Integrand size = 33, antiderivative size = 138 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^n \, dx=-\frac {2^{\frac {3}{2}+m} \operatorname {AppellF1}\left (\frac {3}{2},-\frac {1}{2}-m,-n,\frac {5}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {d (1-\sin (e+f x))}{c+d}\right ) \cos (e+f x) (1-\sin (e+f x)) (1+\sin (e+f x))^{-\frac {1}{2}-m} (a+a \sin (e+f x))^m (c+d \sin (e+f x))^n \left (\frac {c+d \sin (e+f x)}{c+d}\right )^{-n}}{3 f} \] Output:
-1/3*2^(3/2+m)*AppellF1(3/2,-n,-1/2-m,5/2,d*(1-sin(f*x+e))/(c+d),1/2-1/2*s in(f*x+e))*cos(f*x+e)*(1-sin(f*x+e))*(1+sin(f*x+e))^(-1/2-m)*(a+a*sin(f*x+ e))^m*(c+d*sin(f*x+e))^n/f/(((c+d*sin(f*x+e))/(c+d))^n)
\[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^n \, dx=\int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^n \, dx \] Input:
Integrate[Cos[e + f*x]^2*(a + a*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^n,x]
Output:
Integrate[Cos[e + f*x]^2*(a + a*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^n, x]
Time = 0.44 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.98, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {3042, 3397, 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.
\(\displaystyle \int \cos ^2(e+f x) (a \sin (e+f x)+a)^m (c+d \sin (e+f x))^n \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \cos (e+f x)^2 (a \sin (e+f x)+a)^m (c+d \sin (e+f x))^ndx\) |
\(\Big \downarrow \) 3397 |
\(\displaystyle \frac {\cos (e+f x) \int \sqrt {a-a \sin (e+f x)} (\sin (e+f x) a+a)^{m+\frac {1}{2}} (c+d \sin (e+f x))^nd\sin (e+f x)}{f \sqrt {a-a \sin (e+f x)} \sqrt {a \sin (e+f x)+a}}\) |
\(\Big \downarrow \) 157 |
\(\displaystyle \frac {\sqrt {2} \cos (e+f x) \int \frac {\sqrt {1-\sin (e+f x)} (\sin (e+f x) a+a)^{m+\frac {1}{2}} (c+d \sin (e+f x))^n}{\sqrt {2}}d\sin (e+f x)}{f \sqrt {1-\sin (e+f x)} \sqrt {a \sin (e+f x)+a}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\cos (e+f x) \int \sqrt {1-\sin (e+f x)} (\sin (e+f x) a+a)^{m+\frac {1}{2}} (c+d \sin (e+f x))^nd\sin (e+f x)}{f \sqrt {1-\sin (e+f x)} \sqrt {a \sin (e+f x)+a}}\) |
\(\Big \downarrow \) 156 |
\(\displaystyle \frac {\cos (e+f x) (c+d \sin (e+f x))^n \left (\frac {c+d \sin (e+f x)}{c-d}\right )^{-n} \int \sqrt {1-\sin (e+f x)} (\sin (e+f x) a+a)^{m+\frac {1}{2}} \left (\frac {c}{c-d}+\frac {d \sin (e+f x)}{c-d}\right )^nd\sin (e+f x)}{f \sqrt {1-\sin (e+f x)} \sqrt {a \sin (e+f x)+a}}\) |
\(\Big \downarrow \) 155 |
\(\displaystyle \frac {2 \sqrt {2} \cos (e+f x) (a \sin (e+f x)+a)^{m+1} (c+d \sin (e+f x))^n \left (\frac {c+d \sin (e+f x)}{c-d}\right )^{-n} \operatorname {AppellF1}\left (m+\frac {3}{2},-\frac {1}{2},-n,m+\frac {5}{2},\frac {1}{2} (\sin (e+f x)+1),-\frac {d (\sin (e+f x)+1)}{c-d}\right )}{a f (2 m+3) \sqrt {1-\sin (e+f x)}}\) |
Input:
Int[Cos[e + f*x]^2*(a + a*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^n,x]
Output:
