\(\int (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{-2-m} (A+B \sin (e+f x)+C \sin ^2(e+f x)) \, dx\) [25]

Optimal result
Mathematica [F]
Rubi [A] (warning: unable to verify)
Maple [F]
Fricas [F]
Sympy [F(-1)]
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 49, antiderivative size = 400 \[ \int (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{-2-m} \left (A+B \sin (e+f x)+C \sin ^2(e+f x)\right ) \, dx=\frac {\left (c^2 C-B c d+A d^2\right ) \cos (e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{-1-m}}{d \left (c^2-d^2\right ) f (1+m)}-\frac {2^{\frac {1}{2}+m} a \left (c d (A+C+A m+B m+C m)-c^2 (C+2 C m)-d^2 (A m+B (1+m)-C (1+m))\right ) \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2}-m,\frac {3}{2},\frac {(c-d) (1-\sin (e+f x))}{2 (c+d \sin (e+f x))}\right ) (a+a \sin (e+f x))^{-1+m} \left (\frac {(c+d) (1+\sin (e+f x))}{c+d \sin (e+f x)}\right )^{\frac {1}{2}-m} (c+d \sin (e+f x))^{-m}}{(c-d) d (c+d)^2 f (1+m)}-\frac {2^{\frac {3}{2}+m} C \operatorname {AppellF1}\left (\frac {1}{2},-\frac {1}{2}-m,1+m,\frac {3}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {d (1-\sin (e+f x))}{c+d}\right ) \cos (e+f x) (1+\sin (e+f x))^{-\frac {1}{2}-m} (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{-m} \left (\frac {c+d \sin (e+f x)}{c+d}\right )^m}{d (c+d) f} \] Output:

(A*d^2-B*c*d+C*c^2)*cos(f*x+e)*(a+a*sin(f*x+e))^m*(c+d*sin(f*x+e))^(-1-m)/ 
d/(c^2-d^2)/f/(1+m)-2^(1/2+m)*a*(c*d*(A*m+B*m+C*m+A+C)-c^2*(2*C*m+C)-d^2*( 
A*m+B*(1+m)-C*(1+m)))*cos(f*x+e)*hypergeom([1/2, 1/2-m],[3/2],(c-d)*(1-sin 
(f*x+e))/(2*c+2*d*sin(f*x+e)))*(a+a*sin(f*x+e))^(-1+m)*((c+d)*(1+sin(f*x+e 
))/(c+d*sin(f*x+e)))^(1/2-m)/(c-d)/d/(c+d)^2/f/(1+m)/((c+d*sin(f*x+e))^m)- 
2^(3/2+m)*C*AppellF1(1/2,1+m,-1/2-m,3/2,d*(1-sin(f*x+e))/(c+d),1/2-1/2*sin 
(f*x+e))*cos(f*x+e)*(1+sin(f*x+e))^(-1/2-m)*(a+a*sin(f*x+e))^m*((c+d*sin(f 
*x+e))/(c+d))^m/d/(c+d)/f/((c+d*sin(f*x+e))^m)
                                                                                    
                                                                                    
 

Mathematica [F]

\[ \int (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{-2-m} \left (A+B \sin (e+f x)+C \sin ^2(e+f x)\right ) \, dx=\int (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{-2-m} \left (A+B \sin (e+f x)+C \sin ^2(e+f x)\right ) \, dx \] Input:

Integrate[(a + a*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(-2 - m)*(A + B*Sin[ 
e + f*x] + C*Sin[e + f*x]^2),x]
 

Output:

Integrate[(a + a*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(-2 - m)*(A + B*Sin[ 
e + f*x] + C*Sin[e + f*x]^2), x]
 

Rubi [A] (warning: unable to verify)

Time = 1.38 (sec) , antiderivative size = 441, normalized size of antiderivative = 1.10, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.245, Rules used = {3042, 3522, 25, 3042, 3466, 3042, 3267, 142, 157, 27, 156, 155}

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.

