Integrand size = 33, antiderivative size = 173 \[ \int \frac {(a+b \tan (c+d x))^n (A+B \tan (c+d x))}{\sqrt {\cot (c+d x)}} \, dx=\frac {(A+i B) \operatorname {AppellF1}\left (\frac {3}{2},1,-n,\frac {5}{2},-i \tan (c+d x),-\frac {b \tan (c+d x)}{a}\right ) (a+b \tan (c+d x))^n \left (1+\frac {b \tan (c+d x)}{a}\right )^{-n}}{3 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {(A-i B) \operatorname {AppellF1}\left (\frac {3}{2},1,-n,\frac {5}{2},i \tan (c+d x),-\frac {b \tan (c+d x)}{a}\right ) (a+b \tan (c+d x))^n \left (1+\frac {b \tan (c+d x)}{a}\right )^{-n}}{3 d \cot ^{\frac {3}{2}}(c+d x)} \]
1/3*(A+I*B)*AppellF1(3/2,1,-n,5/2,-I*tan(d*x+c),-b*tan(d*x+c)/a)*(a+b*tan( d*x+c))^n/d/cot(d*x+c)^(3/2)/((1+b*tan(d*x+c)/a)^n)+1/3*(A-I*B)*AppellF1(3 /2,1,-n,5/2,I*tan(d*x+c),-b*tan(d*x+c)/a)*(a+b*tan(d*x+c))^n/d/cot(d*x+c)^ (3/2)/((1+b*tan(d*x+c)/a)^n)
\[ \int \frac {(a+b \tan (c+d x))^n (A+B \tan (c+d x))}{\sqrt {\cot (c+d x)}} \, dx=\int \frac {(a+b \tan (c+d x))^n (A+B \tan (c+d x))}{\sqrt {\cot (c+d x)}} \, dx \]
Time = 0.79 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.12, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {3042, 4729, 3042, 4086, 3042, 4085, 148, 395, 394}
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.
\(\displaystyle \int \frac {(A+B \tan (c+d x)) (a+b \tan (c+d x))^n}{\sqrt {\cot (c+d x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(A+B \tan (c+d x)) (a+b \tan (c+d x))^n}{\sqrt {\cot (c+d x)}}dx\) |
\(\Big \downarrow \) 4729 |
\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \int \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^n (A+B \tan (c+d x))dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \int \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^n (A+B \tan (c+d x))dx\) |
\(\Big \downarrow \) 4086 |
\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {1}{2} (A+i B) \int (1-i \tan (c+d x)) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^ndx+\frac {1}{2} (A-i B) \int (i \tan (c+d x)+1) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^ndx\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {1}{2} (A+i B) \int (1-i \tan (c+d x)) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^ndx+\frac {1}{2} (A-i B) \int (i \tan (c+d x)+1) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^ndx\right )\) |
\(\Big \downarrow \) 4085 |
\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {(A-i B) \int \frac {\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^n}{1-i \tan (c+d x)}d\tan (c+d x)}{2 d}+\frac {(A+i B) \int \frac {\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^n}{i \tan (c+d x)+1}d\tan (c+d x)}{2 d}\right )\) |
\(\Big \downarrow \) 148 |
\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {(A-i B) \int \frac {\tan (c+d x) (a+b \tan (c+d x))^n}{1-i \tan (c+d x)}d\sqrt {\tan (c+d x)}}{d}+\frac {(A+i B) \int \frac {\tan (c+d x) (a+b \tan (c+d x))^n}{i \tan (c+d x)+1}d\sqrt {\tan (c+d x)}}{d}\right )\) |
\(\Big \downarrow \) 395 |
\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {(A-i B) (a+b \tan (c+d x))^n \left (\frac {b \tan (c+d x)}{a}+1\right )^{-n} \int \frac {\tan (c+d x) \left (\frac {b \tan (c+d x)}{a}+1\right )^n}{1-i \tan (c+d x)}d\sqrt {\tan (c+d x)}}{d}+\frac {(A+i B) (a+b \tan (c+d x))^n \left (\frac {b \tan (c+d x)}{a}+1\right )^{-n} \int \frac {\tan (c+d x) \left (\frac {b \tan (c+d x)}{a}+1\right )^n}{i \tan (c+d x)+1}d\sqrt {\tan (c+d x)}}{d}\right )\) |
\(\Big \downarrow \) 394 |
\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {(A+i B) \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^n \left (\frac {b \tan (c+d x)}{a}+1\right )^{-n} \operatorname {AppellF1}\left (\frac {3}{2},1,-n,\frac {5}{2},-i \tan (c+d x),-\frac {b \tan (c+d x)}{a}\right )}{3 d}+\frac {(A-i B) \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^n \left (\frac {b \tan (c+d x)}{a}+1\right )^{-n} \operatorname {AppellF1}\left (\frac {3}{2},1,-n,\frac {5}{2},i \tan (c+d x),-\frac {b \tan (c+d x)}{a}\right )}{3 d}\right )\) |
Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]]*(((A + I*B)*AppellF1[3/2, 1, -n, 5/2 , (-I)*Tan[c + d*x], -((b*Tan[c + d*x])/a)]*Tan[c + d*x]^(3/2)*(a + b*Tan[ c + d*x])^n)/(3*d*(1 + (b*Tan[c + d*x])/a)^n) + ((A - I*B)*AppellF1[3/2, 1 , -n, 5/2, I*Tan[c + d*x], -((b*Tan[c + d*x])/a)]*Tan[c + d*x]^(3/2)*(a + b*Tan[c + d*x])^n)/(3*d*(1 + (b*Tan[c + d*x])/a)^n))
3.7.59.3.1 Defintions of rubi rules used
Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))^(p_.), x_] :> With[{k = Denominator[m]}, Simp[k/b Subst[Int[x^(k*(m + 1) - 1)*(c + d*(x^k/b))^n*(e + f*(x^k/b))^p, x], x, (b*x)^(1/k)], x]] /; FreeQ[{b, c, d, e, f, n, p}, x] && FractionQ[m] && IntegerQ[p]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ ), x_Symbol] :> Simp[a^p*c^q*((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m + 1)/2 , -p, -q, 1 + (m + 1)/2, (-b)*(x^2/a), (-d)*(x^2/c)], x] /; FreeQ[{a, b, c, d, e, m, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, 1] && (Int egerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ ), x_Symbol] :> Simp[a^IntPart[p]*((a + b*x^2)^FracPart[p]/(1 + b*(x^2/a))^ FracPart[p]) Int[(e*x)^m*(1 + b*(x^2/a))^p*(c + d*x^2)^q, x], x] /; FreeQ [{a, b, c, d, e, m, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, 1] && !(IntegerQ[p] || GtQ[a, 0])
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[A^2/f Subst[Int[(a + b*x)^m*((c + d*x)^n/(A - B*x)), x], x, Tan[e + f*x ]], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && !IntegerQ[m] && !IntegerQ[n] && !IntegersQ[2*m, 2*n ] && EqQ[A^2 + B^2, 0]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si mp[(A + I*B)/2 Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n*(1 - I*T an[e + f*x]), x], x] + Simp[(A - I*B)/2 Int[(a + b*Tan[e + f*x])^m*(c + d *Tan[e + f*x])^n*(1 + I*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A , B, m, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && !IntegerQ[m] & & !IntegerQ[n] && !IntegersQ[2*m, 2*n] && NeQ[A^2 + B^2, 0]
Int[(cot[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Simp[(c*Cot[a + b*x])^m*(c*Tan[a + b*x])^m Int[ActivateTrig[u]/(c*Tan[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] && !IntegerQ[m] && KnownTangentIntegrandQ[u, x]
\[\int \frac {\left (a +b \tan \left (d x +c \right )\right )^{n} \left (A +B \tan \left (d x +c \right )\right )}{\sqrt {\cot \left (d x +c \right )}}d x\]
\[ \int \frac {(a+b \tan (c+d x))^n (A+B \tan (c+d x))}{\sqrt {\cot (c+d x)}} \, dx=\int { \frac {{\left (B \tan \left (d x + c\right ) + A\right )} {\left (b \tan \left (d x + c\right ) + a\right )}^{n}}{\sqrt {\cot \left (d x + c\right )}} \,d x } \]
\[ \int \frac {(a+b \tan (c+d x))^n (A+B \tan (c+d x))}{\sqrt {\cot (c+d x)}} \, dx=\int \frac {\left (A + B \tan {\left (c + d x \right )}\right ) \left (a + b \tan {\left (c + d x \right )}\right )^{n}}{\sqrt {\cot {\left (c + d x \right )}}}\, dx \]
\[ \int \frac {(a+b \tan (c+d x))^n (A+B \tan (c+d x))}{\sqrt {\cot (c+d x)}} \, dx=\int { \frac {{\left (B \tan \left (d x + c\right ) + A\right )} {\left (b \tan \left (d x + c\right ) + a\right )}^{n}}{\sqrt {\cot \left (d x + c\right )}} \,d x } \]
\[ \int \frac {(a+b \tan (c+d x))^n (A+B \tan (c+d x))}{\sqrt {\cot (c+d x)}} \, dx=\int { \frac {{\left (B \tan \left (d x + c\right ) + A\right )} {\left (b \tan \left (d x + c\right ) + a\right )}^{n}}{\sqrt {\cot \left (d x + c\right )}} \,d x } \]
Timed out. \[ \int \frac {(a+b \tan (c+d x))^n (A+B \tan (c+d x))}{\sqrt {\cot (c+d x)}} \, dx=\int \frac {\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^n}{\sqrt {\mathrm {cot}\left (c+d\,x\right )}} \,d x \]