Integrand size = 12, antiderivative size = 104 \[ \int (a+b \sin (c+d x))^n \, dx=-\frac {\sqrt {2} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-n,\frac {3}{2},\frac {1}{2} (1-\sin (c+d x)),\frac {b (1-\sin (c+d x))}{a+b}\right ) \cos (c+d x) (a+b \sin (c+d x))^n \left (\frac {a+b \sin (c+d x)}{a+b}\right )^{-n}}{d \sqrt {1+\sin (c+d x)}} \]
-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)
Time = 0.17 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.15 \[ \int (a+b \sin (c+d x))^n \, dx=\frac {\operatorname {AppellF1}\left (1+n,\frac {1}{2},\frac {1}{2},2+n,\frac {a+b \sin (c+d x)}{a-b},\frac {a+b \sin (c+d x)}{a+b}\right ) \sec (c+d x) \sqrt {-\frac {b (-1+\sin (c+d x))}{a+b}} \sqrt {\frac {b (1+\sin (c+d x))}{-a+b}} (a+b \sin (c+d x))^{1+n}}{b d (1+n)} \]
(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]*Sqrt[-((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))
Time = 0.24 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3144, 156, 155}
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.
\(\displaystyle \int (a+b \sin (c+d x))^n \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (a+b \sin (c+d x))^ndx\) |
\(\Big \downarrow \) 3144 |
\(\displaystyle \frac {\cos (c+d x) \int \frac {(a+b \sin (c+d x))^n}{\sqrt {1-\sin (c+d x)} \sqrt {\sin (c+d x)+1}}d\sin (c+d x)}{d \sqrt {1-\sin (c+d x)} \sqrt {\sin (c+d x)+1}}\) |
\(\Big \downarrow \) 156 |
\(\displaystyle \frac {\cos (c+d x) (a+b \sin (c+d x))^n \left (\frac {a+b \sin (c+d x)}{a+b}\right )^{-n} \int \frac {\left (\frac {a}{a+b}+\frac {b \sin (c+d x)}{a+b}\right )^n}{\sqrt {1-\sin (c+d x)} \sqrt {\sin (c+d x)+1}}d\sin (c+d x)}{d \sqrt {1-\sin (c+d x)} \sqrt {\sin (c+d x)+1}}\) |
\(\Big \downarrow \) 155 |
\(\displaystyle -\frac {\sqrt {2} \cos (c+d x) (a+b \sin (c+d x))^n \left (\frac {a+b \sin (c+d x)}{a+b}\right )^{-n} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-n,\frac {3}{2},\frac {1}{2} (1-\sin (c+d x)),\frac {b (1-\sin (c+d x))}{a+b}\right )}{d \sqrt {\sin (c+d x)+1}}\) |
-((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))
3.3.25.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) ^(p_), x_] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*Simplify[b/(b*c - a*d)]^n* Simplify[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] && !IntegerQ[n] && !IntegerQ[p] && GtQ[Sim plify[b/(b*c - a*d)], 0] && GtQ[Simplify[b/(b*e - a*f)], 0] && !(GtQ[Simpl ify[d/(d*a - c*b)], 0] && GtQ[Simplify[d/(d*e - c*f)], 0] && SimplerQ[c + d *x, a + b*x]) && !(GtQ[Simplify[f/(f*a - e*b)], 0] && GtQ[Simplify[f/(f*c - e*d)], 0] && SimplerQ[e + f*x, a + b*x])
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) ^(p_), x_] :> Simp[(e + f*x)^FracPart[p]/(Simplify[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*Si mp[b*(e/(b*e - a*f)) + b*f*(x/(b*e - a*f)), x]^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && !IntegerQ[m] && !IntegerQ[n] && !IntegerQ[p] & & GtQ[Simplify[b/(b*c - a*d)], 0] && !GtQ[Simplify[b/(b*e - a*f)], 0]
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[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]
\[\int \left (a +b \sin \left (d x +c \right )\right )^{n}d x\]
\[ \int (a+b \sin (c+d x))^n \, dx=\int { {\left (b \sin \left (d x + c\right ) + a\right )}^{n} \,d x } \]
\[ \int (a+b \sin (c+d x))^n \, dx=\int \left (a + b \sin {\left (c + d x \right )}\right )^{n}\, dx \]
\[ \int (a+b \sin (c+d x))^n \, dx=\int { {\left (b \sin \left (d x + c\right ) + a\right )}^{n} \,d x } \]
\[ \int (a+b \sin (c+d x))^n \, dx=\int { {\left (b \sin \left (d x + c\right ) + a\right )}^{n} \,d x } \]
Timed out. \[ \int (a+b \sin (c+d x))^n \, dx=\int {\left (a+b\,\sin \left (c+d\,x\right )\right )}^n \,d x \]