Integrand size = 21, antiderivative size = 95 \[ \int (a+a \sec (c+d x))^n \sin ^2(c+d x) \, dx=-\frac {\operatorname {AppellF1}\left (1-n,-\frac {1}{2},-\frac {1}{2}-n,2-n,\cos (c+d x),-\cos (c+d x)\right ) \sqrt {1-\cos (c+d x)} (1+\cos (c+d x))^{\frac {1}{2}-n} \cot (c+d x) (a+a \sec (c+d x))^n}{d (1-n)} \]
-AppellF1(1-n,-1/2-n,-1/2,2-n,-cos(d*x+c),cos(d*x+c))*(1+cos(d*x+c))^(1/2- n)*cot(d*x+c)*(a+a*sec(d*x+c))^n*(1-cos(d*x+c))^(1/2)/d/(1-n)
Result contains complex when optimal does not.
Time = 17.34 (sec) , antiderivative size = 4297, normalized size of antiderivative = 45.23 \[ \int (a+a \sec (c+d x))^n \sin ^2(c+d x) \, dx=\text {Result too large to show} \]
(2^(3 + n)*Cos[(c + d*x)/2]^5*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^n*(a*(1 + Sec[c + d*x]))^n*Sin[(c + d*x)/2]*(Cos[2*(c + d*x)]*(-1/4*(1 + Sec[c + d*x ])^n - ((1 + Sec[c + d*x])^n*Sin[c + d*x]^2)/2 - ((1 + Sec[c + d*x])^n*Sin [c + d*x]^4)/4) + (I/4)*(1 + Sec[c + d*x])^n*Sin[2*(c + d*x)] + (I/2)*(1 + Sec[c + d*x])^n*Sin[c + d*x]^2*Sin[2*(c + d*x)] + (I/4)*(1 + Sec[c + d*x] )^n*Sin[c + d*x]^4*Sin[2*(c + d*x)] + Cos[c + d*x]^4*(-1/4*(Cos[2*(c + d*x )]*(1 + Sec[c + d*x])^n) + (I/4)*(1 + Sec[c + d*x])^n*Sin[2*(c + d*x)]) + Cos[c + d*x]^3*((-I)*Cos[2*(c + d*x)]*(1 + Sec[c + d*x])^n*Sin[c + d*x] - (1 + Sec[c + d*x])^n*Sin[c + d*x]*Sin[2*(c + d*x)]) + Cos[c + d*x]^2*(Cos[ 2*(c + d*x)]*((1 + Sec[c + d*x])^n/2 + (3*(1 + Sec[c + d*x])^n*Sin[c + d*x ]^2)/2) - (I/2)*(1 + Sec[c + d*x])^n*Sin[2*(c + d*x)] - ((3*I)/2)*(1 + Sec [c + d*x])^n*Sin[c + d*x]^2*Sin[2*(c + d*x)]) + Cos[c + d*x]*(Cos[2*(c + d *x)]*(I*(1 + Sec[c + d*x])^n*Sin[c + d*x] + I*(1 + Sec[c + d*x])^n*Sin[c + d*x]^3) + (1 + Sec[c + d*x])^n*Sin[c + d*x]*Sin[2*(c + d*x)] + (1 + Sec[c + d*x])^n*Sin[c + d*x]^3*Sin[2*(c + d*x)]))*((3*AppellF1[1/2, n, 2, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Sec[(c + d*x)/2]^2)/(3*AppellF1[1 /2, n, 2, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + 2*(-2*AppellF1[3 /2, n, 3, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + n*AppellF1[3/2, 1 + n, 2, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])*Tan[(c + d*x)/2]^ 2) - AppellF1[1/2, n, 3, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]/...
Time = 0.69 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.27, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {3042, 4364, 3042, 3353, 25, 3042, 3487, 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.
