Integrand size = 25, antiderivative size = 115 \[ \int \frac {(e \sin (c+d x))^m}{\sqrt {a+a \sec (c+d x)}} \, dx=-\frac {2 e \operatorname {AppellF1}\left (\frac {3}{2},\frac {1-m}{2},\frac {2-m}{2},\frac {5}{2},\cos (c+d x),-\cos (c+d x)\right ) (1-\cos (c+d x))^{\frac {1-m}{2}} \cos (c+d x) (1+\cos (c+d x))^{1-\frac {m}{2}} (e \sin (c+d x))^{-1+m}}{3 d \sqrt {a+a \sec (c+d x)}} \]
-2/3*e*AppellF1(3/2,1-1/2*m,-1/2*m+1/2,5/2,-cos(d*x+c),cos(d*x+c))*(1-cos( d*x+c))^(-1/2*m+1/2)*cos(d*x+c)*(1+cos(d*x+c))^(1-1/2*m)*(e*sin(d*x+c))^(- 1+m)/d/(a+a*sec(d*x+c))^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(277\) vs. \(2(115)=230\).
Time = 2.85 (sec) , antiderivative size = 277, normalized size of antiderivative = 2.41 \[ \int \frac {(e \sin (c+d x))^m}{\sqrt {a+a \sec (c+d x)}} \, dx=\frac {4 (3+m) \operatorname {AppellF1}\left (\frac {1+m}{2},-\frac {1}{2},1+m,\frac {3+m}{2},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \cos ^3\left (\frac {1}{2} (c+d x)\right ) \sin \left (\frac {1}{2} (c+d x)\right ) (e \sin (c+d x))^m}{d (1+m) \left (\left (2 (1+m) \operatorname {AppellF1}\left (\frac {3+m}{2},-\frac {1}{2},2+m,\frac {5+m}{2},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )+\operatorname {AppellF1}\left (\frac {3+m}{2},\frac {1}{2},1+m,\frac {5+m}{2},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )\right ) (-1+\cos (c+d x))+(3+m) \operatorname {AppellF1}\left (\frac {1+m}{2},-\frac {1}{2},1+m,\frac {3+m}{2},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) (1+\cos (c+d x))\right ) \sqrt {a (1+\sec (c+d x))}} \]
(4*(3 + m)*AppellF1[(1 + m)/2, -1/2, 1 + m, (3 + m)/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Cos[(c + d*x)/2]^3*Sin[(c + d*x)/2]*(e*Sin[c + d*x]) ^m)/(d*(1 + m)*((2*(1 + m)*AppellF1[(3 + m)/2, -1/2, 2 + m, (5 + m)/2, Tan [(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + AppellF1[(3 + m)/2, 1/2, 1 + m, (5 + m)/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])*(-1 + Cos[c + d*x]) + ( 3 + m)*AppellF1[(1 + m)/2, -1/2, 1 + m, (3 + m)/2, Tan[(c + d*x)/2]^2, -Ta n[(c + d*x)/2]^2]*(1 + Cos[c + d*x]))*Sqrt[a*(1 + Sec[c + d*x])])
Time = 0.61 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.55, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3042, 4364, 3042, 3365, 152, 152, 150}
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.
| \(\displaystyle \int \frac {(e \sin (c+d x))^m}{\sqrt {a \sec (c+d x)+a}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (e \cos \left (c+d x-\frac {\pi }{2}\right )\right )^m}{\sqrt {a-a \csc \left (c+d x-\frac {\pi }{2}\right )}}dx\) |
| \(\Big \downarrow \) 4364 |
| \(\displaystyle \frac {\sqrt {a (-\cos (c+d x))-a} \int \frac {\sqrt {-\cos (c+d x)} (e \sin (c+d x))^m}{\sqrt {-\cos (c+d x) a-a}}dx}{\sqrt {-\cos (c+d x)} \sqrt {a \sec (c+d x)+a}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sqrt {a (-\cos (c+d x))-a} \int \frac {\left (-e \cos \left (c+d x+\frac {\pi }{2}\right )\right )^m \sqrt {-\sin \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {-\sin \left (c+d x+\frac {\pi }{2}\right ) a-a}}dx}{\sqrt {-\cos (c+d x)} \sqrt {a \sec (c+d x)+a}}\) |
| \(\Big \downarrow \) 3365 |
| \(\displaystyle -\frac {e (a (-\cos (c+d x))-a)^{\frac {1-m}{2}+\frac {1}{2}} (a \cos (c+d x)-a)^{\frac {1-m}{2}} (e \sin (c+d x))^{m-1} \int \sqrt {-\cos (c+d x)} (-\cos (c+d x) a-a)^{\frac {m-2}{2}} (a \cos (c+d x)-a)^{\frac {m-1}{2}}d\cos (c+d x)}{d \sqrt {-\cos (c+d x)} \sqrt {a \sec (c+d x)+a}}\) |
| \(\Big \downarrow \) 152 |
