Integrand size = 38, antiderivative size = 134 \[ \int \frac {(a+a \sin (e+f x))^m (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{5/2}} \, dx=\frac {(A+B) \cos (e+f x) (a+a \sin (e+f x))^m}{4 f (c-c \sin (e+f x))^{5/2}}+\frac {(A (3-2 m)-B (5+2 m)) \cos (e+f x) \operatorname {Hypergeometric2F1}\left (2,\frac {1}{2}+m,\frac {3}{2}+m,\frac {1}{2} (1+\sin (e+f x))\right ) (a+a \sin (e+f x))^m}{16 c^2 f (1+2 m) \sqrt {c-c \sin (e+f x)}} \]
1/4*(A+B)*cos(f*x+e)*(a+a*sin(f*x+e))^m/f/(c-c*sin(f*x+e))^(5/2)+1/16*(A*( 3-2*m)-B*(5+2*m))*cos(f*x+e)*hypergeom([2, 1/2+m],[3/2+m],1/2+1/2*sin(f*x+ e))*(a+a*sin(f*x+e))^m/c^2/f/(1+2*m)/(c-c*sin(f*x+e))^(1/2)
Time = 33.23 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.19 \[ \int \frac {(a+a \sin (e+f x))^m (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{5/2}} \, dx=-\frac {\cos (e+f x) \left (B (5+2 m) \operatorname {Hypergeometric2F1}\left (2,\frac {1}{2}+m,\frac {3}{2}+m,\frac {1}{2} (1+\sin (e+f x))\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^4-4 \left (B+2 B m+A \operatorname {Hypergeometric2F1}\left (3,\frac {1}{2}+m,\frac {3}{2}+m,\frac {1}{2} (1+\sin (e+f x))\right ) (-1+\sin (e+f x))^2\right )\right ) (a (1+\sin (e+f x)))^m}{16 c^2 (f+2 f m) (-1+\sin (e+f x))^2 \sqrt {c-c \sin (e+f x)}} \]
-1/16*(Cos[e + f*x]*(B*(5 + 2*m)*Hypergeometric2F1[2, 1/2 + m, 3/2 + m, (1 + Sin[e + f*x])/2]*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^4 - 4*(B + 2*B*m + A*Hypergeometric2F1[3, 1/2 + m, 3/2 + m, (1 + Sin[e + f*x])/2]*(-1 + Si n[e + f*x])^2))*(a*(1 + Sin[e + f*x]))^m)/(c^2*(f + 2*f*m)*(-1 + Sin[e + f *x])^2*Sqrt[c - c*Sin[e + f*x]])
Time = 0.71 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.04, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.184, Rules used = {3042, 3451, 3042, 3224, 3042, 3146, 78}
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.
\(\displaystyle \int \frac {(a \sin (e+f x)+a)^m (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a \sin (e+f x)+a)^m (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{5/2}}dx\) |
\(\Big \downarrow \) 3451 |
\(\displaystyle \frac {(A (3-2 m)-B (2 m+5)) \int \frac {(\sin (e+f x) a+a)^m}{(c-c \sin (e+f x))^{3/2}}dx}{8 c}+\frac {(A+B) \cos (e+f x) (a \sin (e+f x)+a)^m}{4 f (c-c \sin (e+f x))^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {(A (3-2 m)-B (2 m+5)) \int \frac {(\sin (e+f x) a+a)^m}{(c-c \sin (e+f x))^{3/2}}dx}{8 c}+\frac {(A+B) \cos (e+f x) (a \sin (e+f x)+a)^m}{4 f (c-c \sin (e+f x))^{5/2}}\) |
\(\Big \downarrow \) 3224 |
\(\displaystyle \frac {(A (3-2 m)-B (2 m+5)) \cos (e+f x) \int \sec ^3(e+f x) (\sin (e+f x) a+a)^{m+\frac {3}{2}}dx}{8 a c^2 \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}+\frac {(A+B) \cos (e+f x) (a \sin (e+f x)+a)^m}{4 f (c-c \sin (e+f x))^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {(A (3-2 m)-B (2 m+5)) \cos (e+f x) \int \frac {(\sin (e+f x) a+a)^{m+\frac {3}{2}}}{\cos (e+f x)^3}dx}{8 a c^2 \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}+\frac {(A+B) \cos (e+f x) (a \sin (e+f x)+a)^m}{4 f (c-c \sin (e+f x))^{5/2}}\) |
