Integrand size = 23, antiderivative size = 106 \[ \int \frac {(a+a \sec (c+d x))^n}{\sqrt {\sin (c+d x)}} \, dx=\frac {2^{\frac {5}{4}+n} \operatorname {AppellF1}\left (\frac {1}{4},n,\frac {3}{4}-n,\frac {5}{4},1-\cos (c+d x),\frac {1}{2} (1-\cos (c+d x))\right ) (1-\cos (c+d x)) \cos ^n(c+d x) (1+\cos (c+d x))^{\frac {3}{4}-n} (a+a \sec (c+d x))^n}{d \sin ^{\frac {3}{2}}(c+d x)} \] Output:
2^(5/4+n)*AppellF1(1/4,n,3/4-n,5/4,1-cos(d*x+c),1/2-1/2*cos(d*x+c))*(1-cos (d*x+c))*cos(d*x+c)^n*(1+cos(d*x+c))^(3/4-n)*(a+a*sec(d*x+c))^n/d/sin(d*x+ c)^(3/2)
Time = 2.90 (sec) , antiderivative size = 212, normalized size of antiderivative = 2.00 \[ \int \frac {(a+a \sec (c+d x))^n}{\sqrt {\sin (c+d x)}} \, dx=\frac {10 \operatorname {AppellF1}\left (\frac {1}{4},n,\frac {1}{2},\frac {5}{4},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) (1+\cos (c+d x)) (a (1+\sec (c+d x)))^n \sqrt {\sin (c+d x)}}{d \left (2 \left (\operatorname {AppellF1}\left (\frac {5}{4},n,\frac {3}{2},\frac {9}{4},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )-2 n \operatorname {AppellF1}\left (\frac {5}{4},1+n,\frac {1}{2},\frac {9}{4},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )\right ) (-1+\cos (c+d x))+5 \operatorname {AppellF1}\left (\frac {1}{4},n,\frac {1}{2},\frac {5}{4},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) (1+\cos (c+d x))\right )} \] Input:
Integrate[(a + a*Sec[c + d*x])^n/Sqrt[Sin[c + d*x]],x]
Output:
(10*AppellF1[1/4, n, 1/2, 5/4, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*(1 + Cos[c + d*x])*(a*(1 + Sec[c + d*x]))^n*Sqrt[Sin[c + d*x]])/(d*(2*(Appel lF1[5/4, n, 3/2, 9/4, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] - 2*n*Appel lF1[5/4, 1 + n, 1/2, 9/4, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])*(-1 + Cos[c + d*x]) + 5*AppellF1[1/4, n, 1/2, 5/4, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*(1 + Cos[c + d*x])))
Time = 0.55 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.99, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, 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.
\(\displaystyle \int \frac {(a \sec (c+d x)+a)^n}{\sqrt {\sin (c+d x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (a-a \csc \left (c+d x-\frac {\pi }{2}\right )\right )^n}{\sqrt {\cos \left (c+d x-\frac {\pi }{2}\right )}}dx\) |
\(\Big \downarrow \) 4364 |
\(\displaystyle (-\cos (c+d x))^n (a (-\cos (c+d x))-a)^{-n} (a \sec (c+d x)+a)^n \int \frac {(-\cos (c+d x))^{-n} (-\cos (c+d x) a-a)^n}{\sqrt {\sin (c+d x)}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle (-\cos (c+d x))^n (a (-\cos (c+d x))-a)^{-n} (a \sec (c+d x)+a)^n \int \frac {\left (-\sin \left (c+d x+\frac {\pi }{2}\right )\right )^{-n} \left (-\sin \left (c+d x+\frac {\pi }{2}\right ) a-a\right )^n}{\sqrt {-\cos \left (c+d x+\frac {\pi }{2}\right )}}dx\) |
\(\Big \downarrow \) 3365 |
\(\displaystyle -\frac {(a \cos (c+d x)-a)^{3/4} (-\cos (c+d x))^n (a (-\cos (c+d x))-a)^{\frac {3}{4}-n} (a \sec (c+d x)+a)^n \int \frac {(-\cos (c+d x))^{-n} (-\cos (c+d x) a-a)^{n-\frac {3}{4}}}{(a \cos (c+d x)-a)^{3/4}}d\cos (c+d x)}{d \sin ^{\frac {3}{2}}(c+d x)}\) |
\(\Big \downarrow \) 152 |