\(\displaystyle \int (a \sin (e+f x)+a)^m \left (A+B \sin (e+f x)+C \sin ^2(e+f x)\right ) (c+d \sin (e+f x))^{-m-2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int (a \sin (e+f x)+a)^m \left (A+B \sin (e+f x)+C \sin (e+f x)^2\right ) (c+d \sin (e+f x))^{-m-2}dx\)

\(\Big \downarrow \) 3522

\(\displaystyle \frac {\cos (e+f x) \left (A d^2-B c d+c^2 C\right ) (a \sin (e+f x)+a)^m (c+d \sin (e+f x))^{-m-1}}{d f (m+1) \left (c^2-d^2\right )}-\frac {\int -(\sin (e+f x) a+a)^m (c+d \sin (e+f x))^{-m-1} \left (a (A d (m c+c-d m)+(c C-B d) (m d+d-c m))+a C \left (c^2-d^2\right ) (m+1) \sin (e+f x)\right )dx}{a d (m+1) \left (c^2-d^2\right )}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\int (\sin (e+f x) a+a)^m (c+d \sin (e+f x))^{-m-1} \left (a (A d (m c+c-d m)+(c C-B d) (m d+d-c m))+a C \left (c^2-d^2\right ) (m+1) \sin (e+f x)\right )dx}{a d (m+1) \left (c^2-d^2\right )}+\frac {\cos (e+f x) \left (A d^2-B c d+c^2 C\right ) (a \sin (e+f x)+a)^m (c+d \sin (e+f x))^{-m-1}}{d f (m+1) \left (c^2-d^2\right )}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int (\sin (e+f x) a+a)^m (c+d \sin (e+f x))^{-m-1} \left (a (A d (m c+c-d m)+(c C-B d) (m d+d-c m))+a C \left (c^2-d^2\right ) (m+1) \sin (e+f x)\right )dx}{a d (m+1) \left (c^2-d^2\right )}+\frac {\cos (e+f x) \left (A d^2-B c d+c^2 C\right ) (a \sin (e+f x)+a)^m (c+d \sin (e+f x))^{-m-1}}{d f (m+1) \left (c^2-d^2\right )}\)

\(\Big \downarrow \) 3466

\(\displaystyle \frac {a \left (c d (A m+A+B m+C m+C)-d^2 (A m+B (m+1)-C (m+1))-\left (c^2 (2 C m+C)\right )\right ) \int (\sin (e+f x) a+a)^m (c+d \sin (e+f x))^{-m-1}dx+C (m+1) \left (c^2-d^2\right ) \int (\sin (e+f x) a+a)^{m+1} (c+d \sin (e+f x))^{-m-1}dx}{a d (m+1) \left (c^2-d^2\right )}+\frac {\cos (e+f x) \left (A d^2-B c d+c^2 C\right ) (a \sin (e+f x)+a)^m (c+d \sin (e+f x))^{-m-1}}{d f (m+1) \left (c^2-d^2\right )}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {a \left (c d (A m+A+B m+C m+C)-d^2 (A m+B (m+1)-C (m+1))-\left (c^2 (2 C m+C)\right )\right ) \int (\sin (e+f x) a+a)^m (c+d \sin (e+f x))^{-m-1}dx+C (m+1) \left (c^2-d^2\right ) \int (\sin (e+f x) a+a)^{m+1} (c+d \sin (e+f x))^{-m-1}dx}{a d (m+1) \left (c^2-d^2\right )}+\frac {\cos (e+f x) \left (A d^2-B c d+c^2 C\right ) (a \sin (e+f x)+a)^m (c+d \sin (e+f x))^{-m-1}}{d f (m+1) \left (c^2-d^2\right )}\)

\(\Big \downarrow \) 3267

\(\displaystyle \frac {\frac {a^3 \cos (e+f x) \left (c d (A m+A+B m+C m+C)-d^2 (A m+B (m+1)-C (m+1))-\left (c^2 (2 C m+C)\right )\right ) \int \frac {(\sin (e+f x) a+a)^{m-\frac {1}{2}} (c+d \sin (e+f x))^{-m-1}}{\sqrt {a-a \sin (e+f x)}}d\sin (e+f x)}{f \sqrt {a-a \sin (e+f x)} \sqrt {a \sin (e+f x)+a}}+\frac {a^2 C (m+1) \left (c^2-d^2\right ) \cos (e+f x) \int \frac {(\sin (e+f x) a+a)^{m+\frac {1}{2}} (c+d \sin (e+f x))^{-m-1}}{\sqrt {a-a \sin (e+f x)}}d\sin (e+f x)}{f \sqrt {a-a \sin (e+f x)} \sqrt {a \sin (e+f x)+a}}}{a d (m+1) \left (c^2-d^2\right )}+\frac {\cos (e+f x) \left (A d^2-B c d+c^2 C\right ) (a \sin (e+f x)+a)^m (c+d \sin (e+f x))^{-m-1}}{d f (m+1) \left (c^2-d^2\right )}\)

