Integrand size = 30, antiderivative size = 125 \[ \int \frac {(a+i a \tan (e+f x))^m}{(c+d \tan (e+f x))^{5/2}} \, dx=-\frac {i \operatorname {AppellF1}\left (m,\frac {5}{2},1,1+m,-\frac {d (1+i \tan (e+f x))}{i c-d},\frac {1}{2} (1+i \tan (e+f x))\right ) (a+i a \tan (e+f x))^m \sqrt {\frac {c+d \tan (e+f x)}{c+i d}}}{2 (c+i d)^2 f m \sqrt {c+d \tan (e+f x)}} \]
-1/2*I*AppellF1(m,5/2,1,1+m,-d*(1+I*tan(f*x+e))/(I*c-d),1/2+1/2*I*tan(f*x+ e))*((c+d*tan(f*x+e))/(c+I*d))^(1/2)*(a+I*a*tan(f*x+e))^m/(c+I*d)^2/f/m/(c +d*tan(f*x+e))^(1/2)
\[ \int \frac {(a+i a \tan (e+f x))^m}{(c+d \tan (e+f x))^{5/2}} \, dx=\int \frac {(a+i a \tan (e+f x))^m}{(c+d \tan (e+f x))^{5/2}} \, dx \]
Time = 0.32 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.11, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 4047, 25, 27, 154, 153}
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.
| \(\displaystyle \int \frac {(a+i a \tan (e+f x))^m}{(c+d \tan (e+f x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a+i a \tan (e+f x))^m}{(c+d \tan (e+f x))^{5/2}}dx\) |
| \(\Big \downarrow \) 4047 |
| \(\displaystyle \frac {i a^2 \int -\frac {(i \tan (e+f x) a+a)^{m-1}}{a (a-i a \tan (e+f x)) (c+d \tan (e+f x))^{5/2}}d(i a \tan (e+f x))}{f}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {i a^2 \int \frac {(i \tan (e+f x) a+a)^{m-1}}{a (a-i a \tan (e+f x)) (c+d \tan (e+f x))^{5/2}}d(i a \tan (e+f x))}{f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {i a \int \frac {(i \tan (e+f x) a+a)^{m-1}}{(a-i a \tan (e+f x)) (c+d \tan (e+f x))^{5/2}}d(i a \tan (e+f x))}{f}\) |
| \(\Big \downarrow \) 154 |
| \(\displaystyle -\frac {i a \sqrt {\frac {a c+a d \tan (e+f x)}{a (c+i d)}} \int \frac {(i \tan (e+f x) a+a)^{m-1}}{(a-i a \tan (e+f x)) \left (\frac {c}{c+i d}+\frac {i d \tan (e+f x)}{i c-d}\right )^{5/2}}d(i a \tan (e+f x))}{f (c+i d)^2 \sqrt {c+d \tan (e+f x)}}\) |
| \(\Big \downarrow \) 153 |
| \(\displaystyle -\frac {i (a+i a \tan (e+f x))^m \sqrt {\frac {a c+a d \tan (e+f x)}{a (c+i d)}} \operatorname {AppellF1}\left (m,\frac {5}{2},1,m+1,-\frac {d (i \tan (e+f x) a+a)}{a (i c-d)},\frac {i \tan (e+f x) a+a}{2 a}\right )}{2 f m (c+i d)^2 \sqrt {c+d \tan (e+f x)}}\) |
((-1/2*I)*AppellF1[m, 5/2, 1, 1 + m, -((d*(a + I*a*Tan[e + f*x]))/(a*(I*c - d))), (a + I*a*Tan[e + f*x])/(2*a)]*(a + I*a*Tan[e + f*x])^m*Sqrt[(a*c + a*d*Tan[e + f*x])/(a*(c + I*d))])/((c + I*d)^2*f*m*Sqrt[c + d*Tan[e + f*x ]])
3.12.90.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) ^(p_), x_] :> Simp[(b*e - a*f)^p*((a + b*x)^(m + 1)/(b^(p + 1)*(m + 1)*Simp lify[b/(b*c - a*d)]^n))*AppellF1[m + 1, -n, -p, m + 2, (-d)*((a + b*x)/(b*c - a*d)), (-f)*((a + b*x)/(b*e - a*f))], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && !IntegerQ[m] && !IntegerQ[n] && IntegerQ[p] && GtQ[Simplify[b/( b*c - a*d)], 0] && !(GtQ[Simplify[d/(d*a - c*b)], 0] && SimplerQ[c + d*x, a + b*x])
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) ^(p_), x_] :> Simp[(c + d*x)^FracPart[n]/(Simplify[b/(b*c - a*d)]^IntPart[n ]*(b*((c + d*x)/(b*c - a*d)))^FracPart[n]) Int[(a + b*x)^m*Simp[b*(c/(b*c - a*d)) + b*d*(x/(b*c - a*d)), x]^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && !IntegerQ[m] && !IntegerQ[n] && IntegerQ[p] && !G tQ[Simplify[b/(b*c - a*d)], 0] && !SimplerQ[c + d*x, a + b*x]
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[a*(b/f) Subst[Int[(a + x)^(m - 1)*(( c + (d/b)*x)^n/(b^2 + a*x)), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d ^2, 0]
\[\int \frac {\left (a +i a \tan \left (f x +e \right )\right )^{m}}{\left (c +d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}d x\]
\[ \int \frac {(a+i a \tan (e+f x))^m}{(c+d \tan (e+f x))^{5/2}} \, dx=\int { \frac {{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{m}}{{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}}} \,d x } \]
integral((2*a*e^(2*I*f*x + 2*I*e)/(e^(2*I*f*x + 2*I*e) + 1))^m*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*(-I*e^(6*I* f*x + 6*I*e) - 3*I*e^(4*I*f*x + 4*I*e) - 3*I*e^(2*I*f*x + 2*I*e) - I)/(-I* c^3 + 3*c^2*d + 3*I*c*d^2 - d^3 + (-I*c^3 - 3*c^2*d + 3*I*c*d^2 + d^3)*e^( 6*I*f*x + 6*I*e) - 3*(I*c^3 + c^2*d + I*c*d^2 + d^3)*e^(4*I*f*x + 4*I*e) - 3*(I*c^3 - c^2*d + I*c*d^2 - d^3)*e^(2*I*f*x + 2*I*e)), x)
\[ \int \frac {(a+i a \tan (e+f x))^m}{(c+d \tan (e+f x))^{5/2}} \, dx=\int \frac {\left (i a \left (\tan {\left (e + f x \right )} - i\right )\right )^{m}}{\left (c + d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \]
\[ \int \frac {(a+i a \tan (e+f x))^m}{(c+d \tan (e+f x))^{5/2}} \, dx=\int { \frac {{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{m}}{{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}}} \,d x } \]
Exception generated. \[ \int \frac {(a+i a \tan (e+f x))^m}{(c+d \tan (e+f x))^{5/2}} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Valuesym2poly/r2sym(con st gen &
Timed out. \[ \int \frac {(a+i a \tan (e+f x))^m}{(c+d \tan (e+f x))^{5/2}} \, dx=\int \frac {{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^m}{{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}} \,d x \]