Integrand size = 42, antiderivative size = 123 \[ \int (g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{2-m} \, dx=\frac {2^{\frac {17}{4}-m} c^3 (g \cos (e+f x))^{5/2} \operatorname {Hypergeometric2F1}\left (\frac {1}{4} (-9+4 m),\frac {1}{4} (5+4 m),\frac {1}{4} (9+4 m),\frac {1}{2} (1+\sin (e+f x))\right ) (1-\sin (e+f x))^{-\frac {1}{4}+m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-1-m}}{f g (5+4 m)} \]
2^(17/4-m)*c^3*(g*cos(f*x+e))^(5/2)*hypergeom([5/4+m, -9/4+m],[9/4+m],1/2+ 1/2*sin(f*x+e))*(1-sin(f*x+e))^(-1/4+m)*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e) )^(-1-m)/f/g/(5+4*m)
\[ \int (g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{2-m} \, dx=\int (g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{2-m} \, dx \]
Time = 0.63 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.31, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, 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.
| \(\displaystyle \int (g \cos (e+f x))^{3/2} (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{2-m} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int (g \cos (e+f x))^{3/2} (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{2-m}dx\) |
| \(\Big \downarrow \) 3332 |
| \(\displaystyle (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^m (g \cos (e+f x))^{-2 m} \int (g \cos (e+f x))^{2 m+\frac {3}{2}} (c-c \sin (e+f x))^{2-2 m}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^m (g \cos (e+f x))^{-2 m} \int (g \cos (e+f x))^{2 m+\frac {3}{2}} (c-c \sin (e+f x))^{2-2 m}dx\) |
| \(\Big \downarrow \) 3168 |
| \(\displaystyle \frac {c^2 (g \cos (e+f x))^{5/2} (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{\frac {1}{4} (-4 m-5)+m} (c \sin (e+f x)+c)^{\frac {1}{4} (-4 m-5)} \int (c-c \sin (e+f x))^{\frac {1}{4} (9-4 m)} (\sin (e+f x) c+c)^{\frac {1}{4} (4 m+1)}d\sin (e+f x)}{f g}\) |
| \(\Big \downarrow \) 80 |
| \(\displaystyle \frac {c^4 2^{\frac {9}{4}-m} (g \cos (e+f x))^{5/2} (1-\sin (e+f x))^{m-\frac {1}{4}} (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{\frac {1}{4} (-4 m-5)+\frac {1}{4}} (c \sin (e+f x)+c)^{\frac {1}{4} (-4 m-5)} \int \left (\frac {1}{2}-\frac {1}{2} \sin (e+f x)\right )^{\frac {1}{4} (9-4 m)} (\sin (e+f x) c+c)^{\frac {1}{4} (4 m+1)}d\sin (e+f x)}{f g}\) |
| \(\Big \downarrow \) 79 |
| \(\displaystyle \frac {c^3 2^{\frac {17}{4}-m} (g \cos (e+f x))^{5/2} (1-\sin (e+f x))^{m-\frac {1}{4}} (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{\frac {1}{4} (-4 m-5)+\frac {1}{4}} (c \sin (e+f x)+c)^{\frac {1}{4} (-4 m-5)+\frac {1}{4} (4 m+5)} \operatorname {Hypergeometric2F1}\left (\frac {1}{4} (4 m-9),\frac {1}{4} (4 m+5),\frac {1}{4} (4 m+9),\frac {1}{2} (\sin (e+f x)+1)\right )}{f g (4 m+5)}\) |
(2^(17/4 - m)*c^3*(g*Cos[e + f*x])^(5/2)*Hypergeometric2F1[(-9 + 4*m)/4, ( 5 + 4*m)/4, (9 + 4*m)/4, (1 + Sin[e + f*x])/2]*(1 - Sin[e + f*x])^(-1/4 + m)*(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^(1/4 + (-5 - 4*m)/4)*(c + c *Sin[e + f*x])^((-5 - 4*m)/4 + (5 + 4*m)/4))/(f*g*(5 + 4*m))
3.2.72.3.1 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])
\[\int \left (g \cos \left (f x +e \right )\right )^{\frac {3}{2}} \left (a +a \sin \left (f x +e \right )\right )^{m} \left (c -c \sin \left (f x +e \right )\right )^{2-m}d x\]
\[ \int (g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{2-m} \, dx=\int { \left (g \cos \left (f x + e\right )\right )^{\frac {3}{2}} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{-m + 2} \,d x } \]
integrate((g*cos(f*x+e))^(3/2)*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(2-m),x , algorithm="fricas")
integral(sqrt(g*cos(f*x + e))*(a*sin(f*x + e) + a)^m*(-c*sin(f*x + e) + c) ^(-m + 2)*g*cos(f*x + e), x)
Timed out. \[ \int (g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{2-m} \, dx=\text {Timed out} \]
\[ \int (g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{2-m} \, dx=\int { \left (g \cos \left (f x + e\right )\right )^{\frac {3}{2}} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{-m + 2} \,d x } \]
integrate((g*cos(f*x+e))^(3/2)*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(2-m),x , algorithm="maxima")
\[ \int (g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{2-m} \, dx=\int { \left (g \cos \left (f x + e\right )\right )^{\frac {3}{2}} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{-m + 2} \,d x } \]
integrate((g*cos(f*x+e))^(3/2)*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(2-m),x , algorithm="giac")
Timed out. \[ \int (g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{2-m} \, dx=\int {\left (g\,\cos \left (e+f\,x\right )\right )}^{3/2}\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{2-m} \,d x \]