Integrand size = 27, antiderivative size = 311 \[ \int (e \cos (c+d x))^{-3-m} (a+b \sin (c+d x))^m \, dx=\frac {(e \cos (c+d x))^{-m} \sec ^4(c+d x) (-1+\sin (c+d x)) (1+\sin (c+d x)) (a+b \sin (c+d x))^{1+m}}{(a-b) d e^3 (2+m)}+\frac {(-2 b+a (2+m)) (e \cos (c+d x))^{-m} \sec ^4(c+d x) (-1+\sin (c+d x)) (1+\sin (c+d x))^2 (a+b \sin (c+d x))^{1+m}}{(a-b)^2 d e^3 m (2+m)}-\frac {\left (-b^2+a^2 (1+m)\right ) (e \cos (c+d x))^{-m} \operatorname {Hypergeometric2F1}\left (\frac {m}{2},1+m,2+m,-\frac {2 (a+b \sin (c+d x))}{(a-b) (-1+\sin (c+d x))}\right ) \sec ^4(c+d x) (1+\sin (c+d x))^3 \left (\frac {(a+b) (1+\sin (c+d x))}{(a-b) (-1+\sin (c+d x))}\right )^{\frac {1}{2} (-2+m)} (a+b \sin (c+d x))^{1+m}}{(a-b)^3 d e^3 m (1+m)} \]
sec(d*x+c)^4*(-1+sin(d*x+c))*(1+sin(d*x+c))*(a+b*sin(d*x+c))^(1+m)/(a-b)/d /e^3/(2+m)/((e*cos(d*x+c))^m)+(-2*b+a*(2+m))*sec(d*x+c)^4*(-1+sin(d*x+c))* (1+sin(d*x+c))^2*(a+b*sin(d*x+c))^(1+m)/(a-b)^2/d/e^3/m/(2+m)/((e*cos(d*x+ c))^m)-(-b^2+a^2*(1+m))*hypergeom([1/2*m, 1+m],[2+m],-2*(a+b*sin(d*x+c))/( a-b)/(-1+sin(d*x+c)))*sec(d*x+c)^4*(1+sin(d*x+c))^3*((a+b)*(1+sin(d*x+c))/ (a-b)/(-1+sin(d*x+c)))^(-1+1/2*m)*(a+b*sin(d*x+c))^(1+m)/(a-b)^3/d/e^3/m/( 1+m)/((e*cos(d*x+c))^m)
Time = 3.30 (sec) , antiderivative size = 319, normalized size of antiderivative = 1.03 \[ \int (e \cos (c+d x))^{-3-m} (a+b \sin (c+d x))^m \, dx=\frac {(e \cos (c+d x))^{-m} \sec ^2(c+d x) (a+b \sin (c+d x))^m \left (-a-b \sin (c+d x)+\frac {b \operatorname {Hypergeometric2F1}\left (1+m,\frac {2+m}{2},2+m,-\frac {2 (a+b \sin (c+d x))}{(a-b) (-1+\sin (c+d x))}\right ) (1+\sin (c+d x)) \left (\frac {(a+b) (1+\sin (c+d x))}{(a-b) (-1+\sin (c+d x))}\right )^{m/2} (a+b \sin (c+d x))}{(a-b) (1+m)}+\frac {a (1-\sin (c+d x)) (1+\sin (c+d x)) \left (2^{-m/2} (a+b+a m) \operatorname {Hypergeometric2F1}\left (-\frac {m}{2},\frac {2+m}{2},1-\frac {m}{2},-\frac {(a-b) (-1+\sin (c+d x))}{2 (a+b \sin (c+d x))}\right ) \left (\frac {(a+b) (1+\sin (c+d x))}{a+b \sin (c+d x)}\right )^{m/2}-\frac {m (a+b \sin (c+d x))}{-1+\sin (c+d x)}\right )}{(a+b) m}\right )}{(a-b) d e^3 (2+m)} \]
(Sec[c + d*x]^2*(a + b*Sin[c + d*x])^m*(-a - b*Sin[c + d*x] + (b*Hypergeom etric2F1[1 + m, (2 + m)/2, 2 + m, (-2*(a + b*Sin[c + d*x]))/((a - b)*(-1 + Sin[c + d*x]))]*(1 + Sin[c + d*x])*(((a + b)*(1 + Sin[c + d*x]))/((a - b) *(-1 + Sin[c + d*x])))^(m/2)*(a + b*Sin[c + d*x]))/((a - b)*(1 + m)) + (a* (1 - Sin[c + d*x])*(1 + Sin[c + d*x])*(((a + b + a*m)*Hypergeometric2F1[-1 /2*m, (2 + m)/2, 1 - m/2, -1/2*((a - b)*(-1 + Sin[c + d*x]))/(a + b*Sin[c + d*x])]*(((a + b)*(1 + Sin[c + d*x]))/(a + b*Sin[c + d*x]))^(m/2))/2^(m/2 ) - (m*(a + b*Sin[c + d*x]))/(-1 + Sin[c + d*x])))/((a + b)*m)))/((a - b)* d*e^3*(2 + m)*(e*Cos[c + d*x])^m)
Time = 0.98 (sec) , antiderivative size = 480, normalized size of antiderivative = 1.54, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {3042, 3179, 3042, 3177, 3399, 107, 142}
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.
