Integrand size = 25, antiderivative size = 85 \[ \int \frac {(d \sin (e+f x))^n}{(a+a \sin (e+f x))^{3/2}} \, dx=-\frac {\operatorname {AppellF1}\left (\frac {1}{2},-n,2,\frac {3}{2},1-\sin (e+f x),\frac {1}{2} (1-\sin (e+f x))\right ) \cos (e+f x) \sin ^{-n}(e+f x) (d \sin (e+f x))^n}{2 a f \sqrt {a+a \sin (e+f x)}} \]
-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/a/f/(sin(f*x+e)^n)/(a+a*sin(f*x+e))^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(276\) vs. \(2(85)=170\).
Time = 0.94 (sec) , antiderivative size = 276, normalized size of antiderivative = 3.25 \[ \int \frac {(d \sin (e+f x))^n}{(a+a \sin (e+f x))^{3/2}} \, dx=\frac {\sec (e+f x) (d \sin (e+f x))^n \left (a^2 \operatorname {AppellF1}\left (1,\frac {1}{2},-n,2,\frac {1}{2} (1+\sin (e+f x)),1+\sin (e+f x)\right ) \sqrt {2-2 \sin (e+f x)} (-\sin (e+f x))^{-n} (1+\sin (e+f x))^2-\frac {4 a (-1+\sin (e+f x)) \left (1-\frac {1}{1+\sin (e+f x)}\right )^{-n} \left (2 a (1+2 n) \operatorname {AppellF1}\left (\frac {1}{2}-n,-\frac {1}{2},-n,\frac {3}{2}-n,\frac {2}{1+\sin (e+f x)},\frac {1}{1+\sin (e+f x)}\right )+a (-1+2 n) \operatorname {AppellF1}\left (-\frac {1}{2}-n,-\frac {1}{2},-n,\frac {1}{2}-n,\frac {2}{1+\sin (e+f x)},\frac {1}{1+\sin (e+f x)}\right ) (1+\sin (e+f x))\right )}{\left (-1+4 n^2\right ) \sqrt {1-\frac {2}{1+\sin (e+f x)}}}\right )}{8 a^3 f \sqrt {a (1+\sin (e+f x))}} \]
(Sec[e + f*x]*(d*Sin[e + f*x])^n*((a^2*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*a*(-1 + Sin[e + f*x])*(2*a*(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)] + a*(-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)*Sqrt[1 - 2/(1 + Sin[e + f*x])]*(1 - (1 + Sin[e + f*x])^(-1))^n)))/(8*a^3*f*Sqrt[a*(1 + Si n[e + f*x])])
Time = 0.48 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 3266, 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.
\(\displaystyle \int \frac {(d \sin (e+f x))^n}{(a \sin (e+f x)+a)^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(d \sin (e+f x))^n}{(a \sin (e+f x)+a)^{3/2}}dx\) |
\(\Big \downarrow \) 3266 |
\(\displaystyle \frac {\sqrt {\sin (e+f x)+1} \int \frac {(d \sin (e+f x))^n}{(\sin (e+f x)+1)^{3/2}}dx}{a \sqrt {a \sin (e+f x)+a}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sqrt {\sin (e+f x)+1} \int \frac {(d \sin (e+f x))^n}{(\sin (e+f x)+1)^{3/2}}dx}{a \sqrt {a \sin (e+f x)+a}}\) |
\(\Big \downarrow \) 3265 |
\(\displaystyle \frac {\sqrt {\sin (e+f x)+1} \sin ^{-n}(e+f x) (d \sin (e+f x))^n \int \frac {\sin ^n(e+f x)}{(\sin (e+f x)+1)^{3/2}}dx}{a \sqrt {a \sin (e+f x)+a}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sqrt {\sin (e+f x)+1} \sin ^{-n}(e+f x) (d \sin (e+f x))^n \int \frac {\sin (e+f x)^n}{(\sin (e+f x)+1)^{3/2}}dx}{a \sqrt {a \sin (e+f x)+a}}\) |
\(\Big \downarrow \) 3264 |
\(\displaystyle -\frac {\cos (e+f x) \sin ^{-n}(e+f x) (d \sin (e+f x))^n \int \frac {\sin ^n(e+f x)}{\sqrt {1-\sin (e+f x)} (\sin (e+f x)+1)^2}d(1-\sin (e+f x))}{a f \sqrt {1-\sin (e+f x)} \sqrt {a \sin (e+f x)+a}}\) |
\(\Big \downarrow \) 148 |
\(\displaystyle -\frac {2 \cos (e+f x) \sin ^{-n}(e+f x) (d \sin (e+f x))^n \int \frac {\sin ^n(e+f x)}{(\sin (e+f x)+1)^2}d\sqrt {1-\sin (e+f x)}}{a f \sqrt {1-\sin (e+f x)} \sqrt {a \sin (e+f x)+a}}\) |
\(\Big \downarrow \) 333 |
\(\displaystyle -\frac {\cos (e+f x) \sin ^{-n}(e+f x) \operatorname {AppellF1}\left (\frac {1}{2},-n,2,\frac {3}{2},1-\sin (e+f x),\frac {1}{2} (1-\sin (e+f x))\right ) (d \sin (e+f x))^n}{2 a f \sqrt {a \sin (e+f x)+a}}\) |
-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)/(a*f*Sin[e + f*x]^n*Sqrt[a + a*Sin[e + f*x] ])
3.2.31.3.1 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[((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*( x_)])^(m_), x_Symbol] :> Simp[a^IntPart[m]*((a + b*Sin[e + f*x])^FracPart[m ]/(1 + (b/a)*Sin[e + f*x])^FracPart[m]) Int[(1 + (b/a)*Sin[e + f*x])^m*(d *Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^ 2, 0] && !IntegerQ[m] && !GtQ[a, 0]
\[\int \frac {\left (d \sin \left (f x +e \right )\right )^{n}}{\left (a +a \sin \left (f x +e \right )\right )^{\frac {3}{2}}}d x\]
\[ \int \frac {(d \sin (e+f x))^n}{(a+a \sin (e+f x))^{3/2}} \, dx=\int { \frac {\left (d \sin \left (f x + e\right )\right )^{n}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]
integral(-sqrt(a*sin(f*x + e) + a)*(d*sin(f*x + e))^n/(a^2*cos(f*x + e)^2 - 2*a^2*sin(f*x + e) - 2*a^2), x)
\[ \int \frac {(d \sin (e+f x))^n}{(a+a \sin (e+f x))^{3/2}} \, dx=\int \frac {\left (d \sin {\left (e + f x \right )}\right )^{n}}{\left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {(d \sin (e+f x))^n}{(a+a \sin (e+f x))^{3/2}} \, dx=\int { \frac {\left (d \sin \left (f x + e\right )\right )^{n}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {(d \sin (e+f x))^n}{(a+a \sin (e+f x))^{3/2}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {(d \sin (e+f x))^n}{(a+a \sin (e+f x))^{3/2}} \, dx=\int \frac {{\left (d\,\sin \left (e+f\,x\right )\right )}^n}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{3/2}} \,d x \]