))^(5/2)/(a+a*sin(f*x+e))^(1/2)+3/8*cos(f*x+e)/a^2/c/f/(c- c*sin(f*x+e))^(3/2)/(a+a*sin(f*x+e))^(1/2)+3/8*arctanh(sin(f*x+e))*cos(f*x +e)/a^2/c^2/f/(a+a*sin(f*x+e))^(1/2)/(c-c*sin(f*x+e))^(1/2)
Time = 1.33 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.02 \[ \int \frac {1}{(3+3 \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{5/2}} \, dx=\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (-9 \log \left (1-\tan \left (\frac {1}{2} (e+f x)\right )\right )-12 \cos (2 (e+f x)) \left (\log \left (1-\tan \left (\frac {1}{2} (e+f x)\right )\right )-\log \left (1+\tan \left (\frac {1}{2} (e+f x)\right )\right )\right )-3 \cos (4 (e+f x)) \left (\log \left (1-\tan \left (\frac {1}{2} (e+f x)\right )\right )-\log \left (1+\tan \left (\frac {1}{2} (e+f x)\right )\right )\right )+9 \log \left (1+\tan \left (\frac {1}{2} (e+f x)\right )\right )+22 \sin (e+f x)+6 \sin (3 (e+f x))\right )}{576 \sqrt {3} c^2 f (-1+\sin (e+f x))^2 (1+\sin (e+f x))^{5/2} \sqrt {c-c \sin (e+f x)}} \]
((Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2 ])*(-9*Log[1 - Tan[(e + f*x)/2]] - 12*Cos[2*(e + f*x)]*(Log[1 - Tan[(e + f *x)/2]] - Log[1 + Tan[(e + f*x)/2]]) - 3*Cos[4*(e + f*x)]*(Log[1 - Tan[(e + f*x)/2]] - Log[1 + Tan[(e + f*x)/2]]) + 9*Log[1 + Tan[(e + f*x)/2]] + 22 *Sin[e + f*x] + 6*Sin[3*(e + f*x)]))/(576*Sqrt[3]*c^2*f*(-1 + Sin[e + f*x] )^2*(1 + Sin[e + f*x])^(5/2)*Sqrt[c - c*Sin[e + f*x]])
Time = 1.20 (sec) , antiderivative size = 242, normalized size of antiderivative = 1.08, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 3222, 3042, 3222, 3042, 3222, 3042, 3222, 3042, 3220, 3042, 4257}
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.
| \(\displaystyle \int \frac {1}{(a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{(a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{5/2}}dx\) |
| \(\Big \downarrow \) 3222 |
| \(\displaystyle \frac {\int \frac {1}{(\sin (e+f x) a+a)^{3/2} (c-c \sin (e+f x))^{5/2}}dx}{a}-\frac {\cos (e+f x)}{4 f (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {1}{(\sin (e+f x) a+a)^{3/2} (c-c \sin (e+f x))^{5/2}}dx}{a}-\frac {\cos (e+f x)}{4 f (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 3222 |
| \(\displaystyle \frac {\frac {3 \int \frac {1}{\sqrt {\sin (e+f x) a+a} (c-c \sin (e+f x))^{5/2}}dx}{2 a}-\frac {\cos (e+f x)}{2 f (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{5/2}}}{a}-\frac {\cos (e+f x)}{4 f (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {3 \int \frac {1}{\sqrt {\sin (e+f x) a+a} (c-c \sin (e+f x))^{5/2}}dx}{2 a}-\frac {\cos (e+f x)}{2 f (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{5/2}}}{a}-\frac {\cos (e+f x)}{4 f (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 3222 |
| \(\displaystyle \frac {\frac {3 \left (\frac {\int \frac {1}{\sqrt {\sin (e+f x) a+a} (c-c \sin (e+f x))^{3/2}}dx}{2 c}+\frac {\cos (e+f x)}{4 f \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{5/2}}\right )}{2 a}-\frac {\cos (e+f x)}{2 f (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{5/2}}}{a}-\frac {\cos (e+f x)}{4 f (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {3 \left (\frac {\int \frac {1}{\sqrt {\sin (e+f x) a+a} (c-c \sin (e+f x))^{3/2}}dx}{2 c}+\frac {\cos (e+f x)}{4 f \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{5/2}}\right )}{2 a}-\frac {\cos (e+f x)}{2 f (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{5/2}}}{a}-\frac {\cos (e+f x)}{4 f (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 3222 |