| \(\displaystyle \int (e \cos (c+d x))^{-m-3} (a+b \sin (c+d x))^m \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int (e \cos (c+d x))^{-m-3} (a+b \sin (c+d x))^mdx\) |
| \(\Big \downarrow \) 3179 |
| \(\displaystyle -\frac {b \int (e \cos (c+d x))^{-m-1} (a+b \sin (c+d x))^mdx}{e^2 (m+2) (a-b)}+\frac {a \int \frac {(e \cos (c+d x))^{-m-1} (a+b \sin (c+d x))^m}{1-\sin (c+d x)}dx}{e^2 (a-b)}-\frac {(e \cos (c+d x))^{-m-2} (a+b \sin (c+d x))^{m+1}}{d e (m+2) (a-b)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {b \int (e \cos (c+d x))^{-m-1} (a+b \sin (c+d x))^mdx}{e^2 (m+2) (a-b)}+\frac {a \int \frac {(e \cos (c+d x))^{-m-1} (a+b \sin (c+d x))^m}{1-\sin (c+d x)}dx}{e^2 (a-b)}-\frac {(e \cos (c+d x))^{-m-2} (a+b \sin (c+d x))^{m+1}}{d e (m+2) (a-b)}\) |
| \(\Big \downarrow \) 3177 |
| \(\displaystyle \frac {a \int \frac {(e \cos (c+d x))^{-m-1} (a+b \sin (c+d x))^m}{1-\sin (c+d x)}dx}{e^2 (a-b)}-\frac {b (1-\sin (c+d x)) (e \cos (c+d x))^{-m-2} \left (-\frac {(a-b) (1-\sin (c+d x))}{(a+b) (\sin (c+d x)+1)}\right )^{m/2} (a+b \sin (c+d x))^{m+1} \operatorname {Hypergeometric2F1}\left (m+1,\frac {m+2}{2},m+2,\frac {2 (a+b \sin (c+d x))}{(a+b) (\sin (c+d x)+1)}\right )}{d e (m+1) (m+2) (a-b) (a+b)}-\frac {(e \cos (c+d x))^{-m-2} (a+b \sin (c+d x))^{m+1}}{d e (m+2) (a-b)}\) |
| \(\Big \downarrow \) 3399 |
| \(\displaystyle \frac {a (1-\sin (c+d x))^{\frac {m+2}{2}} (\sin (c+d x)+1)^{\frac {m+2}{2}} (e \cos (c+d x))^{-m-2} \int (1-\sin (c+d x))^{\frac {1}{2} (-m-4)} (\sin (c+d x)+1)^{\frac {1}{2} (-m-2)} (a+b \sin (c+d x))^md\sin (c+d x)}{d e (a-b)}-\frac {b (1-\sin (c+d x)) (e \cos (c+d x))^{-m-2} \left (-\frac {(a-b) (1-\sin (c+d x))}{(a+b) (\sin (c+d x)+1)}\right )^{m/2} (a+b \sin (c+d x))^{m+1} \operatorname {Hypergeometric2F1}\left (m+1,\frac {m+2}{2},m+2,\frac {2 (a+b \sin (c+d x))}{(a+b) (\sin (c+d x)+1)}\right )}{d e (m+1) (m+2) (a-b) (a+b)}-\frac {(e \cos (c+d x))^{-m-2} (a+b \sin (c+d x))^{m+1}}{d e (m+2) (a-b)}\) |
| \(\Big \downarrow \) 107 |
| \(\displaystyle \frac {a (1-\sin (c+d x))^{\frac {m+2}{2}} (\sin (c+d x)+1)^{\frac {m+2}{2}} (e \cos (c+d x))^{-m-2} \left (\frac {(a m+a+b) \int (1-\sin (c+d x))^{\frac {1}{2} (-m-2)} (\sin (c+d x)+1)^{\frac {1}{2} (-m-2)} (a+b \sin (c+d x))^md\sin (c+d x)}{(m+2) (a+b)}+\frac {(\sin (c+d x)+1)^{-m/2} (1-\sin (c+d x))^{\frac {1}{2} (-m-2)} (a+b \sin (c+d x))^{m+1}}{(m+2) (a+b)}\right )}{d e (a-b)}-\frac {b (1-\sin (c+d x)) (e \cos (c+d x))^{-m-2} \left (-\frac {(a-b) (1-\sin (c+d x))}{(a+b) (\sin (c+d x)+1)}\right )^{m/2} (a+b \sin (c+d x))^{m+1} \operatorname {Hypergeometric2F1}\left (m+1,\frac {m+2}{2},m+2,\frac {2 (a+b \sin (c+d x))}{(a+b) (\sin (c+d x)+1)}\right )}{d e (m+1) (m+2) (a-b) (a+b)}-\frac {(e \cos (c+d x))^{-m-2} (a+b \sin (c+d x))^{m+1}}{d e (m+2) (a-b)}\) |
