Integrand size = 30, antiderivative size = 118 \[ \int (e \sec (c+d x))^{1-n} (a+i a \tan (c+d x))^n \, dx=\frac {i 2^{\frac {1+n}{2}} \operatorname {Hypergeometric2F1}\left (\frac {1-n}{2},\frac {1-n}{2},\frac {3-n}{2},\frac {1}{2} (1-i \tan (c+d x))\right ) (e \sec (c+d x))^{1-n} (1+i \tan (c+d x))^{\frac {1}{2} (-1-n)} (a+i a \tan (c+d x))^n}{d (1-n)} \]
I*2^(1/2+1/2*n)*hypergeom([1/2-1/2*n, 1/2-1/2*n],[3/2-1/2*n],1/2-1/2*I*tan (d*x+c))*(e*sec(d*x+c))^(1-n)*(1+I*tan(d*x+c))^(-1/2-1/2*n)*(a+I*a*tan(d*x +c))^n/d/(1-n)
Time = 5.84 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.74 \[ \int (e \sec (c+d x))^{1-n} (a+i a \tan (c+d x))^n \, dx=-\frac {e (\operatorname {Hypergeometric2F1}(1,n,1+n,i \cos (c+d x)-\sin (c+d x))-\operatorname {Hypergeometric2F1}(1,n,1+n,-i \cos (c+d x)+\sin (c+d x))) (e \sec (c+d x))^{-n} (a+i a \tan (c+d x))^n}{d n} \]
-((e*(Hypergeometric2F1[1, n, 1 + n, I*Cos[c + d*x] - Sin[c + d*x]] - Hype rgeometric2F1[1, n, 1 + n, (-I)*Cos[c + d*x] + Sin[c + d*x]])*(a + I*a*Tan [c + d*x])^n)/(d*n*(e*Sec[c + d*x])^n))
Time = 0.55 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.32, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, 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.
| \(\displaystyle \int (a+i a \tan (c+d x))^n (e \sec (c+d x))^{1-n} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int (a+i a \tan (c+d x))^n (e \sec (c+d x))^{1-n}dx\) |
| \(\Big \downarrow \) 3986 |
| \(\displaystyle (a-i a \tan (c+d x))^{\frac {n-1}{2}} (a+i a \tan (c+d x))^{\frac {n-1}{2}} (e \sec (c+d x))^{1-n} \int (a-i a \tan (c+d x))^{\frac {1-n}{2}} (i \tan (c+d x) a+a)^{\frac {n+1}{2}}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle (a-i a \tan (c+d x))^{\frac {n-1}{2}} (a+i a \tan (c+d x))^{\frac {n-1}{2}} (e \sec (c+d x))^{1-n} \int (a-i a \tan (c+d x))^{\frac {1-n}{2}} (i \tan (c+d x) a+a)^{\frac {n+1}{2}}dx\) |
| \(\Big \downarrow \) 4006 |
| \(\displaystyle \frac {a^2 (a-i a \tan (c+d x))^{\frac {n-1}{2}} (a+i a \tan (c+d x))^{\frac {n-1}{2}} (e \sec (c+d x))^{1-n} \int (a-i a \tan (c+d x))^{\frac {1}{2} (-n-1)} (i \tan (c+d x) a+a)^{\frac {n-1}{2}}d\tan (c+d x)}{d}\) |
| \(\Big \downarrow \) 80 |
| \(\displaystyle \frac {a^2 2^{-\frac {n}{2}-\frac {1}{2}} (1-i \tan (c+d x))^{\frac {n+1}{2}} (a-i a \tan (c+d x))^{\frac {1}{2} (-n-1)+\frac {n-1}{2}} (a+i a \tan (c+d x))^{\frac {n-1}{2}} (e \sec (c+d x))^{1-n} \int \left (\frac {1}{2}-\frac {1}{2} i \tan (c+d x)\right )^{\frac {1}{2} (-n-1)} (i \tan (c+d x) a+a)^{\frac {n-1}{2}}d\tan (c+d x)}{d}\) |
| \(\Big \downarrow \) 79 |
| \(\displaystyle -\frac {i a 2^{\frac {1}{2}-\frac {n}{2}} (1-i \tan (c+d x))^{\frac {n+1}{2}} (a-i a \tan (c+d x))^{\frac {1}{2} (-n-1)+\frac {n-1}{2}} (a+i a \tan (c+d x))^{\frac {n-1}{2}+\frac {n+1}{2}} (e \sec (c+d x))^{1-n} \operatorname {Hypergeometric2F1}\left (\frac {n+1}{2},\frac {n+1}{2},\frac {n+3}{2},\frac {1}{2} (i \tan (c+d x)+1)\right )}{d (n+1)}\) |
