Integrand size = 37, antiderivative size = 284 \[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c+d \sin (e+f x))^{3/2} \, dx=\frac {\sqrt {2} (A-B) (c-d) \operatorname {AppellF1}\left (\frac {1}{2}+m,\frac {1}{2},-\frac {3}{2},\frac {3}{2}+m,\frac {1}{2} (1+\sin (e+f x)),-\frac {d (1+\sin (e+f x))}{c-d}\right ) \cos (e+f x) (a+a \sin (e+f x))^m \sqrt {c+d \sin (e+f x)}}{f (1+2 m) \sqrt {1-\sin (e+f x)} \sqrt {\frac {c+d \sin (e+f x)}{c-d}}}+\frac {\sqrt {2} B (c-d) \operatorname {AppellF1}\left (\frac {3}{2}+m,\frac {1}{2},-\frac {3}{2},\frac {5}{2}+m,\frac {1}{2} (1+\sin (e+f x)),-\frac {d (1+\sin (e+f x))}{c-d}\right ) \cos (e+f x) (a+a \sin (e+f x))^{1+m} \sqrt {c+d \sin (e+f x)}}{a f (3+2 m) \sqrt {1-\sin (e+f x)} \sqrt {\frac {c+d \sin (e+f x)}{c-d}}} \]
(A-B)*(c-d)*AppellF1(1/2+m,-3/2,1/2,3/2+m,-d*(1+sin(f*x+e))/(c-d),1/2+1/2* sin(f*x+e))*cos(f*x+e)*(a+a*sin(f*x+e))^m*2^(1/2)*(c+d*sin(f*x+e))^(1/2)/f /(1+2*m)/(1-sin(f*x+e))^(1/2)/((c+d*sin(f*x+e))/(c-d))^(1/2)+B*(c-d)*Appel lF1(3/2+m,-3/2,1/2,5/2+m,-d*(1+sin(f*x+e))/(c-d),1/2+1/2*sin(f*x+e))*cos(f *x+e)*(a+a*sin(f*x+e))^(1+m)*2^(1/2)*(c+d*sin(f*x+e))^(1/2)/a/f/(3+2*m)/(1 -sin(f*x+e))^(1/2)/((c+d*sin(f*x+e))/(c-d))^(1/2)
\[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c+d \sin (e+f x))^{3/2} \, dx=\int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c+d \sin (e+f x))^{3/2} \, dx \]
Time = 0.75 (sec) , antiderivative size = 308, normalized size of antiderivative = 1.08, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.216, Rules used = {3042, 3466, 3042, 3267, 157, 27, 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.
\(\displaystyle \int (a \sin (e+f x)+a)^m (A+B \sin (e+f x)) (c+d \sin (e+f x))^{3/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (a \sin (e+f x)+a)^m (A+B \sin (e+f x)) (c+d \sin (e+f x))^{3/2}dx\) |
\(\Big \downarrow \) 3466 |
\(\displaystyle (A-B) \int (\sin (e+f x) a+a)^m (c+d \sin (e+f x))^{3/2}dx+\frac {B \int (\sin (e+f x) a+a)^{m+1} (c+d \sin (e+f x))^{3/2}dx}{a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle (A-B) \int (\sin (e+f x) a+a)^m (c+d \sin (e+f x))^{3/2}dx+\frac {B \int (\sin (e+f x) a+a)^{m+1} (c+d \sin (e+f x))^{3/2}dx}{a}\) |
\(\Big \downarrow \) 3267 |
\(\displaystyle \frac {a^2 (A-B) \cos (e+f x) \int \frac {(\sin (e+f x) a+a)^{m-\frac {1}{2}} (c+d \sin (e+f x))^{3/2}}{\sqrt {a-a \sin (e+f x)}}d\sin (e+f x)}{f \sqrt {a-a \sin (e+f x)} \sqrt {a \sin (e+f x)+a}}+\frac {a B \cos (e+f x) \int \frac {(\sin (e+f x) a+a)^{m+\frac {1}{2}} (c+d \sin (e+f x))^{3/2}}{\sqrt {a-a \sin (e+f x)}}d\sin (e+f x)}{f \sqrt {a-a \sin (e+f x)} \sqrt {a \sin (e+f x)+a}}\) |
