Integrand size = 33, antiderivative size = 145 \[ \int (g \cos (e+f x))^p (a+a \sin (e+f x)) (c+d \sin (e+f x))^n \, dx=-\frac {2^{\frac {3}{2}+\frac {p}{2}} a \operatorname {AppellF1}\left (\frac {1+p}{2},\frac {1}{2} (-1-p),-n,\frac {3+p}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {d (1-\sin (e+f x))}{c+d}\right ) (g \cos (e+f x))^{1+p} (1+\sin (e+f x))^{\frac {1}{2} (-1-p)} (c+d \sin (e+f x))^n \left (\frac {c+d \sin (e+f x)}{c+d}\right )^{-n}}{f g (1+p)} \]
-2^(3/2+1/2*p)*a*AppellF1(1/2+1/2*p,-n,-1/2-1/2*p,3/2+1/2*p,d*(1-sin(f*x+e ))/(c+d),1/2-1/2*sin(f*x+e))*(g*cos(f*x+e))^(p+1)*(1+sin(f*x+e))^(-1/2-1/2 *p)*(c+d*sin(f*x+e))^n/f/g/(p+1)/(((c+d*sin(f*x+e))/(c+d))^n)
\[ \int (g \cos (e+f x))^p (a+a \sin (e+f x)) (c+d \sin (e+f x))^n \, dx=\int (g \cos (e+f x))^p (a+a \sin (e+f x)) (c+d \sin (e+f x))^n \, dx \]
Time = 0.35 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.17, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {3042, 3347, 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.
| \(\displaystyle \int (a \sin (e+f x)+a) (g \cos (e+f x))^p (c+d \sin (e+f x))^n \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int (a \sin (e+f x)+a) (g \cos (e+f x))^p (c+d \sin (e+f x))^ndx\) |
| \(\Big \downarrow \) 3347 |
| \(\displaystyle \frac {a g (1-\sin (e+f x))^{\frac {1-p}{2}} (\sin (e+f x)+1)^{\frac {1-p}{2}} (g \cos (e+f x))^{p-1} \int (1-\sin (e+f x))^{\frac {p-1}{2}} (\sin (e+f x)+1)^{\frac {p+1}{2}} (c+d \sin (e+f x))^nd\sin (e+f x)}{f}\) |
| \(\Big \downarrow \) 156 |
| \(\displaystyle \frac {a g (1-\sin (e+f x))^{\frac {1-p}{2}} (\sin (e+f x)+1)^{\frac {1-p}{2}} (g \cos (e+f x))^{p-1} (c+d \sin (e+f x))^n \left (\frac {c+d \sin (e+f x)}{c+d}\right )^{-n} \int (1-\sin (e+f x))^{\frac {p-1}{2}} (\sin (e+f x)+1)^{\frac {p+1}{2}} \left (\frac {c}{c+d}+\frac {d \sin (e+f x)}{c+d}\right )^nd\sin (e+f x)}{f}\) |
| \(\Big \downarrow \) 155 |
| \(\displaystyle -\frac {a g 2^{\frac {p+3}{2}} (1-\sin (e+f x))^{\frac {1-p}{2}+\frac {p+1}{2}} (\sin (e+f x)+1)^{\frac {1-p}{2}} (g \cos (e+f x))^{p-1} (c+d \sin (e+f x))^n \left (\frac {c+d \sin (e+f x)}{c+d}\right )^{-n} \operatorname {AppellF1}\left (\frac {p+1}{2},\frac {1}{2} (-p-1),-n,\frac {p+3}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {d (1-\sin (e+f x))}{c+d}\right )}{f (p+1)}\) |
-((2^((3 + p)/2)*a*g*AppellF1[(1 + p)/2, (-1 - p)/2, -n, (3 + p)/2, (1 - S in[e + f*x])/2, (d*(1 - Sin[e + f*x]))/(c + d)]*(g*Cos[e + f*x])^(-1 + p)* (1 - Sin[e + f*x])^((1 - p)/2 + (1 + p)/2)*(1 + Sin[e + f*x])^((1 - p)/2)* (c + d*Sin[e + f*x])^n)/(f*(1 + p)*((c + d*Sin[e + f*x])/(c + d))^n))
3.11.44.3.1 Defintions of rubi rules used
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])
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]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[c*g*((g *Cos[e + f*x])^(p - 1)/(f*(1 + Sin[e + f*x])^((p - 1)/2)*(1 - Sin[e + f*x]) ^((p - 1)/2))) Subst[Int[(1 + (d/c)*x)^((p + 1)/2)*(1 - (d/c)*x)^((p - 1) /2)*(a + b*x)^m, x], x, Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, p} , x] && NeQ[a^2 - b^2, 0] && EqQ[c^2 - d^2, 0]
\[\int \left (g \cos \left (f x +e \right )\right )^{p} \left (a +a \sin \left (f x +e \right )\right ) \left (c +d \sin \left (f x +e \right )\right )^{n}d x\]
\[ \int (g \cos (e+f x))^p (a+a \sin (e+f x)) (c+d \sin (e+f x))^n \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )} \left (g \cos \left (f x + e\right )\right )^{p} {\left (d \sin \left (f x + e\right ) + c\right )}^{n} \,d x } \]
\[ \int (g \cos (e+f x))^p (a+a \sin (e+f x)) (c+d \sin (e+f x))^n \, dx=a \left (\int \left (g \cos {\left (e + f x \right )}\right )^{p} \left (c + d \sin {\left (e + f x \right )}\right )^{n}\, dx + \int \left (g \cos {\left (e + f x \right )}\right )^{p} \left (c + d \sin {\left (e + f x \right )}\right )^{n} \sin {\left (e + f x \right )}\, dx\right ) \]
a*(Integral((g*cos(e + f*x))**p*(c + d*sin(e + f*x))**n, x) + Integral((g* cos(e + f*x))**p*(c + d*sin(e + f*x))**n*sin(e + f*x), x))
\[ \int (g \cos (e+f x))^p (a+a \sin (e+f x)) (c+d \sin (e+f x))^n \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )} \left (g \cos \left (f x + e\right )\right )^{p} {\left (d \sin \left (f x + e\right ) + c\right )}^{n} \,d x } \]
\[ \int (g \cos (e+f x))^p (a+a \sin (e+f x)) (c+d \sin (e+f x))^n \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )} \left (g \cos \left (f x + e\right )\right )^{p} {\left (d \sin \left (f x + e\right ) + c\right )}^{n} \,d x } \]
Timed out. \[ \int (g \cos (e+f x))^p (a+a \sin (e+f x)) (c+d \sin (e+f x))^n \, dx=\int {\left (g\,\cos \left (e+f\,x\right )\right )}^p\,\left (a+a\,\sin \left (e+f\,x\right )\right )\,{\left (c+d\,\sin \left (e+f\,x\right )\right )}^n \,d x \]