Integrand size = 45, antiderivative size = 134 \[ \int (g \cos (e+f x))^{-1-m-n} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{3+n} \, dx=\frac {2^{3-\frac {m}{2}+\frac {n}{2}} c^3 (g \cos (e+f x))^{-m-n} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (-4+m-n),\frac {m-n}{2},\frac {1}{2} (2+m-n),\frac {1}{2} (1+\sin (e+f x))\right ) (1-\sin (e+f x))^{\frac {m-n}{2}} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n}{f g (m-n)} \]
2^(3-1/2*m+1/2*n)*c^3*(g*cos(f*x+e))^(-n-m)*hypergeom([1/2*m-1/2*n, -2+1/2 *m-1/2*n],[1+1/2*m-1/2*n],1/2+1/2*sin(f*x+e))*(1-sin(f*x+e))^(1/2*m-1/2*n) *(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^n/f/g/(m-n)
\[ \int (g \cos (e+f x))^{-1-m-n} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{3+n} \, dx=\int (g \cos (e+f x))^{-1-m-n} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{3+n} \, dx \]
Integrate[(g*Cos[e + f*x])^(-1 - m - n)*(a + a*Sin[e + f*x])^m*(c - c*Sin[ e + f*x])^(3 + n),x]
Integrate[(g*Cos[e + f*x])^(-1 - m - n)*(a + a*Sin[e + f*x])^m*(c - c*Sin[ e + f*x])^(3 + n), x]
Time = 0.65 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.22, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, 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.
| \(\displaystyle \int (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{n+3} (g \cos (e+f x))^{-m-n-1} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{n+3} (g \cos (e+f x))^{-m-n-1}dx\) |
| \(\Big \downarrow \) 3332 |
| \(\displaystyle (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^m (g \cos (e+f x))^{-2 m} \int (g \cos (e+f x))^{m-n-1} (c-c \sin (e+f x))^{-m+n+3}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^m (g \cos (e+f x))^{-2 m} \int (g \cos (e+f x))^{m-n-1} (c-c \sin (e+f x))^{-m+n+3}dx\) |
| \(\Big \downarrow \) 3168 |
| \(\displaystyle \frac {c^2 (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{\frac {n-m}{2}+m} (c \sin (e+f x)+c)^{\frac {n-m}{2}} (g \cos (e+f x))^{-m-n} \int (c-c \sin (e+f x))^{\frac {1}{2} (-m+n+4)} (\sin (e+f x) c+c)^{\frac {1}{2} (m-n-2)}d\sin (e+f x)}{f g}\) |
| \(\Big \downarrow \) 80 |
| \(\displaystyle \frac {c^4 2^{\frac {1}{2} (-m+n+4)} (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^n (1-\sin (e+f x))^{\frac {m-n}{2}} (c \sin (e+f x)+c)^{\frac {n-m}{2}} (g \cos (e+f x))^{-m-n} \int \left (\frac {1}{2}-\frac {1}{2} \sin (e+f x)\right )^{\frac {1}{2} (-m+n+4)} (\sin (e+f x) c+c)^{\frac {1}{2} (m-n-2)}d\sin (e+f x)}{f g}\) |
| \(\Big \downarrow \) 79 |
| \(\displaystyle \frac {c^3 2^{\frac {1}{2} (-m+n+4)+1} (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^n (1-\sin (e+f x))^{\frac {m-n}{2}} (c \sin (e+f x)+c)^{\frac {m-n}{2}+\frac {n-m}{2}} (g \cos (e+f x))^{-m-n} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (m-n-4),\frac {m-n}{2},\frac {1}{2} (m-n+2),\frac {1}{2} (\sin (e+f x)+1)\right )}{f g (m-n)}\) |
(2^(1 + (4 - m + n)/2)*c^3*(g*Cos[e + f*x])^(-m - n)*Hypergeometric2F1[(-4 + m - n)/2, (m - n)/2, (2 + m - n)/2, (1 + Sin[e + f*x])/2]*(1 - Sin[e + f*x])^((m - n)/2)*(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^n*(c + c*Sin [e + f*x])^((m - n)/2 + (-m + n)/2))/(f*g*(m - n))
3.2.82.3.1 Defintions of rubi rules used
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])
\[\int \left (g \cos \left (f x +e \right )\right )^{-1-m -n} \left (a +a \sin \left (f x +e \right )\right )^{m} \left (c -c \sin \left (f x +e \right )\right )^{3+n}d x\]
\[ \int (g \cos (e+f x))^{-1-m-n} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{3+n} \, dx=\int { \left (g \cos \left (f x + e\right )\right )^{-m - n - 1} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{n + 3} \,d x } \]
integrate((g*cos(f*x+e))^(-1-m-n)*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(3+n ),x, algorithm="fricas")
integral((g*cos(f*x + e))^(-m - n - 1)*(a*sin(f*x + e) + a)^m*(-c*sin(f*x + e) + c)^(n + 3), x)
Timed out. \[ \int (g \cos (e+f x))^{-1-m-n} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{3+n} \, dx=\text {Timed out} \]
\[ \int (g \cos (e+f x))^{-1-m-n} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{3+n} \, dx=\int { \left (g \cos \left (f x + e\right )\right )^{-m - n - 1} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{n + 3} \,d x } \]
integrate((g*cos(f*x+e))^(-1-m-n)*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(3+n ),x, algorithm="maxima")
integrate((g*cos(f*x + e))^(-m - n - 1)*(a*sin(f*x + e) + a)^m*(-c*sin(f*x + e) + c)^(n + 3), x)
\[ \int (g \cos (e+f x))^{-1-m-n} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{3+n} \, dx=\int { \left (g \cos \left (f x + e\right )\right )^{-m - n - 1} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{n + 3} \,d x } \]
integrate((g*cos(f*x+e))^(-1-m-n)*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(3+n ),x, algorithm="giac")
integrate((g*cos(f*x + e))^(-m - n - 1)*(a*sin(f*x + e) + a)^m*(-c*sin(f*x + e) + c)^(n + 3), x)
Timed out. \[ \int (g \cos (e+f x))^{-1-m-n} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{3+n} \, dx=\int \frac {{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{n+3}}{{\left (g\,\cos \left (e+f\,x\right )\right )}^{m+n+1}} \,d x \]