Integrand size = 21, antiderivative size = 116 \[ \int \cos ^2(e+f x) (c+d \sin (e+f x))^n \, dx=-\frac {2 \sqrt {2} \operatorname {AppellF1}\left (\frac {3}{2},-\frac {1}{2},-n,\frac {5}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {d (1-\sin (e+f x))}{c+d}\right ) \cos (e+f x) (1-\sin (e+f x)) (c+d \sin (e+f x))^n \left (\frac {c+d \sin (e+f x)}{c+d}\right )^{-n}}{3 f \sqrt {1+\sin (e+f x)}} \] Output:
-2/3*2^(1/2)*AppellF1(3/2,-n,-1/2,5/2,d*(1-sin(f*x+e))/(c+d),1/2-1/2*sin(f *x+e))*cos(f*x+e)*(1-sin(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)
\[ \int \cos ^2(e+f x) (c+d \sin (e+f x))^n \, dx=\int \cos ^2(e+f x) (c+d \sin (e+f x))^n \, dx \] Input:
Integrate[Cos[e + f*x]^2*(c + d*Sin[e + f*x])^n,x]
Output:
Integrate[Cos[e + f*x]^2*(c + d*Sin[e + f*x])^n, x]
Time = 0.30 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.09, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3042, 3183, 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.
\(\displaystyle \int \cos ^2(e+f x) (c+d \sin (e+f x))^n \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \cos (e+f x)^2 (c+d \sin (e+f x))^ndx\) |
\(\Big \downarrow \) 3183 |
\(\displaystyle \frac {\cos (e+f x) \int (c+d \sin (e+f x))^n \sqrt {-\frac {\sin (e+f x) d}{c-d}-\frac {d}{c-d}} \sqrt {\frac {d}{c+d}-\frac {d \sin (e+f x)}{c+d}}d\sin (e+f x)}{f \sqrt {1-\frac {c+d \sin (e+f x)}{c-d}} \sqrt {1-\frac {c+d \sin (e+f x)}{c+d}}}\) |
\(\Big \downarrow \) 155 |
\(\displaystyle \frac {\cos (e+f x) (c+d \sin (e+f x))^{n+1} \operatorname {AppellF1}\left (n+1,-\frac {1}{2},-\frac {1}{2},n+2,\frac {c+d \sin (e+f x)}{c-d},\frac {c+d \sin (e+f x)}{c+d}\right )}{d f (n+1) \sqrt {1-\frac {c+d \sin (e+f x)}{c-d}} \sqrt {1-\frac {c+d \sin (e+f x)}{c+d}}}\) |
Input:
Int[Cos[e + f*x]^2*(c + d*Sin[e + f*x])^n,x]
Output:
(AppellF1[1 + n, -1/2, -1/2, 2 + n, (c + d*Sin[e + f*x])/(c - d), (c + d*S in[e + f*x])/(c + d)]*Cos[e + f*x]*(c + d*Sin[e + f*x])^(1 + n))/(d*f*(1 + n)*Sqrt[1 - (c + d*Sin[e + f*x])/(c - d)]*Sqrt[1 - (c + d*Sin[e + f*x])/( c + d)])
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[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_), x_Symbol] :> Simp[g*((g*Cos[e + f*x])^(p - 1)/(f*(1 - (a + b*Sin [e + f*x])/(a - b))^((p - 1)/2)*(1 - (a + b*Sin[e + f*x])/(a + b))^((p - 1) /2))) Subst[Int[(-b/(a - b) - b*(x/(a - b)))^((p - 1)/2)*(b/(a + b) - b*( x/(a + b)))^((p - 1)/2)*(a + b*x)^m, x], x, Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, g, m, p}, x] && NeQ[a^2 - b^2, 0] && !IGtQ[m, 0]
\[\int \cos \left (f x +e \right )^{2} \left (c +d \sin \left (f x +e \right )\right )^{n}d x\]
Input:
int(cos(f*x+e)^2*(c+d*sin(f*x+e))^n,x)
Output:
int(cos(f*x+e)^2*(c+d*sin(f*x+e))^n,x)
\[ \int \cos ^2(e+f x) (c+d \sin (e+f x))^n \, dx=\int { {\left (d \sin \left (f x + e\right ) + c\right )}^{n} \cos \left (f x + e\right )^{2} \,d x } \] Input:
integrate(cos(f*x+e)^2*(c+d*sin(f*x+e))^n,x, algorithm="fricas")
Output:
integral((d*sin(f*x + e) + c)^n*cos(f*x + e)^2, x)
Timed out. \[ \int \cos ^2(e+f x) (c+d \sin (e+f x))^n \, dx=\text {Timed out} \] Input:
integrate(cos(f*x+e)**2*(c+d*sin(f*x+e))**n,x)
Output:
Timed out
\[ \int \cos ^2(e+f x) (c+d \sin (e+f x))^n \, dx=\int { {\left (d \sin \left (f x + e\right ) + c\right )}^{n} \cos \left (f x + e\right )^{2} \,d x } \] Input:
integrate(cos(f*x+e)^2*(c+d*sin(f*x+e))^n,x, algorithm="maxima")
Output:
integrate((d*sin(f*x + e) + c)^n*cos(f*x + e)^2, x)
\[ \int \cos ^2(e+f x) (c+d \sin (e+f x))^n \, dx=\int { {\left (d \sin \left (f x + e\right ) + c\right )}^{n} \cos \left (f x + e\right )^{2} \,d x } \] Input:
integrate(cos(f*x+e)^2*(c+d*sin(f*x+e))^n,x, algorithm="giac")
Output:
integrate((d*sin(f*x + e) + c)^n*cos(f*x + e)^2, x)
Timed out. \[ \int \cos ^2(e+f x) (c+d \sin (e+f x))^n \, dx=\int {\cos \left (e+f\,x\right )}^2\,{\left (c+d\,\sin \left (e+f\,x\right )\right )}^n \,d x \] Input:
int(cos(e + f*x)^2*(c + d*sin(e + f*x))^n,x)
Output:
int(cos(e + f*x)^2*(c + d*sin(e + f*x))^n, x)
\[ \int \cos ^2(e+f x) (c+d \sin (e+f x))^n \, dx=\int \left (\sin \left (f x +e \right ) d +c \right )^{n} \cos \left (f x +e \right )^{2}d x \] Input:
int(cos(f*x+e)^2*(c+d*sin(f*x+e))^n,x)
Output:
int((sin(e + f*x)*d + c)**n*cos(e + f*x)**2,x)