| \(\Big \downarrow \) 142 |
| \(\displaystyle -\frac {b (1-\sin (c+d x)) (e \cos (c+d x))^{-m-2} \left (-\frac {(a-b) (1-\sin (c+d x))}{(a+b) (\sin (c+d x)+1)}\right )^{m/2} (a+b \sin (c+d x))^{m+1} \operatorname {Hypergeometric2F1}\left (m+1,\frac {m+2}{2},m+2,\frac {2 (a+b \sin (c+d x))}{(a+b) (\sin (c+d x)+1)}\right )}{d e (m+1) (m+2) (a-b) (a+b)}+\frac {a (1-\sin (c+d x))^{\frac {m+2}{2}} (\sin (c+d x)+1)^{\frac {m+2}{2}} (e \cos (c+d x))^{-m-2} \left (\frac {2^{-m/2} (a m+a+b) (\sin (c+d x)+1)^{\frac {1}{2} (-m-2)} (1-\sin (c+d x))^{-m/2} \left (\frac {(a+b) (\sin (c+d x)+1)}{a+b \sin (c+d x)}\right )^{\frac {m+2}{2}} (a+b \sin (c+d x))^{m+1} \operatorname {Hypergeometric2F1}\left (-\frac {m}{2},\frac {m+2}{2},\frac {2-m}{2},\frac {(a-b) (1-\sin (c+d x))}{2 (a+b \sin (c+d x))}\right )}{m (m+2) (a+b)^2}+\frac {(\sin (c+d x)+1)^{-m/2} (1-\sin (c+d x))^{\frac {1}{2} (-m-2)} (a+b \sin (c+d x))^{m+1}}{(m+2) (a+b)}\right )}{d e (a-b)}-\frac {(e \cos (c+d x))^{-m-2} (a+b \sin (c+d x))^{m+1}}{d e (m+2) (a-b)}\) |
-(((e*Cos[c + d*x])^(-2 - m)*(a + b*Sin[c + d*x])^(1 + m))/((a - b)*d*e*(2 + m))) - (b*(e*Cos[c + d*x])^(-2 - m)*Hypergeometric2F1[1 + m, (2 + m)/2, 2 + m, (2*(a + b*Sin[c + d*x]))/((a + b)*(1 + Sin[c + d*x]))]*(1 - Sin[c + d*x])*(-(((a - b)*(1 - Sin[c + d*x]))/((a + b)*(1 + Sin[c + d*x]))))^(m/ 2)*(a + b*Sin[c + d*x])^(1 + m))/((a - b)*(a + b)*d*e*(1 + m)*(2 + m)) + ( a*(e*Cos[c + d*x])^(-2 - m)*(1 - Sin[c + d*x])^((2 + m)/2)*(1 + Sin[c + d* x])^((2 + m)/2)*(((1 - Sin[c + d*x])^((-2 - m)/2)*(a + b*Sin[c + d*x])^(1 + m))/((a + b)*(2 + m)*(1 + Sin[c + d*x])^(m/2)) + ((a + b + a*m)*Hypergeo metric2F1[-1/2*m, (2 + m)/2, (2 - m)/2, ((a - b)*(1 - Sin[c + d*x]))/(2*(a + b*Sin[c + d*x]))]*(1 + Sin[c + d*x])^((-2 - m)/2)*(((a + b)*(1 + Sin[c + d*x]))/(a + b*Sin[c + d*x]))^((2 + m)/2)*(a + b*Sin[c + d*x])^(1 + m))/( 2^(m/2)*(a + b)^2*m*(2 + m)*(1 - Sin[c + d*x])^(m/2))))/((a - b)*d*e)
3.7.48.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Simp[b*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1 )/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + Simp[(a*d*f*(m + 1) + b*c*f*(n + 1) + b*d*e*(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)) Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x ] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || SumSimplerQ[m, 1])
