Integrand size = 27, antiderivative size = 101 \[ \int \frac {(c+d \sin (e+f x))^n}{(3+3 \sin (e+f x))^{5/2}} \, dx=-\frac {\operatorname {AppellF1}\left (\frac {1}{2},-n,3,\frac {3}{2},\frac {d (1-\sin (e+f x))}{c+d},\frac {1}{2} (1-\sin (e+f x))\right ) \cos (e+f x) (c+d \sin (e+f x))^n \left (\frac {c+d \sin (e+f x)}{c+d}\right )^{-n}}{36 f \sqrt {3+3 \sin (e+f x)}} \]
-1/4*AppellF1(1/2,-n,3,3/2,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^2/f/(((c+d*sin(f*x+e))/(c+d))^n)/(a+a*sin(f*x+ e))^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(406\) vs. \(2(101)=202\).
Time = 7.70 (sec) , antiderivative size = 406, normalized size of antiderivative = 4.02 \[ \int \frac {(c+d \sin (e+f x))^n}{(3+3 \sin (e+f x))^{5/2}} \, dx=\frac {\sec (e+f x) (c+d \sin (e+f x))^n \left (\operatorname {AppellF1}\left (1,\frac {1}{2},-n,2,\frac {1}{2} (1+\sin (e+f x)),-\frac {d (1+\sin (e+f x))}{c-d}\right ) \sqrt {2-2 \sin (e+f x)} (1+\sin (e+f x))^3 \left (\frac {c+d \sin (e+f x)}{c-d}\right )^{-n}-\frac {4 (1+\sin (e+f x)) \sqrt {1-\frac {2}{1+\sin (e+f x)}} \left (1+\frac {-1+\frac {c}{d}}{1+\sin (e+f x)}\right )^{-n} \left (\left (3-8 n+4 n^2\right ) \operatorname {AppellF1}\left (-\frac {1}{2}-n,-\frac {1}{2},-n,\frac {1}{2}-n,\frac {2}{1+\sin (e+f x)},\frac {-c+d}{d+d \sin (e+f x)}\right ) (1+\sin (e+f x))^2+2 (1+2 n) \left (2 (-1+2 n) \operatorname {AppellF1}\left (\frac {3}{2}-n,-\frac {1}{2},-n,\frac {5}{2}-n,\frac {2}{1+\sin (e+f x)},\frac {-c+d}{d+d \sin (e+f x)}\right )+(-3+2 n) \operatorname {AppellF1}\left (\frac {1}{2}-n,-\frac {1}{2},-n,\frac {3}{2}-n,\frac {2}{1+\sin (e+f x)},\frac {-c+d}{d+d \sin (e+f x)}\right ) (1+\sin (e+f x))\right )\right )}{(-3+2 n) (-1+2 n) (1+2 n)}\right )}{144 \sqrt {3} f (1+\sin (e+f x))^{3/2}} \]
(Sec[e + f*x]*(c + d*Sin[e + f*x])^n*((AppellF1[1, 1/2, -n, 2, (1 + Sin[e + f*x])/2, -((d*(1 + Sin[e + f*x]))/(c - d))]*Sqrt[2 - 2*Sin[e + f*x]]*(1 + Sin[e + f*x])^3)/((c + d*Sin[e + f*x])/(c - d))^n - (4*(1 + Sin[e + f*x] )*Sqrt[1 - 2/(1 + Sin[e + f*x])]*((3 - 8*n + 4*n^2)*AppellF1[-1/2 - n, -1/ 2, -n, 1/2 - n, 2/(1 + Sin[e + f*x]), (-c + d)/(d + d*Sin[e + f*x])]*(1 + Sin[e + f*x])^2 + 2*(1 + 2*n)*(2*(-1 + 2*n)*AppellF1[3/2 - n, -1/2, -n, 5/ 2 - n, 2/(1 + Sin[e + f*x]), (-c + d)/(d + d*Sin[e + f*x])] + (-3 + 2*n)*A ppellF1[1/2 - n, -1/2, -n, 3/2 - n, 2/(1 + Sin[e + f*x]), (-c + d)/(d + d* Sin[e + f*x])]*(1 + Sin[e + f*x]))))/((-3 + 2*n)*(-1 + 2*n)*(1 + 2*n)*(1 + (-1 + c/d)/(1 + Sin[e + f*x]))^n)))/(144*Sqrt[3]*f*(1 + Sin[e + f*x])^(3/ 2))
Time = 0.35 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.35, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {3042, 3267, 27, 154, 153}
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.
