Integrand size = 24, antiderivative size = 94 \[ \int \cos ^5(c+d x) (a+i a \tan (c+d x))^n \, dx=-\frac {i 2^{-\frac {5}{2}+n} \cos ^5(c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},\frac {7}{2}-n,-\frac {3}{2},\frac {1}{2} (1-i \tan (c+d x))\right ) (1+i \tan (c+d x))^{\frac {1}{2}-n} (a+i a \tan (c+d x))^{2+n}}{5 a^2 d} \]
-1/5*I*2^(-5/2+n)*cos(d*x+c)^5*hypergeom([-5/2, 7/2-n],[-3/2],1/2-1/2*I*ta n(d*x+c))*(1+I*tan(d*x+c))^(1/2-n)*(a+I*a*tan(d*x+c))^(2+n)/a^2/d
Time = 15.68 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.59 \[ \int \cos ^5(c+d x) (a+i a \tan (c+d x))^n \, dx=-\frac {i 2^{-5+n} e^{-5 i (c+d x)} \left (e^{i d x}\right )^n \left (\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^n \left (1+e^{2 i (c+d x)}\right )^6 \operatorname {Hypergeometric2F1}\left (1,\frac {7}{2},-\frac {3}{2}+n,-e^{2 i (c+d x)}\right ) \sec ^{-n}(c+d x) (\cos (d x)+i \sin (d x))^{-n} (a+i a \tan (c+d x))^n}{d (-5+2 n)} \]
((-I)*2^(-5 + n)*(E^(I*d*x))^n*(E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))) ^n*(1 + E^((2*I)*(c + d*x)))^6*Hypergeometric2F1[1, 7/2, -3/2 + n, -E^((2* I)*(c + d*x))]*(a + I*a*Tan[c + d*x])^n)/(d*E^((5*I)*(c + d*x))*(-5 + 2*n) *Sec[c + d*x]^n*(Cos[d*x] + I*Sin[d*x])^n)
Time = 0.48 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, 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.
\(\displaystyle \int \cos ^5(c+d x) (a+i a \tan (c+d x))^n \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a+i a \tan (c+d x))^n}{\sec (c+d x)^5}dx\) |
\(\Big \downarrow \) 3986 |
\(\displaystyle \cos ^5(c+d x) (a-i a \tan (c+d x))^{5/2} (a+i a \tan (c+d x))^{5/2} \int \frac {(i \tan (c+d x) a+a)^{n-\frac {5}{2}}}{(a-i a \tan (c+d x))^{5/2}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \cos ^5(c+d x) (a-i a \tan (c+d x))^{5/2} (a+i a \tan (c+d x))^{5/2} \int \frac {(i \tan (c+d x) a+a)^{n-\frac {5}{2}}}{(a-i a \tan (c+d x))^{5/2}}dx\) |
\(\Big \downarrow \) 4006 |
\(\displaystyle \frac {a^2 \cos ^5(c+d x) (a-i a \tan (c+d x))^{5/2} (a+i a \tan (c+d x))^{5/2} \int \frac {(i \tan (c+d x) a+a)^{n-\frac {7}{2}}}{(a-i a \tan (c+d x))^{7/2}}d\tan (c+d x)}{d}\) |
\(\Big \downarrow \) 80 |
\(\displaystyle \frac {2^{n-\frac {7}{2}} \cos ^5(c+d x) (a-i a \tan (c+d x))^{5/2} (1+i \tan (c+d x))^{\frac {1}{2}-n} (a+i a \tan (c+d x))^{n+2} \int \frac {\left (\frac {1}{2} i \tan (c+d x)+\frac {1}{2}\right )^{n-\frac {7}{2}}}{(a-i a \tan (c+d x))^{7/2}}d\tan (c+d x)}{a d}\) |
\(\Big \downarrow \) 79 |
\(\displaystyle -\frac {i 2^{n-\frac {5}{2}} \cos ^5(c+d x) (1+i \tan (c+d x))^{\frac {1}{2}-n} (a+i a \tan (c+d x))^{n+2} \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},\frac {7}{2}-n,-\frac {3}{2},\frac {1}{2} (1-i \tan (c+d x))\right )}{5 a^2 d}\) |
((-1/5*I)*2^(-5/2 + n)*Cos[c + d*x]^5*Hypergeometric2F1[-5/2, 7/2 - n, -3/ 2, (1 - I*Tan[c + d*x])/2]*(1 + I*Tan[c + d*x])^(1/2 - n)*(a + I*a*Tan[c + d*x])^(2 + n))/(a^2*d)
3.5.76.3.1 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]
\[\int \left (\cos ^{5}\left (d x +c \right )\right ) \left (a +i a \tan \left (d x +c \right )\right )^{n}d x\]
\[ \int \cos ^5(c+d x) (a+i a \tan (c+d x))^n \, dx=\int { {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n} \cos \left (d x + c\right )^{5} \,d x } \]
integral(1/32*(2*a*e^(2*I*d*x + 2*I*c)/(e^(2*I*d*x + 2*I*c) + 1))^n*(e^(10 *I*d*x + 10*I*c) + 5*e^(8*I*d*x + 8*I*c) + 10*e^(6*I*d*x + 6*I*c) + 10*e^( 4*I*d*x + 4*I*c) + 5*e^(2*I*d*x + 2*I*c) + 1)*e^(-5*I*d*x - 5*I*c), x)
Timed out. \[ \int \cos ^5(c+d x) (a+i a \tan (c+d x))^n \, dx=\text {Timed out} \]
\[ \int \cos ^5(c+d x) (a+i a \tan (c+d x))^n \, dx=\int { {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n} \cos \left (d x + c\right )^{5} \,d x } \]
\[ \int \cos ^5(c+d x) (a+i a \tan (c+d x))^n \, dx=\int { {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n} \cos \left (d x + c\right )^{5} \,d x } \]
Timed out. \[ \int \cos ^5(c+d x) (a+i a \tan (c+d x))^n \, dx=\int {\cos \left (c+d\,x\right )}^5\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^n \,d x \]