3.1.14 \(\int (a+b \sin (e+f x))^m (A-A \sin ^2(e+f x)) \, dx\) [14]

3.1.14.1 Optimal result

Integrand size = 26, antiderivative size = 211 \[ \int (a+b \sin (e+f x))^m \left (A-A \sin ^2(e+f x)\right ) \, dx=\frac {4 \sqrt {2} A \operatorname {AppellF1}\left (\frac {1}{2},-\frac {3}{2},-m,\frac {3}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {b (1-\sin (e+f x))}{a+b}\right ) \cos (e+f x) (a+b \sin (e+f x))^m \left (\frac {a+b \sin (e+f x)}{a+b}\right )^{-m}}{f \sqrt {1+\sin (e+f x)}}-\frac {4 \sqrt {2} A \operatorname {AppellF1}\left (\frac {1}{2},-\frac {1}{2},-m,\frac {3}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {b (1-\sin (e+f x))}{a+b}\right ) \cos (e+f x) (a+b \sin (e+f x))^m \left (\frac {a+b \sin (e+f x)}{a+b}\right )^{-m}}{f \sqrt {1+\sin (e+f x)}} \]

output

4*A*AppellF1(1/2,-m,-3/2,3/2,b*(1-sin(f*x+e))/(a+b),1/2-1/2*sin(f*x+e))*co 
s(f*x+e)*(a+b*sin(f*x+e))^m*2^(1/2)/f/(((a+b*sin(f*x+e))/(a+b))^m)/(1+sin( 
f*x+e))^(1/2)-4*A*AppellF1(1/2,-m,-1/2,3/2,b*(1-sin(f*x+e))/(a+b),1/2-1/2* 
sin(f*x+e))*cos(f*x+e)*(a+b*sin(f*x+e))^m*2^(1/2)/f/(((a+b*sin(f*x+e))/(a+ 
b))^m)/(1+sin(f*x+e))^(1/2)
 

3.1.14.2 Mathematica [F]

\[ \int (a+b \sin (e+f x))^m \left (A-A \sin ^2(e+f x)\right ) \, dx=\int (a+b \sin (e+f x))^m \left (A-A \sin ^2(e+f x)\right ) \, dx \]

input

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

output

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

3.1.14.3 Rubi [A] (verified)

Time = 0.56 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.346, Rules used = {3042, 3497, 3042, 3234, 156, 155, 3263, 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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \left (A-A \sin (e+f x)^2\right ) (a+b \sin (e+f x))^mdx\)

\(\Big \downarrow \) 3497

\(\displaystyle 2 A \int (\sin (e+f x)+1) (a+b \sin (e+f x))^mdx-A \int (\sin (e+f x)+1)^2 (a+b \sin (e+f x))^mdx\)

\(\Big \downarrow \) 3042

\(\displaystyle 2 A \int (\sin (e+f x)+1) (a+b \sin (e+f x))^mdx-A \int (\sin (e+f x)+1)^2 (a+b \sin (e+f x))^mdx\)

\(\Big \downarrow \) 3234

\(\displaystyle \frac {2 A \cos (e+f x) \int \frac {\sqrt {\sin (e+f x)+1} (a+b \sin (e+f x))^m}{\sqrt {1-\sin (e+f x)}}d\sin (e+f x)}{f \sqrt {1-\sin (e+f x)} \sqrt {\sin (e+f x)+1}}-A \int (\sin (e+f x)+1)^2 (a+b \sin (e+f x))^mdx\)

\(\Big \downarrow \) 156

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

\(\Big \downarrow \) 155

\(\displaystyle -A \int (\sin (e+f x)+1)^2 (a+b \sin (e+f x))^mdx-\frac {4 \sqrt {2} A \cos (e+f x) (a+b \sin (e+f x))^m \left (\frac {a+b \sin (e+f x)}{a+b}\right )^{-m} \operatorname {AppellF1}\left (\frac {1}{2},-\frac {1}{2},-m,\frac {3}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {b (1-\sin (e+f x))}{a+b}\right )}{f \sqrt {\sin (e+f x)+1}}\)

\(\Big \downarrow \) 3263

\(\displaystyle -\frac {A \cos (e+f x) \int \frac {(\sin (e+f x)+1)^{3/2} (a+b \sin (e+f x))^m}{\sqrt {1-\sin (e+f x)}}d\sin (e+f x)}{f \sqrt {1-\sin (e+f x)} \sqrt {\sin (e+f x)+1}}-\frac {4 \sqrt {2} A \cos (e+f x) (a+b \sin (e+f x))^m \left (\frac {a+b \sin (e+f x)}{a+b}\right )^{-m} \operatorname {AppellF1}\left (\frac {1}{2},-\frac {1}{2},-m,\frac {3}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {b (1-\sin (e+f x))}{a+b}\right )}{f \sqrt {\sin (e+f x)+1}}\)