\(\Big \downarrow \) 142

\(\displaystyle \frac {\frac {a^2 C (m+1) \left (c^2-d^2\right ) \cos (e+f x) \int \frac {(\sin (e+f x) a+a)^{m+\frac {1}{2}} (c+d \sin (e+f x))^{-m-1}}{\sqrt {a-a \sin (e+f x)}}d\sin (e+f x)}{f \sqrt {a-a \sin (e+f x)} \sqrt {a \sin (e+f x)+a}}-\frac {a^2 2^{m+\frac {1}{2}} \cos (e+f x) (a \sin (e+f x)+a)^{m-1} \left (c d (A m+A+B m+C m+C)-d^2 (A m+B (m+1)-C (m+1))-\left (c^2 (2 C m+C)\right )\right ) \left (\frac {(c+d) (\sin (e+f x)+1)}{c+d \sin (e+f x)}\right )^{\frac {1}{2}-m} (c+d \sin (e+f x))^{-m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2}-m,\frac {3}{2},\frac {(c-d) (1-\sin (e+f x))}{2 (c+d \sin (e+f x))}\right )}{f (c+d)}}{a d (m+1) \left (c^2-d^2\right )}+\frac {\cos (e+f x) \left (A d^2-B c d+c^2 C\right ) (a \sin (e+f x)+a)^m (c+d \sin (e+f x))^{-m-1}}{d f (m+1) \left (c^2-d^2\right )}\)

\(\Big \downarrow \) 157

\(\displaystyle \frac {\frac {a^2 C (m+1) \left (c^2-d^2\right ) \sqrt {1-\sin (e+f x)} \cos (e+f x) \int \frac {\sqrt {2} (\sin (e+f x) a+a)^{m+\frac {1}{2}} (c+d \sin (e+f x))^{-m-1}}{\sqrt {1-\sin (e+f x)}}d\sin (e+f x)}{\sqrt {2} f (a-a \sin (e+f x)) \sqrt {a \sin (e+f x)+a}}-\frac {a^2 2^{m+\frac {1}{2}} \cos (e+f x) (a \sin (e+f x)+a)^{m-1} \left (c d (A m+A+B m+C m+C)-d^2 (A m+B (m+1)-C (m+1))-\left (c^2 (2 C m+C)\right )\right ) \left (\frac {(c+d) (\sin (e+f x)+1)}{c+d \sin (e+f x)}\right )^{\frac {1}{2}-m} (c+d \sin (e+f x))^{-m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2}-m,\frac {3}{2},\frac {(c-d) (1-\sin (e+f x))}{2 (c+d \sin (e+f x))}\right )}{f (c+d)}}{a d (m+1) \left (c^2-d^2\right )}+\frac {\cos (e+f x) \left (A d^2-B c d+c^2 C\right ) (a \sin (e+f x)+a)^m (c+d \sin (e+f x))^{-m-1}}{d f (m+1) \left (c^2-d^2\right )}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {a^2 C (m+1) \left (c^2-d^2\right ) \sqrt {1-\sin (e+f x)} \cos (e+f x) \int \frac {(\sin (e+f x) a+a)^{m+\frac {1}{2}} (c+d \sin (e+f x))^{-m-1}}{\sqrt {1-\sin (e+f x)}}d\sin (e+f x)}{f (a-a \sin (e+f x)) \sqrt {a \sin (e+f x)+a}}-\frac {a^2 2^{m+\frac {1}{2}} \cos (e+f x) (a \sin (e+f x)+a)^{m-1} \left (c d (A m+A+B m+C m+C)-d^2 (A m+B (m+1)-C (m+1))-\left (c^2 (2 C m+C)\right )\right ) \left (\frac {(c+d) (\sin (e+f x)+1)}{c+d \sin (e+f x)}\right )^{\frac {1}{2}-m} (c+d \sin (e+f x))^{-m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2}-m,\frac {3}{2},\frac {(c-d) (1-\sin (e+f x))}{2 (c+d \sin (e+f x))}\right )}{f (c+d)}}{a d (m+1) \left (c^2-d^2\right )}+\frac {\cos (e+f x) \left (A d^2-B c d+c^2 C\right ) (a \sin (e+f x)+a)^m (c+d \sin (e+f x))^{-m-1}}{d f (m+1) \left (c^2-d^2\right )}\)

