Integrand size = 48, antiderivative size = 230 \[ \int \frac {(a+a \sin (e+f x))^m \left (A+B \sin (e+f x)+C \sin ^2(e+f x)\right )}{(c-c \sin (e+f x))^{5/2}} \, dx=\frac {(A+B+C) \cos (e+f x) (a+a \sin (e+f x))^{1+m}}{8 a f (c-c \sin (e+f x))^{5/2}}+\frac {(A (5-2 m)-B (3+2 m)-C (11+2 m)) \cos (e+f x) (a+a \sin (e+f x))^m}{16 c f (c-c \sin (e+f x))^{3/2}}-\frac {\left (B \left (5-8 m-4 m^2\right )-A \left (3-8 m+4 m^2\right )-C \left (19+24 m+4 m^2\right )\right ) \cos (e+f x) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2}+m,\frac {3}{2}+m,\frac {1}{2} (1+\sin (e+f x))\right ) (a+a \sin (e+f x))^m}{32 c^2 f (1+2 m) \sqrt {c-c \sin (e+f x)}} \]
1/8*(A+B+C)*cos(f*x+e)*(a+a*sin(f*x+e))^(1+m)/a/f/(c-c*sin(f*x+e))^(5/2)+1 /16*(A*(5-2*m)-B*(3+2*m)-C*(11+2*m))*cos(f*x+e)*(a+a*sin(f*x+e))^m/c/f/(c- c*sin(f*x+e))^(3/2)-1/32*(B*(-4*m^2-8*m+5)-A*(4*m^2-8*m+3)-C*(4*m^2+24*m+1 9))*cos(f*x+e)*hypergeom([1, 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 = 29.42 (sec) , antiderivative size = 228, normalized size of antiderivative = 0.99 \[ \int \frac {(a+a \sin (e+f x))^m \left (A+B \sin (e+f x)+C \sin ^2(e+f x)\right )}{(c-c \sin (e+f x))^{5/2}} \, dx=\frac {\cos (e+f x) (a (1+\sin (e+f x)))^m \left (8 B (1+2 m)-2 (16 C+B (5+2 m)) \operatorname {Hypergeometric2F1}\left (2,\frac {1}{2}+m,\frac {3}{2}+m,\frac {1}{2} (1+\sin (e+f x))\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^4+8 (A+C) \operatorname {Hypergeometric2F1}\left (3,\frac {1}{2}+m,\frac {3}{2}+m,\frac {1}{2} (1+\sin (e+f x))\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^4-16 C \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2}+m,\frac {3}{2}+m,\frac {1}{2} (1+\sin (e+f x))\right ) (-3+\cos (2 (e+f x))+4 \sin (e+f x))\right )}{32 c^2 (f+2 f m) (-1+\sin (e+f x))^2 \sqrt {c-c \sin (e+f x)}} \]
Integrate[((a + a*Sin[e + f*x])^m*(A + B*Sin[e + f*x] + C*Sin[e + f*x]^2)) /(c - c*Sin[e + f*x])^(5/2),x]
(Cos[e + f*x]*(a*(1 + Sin[e + f*x]))^m*(8*B*(1 + 2*m) - 2*(16*C + 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 + 8*(A + C)*Hypergeometric2F1[3, 1/2 + m, 3/2 + m, (1 + Sin[e + f*x])/2]*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^4 - 1 6*C*Hypergeometric2F1[1, 1/2 + m, 3/2 + m, (1 + Sin[e + f*x])/2]*(-3 + Cos [2*(e + f*x)] + 4*Sin[e + f*x])))/(32*c^2*(f + 2*f*m)*(-1 + Sin[e + f*x])^ 2*Sqrt[c - c*Sin[e + f*x]])
Time = 1.25 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.07, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {3042, 3514, 27, 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.
