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

Optimal result

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

-1/2*AppellF1(1/2,-m,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)
 

Mathematica [B] (warning: unable to verify)

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

Time = 6.06 (sec) , antiderivative size = 263, normalized size of antiderivative = 4.38 \[ \int \frac {\sin ^m(e+f x)}{(1+\sin (e+f x))^{3/2}} \, dx=\frac {\sec (e+f x) \sin ^m(e+f x) \left (\operatorname {AppellF1}\left (1,\frac {1}{2},-m,2,\frac {1}{2} (1+\sin (e+f x)),1+\sin (e+f x)\right ) \sqrt {2-2 \sin (e+f x)} (-\sin (e+f x))^{-m} (1+\sin (e+f x))^2-\frac {4 (1+\sin (e+f x)) \sqrt {1-\frac {2}{1+\sin (e+f x)}} \left (1-\frac {1}{1+\sin (e+f x)}\right )^{-m} \left (2 (1+2 m) \operatorname {AppellF1}\left (\frac {1}{2}-m,-\frac {1}{2},-m,\frac {3}{2}-m,\frac {2}{1+\sin (e+f x)},\frac {1}{1+\sin (e+f x)}\right )+(-1+2 m) \operatorname {AppellF1}\left (-\frac {1}{2}-m,-\frac {1}{2},-m,\frac {1}{2}-m,\frac {2}{1+\sin (e+f x)},\frac {1}{1+\sin (e+f x)}\right ) (1+\sin (e+f x))\right )}{-1+4 m^2}\right )}{8 f \sqrt {1+\sin (e+f x)}} \] Input:

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

Output:

(Sec[e + f*x]*Sin[e + f*x]^m*((AppellF1[1, 1/2, -m, 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])^m - (4*(1 + Sin[e + f*x])*Sqrt[1 - 2/(1 + Sin[e + f*x])]*(2*(1 + 
 2*m)*AppellF1[1/2 - m, -1/2, -m, 3/2 - m, 2/(1 + Sin[e + f*x]), (1 + Sin[ 
e + f*x])^(-1)] + (-1 + 2*m)*AppellF1[-1/2 - m, -1/2, -m, 1/2 - m, 2/(1 + 
Sin[e + f*x]), (1 + Sin[e + f*x])^(-1)]*(1 + Sin[e + f*x])))/((-1 + 4*m^2) 
*(1 - (1 + Sin[e + f*x])^(-1))^m)))/(8*f*Sqrt[1 + Sin[e + f*x]])
 

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {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.

\(\displaystyle \int \frac {\sin ^m(e+f x)}{(\sin (e+f x)+1)^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sin (e+f x)^m}{(\sin (e+f x)+1)^{3/2}}dx\)

\(\Big \downarrow \) 3264

\(\displaystyle -\frac {\cos (e+f x) \int \frac {\sin ^m(e+f x)}{\sqrt {1-\sin (e+f x)} (\sin (e+f x)+1)^2}d(1-\sin (e+f x))}{f \sqrt {1-\sin (e+f x)} \sqrt {\sin (e+f x)+1}}\)

\(\Big \downarrow \) 148

\(\displaystyle -\frac {2 \cos (e+f x) \int \frac {\sin ^m(e+f x)}{(\sin (e+f x)+1)^2}d\sqrt {1-\sin (e+f x)}}{f \sqrt {1-\sin (e+f x)} \sqrt {\sin (e+f x)+1}}\)

\(\Big \downarrow \) 333

\(\displaystyle -\frac {\cos (e+f x) \operatorname {AppellF1}\left (\frac {1}{2},-m,2,\frac {3}{2},1-\sin (e+f x),\frac {1}{2} (1-\sin (e+f x))\right )}{2 f \sqrt {\sin (e+f x)+1}}\)

Input:

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

Output:

-1/2*(AppellF1[1/2, -m, 2, 3/2, 1 - Sin[e + f*x], (1 - Sin[e + f*x])/2]*Co 
s[e + f*x])/(f*Sqrt[1 + Sin[e + f*x]])
 

Defintions of rubi rules used

rule 148

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]
 

rule 333

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])
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3264

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]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F(-1)]

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

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

Output:

Timed out
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

int((sin(e + f*x)**m*sqrt(sin(e + f*x) + 1))/(sin(e + f*x)**2 + 2*sin(e + 
f*x) + 1),x)