\(\int (a-a \sin (e+f x)) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{-1-m} \, dx\) [349]

Optimal result

Integrand size = 40, antiderivative size = 139 \[ \int (a-a \sin (e+f x)) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{-1-m} \, dx=\frac {2 \sqrt {2} \operatorname {AppellF1}\left (\frac {1}{2}+m,-\frac {1}{2},1+m,\frac {3}{2}+m,\frac {1}{2} (1+\sin (e+f x)),-\frac {d (1+\sin (e+f x))}{c-d}\right ) \sec (e+f x) \sqrt {1-\sin (e+f x)} (a+a \sin (e+f x))^{1+m} (c+d \sin (e+f x))^{-m} \left (\frac {c+d \sin (e+f x)}{c-d}\right )^m}{(c-d) f (1+2 m)} \]

[Out]

2*AppellF1(1/2+m,1+m,-1/2,3/2+m,-d*(1+sin(f*x+e))/(c-d),1/2+1/2*sin(f*x+e))*sec(f*x+e)*(a+a*sin(f*x+e))^(1+m)*
((c+d*sin(f*x+e))/(c-d))^m*2^(1/2)*(1-sin(f*x+e))^(1/2)/(c-d)/f/(1+2*m)/((c+d*sin(f*x+e))^m)

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {3087, 145, 144, 143} \[ \int (a-a \sin (e+f x)) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{-1-m} \, dx=\frac {2 \sqrt {2} \sqrt {1-\sin (e+f x)} \sec (e+f x) (a \sin (e+f x)+a)^{m+1} (c+d \sin (e+f x))^{-m} \left (\frac {c+d \sin (e+f x)}{c-d}\right )^m \operatorname {AppellF1}\left (m+\frac {1}{2},-\frac {1}{2},m+1,m+\frac {3}{2},\frac {1}{2} (\sin (e+f x)+1),-\frac {d (\sin (e+f x)+1)}{c-d}\right )}{f (2 m+1) (c-d)} \]

[In]

Int[(a - a*Sin[e + f*x])*(a + a*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(-1 - m),x]

[Out]

(2*Sqrt[2]*AppellF1[1/2 + m, -1/2, 1 + m, 3/2 + m, (1 + Sin[e + f*x])/2, -((d*(1 + Sin[e + f*x]))/(c - d))]*Se
c[e + f*x]*Sqrt[1 - Sin[e + f*x]]*(a + a*Sin[e + f*x])^(1 + m)*((c + d*Sin[e + f*x])/(c - d))^m)/((c - d)*f*(1
+ 2*m)*(c + d*Sin[e + f*x])^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 145

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

Rule 3087

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

Rubi steps \begin{align*} \text {integral}& = \frac {\left (\sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {a+a \sin (e+f x)}\right ) \text {Subst}\left (\int \sqrt {a-a x} (a+a x)^{-\frac {1}{2}+m} (c+d x)^{-1-m} \, dx,x,\sin (e+f x)\right )}{f} \\ & = \frac {\left (\sqrt {2} \sec (e+f x) (a-a \sin (e+f x)) \sqrt {a+a \sin (e+f x)}\right ) \text {Subst}\left (\int \sqrt {\frac {1}{2}-\frac {x}{2}} (a+a x)^{-\frac {1}{2}+m} (c+d x)^{-1-m} \, dx,x,\sin (e+f x)\right )}{f \sqrt {\frac {a-a \sin (e+f x)}{a}}} \\ & = \frac {\left (\sqrt {2} a \sec (e+f x) (a-a \sin (e+f x)) \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^{-m} \left (\frac {a (c+d \sin (e+f x))}{a c-a d}\right )^m\right ) \text {Subst}\left (\int \sqrt {\frac {1}{2}-\frac {x}{2}} (a+a x)^{-\frac {1}{2}+m} \left (\frac {a c}{a c-a d}+\frac {a d x}{a c-a d}\right )^{-1-m} \, dx,x,\sin (e+f x)\right )}{(a c-a d) f \sqrt {\frac {a-a \sin (e+f x)}{a}}} \\ & = \frac {2 \sqrt {2} \operatorname {AppellF1}\left (\frac {1}{2}+m,-\frac {1}{2},1+m,\frac {3}{2}+m,\frac {1}{2} (1+\sin (e+f x)),-\frac {d (1+\sin (e+f x))}{c-d}\right ) \sec (e+f x) \sqrt {1-\sin (e+f x)} (a+a \sin (e+f x))^{1+m} (c+d \sin (e+f x))^{-m} \left (\frac {c+d \sin (e+f x)}{c-d}\right )^m}{(c-d) f (1+2 m)} \\ \end{align*}

