Integrand size = 40, antiderivative size = 114 \[ \int (g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{5/2} \, dx=-\frac {2^{\frac {9}{4}+m} a^3 c^2 (g \cos (e+f x))^{15/2} \operatorname {Hypergeometric2F1}\left (\frac {15}{4},-\frac {1}{4}-m,\frac {19}{4},\frac {1}{2} (1-\sin (e+f x))\right ) \sec (e+f x) (1+\sin (e+f x))^{-\frac {1}{4}-m} (a+a \sin (e+f x))^{-3+m} \sqrt {c-c \sin (e+f x)}}{15 f g^6} \]
-1/15*2^(9/4+m)*a^3*c^2*(g*cos(f*x+e))^(15/2)*hypergeom([15/4, -1/4-m],[19 /4],1/2-1/2*sin(f*x+e))*sec(f*x+e)*(1+sin(f*x+e))^(-1/4-m)*(a+a*sin(f*x+e) )^(-3+m)*(c-c*sin(f*x+e))^(1/2)/f/g^6
\[ \int (g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{5/2} \, dx=\int (g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{5/2} \, dx \]
Time = 0.62 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, 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.
| \(\displaystyle \int (c-c \sin (e+f x))^{5/2} (g \cos (e+f x))^{3/2} (a \sin (e+f x)+a)^m \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int (c-c \sin (e+f x))^{5/2} (g \cos (e+f x))^{3/2} (a \sin (e+f x)+a)^mdx\) |
| \(\Big \downarrow \) 3332 |
| \(\displaystyle \frac {a^2 c^2 \sec (e+f x) \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)} \int (g \cos (e+f x))^{13/2} (\sin (e+f x) a+a)^{m-\frac {5}{2}}dx}{g^5}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a^2 c^2 \sec (e+f x) \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)} \int (g \cos (e+f x))^{13/2} (\sin (e+f x) a+a)^{m-\frac {5}{2}}dx}{g^5}\) |
| \(\Big \downarrow \) 3168 |
| \(\displaystyle \frac {a^4 c^2 \sec (e+f x) \sqrt {c-c \sin (e+f x)} (g \cos (e+f x))^{15/2} \int (a-a \sin (e+f x))^{11/4} (\sin (e+f x) a+a)^{m+\frac {1}{4}}d\sin (e+f x)}{f g^6 (a-a \sin (e+f x))^{15/4} (a \sin (e+f x)+a)^{13/4}}\) |
| \(\Big \downarrow \) 80 |
| \(\displaystyle \frac {a^4 c^2 2^{m+\frac {1}{4}} \sec (e+f x) \sqrt {c-c \sin (e+f x)} (g \cos (e+f x))^{15/2} (\sin (e+f x)+1)^{-m-\frac {1}{4}} (a \sin (e+f x)+a)^{m-3} \int \left (\frac {1}{2} \sin (e+f x)+\frac {1}{2}\right )^{m+\frac {1}{4}} (a-a \sin (e+f x))^{11/4}d\sin (e+f x)}{f g^6 (a-a \sin (e+f x))^{15/4}}\) |
| \(\Big \downarrow \) 79 |
| \(\displaystyle -\frac {a^3 c^2 2^{m+\frac {9}{4}} \sec (e+f x) \sqrt {c-c \sin (e+f x)} (g \cos (e+f x))^{15/2} (\sin (e+f x)+1)^{-m-\frac {1}{4}} (a \sin (e+f x)+a)^{m-3} \operatorname {Hypergeometric2F1}\left (\frac {15}{4},-m-\frac {1}{4},\frac {19}{4},\frac {1}{2} (1-\sin (e+f x))\right )}{15 f g^6}\) |
-1/15*(2^(9/4 + m)*a^3*c^2*(g*Cos[e + f*x])^(15/2)*Hypergeometric2F1[15/4, -1/4 - m, 19/4, (1 - Sin[e + f*x])/2]*Sec[e + f*x]*(1 + Sin[e + f*x])^(-1 /4 - m)*(a + a*Sin[e + f*x])^(-3 + m)*Sqrt[c - c*Sin[e + f*x]])/(f*g^6)
3.2.59.3.1 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])
\[\int \left (g \cos \left (f x +e \right )\right )^{\frac {3}{2}} \left (a +a \sin \left (f x +e \right )\right )^{m} \left (c -c \sin \left (f x +e \right )\right )^{\frac {5}{2}}d x\]
\[ \int (g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{5/2} \, dx=\int { \left (g \cos \left (f x + e\right )\right )^{\frac {3}{2}} {\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \,d x } \]
integrate((g*cos(f*x+e))^(3/2)*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(5/2),x , algorithm="fricas")
integral(-(c^2*g*cos(f*x + e)^3 + 2*c^2*g*cos(f*x + e)*sin(f*x + e) - 2*c^ 2*g*cos(f*x + e))*sqrt(g*cos(f*x + e))*sqrt(-c*sin(f*x + e) + c)*(a*sin(f* x + e) + a)^m, x)
Timed out. \[ \int (g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{5/2} \, dx=\text {Timed out} \]
\[ \int (g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{5/2} \, dx=\int { \left (g \cos \left (f x + e\right )\right )^{\frac {3}{2}} {\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \,d x } \]
integrate((g*cos(f*x+e))^(3/2)*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(5/2),x , algorithm="maxima")
\[ \int (g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{5/2} \, dx=\int { \left (g \cos \left (f x + e\right )\right )^{\frac {3}{2}} {\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \,d x } \]
integrate((g*cos(f*x+e))^(3/2)*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(5/2),x , algorithm="giac")
Timed out. \[ \int (g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{5/2} \, dx=\int {\left (g\,\cos \left (e+f\,x\right )\right )}^{3/2}\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{5/2} \,d x \]