Integrand size = 33, antiderivative size = 274 \[ \int \frac {1}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^{5/2}} \, dx=\frac {231 i \text {arctanh}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{512 \sqrt {2} a^3 c^{5/2} f}-\frac {231 i}{640 a^3 f (c-i c \tan (e+f x))^{5/2}}+\frac {i}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{5/2}}+\frac {11 i}{48 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}+\frac {33 i}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}-\frac {77 i}{256 a^3 c f (c-i c \tan (e+f x))^{3/2}}-\frac {231 i}{512 a^3 c^2 f \sqrt {c-i c \tan (e+f x)}} \]
231/1024*I*arctanh(1/2*(c-I*c*tan(f*x+e))^(1/2)*2^(1/2)/c^(1/2))/a^3/c^(5/ 2)/f*2^(1/2)-231/512*I/a^3/c^2/f/(c-I*c*tan(f*x+e))^(1/2)-231/640*I/a^3/f/ (c-I*c*tan(f*x+e))^(5/2)+1/6*I/a^3/f/(1+I*tan(f*x+e))^3/(c-I*c*tan(f*x+e)) ^(5/2)+11/48*I/a^3/f/(1+I*tan(f*x+e))^2/(c-I*c*tan(f*x+e))^(5/2)+33/64*I/a ^3/f/(1+I*tan(f*x+e))/(c-I*c*tan(f*x+e))^(5/2)-77/256*I/a^3/c/f/(c-I*c*tan (f*x+e))^(3/2)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 2.37 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.19 \[ \int \frac {1}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^{5/2}} \, dx=-\frac {i \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},4,-\frac {3}{2},-\frac {1}{2} i (i+\tan (e+f x))\right )}{40 a^3 f (c-i c \tan (e+f x))^{5/2}} \]
((-1/40*I)*Hypergeometric2F1[-5/2, 4, -3/2, (-1/2*I)*(I + Tan[e + f*x])])/ (a^3*f*(c - I*c*Tan[e + f*x])^(5/2))
Time = 0.47 (sec) , antiderivative size = 273, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {3042, 4005, 3042, 3968, 52, 52, 52, 61, 61, 61, 73, 219}
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.
| \(\displaystyle \int \frac {1}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^{5/2}}dx\) |
| \(\Big \downarrow \) 4005 |
| \(\displaystyle \frac {\int \cos ^6(e+f x) \sqrt {c-i c \tan (e+f x)}dx}{a^3 c^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\sqrt {c-i c \tan (e+f x)}}{\sec (e+f x)^6}dx}{a^3 c^3}\) |
| \(\Big \downarrow \) 3968 |
| \(\displaystyle \frac {i c^4 \int \frac {1}{(c-i c \tan (e+f x))^{7/2} (i \tan (e+f x) c+c)^4}d(-i c \tan (e+f x))}{a^3 f}\) |
| \(\Big \downarrow \) 52 |
| \(\displaystyle \frac {i c^4 \left (\frac {11 \int \frac {1}{(c-i c \tan (e+f x))^{7/2} (i \tan (e+f x) c+c)^3}d(-i c \tan (e+f x))}{12 c}+\frac {1}{6 c (c-i c \tan (e+f x))^{5/2} (c+i c \tan (e+f x))^3}\right )}{a^3 f}\) |
| \(\Big \downarrow \) 52 |
| \(\displaystyle \frac {i c^4 \left (\frac {11 \left (\frac {9 \int \frac {1}{(c-i c \tan (e+f x))^{7/2} (i \tan (e+f x) c+c)^2}d(-i c \tan (e+f x))}{8 c}+\frac {1}{4 c (c-i c \tan (e+f x))^{5/2} (c+i c \tan (e+f x))^2}\right )}{12 c}+\frac {1}{6 c (c-i c \tan (e+f x))^{5/2} (c+i c \tan (e+f x))^3}\right )}{a^3 f}\) |
| \(\Big \downarrow \) 52 |
| \(\displaystyle \frac {i c^4 \left (\frac {11 \left (\frac {9 \left (\frac {7 \int \frac {1}{(c-i c \tan (e+f x))^{7/2} (i \tan (e+f x) c+c)}d(-i c \tan (e+f x))}{4 c}+\frac {1}{2 c (c-i c \tan (e+f x))^{5/2} (c+i c \tan (e+f x))}\right )}{8 c}+\frac {1}{4 c (c-i c \tan (e+f x))^{5/2} (c+i c \tan (e+f x))^2}\right )}{12 c}+\frac {1}{6 c (c-i c \tan (e+f x))^{5/2} (c+i c \tan (e+f x))^3}\right )}{a^3 f}\) |