\(\Big \downarrow \) 157 |
\(\displaystyle \frac {a^2 (A-B) \sqrt {1-\sin (e+f x)} \cos (e+f x) \int \frac {\sqrt {2} (\sin (e+f x) a+a)^{m-\frac {1}{2}} (c+d \sin (e+f x))^{3/2}}{\sqrt {1-\sin (e+f x)}}d\sin (e+f x)}{\sqrt {2} f (a-a \sin (e+f x)) \sqrt {a \sin (e+f x)+a}}+\frac {a B \sqrt {1-\sin (e+f x)} \cos (e+f x) \int \frac {\sqrt {2} (\sin (e+f x) a+a)^{m+\frac {1}{2}} (c+d \sin (e+f x))^{3/2}}{\sqrt {1-\sin (e+f x)}}d\sin (e+f x)}{\sqrt {2} f (a-a \sin (e+f x)) \sqrt {a \sin (e+f x)+a}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {a^2 (A-B) \sqrt {1-\sin (e+f x)} \cos (e+f x) \int \frac {(\sin (e+f x) a+a)^{m-\frac {1}{2}} (c+d \sin (e+f x))^{3/2}}{\sqrt {1-\sin (e+f x)}}d\sin (e+f x)}{f (a-a \sin (e+f x)) \sqrt {a \sin (e+f x)+a}}+\frac {a B \sqrt {1-\sin (e+f x)} \cos (e+f x) \int \frac {(\sin (e+f x) a+a)^{m+\frac {1}{2}} (c+d \sin (e+f x))^{3/2}}{\sqrt {1-\sin (e+f x)}}d\sin (e+f x)}{f (a-a \sin (e+f x)) \sqrt {a \sin (e+f x)+a}}\) |
\(\Big \downarrow \) 156 |
\(\displaystyle \frac {a^2 (A-B) (c-d) \sqrt {1-\sin (e+f x)} \cos (e+f x) \sqrt {c+d \sin (e+f x)} \int \frac {(\sin (e+f x) a+a)^{m-\frac {1}{2}} \left (\frac {c}{c-d}+\frac {d \sin (e+f x)}{c-d}\right )^{3/2}}{\sqrt {1-\sin (e+f x)}}d\sin (e+f x)}{f (a-a \sin (e+f x)) \sqrt {a \sin (e+f x)+a} \sqrt {\frac {c+d \sin (e+f x)}{c-d}}}+\frac {a B (c-d) \sqrt {1-\sin (e+f x)} \cos (e+f x) \sqrt {c+d \sin (e+f x)} \int \frac {(\sin (e+f x) a+a)^{m+\frac {1}{2}} \left (\frac {c}{c-d}+\frac {d \sin (e+f x)}{c-d}\right )^{3/2}}{\sqrt {1-\sin (e+f x)}}d\sin (e+f x)}{f (a-a \sin (e+f x)) \sqrt {a \sin (e+f x)+a} \sqrt {\frac {c+d \sin (e+f x)}{c-d}}}\) |
\(\Big \downarrow \) 155 |
\(\displaystyle \frac {\sqrt {2} a (A-B) (c-d) \sqrt {1-\sin (e+f x)} \cos (e+f x) (a \sin (e+f x)+a)^m \sqrt {c+d \sin (e+f x)} \operatorname {AppellF1}\left (m+\frac {1}{2},\frac {1}{2},-\frac {3}{2},m+\frac {3}{2},\frac {1}{2} (\sin (e+f x)+1),-\frac {d (\sin (e+f x)+1)}{c-d}\right )}{f (2 m+1) (a-a \sin (e+f x)) \sqrt {\frac {c+d \sin (e+f x)}{c-d}}}+\frac {\sqrt {2} B (c-d) \sqrt {1-\sin (e+f x)} \cos (e+f x) (a \sin (e+f x)+a)^{m+1} \sqrt {c+d \sin (e+f x)} \operatorname {AppellF1}\left (m+\frac {3}{2},\frac {1}{2},-\frac {3}{2},m+\frac {5}{2},\frac {1}{2} (\sin (e+f x)+1),-\frac {d (\sin (e+f x)+1)}{c-d}\right )}{f (2 m+3) (a-a \sin (e+f x)) \sqrt {\frac {c+d \sin (e+f x)}{c-d}}}\) |
