\(\int (a+b \sin (c+d x))^n \, dx\) [225]

Optimal result

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

[Out]

-AppellF1(1/2,-n,1/2,3/2,b*(1-sin(d*x+c))/(a+b),1/2-1/2*sin(d*x+c))*cos(d*x+c)*(a+b*sin(d*x+c))^n*2^(1/2)/d/((
(a+b*sin(d*x+c))/(a+b))^n)/(1+sin(d*x+c))^(1/2)

Rubi [A] (verified)

Time = 0.04 (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} \[ \int (a+b \sin (c+d x))^n \, dx=-\frac {\sqrt {2} \cos (c+d x) (a+b \sin (c+d x))^n \left (\frac {a+b \sin (c+d x)}{a+b}\right )^{-n} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-n,\frac {3}{2},\frac {1}{2} (1-\sin (c+d x)),\frac {b (1-\sin (c+d x))}{a+b}\right )}{d \sqrt {\sin (c+d x)+1}} \]

[In]

Int[(a + b*Sin[c + d*x])^n,x]

[Out]

-((Sqrt[2]*AppellF1[1/2, 1/2, -n, 3/2, (1 - Sin[c + d*x])/2, (b*(1 - Sin[c + d*x]))/(a + b)]*Cos[c + d*x]*(a +
b*Sin[c + d*x])^n)/(d*Sqrt[1 + Sin[c + d*x]]*((a + b*Sin[c + d*x])/(a + b))^n))

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 (c+d x) \text {Subst}\left (\int \frac {(a+b x)^n}{\sqrt {1-x} \sqrt {1+x}} \, dx,x,\sin (c+d x)\right )}{d \sqrt {1-\sin (c+d x)} \sqrt {1+\sin (c+d x)}} \\ & = \frac {\left (\cos (c+d x) (a+b \sin (c+d x))^n \left (-\frac {a+b \sin (c+d x)}{-a-b}\right )^{-n}\right ) \text {Subst}\left (\int \frac {\left (-\frac {a}{-a-b}-\frac {b x}{-a-b}\right )^n}{\sqrt {1-x} \sqrt {1+x}} \, dx,x,\sin (c+d x)\right )}{d \sqrt {1-\sin (c+d x)} \sqrt {1+\sin (c+d x)}} \\ & = -\frac {\sqrt {2} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-n,\frac {3}{2},\frac {1}{2} (1-\sin (c+d x)),\frac {b (1-\sin (c+d x))}{a+b}\right ) \cos (c+d x) (a+b \sin (c+d x))^n \left (\frac {a+b \sin (c+d x)}{a+b}\right )^{-n}}{d \sqrt {1+\sin (c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.15 \[ \int (a+b \sin (c+d x))^n \, dx=\frac {\operatorname {AppellF1}\left (1+n,\frac {1}{2},\frac {1}{2},2+n,\frac {a+b \sin (c+d x)}{a-b},\frac {a+b \sin (c+d x)}{a+b}\right ) \sec (c+d x) \sqrt {-\frac {b (-1+\sin (c+d x))}{a+b}} \sqrt {\frac {b (1+\sin (c+d x))}{-a+b}} (a+b \sin (c+d x))^{1+n}}{b d (1+n)} \]

[In]

Integrate[(a + b*Sin[c + d*x])^n,x]

[Out]

(AppellF1[1 + n, 1/2, 1/2, 2 + n, (a + b*Sin[c + d*x])/(a - b), (a + b*Sin[c + d*x])/(a + b)]*Sec[c + d*x]*Sqr
t[-((b*(-1 + Sin[c + d*x]))/(a + b))]*Sqrt[(b*(1 + Sin[c + d*x]))/(-a + b)]*(a + b*Sin[c + d*x])^(1 + n))/(b*d
*(1 + n))

Maple [F]

\[\int \left (a +b \sin \left (d x +c \right )\right )^{n}d x\]

[In]

int((a+b*sin(d*x+c))^n,x)

[Out]

int((a+b*sin(d*x+c))^n,x)

Fricas [F]

\[ \int (a+b \sin (c+d x))^n \, dx=\int { {\left (b \sin \left (d x + c\right ) + a\right )}^{n} \,d x } \]

[In]

integrate((a+b*sin(d*x+c))^n,x, algorithm="fricas")

[Out]

integral((b*sin(d*x + c) + a)^n, x)

Sympy [F]

\[ \int (a+b \sin (c+d x))^n \, dx=\int \left (a + b \sin {\left (c + d x \right )}\right )^{n}\, dx \]

[In]

integrate((a+b*sin(d*x+c))**n,x)

[Out]

Integral((a + b*sin(c + d*x))**n, x)

Maxima [F]

\[ \int (a+b \sin (c+d x))^n \, dx=\int { {\left (b \sin \left (d x + c\right ) + a\right )}^{n} \,d x } \]

[In]

integrate((a+b*sin(d*x+c))^n,x, algorithm="maxima")

[Out]

integrate((b*sin(d*x + c) + a)^n, x)

Giac [F]

\[ \int (a+b \sin (c+d x))^n \, dx=\int { {\left (b \sin \left (d x + c\right ) + a\right )}^{n} \,d x } \]

[In]

integrate((a+b*sin(d*x+c))^n,x, algorithm="giac")

[Out]

integrate((b*sin(d*x + c) + a)^n, x)

Mupad [F(-1)]

Timed out. \[ \int (a+b \sin (c+d x))^n \, dx=\int {\left (a+b\,\sin \left (c+d\,x\right )\right )}^n \,d x \]

[In]

int((a + b*sin(c + d*x))^n,x)

[Out]

int((a + b*sin(c + d*x))^n, x)