Mathematica [F]

\[ \int (a-a \sin (e+f x)) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{-1-m} \, dx=\int (a-a \sin (e+f x)) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{-1-m} \, dx \]

[In]

Integrate[(a - a*Sin[e + f*x])*(a + a*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(-1 - m),x]

[Out]

Integrate[(a - a*Sin[e + f*x])*(a + a*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(-1 - m), x]

Maple [F]

\[\int \left (a -a \sin \left (f x +e \right )\right ) \left (a +a \sin \left (f x +e \right )\right )^{m} \left (c +d \sin \left (f x +e \right )\right )^{-1-m}d x\]

[In]

int((a-a*sin(f*x+e))*(a+a*sin(f*x+e))^m*(c+d*sin(f*x+e))^(-1-m),x)

[Out]

int((a-a*sin(f*x+e))*(a+a*sin(f*x+e))^m*(c+d*sin(f*x+e))^(-1-m),x)

Fricas [F]

\[ \int (a-a \sin (e+f x)) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{-1-m} \, dx=\int { -{\left (a \sin \left (f x + e\right ) - a\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (d \sin \left (f x + e\right ) + c\right )}^{-m - 1} \,d x } \]

[In]

integrate((a-a*sin(f*x+e))*(a+a*sin(f*x+e))^m*(c+d*sin(f*x+e))^(-1-m),x, algorithm="fricas")

[Out]

integral(-(a*sin(f*x + e) - a)*(a*sin(f*x + e) + a)^m*(d*sin(f*x + e) + c)^(-m - 1), x)

Sympy [F(-1)]

Timed out. \[ \int (a-a \sin (e+f x)) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{-1-m} \, dx=\text {Timed out} \]

[In]

integrate((a-a*sin(f*x+e))*(a+a*sin(f*x+e))**m*(c+d*sin(f*x+e))**(-1-m),x)

[Out]

Timed out

Maxima [F]

\[ \int (a-a \sin (e+f x)) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{-1-m} \, dx=\int { -{\left (a \sin \left (f x + e\right ) - a\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (d \sin \left (f x + e\right ) + c\right )}^{-m - 1} \,d x } \]

[In]

integrate((a-a*sin(f*x+e))*(a+a*sin(f*x+e))^m*(c+d*sin(f*x+e))^(-1-m),x, algorithm="maxima")

[Out]

-integrate((a*sin(f*x + e) - a)*(a*sin(f*x + e) + a)^m*(d*sin(f*x + e) + c)^(-m - 1), x)

Giac [F]

\[ \int (a-a \sin (e+f x)) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{-1-m} \, dx=\int { -{\left (a \sin \left (f x + e\right ) - a\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (d \sin \left (f x + e\right ) + c\right )}^{-m - 1} \,d x } \]

[In]

integrate((a-a*sin(f*x+e))*(a+a*sin(f*x+e))^m*(c+d*sin(f*x+e))^(-1-m),x, algorithm="giac")

[Out]

integrate(-(a*sin(f*x + e) - a)*(a*sin(f*x + e) + a)^m*(d*sin(f*x + e) + c)^(-m - 1), x)

Mupad [F(-1)]

Timed out. \[ \int (a-a \sin (e+f x)) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{-1-m} \, dx=\int \frac {{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,\left (a-a\,\sin \left (e+f\,x\right )\right )}{{\left (c+d\,\sin \left (e+f\,x\right )\right )}^{m+1}} \,d x \]

[In]

int(((a + a*sin(e + f*x))^m*(a - a*sin(e + f*x)))/(c + d*sin(e + f*x))^(m + 1),x)

[Out]

int(((a + a*sin(e + f*x))^m*(a - a*sin(e + f*x)))/(c + d*sin(e + f*x))^(m + 1), x)