(Sqrt[2]*a*(A - B)*(c - d)*AppellF1[1/2 + m, 1/2, -3/2, 3/2 + m, (1 + Sin[ e + f*x])/2, -((d*(1 + Sin[e + f*x]))/(c - d))]*Cos[e + f*x]*Sqrt[1 - Sin[ e + f*x]]*(a + a*Sin[e + f*x])^m*Sqrt[c + d*Sin[e + f*x]])/(f*(1 + 2*m)*(a - a*Sin[e + f*x])*Sqrt[(c + d*Sin[e + f*x])/(c - d)]) + (Sqrt[2]*B*(c - d )*AppellF1[3/2 + m, 1/2, -3/2, 5/2 + m, (1 + Sin[e + f*x])/2, -((d*(1 + Si n[e + f*x]))/(c - d))]*Cos[e + f*x]*Sqrt[1 - Sin[e + f*x]]*(a + a*Sin[e + f*x])^(1 + m)*Sqrt[c + d*Sin[e + f*x]])/(f*(3 + 2*m)*(a - a*Sin[e + f*x])* Sqrt[(c + d*Sin[e + f*x])/(c - d)])
3.4.42.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) ^(p_), x_] :> Simp[(c + d*x)^FracPart[n]/(Simplify[b/(b*c - a*d)]^IntPart[n ]*(b*((c + d*x)/(b*c - a*d)))^FracPart[n]) Int[(a + b*x)^m*Simp[b*(c/(b*c - a*d)) + b*d*(x/(b*c - a*d)), x]^n*(e + 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] && !SimplerQ[c + d*x, a + b*x] && !Si mplerQ[e + f*x, a + b*x]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + ( f_.)*(x_)])^(n_.), x_Symbol] :> Simp[a^2*(Cos[e + f*x]/(f*Sqrt[a + b*Sin[e + f*x]]*Sqrt[a - b*Sin[e + f*x]])) Subst[Int[(a + b*x)^(m - 1/2)*((c + d* x)^n/Sqrt[a - b*x]), x], x, Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m , 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_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si mp[(A*b - a*B)/b Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^n, x], x ] + Simp[B/b Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ [a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && NeQ[A*b + a*B, 0]
\[\int \left (a +a \sin \left (f x +e \right )\right )^{m} \left (A +B \sin \left (f x +e \right )\right ) \left (c +d \sin \left (f x +e \right )\right )^{\frac {3}{2}}d x\]
\[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c+d \sin (e+f x))^{3/2} \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \,d x } \]
integral(-(B*d*cos(f*x + e)^2 - A*c - B*d - (B*c + A*d)*sin(f*x + e))*sqrt (d*sin(f*x + e) + c)*(a*sin(f*x + e) + a)^m, x)
Timed out. \[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c+d \sin (e+f x))^{3/2} \, dx=\text {Timed out} \]
\[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c+d \sin (e+f x))^{3/2} \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \,d x } \]
\[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c+d \sin (e+f x))^{3/2} \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \,d x } \]
Timed out. \[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c+d \sin (e+f x))^{3/2} \, dx=\int \left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,{\left (c+d\,\sin \left (e+f\,x\right )\right )}^{3/2} \,d x \]