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n*((e + f*x)^(p + 1)/((b*e - a*f)*(m + 1)))*Hypergeometric2F1[m + 1, -n, m + 2, (-(d*e - c*f))*((a + b*x)/((b*c - a*d)*(e + f*x)))])/((b*e - a*f)*((c + d*x)/((b*c - a*d)*(e + f *x))))^n, x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[m + n + p + 2, 0] && !IntegerQ[n]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_), x_Symbol] :> Simp[g*(g*Cos[e + f*x])^(p - 1)*(1 - Sin[e + f*x])* (a + b*Sin[e + f*x])^(m + 1)*(((-(a - b))*((1 - Sin[e + f*x])/((a + b)*(1 + Sin[e + f*x]))))^(m/2)/(f*(a + b)*(m + 1)))*Hypergeometric2F1[m + 1, m/2 + 1, m + 2, 2*((a + b*Sin[e + f*x])/((a + b)*(1 + Sin[e + f*x])))], x] /; Fr eeQ[{a, b, e, f, g, m, p}, x] && NeQ[a^2 - b^2, 0] && EqQ[m + p + 1, 0]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_), x_Symbol] :> Simp[(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x]) ^(m + 1)/(f*g*(a - b)*(p + 1))), x] + (-Simp[b*((m + p + 2)/(g^2*(a - b)*(p + 1))) Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^m, x], x] + Sim p[a/(g^2*(a - b)) Int[(g*Cos[e + f*x])^(p + 2)*((a + b*Sin[e + f*x])^m/(1 - Sin[e + f*x])), x], x]) /; FreeQ[{a, b, e, f, g, m, p}, x] && NeQ[a^2 - b^2, 0] && ILtQ[m + p + 2, 0]
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^ m*g*((g*Cos[e + f*x])^(p - 1)/(f*(1 + Sin[e + f*x])^((p - 1)/2)*(1 - Sin[e + f*x])^((p - 1)/2))) Subst[Int[(1 + (b/a)*x)^(m + (p - 1)/2)*(1 - (b/a)* x)^((p - 1)/2)*(c + d*x)^n, x], x, Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && EqQ[a^2 - b^2, 0] && IntegerQ[m]
\[\int \left (e \cos \left (d x +c \right )\right )^{-3-m} \left (a +b \sin \left (d x +c \right )\right )^{m}d x\]
\[ \int (e \cos (c+d x))^{-3-m} (a+b \sin (c+d x))^m \, dx=\int { \left (e \cos \left (d x + c\right )\right )^{-m - 3} {\left (b \sin \left (d x + c\right ) + a\right )}^{m} \,d x } \]
\[ \int (e \cos (c+d x))^{-3-m} (a+b \sin (c+d x))^m \, dx=\int \left (e \cos {\left (c + d x \right )}\right )^{- m - 3} \left (a + b \sin {\left (c + d x \right )}\right )^{m}\, dx \]
\[ \int (e \cos (c+d x))^{-3-m} (a+b \sin (c+d x))^m \, dx=\int { \left (e \cos \left (d x + c\right )\right )^{-m - 3} {\left (b \sin \left (d x + c\right ) + a\right )}^{m} \,d x } \]
\[ \int (e \cos (c+d x))^{-3-m} (a+b \sin (c+d x))^m \, dx=\int { \left (e \cos \left (d x + c\right )\right )^{-m - 3} {\left (b \sin \left (d x + c\right ) + a\right )}^{m} \,d x } \]
Timed out. \[ \int (e \cos (c+d x))^{-3-m} (a+b \sin (c+d x))^m \, dx=\int \frac {{\left (a+b\,\sin \left (c+d\,x\right )\right )}^m}{{\left (e\,\cos \left (c+d\,x\right )\right )}^{m+3}} \,d x \]