| \(\displaystyle \int \frac {(c+d \sin (e+f x))^n}{(a \sin (e+f x)+a)^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(c+d \sin (e+f x))^n}{(a \sin (e+f x)+a)^{5/2}}dx\) |
| \(\Big \downarrow \) 3267 |
| \(\displaystyle \frac {a^2 \cos (e+f x) \int \frac {(c+d \sin (e+f x))^n}{a^3 (\sin (e+f x)+1)^3 \sqrt {a-a \sin (e+f x)}}d\sin (e+f x)}{f \sqrt {a-a \sin (e+f x)} \sqrt {a \sin (e+f x)+a}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\cos (e+f x) \int \frac {(c+d \sin (e+f x))^n}{(\sin (e+f x)+1)^3 \sqrt {a-a \sin (e+f x)}}d\sin (e+f x)}{a f \sqrt {a-a \sin (e+f x)} \sqrt {a \sin (e+f x)+a}}\) |
| \(\Big \downarrow \) 154 |
| \(\displaystyle \frac {\cos (e+f x) \sqrt {\frac {d (1-\sin (e+f x))}{c+d}} \int \frac {(c+d \sin (e+f x))^n}{(\sin (e+f x)+1)^3 \sqrt {\frac {d}{c+d}-\frac {d \sin (e+f x)}{c+d}}}d\sin (e+f x)}{a f (a-a \sin (e+f x)) \sqrt {a \sin (e+f x)+a}}\) |
| \(\Big \downarrow \) 153 |
| \(\displaystyle -\frac {d^2 \cos (e+f x) \sqrt {\frac {d (1-\sin (e+f x))}{c+d}} (c+d \sin (e+f x))^{n+1} \operatorname {AppellF1}\left (n+1,\frac {1}{2},3,n+2,\frac {c+d \sin (e+f x)}{c+d},\frac {c+d \sin (e+f x)}{c-d}\right )}{a f (n+1) (c-d)^3 (a-a \sin (e+f x)) \sqrt {a \sin (e+f x)+a}}\) |
-((d^2*AppellF1[1 + n, 1/2, 3, 2 + n, (c + d*Sin[e + f*x])/(c + d), (c + d *Sin[e + f*x])/(c - d)]*Cos[e + f*x]*Sqrt[(d*(1 - Sin[e + f*x]))/(c + d)]* (c + d*Sin[e + f*x])^(1 + n))/(a*(c - d)^3*f*(1 + n)*(a - a*Sin[e + f*x])* Sqrt[a + a*Sin[e + f*x]]))
3.7.67.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) ^(p_), x_] :> Simp[(b*e - a*f)^p*((a + b*x)^(m + 1)/(b^(p + 1)*(m + 1)*Simp lify[b/(b*c - a*d)]^n))*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}, x] && !IntegerQ[m] && !IntegerQ[n] && IntegerQ[p] && GtQ[Simplify[b/( b*c - a*d)], 0] && !(GtQ[Simplify[d/(d*a - c*b)], 0] && SimplerQ[c + d*x, a + b*x])
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) ^(p_), x_] :> Simp[(c + d*x)^FracPart[n]/(Simplify[b/(b*c - a*d)]^IntPart[n ]*(b*((c + d*x)/(b*c - a*d)))^FracPart[n]) Int[(a + b*x)^m*Simp[b*(c/(b*c - a*d)) + b*d*(x/(b*c - a*d)), x]^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && !IntegerQ[m] && !IntegerQ[n] && IntegerQ[p] && !G tQ[Simplify[b/(b*c - a*d)], 0] && !SimplerQ[c + d*x, a + b*x]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + ( f_.)*(x_)])^(n_.), x_Symbol] :> Simp[a^2*(Cos[e + f*x]/(f*Sqrt[a + b*Sin[e + f*x]]*Sqrt[a - b*Sin[e + f*x]])) Subst[Int[(a + b*x)^(m - 1/2)*((c + d* x)^n/Sqrt[a - b*x]), x], x, Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m , n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && !IntegerQ[m]
\[\int \frac {\left (c +d \sin \left (f x +e \right )\right )^{n}}{\left (a +a \sin \left (f x +e \right )\right )^{\frac {5}{2}}}d x\]
\[ \int \frac {(c+d \sin (e+f x))^n}{(3+3 \sin (e+f x))^{5/2}} \, dx=\int { \frac {{\left (d \sin \left (f x + e\right ) + c\right )}^{n}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
integral(-sqrt(a*sin(f*x + e) + a)*(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)
\[ \int \frac {(c+d \sin (e+f x))^n}{(3+3 \sin (e+f x))^{5/2}} \, dx=\int \frac {\left (c + d \sin {\left (e + f x \right )}\right )^{n}}{\left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{\frac {5}{2}}}\, dx \]
\[ \int \frac {(c+d \sin (e+f x))^n}{(3+3 \sin (e+f x))^{5/2}} \, dx=\int { \frac {{\left (d \sin \left (f x + e\right ) + c\right )}^{n}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {(c+d \sin (e+f x))^n}{(3+3 \sin (e+f x))^{5/2}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {(c+d \sin (e+f x))^n}{(3+3 \sin (e+f x))^{5/2}} \, dx=\int \frac {{\left (c+d\,\sin \left (e+f\,x\right )\right )}^n}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{5/2}} \,d x \]