| \(\Big \downarrow \) 61 |
| \(\displaystyle \frac {i c^4 \left (\frac {11 \left (\frac {9 \left (\frac {7 \left (\frac {\int \frac {1}{(c-i c \tan (e+f x))^{5/2} (i \tan (e+f x) c+c)}d(-i c \tan (e+f x))}{2 c}-\frac {1}{5 c (c-i c \tan (e+f x))^{5/2}}\right )}{4 c}+\frac {1}{2 c (c-i c \tan (e+f x))^{5/2} (c+i c \tan (e+f x))}\right )}{8 c}+\frac {1}{4 c (c-i c \tan (e+f x))^{5/2} (c+i c \tan (e+f x))^2}\right )}{12 c}+\frac {1}{6 c (c-i c \tan (e+f x))^{5/2} (c+i c \tan (e+f x))^3}\right )}{a^3 f}\) |
| \(\Big \downarrow \) 61 |
| \(\displaystyle \frac {i c^4 \left (\frac {11 \left (\frac {9 \left (\frac {7 \left (\frac {\frac {\int \frac {1}{(c-i c \tan (e+f x))^{3/2} (i \tan (e+f x) c+c)}d(-i c \tan (e+f x))}{2 c}-\frac {1}{3 c (c-i c \tan (e+f x))^{3/2}}}{2 c}-\frac {1}{5 c (c-i c \tan (e+f x))^{5/2}}\right )}{4 c}+\frac {1}{2 c (c-i c \tan (e+f x))^{5/2} (c+i c \tan (e+f x))}\right )}{8 c}+\frac {1}{4 c (c-i c \tan (e+f x))^{5/2} (c+i c \tan (e+f x))^2}\right )}{12 c}+\frac {1}{6 c (c-i c \tan (e+f x))^{5/2} (c+i c \tan (e+f x))^3}\right )}{a^3 f}\) |
| \(\Big \downarrow \) 61 |
| \(\displaystyle \frac {i c^4 \left (\frac {11 \left (\frac {9 \left (\frac {7 \left (\frac {\frac {\frac {\int \frac {1}{\sqrt {c-i c \tan (e+f x)} (i \tan (e+f x) c+c)}d(-i c \tan (e+f x))}{2 c}-\frac {1}{c \sqrt {c-i c \tan (e+f x)}}}{2 c}-\frac {1}{3 c (c-i c \tan (e+f x))^{3/2}}}{2 c}-\frac {1}{5 c (c-i c \tan (e+f x))^{5/2}}\right )}{4 c}+\frac {1}{2 c (c-i c \tan (e+f x))^{5/2} (c+i c \tan (e+f x))}\right )}{8 c}+\frac {1}{4 c (c-i c \tan (e+f x))^{5/2} (c+i c \tan (e+f x))^2}\right )}{12 c}+\frac {1}{6 c (c-i c \tan (e+f x))^{5/2} (c+i c \tan (e+f x))^3}\right )}{a^3 f}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {i c^4 \left (\frac {11 \left (\frac {9 \left (\frac {7 \left (\frac {\frac {\frac {\int \frac {1}{c^2 \tan ^2(e+f x)+2 c}d\sqrt {c-i c \tan (e+f x)}}{c}-\frac {1}{c \sqrt {c-i c \tan (e+f x)}}}{2 c}-\frac {1}{3 c (c-i c \tan (e+f x))^{3/2}}}{2 c}-\frac {1}{5 c (c-i c \tan (e+f x))^{5/2}}\right )}{4 c}+\frac {1}{2 c (c-i c \tan (e+f x))^{5/2} (c+i c \tan (e+f x))}\right )}{8 c}+\frac {1}{4 c (c-i c \tan (e+f x))^{5/2} (c+i c \tan (e+f x))^2}\right )}{12 c}+\frac {1}{6 c (c-i c \tan (e+f x))^{5/2} (c+i c \tan (e+f x))^3}\right )}{a^3 f}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {i c^4 \left (\frac {11 \left (\frac {9 \left (\frac {7 \left (\frac {\frac {-\frac {i \arctan \left (\frac {\sqrt {c} \tan (e+f x)}{\sqrt {2}}\right )}{\sqrt {2} c^{3/2}}-\frac {1}{c \sqrt {c-i c \tan (e+f x)}}}{2 c}-\frac {1}{3 c (c-i c \tan (e+f x))^{3/2}}}{2 c}-\frac {1}{5 c (c-i c \tan (e+f x))^{5/2}}\right )}{4 c}+\frac {1}{2 c (c-i c \tan (e+f x))^{5/2} (c+i c \tan (e+f x))}\right )}{8 c}+\frac {1}{4 c (c-i c \tan (e+f x))^{5/2} (c+i c \tan (e+f x))^2}\right )}{12 c}+\frac {1}{6 c (c-i c \tan (e+f x))^{5/2} (c+i c \tan (e+f x))^3}\right )}{a^3 f}\) |
(I*c^4*(1/(6*c*(c - I*c*Tan[e + f*x])^(5/2)*(c + I*c*Tan[e + f*x])^3) + (1 1*(1/(4*c*(c - I*c*Tan[e + f*x])^(5/2)*(c + I*c*Tan[e + f*x])^2) + (9*(1/( 2*c*(c - I*c*Tan[e + f*x])^(5/2)*(c + I*c*Tan[e + f*x])) + (7*(-1/5*1/(c*( c - I*c*Tan[e + f*x])^(5/2)) + (-1/3*1/(c*(c - I*c*Tan[e + f*x])^(3/2)) + (((-I)*ArcTan[(Sqrt[c]*Tan[e + f*x])/Sqrt[2]])/(Sqrt[2]*c^(3/2)) - 1/(c*Sq rt[c - I*c*Tan[e + f*x]]))/(2*c))/(2*c)))/(4*c)))/(8*c)))/(12*c)))/(a^3*f)