\(\Big \downarrow \) 156

\(\displaystyle \frac {\frac {a^2 C (m+1) \left (c^2-d^2\right ) \sqrt {1-\sin (e+f x)} \cos (e+f x) (c+d \sin (e+f x))^{-m} \left (\frac {c+d \sin (e+f x)}{c-d}\right )^m \int \frac {(\sin (e+f x) a+a)^{m+\frac {1}{2}} \left (\frac {c}{c-d}+\frac {d \sin (e+f x)}{c-d}\right )^{-m-1}}{\sqrt {1-\sin (e+f x)}}d\sin (e+f x)}{f (c-d) (a-a \sin (e+f x)) \sqrt {a \sin (e+f x)+a}}-\frac {a^2 2^{m+\frac {1}{2}} \cos (e+f x) (a \sin (e+f x)+a)^{m-1} \left (c d (A m+A+B m+C m+C)-d^2 (A m+B (m+1)-C (m+1))-\left (c^2 (2 C m+C)\right )\right ) \left (\frac {(c+d) (\sin (e+f x)+1)}{c+d \sin (e+f x)}\right )^{\frac {1}{2}-m} (c+d \sin (e+f x))^{-m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2}-m,\frac {3}{2},\frac {(c-d) (1-\sin (e+f x))}{2 (c+d \sin (e+f x))}\right )}{f (c+d)}}{a d (m+1) \left (c^2-d^2\right )}+\frac {\cos (e+f x) \left (A d^2-B c d+c^2 C\right ) (a \sin (e+f x)+a)^m (c+d \sin (e+f x))^{-m-1}}{d f (m+1) \left (c^2-d^2\right )}\)

\(\Big \downarrow \) 155

\(\displaystyle \frac {\frac {\sqrt {2} a C (m+1) \left (c^2-d^2\right ) \sqrt {1-\sin (e+f x)} \cos (e+f x) (a \sin (e+f x)+a)^{m+1} (c+d \sin (e+f x))^{-m} \left (\frac {c+d \sin (e+f x)}{c-d}\right )^m \operatorname {AppellF1}\left (m+\frac {3}{2},\frac {1}{2},m+1,m+\frac {5}{2},\frac {1}{2} (\sin (e+f x)+1),-\frac {d (\sin (e+f x)+1)}{c-d}\right )}{f (2 m+3) (c-d) (a-a \sin (e+f x))}-\frac {a^2 2^{m+\frac {1}{2}} \cos (e+f x) (a \sin (e+f x)+a)^{m-1} \left (c d (A m+A+B m+C m+C)-d^2 (A m+B (m+1)-C (m+1))-\left (c^2 (2 C m+C)\right )\right ) \left (\frac {(c+d) (\sin (e+f x)+1)}{c+d \sin (e+f x)}\right )^{\frac {1}{2}-m} (c+d \sin (e+f x))^{-m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2}-m,\frac {3}{2},\frac {(c-d) (1-\sin (e+f x))}{2 (c+d \sin (e+f x))}\right )}{f (c+d)}}{a d (m+1) \left (c^2-d^2\right )}+\frac {\cos (e+f x) \left (A d^2-B c d+c^2 C\right ) (a \sin (e+f x)+a)^m (c+d \sin (e+f x))^{-m-1}}{d f (m+1) \left (c^2-d^2\right )}\)

Input:

Int[(a + a*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(-2 - m)*(A + B*Sin[e + f* 
x] + C*Sin[e + f*x]^2),x]
 

Output:

