Integrand size = 25, antiderivative size = 286 \[ \int (a+b \sin (e+f x))^m \left (A+C \sin ^2(e+f x)\right ) \, dx=-\frac {C \cos (e+f x) (a+b \sin (e+f x))^{1+m}}{b f (2+m)}+\frac {\sqrt {2} a (a+b) C \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-1-m,\frac {3}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {b (1-\sin (e+f x))}{a+b}\right ) \cos (e+f x) (a+b \sin (e+f x))^m \left (\frac {a+b \sin (e+f x)}{a+b}\right )^{-m}}{b^2 f (2+m) \sqrt {1+\sin (e+f x)}}-\frac {\sqrt {2} \left (a^2 C+b^2 (C (1+m)+A (2+m))\right ) \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-m,\frac {3}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {b (1-\sin (e+f x))}{a+b}\right ) \cos (e+f x) (a+b \sin (e+f x))^m \left (\frac {a+b \sin (e+f x)}{a+b}\right )^{-m}}{b^2 f (2+m) \sqrt {1+\sin (e+f x)}} \]
-C*cos(f*x+e)*(a+b*sin(f*x+e))^(1+m)/b/f/(2+m)+a*(a+b)*C*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^2/f/(2+m)/(((a+b*sin(f*x+e))/(a+b))^m)/(1+sin(f*x+e))^(1 /2)-(a^2*C+b^2*(C*(1+m)+A*(2+m)))*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^2/f/(2+ m)/(((a+b*sin(f*x+e))/(a+b))^m)/(1+sin(f*x+e))^(1/2)
\[ \int (a+b \sin (e+f x))^m \left (A+C \sin ^2(e+f x)\right ) \, dx=\int (a+b \sin (e+f x))^m \left (A+C \sin ^2(e+f x)\right ) \, dx \]
Time = 0.63 (sec) , antiderivative size = 286, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 3503, 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.
\(\displaystyle \int \left (A+C \sin ^2(e+f x)\right ) (a+b \sin (e+f x))^m \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \left (A+C \sin (e+f x)^2\right ) (a+b \sin (e+f x))^mdx\) |
\(\Big \downarrow \) 3503 |
\(\displaystyle \frac {\int (a+b \sin (e+f x))^m (b (C (m+1)+A (m+2))-a C \sin (e+f x))dx}{b (m+2)}-\frac {C \cos (e+f x) (a+b \sin (e+f x))^{m+1}}{b f (m+2)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int (a+b \sin (e+f x))^m (b (C (m+1)+A (m+2))-a C \sin (e+f x))dx}{b (m+2)}-\frac {C \cos (e+f x) (a+b \sin (e+f x))^{m+1}}{b f (m+2)}\) |
\(\Big \downarrow \) 3235 |
\(\displaystyle \frac {\frac {\left (a^2 C+b^2 (A (m+2)+C (m+1))\right ) \int (a+b \sin (e+f x))^mdx}{b}-\frac {a C \int (a+b \sin (e+f x))^{m+1}dx}{b}}{b (m+2)}-\frac {C \cos (e+f x) (a+b \sin (e+f x))^{m+1}}{b f (m+2)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\left (a^2 C+b^2 (A (m+2)+C (m+1))\right ) \int (a+b \sin (e+f x))^mdx}{b}-\frac {a C \int (a+b \sin (e+f x))^{m+1}dx}{b}}{b (m+2)}-\frac {C \cos (e+f x) (a+b \sin (e+f x))^{m+1}}{b f (m+2)}\) |
\(\Big \downarrow \) 3144 |
\(\displaystyle \frac {\frac {\cos (e+f x) \left (a^2 C+b^2 (A (m+2)+C (m+1))\right ) \int \frac {(a+b \sin (e+f x))^m}{\sqrt {1-\sin (e+f x)} \sqrt {\sin (e+f x)+1}}d\sin (e+f x)}{b f \sqrt {1-\sin (e+f x)} \sqrt {\sin (e+f x)+1}}-\frac {a C \cos (e+f x) \int \frac {(a+b \sin (e+f x))^{m+1}}{\sqrt {1-\sin (e+f x)} \sqrt {\sin (e+f x)+1}}d\sin (e+f x)}{b f \sqrt {1-\sin (e+f x)} \sqrt {\sin (e+f x)+1}}}{b (m+2)}-\frac {C \cos (e+f x) (a+b \sin (e+f x))^{m+1}}{b f (m+2)}\) |
\(\Big \downarrow \) 156 |
