Integrand size = 21, antiderivative size = 78 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m \, dx=-\frac {2^{\frac {3}{2}+m} \cos ^3(e+f x) \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-\frac {1}{2}-m,\frac {5}{2},\frac {1}{2} (1-\sin (e+f x))\right ) (1+\sin (e+f x))^{-\frac {3}{2}-m} (a+a \sin (e+f x))^m}{3 f} \] Output:
-1/3*2^(3/2+m)*cos(f*x+e)^3*hypergeom([3/2, -1/2-m],[5/2],1/2-1/2*sin(f*x+ e))*(1+sin(f*x+e))^(-3/2-m)*(a+a*sin(f*x+e))^m/f
Time = 0.09 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.00 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m \, dx=-\frac {2^{\frac {3}{2}+m} \cos ^3(e+f x) \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-\frac {1}{2}-m,\frac {5}{2},\frac {1}{2} (1-\sin (e+f x))\right ) (1+\sin (e+f x))^{-\frac {3}{2}-m} (a (1+\sin (e+f x)))^m}{3 f} \] Input:
Integrate[Cos[e + f*x]^2*(a + a*Sin[e + f*x])^m,x]
Output:
-1/3*(2^(3/2 + m)*Cos[e + f*x]^3*Hypergeometric2F1[3/2, -1/2 - m, 5/2, (1 - Sin[e + f*x])/2]*(1 + Sin[e + f*x])^(-3/2 - m)*(a*(1 + Sin[e + f*x]))^m) /f
Time = 0.27 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.04, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3042, 3168, 80, 79}
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.
\(\displaystyle \int \cos ^2(e+f x) (a \sin (e+f x)+a)^m \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \cos (e+f x)^2 (a \sin (e+f x)+a)^mdx\) |
\(\Big \downarrow \) 3168 |
\(\displaystyle \frac {a^2 \cos ^3(e+f x) \int \sqrt {a-a \sin (e+f x)} (\sin (e+f x) a+a)^{m+\frac {1}{2}}d\sin (e+f x)}{f (a-a \sin (e+f x))^{3/2} (a \sin (e+f x)+a)^{3/2}}\) |
\(\Big \downarrow \) 80 |
\(\displaystyle \frac {a^2 2^{m+\frac {1}{2}} \cos ^3(e+f x) (\sin (e+f x)+1)^{-m-\frac {1}{2}} (a \sin (e+f x)+a)^{m-1} \int \left (\frac {1}{2} \sin (e+f x)+\frac {1}{2}\right )^{m+\frac {1}{2}} \sqrt {a-a \sin (e+f x)}d\sin (e+f x)}{f (a-a \sin (e+f x))^{3/2}}\) |
\(\Big \downarrow \) 79 |
\(\displaystyle -\frac {a 2^{m+\frac {3}{2}} \cos ^3(e+f x) (\sin (e+f x)+1)^{-m-\frac {1}{2}} (a \sin (e+f x)+a)^{m-1} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-m-\frac {1}{2},\frac {5}{2},\frac {1}{2} (1-\sin (e+f x))\right )}{3 f}\) |
Input:
Int[Cos[e + f*x]^2*(a + a*Sin[e + f*x])^m,x]
Output:
-1/3*(2^(3/2 + m)*a*Cos[e + f*x]^3*Hypergeometric2F1[3/2, -1/2 - m, 5/2, ( 1 - Sin[e + f*x])/2]*(1 + Sin[e + f*x])^(-1/2 - m)*(a + a*Sin[e + f*x])^(- 1 + m))/f
Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(( a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c - a*d))^n))*Hypergeometric2F1[-n, m + 1 , m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m, n}, x] && !IntegerQ[m] && !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] || !(RationalQ[n] && GtQ[-d/(b*c - a*d), 0]))
Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(c + d*x)^FracPart[n]/((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, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && !IntegerQ[m] && !Integ erQ[n] && (RationalQ[m] || !SimplerQ[n + 1, m + 1])
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_.), x_Symbol] :> Simp[a^2*((g*Cos[e + f*x])^(p + 1)/(f*g*(a + b*Sin [e + f*x])^((p + 1)/2)*(a - b*Sin[e + f*x])^((p + 1)/2))) Subst[Int[(a + b*x)^(m + (p - 1)/2)*(a - b*x)^((p - 1)/2), x], x, Sin[e + f*x]], x] /; Fre eQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && !IntegerQ[m]
\[\int \cos \left (f x +e \right )^{2} \left (a +a \sin \left (f x +e \right )\right )^{m}d x\]
Input:
int(cos(f*x+e)^2*(a+a*sin(f*x+e))^m,x)
Output:
int(cos(f*x+e)^2*(a+a*sin(f*x+e))^m,x)
\[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \cos \left (f x + e\right )^{2} \,d x } \] Input:
integrate(cos(f*x+e)^2*(a+a*sin(f*x+e))^m,x, algorithm="fricas")
Output:
integral((a*sin(f*x + e) + a)^m*cos(f*x + e)^2, x)
\[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m \, dx=\int \left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{m} \cos ^{2}{\left (e + f x \right )}\, dx \] Input:
integrate(cos(f*x+e)**2*(a+a*sin(f*x+e))**m,x)
Output:
Integral((a*(sin(e + f*x) + 1))**m*cos(e + f*x)**2, x)
\[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \cos \left (f x + e\right )^{2} \,d x } \] Input:
integrate(cos(f*x+e)^2*(a+a*sin(f*x+e))^m,x, algorithm="maxima")
Output:
integrate((a*sin(f*x + e) + a)^m*cos(f*x + e)^2, x)
\[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \cos \left (f x + e\right )^{2} \,d x } \] Input:
integrate(cos(f*x+e)^2*(a+a*sin(f*x+e))^m,x, algorithm="giac")
Output:
integrate((a*sin(f*x + e) + a)^m*cos(f*x + e)^2, x)
Timed out. \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m \, dx=\int {\cos \left (e+f\,x\right )}^2\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m \,d x \] Input:
int(cos(e + f*x)^2*(a + a*sin(e + f*x))^m,x)
Output:
int(cos(e + f*x)^2*(a + a*sin(e + f*x))^m, x)
\[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m \, dx=\int \left (a +a \sin \left (f x +e \right )\right )^{m} \cos \left (f x +e \right )^{2}d x \] Input:
int(cos(f*x+e)^2*(a+a*sin(f*x+e))^m,x)
Output:
int((sin(e + f*x)*a + a)**m*cos(e + f*x)**2,x)