((c^2*C - B*c*d + A*d^2)*Cos[e + f*x]*(a + a*Sin[e + f*x])^m*(c + d*Sin[e 
+ f*x])^(-1 - m))/(d*(c^2 - d^2)*f*(1 + m)) + (-((2^(1/2 + m)*a^2*(c*d*(A 
+ C + A*m + B*m + C*m) - c^2*(C + 2*C*m) - d^2*(A*m + B*(1 + m) - C*(1 + m 
)))*Cos[e + f*x]*Hypergeometric2F1[1/2, 1/2 - m, 3/2, ((c - d)*(1 - Sin[e 
+ f*x]))/(2*(c + d*Sin[e + f*x]))]*(a + a*Sin[e + f*x])^(-1 + m)*(((c + d) 
*(1 + Sin[e + f*x]))/(c + d*Sin[e + f*x]))^(1/2 - m))/((c + d)*f*(c + d*Si 
n[e + f*x])^m)) + (Sqrt[2]*a*C*(c^2 - d^2)*(1 + m)*AppellF1[3/2 + m, 1/2, 
1 + m, 5/2 + m, (1 + Sin[e + f*x])/2, -((d*(1 + Sin[e + f*x]))/(c - d))]*C 
os[e + f*x]*Sqrt[1 - Sin[e + f*x]]*(a + a*Sin[e + f*x])^(1 + m)*((c + d*Si 
n[e + f*x])/(c - d))^m)/((c - d)*f*(3 + 2*m)*(a - a*Sin[e + f*x])*(c + d*S 
in[e + f*x])^m))/(a*d*(c^2 - d^2)*(1 + m))
 

Defintions of rubi rules used

rule 25
Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27
Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 142
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n*((e + f*x)^(p + 1)/((b*e 
 - a*f)*(m + 1)))*Hypergeometric2F1[m + 1, -n, m + 2, (-(d*e - c*f))*((a + 
b*x)/((b*c - a*d)*(e + f*x)))])/((b*e - a*f)*((c + d*x)/((b*c - a*d)*(e + f 
*x))))^n, x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[m + n + p + 2, 
 0] &&  !IntegerQ[n]
 

rule 155
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) 
^(p_), x_] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*Simplify[b/(b*c - a*d)]^n* 
Simplify[b/(b*e - a*f)]^p))*AppellF1[m + 1, -n, -p, m + 2, (-d)*((a + b*x)/ 
(b*c - a*d)), (-f)*((a + b*x)/(b*e - a*f))], x] /; FreeQ[{a, b, c, d, e, f, 
 m, n, p}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] &&  !IntegerQ[p] && GtQ[Sim 
plify[b/(b*c - a*d)], 0] && GtQ[Simplify[b/(b*e - a*f)], 0] &&  !(GtQ[Simpl 
ify[d/(d*a - c*b)], 0] && GtQ[Simplify[d/(d*e - c*f)], 0] && SimplerQ[c + d 
*x, a + b*x]) &&  !(GtQ[Simplify[f/(f*a - e*b)], 0] && GtQ[Simplify[f/(f*c 
- e*d)], 0] && SimplerQ[e + f*x, a + b*x])
 

rule 156
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) 
^(p_), x_] :> Simp[(e + f*x)^FracPart[p]/(Simplify[b/(b*e - a*f)]^IntPart[p 
]*(b*((e + f*x)/(b*e - a*f)))^FracPart[p])   Int[(a + b*x)^m*(c + d*x)^n*Si 
mp[b*(e/(b*e - a*f)) + b*f*(x/(b*e - a*f)), x]^p, x], x] /; FreeQ[{a, b, c, 
 d, e, f, m, n, p}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] &&  !IntegerQ[p] & 
& GtQ[Simplify[b/(b*c - a*d)], 0] &&  !GtQ[Simplify[b/(b*e - a*f)], 0]
 

