Integrand size = 35, antiderivative size = 215 \[ \int (a+b \sin (e+f x))^m \left (A+(A+C) \sin (e+f x)+C \sin ^2(e+f x)\right ) \, dx=-\frac {4 \sqrt {2} C \operatorname {AppellF1}\left (\frac {1}{2},-\frac {3}{2},-m,\frac {3}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {b (1-\sin (e+f x))}{a+b}\right ) \cos (e+f x) (a+b \sin (e+f x))^m \left (\frac {a+b \sin (e+f x)}{a+b}\right )^{-m}}{f \sqrt {1+\sin (e+f x)}}-\frac {2 \sqrt {2} (A-C) \operatorname {AppellF1}\left (\frac {1}{2},-\frac {1}{2},-m,\frac {3}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {b (1-\sin (e+f x))}{a+b}\right ) \cos (e+f x) (a+b \sin (e+f x))^m \left (\frac {a+b \sin (e+f x)}{a+b}\right )^{-m}}{f \sqrt {1+\sin (e+f x)}} \]
-4*C*AppellF1(1/2,-m,-3/2,3/2,b*(1-sin(f*x+e))/(a+b),1/2-1/2*sin(f*x+e))*c os(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)-2*(A-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)/f/(((a+b*sin(f*x+e) )/(a+b))^m)/(1+sin(f*x+e))^(1/2)
\[ \int (a+b \sin (e+f x))^m \left (A+(A+C) \sin (e+f x)+C \sin ^2(e+f x)\right ) \, dx=\int (a+b \sin (e+f x))^m \left (A+(A+C) \sin (e+f x)+C \sin ^2(e+f x)\right ) \, dx \]
Time = 0.55 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.257, Rules used = {3042, 3496, 3042, 3234, 156, 155, 3263, 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.
| \(\displaystyle \int \left ((A+C) \sin (e+f x)+A+C \sin ^2(e+f x)\right ) (a+b \sin (e+f x))^m \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \left ((A+C) \sin (e+f x)+A+C \sin (e+f x)^2\right ) (a+b \sin (e+f x))^mdx\) |
| \(\Big \downarrow \) 3496 |
| \(\displaystyle (A-C) \int (\sin (e+f x)+1) (a+b \sin (e+f x))^mdx+C \int (\sin (e+f x)+1)^2 (a+b \sin (e+f x))^mdx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle (A-C) \int (\sin (e+f x)+1) (a+b \sin (e+f x))^mdx+C \int (\sin (e+f x)+1)^2 (a+b \sin (e+f x))^mdx\) |
| \(\Big \downarrow \) 3234 |
| \(\displaystyle \frac {(A-C) \cos (e+f x) \int \frac {\sqrt {\sin (e+f x)+1} (a+b \sin (e+f x))^m}{\sqrt {1-\sin (e+f x)}}d\sin (e+f x)}{f \sqrt {1-\sin (e+f x)} \sqrt {\sin (e+f x)+1}}+C \int (\sin (e+f x)+1)^2 (a+b \sin (e+f x))^mdx\) |
| \(\Big \downarrow \) 156 |
| \(\displaystyle \frac {(A-C) \cos (e+f x) (a+b \sin (e+f x))^m \left (\frac {a+b \sin (e+f x)}{a+b}\right )^{-m} \int \frac {\sqrt {\sin (e+f x)+1} \left (\frac {a}{a+b}+\frac {b \sin (e+f x)}{a+b}\right )^m}{\sqrt {1-\sin (e+f x)}}d\sin (e+f x)}{f \sqrt {1-\sin (e+f x)} \sqrt {\sin (e+f x)+1}}+C \int (\sin (e+f x)+1)^2 (a+b \sin (e+f x))^mdx\) |
| \(\Big \downarrow \) 155 |
| \(\displaystyle C \int (\sin (e+f x)+1)^2 (a+b \sin (e+f x))^mdx-\frac {2 \sqrt {2} (A-C) \cos (e+f x) (a+b \sin (e+f x))^m \left (\frac {a+b \sin (e+f x)}{a+b}\right )^{-m} \operatorname {AppellF1}\left (\frac {1}{2},-\frac {1}{2},-m,\frac {3}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {b (1-\sin (e+f x))}{a+b}\right )}{f \sqrt {\sin (e+f x)+1}}\) |
| \(\Big \downarrow \) 3263 |
| \(\displaystyle \frac {C \cos (e+f x) \int \frac {(\sin (e+f x)+1)^{3/2} (a+b \sin (e+f x))^m}{\sqrt {1-\sin (e+f x)}}d\sin (e+f x)}{f \sqrt {1-\sin (e+f x)} \sqrt {\sin (e+f x)+1}}-\frac {2 \sqrt {2} (A-C) \cos (e+f x) (a+b \sin (e+f x))^m \left (\frac {a+b \sin (e+f x)}{a+b}\right )^{-m} \operatorname {AppellF1}\left (\frac {1}{2},-\frac {1}{2},-m,\frac {3}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {b (1-\sin (e+f x))}{a+b}\right )}{f \sqrt {\sin (e+f x)+1}}\) |
| \(\Big \downarrow \) 156 |