| \(\displaystyle \frac {\frac {3 \left (\frac {\frac {\int \frac {1}{\sqrt {\sin (e+f x) a+a} \sqrt {c-c \sin (e+f x)}}dx}{2 c}+\frac {\cos (e+f x)}{2 f \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{3/2}}}{2 c}+\frac {\cos (e+f x)}{4 f \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{5/2}}\right )}{2 a}-\frac {\cos (e+f x)}{2 f (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{5/2}}}{a}-\frac {\cos (e+f x)}{4 f (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {3 \left (\frac {\frac {\int \frac {1}{\sqrt {\sin (e+f x) a+a} \sqrt {c-c \sin (e+f x)}}dx}{2 c}+\frac {\cos (e+f x)}{2 f \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{3/2}}}{2 c}+\frac {\cos (e+f x)}{4 f \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{5/2}}\right )}{2 a}-\frac {\cos (e+f x)}{2 f (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{5/2}}}{a}-\frac {\cos (e+f x)}{4 f (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 3220 |
| \(\displaystyle \frac {\frac {3 \left (\frac {\frac {\cos (e+f x) \int \sec (e+f x)dx}{2 c \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}+\frac {\cos (e+f x)}{2 f \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{3/2}}}{2 c}+\frac {\cos (e+f x)}{4 f \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{5/2}}\right )}{2 a}-\frac {\cos (e+f x)}{2 f (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{5/2}}}{a}-\frac {\cos (e+f x)}{4 f (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {3 \left (\frac {\frac {\cos (e+f x) \int \csc \left (e+f x+\frac {\pi }{2}\right )dx}{2 c \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}+\frac {\cos (e+f x)}{2 f \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{3/2}}}{2 c}+\frac {\cos (e+f x)}{4 f \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{5/2}}\right )}{2 a}-\frac {\cos (e+f x)}{2 f (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{5/2}}}{a}-\frac {\cos (e+f x)}{4 f (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {\frac {3 \left (\frac {\frac {\cos (e+f x) \text {arctanh}(\sin (e+f x))}{2 c f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}+\frac {\cos (e+f x)}{2 f \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{3/2}}}{2 c}+\frac {\cos (e+f x)}{4 f \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{5/2}}\right )}{2 a}-\frac {\cos (e+f x)}{2 f (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{5/2}}}{a}-\frac {\cos (e+f x)}{4 f (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{5/2}}\) |
-1/4*Cos[e + f*x]/(f*(a + a*Sin[e + f*x])^(5/2)*(c - c*Sin[e + f*x])^(5/2) ) + (-1/2*Cos[e + f*x]/(f*(a + a*Sin[e + f*x])^(3/2)*(c - c*Sin[e + f*x])^ (5/2)) + (3*(Cos[e + f*x]/(4*f*Sqrt[a + a*Sin[e + f*x]]*(c - c*Sin[e + f*x ])^(5/2)) + (Cos[e + f*x]/(2*f*Sqrt[a + a*Sin[e + f*x]]*(c - c*Sin[e + f*x ])^(3/2)) + (ArcTanh[Sin[e + f*x]]*Cos[e + f*x])/(2*c*f*Sqrt[a + a*Sin[e + f*x]]*Sqrt[c - c*Sin[e + f*x]]))/(2*c)))/(2*a))/a
3.5.3.3.1 Defintions of rubi rules used
Int[1/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_) + (d_.)*sin[(e_ .) + (f_.)*(x_)]]), x_Symbol] :> Simp[Cos[e + f*x]/(Sqrt[a + b*Sin[e + f*x] ]*Sqrt[c + d*Sin[e + f*x]]) Int[1/Cos[e + f*x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + ( f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*( (c + d*Sin[e + f*x])^n/(a*f*(2*m + 1))), x] + Simp[(m + n + 1)/(a*(2*m + 1) ) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n, x], x] /; Free Q[{a, b, c, d, e, f, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && ILtQ[Simplify[m + n + 1], 0] && NeQ[m, -2^(-1)] && (SumSimplerQ[m, 1] || !SumSimplerQ[n, 1])
