Integrand size = 36, antiderivative size = 132 \[ \int (g \cos (e+f x))^p (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n \, dx=\frac {2^{\frac {1}{2}+n+\frac {p}{2}} (g \cos (e+f x))^{1+p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (1-2 n-p),\frac {1}{2} (1+2 m+p),\frac {1}{2} (3+2 m+p),\frac {1}{2} (1+\sin (e+f x))\right ) (1-\sin (e+f x))^{\frac {1}{2} (-1-2 n-p)} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n}{f g (1+2 m+p)} \] Output:
2^(1/2+n+1/2*p)*(g*cos(f*x+e))^(p+1)*hypergeom([1/2-n-1/2*p, 1/2+m+1/2*p], [3/2+m+1/2*p],1/2+1/2*sin(f*x+e))*(1-sin(f*x+e))^(-1/2-n-1/2*p)*(a+a*sin(f *x+e))^m*(c-c*sin(f*x+e))^n/f/g/(1+2*m+p)
\[ \int (g \cos (e+f x))^p (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n \, dx=\int (g \cos (e+f x))^p (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n \, dx \] Input:
Integrate[(g*Cos[e + f*x])^p*(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^n ,x]
Output:
Integrate[(g*Cos[e + f*x])^p*(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^n , x]
Time = 0.61 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.45, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3042, 3332, 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.
\(\displaystyle \int (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^n (g \cos (e+f x))^p \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^n (g \cos (e+f x))^pdx\) |
\(\Big \downarrow \) 3332 |
\(\displaystyle (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^m (g \cos (e+f x))^{-2 m} \int (g \cos (e+f x))^{2 m+p} (c-c \sin (e+f x))^{n-m}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^m (g \cos (e+f x))^{-2 m} \int (g \cos (e+f x))^{2 m+p} (c-c \sin (e+f x))^{n-m}dx\) |
\(\Big \downarrow \) 3168 |
\(\displaystyle \frac {c^2 (a \sin (e+f x)+a)^m (g \cos (e+f x))^{p+1} (c-c \sin (e+f x))^{\frac {1}{2} (-2 m-p-1)+m} (c \sin (e+f x)+c)^{\frac {1}{2} (-2 m-p-1)} \int (c-c \sin (e+f x))^{\frac {1}{2} (2 n+p-1)} (\sin (e+f x) c+c)^{\frac {1}{2} (2 m+p-1)}d\sin (e+f x)}{f g}\) |
\(\Big \downarrow \) 80 |
\(\displaystyle \frac {c^2 2^{\frac {1}{2} (2 n+p-1)} (a \sin (e+f x)+a)^m (g \cos (e+f x))^{p+1} (1-\sin (e+f x))^{\frac {1}{2} (-2 n-p+1)} (c \sin (e+f x)+c)^{\frac {1}{2} (-2 m-p-1)} (c-c \sin (e+f x))^{\frac {1}{2} (-2 m-p-1)+m+\frac {1}{2} (2 n+p-1)} \int \left (\frac {1}{2}-\frac {1}{2} \sin (e+f x)\right )^{\frac {1}{2} (2 n+p-1)} (\sin (e+f x) c+c)^{\frac {1}{2} (2 m+p-1)}d\sin (e+f x)}{f g}\) |
\(\Big \downarrow \) 79 |
\(\displaystyle \frac {c 2^{\frac {1}{2} (2 n+p-1)+1} (a \sin (e+f x)+a)^m (g \cos (e+f x))^{p+1} (1-\sin (e+f x))^{\frac {1}{2} (-2 n-p+1)} (c \sin (e+f x)+c)^{\frac {1}{2} (-2 m-p-1)+\frac {1}{2} (2 m+p+1)} (c-c \sin (e+f x))^{\frac {1}{2} (-2 m-p-1)+m+\frac {1}{2} (2 n+p-1)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (-2 n-p+1),\frac {1}{2} (2 m+p+1),\frac {1}{2} (2 m+p+3),\frac {1}{2} (\sin (e+f x)+1)\right )}{f g (2 m+p+1)}\) |
Input:
Int[(g*Cos[e + f*x])^p*(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^n,x]
Output:
(2^(1 + (-1 + 2*n + p)/2)*c*(g*Cos[e + f*x])^(1 + p)*Hypergeometric2F1[(1 - 2*n - p)/2, (1 + 2*m + p)/2, (3 + 2*m + p)/2, (1 + Sin[e + f*x])/2]*(1 - Sin[e + f*x])^((1 - 2*n - p)/2)*(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x ])^(m + (-1 - 2*m - p)/2 + (-1 + 2*n + p)/2)*(c + c*Sin[e + f*x])^((-1 - 2 *m - p)/2 + (1 + 2*m + p)/2))/(f*g*(1 + 2*m + p))
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[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((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]/(g^(2*IntPart[m])*(g*Cos[e + f*x])^(2*FracPart[m]))) Int[ (g*Cos[e + f*x])^(2*m + p)*(c + d*Sin[e + f*x])^(n - m), x], x] /; FreeQ[{a , b, c, d, e, f, g, m, n, p}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && (FractionQ[m] || !FractionQ[n])
\[\int \left (g \cos \left (f x +e \right )\right )^{p} \left (a +a \sin \left (f x +e \right )\right )^{m} \left (c -c \sin \left (f x +e \right )\right )^{n}d x\]
Input:
int((g*cos(f*x+e))^p*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^n,x)
Output:
int((g*cos(f*x+e))^p*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^n,x)
\[ \int (g \cos (e+f x))^p (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n \, dx=\int { \left (g \cos \left (f x + e\right )\right )^{p} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{n} \,d x } \] Input:
integrate((g*cos(f*x+e))^p*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^n,x, algori thm="fricas")
Output:
integral((g*cos(f*x + e))^p*(a*sin(f*x + e) + a)^m*(-c*sin(f*x + e) + c)^n , x)
Timed out. \[ \int (g \cos (e+f x))^p (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n \, dx=\text {Timed out} \] Input:
integrate((g*cos(f*x+e))**p*(a+a*sin(f*x+e))**m*(c-c*sin(f*x+e))**n,x)
Output:
Timed out
\[ \int (g \cos (e+f x))^p (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n \, dx=\int { \left (g \cos \left (f x + e\right )\right )^{p} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{n} \,d x } \] Input:
integrate((g*cos(f*x+e))^p*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^n,x, algori thm="maxima")
Output:
integrate((g*cos(f*x + e))^p*(a*sin(f*x + e) + a)^m*(-c*sin(f*x + e) + c)^ n, x)
\[ \int (g \cos (e+f x))^p (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n \, dx=\int { \left (g \cos \left (f x + e\right )\right )^{p} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{n} \,d x } \] Input:
integrate((g*cos(f*x+e))^p*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^n,x, algori thm="giac")
Output:
integrate((g*cos(f*x + e))^p*(a*sin(f*x + e) + a)^m*(-c*sin(f*x + e) + c)^ n, x)
Timed out. \[ \int (g \cos (e+f x))^p (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n \, dx=\int {\left (g\,\cos \left (e+f\,x\right )\right )}^p\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^n \,d x \] Input:
int((g*cos(e + f*x))^p*(a + a*sin(e + f*x))^m*(c - c*sin(e + f*x))^n,x)
Output:
int((g*cos(e + f*x))^p*(a + a*sin(e + f*x))^m*(c - c*sin(e + f*x))^n, x)
\[ \int (g \cos (e+f x))^p (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n \, dx=g^{p} \left (\int \left (a +a \sin \left (f x +e \right )\right )^{m} \left (-\sin \left (f x +e \right ) c +c \right )^{n} \cos \left (f x +e \right )^{p}d x \right ) \] Input:
int((g*cos(f*x+e))^p*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^n,x)
Output:
g**p*int((sin(e + f*x)*a + a)**m*( - sin(e + f*x)*c + c)**n*cos(e + f*x)** p,x)