Integrand size = 28, antiderivative size = 96 \[ \int (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))^n \, dx=\frac {i 2^{\frac {9}{4}+n} a \operatorname {Hypergeometric2F1}\left (\frac {5}{4},-\frac {1}{4}-n,\frac {9}{4},\frac {1}{2} (1-i \tan (c+d x))\right ) (e \sec (c+d x))^{5/2} (1+i \tan (c+d x))^{-\frac {1}{4}-n} (a+i a \tan (c+d x))^{-1+n}}{5 d} \]
1/5*I*2^(9/4+n)*a*hypergeom([5/4, -1/4-n],[9/4],1/2-1/2*I*tan(d*x+c))*(e*s ec(d*x+c))^(5/2)*(1+I*tan(d*x+c))^(-1/4-n)*(a+I*a*tan(d*x+c))^(-1+n)/d
Time = 10.90 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.89 \[ \int (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))^n \, dx=-\frac {i 2^{\frac {7}{2}+n} e^{2 i (c+d x)} \left (e^{i d x}\right )^n \left (\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{\frac {1}{2}+n} \left (1+e^{2 i (c+d x)}\right )^{\frac {1}{2}+n} \operatorname {Hypergeometric2F1}\left (\frac {5}{4}+n,\frac {5}{2}+n,\frac {9}{4}+n,-e^{2 i (c+d x)}\right ) \sec ^{-\frac {5}{2}-n}(c+d x) (e \sec (c+d x))^{5/2} (\cos (d x)+i \sin (d x))^{-n} (a+i a \tan (c+d x))^n}{d (5+4 n)} \]
((-I)*2^(7/2 + n)*E^((2*I)*(c + d*x))*(E^(I*d*x))^n*(E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x))))^(1/2 + n)*(1 + E^((2*I)*(c + d*x)))^(1/2 + n)*Hyperg eometric2F1[5/4 + n, 5/2 + n, 9/4 + n, -E^((2*I)*(c + d*x))]*Sec[c + d*x]^ (-5/2 - n)*(e*Sec[c + d*x])^(5/2)*(a + I*a*Tan[c + d*x])^n)/(d*(5 + 4*n)*( Cos[d*x] + I*Sin[d*x])^n)
Time = 0.49 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {3042, 3986, 3042, 4006, 80, 79}
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.
\(\displaystyle \int (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))^n \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))^ndx\) |
\(\Big \downarrow \) 3986 |
\(\displaystyle \frac {(e \sec (c+d x))^{5/2} \int (a-i a \tan (c+d x))^{5/4} (i \tan (c+d x) a+a)^{n+\frac {5}{4}}dx}{(a-i a \tan (c+d x))^{5/4} (a+i a \tan (c+d x))^{5/4}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {(e \sec (c+d x))^{5/2} \int (a-i a \tan (c+d x))^{5/4} (i \tan (c+d x) a+a)^{n+\frac {5}{4}}dx}{(a-i a \tan (c+d x))^{5/4} (a+i a \tan (c+d x))^{5/4}}\) |
\(\Big \downarrow \) 4006 |
\(\displaystyle \frac {a^2 (e \sec (c+d x))^{5/2} \int \sqrt [4]{a-i a \tan (c+d x)} (i \tan (c+d x) a+a)^{n+\frac {1}{4}}d\tan (c+d x)}{d (a-i a \tan (c+d x))^{5/4} (a+i a \tan (c+d x))^{5/4}}\) |
\(\Big \downarrow \) 80 |
\(\displaystyle \frac {a^2 2^{n+\frac {1}{4}} (e \sec (c+d x))^{5/2} (1+i \tan (c+d x))^{-n-\frac {1}{4}} (a+i a \tan (c+d x))^{n-1} \int \left (\frac {1}{2} i \tan (c+d x)+\frac {1}{2}\right )^{n+\frac {1}{4}} \sqrt [4]{a-i a \tan (c+d x)}d\tan (c+d x)}{d (a-i a \tan (c+d x))^{5/4}}\) |
\(\Big \downarrow \) 79 |
\(\displaystyle \frac {i a 2^{n+\frac {9}{4}} (e \sec (c+d x))^{5/2} (1+i \tan (c+d x))^{-n-\frac {1}{4}} (a+i a \tan (c+d x))^{n-1} \operatorname {Hypergeometric2F1}\left (\frac {5}{4},-n-\frac {1}{4},\frac {9}{4},\frac {1}{2} (1-i \tan (c+d x))\right )}{5 d}\) |
((I/5)*2^(9/4 + n)*a*Hypergeometric2F1[5/4, -1/4 - n, 9/4, (1 - I*Tan[c + d*x])/2]*(e*Sec[c + d*x])^(5/2)*(1 + I*Tan[c + d*x])^(-1/4 - n)*(a + I*a*T an[c + d*x])^(-1 + n))/d
3.5.77.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(( a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c - a*d))^n))*Hypergeometric2F1[-n, m + 1 , m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m, n}, x] && !IntegerQ[m] && !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] || !(RationalQ[n] && GtQ[-d/(b*c - a*d), 0]))
Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(c + d*x)^FracPart[n]/((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, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && !IntegerQ[m] && !Integ erQ[n] && (RationalQ[m] || !SimplerQ[n + 1, m + 1])
Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*( x_)])^(n_.), x_Symbol] :> Simp[(d*Sec[e + f*x])^m/((a + b*Tan[e + f*x])^(m/ 2)*(a - b*Tan[e + f*x])^(m/2)) Int[(a + b*Tan[e + f*x])^(m/2 + n)*(a - b* Tan[e + f*x])^(m/2), x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 + b^2, 0]
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + ( f_.)*(x_)])^(n_), x_Symbol] :> Simp[a*(c/f) Subst[Int[(a + b*x)^(m - 1)*( c + d*x)^(n - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n }, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 + b^2, 0]
\[\int \left (e \sec \left (d x +c \right )\right )^{\frac {5}{2}} \left (a +i a \tan \left (d x +c \right )\right )^{n}d x\]
\[ \int (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))^n \, dx=\int { \left (e \sec \left (d x + c\right )\right )^{\frac {5}{2}} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n} \,d x } \]
integral(4*sqrt(2)*e^2*(2*a*e^(2*I*d*x + 2*I*c)/(e^(2*I*d*x + 2*I*c) + 1)) ^n*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(5/2*I*d*x + 5/2*I*c)/(e^(4*I*d*x + 4*I*c) + 2*e^(2*I*d*x + 2*I*c) + 1), x)
Timed out. \[ \int (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))^n \, dx=\text {Timed out} \]
\[ \int (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))^n \, dx=\int { \left (e \sec \left (d x + c\right )\right )^{\frac {5}{2}} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n} \,d x } \]
\[ \int (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))^n \, dx=\int { \left (e \sec \left (d x + c\right )\right )^{\frac {5}{2}} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n} \,d x } \]
Timed out. \[ \int (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))^n \, dx=\int {\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{5/2}\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^n \,d x \]