Integrand size = 23, antiderivative size = 173 \[ \int \tan ^m(c+d x) \sqrt {a+b \tan (c+d x)} \, dx=\frac {\operatorname {AppellF1}\left (1+m,-\frac {1}{2},1,2+m,-\frac {b \tan (c+d x)}{a},-i \tan (c+d x)\right ) \tan ^{1+m}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d (1+m) \sqrt {1+\frac {b \tan (c+d x)}{a}}}+\frac {\operatorname {AppellF1}\left (1+m,-\frac {1}{2},1,2+m,-\frac {b \tan (c+d x)}{a},i \tan (c+d x)\right ) \tan ^{1+m}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d (1+m) \sqrt {1+\frac {b \tan (c+d x)}{a}}} \]
1/2*AppellF1(1+m,1,-1/2,2+m,-I*tan(d*x+c),-b*tan(d*x+c)/a)*(a+b*tan(d*x+c) )^(1/2)*tan(d*x+c)^(1+m)/d/(1+m)/(1+b*tan(d*x+c)/a)^(1/2)+1/2*AppellF1(1+m ,1,-1/2,2+m,I*tan(d*x+c),-b*tan(d*x+c)/a)*(a+b*tan(d*x+c))^(1/2)*tan(d*x+c )^(1+m)/d/(1+m)/(1+b*tan(d*x+c)/a)^(1/2)
\[ \int \tan ^m(c+d x) \sqrt {a+b \tan (c+d x)} \, dx=\int \tan ^m(c+d x) \sqrt {a+b \tan (c+d x)} \, dx \]
Time = 0.37 (sec) , antiderivative size = 171, 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.
\(\displaystyle \int \tan ^m(c+d x) \sqrt {a+b \tan (c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \tan (c+d x)^m \sqrt {a+b \tan (c+d x)}dx\) |
\(\Big \downarrow \) 4058 |
\(\displaystyle \frac {\int \frac {\tan ^m(c+d x) \sqrt {a+b \tan (c+d x)}}{\tan ^2(c+d x)+1}d\tan (c+d x)}{d}\) |
\(\Big \downarrow \) 615 |
\(\displaystyle \frac {\int \left (\frac {i \sqrt {a+b \tan (c+d x)} \tan ^m(c+d x)}{2 (i-\tan (c+d x))}+\frac {i \sqrt {a+b \tan (c+d x)} \tan ^m(c+d x)}{2 (\tan (c+d x)+i)}\right )d\tan (c+d x)}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {\tan ^{m+1}(c+d x) \sqrt {a+b \tan (c+d x)} \operatorname {AppellF1}\left (m+1,-\frac {1}{2},1,m+2,-\frac {b \tan (c+d x)}{a},-i \tan (c+d x)\right )}{2 (m+1) \sqrt {\frac {b \tan (c+d x)}{a}+1}}+\frac {\tan ^{m+1}(c+d x) \sqrt {a+b \tan (c+d x)} \operatorname {AppellF1}\left (m+1,-\frac {1}{2},1,m+2,-\frac {b \tan (c+d x)}{a},i \tan (c+d x)\right )}{2 (m+1) \sqrt {\frac {b \tan (c+d x)}{a}+1}}}{d}\) |
((AppellF1[1 + m, -1/2, 1, 2 + m, -((b*Tan[c + d*x])/a), (-I)*Tan[c + d*x] ]*Tan[c + d*x]^(1 + m)*Sqrt[a + b*Tan[c + d*x]])/(2*(1 + m)*Sqrt[1 + (b*Ta n[c + d*x])/a]) + (AppellF1[1 + m, -1/2, 1, 2 + m, -((b*Tan[c + d*x])/a), I*Tan[c + d*x]]*Tan[c + d*x]^(1 + m)*Sqrt[a + b*Tan[c + d*x]])/(2*(1 + m)* Sqrt[1 + (b*Tan[c + d*x])/a]))/d
3.8.3.3.1 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]
\[\int \left (\tan ^{m}\left (d x +c \right )\right ) \sqrt {a +b \tan \left (d x +c \right )}d x\]
\[ \int \tan ^m(c+d x) \sqrt {a+b \tan (c+d x)} \, dx=\int { \sqrt {b \tan \left (d x + c\right ) + a} \tan \left (d x + c\right )^{m} \,d x } \]
\[ \int \tan ^m(c+d x) \sqrt {a+b \tan (c+d x)} \, dx=\int \sqrt {a + b \tan {\left (c + d x \right )}} \tan ^{m}{\left (c + d x \right )}\, dx \]
\[ \int \tan ^m(c+d x) \sqrt {a+b \tan (c+d x)} \, dx=\int { \sqrt {b \tan \left (d x + c\right ) + a} \tan \left (d x + c\right )^{m} \,d x } \]
\[ \int \tan ^m(c+d x) \sqrt {a+b \tan (c+d x)} \, dx=\int { \sqrt {b \tan \left (d x + c\right ) + a} \tan \left (d x + c\right )^{m} \,d x } \]
Timed out. \[ \int \tan ^m(c+d x) \sqrt {a+b \tan (c+d x)} \, dx=\int {\mathrm {tan}\left (c+d\,x\right )}^m\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )} \,d x \]