Integrand size = 23, antiderivative size = 228 \[ \int (3+b \sin (e+f x))^m (c+d \sin (e+f x)) \, dx=-\frac {\sqrt {2} (3+b) d \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-1-m,\frac {3}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {b (1-\sin (e+f x))}{3+b}\right ) \cos (e+f x) (3+b \sin (e+f x))^m \left (\frac {3+b \sin (e+f x)}{3+b}\right )^{-m}}{b f \sqrt {1+\sin (e+f x)}}-\frac {\sqrt {2} (b c-3 d) \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-m,\frac {3}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {b (1-\sin (e+f x))}{3+b}\right ) \cos (e+f x) (3+b \sin (e+f x))^m \left (\frac {3+b \sin (e+f x)}{3+b}\right )^{-m}}{b f \sqrt {1+\sin (e+f x)}} \]
-(a+b)*d*AppellF1(1/2,-1-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)/b/f/(((a+b*sin(f*x+e))/(a+b))^m) /(1+sin(f*x+e))^(1/2)-(-a*d+b*c)*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)/b/f/(((a+b *sin(f*x+e))/(a+b))^m)/(1+sin(f*x+e))^(1/2)
Time = 0.41 (sec) , antiderivative size = 195, normalized size of antiderivative = 0.86 \[ \int (3+b \sin (e+f x))^m (c+d \sin (e+f x)) \, dx=\frac {\sec (e+f x) \sqrt {-\frac {b (-1+\sin (e+f x))}{3+b}} \sqrt {\frac {b (1+\sin (e+f x))}{-3+b}} (3+b \sin (e+f x))^{1+m} \left ((b c-3 d) (2+m) \operatorname {AppellF1}\left (1+m,\frac {1}{2},\frac {1}{2},2+m,-\frac {3+b \sin (e+f x)}{-3+b},\frac {3+b \sin (e+f x)}{3+b}\right )+d (1+m) \operatorname {AppellF1}\left (2+m,\frac {1}{2},\frac {1}{2},3+m,-\frac {3+b \sin (e+f x)}{-3+b},\frac {3+b \sin (e+f x)}{3+b}\right ) (3+b \sin (e+f x))\right )}{b^2 f (1+m) (2+m)} \]
(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*c - 3*d)*(2 + m)*Appel lF1[1 + m, 1/2, 1/2, 2 + m, -((3 + b*Sin[e + f*x])/(-3 + b)), (3 + b*Sin[e + f*x])/(3 + b)] + d*(1 + m)*AppellF1[2 + m, 1/2, 1/2, 3 + m, -((3 + b*Si n[e + f*x])/(-3 + b)), (3 + b*Sin[e + f*x])/(3 + b)]*(3 + b*Sin[e + f*x])) )/(b^2*f*(1 + m)*(2 + m))
Time = 0.41 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3042, 3235, 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.
