Integrand size = 23, antiderivative size = 80 \[ \int \frac {(d \sin (e+f x))^n}{(1+\sin (e+f x))^{3/2}} \, dx=-\frac {\operatorname {AppellF1}\left (\frac {1}{2},-n,2,\frac {3}{2},1-\sin (e+f x),\frac {1}{2} (1-\sin (e+f x))\right ) \cos (e+f x) \sin ^{-n}(e+f x) (d \sin (e+f x))^n}{2 f \sqrt {1+\sin (e+f x)}} \]
-1/2*AppellF1(1/2,-n,2,3/2,1-sin(f*x+e),1/2-1/2*sin(f*x+e))*cos(f*x+e)*(d* sin(f*x+e))^n/f/(sin(f*x+e)^n)/(1+sin(f*x+e))^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(265\) vs. \(2(80)=160\).
Time = 0.58 (sec) , antiderivative size = 265, normalized size of antiderivative = 3.31 \[ \int \frac {(d \sin (e+f x))^n}{(1+\sin (e+f x))^{3/2}} \, dx=\frac {\sec (e+f x) (d \sin (e+f x))^n \left (\operatorname {AppellF1}\left (1,\frac {1}{2},-n,2,\frac {1}{2} (1+\sin (e+f x)),1+\sin (e+f x)\right ) \sqrt {2-2 \sin (e+f x)} (-\sin (e+f x))^{-n} (1+\sin (e+f x))^2-\frac {4 (1+\sin (e+f x)) \sqrt {1-\frac {2}{1+\sin (e+f x)}} \left (1-\frac {1}{1+\sin (e+f x)}\right )^{-n} \left (2 (1+2 n) \operatorname {AppellF1}\left (\frac {1}{2}-n,-\frac {1}{2},-n,\frac {3}{2}-n,\frac {2}{1+\sin (e+f x)},\frac {1}{1+\sin (e+f x)}\right )+(-1+2 n) \operatorname {AppellF1}\left (-\frac {1}{2}-n,-\frac {1}{2},-n,\frac {1}{2}-n,\frac {2}{1+\sin (e+f x)},\frac {1}{1+\sin (e+f x)}\right ) (1+\sin (e+f x))\right )}{-1+4 n^2}\right )}{8 f \sqrt {1+\sin (e+f x)}} \]
(Sec[e + f*x]*(d*Sin[e + f*x])^n*((AppellF1[1, 1/2, -n, 2, (1 + Sin[e + f* x])/2, 1 + Sin[e + f*x]]*Sqrt[2 - 2*Sin[e + f*x]]*(1 + Sin[e + f*x])^2)/(- Sin[e + f*x])^n - (4*(1 + Sin[e + f*x])*Sqrt[1 - 2/(1 + Sin[e + f*x])]*(2* (1 + 2*n)*AppellF1[1/2 - n, -1/2, -n, 3/2 - n, 2/(1 + Sin[e + f*x]), (1 + Sin[e + f*x])^(-1)] + (-1 + 2*n)*AppellF1[-1/2 - n, -1/2, -n, 1/2 - n, 2/( 1 + Sin[e + f*x]), (1 + Sin[e + f*x])^(-1)]*(1 + Sin[e + f*x])))/((-1 + 4* n^2)*(1 - (1 + Sin[e + f*x])^(-1))^n)))/(8*f*Sqrt[1 + Sin[e + f*x]])
Time = 0.37 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3042, 3265, 3042, 3264, 148, 333}
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.