\(\displaystyle \frac {\frac {\cos (e+f x) \left (a^2 C+b^2 (A (m+2)+C (m+1))\right ) (a+b \sin (e+f x))^m \left (\frac {a+b \sin (e+f x)}{a+b}\right )^{-m} \int \frac {\left (\frac {a}{a+b}+\frac {b \sin (e+f x)}{a+b}\right )^m}{\sqrt {1-\sin (e+f x)} \sqrt {\sin (e+f x)+1}}d\sin (e+f x)}{b f \sqrt {1-\sin (e+f x)} \sqrt {\sin (e+f x)+1}}-\frac {a C (a+b) \cos (e+f x) (a+b \sin (e+f x))^m \left (\frac {a+b \sin (e+f x)}{a+b}\right )^{-m} \int \frac {\left (\frac {a}{a+b}+\frac {b \sin (e+f x)}{a+b}\right )^{m+1}}{\sqrt {1-\sin (e+f x)} \sqrt {\sin (e+f x)+1}}d\sin (e+f x)}{b f \sqrt {1-\sin (e+f x)} \sqrt {\sin (e+f x)+1}}}{b (m+2)}-\frac {C \cos (e+f x) (a+b \sin (e+f x))^{m+1}}{b f (m+2)}\) |
\(\Big \downarrow \) 155 |
\(\displaystyle \frac {\frac {\sqrt {2} a C (a+b) \cos (e+f x) (a+b \sin (e+f x))^m \left (\frac {a+b \sin (e+f x)}{a+b}\right )^{-m} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-m-1,\frac {3}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {b (1-\sin (e+f x))}{a+b}\right )}{b f \sqrt {\sin (e+f x)+1}}-\frac {\sqrt {2} \cos (e+f x) \left (a^2 C+b^2 (A (m+2)+C (m+1))\right ) (a+b \sin (e+f x))^m \left (\frac {a+b \sin (e+f x)}{a+b}\right )^{-m} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-m,\frac {3}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {b (1-\sin (e+f x))}{a+b}\right )}{b f \sqrt {\sin (e+f x)+1}}}{b (m+2)}-\frac {C \cos (e+f x) (a+b \sin (e+f x))^{m+1}}{b f (m+2)}\) |
-((C*Cos[e + f*x]*(a + b*Sin[e + f*x])^(1 + m))/(b*f*(2 + m))) + ((Sqrt[2] *a*(a + b)*C*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]*(a^2*C + b^2* (C*(1 + m) + A*(2 + m)))*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))/(b*(2 + m))
3.1.15.3.1 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[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*((a + b*Sin[e + f*x])^ (m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m + 2)) Int[(a + b*Sin[e + f*x])^ m*Simp[A*b*(m + 2) + b*C*(m + 1) - a*C*Sin[e + f*x], x], x], x] /; FreeQ[{a , b, e, f, A, C, m}, x] && !LtQ[m, -1]
\[\int \left (a +b \sin \left (f x +e \right )\right )^{m} \left (A +C \left (\sin ^{2}\left (f x +e \right )\right )\right )d x\]
\[ \int (a+b \sin (e+f x))^m \left (A+C \sin ^2(e+f x)\right ) \, dx=\int { {\left (C \sin \left (f x + e\right )^{2} + A\right )} {\left (b \sin \left (f x + e\right ) + a\right )}^{m} \,d x } \]
Timed out. \[ \int (a+b \sin (e+f x))^m \left (A+C \sin ^2(e+f x)\right ) \, dx=\text {Timed out} \]
\[ \int (a+b \sin (e+f x))^m \left (A+C \sin ^2(e+f x)\right ) \, dx=\int { {\left (C \sin \left (f x + e\right )^{2} + A\right )} {\left (b \sin \left (f x + e\right ) + a\right )}^{m} \,d x } \]
\[ \int (a+b \sin (e+f x))^m \left (A+C \sin ^2(e+f x)\right ) \, dx=\int { {\left (C \sin \left (f x + e\right )^{2} + A\right )} {\left (b \sin \left (f x + e\right ) + a\right )}^{m} \,d x } \]
Timed out. \[ \int (a+b \sin (e+f x))^m \left (A+C \sin ^2(e+f x)\right ) \, dx=\int \left (C\,{\sin \left (e+f\,x\right )}^2+A\right )\,{\left (a+b\,\sin \left (e+f\,x\right )\right )}^m \,d x \]