| \(\displaystyle \frac {C \cos (e+f x) (a+b \sin (e+f x))^m \left (\frac {a+b \sin (e+f x)}{a+b}\right )^{-m} \int \frac {(\sin (e+f x)+1)^{3/2} \left (\frac {a}{a+b}+\frac {b \sin (e+f x)}{a+b}\right )^m}{\sqrt {1-\sin (e+f x)}}d\sin (e+f x)}{f \sqrt {1-\sin (e+f x)} \sqrt {\sin (e+f x)+1}}-\frac {2 \sqrt {2} (A-C) \cos (e+f x) (a+b \sin (e+f x))^m \left (\frac {a+b \sin (e+f x)}{a+b}\right )^{-m} \operatorname {AppellF1}\left (\frac {1}{2},-\frac {1}{2},-m,\frac {3}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {b (1-\sin (e+f x))}{a+b}\right )}{f \sqrt {\sin (e+f x)+1}}\) |
| \(\Big \downarrow \) 155 |
| \(\displaystyle -\frac {2 \sqrt {2} (A-C) \cos (e+f x) (a+b \sin (e+f x))^m \left (\frac {a+b \sin (e+f x)}{a+b}\right )^{-m} \operatorname {AppellF1}\left (\frac {1}{2},-\frac {1}{2},-m,\frac {3}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {b (1-\sin (e+f x))}{a+b}\right )}{f \sqrt {\sin (e+f x)+1}}-\frac {4 \sqrt {2} C \cos (e+f x) (a+b \sin (e+f x))^m \left (\frac {a+b \sin (e+f x)}{a+b}\right )^{-m} \operatorname {AppellF1}\left (\frac {1}{2},-\frac {3}{2},-m,\frac {3}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {b (1-\sin (e+f x))}{a+b}\right )}{f \sqrt {\sin (e+f x)+1}}\) |
(-4*Sqrt[2]*C*AppellF1[1/2, -3/2, -m, 3/2, (1 - Sin[e + f*x])/2, (b*(1 - S in[e + f*x]))/(a + b)]*Cos[e + f*x]*(a + b*Sin[e + f*x])^m)/(f*Sqrt[1 + Si n[e + f*x]]*((a + b*Sin[e + f*x])/(a + b))^m) - (2*Sqrt[2]*(A - C)*AppellF 1[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*S in[e + f*x])/(a + b))^m)
3.1.18.3.1 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[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + ( f_.)*(x_)]), x_Symbol] :> Simp[c*(Cos[e + f*x]/(f*Sqrt[1 + Sin[e + f*x]]*Sq rt[1 - Sin[e + f*x]])) Subst[Int[(a + b*x)^m*(Sqrt[1 + (d/c)*x]/Sqrt[1 - (d/c)*x]), x], x, Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && N eQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && !IntegerQ[2*m] && EqQ[c^2 - d^2, 0]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[a^m*(Cos[e + f*x]/(f*Sqrt[1 + Sin[e + f*x]]*Sqrt[1 - Sin[e + f*x]])) Subst[Int[(1 + (b/a)*x)^(m - 1/2)*((c + d *x)^n/Sqrt[1 - (b/a)*x]), x], x, Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] & & IntegerQ[m]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(A - C) Int[(a + b*Sin[e + f*x])^m*(1 + Sin[e + f*x]), x], x] + Simp[C Int[(a + b*Sin[e + f*x])^m*(1 + Sin[e + f*x])^2, x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && EqQ[A - B + C, 0] && !IntegerQ[2*m]
\[\int \left (a +b \sin \left (f x +e \right )\right )^{m} \left (A +\left (A +C \right ) \sin \left (f x +e \right )+C \left (\sin ^{2}\left (f x +e \right )\right )\right )d x\]
\[ \int (a+b \sin (e+f x))^m \left (A+(A+C) \sin (e+f x)+C \sin ^2(e+f x)\right ) \, dx=\int { {\left (C \sin \left (f x + e\right )^{2} + {\left (A + C\right )} \sin \left (f x + e\right ) + A\right )} {\left (b \sin \left (f x + e\right ) + a\right )}^{m} \,d x } \]
Timed out. \[ \int (a+b \sin (e+f x))^m \left (A+(A+C) \sin (e+f x)+C \sin ^2(e+f x)\right ) \, dx=\text {Timed out} \]
\[ \int (a+b \sin (e+f x))^m \left (A+(A+C) \sin (e+f x)+C \sin ^2(e+f x)\right ) \, dx=\int { {\left (C \sin \left (f x + e\right )^{2} + {\left (A + C\right )} \sin \left (f x + e\right ) + A\right )} {\left (b \sin \left (f x + e\right ) + a\right )}^{m} \,d x } \]
\[ \int (a+b \sin (e+f x))^m \left (A+(A+C) \sin (e+f x)+C \sin ^2(e+f x)\right ) \, dx=\int { {\left (C \sin \left (f x + e\right )^{2} + {\left (A + C\right )} \sin \left (f x + e\right ) + A\right )} {\left (b \sin \left (f x + e\right ) + a\right )}^{m} \,d x } \]
Timed out. \[ \int (a+b \sin (e+f x))^m \left (A+(A+C) \sin (e+f x)+C \sin ^2(e+f x)\right ) \, dx=\int {\left (a+b\,\sin \left (e+f\,x\right )\right )}^m\,\left (C\,{\sin \left (e+f\,x\right )}^2+\left (A+C\right )\,\sin \left (e+f\,x\right )+A\right ) \,d x \]