rule 157
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) 
^(p_), x_] :> Simp[(c + d*x)^FracPart[n]/(Simplify[b/(b*c - a*d)]^IntPart[n 
]*(b*((c + d*x)/(b*c - a*d)))^FracPart[n])   Int[(a + b*x)^m*Simp[b*(c/(b*c 
 - a*d)) + b*d*(x/(b*c - a*d)), x]^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, 
 d, e, f, m, n, p}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] &&  !IntegerQ[p] & 
&  !GtQ[Simplify[b/(b*c - a*d)], 0] &&  !SimplerQ[c + d*x, a + b*x] &&  !Si 
mplerQ[e + f*x, a + b*x]
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3267
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + ( 
f_.)*(x_)])^(n_.), x_Symbol] :> Simp[a^2*(Cos[e + f*x]/(f*Sqrt[a + b*Sin[e 
+ f*x]]*Sqrt[a - b*Sin[e + f*x]]))   Subst[Int[(a + b*x)^(m - 1/2)*((c + d* 
x)^n/Sqrt[a - b*x]), x], x, Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m 
, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && 
 !IntegerQ[m]
 

rule 3466
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + 
 (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si 
mp[(A*b - a*B)/b   Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^n, x], x 
] + Simp[B/b   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] && NeQ[b*c - a*d, 0] && EqQ 
[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && NeQ[A*b + a*B, 0]
 

rule 3522
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[(-(c^2*C - B*c*d + A*d^2))*Cos[e + f*x 
]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 - 
d^2))), x] + Simp[1/(b*d*(n + 1)*(c^2 - d^2))   Int[(a + b*Sin[e + f*x])^m* 
(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(a*d*m + b*c*(n + 1)) + (c*C - B*d)*( 
a*c*m + b*d*(n + 1)) + b*(d*(B*c - A*d)*(m + n + 2) - C*(c^2*(m + 1) + d^2* 
(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m}, 
 x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &&  !LtQ 
[m, -2^(-1)] && (LtQ[n, -1] || EqQ[m + n + 2, 0])
 
Maple [F]

\[\int \left (a +a \sin \left (f x +e \right )\right )^{m} \left (c +d \sin \left (f x +e \right )\right )^{-2-m} \left (A +B \sin \left (f x +e \right )+C \sin \left (f x +e \right )^{2}\right )d x\]

Input:

int((a+a*sin(f*x+e))^m*(c+d*sin(f*x+e))^(-2-m)*(A+B*sin(f*x+e)+C*sin(f*x+e 
)^2),x)
 

Output:

int((a+a*sin(f*x+e))^m*(c+d*sin(f*x+e))^(-2-m)*(A+B*sin(f*x+e)+C*sin(f*x+e 
)^2),x)
 

Fricas [F]

\[ \int (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{-2-m} \left (A+B \sin (e+f x)+C \sin ^2(e+f x)\right ) \, dx=\int { {\left (C \sin \left (f x + e\right )^{2} + B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (d \sin \left (f x + e\right ) + c\right )}^{-m - 2} \,d x } \] Input:

integrate((a+a*sin(f*x+e))^m*(c+d*sin(f*x+e))^(-2-m)*(A+B*sin(f*x+e)+C*sin 
(f*x+e)^2),x, algorithm="fricas")
 

Output:

integral(-(C*cos(f*x + e)^2 - B*sin(f*x + e) - A - C)*(a*sin(f*x + e) + a) 
^m*(d*sin(f*x + e) + c)^(-m - 2), x)
 

Sympy [F(-1)]

Timed out. \[ \int (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{-2-m} \left (A+B \sin (e+f x)+C \sin ^2(e+f x)\right ) \, dx=\text {Timed out} \] Input:

integrate((a+a*sin(f*x+e))**m*(c+d*sin(f*x+e))**(-2-m)*(A+B*sin(f*x+e)+C*s 
in(f*x+e)**2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{-2-m} \left (A+B \sin (e+f x)+C \sin ^2(e+f x)\right ) \, dx=\int { {\left (C \sin \left (f x + e\right )^{2} + B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (d \sin \left (f x + e\right ) + c\right )}^{-m - 2} \,d x } \] Input:

integrate((a+a*sin(f*x+e))^m*(c+d*sin(f*x+e))^(-2-m)*(A+B*sin(f*x+e)+C*sin 
(f*x+e)^2),x, algorithm="maxima")
 

Output:

integrate((C*sin(f*x + e)^2 + B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)^m*( 
d*sin(f*x + e) + c)^(-m - 2), x)
 

