Integrand size = 40, antiderivative size = 210 \[ \int \frac {(a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{3/2}} \, dx=\frac {(A+B) \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{2 f (c-c \sin (e+f x))^{3/2}}+\frac {4 a^3 (A+2 B) \cos (e+f x) \log (1-\sin (e+f x))}{c f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}+\frac {2 a^2 (A+2 B) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{c f \sqrt {c-c \sin (e+f x)}}+\frac {a (A+2 B) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{2 c f \sqrt {c-c \sin (e+f x)}} \]
1/2*(A+B)*cos(f*x+e)*(a+a*sin(f*x+e))^(5/2)/f/(c-c*sin(f*x+e))^(3/2)+1/2*a *(A+2*B)*cos(f*x+e)*(a+a*sin(f*x+e))^(3/2)/c/f/(c-c*sin(f*x+e))^(1/2)+4*a^ 3*(A+2*B)*cos(f*x+e)*ln(1-sin(f*x+e))/c/f/(a+a*sin(f*x+e))^(1/2)/(c-c*sin( f*x+e))^(1/2)+2*a^2*(A+2*B)*cos(f*x+e)*(a+a*sin(f*x+e))^(1/2)/c/f/(c-c*sin (f*x+e))^(1/2)
Time = 11.77 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.10 \[ \int \frac {(a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{3/2}} \, dx=-\frac {a^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {a (1+\sin (e+f x))} \left (28 A+16 B+2 (2 A+7 B) \cos (2 (e+f x))+64 A \log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )+128 B \log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )+\left (8 A+31 B-64 (A+2 B) \log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )\right ) \sin (e+f x)+B \sin (3 (e+f x))\right )}{8 c f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) (-1+\sin (e+f x)) \sqrt {c-c \sin (e+f x)}} \]
-1/8*(a^2*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*Sqrt[a*(1 + Sin[e + f*x])] *(28*A + 16*B + 2*(2*A + 7*B)*Cos[2*(e + f*x)] + 64*A*Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]] + 128*B*Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]] + (8 *A + 31*B - 64*(A + 2*B)*Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]])*Sin[e + f*x] + B*Sin[3*(e + f*x)]))/(c*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(- 1 + Sin[e + f*x])*Sqrt[c - c*Sin[e + f*x]])
Time = 1.13 (sec) , antiderivative size = 200, normalized size of antiderivative = 0.95, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.275, Rules used = {3042, 3451, 3042, 3219, 3042, 3219, 3042, 3216, 3042, 3146, 16}
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.
| \(\displaystyle \int \frac {(a \sin (e+f x)+a)^{5/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a \sin (e+f x)+a)^{5/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{3/2}}dx\) |
| \(\Big \downarrow \) 3451 |
| \(\displaystyle \frac {(A+B) \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{2 f (c-c \sin (e+f x))^{3/2}}-\frac {(A+2 B) \int \frac {(\sin (e+f x) a+a)^{5/2}}{\sqrt {c-c \sin (e+f x)}}dx}{c}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {(A+B) \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{2 f (c-c \sin (e+f x))^{3/2}}-\frac {(A+2 B) \int \frac {(\sin (e+f x) a+a)^{5/2}}{\sqrt {c-c \sin (e+f x)}}dx}{c}\) |
| \(\Big \downarrow \) 3219 |
| \(\displaystyle \frac {(A+B) \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{2 f (c-c \sin (e+f x))^{3/2}}-\frac {(A+2 B) \left (2 a \int \frac {(\sin (e+f x) a+a)^{3/2}}{\sqrt {c-c \sin (e+f x)}}dx-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{2 f \sqrt {c-c \sin (e+f x)}}\right )}{c}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {(A+B) \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{2 f (c-c \sin (e+f x))^{3/2}}-\frac {(A+2 B) \left (2 a \int \frac {(\sin (e+f x) a+a)^{3/2}}{\sqrt {c-c \sin (e+f x)}}dx-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{2 f \sqrt {c-c \sin (e+f x)}}\right )}{c}\) |
| \(\Big \downarrow \) 3219 |
| \(\displaystyle \frac {(A+B) \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{2 f (c-c \sin (e+f x))^{3/2}}-\frac {(A+2 B) \left (2 a \left (2 a \int \frac {\sqrt {\sin (e+f x) a+a}}{\sqrt {c-c \sin (e+f x)}}dx-\frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{f \sqrt {c-c \sin (e+f x)}}\right )-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{2 f \sqrt {c-c \sin (e+f x)}}\right )}{c}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {(A+B) \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{2 f (c-c \sin (e+f x))^{3/2}}-\frac {(A+2 B) \left (2 a \left (2 a \int \frac {\sqrt {\sin (e+f x) a+a}}{\sqrt {c-c \sin (e+f x)}}dx-\frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{f \sqrt {c-c \sin (e+f x)}}\right )-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{2 f \sqrt {c-c \sin (e+f x)}}\right )}{c}\) |
