Integrand size = 23, antiderivative size = 105 \[ \int (a+a \sin (e+f x)) (c+d \sin (e+f x))^n \, dx=-\frac {2 \sqrt {2} a \operatorname {AppellF1}\left (\frac {1}{2},-\frac {1}{2},-n,\frac {3}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {d (1-\sin (e+f x))}{c+d}\right ) \cos (e+f x) (c+d \sin (e+f x))^n \left (\frac {c+d \sin (e+f x)}{c+d}\right )^{-n}}{f \sqrt {1+\sin (e+f x)}} \] Output:
-2*2^(1/2)*a*AppellF1(1/2,-n,-1/2,3/2,d*(1-sin(f*x+e))/(c+d),1/2-1/2*sin(f *x+e))*cos(f*x+e)*(c+d*sin(f*x+e))^n/f/(1+sin(f*x+e))^(1/2)/(((c+d*sin(f*x +e))/(c+d))^n)
\[ \int (a+a \sin (e+f x)) (c+d \sin (e+f x))^n \, dx=\int (a+a \sin (e+f x)) (c+d \sin (e+f x))^n \, dx \] Input:
Integrate[(a + a*Sin[e + f*x])*(c + d*Sin[e + f*x])^n,x]
Output:
Integrate[(a + a*Sin[e + f*x])*(c + d*Sin[e + f*x])^n, x]
Time = 0.26 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3042, 3234, 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.
\(\displaystyle \int (a \sin (e+f x)+a) (c+d \sin (e+f x))^n \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (a \sin (e+f x)+a) (c+d \sin (e+f x))^ndx\) |
\(\Big \downarrow \) 3234 |
\(\displaystyle \frac {a \cos (e+f x) \int \frac {\sqrt {\sin (e+f x)+1} (c+d \sin (e+f x))^n}{\sqrt {1-\sin (e+f x)}}d\sin (e+f x)}{f \sqrt {1-\sin (e+f x)} \sqrt {\sin (e+f x)+1}}\) |
\(\Big \downarrow \) 156 |
\(\displaystyle \frac {a \cos (e+f x) (c+d \sin (e+f x))^n \left (\frac {c+d \sin (e+f x)}{c+d}\right )^{-n} \int \frac {\sqrt {\sin (e+f x)+1} \left (\frac {c}{c+d}+\frac {d \sin (e+f x)}{c+d}\right )^n}{\sqrt {1-\sin (e+f x)}}d\sin (e+f x)}{f \sqrt {1-\sin (e+f x)} \sqrt {\sin (e+f x)+1}}\) |
\(\Big \downarrow \) 155 |
\(\displaystyle -\frac {2 \sqrt {2} a \cos (e+f x) (c+d \sin (e+f x))^n \left (\frac {c+d \sin (e+f x)}{c+d}\right )^{-n} \operatorname {AppellF1}\left (\frac {1}{2},-\frac {1}{2},-n,\frac {3}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {d (1-\sin (e+f x))}{c+d}\right )}{f \sqrt {\sin (e+f x)+1}}\) |
Input:
Int[(a + a*Sin[e + f*x])*(c + d*Sin[e + f*x])^n,x]
Output:
(-2*Sqrt[2]*a*AppellF1[1/2, -1/2, -n, 3/2, (1 - Sin[e + f*x])/2, (d*(1 - S in[e + f*x]))/(c + d)]*Cos[e + f*x]*(c + d*Sin[e + f*x])^n)/(f*Sqrt[1 + Si n[e + f*x]]*((c + d*Sin[e + f*x])/(c + d))^n)
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]
\[\int \left (a +a \sin \left (f x +e \right )\right ) \left (c +d \sin \left (f x +e \right )\right )^{n}d x\]
Input:
int((a+a*sin(f*x+e))*(c+d*sin(f*x+e))^n,x)
Output:
int((a+a*sin(f*x+e))*(c+d*sin(f*x+e))^n,x)
\[ \int (a+a \sin (e+f x)) (c+d \sin (e+f x))^n \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )} {\left (d \sin \left (f x + e\right ) + c\right )}^{n} \,d x } \] Input:
integrate((a+a*sin(f*x+e))*(c+d*sin(f*x+e))^n,x, algorithm="fricas")
Output:
integral((a*sin(f*x + e) + a)*(d*sin(f*x + e) + c)^n, x)
Timed out. \[ \int (a+a \sin (e+f x)) (c+d \sin (e+f x))^n \, dx=\text {Timed out} \] Input:
integrate((a+a*sin(f*x+e))*(c+d*sin(f*x+e))**n,x)
Output:
Timed out
\[ \int (a+a \sin (e+f x)) (c+d \sin (e+f x))^n \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )} {\left (d \sin \left (f x + e\right ) + c\right )}^{n} \,d x } \] Input:
integrate((a+a*sin(f*x+e))*(c+d*sin(f*x+e))^n,x, algorithm="maxima")
Output:
integrate((a*sin(f*x + e) + a)*(d*sin(f*x + e) + c)^n, x)
\[ \int (a+a \sin (e+f x)) (c+d \sin (e+f x))^n \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )} {\left (d \sin \left (f x + e\right ) + c\right )}^{n} \,d x } \] Input:
integrate((a+a*sin(f*x+e))*(c+d*sin(f*x+e))^n,x, algorithm="giac")
Output:
integrate((a*sin(f*x + e) + a)*(d*sin(f*x + e) + c)^n, x)
Timed out. \[ \int (a+a \sin (e+f x)) (c+d \sin (e+f x))^n \, dx=\int \left (a+a\,\sin \left (e+f\,x\right )\right )\,{\left (c+d\,\sin \left (e+f\,x\right )\right )}^n \,d x \] Input:
int((a + a*sin(e + f*x))*(c + d*sin(e + f*x))^n,x)
Output:
int((a + a*sin(e + f*x))*(c + d*sin(e + f*x))^n, x)
\[ \int (a+a \sin (e+f x)) (c+d \sin (e+f x))^n \, dx=a \left (\int \left (\sin \left (f x +e \right ) d +c \right )^{n}d x +\int \left (\sin \left (f x +e \right ) d +c \right )^{n} \sin \left (f x +e \right )d x \right ) \] Input:
int((a+a*sin(f*x+e))*(c+d*sin(f*x+e))^n,x)
Output:
a*(int((sin(e + f*x)*d + c)**n,x) + int((sin(e + f*x)*d + c)**n*sin(e + f* x),x))