| \(\displaystyle \int \frac {(a \sin (e+f x)+a)^m \left (A+B \sin (e+f x)+C \sin ^2(e+f x)\right )}{(c-c \sin (e+f x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a \sin (e+f x)+a)^m \left (A+B \sin (e+f x)+C \sin (e+f x)^2\right )}{(c-c \sin (e+f x))^{5/2}}dx\) |
| \(\Big \downarrow \) 3514 |
| \(\displaystyle \frac {(A+B+C) \cos (e+f x) (a \sin (e+f x)+a)^{m+1}}{8 a f (c-c \sin (e+f x))^{5/2}}-\frac {\int -\frac {(\sin (e+f x) a+a)^m \left ((A (9-2 m)-(B+C) (2 m+7)) a^2+((A+B) (1-2 m)-C (2 m+15)) \sin (e+f x) a^2\right )}{2 (c-c \sin (e+f x))^{3/2}}dx}{8 a^2 c}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {(\sin (e+f x) a+a)^m \left ((A (9-2 m)-(B+C) (2 m+7)) a^2+((A+B) (1-2 m)-C (2 m+15)) \sin (e+f x) a^2\right )}{(c-c \sin (e+f x))^{3/2}}dx}{16 a^2 c}+\frac {(A+B+C) \cos (e+f x) (a \sin (e+f x)+a)^{m+1}}{8 a f (c-c \sin (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {(\sin (e+f x) a+a)^m \left ((A (9-2 m)-(B+C) (2 m+7)) a^2+((A+B) (1-2 m)-C (2 m+15)) \sin (e+f x) a^2\right )}{(c-c \sin (e+f x))^{3/2}}dx}{16 a^2 c}+\frac {(A+B+C) \cos (e+f x) (a \sin (e+f x)+a)^{m+1}}{8 a f (c-c \sin (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 3451 |
| \(\displaystyle \frac {\frac {a^2 (A (5-2 m)-B (2 m+3)-C (2 m+11)) \cos (e+f x) (a \sin (e+f x)+a)^m}{f (c-c \sin (e+f x))^{3/2}}-\frac {a^2 \left (-A \left (4 m^2-8 m+3\right )+B \left (-4 m^2-8 m+5\right )-C \left (4 m^2+24 m+19\right )\right ) \int \frac {(\sin (e+f x) a+a)^m}{\sqrt {c-c \sin (e+f x)}}dx}{2 c}}{16 a^2 c}+\frac {(A+B+C) \cos (e+f x) (a \sin (e+f x)+a)^{m+1}}{8 a f (c-c \sin (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {a^2 (A (5-2 m)-B (2 m+3)-C (2 m+11)) \cos (e+f x) (a \sin (e+f x)+a)^m}{f (c-c \sin (e+f x))^{3/2}}-\frac {a^2 \left (-A \left (4 m^2-8 m+3\right )+B \left (-4 m^2-8 m+5\right )-C \left (4 m^2+24 m+19\right )\right ) \int \frac {(\sin (e+f x) a+a)^m}{\sqrt {c-c \sin (e+f x)}}dx}{2 c}}{16 a^2 c}+\frac {(A+B+C) \cos (e+f x) (a \sin (e+f x)+a)^{m+1}}{8 a f (c-c \sin (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 3224 |
| \(\displaystyle \frac {\frac {a^2 (A (5-2 m)-B (2 m+3)-C (2 m+11)) \cos (e+f x) (a \sin (e+f x)+a)^m}{f (c-c \sin (e+f x))^{3/2}}-\frac {a^2 \left (-A \left (4 m^2-8 m+3\right )+B \left (-4 m^2-8 m+5\right )-C \left (4 m^2+24 m+19\right )\right ) \cos (e+f x) \int \sec (e+f x) (\sin (e+f x) a+a)^{m+\frac {1}{2}}dx}{2 c \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}}{16 a^2 c}+\frac {(A+B+C) \cos (e+f x) (a \sin (e+f x)+a)^{m+1}}{8 a f (c-c \sin (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {a^2 (A (5-2 m)-B (2 m+3)-C (2 m+11)) \cos (e+f x) (a \sin (e+f x)+a)^m}{f (c-c \sin (e+f x))^{3/2}}-\frac {a^2 \left (-A \left (4 m^2-8 m+3\right )+B \left (-4 m^2-8 m+5\right )-C \left (4 m^2+24 m+19\right )\right ) \cos (e+f x) \int \frac {(\sin (e+f x) a+a)^{m+\frac {1}{2}}}{\cos (e+f x)}dx}{2 c \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}}{16 a^2 c}+\frac {(A+B+C) \cos (e+f x) (a \sin (e+f x)+a)^{m+1}}{8 a f (c-c \sin (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 3146 |
| \(\displaystyle \frac {\frac {a^2 (A (5-2 m)-B (2 m+3)-C (2 m+11)) \cos (e+f x) (a \sin (e+f x)+a)^m}{f (c-c \sin (e+f x))^{3/2}}-\frac {a^3 \left (-A \left (4 m^2-8 m+3\right )+B \left (-4 m^2-8 m+5\right )-C \left (4 m^2+24 m+19\right )\right ) \cos (e+f x) \int \frac {(\sin (e+f x) a+a)^{m-\frac {1}{2}}}{a-a \sin (e+f x)}d(a \sin (e+f x))}{2 c f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}}{16 a^2 c}+\frac {(A+B+C) \cos (e+f x) (a \sin (e+f x)+a)^{m+1}}{8 a f (c-c \sin (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 78 |
