Integrand size = 27, antiderivative size = 115 \[ \int (e \cos (c+d x))^{2-m} (a+a \sin (c+d x))^m \, dx=-\frac {2^{\frac {3}{2}+\frac {m}{2}} a (e \cos (c+d x))^{3-m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (-1-m),\frac {3-m}{2},\frac {5-m}{2},\frac {1}{2} (1-\sin (c+d x))\right ) (1+\sin (c+d x))^{\frac {1}{2} (-1-m)} (a+a \sin (c+d x))^{-1+m}}{d e (3-m)} \]
-2^(3/2+1/2*m)*a*(e*cos(d*x+c))^(3-m)*hypergeom([-1/2-1/2*m, 3/2-1/2*m],[5 /2-1/2*m],1/2-1/2*sin(d*x+c))*(1+sin(d*x+c))^(-1/2-1/2*m)*(a+a*sin(d*x+c)) ^(-1+m)/d/e/(3-m)
Time = 0.18 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.98 \[ \int (e \cos (c+d x))^{2-m} (a+a \sin (c+d x))^m \, dx=\frac {2^{\frac {3+m}{2}} e^2 \cos ^3(c+d x) (e \cos (c+d x))^{-m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (-1-m),\frac {3-m}{2},\frac {5-m}{2},\frac {1}{2} (1-\sin (c+d x))\right ) (1+\sin (c+d x))^{\frac {1}{2} (-3-m)} (a (1+\sin (c+d x)))^m}{d (-3+m)} \]
(2^((3 + m)/2)*e^2*Cos[c + d*x]^3*Hypergeometric2F1[(-1 - m)/2, (3 - m)/2, (5 - m)/2, (1 - Sin[c + d*x])/2]*(1 + Sin[c + d*x])^((-3 - m)/2)*(a*(1 + Sin[c + d*x]))^m)/(d*(-3 + m)*(e*Cos[c + d*x])^m)
Time = 0.36 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.26, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {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.
\(\displaystyle \int (a \sin (c+d x)+a)^m (e \cos (c+d x))^{2-m} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (a \sin (c+d x)+a)^m (e \cos (c+d x))^{2-m}dx\) |
\(\Big \downarrow \) 3168 |
\(\displaystyle \frac {a^2 (a-a \sin (c+d x))^{\frac {m-3}{2}} (a \sin (c+d x)+a)^{\frac {m-3}{2}} (e \cos (c+d x))^{3-m} \int (a-a \sin (c+d x))^{\frac {1-m}{2}} (\sin (c+d x) a+a)^{\frac {m+1}{2}}d\sin (c+d x)}{d e}\) |
\(\Big \downarrow \) 80 |
\(\displaystyle \frac {a^2 2^{\frac {1}{2}-\frac {m}{2}} (1-\sin (c+d x))^{\frac {m-1}{2}} (a-a \sin (c+d x))^{\frac {1-m}{2}+\frac {m-3}{2}} (a \sin (c+d x)+a)^{\frac {m-3}{2}} (e \cos (c+d x))^{3-m} \int \left (\frac {1}{2}-\frac {1}{2} \sin (c+d x)\right )^{\frac {1-m}{2}} (\sin (c+d x) a+a)^{\frac {m+1}{2}}d\sin (c+d x)}{d e}\) |
\(\Big \downarrow \) 79 |
\(\displaystyle \frac {a 2^{\frac {3}{2}-\frac {m}{2}} (1-\sin (c+d x))^{\frac {m-1}{2}} (a-a \sin (c+d x))^{\frac {1-m}{2}+\frac {m-3}{2}} (a \sin (c+d x)+a)^{\frac {m-3}{2}+\frac {m+3}{2}} (e \cos (c+d x))^{3-m} \operatorname {Hypergeometric2F1}\left (\frac {m-1}{2},\frac {m+3}{2},\frac {m+5}{2},\frac {1}{2} (\sin (c+d x)+1)\right )}{d e (m+3)}\) |
(2^(3/2 - m/2)*a*(e*Cos[c + d*x])^(3 - m)*Hypergeometric2F1[(-1 + m)/2, (3 + m)/2, (5 + m)/2, (1 + Sin[c + d*x])/2]*(1 - Sin[c + d*x])^((-1 + m)/2)* (a - a*Sin[c + d*x])^((1 - m)/2 + (-3 + m)/2)*(a + a*Sin[c + d*x])^((-3 + m)/2 + (3 + m)/2))/(d*e*(3 + m))
3.4.66.3.1 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]
\[\int \left (e \cos \left (d x +c \right )\right )^{2-m} \left (a +a \sin \left (d x +c \right )\right )^{m}d x\]
\[ \int (e \cos (c+d x))^{2-m} (a+a \sin (c+d x))^m \, dx=\int { \left (e \cos \left (d x + c\right )\right )^{-m + 2} {\left (a \sin \left (d x + c\right ) + a\right )}^{m} \,d x } \]
\[ \int (e \cos (c+d x))^{2-m} (a+a \sin (c+d x))^m \, dx=\int \left (a \left (\sin {\left (c + d x \right )} + 1\right )\right )^{m} \left (e \cos {\left (c + d x \right )}\right )^{2 - m}\, dx \]
\[ \int (e \cos (c+d x))^{2-m} (a+a \sin (c+d x))^m \, dx=\int { \left (e \cos \left (d x + c\right )\right )^{-m + 2} {\left (a \sin \left (d x + c\right ) + a\right )}^{m} \,d x } \]
\[ \int (e \cos (c+d x))^{2-m} (a+a \sin (c+d x))^m \, dx=\int { \left (e \cos \left (d x + c\right )\right )^{-m + 2} {\left (a \sin \left (d x + c\right ) + a\right )}^{m} \,d x } \]
Timed out. \[ \int (e \cos (c+d x))^{2-m} (a+a \sin (c+d x))^m \, dx=\int {\left (e\,\cos \left (c+d\,x\right )\right )}^{2-m}\,{\left (a+a\,\sin \left (c+d\,x\right )\right )}^m \,d x \]