Integrand size = 35, antiderivative size = 149 \[ \int \frac {(g \cos (e+f x))^p (c+d \sin (e+f x))^n}{(a+a \sin (e+f x))^3} \, dx=-\frac {2^{-\frac {5}{2}+\frac {p}{2}} \operatorname {AppellF1}\left (\frac {1+p}{2},\frac {7-p}{2},-n,\frac {3+p}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {d (1-\sin (e+f x))}{c+d}\right ) (g \cos (e+f x))^{1+p} (1+\sin (e+f x))^{-3+\frac {5-p}{2}} (c+d \sin (e+f x))^n \left (\frac {c+d \sin (e+f x)}{c+d}\right )^{-n}}{a^3 f g (1+p)} \]
-2^(-5/2+1/2*p)*AppellF1(1/2+1/2*p,-n,7/2-1/2*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/a^3/f/g/(p+1)/(((c+d*sin(f*x+e))/(c+d))^n)
\[ \int \frac {(g \cos (e+f x))^p (c+d \sin (e+f x))^n}{(a+a \sin (e+f x))^3} \, dx=\int \frac {(g \cos (e+f x))^p (c+d \sin (e+f x))^n}{(a+a \sin (e+f x))^3} \, dx \]
Time = 0.40 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.15, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.114, Rules used = {3042, 3399, 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.
\(\displaystyle \int \frac {(g \cos (e+f x))^p (c+d \sin (e+f x))^n}{(a \sin (e+f x)+a)^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(g \cos (e+f x))^p (c+d \sin (e+f x))^n}{(a \sin (e+f x)+a)^3}dx\) |
\(\Big \downarrow \) 3399 |
\(\displaystyle \frac {g (1-\sin (e+f x))^{\frac {1-p}{2}} (\sin (e+f x)+1)^{\frac {1-p}{2}} (g \cos (e+f x))^{p-1} \int (1-\sin (e+f x))^{\frac {p-1}{2}} (\sin (e+f x)+1)^{\frac {p-7}{2}} (c+d \sin (e+f x))^nd\sin (e+f x)}{a^3 f}\) |
\(\Big \downarrow \) 156 |
\(\displaystyle \frac {g (1-\sin (e+f x))^{\frac {1-p}{2}} (\sin (e+f x)+1)^{\frac {1-p}{2}} (g \cos (e+f x))^{p-1} (c+d \sin (e+f x))^n \left (\frac {c+d \sin (e+f x)}{c+d}\right )^{-n} \int (1-\sin (e+f x))^{\frac {p-1}{2}} (\sin (e+f x)+1)^{\frac {p-7}{2}} \left (\frac {c}{c+d}+\frac {d \sin (e+f x)}{c+d}\right )^nd\sin (e+f x)}{a^3 f}\) |
\(\Big \downarrow \) 155 |
\(\displaystyle -\frac {g 2^{\frac {p-5}{2}} (1-\sin (e+f x))^{\frac {1-p}{2}+\frac {p+1}{2}} (\sin (e+f x)+1)^{\frac {1-p}{2}} (g \cos (e+f x))^{p-1} (c+d \sin (e+f x))^n \left (\frac {c+d \sin (e+f x)}{c+d}\right )^{-n} \operatorname {AppellF1}\left (\frac {p+1}{2},\frac {7-p}{2},-n,\frac {p+3}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {d (1-\sin (e+f x))}{c+d}\right )}{a^3 f (p+1)}\) |
-((2^((-5 + p)/2)*g*AppellF1[(1 + p)/2, (7 - p)/2, -n, (3 + p)/2, (1 - Sin [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)/(a^3*f*(1 + p)*((c + d*Sin[e + f*x])/(c + d))^n))
3.11.47.3.1 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_)])^(n_), x_Symbol] :> Simp[a^ m*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 + (b/a)*x)^(m + (p - 1)/2)*(1 - (b/a)* x)^((p - 1)/2)*(c + d*x)^n, x], x, Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && EqQ[a^2 - b^2, 0] && IntegerQ[m]
\[\int \frac {\left (g \cos \left (f x +e \right )\right )^{p} \left (c +d \sin \left (f x +e \right )\right )^{n}}{\left (a +a \sin \left (f x +e \right )\right )^{3}}d x\]
\[ \int \frac {(g \cos (e+f x))^p (c+d \sin (e+f x))^n}{(a+a \sin (e+f x))^3} \, dx=\int { \frac {\left (g \cos \left (f x + e\right )\right )^{p} {\left (d \sin \left (f x + e\right ) + c\right )}^{n}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{3}} \,d x } \]
integral(-(g*cos(f*x + e))^p*(d*sin(f*x + e) + c)^n/(3*a^3*cos(f*x + e)^2 - 4*a^3 + (a^3*cos(f*x + e)^2 - 4*a^3)*sin(f*x + e)), x)
Timed out. \[ \int \frac {(g \cos (e+f x))^p (c+d \sin (e+f x))^n}{(a+a \sin (e+f x))^3} \, dx=\text {Timed out} \]
\[ \int \frac {(g \cos (e+f x))^p (c+d \sin (e+f x))^n}{(a+a \sin (e+f x))^3} \, dx=\int { \frac {\left (g \cos \left (f x + e\right )\right )^{p} {\left (d \sin \left (f x + e\right ) + c\right )}^{n}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{3}} \,d x } \]
\[ \int \frac {(g \cos (e+f x))^p (c+d \sin (e+f x))^n}{(a+a \sin (e+f x))^3} \, dx=\int { \frac {\left (g \cos \left (f x + e\right )\right )^{p} {\left (d \sin \left (f x + e\right ) + c\right )}^{n}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{3}} \,d x } \]
Timed out. \[ \int \frac {(g \cos (e+f x))^p (c+d \sin (e+f x))^n}{(a+a \sin (e+f x))^3} \, dx=\int \frac {{\left (g\,\cos \left (e+f\,x\right )\right )}^p\,{\left (c+d\,\sin \left (e+f\,x\right )\right )}^n}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^3} \,d x \]