| \(\displaystyle \int \sin ^2(c+d x) (a \sec (c+d x)+a)^n \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \cos \left (c+d x-\frac {\pi }{2}\right )^2 \left (a-a \csc \left (c+d x-\frac {\pi }{2}\right )\right )^ndx\) |
| \(\Big \downarrow \) 4364 |
| \(\displaystyle (-\cos (c+d x))^n (a (-\cos (c+d x))-a)^{-n} (a \sec (c+d x)+a)^n \int (-\cos (c+d x))^{-n} (-\cos (c+d x) a-a)^n \sin ^2(c+d x)dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle (-\cos (c+d x))^n (a (-\cos (c+d x))-a)^{-n} (a \sec (c+d x)+a)^n \int \cos \left (c+d x+\frac {\pi }{2}\right )^2 \left (-\sin \left (c+d x+\frac {\pi }{2}\right )\right )^{-n} \left (-\sin \left (c+d x+\frac {\pi }{2}\right ) a-a\right )^ndx\) |
| \(\Big \downarrow \) 3353 |
| \(\displaystyle \frac {(-\cos (c+d x))^n (a (-\cos (c+d x))-a)^{-n} (a \sec (c+d x)+a)^n \int -(-\cos (c+d x))^{-n} (-\cos (c+d x) a-a)^{n+1} (a-a \cos (c+d x))dx}{a^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {(-\cos (c+d x))^n (a (-\cos (c+d x))-a)^{-n} (a \sec (c+d x)+a)^n \int (-\cos (c+d x))^{-n} (-\cos (c+d x) a-a)^{n+1} (a-a \cos (c+d x))dx}{a^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {(-\cos (c+d x))^n (a (-\cos (c+d x))-a)^{-n} (a \sec (c+d x)+a)^n \int \left (-\sin \left (c+d x+\frac {\pi }{2}\right )\right )^{-n} \left (-\sin \left (c+d x+\frac {\pi }{2}\right ) a-a\right )^{n+1} \left (a-a \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx}{a^2}\) |
| \(\Big \downarrow \) 3487 |
| \(\displaystyle \frac {\csc (c+d x) \sqrt {a-a \cos (c+d x)} (-\cos (c+d x))^n (a (-\cos (c+d x))-a)^{\frac {1}{2}-n} (a \sec (c+d x)+a)^n \int (-\cos (c+d x))^{-n} (-\cos (c+d x) a-a)^{n+\frac {1}{2}} \sqrt {a-a \cos (c+d x)}d\cos (c+d x)}{a^2 d}\) |
| \(\Big \downarrow \) 152 |
| \(\displaystyle \frac {\csc (c+d x) (a (-\cos (c+d x))-a) \sqrt {a-a \cos (c+d x)} (-\cos (c+d x))^n (\cos (c+d x)+1)^{-n-\frac {1}{2}} (a \sec (c+d x)+a)^n \int (-\cos (c+d x))^{-n} (\cos (c+d x)+1)^{n+\frac {1}{2}} \sqrt {a-a \cos (c+d x)}d\cos (c+d x)}{a^2 d}\) |
| \(\Big \downarrow \) 152 |
| \(\displaystyle \frac {\csc (c+d x) (a (-\cos (c+d x))-a) (a-a \cos (c+d x)) (-\cos (c+d x))^n (\cos (c+d x)+1)^{-n-\frac {1}{2}} (a \sec (c+d x)+a)^n \int \sqrt {1-\cos (c+d x)} (-\cos (c+d x))^{-n} (\cos (c+d x)+1)^{n+\frac {1}{2}}d\cos (c+d x)}{a^2 d \sqrt {1-\cos (c+d x)}}\) |
| \(\Big \downarrow \) 150 |
| \(\displaystyle \frac {\cot (c+d x) (a (-\cos (c+d x))-a) (a-a \cos (c+d x)) (\cos (c+d x)+1)^{-n-\frac {1}{2}} (a \sec (c+d x)+a)^n \operatorname {AppellF1}\left (1-n,-\frac {1}{2},-n-\frac {1}{2},2-n,\cos (c+d x),-\cos (c+d x)\right )}{a^2 d (1-n) \sqrt {1-\cos (c+d x)}}\) |
(AppellF1[1 - n, -1/2, -1/2 - n, 2 - n, Cos[c + d*x], -Cos[c + d*x]]*(1 + Cos[c + d*x])^(-1/2 - n)*(-a - a*Cos[c + d*x])*(a - a*Cos[c + d*x])*Cot[c + d*x]*(a + a*Sec[c + d*x])^n)/(a^2*d*(1 - n)*Sqrt[1 - Cos[c + d*x]])
3.2.53.3.1 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_)]^2*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[1/b^2 Int[(d*Sin[e + f*x])^n*(a + b*Sin[e + f*x])^(m + 1)*(a - b*Sin[e + f*x]), x], x] /; Fre eQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] && (ILtQ[m, 0] || !IGtQ[ n, 0])
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])^(p_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[Sqrt[a + b*Sin[e + f*x]]*(Sqrt[c + d*Sin[e + f*x]]/(f*Cos[e + f*x]) ) Subst[Int[(a + b*x)^(m - 1/2)*(c + d*x)^(n - 1/2)*(A + B*x)^p, x], x, S in[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n, p}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0]
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])
\[\int \left (a +a \sec \left (d x +c \right )\right )^{n} \sin \left (d x +c \right )^{2}d x\]
\[ \int (a+a \sec (c+d x))^n \sin ^2(c+d x) \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{n} \sin \left (d x + c\right )^{2} \,d x } \]
\[ \int (a+a \sec (c+d x))^n \sin ^2(c+d x) \, dx=\int \left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{n} \sin ^{2}{\left (c + d x \right )}\, dx \]
\[ \int (a+a \sec (c+d x))^n \sin ^2(c+d x) \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{n} \sin \left (d x + c\right )^{2} \,d x } \]
\[ \int (a+a \sec (c+d x))^n \sin ^2(c+d x) \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{n} \sin \left (d x + c\right )^{2} \,d x } \]
Timed out. \[ \int (a+a \sec (c+d x))^n \sin ^2(c+d x) \, dx=\int {\sin \left (c+d\,x\right )}^2\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^n \,d x \]