(2*Sqrt[2]*AppellF1[3/2 + m, -1/2, -n, 5/2 + m, (1 + Sin[e + f*x])/2, -((d *(1 + Sin[e + f*x]))/(c - d))]*Cos[e + f*x]*(a + a*Sin[e + f*x])^(1 + m)*( c + d*Sin[e + f*x])^n)/(a*f*(3 + 2*m)*Sqrt[1 - Sin[e + f*x]]*((c + d*Sin[e + f*x])/(c - d))^n)
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[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_ )*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[e + f* x]/(a^(p - 2)*f*Sqrt[a + b*Sin[e + f*x]]*Sqrt[a - b*Sin[e + f*x]]) Subst[ Int[(a + b*x)^(m + p/2 - 1/2)*(a - b*x)^(p/2 - 1/2)*(c + d*x)^n, x], x, Sin [e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] & & IntegerQ[p/2] && !IntegerQ[m]
\[\int \cos \left (f x +e \right )^{2} \left (a +a \sin \left (f x +e \right )\right )^{m} \left (c +d \sin \left (f x +e \right )\right )^{n}d x\]
Input:
int(cos(f*x+e)^2*(a+a*sin(f*x+e))^m*(c+d*sin(f*x+e))^n,x)
Output:
int(cos(f*x+e)^2*(a+a*sin(f*x+e))^m*(c+d*sin(f*x+e))^n,x)
\[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^n \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (d \sin \left (f x + e\right ) + c\right )}^{n} \cos \left (f x + e\right )^{2} \,d x } \] Input:
integrate(cos(f*x+e)^2*(a+a*sin(f*x+e))^m*(c+d*sin(f*x+e))^n,x, algorithm= "fricas")
Output:
integral((a*sin(f*x + e) + a)^m*(d*sin(f*x + e) + c)^n*cos(f*x + e)^2, x)
Timed out. \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^n \, dx=\text {Timed out} \] Input:
integrate(cos(f*x+e)**2*(a+a*sin(f*x+e))**m*(c+d*sin(f*x+e))**n,x)
Output:
Timed out
\[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^n \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (d \sin \left (f x + e\right ) + c\right )}^{n} \cos \left (f x + e\right )^{2} \,d x } \] Input:
integrate(cos(f*x+e)^2*(a+a*sin(f*x+e))^m*(c+d*sin(f*x+e))^n,x, algorithm= "maxima")
Output:
integrate((a*sin(f*x + e) + a)^m*(d*sin(f*x + e) + c)^n*cos(f*x + e)^2, x)
\[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^n \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (d \sin \left (f x + e\right ) + c\right )}^{n} \cos \left (f x + e\right )^{2} \,d x } \] Input:
integrate(cos(f*x+e)^2*(a+a*sin(f*x+e))^m*(c+d*sin(f*x+e))^n,x, algorithm= "giac")
Output:
integrate((a*sin(f*x + e) + a)^m*(d*sin(f*x + e) + c)^n*cos(f*x + e)^2, x)
Timed out. \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^n \, dx=\int {\cos \left (e+f\,x\right )}^2\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,{\left (c+d\,\sin \left (e+f\,x\right )\right )}^n \,d x \] Input:
int(cos(e + f*x)^2*(a + a*sin(e + f*x))^m*(c + d*sin(e + f*x))^n,x)
Output:
int(cos(e + f*x)^2*(a + a*sin(e + f*x))^m*(c + d*sin(e + f*x))^n, x)
\[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^n \, dx=\int \cos \left (f x +e \right )^{2} \left (a +a \sin \left (f x +e \right )\right )^{m} \left (\sin \left (f x +e \right ) d +c \right )^{n}d x \] Input:
int(cos(f*x+e)^2*(a+a*sin(f*x+e))^m*(c+d*sin(f*x+e))^n,x)
Output:
int(cos(f*x+e)^2*(a+a*sin(f*x+e))^m*(c+d*sin(f*x+e))^n,x)