Integrand size = 18, antiderivative size = 131 \[ \int (a+b \cos (c+d x) \sin (c+d x))^m \, dx=-\frac {\operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-m,\frac {3}{2},\frac {1}{2} (1-\sin (2 c+2 d x)),\frac {b (1-\sin (2 c+2 d x))}{2 a+b}\right ) \cos (2 c+2 d x) \left (a+\frac {1}{2} b \sin (2 c+2 d x)\right )^m \left (\frac {2 a+b \sin (2 c+2 d x)}{2 a+b}\right )^{-m}}{\sqrt {2} d \sqrt {1+\sin (2 c+2 d x)}} \]
-1/2*AppellF1(1/2,-m,1/2,3/2,b*(1-sin(2*d*x+2*c))/(2*a+b),1/2-1/2*sin(2*d* x+2*c))*cos(2*d*x+2*c)*(a+1/2*b*sin(2*d*x+2*c))^m/d/(((2*a+b*sin(2*d*x+2*c ))/(2*a+b))^m)*2^(1/2)/(1+sin(2*d*x+2*c))^(1/2)
Time = 0.68 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.11 \[ \int (a+b \cos (c+d x) \sin (c+d x))^m \, dx=\frac {\operatorname {AppellF1}\left (1+m,\frac {1}{2},\frac {1}{2},2+m,\frac {2 a+b \sin (2 (c+d x))}{2 a-b},\frac {2 a+b \sin (2 (c+d x))}{2 a+b}\right ) \sec (2 (c+d x)) \sqrt {-\frac {b (-1+\sin (2 (c+d x)))}{2 a+b}} \sqrt {\frac {b (1+\sin (2 (c+d x)))}{-2 a+b}} \left (a+\frac {1}{2} b \sin (2 (c+d x))\right )^{1+m}}{b d (1+m)} \]
(AppellF1[1 + m, 1/2, 1/2, 2 + m, (2*a + b*Sin[2*(c + d*x)])/(2*a - b), (2 *a + b*Sin[2*(c + d*x)])/(2*a + b)]*Sec[2*(c + d*x)]*Sqrt[-((b*(-1 + Sin[2 *(c + d*x)]))/(2*a + b))]*Sqrt[(b*(1 + Sin[2*(c + d*x)]))/(-2*a + b)]*(a + (b*Sin[2*(c + d*x)])/2)^(1 + m))/(b*d*(1 + m))
Time = 0.33 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3145, 3042, 3144, 156, 155}
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.
\(\displaystyle \int (a+b \sin (c+d x) \cos (c+d x))^m \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (a+b \sin (c+d x) \cos (c+d x))^mdx\) |
\(\Big \downarrow \) 3145 |
\(\displaystyle \int \left (a+\frac {1}{2} b \sin (2 c+2 d x)\right )^mdx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \left (a+\frac {1}{2} b \sin (2 c+2 d x)\right )^mdx\) |
\(\Big \downarrow \) 3144 |
\(\displaystyle \frac {\cos (2 c+2 d x) \int \frac {\left (a+\frac {1}{2} b \sin (2 c+2 d x)\right )^m}{\sqrt {1-\sin (2 c+2 d x)} \sqrt {\sin (2 c+2 d x)+1}}d\sin (2 c+2 d x)}{2 d \sqrt {1-\sin (2 c+2 d x)} \sqrt {\sin (2 c+2 d x)+1}}\) |
\(\Big \downarrow \) 156 |
\(\displaystyle \frac {\cos (2 c+2 d x) \left (a+\frac {1}{2} b \sin (2 c+2 d x)\right )^m \left (\frac {2 a+b \sin (2 c+2 d x)}{2 a+b}\right )^{-m} \int \frac {\left (\frac {2 a}{2 a+b}+\frac {b \sin (2 c+2 d x)}{2 a+b}\right )^m}{\sqrt {1-\sin (2 c+2 d x)} \sqrt {\sin (2 c+2 d x)+1}}d\sin (2 c+2 d x)}{2 d \sqrt {1-\sin (2 c+2 d x)} \sqrt {\sin (2 c+2 d x)+1}}\) |
\(\Big \downarrow \) 155 |
