Integrand size = 23, antiderivative size = 153 \[ \int \frac {(a+b \tan (c+d x))^n}{\sqrt {\tan (c+d x)}} \, dx=\frac {\operatorname {AppellF1}\left (\frac {1}{2},1,-n,\frac {3}{2},-i \tan (c+d x),-\frac {b \tan (c+d x)}{a}\right ) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^n \left (1+\frac {b \tan (c+d x)}{a}\right )^{-n}}{d}+\frac {\operatorname {AppellF1}\left (\frac {1}{2},1,-n,\frac {3}{2},i \tan (c+d x),-\frac {b \tan (c+d x)}{a}\right ) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^n \left (1+\frac {b \tan (c+d x)}{a}\right )^{-n}}{d} \]
AppellF1(1/2,1,-n,3/2,-I*tan(d*x+c),-b*tan(d*x+c)/a)*tan(d*x+c)^(1/2)*(a+b *tan(d*x+c))^n/d/((1+b*tan(d*x+c)/a)^n)+AppellF1(1/2,1,-n,3/2,I*tan(d*x+c) ,-b*tan(d*x+c)/a)*tan(d*x+c)^(1/2)*(a+b*tan(d*x+c))^n/d/((1+b*tan(d*x+c)/a )^n)
\[ \int \frac {(a+b \tan (c+d x))^n}{\sqrt {\tan (c+d x)}} \, dx=\int \frac {(a+b \tan (c+d x))^n}{\sqrt {\tan (c+d x)}} \, dx \]
Time = 0.37 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.99, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3042, 4058, 615, 2009}
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.
\(\displaystyle \int \frac {(a+b \tan (c+d x))^n}{\sqrt {\tan (c+d x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a+b \tan (c+d x))^n}{\sqrt {\tan (c+d x)}}dx\) |
\(\Big \downarrow \) 4058 |
\(\displaystyle \frac {\int \frac {(a+b \tan (c+d x))^n}{\sqrt {\tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}d\tan (c+d x)}{d}\) |
\(\Big \downarrow \) 615 |
\(\displaystyle \frac {\int \left (\frac {i (a+b \tan (c+d x))^n}{2 (i-\tan (c+d x)) \sqrt {\tan (c+d x)}}+\frac {i (a+b \tan (c+d x))^n}{2 \sqrt {\tan (c+d x)} (\tan (c+d x)+i)}\right )d\tan (c+d x)}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^n \left (\frac {b \tan (c+d x)}{a}+1\right )^{-n} \operatorname {AppellF1}\left (\frac {1}{2},1,-n,\frac {3}{2},-i \tan (c+d x),-\frac {b \tan (c+d x)}{a}\right )+\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^n \left (\frac {b \tan (c+d x)}{a}+1\right )^{-n} \operatorname {AppellF1}\left (\frac {1}{2},1,-n,\frac {3}{2},i \tan (c+d x),-\frac {b \tan (c+d x)}{a}\right )}{d}\) |
((AppellF1[1/2, 1, -n, 3/2, (-I)*Tan[c + d*x], -((b*Tan[c + d*x])/a)]*Sqrt [Tan[c + d*x]]*(a + b*Tan[c + d*x])^n)/(1 + (b*Tan[c + d*x])/a)^n + (Appel lF1[1/2, 1, -n, 3/2, I*Tan[c + d*x], -((b*Tan[c + d*x])/a)]*Sqrt[Tan[c + d *x]]*(a + b*Tan[c + d*x])^n)/(1 + (b*Tan[c + d*x])/a)^n)/d
3.8.17.3.1 Defintions of rubi rules used
Int[((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(c + d*x)^n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && ILtQ[p, 0]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, S imp[ff/f Subst[Int[(a + b*ff*x)^m*((c + d*ff*x)^n/(1 + ff^2*x^2)), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && NeQ[b*c - a *d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]
\[\int \frac {\left (a +b \tan \left (d x +c \right )\right )^{n}}{\sqrt {\tan \left (d x +c \right )}}d x\]
\[ \int \frac {(a+b \tan (c+d x))^n}{\sqrt {\tan (c+d x)}} \, dx=\int { \frac {{\left (b \tan \left (d x + c\right ) + a\right )}^{n}}{\sqrt {\tan \left (d x + c\right )}} \,d x } \]
\[ \int \frac {(a+b \tan (c+d x))^n}{\sqrt {\tan (c+d x)}} \, dx=\int \frac {\left (a + b \tan {\left (c + d x \right )}\right )^{n}}{\sqrt {\tan {\left (c + d x \right )}}}\, dx \]
\[ \int \frac {(a+b \tan (c+d x))^n}{\sqrt {\tan (c+d x)}} \, dx=\int { \frac {{\left (b \tan \left (d x + c\right ) + a\right )}^{n}}{\sqrt {\tan \left (d x + c\right )}} \,d x } \]
\[ \int \frac {(a+b \tan (c+d x))^n}{\sqrt {\tan (c+d x)}} \, dx=\int { \frac {{\left (b \tan \left (d x + c\right ) + a\right )}^{n}}{\sqrt {\tan \left (d x + c\right )}} \,d x } \]
Timed out. \[ \int \frac {(a+b \tan (c+d x))^n}{\sqrt {\tan (c+d x)}} \, dx=\int \frac {{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^n}{\sqrt {\mathrm {tan}\left (c+d\,x\right )}} \,d x \]