((-I)*2^(1/2 - n/2)*a*Hypergeometric2F1[(1 + n)/2, (1 + n)/2, (3 + n)/2, ( 1 + I*Tan[c + d*x])/2]*(e*Sec[c + d*x])^(1 - n)*(1 - I*Tan[c + d*x])^((1 + n)/2)*(a - I*a*Tan[c + d*x])^((-1 - n)/2 + (-1 + n)/2)*(a + I*a*Tan[c + d *x])^((-1 + n)/2 + (1 + n)/2))/(d*(1 + n))
3.5.88.3.1 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]
\[\int \left (e \sec \left (d x +c \right )\right )^{1-n} \left (a +i a \tan \left (d x +c \right )\right )^{n}d x\]
\[ \int (e \sec (c+d x))^{1-n} (a+i a \tan (c+d x))^n \, dx=\int { \left (e \sec \left (d x + c\right )\right )^{-n + 1} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n} \,d x } \]
integral((2*e*e^(I*d*x + I*c)/(e^(2*I*d*x + 2*I*c) + 1))^(-n + 1)*e^(I*d*n *x + I*c*n + n*log(2*e*e^(I*d*x + I*c)/(e^(2*I*d*x + 2*I*c) + 1)) + n*log( a/e)), x)
\[ \int (e \sec (c+d x))^{1-n} (a+i a \tan (c+d x))^n \, dx=\int \left (e \sec {\left (c + d x \right )}\right )^{1 - n} \left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{n}\, dx \]
\[ \int (e \sec (c+d x))^{1-n} (a+i a \tan (c+d x))^n \, dx=\int { \left (e \sec \left (d x + c\right )\right )^{-n + 1} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n} \,d x } \]
-2*(a^n*e*cos(c*n + (d*n + d)*x + c) + I*a^n*e*sin(c*n + (d*n + d)*x + c) - 2*(I*a^n*d*e^(n + 1)*n - I*a^n*d*e^(n + 1) + (I*a^n*d*e^(n + 1)*n - I*a^ n*d*e^(n + 1))*cos(2*d*x + 2*c) - (a^n*d*e^(n + 1)*n - a^n*d*e^(n + 1))*si n(2*d*x + 2*c))*integrate(((cos(4*d*x + 4*c) + 2*cos(2*d*x + 2*c) + 1)*cos (c*n + (d*n + d)*x + c) + (sin(4*d*x + 4*c) + 2*sin(2*d*x + 2*c))*sin(c*n + (d*n + d)*x + c))/((e^n*n - e^n)*cos(4*d*x + 4*c)^2 + 4*(e^n*n - e^n)*co s(2*d*x + 2*c)^2 + (e^n*n - e^n)*sin(4*d*x + 4*c)^2 + 4*(e^n*n - e^n)*sin( 4*d*x + 4*c)*sin(2*d*x + 2*c) + 4*(e^n*n - e^n)*sin(2*d*x + 2*c)^2 + e^n*n + 2*(e^n*n + 2*(e^n*n - e^n)*cos(2*d*x + 2*c) - e^n)*cos(4*d*x + 4*c) + 4 *(e^n*n - e^n)*cos(2*d*x + 2*c) - e^n), x) + 2*(a^n*d*e^(n + 1)*n - a^n*d* e^(n + 1) + (a^n*d*e^(n + 1)*n - a^n*d*e^(n + 1))*cos(2*d*x + 2*c) - (-I*a ^n*d*e^(n + 1)*n + I*a^n*d*e^(n + 1))*sin(2*d*x + 2*c))*integrate(-((sin(4 *d*x + 4*c) + 2*sin(2*d*x + 2*c))*cos(c*n + (d*n + d)*x + c) - (cos(4*d*x + 4*c) + 2*cos(2*d*x + 2*c) + 1)*sin(c*n + (d*n + d)*x + c))/((e^n*n - e^n )*cos(4*d*x + 4*c)^2 + 4*(e^n*n - e^n)*cos(2*d*x + 2*c)^2 + (e^n*n - e^n)* sin(4*d*x + 4*c)^2 + 4*(e^n*n - e^n)*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 4 *(e^n*n - e^n)*sin(2*d*x + 2*c)^2 + e^n*n + 2*(e^n*n + 2*(e^n*n - e^n)*cos (2*d*x + 2*c) - e^n)*cos(4*d*x + 4*c) + 4*(e^n*n - e^n)*cos(2*d*x + 2*c) - e^n), x))/(-I*d*e^n*n + I*d*e^n + (-I*d*e^n*n + I*d*e^n)*cos(2*d*x + 2*c) + (d*e^n*n - d*e^n)*sin(2*d*x + 2*c))
\[ \int (e \sec (c+d x))^{1-n} (a+i a \tan (c+d x))^n \, dx=\int { \left (e \sec \left (d x + c\right )\right )^{-n + 1} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n} \,d x } \]
Timed out. \[ \int (e \sec (c+d x))^{1-n} (a+i a \tan (c+d x))^n \, dx=\int {\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{1-n}\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^n \,d x \]