\(\int (3+b \sin (e+f x))^m \, dx\) [804]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [F]
   Fricas [F]
   Sympy [F]
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 12, antiderivative size = 104 \[ \int (3+b \sin (e+f x))^m \, dx=-\frac {\sqrt {2} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-m,\frac {3}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {b (1-\sin (e+f x))}{3+b}\right ) \cos (e+f x) (3+b \sin (e+f x))^m \left (\frac {3+b \sin (e+f x)}{3+b}\right )^{-m}}{f \sqrt {1+\sin (e+f x)}} \]

[Out]

-AppellF1(1/2,-m,1/2,3/2,b*(1-sin(f*x+e))/(a+b),1/2-1/2*sin(f*x+e))*cos(f*x+e)*(a+b*sin(f*x+e))^m*2^(1/2)/f/((
(a+b*sin(f*x+e))/(a+b))^m)/(1+sin(f*x+e))^(1/2)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2744, 144, 143} \[ \int (3+b \sin (e+f x))^m \, dx=-\frac {\sqrt {2} \cos (e+f x) (a+b \sin (e+f x))^m \left (\frac {a+b \sin (e+f x)}{a+b}\right )^{-m} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-m,\frac {3}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {b (1-\sin (e+f x))}{a+b}\right )}{f \sqrt {\sin (e+f x)+1}} \]

[In]

Int[(a + b*Sin[e + f*x])^m,x]

[Out]

-((Sqrt[2]*AppellF1[1/2, 1/2, -m, 3/2, (1 - Sin[e + f*x])/2, (b*(1 - Sin[e + f*x]))/(a + b)]*Cos[e + f*x]*(a +
 b*Sin[e + f*x])^m)/(f*Sqrt[1 + Sin[e + f*x]]*((a + b*Sin[e + f*x])/(a + b))^m))

Rule 143

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((a + b*x)
^(m + 1)/(b*(m + 1)*(b/(b*c - a*d))^n*(b/(b*e - a*f))^p))*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, p}, x] &&  !IntegerQ[m] &&  !Inte
gerQ[n] &&  !IntegerQ[p] && GtQ[b/(b*c - a*d), 0] && GtQ[b/(b*e - a*f), 0] &&  !(GtQ[d/(d*a - c*b), 0] && GtQ[
d/(d*e - c*f), 0] && SimplerQ[c + d*x, a + b*x]) &&  !(GtQ[f/(f*a - e*b), 0] && GtQ[f/(f*c - e*d), 0] && Simpl
erQ[e + f*x, a + b*x])

Rule 144

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Dist[(e + f*x)^
FracPart[p]/((b/(b*e - a*f))^IntPart[p]*(b*((e + f*x)/(b*e - a*f)))^FracPart[p]), Int[(a + b*x)^m*(c + d*x)^n*
(b*(e/(b*e - a*f)) + b*f*(x/(b*e - a*f)))^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] &&  !IntegerQ[m]
&&  !IntegerQ[n] &&  !IntegerQ[p] && GtQ[b/(b*c - a*d), 0] &&  !GtQ[b/(b*e - a*f), 0]

Rule 2744

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[Cos[c + d*x]/(d*Sqrt[1 + Sin[c + d*x]]*Sqrt
[1 - Sin[c + d*x]]), Subst[Int[(a + b*x)^n/(Sqrt[1 + x]*Sqrt[1 - x]), x], x, Sin[c + d*x]], x] /; FreeQ[{a, b,
 c, d, n}, x] && NeQ[a^2 - b^2, 0] &&  !IntegerQ[2*n]

Rubi steps \begin{align*} \text {integral}& = \frac {\cos (e+f x) \text {Subst}\left (\int \frac {(a+b x)^m}{\sqrt {1-x} \sqrt {1+x}} \, dx,x,\sin (e+f x)\right )}{f \sqrt {1-\sin (e+f x)} \sqrt {1+\sin (e+f x)}} \\ & = \frac {\left (\cos (e+f x) (a+b \sin (e+f x))^m \left (-\frac {a+b \sin (e+f x)}{-a-b}\right )^{-m}\right ) \text {Subst}\left (\int \frac {\left (-\frac {a}{-a-b}-\frac {b x}{-a-b}\right )^m}{\sqrt {1-x} \sqrt {1+x}} \, dx,x,\sin (e+f x)\right )}{f \sqrt {1-\sin (e+f x)} \sqrt {1+\sin (e+f x)}} \\ & = -\frac {\sqrt {2} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-m,\frac {3}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {b (1-\sin (e+f x))}{a+b}\right ) \cos (e+f x) (a+b \sin (e+f x))^m \left (\frac {a+b \sin (e+f x)}{a+b}\right )^{-m}}{f \sqrt {1+\sin (e+f x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.12 \[ \int (3+b \sin (e+f x))^m \, dx=\frac {\operatorname {AppellF1}\left (1+m,\frac {1}{2},\frac {1}{2},2+m,-\frac {3+b \sin (e+f x)}{-3+b},\frac {3+b \sin (e+f x)}{3+b}\right ) \sec (e+f x) \sqrt {-\frac {b (-1+\sin (e+f x))}{3+b}} \sqrt {\frac {b (1+\sin (e+f x))}{-3+b}} (3+b \sin (e+f x))^{1+m}}{b f (1+m)} \]

[In]

Integrate[(3 + b*Sin[e + f*x])^m,x]

[Out]

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

Maple [F]

\[\int \left (a +b \sin \left (f x +e \right )\right )^{m}d x\]

[In]

int((a+b*sin(f*x+e))^m,x)

[Out]

int((a+b*sin(f*x+e))^m,x)

Fricas [F]

\[ \int (3+b \sin (e+f x))^m \, dx=\int { {\left (b \sin \left (f x + e\right ) + a\right )}^{m} \,d x } \]

[In]

integrate((a+b*sin(f*x+e))^m,x, algorithm="fricas")

[Out]

integral((b*sin(f*x + e) + a)^m, x)

Sympy [F]

\[ \int (3+b \sin (e+f x))^m \, dx=\int \left (a + b \sin {\left (e + f x \right )}\right )^{m}\, dx \]

[In]

integrate((a+b*sin(f*x+e))**m,x)

[Out]

Integral((a + b*sin(e + f*x))**m, x)

Maxima [F]

\[ \int (3+b \sin (e+f x))^m \, dx=\int { {\left (b \sin \left (f x + e\right ) + a\right )}^{m} \,d x } \]

[In]

integrate((a+b*sin(f*x+e))^m,x, algorithm="maxima")

[Out]

integrate((b*sin(f*x + e) + a)^m, x)

Giac [F]

\[ \int (3+b \sin (e+f x))^m \, dx=\int { {\left (b \sin \left (f x + e\right ) + a\right )}^{m} \,d x } \]

[In]

integrate((a+b*sin(f*x+e))^m,x, algorithm="giac")

[Out]

integrate((b*sin(f*x + e) + a)^m, x)

Mupad [F(-1)]

Timed out. \[ \int (3+b \sin (e+f x))^m \, dx=\int {\left (a+b\,\sin \left (e+f\,x\right )\right )}^m \,d x \]

[In]

int((a + b*sin(e + f*x))^m,x)

[Out]

int((a + b*sin(e + f*x))^m, x)