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 3.37 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.50
| method | result | size |
| default | \(-\frac {3 \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )-1\right ) \cos \left (f x +e \right )-3 \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1\right ) \cos \left (f x +e \right )-3 \tan \left (f x +e \right )-2 \tan \left (f x +e \right ) \left (\sec ^{2}\left (f x +e \right )\right )}{8 f \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, \sqrt {a \left (\sin \left (f x +e \right )+1\right )}\, a^{2} c^{2}}\) | \(112\) |
-1/8/f/(-c*(sin(f*x+e)-1))^(1/2)/(a*(sin(f*x+e)+1))^(1/2)/a^2/c^2*(3*ln(-c ot(f*x+e)+csc(f*x+e)-1)*cos(f*x+e)-3*ln(-cot(f*x+e)+csc(f*x+e)+1)*cos(f*x+ e)-3*tan(f*x+e)-2*tan(f*x+e)*sec(f*x+e)^2)
Time = 0.33 (sec) , antiderivative size = 288, normalized size of antiderivative = 1.29 \[ \int \frac {1}{(3+3 \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{5/2}} \, dx=\left [\frac {3 \, \sqrt {a c} \cos \left (f x + e\right )^{5} \log \left (-\frac {a c \cos \left (f x + e\right )^{3} - 2 \, a c \cos \left (f x + e\right ) - 2 \, \sqrt {a c} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c} \sin \left (f x + e\right )}{\cos \left (f x + e\right )^{3}}\right ) + 2 \, {\left (3 \, \cos \left (f x + e\right )^{2} + 2\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c} \sin \left (f x + e\right )}{16 \, a^{3} c^{3} f \cos \left (f x + e\right )^{5}}, -\frac {3 \, \sqrt {-a c} \arctan \left (\frac {\sqrt {-a c} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{a c \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right ) \cos \left (f x + e\right )^{5} - {\left (3 \, \cos \left (f x + e\right )^{2} + 2\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c} \sin \left (f x + e\right )}{8 \, a^{3} c^{3} f \cos \left (f x + e\right )^{5}}\right ] \]
[1/16*(3*sqrt(a*c)*cos(f*x + e)^5*log(-(a*c*cos(f*x + e)^3 - 2*a*c*cos(f*x + e) - 2*sqrt(a*c)*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c)*sin (f*x + e))/cos(f*x + e)^3) + 2*(3*cos(f*x + e)^2 + 2)*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c)*sin(f*x + e))/(a^3*c^3*f*cos(f*x + e)^5), - 1/8*(3*sqrt(-a*c)*arctan(sqrt(-a*c)*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f *x + e) + c)/(a*c*cos(f*x + e)*sin(f*x + e)))*cos(f*x + e)^5 - (3*cos(f*x + e)^2 + 2)*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c)*sin(f*x + e ))/(a^3*c^3*f*cos(f*x + e)^5)]
Timed out. \[ \int \frac {1}{(3+3 \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{5/2}} \, dx=\text {Timed out} \]
\[ \int \frac {1}{(3+3 \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{5/2}} \, dx=\int { \frac {1}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}} {\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}}} \,d x } \]
Exception generated. \[ \int \frac {1}{(3+3 \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{5/2}} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:index.cc index_m i_lex_is_greater E rror: Bad Argument Value
Timed out. \[ \int \frac {1}{(3+3 \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{5/2}} \, dx=\int \frac {1}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{5/2}\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{5/2}} \,d x \]