\(\int \frac {\sin ^n(e+f x)}{(1+\sin (e+f x))^{3/2}} \, dx\) [119]

Optimal result

Integrand size = 21, antiderivative size = 60 \[ \int \frac {\sin ^n(e+f x)}{(1+\sin (e+f x))^{3/2}} \, dx=-\frac {\operatorname {AppellF1}\left (\frac {1}{2},-n,2,\frac {3}{2},1-\sin (e+f x),\frac {1}{2} (1-\sin (e+f x))\right ) \cos (e+f x)}{2 f \sqrt {1+\sin (e+f x)}} \]

[Out]

-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)/f/(1+sin(f*x+e))^(1/2)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2864, 129, 440} \[ \int \frac {\sin ^n(e+f x)}{(1+\sin (e+f x))^{3/2}} \, dx=-\frac {\cos (e+f x) \operatorname {AppellF1}\left (\frac {1}{2},-n,2,\frac {3}{2},1-\sin (e+f x),\frac {1}{2} (1-\sin (e+f x))\right )}{2 f \sqrt {\sin (e+f x)+1}} \]

[In]

Int[Sin[e + f*x]^n/(1 + Sin[e + f*x])^(3/2),x]

[Out]

-1/2*(AppellF1[1/2, -n, 2, 3/2, 1 - Sin[e + f*x], (1 - Sin[e + f*x])/2]*Cos[e + f*x])/(f*Sqrt[1 + Sin[e + f*x]
])

Rule 129

Int[((e_.)*(x_))^(p_)*((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> With[{k = Denominator[p]
}, Dist[k/e, Subst[Int[x^(k*(p + 1) - 1)*(a + b*(x^k/e))^m*(c + d*(x^k/e))^n, x], x, (e*x)^(1/k)], x]] /; Free
Q[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] && FractionQ[p] && IntegerQ[m]

Rule 440

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*x*AppellF1[1/n, -p,
-q, 1 + 1/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; FreeQ[{a, b, c, d, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n
, -1] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rule 2864

Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[(-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]

Rubi steps \begin{align*} \text {integral}& = -\frac {\cos (e+f x) \text {Subst}\left (\int \frac {(1-x)^n}{(2-x)^2 \sqrt {x}} \, dx,x,1-\sin (e+f x)\right )}{f \sqrt {1-\sin (e+f x)} \sqrt {1+\sin (e+f x)}} \\ & = -\frac {(2 \cos (e+f x)) \text {Subst}\left (\int \frac {\left (1-x^2\right )^n}{\left (2-x^2\right )^2} \, dx,x,\sqrt {1-\sin (e+f x)}\right )}{f \sqrt {1-\sin (e+f x)} \sqrt {1+\sin (e+f x)}} \\ & = -\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)}{2 f \sqrt {1+\sin (e+f x)}} \\ \end{align*}

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(263\) vs. \(2(60)=120\).

Time = 5.33 (sec) , antiderivative size = 263, normalized size of antiderivative = 4.38 \[ \int \frac {\sin ^n(e+f x)}{(1+\sin (e+f x))^{3/2}} \, dx=\frac {\sec (e+f x) \sin ^n(e+f x) \left (\operatorname {AppellF1}\left (1,\frac {1}{2},-n,2,\frac {1}{2} (1+\sin (e+f x)),1+\sin (e+f x)\right ) \sqrt {2-2 \sin (e+f x)} (-\sin (e+f x))^{-n} (1+\sin (e+f x))^2-\frac {4 (1+\sin (e+f x)) \sqrt {1-\frac {2}{1+\sin (e+f x)}} \left (1-\frac {1}{1+\sin (e+f x)}\right )^{-n} \left (2 (1+2 n) \operatorname {AppellF1}\left (\frac {1}{2}-n,-\frac {1}{2},-n,\frac {3}{2}-n,\frac {2}{1+\sin (e+f x)},\frac {1}{1+\sin (e+f x)}\right )+(-1+2 n) \operatorname {AppellF1}\left (-\frac {1}{2}-n,-\frac {1}{2},-n,\frac {1}{2}-n,\frac {2}{1+\sin (e+f x)},\frac {1}{1+\sin (e+f x)}\right ) (1+\sin (e+f x))\right )}{-1+4 n^2}\right )}{8 f \sqrt {1+\sin (e+f x)}} \]

[In]

Integrate[Sin[e + f*x]^n/(1 + Sin[e + f*x])^(3/2),x]

[Out]

(Sec[e + f*x]*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)*Appel
lF1[-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]])

Maple [F]

\[\int \frac {\sin ^{n}\left (f x +e \right )}{\left (\sin \left (f x +e \right )+1\right )^{\frac {3}{2}}}d x\]

[In]

int(sin(f*x+e)^n/(sin(f*x+e)+1)^(3/2),x)

[Out]

int(sin(f*x+e)^n/(sin(f*x+e)+1)^(3/2),x)

Fricas [F]

\[ \int \frac {\sin ^n(e+f x)}{(1+\sin (e+f x))^{3/2}} \, dx=\int { \frac {\sin \left (f x + e\right )^{n}}{{\left (\sin \left (f x + e\right ) + 1\right )}^{\frac {3}{2}}} \,d x } \]

[In]

integrate(sin(f*x+e)^n/(1+sin(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

integral(-sin(f*x + e)^n*sqrt(sin(f*x + e) + 1)/(cos(f*x + e)^2 - 2*sin(f*x + e) - 2), x)

Sympy [F]

\[ \int \frac {\sin ^n(e+f x)}{(1+\sin (e+f x))^{3/2}} \, dx=\int \frac {\sin ^{n}{\left (e + f x \right )}}{\left (\sin {\left (e + f x \right )} + 1\right )^{\frac {3}{2}}}\, dx \]

[In]

integrate(sin(f*x+e)**n/(1+sin(f*x+e))**(3/2),x)

[Out]

Integral(sin(e + f*x)**n/(sin(e + f*x) + 1)**(3/2), x)

Maxima [F]

\[ \int \frac {\sin ^n(e+f x)}{(1+\sin (e+f x))^{3/2}} \, dx=\int { \frac {\sin \left (f x + e\right )^{n}}{{\left (\sin \left (f x + e\right ) + 1\right )}^{\frac {3}{2}}} \,d x } \]

[In]

integrate(sin(f*x+e)^n/(1+sin(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

integrate(sin(f*x + e)^n/(sin(f*x + e) + 1)^(3/2), x)

Giac [F(-1)]

Timed out. \[ \int \frac {\sin ^n(e+f x)}{(1+\sin (e+f x))^{3/2}} \, dx=\text {Timed out} \]

[In]

integrate(sin(f*x+e)^n/(1+sin(f*x+e))^(3/2),x, algorithm="giac")

[Out]

Timed out

Mupad [F(-1)]

Timed out. \[ \int \frac {\sin ^n(e+f x)}{(1+\sin (e+f x))^{3/2}} \, dx=\int \frac {{\sin \left (e+f\,x\right )}^n}{{\left (\sin \left (e+f\,x\right )+1\right )}^{3/2}} \,d x \]

[In]

int(sin(e + f*x)^n/(sin(e + f*x) + 1)^(3/2),x)

[Out]

int(sin(e + f*x)^n/(sin(e + f*x) + 1)^(3/2), x)