\(\displaystyle -\frac {(a \cos (c+d x)-a)^{3/4} (-\cos (c+d x))^n (\cos (c+d x)+1)^{\frac {3}{4}-n} (a \sec (c+d x)+a)^n \int \frac {(-\cos (c+d x))^{-n} (\cos (c+d x)+1)^{n-\frac {3}{4}}}{(a \cos (c+d x)-a)^{3/4}}d\cos (c+d x)}{d \sin ^{\frac {3}{2}}(c+d x)}\) |
\(\Big \downarrow \) 152 |
\(\displaystyle -\frac {(1-\cos (c+d x))^{3/4} (-\cos (c+d x))^n (\cos (c+d x)+1)^{\frac {3}{4}-n} (a \sec (c+d x)+a)^n \int \frac {(-\cos (c+d x))^{-n} (\cos (c+d x)+1)^{n-\frac {3}{4}}}{(1-\cos (c+d x))^{3/4}}d\cos (c+d x)}{d \sin ^{\frac {3}{2}}(c+d x)}\) |
\(\Big \downarrow \) 150 |
\(\displaystyle -\frac {(1-\cos (c+d x))^{3/4} \cos (c+d x) (\cos (c+d x)+1)^{\frac {3}{4}-n} (a \sec (c+d x)+a)^n \operatorname {AppellF1}\left (1-n,\frac {3}{4},\frac {3}{4}-n,2-n,\cos (c+d x),-\cos (c+d x)\right )}{d (1-n) \sin ^{\frac {3}{2}}(c+d x)}\) |
Input:
Int[(a + a*Sec[c + d*x])^n/Sqrt[Sin[c + d*x]],x]
Output:
-((AppellF1[1 - n, 3/4, 3/4 - n, 2 - n, Cos[c + d*x], -Cos[c + d*x]]*(1 - Cos[c + d*x])^(3/4)*Cos[c + d*x]*(1 + Cos[c + d*x])^(3/4 - n)*(a + a*Sec[c + d*x])^n)/(d*(1 - n)*Sin[c + d*x]^(3/2)))
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])
\[\int \frac {\left (a +a \sec \left (d x +c \right )\right )^{n}}{\sqrt {\sin \left (d x +c \right )}}d x\]
Input:
int((a+a*sec(d*x+c))^n/sin(d*x+c)^(1/2),x)
Output:
int((a+a*sec(d*x+c))^n/sin(d*x+c)^(1/2),x)
\[ \int \frac {(a+a \sec (c+d x))^n}{\sqrt {\sin (c+d x)}} \, dx=\int { \frac {{\left (a \sec \left (d x + c\right ) + a\right )}^{n}}{\sqrt {\sin \left (d x + c\right )}} \,d x } \] Input:
integrate((a+a*sec(d*x+c))^n/sin(d*x+c)^(1/2),x, algorithm="fricas")
Output:
integral((a*sec(d*x + c) + a)^n/sqrt(sin(d*x + c)), x)
\[ \int \frac {(a+a \sec (c+d x))^n}{\sqrt {\sin (c+d x)}} \, dx=\int \frac {\left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{n}}{\sqrt {\sin {\left (c + d x \right )}}}\, dx \] Input:
integrate((a+a*sec(d*x+c))**n/sin(d*x+c)**(1/2),x)
Output:
Integral((a*(sec(c + d*x) + 1))**n/sqrt(sin(c + d*x)), x)
\[ \int \frac {(a+a \sec (c+d x))^n}{\sqrt {\sin (c+d x)}} \, dx=\int { \frac {{\left (a \sec \left (d x + c\right ) + a\right )}^{n}}{\sqrt {\sin \left (d x + c\right )}} \,d x } \] Input:
integrate((a+a*sec(d*x+c))^n/sin(d*x+c)^(1/2),x, algorithm="maxima")
Output:
integrate((a*sec(d*x + c) + a)^n/sqrt(sin(d*x + c)), x)
\[ \int \frac {(a+a \sec (c+d x))^n}{\sqrt {\sin (c+d x)}} \, dx=\int { \frac {{\left (a \sec \left (d x + c\right ) + a\right )}^{n}}{\sqrt {\sin \left (d x + c\right )}} \,d x } \] Input:
integrate((a+a*sec(d*x+c))^n/sin(d*x+c)^(1/2),x, algorithm="giac")
Output:
integrate((a*sec(d*x + c) + a)^n/sqrt(sin(d*x + c)), x)
Timed out. \[ \int \frac {(a+a \sec (c+d x))^n}{\sqrt {\sin (c+d x)}} \, dx=\int \frac {{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^n}{\sqrt {\sin \left (c+d\,x\right )}} \,d x \] Input:
int((a + a/cos(c + d*x))^n/sin(c + d*x)^(1/2),x)
Output:
int((a + a/cos(c + d*x))^n/sin(c + d*x)^(1/2), x)
\[ \int \frac {(a+a \sec (c+d x))^n}{\sqrt {\sin (c+d x)}} \, dx=\int \frac {\sqrt {\sin \left (d x +c \right )}\, \left (\sec \left (d x +c \right ) a +a \right )^{n}}{\sin \left (d x +c \right )}d x \] Input:
int((a+a*sec(d*x+c))^n/sin(d*x+c)^(1/2),x)
Output:
int((sqrt(sin(c + d*x))*(sec(c + d*x)*a + a)**n)/sin(c + d*x),x)