\(\displaystyle \int (c+d \sin (e+f x)) (a+b \sin (e+f x))^m \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (c+d \sin (e+f x)) (a+b \sin (e+f x))^mdx\) |
\(\Big \downarrow \) 3235 |
\(\displaystyle \frac {(b c-a d) \int (a+b \sin (e+f x))^mdx}{b}+\frac {d \int (a+b \sin (e+f x))^{m+1}dx}{b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {(b c-a d) \int (a+b \sin (e+f x))^mdx}{b}+\frac {d \int (a+b \sin (e+f x))^{m+1}dx}{b}\) |
\(\Big \downarrow \) 3144 |
\(\displaystyle \frac {(b c-a d) \cos (e+f x) \int \frac {(a+b \sin (e+f x))^m}{\sqrt {1-\sin (e+f x)} \sqrt {\sin (e+f x)+1}}d\sin (e+f x)}{b f \sqrt {1-\sin (e+f x)} \sqrt {\sin (e+f x)+1}}+\frac {d \cos (e+f x) \int \frac {(a+b \sin (e+f x))^{m+1}}{\sqrt {1-\sin (e+f x)} \sqrt {\sin (e+f x)+1}}d\sin (e+f x)}{b f \sqrt {1-\sin (e+f x)} \sqrt {\sin (e+f x)+1}}\) |
\(\Big \downarrow \) 156 |
\(\displaystyle \frac {(b c-a d) \cos (e+f x) (a+b \sin (e+f x))^m \left (\frac {a+b \sin (e+f x)}{a+b}\right )^{-m} \int \frac {\left (\frac {a}{a+b}+\frac {b \sin (e+f x)}{a+b}\right )^m}{\sqrt {1-\sin (e+f x)} \sqrt {\sin (e+f x)+1}}d\sin (e+f x)}{b f \sqrt {1-\sin (e+f x)} \sqrt {\sin (e+f x)+1}}+\frac {d (a+b) \cos (e+f x) (a+b \sin (e+f x))^m \left (\frac {a+b \sin (e+f x)}{a+b}\right )^{-m} \int \frac {\left (\frac {a}{a+b}+\frac {b \sin (e+f x)}{a+b}\right )^{m+1}}{\sqrt {1-\sin (e+f x)} \sqrt {\sin (e+f x)+1}}d\sin (e+f x)}{b f \sqrt {1-\sin (e+f x)} \sqrt {\sin (e+f x)+1}}\) |
\(\Big \downarrow \) 155 |
\(\displaystyle -\frac {\sqrt {2} (b c-a d) \cos (e+f x) (a+b \sin (e+f x))^m \left (\frac {a+b \sin (e+f x)}{a+b}\right )^{-m} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-m,\frac {3}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {b (1-\sin (e+f x))}{a+b}\right )}{b f \sqrt {\sin (e+f x)+1}}-\frac {\sqrt {2} d (a+b) \cos (e+f x) (a+b \sin (e+f x))^m \left (\frac {a+b \sin (e+f x)}{a+b}\right )^{-m} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-m-1,\frac {3}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {b (1-\sin (e+f x))}{a+b}\right )}{b f \sqrt {\sin (e+f x)+1}}\) |
-((Sqrt[2]*(a + b)*d*AppellF1[1/2, 1/2, -1 - 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)/(b*f *Sqrt[1 + Sin[e + f*x]]*((a + b*Sin[e + f*x])/(a + b))^m)) - (Sqrt[2]*(b*c - a*d)*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)/(b*f*Sqrt[1 + Sin[e + f*x]]*((a + b*Sin[e + f*x])/(a + b))^m)
3.9.3.3.1 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[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*c - a*d)/b Int[(a + b*Sin[e + f*x])^m, x], x] + Simp[d/b Int[(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]
\[\int \left (a +b \sin \left (f x +e \right )\right )^{m} \left (c +d \sin \left (f x +e \right )\right )d x\]
\[ \int (3+b \sin (e+f x))^m (c+d \sin (e+f x)) \, dx=\int { {\left (d \sin \left (f x + e\right ) + c\right )} {\left (b \sin \left (f x + e\right ) + a\right )}^{m} \,d x } \]
Timed out. \[ \int (3+b \sin (e+f x))^m (c+d \sin (e+f x)) \, dx=\text {Timed out} \]
\[ \int (3+b \sin (e+f x))^m (c+d \sin (e+f x)) \, dx=\int { {\left (d \sin \left (f x + e\right ) + c\right )} {\left (b \sin \left (f x + e\right ) + a\right )}^{m} \,d x } \]
\[ \int (3+b \sin (e+f x))^m (c+d \sin (e+f x)) \, dx=\int { {\left (d \sin \left (f x + e\right ) + c\right )} {\left (b \sin \left (f x + e\right ) + a\right )}^{m} \,d x } \]
Timed out. \[ \int (3+b \sin (e+f x))^m (c+d \sin (e+f x)) \, dx=\int {\left (a+b\,\sin \left (e+f\,x\right )\right )}^m\,\left (c+d\,\sin \left (e+f\,x\right )\right ) \,d x \]