3.10.91.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( m + n + 2)/((b*c - a*d)*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( m + n + 2)/((b*c - a*d)*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[m, -1] && !(LtQ[n, -1] && (EqQ[a, 0 ] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d , m, n, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_ ), x_Symbol] :> Simp[1/(a^(m - 2)*b*f) Subst[Int[(a - x)^(m/2 - 1)*(a + x )^(n + m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x] && EqQ[a^2 + b^2, 0] && IntegerQ[m/2]
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[a^m*c^m Int[Sec[e + f*x]^(2*m)*(c + d*Tan[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && EqQ[ b*c + a*d, 0] && EqQ[a^2 + b^2, 0] && IntegerQ[m] && !(IGtQ[n, 0] && (LtQ[ m, 0] || GtQ[m, n]))
Time = 0.57 (sec) , antiderivative size = 177, normalized size of antiderivative = 0.65
| method | result | size |
| derivativedivides | \(\frac {2 i c^{4} \left (-\frac {5}{32 c^{6} \sqrt {c -i c \tan \left (f x +e \right )}}-\frac {1}{24 c^{5} \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}-\frac {1}{80 c^{4} \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}+\frac {\frac {\frac {71 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{32}-\frac {59 c \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{6}+\frac {89 c^{2} \sqrt {c -i c \tan \left (f x +e \right )}}{8}}{\left (c +i c \tan \left (f x +e \right )\right )^{3}}+\frac {231 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{64 \sqrt {c}}}{32 c^{6}}\right )}{f \,a^{3}}\) | \(177\) |
| default | \(\frac {2 i c^{4} \left (-\frac {5}{32 c^{6} \sqrt {c -i c \tan \left (f x +e \right )}}-\frac {1}{24 c^{5} \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}-\frac {1}{80 c^{4} \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}+\frac {\frac {\frac {71 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{32}-\frac {59 c \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{6}+\frac {89 c^{2} \sqrt {c -i c \tan \left (f x +e \right )}}{8}}{\left (c +i c \tan \left (f x +e \right )\right )^{3}}+\frac {231 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{64 \sqrt {c}}}{32 c^{6}}\right )}{f \,a^{3}}\) | \(177\) |
2*I/f/a^3*c^4*(-5/32/c^6/(c-I*c*tan(f*x+e))^(1/2)-1/24/c^5/(c-I*c*tan(f*x+ e))^(3/2)-1/80/c^4/(c-I*c*tan(f*x+e))^(5/2)+1/32/c^6*(8*(71/256*(c-I*c*tan (f*x+e))^(5/2)-59/48*c*(c-I*c*tan(f*x+e))^(3/2)+89/64*c^2*(c-I*c*tan(f*x+e ))^(1/2))/(c+I*c*tan(f*x+e))^3+231/64*2^(1/2)/c^(1/2)*arctanh(1/2*(c-I*c*t an(f*x+e))^(1/2)*2^(1/2)/c^(1/2))))
