\(\int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^n}{a+a \sin (e+f x)} \, dx\) [330]

Optimal result

Integrand size = 35, antiderivative size = 221 \[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^n}{a+a \sin (e+f x)} \, dx=-\frac {\sqrt {2} B \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-n,\frac {3}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {d (1-\sin (e+f x))}{c+d}\right ) \cos (e+f x) (c+d \sin (e+f x))^n \left (\frac {c+d \sin (e+f x)}{c+d}\right )^{-n}}{a f \sqrt {1+\sin (e+f x)}}-\frac {(A-B) \operatorname {AppellF1}\left (\frac {1}{2},\frac {3}{2},-n,\frac {3}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {d (1-\sin (e+f x))}{c+d}\right ) \cos (e+f x) (c+d \sin (e+f x))^n \left (\frac {c+d \sin (e+f x)}{c+d}\right )^{-n}}{\sqrt {2} a f \sqrt {1+\sin (e+f x)}} \]

[Out]

-1/2*(A-B)*AppellF1(1/2,-n,3/2,3/2,d*(1-sin(f*x+e))/(c+d),1/2-1/2*sin(f*x+e))*cos(f*x+e)*(c+d*sin(f*x+e))^n/a/
f/(((c+d*sin(f*x+e))/(c+d))^n)*2^(1/2)/(1+sin(f*x+e))^(1/2)-B*AppellF1(1/2,-n,1/2,3/2,d*(1-sin(f*x+e))/(c+d),1
/2-1/2*sin(f*x+e))*cos(f*x+e)*(c+d*sin(f*x+e))^n*2^(1/2)/a/f/(((c+d*sin(f*x+e))/(c+d))^n)/(1+sin(f*x+e))^(1/2)

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3066, 2863, 144, 143, 2744} \[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^n}{a+a \sin (e+f x)} \, dx=-\frac {(A-B) \cos (e+f x) (c+d \sin (e+f x))^n \left (\frac {c+d \sin (e+f x)}{c+d}\right )^{-n} \operatorname {AppellF1}\left (\frac {1}{2},\frac {3}{2},-n,\frac {3}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {d (1-\sin (e+f x))}{c+d}\right )}{\sqrt {2} a f \sqrt {\sin (e+f x)+1}}-\frac {\sqrt {2} B \cos (e+f x) (c+d \sin (e+f x))^n \left (\frac {c+d \sin (e+f x)}{c+d}\right )^{-n} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-n,\frac {3}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {d (1-\sin (e+f x))}{c+d}\right )}{a f \sqrt {\sin (e+f x)+1}} \]

[In]

Int[((A + B*Sin[e + f*x])*(c + d*Sin[e + f*x])^n)/(a + a*Sin[e + f*x]),x]

[Out]

-((Sqrt[2]*B*AppellF1[1/2, 1/2, -n, 3/2, (1 - Sin[e + f*x])/2, (d*(1 - Sin[e + f*x]))/(c + d)]*Cos[e + f*x]*(c
+ d*Sin[e + f*x])^n)/(a*f*Sqrt[1 + Sin[e + f*x]]*((c + d*Sin[e + f*x])/(c + d))^n)) - ((A - B)*AppellF1[1/2,
3/2, -n, 3/2, (1 - Sin[e + f*x])/2, (d*(1 - Sin[e + f*x]))/(c + d)]*Cos[e + f*x]*(c + d*Sin[e + f*x])^n)/(Sqrt
[2]*a*f*Sqrt[1 + Sin[e + f*x]]*((c + d*Sin[e + f*x])/(c + d))^n)

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 2744

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

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 3066

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e
_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(A*b - a*B)/b, Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^n, x
], x] + Dist[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]

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

Mathematica [F]

\[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^n}{a+a \sin (e+f x)} \, dx=\int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^n}{a+a \sin (e+f x)} \, dx \]

[In]

Integrate[((A + B*Sin[e + f*x])*(c + d*Sin[e + f*x])^n)/(a + a*Sin[e + f*x]),x]

[Out]

Integrate[((A + B*Sin[e + f*x])*(c + d*Sin[e + f*x])^n)/(a + a*Sin[e + f*x]), x]

Maple [F]

\[\int \frac {\left (A +B \sin \left (f x +e \right )\right ) \left (c +d \sin \left (f x +e \right )\right )^{n}}{a +a \sin \left (f x +e \right )}d x\]

[In]

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

[Out]

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

Fricas [F]

\[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^n}{a+a \sin (e+f x)} \, dx=\int { \frac {{\left (B \sin \left (f x + e\right ) + A\right )} {\left (d \sin \left (f x + e\right ) + c\right )}^{n}}{a \sin \left (f x + e\right ) + a} \,d x } \]

[In]

integrate((A+B*sin(f*x+e))*(c+d*sin(f*x+e))^n/(a+a*sin(f*x+e)),x, algorithm="fricas")

[Out]

integral((B*sin(f*x + e) + A)*(d*sin(f*x + e) + c)^n/(a*sin(f*x + e) + a), x)

Sympy [F(-1)]

Timed out. \[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^n}{a+a \sin (e+f x)} \, dx=\text {Timed out} \]

[In]

integrate((A+B*sin(f*x+e))*(c+d*sin(f*x+e))**n/(a+a*sin(f*x+e)),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^n}{a+a \sin (e+f x)} \, dx=\int { \frac {{\left (B \sin \left (f x + e\right ) + A\right )} {\left (d \sin \left (f x + e\right ) + c\right )}^{n}}{a \sin \left (f x + e\right ) + a} \,d x } \]

[In]

integrate((A+B*sin(f*x+e))*(c+d*sin(f*x+e))^n/(a+a*sin(f*x+e)),x, algorithm="maxima")

[Out]

integrate((B*sin(f*x + e) + A)*(d*sin(f*x + e) + c)^n/(a*sin(f*x + e) + a), x)

Giac [F]

\[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^n}{a+a \sin (e+f x)} \, dx=\int { \frac {{\left (B \sin \left (f x + e\right ) + A\right )} {\left (d \sin \left (f x + e\right ) + c\right )}^{n}}{a \sin \left (f x + e\right ) + a} \,d x } \]

[In]

integrate((A+B*sin(f*x+e))*(c+d*sin(f*x+e))^n/(a+a*sin(f*x+e)),x, algorithm="giac")

[Out]

integrate((B*sin(f*x + e) + A)*(d*sin(f*x + e) + c)^n/(a*sin(f*x + e) + a), x)

Mupad [F(-1)]

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

[In]

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

[Out]

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