| \(\displaystyle \frac {\frac {a^2 (A (5-2 m)-B (2 m+3)-C (2 m+11)) \cos (e+f x) (a \sin (e+f x)+a)^m}{f (c-c \sin (e+f x))^{3/2}}-\frac {a^2 \left (-A \left (4 m^2-8 m+3\right )+B \left (-4 m^2-8 m+5\right )-C \left (4 m^2+24 m+19\right )\right ) \cos (e+f x) (a \sin (e+f x)+a)^m \operatorname {Hypergeometric2F1}\left (1,m+\frac {1}{2},m+\frac {3}{2},\frac {\sin (e+f x) a+a}{2 a}\right )}{2 c f (2 m+1) \sqrt {c-c \sin (e+f x)}}}{16 a^2 c}+\frac {(A+B+C) \cos (e+f x) (a \sin (e+f x)+a)^{m+1}}{8 a f (c-c \sin (e+f x))^{5/2}}\) |
Int[((a + a*Sin[e + f*x])^m*(A + B*Sin[e + f*x] + C*Sin[e + f*x]^2))/(c - c*Sin[e + f*x])^(5/2),x]
((A + B + C)*Cos[e + f*x]*(a + a*Sin[e + f*x])^(1 + m))/(8*a*f*(c - c*Sin[ e + f*x])^(5/2)) + ((a^2*(A*(5 - 2*m) - B*(3 + 2*m) - C*(11 + 2*m))*Cos[e + f*x]*(a + a*Sin[e + f*x])^m)/(f*(c - c*Sin[e + f*x])^(3/2)) - (a^2*(B*(5 - 8*m - 4*m^2) - A*(3 - 8*m + 4*m^2) - C*(19 + 24*m + 4*m^2))*Cos[e + f*x ]*Hypergeometric2F1[1, 1/2 + m, 3/2 + m, (a + a*Sin[e + f*x])/(2*a)]*(a + a*Sin[e + f*x])^m)/(2*c*f*(1 + 2*m)*Sqrt[c - c*Sin[e + f*x]]))/(16*a^2*c)
3.1.23.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(a*A - b*B + a*C)*Cos[e + f*x]*(a + b* Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(2*b*c*f*(2*m + 1))), x] - Si mp[1/(2*b*c*d*(2*m + 1)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[A*(c^2*(m + 1) + d^2*(2*m + n + 2)) - B*c*d*(m - n - 1) - C*(c ^2*m - d^2*(n + 1)) + d*((A*c + B*d)*(m + n + 2) - c*C*(3*m - n))*Sin[e + f *x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && (LtQ[m, -2^(-1)] || (EqQ[m + n + 2, 0] && NeQ[2*m + 1, 0]))
\[\int \frac {\left (a +a \sin \left (f x +e \right )\right )^{m} \left (A +B \sin \left (f x +e \right )+C \left (\sin ^{2}\left (f x +e \right )\right )\right )}{\left (c -c \sin \left (f x +e \right )\right )^{\frac {5}{2}}}d x\]
\[ \int \frac {(a+a \sin (e+f x))^m \left (A+B \sin (e+f x)+C \sin ^2(e+f x)\right )}{(c-c \sin (e+f x))^{5/2}} \, dx=\int { \frac {{\left (C \sin \left (f x + e\right )^{2} + B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}}} \,d x } \]
integrate((a+a*sin(f*x+e))^m*(A+B*sin(f*x+e)+C*sin(f*x+e)^2)/(c-c*sin(f*x+ e))^(5/2),x, algorithm="fricas")
integral((C*cos(f*x + e)^2 - B*sin(f*x + e) - A - C)*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. \[ \int \frac {(a+a \sin (e+f x))^m \left (A+B \sin (e+f x)+C \sin ^2(e+f x)\right )}{(c-c \sin (e+f x))^{5/2}} \, dx=\text {Timed out} \]
\[ \int \frac {(a+a \sin (e+f x))^m \left (A+B \sin (e+f x)+C \sin ^2(e+f x)\right )}{(c-c \sin (e+f x))^{5/2}} \, dx=\int { \frac {{\left (C \sin \left (f x + e\right )^{2} + B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}}} \,d x } \]
integrate((a+a*sin(f*x+e))^m*(A+B*sin(f*x+e)+C*sin(f*x+e)^2)/(c-c*sin(f*x+ e))^(5/2),x, algorithm="maxima")
integrate((C*sin(f*x + e)^2 + B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)^m/( -c*sin(f*x + e) + c)^(5/2), x)
Exception generated. \[ \int \frac {(a+a \sin (e+f x))^m \left (A+B \sin (e+f x)+C \sin ^2(e+f x)\right )}{(c-c \sin (e+f x))^{5/2}} \, dx=\text {Exception raised: TypeError} \]
integrate((a+a*sin(f*x+e))^m*(A+B*sin(f*x+e)+C*sin(f*x+e)^2)/(c-c*sin(f*x+ e))^(5/2),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. \[ \int \frac {(a+a \sin (e+f x))^m \left (A+B \sin (e+f x)+C \sin ^2(e+f x)\right )}{(c-c \sin (e+f x))^{5/2}} \, dx=\int \frac {{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,\left (C\,{\sin \left (e+f\,x\right )}^2+B\,\sin \left (e+f\,x\right )+A\right )}{{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{5/2}} \,d x \]
int(((a + a*sin(e + f*x))^m*(A + B*sin(e + f*x) + C*sin(e + f*x)^2))/(c - c*sin(e + f*x))^(5/2),x)