Integrand size = 28, antiderivative size = 109 \[ \int (e \sec (c+d x))^m (a+i a \tan (c+d x))^{5/2} \, dx=\frac {i 2^{\frac {5+m}{2}} a^2 \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (-3-m),\frac {m}{2},\frac {2+m}{2},\frac {1}{2} (1-i \tan (c+d x))\right ) (e \sec (c+d x))^m (1+i \tan (c+d x))^{\frac {1}{2} (-1-m)} \sqrt {a+i a \tan (c+d x)}}{d m} \]
I*2^(5/2+1/2*m)*a^2*hypergeom([1/2*m, -3/2-1/2*m],[1+1/2*m],1/2-1/2*I*tan( d*x+c))*(e*sec(d*x+c))^m*(a+I*a*tan(d*x+c))^(1/2)*(1+I*tan(d*x+c))^(-1/2-1 /2*m)/d/m
Time = 5.21 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.71 \[ \int (e \sec (c+d x))^m (a+i a \tan (c+d x))^{5/2} \, dx=-\frac {i 2^{\frac {5}{2}+m} e^{2 i (c+2 d x)} \sqrt {e^{i d x}} \left (\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{\frac {1}{2}+m} \left (1+e^{2 i (c+d x)}\right )^{\frac {1}{2}+m} \operatorname {Hypergeometric2F1}\left (\frac {5}{2}+m,\frac {5+m}{2},\frac {7+m}{2},-e^{2 i (c+d x)}\right ) \sec ^{-\frac {5}{2}-m}(c+d x) (e \sec (c+d x))^m (a+i a \tan (c+d x))^{5/2}}{d (5+m) (\cos (d x)+i \sin (d x))^{5/2}} \]
((-I)*2^(5/2 + m)*E^((2*I)*(c + 2*d*x))*Sqrt[E^(I*d*x)]*(E^(I*(c + d*x))/( 1 + E^((2*I)*(c + d*x))))^(1/2 + m)*(1 + E^((2*I)*(c + d*x)))^(1/2 + m)*Hy pergeometric2F1[5/2 + m, (5 + m)/2, (7 + m)/2, -E^((2*I)*(c + d*x))]*Sec[c + d*x]^(-5/2 - m)*(e*Sec[c + d*x])^m*(a + I*a*Tan[c + d*x])^(5/2))/(d*(5 + m)*(Cos[d*x] + I*Sin[d*x])^(5/2))
Time = 0.48 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.11, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, 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.
\(\displaystyle \int (a+i a \tan (c+d x))^{5/2} (e \sec (c+d x))^m \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (a+i a \tan (c+d x))^{5/2} (e \sec (c+d x))^mdx\) |
\(\Big \downarrow \) 3986 |
\(\displaystyle (a-i a \tan (c+d x))^{-m/2} (a+i a \tan (c+d x))^{-m/2} (e \sec (c+d x))^m \int (a-i a \tan (c+d x))^{m/2} (i \tan (c+d x) a+a)^{\frac {m+5}{2}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle (a-i a \tan (c+d x))^{-m/2} (a+i a \tan (c+d x))^{-m/2} (e \sec (c+d x))^m \int (a-i a \tan (c+d x))^{m/2} (i \tan (c+d x) a+a)^{\frac {m+5}{2}}dx\) |
\(\Big \downarrow \) 4006 |
\(\displaystyle \frac {a^2 (a-i a \tan (c+d x))^{-m/2} (a+i a \tan (c+d x))^{-m/2} (e \sec (c+d x))^m \int (a-i a \tan (c+d x))^{\frac {m-2}{2}} (i \tan (c+d x) a+a)^{\frac {m+3}{2}}d\tan (c+d x)}{d}\) |
\(\Big \downarrow \) 80 |
\(\displaystyle \frac {a^3 2^{\frac {m+3}{2}} (1+i \tan (c+d x))^{\frac {1}{2} (-m-1)} (a-i a \tan (c+d x))^{-m/2} (a+i a \tan (c+d x))^{\frac {m+1}{2}-\frac {m}{2}} (e \sec (c+d x))^m \int \left (\frac {1}{2} i \tan (c+d x)+\frac {1}{2}\right )^{\frac {m+3}{2}} (a-i a \tan (c+d x))^{\frac {m-2}{2}}d\tan (c+d x)}{d}\) |
\(\Big \downarrow \) 79 |
\(\displaystyle \frac {i a^2 2^{\frac {m+3}{2}+1} (1+i \tan (c+d x))^{\frac {1}{2} (-m-1)} (a+i a \tan (c+d x))^{\frac {m+1}{2}-\frac {m}{2}} (e \sec (c+d x))^m \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (-m-3),\frac {m}{2},\frac {m+2}{2},\frac {1}{2} (1-i \tan (c+d x))\right )}{d m}\) |
(I*2^(1 + (3 + m)/2)*a^2*Hypergeometric2F1[(-3 - m)/2, m/2, (2 + m)/2, (1 - I*Tan[c + d*x])/2]*(e*Sec[c + d*x])^m*(1 + I*Tan[c + d*x])^((-1 - m)/2)* (a + I*a*Tan[c + d*x])^(-1/2*m + (1 + m)/2))/(d*m)
3.5.58.3.1 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]
\[\int \left (e \sec \left (d x +c \right )\right )^{m} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}d x\]
\[ \int (e \sec (c+d x))^m (a+i a \tan (c+d x))^{5/2} \, dx=\int { {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \left (e \sec \left (d x + c\right )\right )^{m} \,d x } \]
integral(4*sqrt(2)*a^2*(2*e*e^(I*d*x + I*c)/(e^(2*I*d*x + 2*I*c) + 1))^m*s qrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(5*I*d*x + 5*I*c)/(e^(4*I*d*x + 4*I*c) + 2*e^(2*I*d*x + 2*I*c) + 1), x)
Timed out. \[ \int (e \sec (c+d x))^m (a+i a \tan (c+d x))^{5/2} \, dx=\text {Timed out} \]
\[ \int (e \sec (c+d x))^m (a+i a \tan (c+d x))^{5/2} \, dx=\int { {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \left (e \sec \left (d x + c\right )\right )^{m} \,d x } \]
\[ \int (e \sec (c+d x))^m (a+i a \tan (c+d x))^{5/2} \, dx=\int { {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \left (e \sec \left (d x + c\right )\right )^{m} \,d x } \]
Timed out. \[ \int (e \sec (c+d x))^m (a+i a \tan (c+d x))^{5/2} \, dx=\int {\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^m\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2} \,d x \]