Time = 0.27 (sec) , antiderivative size = 350, normalized size of antiderivative = 1.28 \[ \int \frac {1}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^{5/2}} \, dx=\frac {{\left (-3465 i \, \sqrt {\frac {1}{2}} a^{3} c^{3} f \sqrt {\frac {1}{a^{6} c^{5} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (-\frac {231 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (i \, a^{3} c^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + i \, a^{3} c^{2} f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {1}{a^{6} c^{5} f^{2}}} - i\right )} e^{\left (-i \, f x - i \, e\right )}}{256 \, a^{3} c^{2} f}\right ) + 3465 i \, \sqrt {\frac {1}{2}} a^{3} c^{3} f \sqrt {\frac {1}{a^{6} c^{5} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (-\frac {231 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (-i \, a^{3} c^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} - i \, a^{3} c^{2} f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {1}{a^{6} c^{5} f^{2}}} - i\right )} e^{\left (-i \, f x - i \, e\right )}}{256 \, a^{3} c^{2} f}\right ) + \sqrt {2} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} {\left (-48 i \, e^{\left (12 i \, f x + 12 i \, e\right )} - 464 i \, e^{\left (10 i \, f x + 10 i \, e\right )} - 3184 i \, e^{\left (8 i \, f x + 8 i \, e\right )} - 1433 i \, e^{\left (6 i \, f x + 6 i \, e\right )} + 1645 i \, e^{\left (4 i \, f x + 4 i \, e\right )} + 350 i \, e^{\left (2 i \, f x + 2 i \, e\right )} + 40 i\right )}\right )} e^{\left (-6 i \, f x - 6 i \, e\right )}}{15360 \, a^{3} c^{3} f} \]
1/15360*(-3465*I*sqrt(1/2)*a^3*c^3*f*sqrt(1/(a^6*c^5*f^2))*e^(6*I*f*x + 6* I*e)*log(-231/256*(sqrt(2)*sqrt(1/2)*(I*a^3*c^2*f*e^(2*I*f*x + 2*I*e) + I* a^3*c^2*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(1/(a^6*c^5*f^2)) - I)*e^ (-I*f*x - I*e)/(a^3*c^2*f)) + 3465*I*sqrt(1/2)*a^3*c^3*f*sqrt(1/(a^6*c^5*f ^2))*e^(6*I*f*x + 6*I*e)*log(-231/256*(sqrt(2)*sqrt(1/2)*(-I*a^3*c^2*f*e^( 2*I*f*x + 2*I*e) - I*a^3*c^2*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(1/( a^6*c^5*f^2)) - I)*e^(-I*f*x - I*e)/(a^3*c^2*f)) + sqrt(2)*sqrt(c/(e^(2*I* f*x + 2*I*e) + 1))*(-48*I*e^(12*I*f*x + 12*I*e) - 464*I*e^(10*I*f*x + 10*I *e) - 3184*I*e^(8*I*f*x + 8*I*e) - 1433*I*e^(6*I*f*x + 6*I*e) + 1645*I*e^( 4*I*f*x + 4*I*e) + 350*I*e^(2*I*f*x + 2*I*e) + 40*I))*e^(-6*I*f*x - 6*I*e) /(a^3*c^3*f)
\[ \int \frac {1}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^{5/2}} \, dx=\frac {i \int \frac {1}{- c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{5}{\left (e + f x \right )} + i c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{4}{\left (e + f x \right )} - 2 c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{3}{\left (e + f x \right )} + 2 i c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )} - c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + i c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c}}\, dx}{a^{3}} \]
I*Integral(1/(-c**2*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**5 + I*c**2*s qrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**4 - 2*c**2*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**3 + 2*I*c**2*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)* *2 - c**2*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x) + I*c**2*sqrt(-I*c*tan( e + f*x) + c)), x)/a**3
