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

Optimal result

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

[Out]

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))*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)-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)

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {3097, 2834, 144, 143, 2863} \[ \int (a+b \sin (e+f x))^m \left (A-A \sin ^2(e+f x)\right ) \, dx=\frac {4 \sqrt {2} A \cos (e+f x) (a+b \sin (e+f x))^m \left (\frac {a+b \sin (e+f x)}{a+b}\right )^{-m} \operatorname {AppellF1}\left (\frac {1}{2},-\frac {3}{2},-m,\frac {3}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {b (1-\sin (e+f x))}{a+b}\right )}{f \sqrt {\sin (e+f x)+1}}-\frac {4 \sqrt {2} A \cos (e+f x) (a+b \sin (e+f x))^m \left (\frac {a+b \sin (e+f x)}{a+b}\right )^{-m} \operatorname {AppellF1}\left (\frac {1}{2},-\frac {1}{2},-m,\frac {3}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {b (1-\sin (e+f x))}{a+b}\right )}{f \sqrt {\sin (e+f x)+1}} \]

[In]

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

[Out]

(4*Sqrt[2]*A*AppellF1[1/2, -3/2, -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) - (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)

Rule 143

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((a + b*x)
^(m + 1)/(b*(m + 1)*(b/(b*c - a*d))^n*(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] &&  !Inte
gerQ[n] &&  !IntegerQ[p] && GtQ[b/(b*c - a*d), 0] && GtQ[b/(b*e - a*f), 0] &&  !(GtQ[d/(d*a - c*b), 0] && GtQ[
d/(d*e - c*f), 0] && SimplerQ[c + d*x, a + b*x]) &&  !(GtQ[f/(f*a - e*b), 0] && GtQ[f/(f*c - e*d), 0] && Simpl
erQ[e + f*x, a + b*x])

Rule 144

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

Rule 2834

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c*(C
os[e + f*x]/(f*Sqrt[1 + Sin[e + f*x]]*Sqrt[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] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b
^2, 0] &&  !IntegerQ[2*m] && EqQ[c^2 - d^2, 0]

Rule 2863

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Di
st[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 3097

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[
A - C, Int[(a + b*Sin[e + f*x])^m*(1 + Sin[e + f*x]), x], x] + Dist[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] &&  !IntegerQ[2*m]

Rubi steps \begin{align*} \text {integral}& = -\left (A \int (1+\sin (e+f x))^2 (a+b \sin (e+f x))^m \, dx\right )+(2 A) \int (1+\sin (e+f x)) (a+b \sin (e+f x))^m \, dx \\ & = -\frac {(A \cos (e+f x)) \text {Subst}\left (\int \frac {(1+x)^{3/2} (a+b x)^m}{\sqrt {1-x}} \, dx,x,\sin (e+f x)\right )}{f \sqrt {1-\sin (e+f x)} \sqrt {1+\sin (e+f x)}}+\frac {(2 A \cos (e+f x)) \text {Subst}\left (\int \frac {\sqrt {1+x} (a+b x)^m}{\sqrt {1-x}} \, dx,x,\sin (e+f x)\right )}{f \sqrt {1-\sin (e+f x)} \sqrt {1+\sin (e+f x)}} \\ & = -\frac {\left (A \cos (e+f x) (a+b \sin (e+f x))^m \left (-\frac {a+b \sin (e+f x)}{-a-b}\right )^{-m}\right ) \text {Subst}\left (\int \frac {(1+x)^{3/2} \left (-\frac {a}{-a-b}-\frac {b x}{-a-b}\right )^m}{\sqrt {1-x}} \, dx,x,\sin (e+f x)\right )}{f \sqrt {1-\sin (e+f x)} \sqrt {1+\sin (e+f x)}}+\frac {\left (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}\right ) \text {Subst}\left (\int \frac {\sqrt {1+x} \left (-\frac {a}{-a-b}-\frac {b x}{-a-b}\right )^m}{\sqrt {1-x}} \, dx,x,\sin (e+f x)\right )}{f \sqrt {1-\sin (e+f x)} \sqrt {1+\sin (e+f x)}} \\ & = \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)}} \\ \end{align*}

Mathematica [F]

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

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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