\(\displaystyle -\frac {\cos (2 c+2 d x) \left (a+\frac {1}{2} b \sin (2 c+2 d x)\right )^m \left (\frac {2 a+b \sin (2 c+2 d x)}{2 a+b}\right )^{-m} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-m,\frac {3}{2},\frac {1}{2} (1-\sin (2 c+2 d x)),\frac {b (1-\sin (2 c+2 d x))}{2 a+b}\right )}{\sqrt {2} d \sqrt {\sin (2 c+2 d x)+1}}\) |
-((AppellF1[1/2, 1/2, -m, 3/2, (1 - Sin[2*c + 2*d*x])/2, (b*(1 - Sin[2*c + 2*d*x]))/(2*a + b)]*Cos[2*c + 2*d*x]*(a + (b*Sin[2*c + 2*d*x])/2)^m)/(Sqr t[2]*d*Sqrt[1 + Sin[2*c + 2*d*x]]*((2*a + b*Sin[2*c + 2*d*x])/(2*a + b))^m ))
3.6.66.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) ^(p_), x_] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*Simplify[b/(b*c - a*d)]^n* Simplify[b/(b*e - a*f)]^p))*AppellF1[m + 1, -n, -p, m + 2, (-d)*((a + b*x)/ (b*c - a*d)), (-f)*((a + b*x)/(b*e - a*f))], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && !IntegerQ[m] && !IntegerQ[n] && !IntegerQ[p] && GtQ[Sim plify[b/(b*c - a*d)], 0] && GtQ[Simplify[b/(b*e - a*f)], 0] && !(GtQ[Simpl ify[d/(d*a - c*b)], 0] && GtQ[Simplify[d/(d*e - c*f)], 0] && SimplerQ[c + d *x, a + b*x]) && !(GtQ[Simplify[f/(f*a - e*b)], 0] && GtQ[Simplify[f/(f*c - e*d)], 0] && SimplerQ[e + f*x, a + b*x])
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) ^(p_), x_] :> Simp[(e + f*x)^FracPart[p]/(Simplify[b/(b*e - a*f)]^IntPart[p ]*(b*((e + f*x)/(b*e - a*f)))^FracPart[p]) Int[(a + b*x)^m*(c + d*x)^n*Si mp[b*(e/(b*e - a*f)) + b*f*(x/(b*e - a*f)), x]^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && !IntegerQ[m] && !IntegerQ[n] && !IntegerQ[p] & & GtQ[Simplify[b/(b*c - a*d)], 0] && !GtQ[Simplify[b/(b*e - a*f)], 0]
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]/(d*Sqrt[1 + Sin[c + d*x]]*Sqrt[1 - Sin[c + d*x]]) Subst[Int[(a + b*x )^n/(Sqrt[1 + x]*Sqrt[1 - x]), x], x, Sin[c + d*x]], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[a^2 - b^2, 0] && !IntegerQ[2*n]
Int[((a_) + cos[(c_.) + (d_.)*(x_)]*(b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_ Symbol] :> Int[(a + b*(Sin[2*c + 2*d*x]/2))^n, x] /; FreeQ[{a, b, c, d, n}, x]
\[\int \left (a +\cos \left (d x +c \right ) \sin \left (d x +c \right ) b \right )^{m}d x\]
\[ \int (a+b \cos (c+d x) \sin (c+d x))^m \, dx=\int { {\left (b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a\right )}^{m} \,d x } \]
Timed out. \[ \int (a+b \cos (c+d x) \sin (c+d x))^m \, dx=\text {Timed out} \]
\[ \int (a+b \cos (c+d x) \sin (c+d x))^m \, dx=\int { {\left (b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a\right )}^{m} \,d x } \]
\[ \int (a+b \cos (c+d x) \sin (c+d x))^m \, dx=\int { {\left (b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a\right )}^{m} \,d x } \]
Timed out. \[ \int (a+b \cos (c+d x) \sin (c+d x))^m \, dx=\int {\left (a+b\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )\right )}^m \,d x \]