\(\Big \downarrow \) 3146 |
\(\displaystyle \frac {a^2 (A (3-2 m)-B (2 m+5)) \cos (e+f x) \int \frac {(\sin (e+f x) a+a)^{m-\frac {1}{2}}}{(a-a \sin (e+f x))^2}d(a \sin (e+f x))}{8 c^2 f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}+\frac {(A+B) \cos (e+f x) (a \sin (e+f x)+a)^m}{4 f (c-c \sin (e+f x))^{5/2}}\) |
\(\Big \downarrow \) 78 |
\(\displaystyle \frac {(A (3-2 m)-B (2 m+5)) \cos (e+f x) (a \sin (e+f x)+a)^m \operatorname {Hypergeometric2F1}\left (2,m+\frac {1}{2},m+\frac {3}{2},\frac {\sin (e+f x) a+a}{2 a}\right )}{16 c^2 f (2 m+1) \sqrt {c-c \sin (e+f x)}}+\frac {(A+B) \cos (e+f x) (a \sin (e+f x)+a)^m}{4 f (c-c \sin (e+f x))^{5/2}}\) |
((A + B)*Cos[e + f*x]*(a + a*Sin[e + f*x])^m)/(4*f*(c - c*Sin[e + f*x])^(5 /2)) + ((A*(3 - 2*m) - B*(5 + 2*m))*Cos[e + f*x]*Hypergeometric2F1[2, 1/2 + m, 3/2 + m, (a + a*Sin[e + f*x])/(2*a)]*(a + a*Sin[e + f*x])^m)/(16*c^2* f*(1 + 2*m)*Sqrt[c - c*Sin[e + f*x]])
3.3.10.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(b *c - a*d)^n*((a + b*x)^(m + 1)/(b^(n + 1)*(m + 1)))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m}, x] && !IntegerQ[m] && IntegerQ[n]
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[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + ( f_.)*(x_)])^(n_), x_Symbol] :> Simp[a^IntPart[m]*c^IntPart[m]*(a + b*Sin[e + f*x])^FracPart[m]*((c + d*Sin[e + f*x])^FracPart[m]/Cos[e + f*x]^(2*FracP art[m])) Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m), x], x] /; F reeQ[{a, b, c, d, e, f, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && (FractionQ[m] || !FractionQ[n])
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]
\[\int \frac {\left (a +a \sin \left (f x +e \right )\right )^{m} \left (A +B \sin \left (f x +e \right )\right )}{\left (c -c \sin \left (f x +e \right )\right )^{\frac {5}{2}}}d x\]
\[ \int \frac {(a+a \sin (e+f x))^m (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{5/2}} \, dx=\int { \frac {{\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}}} \,d x } \]
integral(-(B*sin(f*x + e) + A)*sqrt(-c*sin(f*x + e) + c)*(a*sin(f*x + e) + a)^m/(3*c^3*cos(f*x + e)^2 - 4*c^3 - (c^3*cos(f*x + e)^2 - 4*c^3)*sin(f*x + e)), x)
Timed out. \[ \int \frac {(a+a \sin (e+f x))^m (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{5/2}} \, dx=\text {Timed out} \]
\[ \int \frac {(a+a \sin (e+f x))^m (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{5/2}} \, dx=\int { \frac {{\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}}} \,d x } \]
Exception generated. \[ \int \frac {(a+a \sin (e+f x))^m (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{5/2}} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error index.cc index_gcd Error: Bad Argument Value
Timed out. \[ \int \frac {(a+a \sin (e+f x))^m (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{5/2}} \, dx=\int \frac {\left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m}{{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{5/2}} \,d x \]