3.1.17 \(\int (a+a \sin (e+f x))^m (A+B \sin (e+f x)+C \sin ^2(e+f x)) \, dx\) [17]

3.1.17.1 Optimal result

Integrand size = 33, antiderivative size = 184 \[ \int (a+a \sin (e+f x))^m \left (A+B \sin (e+f x)+C \sin ^2(e+f x)\right ) \, dx=\frac {(C-B (2+m)) \cos (e+f x) (a+a \sin (e+f x))^m}{f (1+m) (2+m)}-\frac {2^{\frac {1}{2}+m} \left (B m (2+m)+C \left (1+m+m^2\right )+A \left (2+3 m+m^2\right )\right ) \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2}-m,\frac {3}{2},\frac {1}{2} (1-\sin (e+f x))\right ) (1+\sin (e+f x))^{-\frac {1}{2}-m} (a+a \sin (e+f x))^m}{f (1+m) (2+m)}-\frac {C \cos (e+f x) (a+a \sin (e+f x))^{1+m}}{a f (2+m)} \]

output

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

3.1.17.2 Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 22.29 (sec) , antiderivative size = 300, normalized size of antiderivative = 1.63 \[ \int (a+a \sin (e+f x))^m \left (A+B \sin (e+f x)+C \sin ^2(e+f x)\right ) \, dx=-\frac {(a (1+\sin (e+f x)))^m (\cos (e+f x)+i (1+\sin (e+f x))) \left (-\frac {2 (2 A+C) \operatorname {Hypergeometric2F1}(1,1+m,1-m,i \cos (e+f x)-\sin (e+f x))}{m}-\frac {2 i B \operatorname {Hypergeometric2F1}(1,m,-m,i \cos (e+f x)-\sin (e+f x)) (\cos (e+f x)-i \sin (e+f x))}{1+m}+\frac {2 i B \operatorname {Hypergeometric2F1}(1,2+m,2-m,i \cos (e+f x)-\sin (e+f x)) (\cos (e+f x)+i \sin (e+f x))}{-1+m}+\frac {C \operatorname {Hypergeometric2F1}(1,-1+m,-1-m,i \cos (e+f x)-\sin (e+f x)) (\cos (2 (e+f x))-i \sin (2 (e+f x)))}{2+m}+\frac {C \operatorname {Hypergeometric2F1}(1,3+m,3-m,i \cos (e+f x)-\sin (e+f x)) (\cos (2 (e+f x))+i \sin (2 (e+f x)))}{-2+m}\right )}{4 f} \]

input

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

output

-1/4*((a*(1 + Sin[e + f*x]))^m*(Cos[e + f*x] + I*(1 + Sin[e + f*x]))*((-2* 
(2*A + C)*Hypergeometric2F1[1, 1 + m, 1 - m, I*Cos[e + f*x] - Sin[e + f*x] 
])/m - ((2*I)*B*Hypergeometric2F1[1, m, -m, I*Cos[e + f*x] - Sin[e + f*x]] 
*(Cos[e + f*x] - I*Sin[e + f*x]))/(1 + m) + ((2*I)*B*Hypergeometric2F1[1, 
2 + m, 2 - m, I*Cos[e + f*x] - Sin[e + f*x]]*(Cos[e + f*x] + I*Sin[e + f*x 
]))/(-1 + m) + (C*Hypergeometric2F1[1, -1 + m, -1 - m, I*Cos[e + f*x] - Si 
n[e + f*x]]*(Cos[2*(e + f*x)] - I*Sin[2*(e + f*x)]))/(2 + m) + (C*Hypergeo 
metric2F1[1, 3 + m, 3 - m, I*Cos[e + f*x] - Sin[e + f*x]]*(Cos[2*(e + f*x) 
] + I*Sin[2*(e + f*x)]))/(-2 + m)))/f
 

3.1.17.3 Rubi [A] (verified)

Time = 0.67 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.01, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.242, Rules used = {3042, 3502, 3042, 3230, 3042, 3131, 3042, 3130}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3502

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3230

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3131

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3130

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

input

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

output

-((C*Cos[e + f*x]*(a + a*Sin[e + f*x])^(1 + m))/(a*f*(2 + m))) + ((a*(C - 
B*(2 + m))*Cos[e + f*x]*(a + a*Sin[e + f*x])^m)/(f*(1 + m)) - (2^(1/2 + m) 
*a*(B*m*(2 + m) + C*(1 + m + m^2) + A*(2 + 3*m + m^2))*Cos[e + f*x]*Hyperg 
eometric2F1[1/2, 1/2 - m, 3/2, (1 - Sin[e + f*x])/2]*(1 + Sin[e + f*x])^(- 
1/2 - m)*(a + a*Sin[e + f*x])^m)/(f*(1 + m)))/(a*(2 + m))
 

3.1.17.3.1 Defintions of rubi rules used

rule 3042

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

rule 3130

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-2^(n + 
 1/2))*a^(n - 1/2)*b*(Cos[c + d*x]/(d*Sqrt[a + b*Sin[c + d*x]]))*Hypergeome 
tric2F1[1/2, 1/2 - n, 3/2, (1/2)*(1 - b*(Sin[c + d*x]/a))], x] /; FreeQ[{a, 
 b, c, d, n}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[2*n] && GtQ[a, 0]
 

rule 3131

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[a^IntPar 
t[n]*((a + b*Sin[c + d*x])^FracPart[n]/(1 + (b/a)*Sin[c + d*x])^FracPart[n] 
)   Int[(1 + (b/a)*Sin[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && 
EqQ[a^2 - b^2, 0] &&  !IntegerQ[2*n] &&  !GtQ[a, 0]
 

rule 3230

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

rule 3502

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) 
+ (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co 
s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m 
+ 2))   Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m 
 + 2) - a*C)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] 
 &&  !LtQ[m, -1]
 

3.1.17.4 Maple [F]

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

input

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

output

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

3.1.17.5 Fricas [F]

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

input

integrate((a+a*sin(f*x+e))^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, x)
 

3.1.17.6 Sympy [F]

\[ \int (a+a \sin (e+f x))^m \left (A+B \sin (e+f x)+C \sin ^2(e+f x)\right ) \, dx=\int \left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{m} \left (A + B \sin {\left (e + f x \right )} + C \sin ^{2}{\left (e + f x \right )}\right )\, dx \]

input

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

output

Integral((a*(sin(e + f*x) + 1))**m*(A + B*sin(e + f*x) + C*sin(e + f*x)**2 
), x)
 

3.1.17.7 Maxima [F]

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

input

integrate((a+a*sin(f*x+e))^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, 
x)
 

3.1.17.8 Giac [F]

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

input

integrate((a+a*sin(f*x+e))^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, 
x)
 

3.1.17.9 Mupad [F(-1)]

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

input

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

output

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