| \(\Big \downarrow \) 3216 |
| \(\displaystyle \frac {(A+B) \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{2 f (c-c \sin (e+f x))^{3/2}}-\frac {(A+2 B) \left (2 a \left (\frac {2 a^2 c \cos (e+f x) \int \frac {\cos (e+f x)}{c-c \sin (e+f x)}dx}{\sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}-\frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{f \sqrt {c-c \sin (e+f x)}}\right )-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{2 f \sqrt {c-c \sin (e+f x)}}\right )}{c}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {(A+B) \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{2 f (c-c \sin (e+f x))^{3/2}}-\frac {(A+2 B) \left (2 a \left (\frac {2 a^2 c \cos (e+f x) \int \frac {\cos (e+f x)}{c-c \sin (e+f x)}dx}{\sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}-\frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{f \sqrt {c-c \sin (e+f x)}}\right )-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{2 f \sqrt {c-c \sin (e+f x)}}\right )}{c}\) |
| \(\Big \downarrow \) 3146 |
| \(\displaystyle \frac {(A+B) \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{2 f (c-c \sin (e+f x))^{3/2}}-\frac {(A+2 B) \left (2 a \left (-\frac {2 a^2 \cos (e+f x) \int \frac {1}{c-c \sin (e+f x)}d(-c \sin (e+f x))}{f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}-\frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{f \sqrt {c-c \sin (e+f x)}}\right )-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{2 f \sqrt {c-c \sin (e+f x)}}\right )}{c}\) |
| \(\Big \downarrow \) 16 |
| \(\displaystyle \frac {(A+B) \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{2 f (c-c \sin (e+f x))^{3/2}}-\frac {(A+2 B) \left (2 a \left (-\frac {2 a^2 \cos (e+f x) \log (c-c \sin (e+f x))}{f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}-\frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{f \sqrt {c-c \sin (e+f x)}}\right )-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{2 f \sqrt {c-c \sin (e+f x)}}\right )}{c}\) |
((A + B)*Cos[e + f*x]*(a + a*Sin[e + f*x])^(5/2))/(2*f*(c - c*Sin[e + f*x] )^(3/2)) - ((A + 2*B)*(-1/2*(a*Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2))/(f *Sqrt[c - c*Sin[e + f*x]]) + 2*a*((-2*a^2*Cos[e + f*x]*Log[c - c*Sin[e + f *x]])/(f*Sqrt[a + a*Sin[e + f*x]]*Sqrt[c - c*Sin[e + f*x]]) - (a*Cos[e + f *x]*Sqrt[a + a*Sin[e + f*x]])/(f*Sqrt[c - c*Sin[e + f*x]]))))/c
3.2.54.3.1 Defintions of rubi rules used
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m _.), x_Symbol] :> Simp[1/(b^p*f) Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x )^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && I ntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] || !IntegerQ[m + 1/ 2])
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Simp[a*c*(Cos[e + f*x]/(Sqrt[a + b*Sin[e + f*x ]]*Sqrt[c + d*Sin[e + f*x]])) Int[Cos[e + f*x]/(c + d*Sin[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 - 1)*((c + d*Sin[e + f*x])^n/(f*(m + n))), x] + Simp[a*((2*m - 1)/(m + n )) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n, x], x] /; Fre eQ[{a, b, c, d, e, f, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && I GtQ[m - 1/2, 0] && !LtQ[n, -1] && !(IGtQ[n - 1/2, 0] && LtQ[n, m]) && !( ILtQ[m + n, 0] && GtQ[2*m + n + 1, 0])
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Sim p[(A*b - a*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[(a*B*(m - n) + A*b*(m + n + 1))/(a*b*(2*m + 1)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[ {a, b, c, d, e, f, A, B, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0 ] && (LtQ[m, -2^(-1)] || (ILtQ[m + n, 0] && !SumSimplerQ[n, 1])) && NeQ[2* m + 1, 0]
Time = 3.57 (sec) , antiderivative size = 262, normalized size of antiderivative = 1.25
| method | result | size |
