Integrand size = 27, antiderivative size = 287 \[ \int \frac {(a+b \tan (e+f x))^m}{(c+d \tan (e+f x))^{5/2}} \, dx=\frac {b^2 \operatorname {AppellF1}\left (1+m,\frac {5}{2},1,2+m,-\frac {d (a+b \tan (e+f x))}{b c-a d},\frac {a+b \tan (e+f x)}{a-i b}\right ) (a+b \tan (e+f x))^{1+m} \sqrt {\frac {b (c+d \tan (e+f x))}{b c-a d}}}{2 (i a+b) (b c-a d)^2 f (1+m) \sqrt {c+d \tan (e+f x)}}-\frac {b^2 \operatorname {AppellF1}\left (1+m,\frac {5}{2},1,2+m,-\frac {d (a+b \tan (e+f x))}{b c-a d},\frac {a+b \tan (e+f x)}{a+i b}\right ) (a+b \tan (e+f x))^{1+m} \sqrt {\frac {b (c+d \tan (e+f x))}{b c-a d}}}{2 (i a-b) (b c-a d)^2 f (1+m) \sqrt {c+d \tan (e+f x)}} \]
1/2*b^2*AppellF1(1+m,1,5/2,2+m,(a+b*tan(f*x+e))/(a-I*b),-d*(a+b*tan(f*x+e) )/(-a*d+b*c))*(b*(c+d*tan(f*x+e))/(-a*d+b*c))^(1/2)*(a+b*tan(f*x+e))^(1+m) /(I*a+b)/(-a*d+b*c)^2/f/(1+m)/(c+d*tan(f*x+e))^(1/2)-1/2*b^2*AppellF1(1+m, 1,5/2,2+m,(a+b*tan(f*x+e))/(a+I*b),-d*(a+b*tan(f*x+e))/(-a*d+b*c))*(b*(c+d *tan(f*x+e))/(-a*d+b*c))^(1/2)*(a+b*tan(f*x+e))^(1+m)/(I*a-b)/(-a*d+b*c)^2 /f/(1+m)/(c+d*tan(f*x+e))^(1/2)
\[ \int \frac {(a+b \tan (e+f x))^m}{(c+d \tan (e+f x))^{5/2}} \, dx=\int \frac {(a+b \tan (e+f x))^m}{(c+d \tan (e+f x))^{5/2}} \, dx \]
Time = 0.48 (sec) , antiderivative size = 285, normalized size of antiderivative = 0.99, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {3042, 4058, 662, 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.
\(\displaystyle \int \frac {(a+b \tan (e+f x))^m}{(c+d \tan (e+f x))^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a+b \tan (e+f x))^m}{(c+d \tan (e+f x))^{5/2}}dx\) |
\(\Big \downarrow \) 4058 |
\(\displaystyle \frac {\int \frac {(a+b \tan (e+f x))^m}{(c+d \tan (e+f x))^{5/2} \left (\tan ^2(e+f x)+1\right )}d\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 662 |
\(\displaystyle \frac {\int \left (\frac {i (a+b \tan (e+f x))^m}{2 (i-\tan (e+f x)) (c+d \tan (e+f x))^{5/2}}+\frac {i (a+b \tan (e+f x))^m}{2 (\tan (e+f x)+i) (c+d \tan (e+f x))^{5/2}}\right )d\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {b^2 (a+b \tan (e+f x))^{m+1} \sqrt {\frac {b (c+d \tan (e+f x))}{b c-a d}} \operatorname {AppellF1}\left (m+1,\frac {5}{2},1,m+2,-\frac {d (a+b \tan (e+f x))}{b c-a d},\frac {a+b \tan (e+f x)}{a-i b}\right )}{2 (m+1) (b+i a) (b c-a d)^2 \sqrt {c+d \tan (e+f x)}}-\frac {b^2 (a+b \tan (e+f x))^{m+1} \sqrt {\frac {b (c+d \tan (e+f x))}{b c-a d}} \operatorname {AppellF1}\left (m+1,\frac {5}{2},1,m+2,-\frac {d (a+b \tan (e+f x))}{b c-a d},\frac {a+b \tan (e+f x)}{a+i b}\right )}{2 (m+1) (-b+i a) (b c-a d)^2 \sqrt {c+d \tan (e+f x)}}}{f}\) |
((b^2*AppellF1[1 + m, 5/2, 1, 2 + m, -((d*(a + b*Tan[e + f*x]))/(b*c - a*d )), (a + b*Tan[e + f*x])/(a - I*b)]*(a + b*Tan[e + f*x])^(1 + m)*Sqrt[(b*( c + d*Tan[e + f*x]))/(b*c - a*d)])/(2*(I*a + b)*(b*c - a*d)^2*(1 + m)*Sqrt [c + d*Tan[e + f*x]]) - (b^2*AppellF1[1 + m, 5/2, 1, 2 + m, -((d*(a + b*Ta n[e + f*x]))/(b*c - a*d)), (a + b*Tan[e + f*x])/(a + I*b)]*(a + b*Tan[e + f*x])^(1 + m)*Sqrt[(b*(c + d*Tan[e + f*x]))/(b*c - a*d)])/(2*(I*a - b)*(b* c - a*d)^2*(1 + m)*Sqrt[c + d*Tan[e + f*x]]))/f
3.14.16.3.1 Defintions of rubi rules used
Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_))/((a_) + (c_.)*(x_ )^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n, 1/(a + c*x^ 2), x], x] /; FreeQ[{a, c, d, e, f, g, m, n}, x] && !IntegerQ[m] && !Inte gerQ[n]
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]
\[\int \frac {\left (a +b \tan \left (f x +e \right )\right )^{m}}{\left (c +d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}d x\]
\[ \int \frac {(a+b \tan (e+f x))^m}{(c+d \tan (e+f x))^{5/2}} \, dx=\int { \frac {{\left (b \tan \left (f x + e\right ) + a\right )}^{m}}{{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}}} \,d x } \]
integral(sqrt(d*tan(f*x + e) + c)*(b*tan(f*x + e) + a)^m/(d^3*tan(f*x + e) ^3 + 3*c*d^2*tan(f*x + e)^2 + 3*c^2*d*tan(f*x + e) + c^3), x)
\[ \int \frac {(a+b \tan (e+f x))^m}{(c+d \tan (e+f x))^{5/2}} \, dx=\int \frac {\left (a + b \tan {\left (e + f x \right )}\right )^{m}}{\left (c + d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \]
\[ \int \frac {(a+b \tan (e+f x))^m}{(c+d \tan (e+f x))^{5/2}} \, dx=\int { \frac {{\left (b \tan \left (f x + e\right ) + a\right )}^{m}}{{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {(a+b \tan (e+f x))^m}{(c+d \tan (e+f x))^{5/2}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {(a+b \tan (e+f x))^m}{(c+d \tan (e+f x))^{5/2}} \, dx=\int \frac {{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^m}{{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}} \,d x \]