Integrand size = 34, antiderivative size = 111 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n \, dx=\frac {2^{\frac {3}{2}+n} \cos ^3(e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (3+2 m),\frac {1}{2} (-1-2 n),\frac {1}{2} (5+2 m),\frac {1}{2} (1+\sin (e+f x))\right ) (1-\sin (e+f x))^{\frac {1}{2} (-3-2 n)} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n}{f (3+2 m)} \] Output:
2^(3/2+n)*cos(f*x+e)^3*hypergeom([-1/2-n, 3/2+m],[5/2+m],1/2+1/2*sin(f*x+e ))*(1-sin(f*x+e))^(-3/2-n)*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^n/f/(3+2*m)
\[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n \, dx=\int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n \, dx \] Input:
Integrate[Cos[e + f*x]^2*(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^n,x]
Output:
Integrate[Cos[e + f*x]^2*(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^n, x]
Time = 0.67 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.39, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {3042, 3320, 3042, 3224, 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.
\(\displaystyle \int \cos ^2(e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^n \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \cos (e+f x)^2 (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^ndx\) |
\(\Big \downarrow \) 3320 |
\(\displaystyle \frac {\int (\sin (e+f x) a+a)^{m+1} (c-c \sin (e+f x))^{n+1}dx}{a c}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int (\sin (e+f x) a+a)^{m+1} (c-c \sin (e+f x))^{n+1}dx}{a c}\) |
\(\Big \downarrow \) 3224 |
\(\displaystyle \cos ^{-2 m}(e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^m \int \cos ^{2 (m+1)}(e+f x) (c-c \sin (e+f x))^{n-m}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \cos ^{-2 m}(e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^m \int \cos (e+f x)^{2 (m+1)} (c-c \sin (e+f x))^{n-m}dx\) |
\(\Big \downarrow \) 3168 |
\(\displaystyle \frac {c^2 \cos ^3(e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{\frac {1}{2} (-2 m-3)+m} (c \sin (e+f x)+c)^{\frac {1}{2} (-2 m-3)} \int (c-c \sin (e+f x))^{\frac {1}{2} (2 n+1)} (\sin (e+f x) c+c)^{\frac {1}{2} (2 m+1)}d\sin (e+f x)}{f}\) |
\(\Big \downarrow \) 80 |
\(\displaystyle \frac {c^2 2^{n+\frac {1}{2}} \cos ^3(e+f x) (1-\sin (e+f x))^{-n-\frac {1}{2}} (a \sin (e+f x)+a)^m (c \sin (e+f x)+c)^{\frac {1}{2} (-2 m-3)} (c-c \sin (e+f x))^{\frac {1}{2} (-2 m-3)+m+n+\frac {1}{2}} \int \left (\frac {1}{2}-\frac {1}{2} \sin (e+f x)\right )^{\frac {1}{2} (2 n+1)} (\sin (e+f x) c+c)^{\frac {1}{2} (2 m+1)}d\sin (e+f x)}{f}\) |
\(\Big \downarrow \) 79 |
\(\displaystyle \frac {c 2^{n+\frac {3}{2}} \cos ^3(e+f x) (1-\sin (e+f x))^{-n-\frac {1}{2}} (a \sin (e+f x)+a)^m (c \sin (e+f x)+c)^{\frac {1}{2} (-2 m-3)+\frac {1}{2} (2 m+3)} (c-c \sin (e+f x))^{\frac {1}{2} (-2 m-3)+m+n+\frac {1}{2}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (2 m+3),\frac {1}{2} (-2 n-1),\frac {1}{2} (2 m+5),\frac {1}{2} (\sin (e+f x)+1)\right )}{f (2 m+3)}\) |
Input:
Int[Cos[e + f*x]^2*(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^n,x]
Output:
(2^(3/2 + n)*c*Cos[e + f*x]^3*Hypergeometric2F1[(3 + 2*m)/2, (-1 - 2*n)/2, (5 + 2*m)/2, (1 + Sin[e + f*x])/2]*(1 - Sin[e + f*x])^(-1/2 - n)*(a + a*S in[e + f*x])^m*(c - c*Sin[e + f*x])^(1/2 + (-3 - 2*m)/2 + m + n)*(c + c*Si n[e + f*x])^((-3 - 2*m)/2 + (3 + 2*m)/2))/(f*(3 + 2*m))
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[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + ( f_.)*(x_)])^(n_), x_Symbol] :> Simp[a^IntPart[m]*c^IntPart[m]*(a + b*Sin[e + f*x])^FracPart[m]*((c + d*Sin[e + f*x])^FracPart[m]/Cos[e + f*x]^(2*FracP art[m])) Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m), x], x] /; F reeQ[{a, b, c, d, e, f, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && (FractionQ[m] || !FractionQ[n])
Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_ .)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[1/(a^(p/ 2)*c^(p/2)) Int[(a + b*Sin[e + f*x])^(m + p/2)*(c + d*Sin[e + f*x])^(n + p/2), x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[p/2]
\[\int \cos \left (f x +e \right )^{2} \left (a +a \sin \left (f x +e \right )\right )^{m} \left (c -c \sin \left (f x +e \right )\right )^{n}d x\]
Input:
int(cos(f*x+e)^2*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^n,x)
Output:
int(cos(f*x+e)^2*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^n,x)
\[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{n} \cos \left (f x + e\right )^{2} \,d x } \] Input:
integrate(cos(f*x+e)^2*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^n,x, algorithm= "fricas")
Output:
integral((a*sin(f*x + e) + a)^m*(-c*sin(f*x + e) + c)^n*cos(f*x + e)^2, x)
\[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n \, dx=\int \left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{m} \left (- c \left (\sin {\left (e + f x \right )} - 1\right )\right )^{n} \cos ^{2}{\left (e + f x \right )}\, dx \] Input:
integrate(cos(f*x+e)**2*(a+a*sin(f*x+e))**m*(c-c*sin(f*x+e))**n,x)
Output:
Integral((a*(sin(e + f*x) + 1))**m*(-c*(sin(e + f*x) - 1))**n*cos(e + f*x) **2, x)
\[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{n} \cos \left (f x + e\right )^{2} \,d x } \] Input:
integrate(cos(f*x+e)^2*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^n,x, algorithm= "maxima")
Output:
integrate((a*sin(f*x + e) + a)^m*(-c*sin(f*x + e) + c)^n*cos(f*x + e)^2, x )
\[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{n} \cos \left (f x + e\right )^{2} \,d x } \] Input:
integrate(cos(f*x+e)^2*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^n,x, algorithm= "giac")
Output:
integrate((a*sin(f*x + e) + a)^m*(-c*sin(f*x + e) + c)^n*cos(f*x + e)^2, x )
Timed out. \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n \, dx=\int {\cos \left (e+f\,x\right )}^2\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^n \,d x \] Input:
int(cos(e + f*x)^2*(a + a*sin(e + f*x))^m*(c - c*sin(e + f*x))^n,x)
Output:
int(cos(e + f*x)^2*(a + a*sin(e + f*x))^m*(c - c*sin(e + f*x))^n, x)
\[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n \, dx=\int \left (a +a \sin \left (f x +e \right )\right )^{m} \left (-\sin \left (f x +e \right ) c +c \right )^{n} \cos \left (f x +e \right )^{2}d x \] Input:
int(cos(f*x+e)^2*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^n,x)
Output:
int((sin(e + f*x)*a + a)**m*( - sin(e + f*x)*c + c)**n*cos(e + f*x)**2,x)