| default | \(-\frac {a^{2} \sec \left (f x +e \right ) \left (B \left (\sin ^{3}\left (f x +e \right )\right )+2 \left (\sin ^{2}\left (f x +e \right )\right ) A -8 A \sin \left (f x +e \right ) \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right )+16 A \sin \left (f x +e \right ) \ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )-1\right )+7 B \left (\sin ^{2}\left (f x +e \right )\right )-16 B \sin \left (f x +e \right ) \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right )+32 B \sin \left (f x +e \right ) \ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )-1\right )-10 A \sin \left (f x +e \right )+8 A \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right )-16 A \ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )-1\right )-16 B \sin \left (f x +e \right )+16 B \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right )-32 B \ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )-1\right )\right ) \sqrt {a \left (1+\sin \left (f x +e \right )\right )}}{2 c f \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}}\) | \(262\) |
| parts | \(-\frac {A \sec \left (f x +e \right ) \left (8 \ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )-1\right ) \sin \left (f x +e \right )-4 \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right ) \sin \left (f x +e \right )-\left (\cos ^{2}\left (f x +e \right )\right )-8 \ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )-1\right )+4 \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right )-5 \sin \left (f x +e \right )+1\right ) \sqrt {a \left (1+\sin \left (f x +e \right )\right )}\, a^{2}}{f c \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}}-\frac {B \sec \left (f x +e \right ) \left (-\left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )+32 \ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )-1\right ) \sin \left (f x +e \right )-16 \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right ) \sin \left (f x +e \right )-7 \left (\cos ^{2}\left (f x +e \right )\right )-32 \ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )-1\right )+16 \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right )-15 \sin \left (f x +e \right )+7\right ) \sqrt {a \left (1+\sin \left (f x +e \right )\right )}\, a^{2}}{2 f \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, c}\) | \(304\) |
-1/2*a^2/c/f*sec(f*x+e)*(B*sin(f*x+e)^3+2*sin(f*x+e)^2*A-8*A*sin(f*x+e)*ln (2/(1+cos(f*x+e)))+16*A*sin(f*x+e)*ln(csc(f*x+e)-cot(f*x+e)-1)+7*B*sin(f*x +e)^2-16*B*sin(f*x+e)*ln(2/(1+cos(f*x+e)))+32*B*sin(f*x+e)*ln(csc(f*x+e)-c ot(f*x+e)-1)-10*A*sin(f*x+e)+8*A*ln(2/(1+cos(f*x+e)))-16*A*ln(csc(f*x+e)-c ot(f*x+e)-1)-16*B*sin(f*x+e)+16*B*ln(2/(1+cos(f*x+e)))-32*B*ln(csc(f*x+e)- cot(f*x+e)-1))*(a*(1+sin(f*x+e)))^(1/2)/(-c*(sin(f*x+e)-1))^(1/2)
\[ \int \frac {(a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{3/2}} \, dx=\int { \frac {{\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}}} \,d x } \]
integrate((a+a*sin(f*x+e))^(5/2)*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(3/2),x , algorithm="fricas")
integral(((A + 2*B)*a^2*cos(f*x + e)^2 - 2*(A + B)*a^2 + (B*a^2*cos(f*x + e)^2 - 2*(A + B)*a^2)*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f *x + e) + c)/(c^2*cos(f*x + e)^2 + 2*c^2*sin(f*x + e) - 2*c^2), x)
Timed out. \[ \int \frac {(a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{3/2}} \, dx=\text {Timed out} \]
\[ \int \frac {(a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{3/2}} \, dx=\int { \frac {{\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}}} \,d x } \]
integrate((a+a*sin(f*x+e))^(5/2)*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(3/2),x , algorithm="maxima")
Exception generated. \[ \int \frac {(a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{3/2}} \, dx=\text {Exception raised: TypeError} \]
integrate((a+a*sin(f*x+e))^(5/2)*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(3/2),x , algorithm="giac")
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error index.cc index_gcd Error: Bad Argument Value
Timed out. \[ \int \frac {(a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{3/2}} \, dx=\int \frac {\left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{5/2}}{{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{3/2}} \,d x \]