Time = 0.29 (sec) , antiderivative size = 244, normalized size of antiderivative = 0.89 \[ \int \frac {1}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^{5/2}} \, dx=-\frac {i \, {\left (\frac {4 \, {\left (3465 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{5} - 18480 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{4} c + 30492 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{3} c^{2} - 12672 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{2} c^{3} - 2816 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )} c^{4} - 1536 \, c^{5}\right )}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {11}{2}} a^{3} c - 6 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {9}{2}} a^{3} c^{2} + 12 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {7}{2}} a^{3} c^{3} - 8 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} a^{3} c^{4}} + \frac {3465 \, \sqrt {2} \log \left (-\frac {\sqrt {2} \sqrt {c} - \sqrt {-i \, c \tan \left (f x + e\right ) + c}}{\sqrt {2} \sqrt {c} + \sqrt {-i \, c \tan \left (f x + e\right ) + c}}\right )}{a^{3} c^{\frac {3}{2}}}\right )}}{30720 \, c f} \]
-1/30720*I*(4*(3465*(-I*c*tan(f*x + e) + c)^5 - 18480*(-I*c*tan(f*x + e) + c)^4*c + 30492*(-I*c*tan(f*x + e) + c)^3*c^2 - 12672*(-I*c*tan(f*x + e) + c)^2*c^3 - 2816*(-I*c*tan(f*x + e) + c)*c^4 - 1536*c^5)/((-I*c*tan(f*x + e) + c)^(11/2)*a^3*c - 6*(-I*c*tan(f*x + e) + c)^(9/2)*a^3*c^2 + 12*(-I*c* tan(f*x + e) + c)^(7/2)*a^3*c^3 - 8*(-I*c*tan(f*x + e) + c)^(5/2)*a^3*c^4) + 3465*sqrt(2)*log(-(sqrt(2)*sqrt(c) - sqrt(-I*c*tan(f*x + e) + c))/(sqrt (2)*sqrt(c) + sqrt(-I*c*tan(f*x + e) + c)))/(a^3*c^(3/2)))/(c*f)
\[ \int \frac {1}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^{5/2}} \, dx=\int { \frac {1}{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{3} {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}}} \,d x } \]
Time = 6.36 (sec) , antiderivative size = 255, normalized size of antiderivative = 0.93 \[ \int \frac {1}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^{5/2}} \, dx=-\frac {-\frac {{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^3\,2541{}\mathrm {i}}{640\,a^3\,f}+\frac {c^3\,1{}\mathrm {i}}{5\,a^3\,f}+\frac {{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^4\,77{}\mathrm {i}}{32\,a^3\,c\,f}-\frac {{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^5\,231{}\mathrm {i}}{512\,a^3\,c^2\,f}+\frac {c\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2\,33{}\mathrm {i}}{20\,a^3\,f}+\frac {c^2\,\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )\,11{}\mathrm {i}}{30\,a^3\,f}}{6\,c\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{9/2}-{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{11/2}+8\,c^3\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}-12\,c^2\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{7/2}}-\frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-c}}\right )\,231{}\mathrm {i}}{1024\,a^3\,{\left (-c\right )}^{5/2}\,f} \]
- ((c^3*1i)/(5*a^3*f) - ((c - c*tan(e + f*x)*1i)^3*2541i)/(640*a^3*f) + (( c - c*tan(e + f*x)*1i)^4*77i)/(32*a^3*c*f) - ((c - c*tan(e + f*x)*1i)^5*23 1i)/(512*a^3*c^2*f) + (c*(c - c*tan(e + f*x)*1i)^2*33i)/(20*a^3*f) + (c^2* (c - c*tan(e + f*x)*1i)*11i)/(30*a^3*f))/(6*c*(c - c*tan(e + f*x)*1i)^(9/2 ) - (c - c*tan(e + f*x)*1i)^(11/2) + 8*c^3*(c - c*tan(e + f*x)*1i)^(5/2) - 12*c^2*(c - c*tan(e + f*x)*1i)^(7/2)) - (2^(1/2)*atan((2^(1/2)*(c - c*tan (e + f*x)*1i)^(1/2))/(2*(-c)^(1/2)))*231i)/(1024*a^3*(-c)^(5/2)*f)