Integrand size = 28, antiderivative size = 106 \[ \int \frac {(e \sec (c+d x))^m}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {i 2^{\frac {1}{2} (-1+m)} \operatorname {Hypergeometric2F1}\left (\frac {3-m}{2},\frac {m}{2},\frac {2+m}{2},\frac {1}{2} (1-i \tan (c+d x))\right ) (e \sec (c+d x))^m (1+i \tan (c+d x))^{\frac {1-m}{2}}}{d m \sqrt {a+i a \tan (c+d x)}} \]
I*2^(-1/2+1/2*m)*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*(1+I*tan(d*x+c))^(-1/2*m+1/2)/d/m/(a+I*a*tan(d*x+c))^ (1/2)
Time = 2.49 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.64 \[ \int \frac {(e \sec (c+d x))^m}{\sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {i 2^{-\frac {1}{2}+m} \left (\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{-\frac {1}{2}+m} \left (1+e^{2 i (c+d x)}\right )^{-\frac {1}{2}+m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (-1+m),-\frac {1}{2}+m,\frac {1+m}{2},-e^{2 i (c+d x)}\right ) \sec ^{\frac {1}{2}-m}(c+d x) (e \sec (c+d x))^m \sqrt {\cos (d x)+i \sin (d x)}}{d \sqrt {e^{i d x}} (-1+m) \sqrt {a+i a \tan (c+d x)}} \]
((-I)*2^(-1/2 + m)*(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)*Hypergeometric2F1[(-1 + m)/2, -1/2 + m, (1 + m)/2, -E^((2*I)*(c + d*x))]*Sec[c + d*x]^(1/2 - m)*(e*Sec[c + d*x] )^m*Sqrt[Cos[d*x] + I*Sin[d*x]])/(d*Sqrt[E^(I*d*x)]*(-1 + m)*Sqrt[a + I*a* Tan[c + d*x]])
Time = 0.49 (sec) , antiderivative size = 118, 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.
| \(\displaystyle \int \frac {(e \sec (c+d x))^m}{\sqrt {a+i a \tan (c+d x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(e \sec (c+d x))^m}{\sqrt {a+i a \tan (c+d x)}}dx\) |
| \(\Big \downarrow \) 3986 |
| \(\displaystyle (a-i a \tan (c+d x))^{-m/2} (a+i a \tan (c+d x))^{-m/2} (e \sec (c+d x))^m \int (a-i a \tan (c+d x))^{m/2} (i \tan (c+d x) a+a)^{\frac {m-1}{2}}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle (a-i a \tan (c+d x))^{-m/2} (a+i a \tan (c+d x))^{-m/2} (e \sec (c+d x))^m \int (a-i a \tan (c+d x))^{m/2} (i \tan (c+d x) a+a)^{\frac {m-1}{2}}dx\) |
| \(\Big \downarrow \) 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 |
| \(\displaystyle \frac {a 2^{\frac {m-3}{2}} (1+i \tan (c+d x))^{\frac {1-m}{2}} (a-i a \tan (c+d x))^{-m/2} (a+i a \tan (c+d x))^{\frac {m-1}{2}-\frac {m}{2}} (e \sec (c+d x))^m \int \left (\frac {1}{2} i \tan (c+d x)+\frac {1}{2}\right )^{\frac {m-3}{2}} (a-i a \tan (c+d x))^{\frac {m-2}{2}}d\tan (c+d x)}{d}\) |
| \(\Big \downarrow \) 79 |
| \(\displaystyle \frac {i 2^{\frac {m-3}{2}+1} (1+i \tan (c+d x))^{\frac {1-m}{2}} (a+i a \tan (c+d x))^{\frac {m-1}{2}-\frac {m}{2}} (e \sec (c+d x))^m \operatorname {Hypergeometric2F1}\left (\frac {3-m}{2},\frac {m}{2},\frac {m+2}{2},\frac {1}{2} (1-i \tan (c+d x))\right )}{d m}\) |
(I*2^(1 + (-3 + m)/2)*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 + m)/2 - m/2))/(d*m)
3.5.61.3.1 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]
\[\int \frac {\left (e \sec \left (d x +c \right )\right )^{m}}{\sqrt {a +i a \tan \left (d x +c \right )}}d x\]
\[ \int \frac {(e \sec (c+d x))^m}{\sqrt {a+i a \tan (c+d x)}} \, dx=\int { \frac {\left (e \sec \left (d x + c\right )\right )^{m}}{\sqrt {i \, a \tan \left (d x + c\right ) + a}} \,d x } \]
integral(1/2*sqrt(2)*(2*e*e^(I*d*x + I*c)/(e^(2*I*d*x + 2*I*c) + 1))^m*sqr t(a/(e^(2*I*d*x + 2*I*c) + 1))*(e^(2*I*d*x + 2*I*c) + 1)*e^(-I*d*x - I*c)/ a, x)
\[ \int \frac {(e \sec (c+d x))^m}{\sqrt {a+i a \tan (c+d x)}} \, dx=\int \frac {\left (e \sec {\left (c + d x \right )}\right )^{m}}{\sqrt {i a \left (\tan {\left (c + d x \right )} - i\right )}}\, dx \]
\[ \int \frac {(e \sec (c+d x))^m}{\sqrt {a+i a \tan (c+d x)}} \, dx=\int { \frac {\left (e \sec \left (d x + c\right )\right )^{m}}{\sqrt {i \, a \tan \left (d x + c\right ) + a}} \,d x } \]
\[ \int \frac {(e \sec (c+d x))^m}{\sqrt {a+i a \tan (c+d x)}} \, dx=\int { \frac {\left (e \sec \left (d x + c\right )\right )^{m}}{\sqrt {i \, a \tan \left (d x + c\right ) + a}} \,d x } \]
Timed out. \[ \int \frac {(e \sec (c+d x))^m}{\sqrt {a+i a \tan (c+d x)}} \, dx=\int \frac {{\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^m}{\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}} \,d x \]