\(\displaystyle \int \frac {(d \sin (e+f x))^n}{(\sin (e+f x)+1)^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(d \sin (e+f x))^n}{(\sin (e+f x)+1)^{3/2}}dx\) |
\(\Big \downarrow \) 3265 |
\(\displaystyle \sin ^{-n}(e+f x) (d \sin (e+f x))^n \int \frac {\sin ^n(e+f x)}{(\sin (e+f x)+1)^{3/2}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sin ^{-n}(e+f x) (d \sin (e+f x))^n \int \frac {\sin (e+f x)^n}{(\sin (e+f x)+1)^{3/2}}dx\) |
\(\Big \downarrow \) 3264 |
\(\displaystyle -\frac {\cos (e+f x) \sin ^{-n}(e+f x) (d \sin (e+f x))^n \int \frac {\sin ^n(e+f x)}{\sqrt {1-\sin (e+f x)} (\sin (e+f x)+1)^2}d(1-\sin (e+f x))}{f \sqrt {1-\sin (e+f x)} \sqrt {\sin (e+f x)+1}}\) |
\(\Big \downarrow \) 148 |
\(\displaystyle -\frac {2 \cos (e+f x) \sin ^{-n}(e+f x) (d \sin (e+f x))^n \int \frac {\sin ^n(e+f x)}{(\sin (e+f x)+1)^2}d\sqrt {1-\sin (e+f x)}}{f \sqrt {1-\sin (e+f x)} \sqrt {\sin (e+f x)+1}}\) |
\(\Big \downarrow \) 333 |
\(\displaystyle -\frac {\cos (e+f x) \sin ^{-n}(e+f x) \operatorname {AppellF1}\left (\frac {1}{2},-n,2,\frac {3}{2},1-\sin (e+f x),\frac {1}{2} (1-\sin (e+f x))\right ) (d \sin (e+f x))^n}{2 f \sqrt {\sin (e+f x)+1}}\) |
-1/2*(AppellF1[1/2, -n, 2, 3/2, 1 - Sin[e + f*x], (1 - Sin[e + f*x])/2]*Co s[e + f*x]*(d*Sin[e + f*x])^n)/(f*Sin[e + f*x]^n*Sqrt[1 + Sin[e + f*x]])
3.2.27.3.1 Defintions of rubi rules used
Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))^(p_.), x_] :> With[{k = Denominator[m]}, Simp[k/b Subst[Int[x^(k*(m + 1) - 1)*(c + d*(x^k/b))^n*(e + f*(x^k/b))^p, x], x, (b*x)^(1/k)], x]] /; FreeQ[{b, c, d, e, f, n, p}, x] && FractionQ[m] && IntegerQ[p]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[a^p*c^q*x*AppellF1[1/2, -p, -q, 3/2, (-b)*(x^2/a), (-d)*(x^2/c)], x] /; F reeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_), x_Symbol] :> Simp[(-b)*(d/b)^n*(Cos[e + f*x]/(f*Sqrt[a + b*Sin[e + f*x]]*Sqrt[a - b*Sin[e + f*x]])) Subst[Int[(a - x)^n*((2*a - x)^(m - 1 /2)/Sqrt[x]), x], x, a - b*Sin[e + f*x]], x] /; FreeQ[{a, b, d, e, f, m, n} , x] && EqQ[a^2 - b^2, 0] && !IntegerQ[m] && GtQ[a, 0] && GtQ[d/b, 0]
Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*( x_)])^(m_), x_Symbol] :> Simp[(d/b)^IntPart[n]*((d*Sin[e + f*x])^FracPart[n ]/(b*Sin[e + f*x])^FracPart[n]) Int[(a + b*Sin[e + f*x])^m*(b*Sin[e + f*x ])^n, x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] && !I ntegerQ[m] && GtQ[a, 0] && !GtQ[d/b, 0]
\[\int \frac {\left (d \sin \left (f x +e \right )\right )^{n}}{\left (\sin \left (f x +e \right )+1\right )^{\frac {3}{2}}}d x\]
\[ \int \frac {(d \sin (e+f x))^n}{(1+\sin (e+f x))^{3/2}} \, dx=\int { \frac {\left (d \sin \left (f x + e\right )\right )^{n}}{{\left (\sin \left (f x + e\right ) + 1\right )}^{\frac {3}{2}}} \,d x } \]
integral(-(d*sin(f*x + e))^n*sqrt(sin(f*x + e) + 1)/(cos(f*x + e)^2 - 2*si n(f*x + e) - 2), x)
\[ \int \frac {(d \sin (e+f x))^n}{(1+\sin (e+f x))^{3/2}} \, dx=\int \frac {\left (d \sin {\left (e + f x \right )}\right )^{n}}{\left (\sin {\left (e + f x \right )} + 1\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {(d \sin (e+f x))^n}{(1+\sin (e+f x))^{3/2}} \, dx=\int { \frac {\left (d \sin \left (f x + e\right )\right )^{n}}{{\left (\sin \left (f x + e\right ) + 1\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {(d \sin (e+f x))^n}{(1+\sin (e+f x))^{3/2}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {(d \sin (e+f x))^n}{(1+\sin (e+f x))^{3/2}} \, dx=\int \frac {{\left (d\,\sin \left (e+f\,x\right )\right )}^n}{{\left (\sin \left (e+f\,x\right )+1\right )}^{3/2}} \,d x \]