Giac [F]

\[ \int (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{-2-m} \left (A+B \sin (e+f x)+C \sin ^2(e+f x)\right ) \, dx=\int { {\left (C \sin \left (f x + e\right )^{2} + B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (d \sin \left (f x + e\right ) + c\right )}^{-m - 2} \,d x } \] Input:

integrate((a+a*sin(f*x+e))^m*(c+d*sin(f*x+e))^(-2-m)*(A+B*sin(f*x+e)+C*sin 
(f*x+e)^2),x, algorithm="giac")
 

Output:

integrate((C*sin(f*x + e)^2 + B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)^m*( 
d*sin(f*x + e) + c)^(-m - 2), x)
 

Mupad [F(-1)]

Timed out. \[ \int (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{-2-m} \left (A+B \sin (e+f x)+C \sin ^2(e+f x)\right ) \, dx=\int \frac {{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,\left (C\,{\sin \left (e+f\,x\right )}^2+B\,\sin \left (e+f\,x\right )+A\right )}{{\left (c+d\,\sin \left (e+f\,x\right )\right )}^{m+2}} \,d x \] Input:

int(((a + a*sin(e + f*x))^m*(A + B*sin(e + f*x) + C*sin(e + f*x)^2))/(c + 
d*sin(e + f*x))^(m + 2),x)
                                                                                    
                                                                                    
 

Output:

int(((a + a*sin(e + f*x))^m*(A + B*sin(e + f*x) + C*sin(e + f*x)^2))/(c + 
d*sin(e + f*x))^(m + 2), x)
 

Reduce [F]

\[ \int (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{-2-m} \left (A+B \sin (e+f x)+C \sin ^2(e+f x)\right ) \, dx=\left (\int \frac {\left (a +a \sin \left (f x +e \right )\right )^{m}}{\left (\sin \left (f x +e \right ) d +c \right )^{m} \sin \left (f x +e \right )^{2} d^{2}+2 \left (\sin \left (f x +e \right ) d +c \right )^{m} \sin \left (f x +e \right ) c d +\left (\sin \left (f x +e \right ) d +c \right )^{m} c^{2}}d x \right ) a +\left (\int \frac {\left (a +a \sin \left (f x +e \right )\right )^{m} \sin \left (f x +e \right )^{2}}{\left (\sin \left (f x +e \right ) d +c \right )^{m} \sin \left (f x +e \right )^{2} d^{2}+2 \left (\sin \left (f x +e \right ) d +c \right )^{m} \sin \left (f x +e \right ) c d +\left (\sin \left (f x +e \right ) d +c \right )^{m} c^{2}}d x \right ) c +\left (\int \frac {\left (a +a \sin \left (f x +e \right )\right )^{m} \sin \left (f x +e \right )}{\left (\sin \left (f x +e \right ) d +c \right )^{m} \sin \left (f x +e \right )^{2} d^{2}+2 \left (\sin \left (f x +e \right ) d +c \right )^{m} \sin \left (f x +e \right ) c d +\left (\sin \left (f x +e \right ) d +c \right )^{m} c^{2}}d x \right ) b \] Input:

int((a+a*sin(f*x+e))^m*(c+d*sin(f*x+e))^(-2-m)*(A+B*sin(f*x+e)+C*sin(f*x+e 
)^2),x)
 

Output:

int((sin(e + f*x)*a + a)**m/((sin(e + f*x)*d + c)**m*sin(e + f*x)**2*d**2 
+ 2*(sin(e + f*x)*d + c)**m*sin(e + f*x)*c*d + (sin(e + f*x)*d + c)**m*c** 
2),x)*a + int(((sin(e + f*x)*a + a)**m*sin(e + f*x)**2)/((sin(e + f*x)*d + 
 c)**m*sin(e + f*x)**2*d**2 + 2*(sin(e + f*x)*d + c)**m*sin(e + f*x)*c*d + 
 (sin(e + f*x)*d + c)**m*c**2),x)*c + int(((sin(e + f*x)*a + a)**m*sin(e + 
 f*x))/((sin(e + f*x)*d + c)**m*sin(e + f*x)**2*d**2 + 2*(sin(e + f*x)*d + 
 c)**m*sin(e + f*x)*c*d + (sin(e + f*x)*d + c)**m*c**2),x)*b