\(\Big \downarrow \) 156

\(\displaystyle -\frac {A \cos (e+f x) (a+b \sin (e+f x))^m \left (\frac {a+b \sin (e+f x)}{a+b}\right )^{-m} \int \frac {(\sin (e+f x)+1)^{3/2} \left (\frac {a}{a+b}+\frac {b \sin (e+f x)}{a+b}\right )^m}{\sqrt {1-\sin (e+f x)}}d\sin (e+f x)}{f \sqrt {1-\sin (e+f x)} \sqrt {\sin (e+f x)+1}}-\frac {4 \sqrt {2} A \cos (e+f x) (a+b \sin (e+f x))^m \left (\frac {a+b \sin (e+f x)}{a+b}\right )^{-m} \operatorname {AppellF1}\left (\frac {1}{2},-\frac {1}{2},-m,\frac {3}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {b (1-\sin (e+f x))}{a+b}\right )}{f \sqrt {\sin (e+f x)+1}}\)

\(\Big \downarrow \) 155

\(\displaystyle \frac {4 \sqrt {2} A \cos (e+f x) (a+b \sin (e+f x))^m \left (\frac {a+b \sin (e+f x)}{a+b}\right )^{-m} \operatorname {AppellF1}\left (\frac {1}{2},-\frac {3}{2},-m,\frac {3}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {b (1-\sin (e+f x))}{a+b}\right )}{f \sqrt {\sin (e+f x)+1}}-\frac {4 \sqrt {2} A \cos (e+f x) (a+b \sin (e+f x))^m \left (\frac {a+b \sin (e+f x)}{a+b}\right )^{-m} \operatorname {AppellF1}\left (\frac {1}{2},-\frac {1}{2},-m,\frac {3}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {b (1-\sin (e+f x))}{a+b}\right )}{f \sqrt {\sin (e+f x)+1}}\)

input

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

output

(4*Sqrt[2]*A*AppellF1[1/2, -3/2, -m, 3/2, (1 - Sin[e + f*x])/2, (b*(1 - Si 
n[e + f*x]))/(a + b)]*Cos[e + f*x]*(a + b*Sin[e + f*x])^m)/(f*Sqrt[1 + Sin 
[e + f*x]]*((a + b*Sin[e + f*x])/(a + b))^m) - (4*Sqrt[2]*A*AppellF1[1/2, 
-1/2, -m, 3/2, (1 - Sin[e + f*x])/2, (b*(1 - Sin[e + f*x]))/(a + b)]*Cos[e 
 + f*x]*(a + b*Sin[e + f*x])^m)/(f*Sqrt[1 + Sin[e + f*x]]*((a + b*Sin[e + 
f*x])/(a + b))^m)
 

3.1.14.3.1 Defintions of rubi rules used

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 3042

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

rule 3234

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

rule 3263

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

rule 3497

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (C_.)*sin[(e_.) + 
 (f_.)*(x_)]^2), x_Symbol] :> Simp[(A - C)   Int[(a + b*Sin[e + f*x])^m*(1 
+ Sin[e + f*x]), x], x] + Simp[C   Int[(a + b*Sin[e + f*x])^m*(1 + Sin[e + 
f*x])^2, x], x] /; FreeQ[{a, b, e, f, A, C, m}, x] && EqQ[A + C, 0] &&  !In 
tegerQ[2*m]
 

3.1.14.4 Maple [F]

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

input

int((a+b*sin(f*x+e))^m*(A-sin(f*x+e)^2*A),x)
 

output

int((a+b*sin(f*x+e))^m*(A-sin(f*x+e)^2*A),x)
 

3.1.14.5 Fricas [F]

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

input

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

output

integral((b*sin(f*x + e) + a)^m*A*cos(f*x + e)^2, x)
 

3.1.14.6 Sympy [F(-1)]

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

input

integrate((a+b*sin(f*x+e))**m*(A-A*sin(f*x+e)**2),x)
 

output

Timed out
 

3.1.14.7 Maxima [F]

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

input

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

output

-integrate((A*sin(f*x + e)^2 - A)*(b*sin(f*x + e) + a)^m, x)
 

3.1.14.8 Giac [F]

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

input

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

output

integrate(-(A*sin(f*x + e)^2 - A)*(b*sin(f*x + e) + a)^m, x)
 

3.1.14.9 Mupad [F(-1)]

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

input

int((A - A*sin(e + f*x)^2)*(a + b*sin(e + f*x))^m,x)
 

output

int((A - A*sin(e + f*x)^2)*(a + b*sin(e + f*x))^m, x)