| \(\displaystyle \frac {e (\cos (c+d x)+1)^{-m/2} (a (-\cos (c+d x))-a)^{\frac {1-m}{2}+\frac {m}{2}+\frac {1}{2}} (a \cos (c+d x)-a)^{\frac {1-m}{2}} (e \sin (c+d x))^{m-1} \int \sqrt {-\cos (c+d x)} (\cos (c+d x)+1)^{\frac {m-2}{2}} (a \cos (c+d x)-a)^{\frac {m-1}{2}}d\cos (c+d x)}{a d \sqrt {-\cos (c+d x)} \sqrt {a \sec (c+d x)+a}}\) |
| \(\Big \downarrow \) 152 |
| \(\displaystyle \frac {e (1-\cos (c+d x))^{\frac {1-m}{2}} (\cos (c+d x)+1)^{-m/2} (a (-\cos (c+d x))-a)^{\frac {1-m}{2}+\frac {m}{2}+\frac {1}{2}} (a \cos (c+d x)-a)^{\frac {1-m}{2}+\frac {m-1}{2}} (e \sin (c+d x))^{m-1} \int (1-\cos (c+d x))^{\frac {m-1}{2}} \sqrt {-\cos (c+d x)} (\cos (c+d x)+1)^{\frac {m-2}{2}}d\cos (c+d x)}{a d \sqrt {-\cos (c+d x)} \sqrt {a \sec (c+d x)+a}}\) |
| \(\Big \downarrow \) 150 |
| \(\displaystyle \frac {2 e \cos (c+d x) (1-\cos (c+d x))^{\frac {1-m}{2}} (\cos (c+d x)+1)^{-m/2} (a (-\cos (c+d x))-a)^{\frac {1-m}{2}+\frac {m}{2}+\frac {1}{2}} (a \cos (c+d x)-a)^{\frac {1-m}{2}+\frac {m-1}{2}} \operatorname {AppellF1}\left (\frac {3}{2},\frac {1-m}{2},\frac {2-m}{2},\frac {5}{2},\cos (c+d x),-\cos (c+d x)\right ) (e \sin (c+d x))^{m-1}}{3 a d \sqrt {a \sec (c+d x)+a}}\) |
(2*e*AppellF1[3/2, (1 - m)/2, (2 - m)/2, 5/2, Cos[c + d*x], -Cos[c + d*x]] *(1 - Cos[c + d*x])^((1 - m)/2)*Cos[c + d*x]*(-a - a*Cos[c + d*x])^(1/2 + (1 - m)/2 + m/2)*(-a + a*Cos[c + d*x])^((1 - m)/2 + (-1 + m)/2)*(e*Sin[c + d*x])^(-1 + m))/(3*a*d*(1 + Cos[c + d*x])^(m/2)*Sqrt[a + a*Sec[c + d*x]])
3.2.42.3.1 Defintions of rubi rules used
Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_ ] :> Simp[c^n*e^p*((b*x)^(m + 1)/(b*(m + 1)))*AppellF1[m + 1, -n, -p, m + 2 , (-d)*(x/c), (-f)*(x/e)], x] /; FreeQ[{b, c, d, e, f, m, n, p}, x] && !In tegerQ[m] && !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])
Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_ ] :> Simp[c^IntPart[n]*((c + d*x)^FracPart[n]/(1 + d*(x/c))^FracPart[n]) Int[(b*x)^m*(1 + d*(x/c))^n*(e + f*x)^p, x], x] /; FreeQ[{b, c, d, e, f, m, n, p}, x] && !IntegerQ[m] && !IntegerQ[n] && !GtQ[c, 0]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n _)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[g*((g*Cos [e + f*x])^(p - 1)/(f*(a + b*Sin[e + f*x])^((p - 1)/2)*(a - b*Sin[e + f*x]) ^((p - 1)/2))) Subst[Int[(d*x)^n*(a + b*x)^(m + (p - 1)/2)*(a - b*x)^((p - 1)/2), x], x, Sin[e + f*x]], x] /; FreeQ[{a, b, d, e, f, m, n, p}, x] && EqQ[a^2 - b^2, 0] && !IntegerQ[m]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[Sin[e + f*x]^FracPart[m]*((a + b*Csc[e + f*x] )^FracPart[m]/(b + a*Sin[e + f*x])^FracPart[m]) Int[(g*Cos[e + f*x])^p*(( b + a*Sin[e + f*x])^m/Sin[e + f*x]^m), x], x] /; FreeQ[{a, b, e, f, g, m, p }, x] && (EqQ[a^2 - b^2, 0] || IntegersQ[2*m, p])
\[\int \frac {\left (e \sin \left (d x +c \right )\right )^{m}}{\sqrt {a +a \sec \left (d x +c \right )}}d x\]
\[ \int \frac {(e \sin (c+d x))^m}{\sqrt {a+a \sec (c+d x)}} \, dx=\int { \frac {\left (e \sin \left (d x + c\right )\right )^{m}}{\sqrt {a \sec \left (d x + c\right ) + a}} \,d x } \]
\[ \int \frac {(e \sin (c+d x))^m}{\sqrt {a+a \sec (c+d x)}} \, dx=\int \frac {\left (e \sin {\left (c + d x \right )}\right )^{m}}{\sqrt {a \left (\sec {\left (c + d x \right )} + 1\right )}}\, dx \]
\[ \int \frac {(e \sin (c+d x))^m}{\sqrt {a+a \sec (c+d x)}} \, dx=\int { \frac {\left (e \sin \left (d x + c\right )\right )^{m}}{\sqrt {a \sec \left (d x + c\right ) + a}} \,d x } \]
\[ \int \frac {(e \sin (c+d x))^m}{\sqrt {a+a \sec (c+d x)}} \, dx=\int { \frac {\left (e \sin \left (d x + c\right )\right )^{m}}{\sqrt {a \sec \left (d x + c\right ) + a}} \,d x } \]
Timed out. \[ \int \frac {(e \sin (c+d x))^m}{\sqrt {a+a \sec (c+d x)}} \, dx=\int \frac {{\left (e\,\sin \left (c+d\,x\right )\right )}^m}{\sqrt {a+\frac {a}{\cos \left (c+d\,x\right )}}} \,d x \]