Integrand size = 33, antiderivative size = 231 \[ \int \frac {1}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}} \, dx=\frac {63 i \text {arctanh}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{128 \sqrt {2} a^2 c^{5/2} f}-\frac {63 i}{160 a^2 f (c-i c \tan (e+f x))^{5/2}}+\frac {i}{4 a^2 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}+\frac {9 i}{16 a^2 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}-\frac {21 i}{64 a^2 c f (c-i c \tan (e+f x))^{3/2}}-\frac {63 i}{128 a^2 c^2 f \sqrt {c-i c \tan (e+f x)}} \]
63/256*I*arctanh(1/2*(c-I*c*tan(f*x+e))^(1/2)*2^(1/2)/c^(1/2))/a^2/c^(5/2) /f*2^(1/2)-63/128*I/a^2/c^2/f/(c-I*c*tan(f*x+e))^(1/2)-63/160*I/a^2/f/(c-I *c*tan(f*x+e))^(5/2)+1/4*I/a^2/f/(1+I*tan(f*x+e))^2/(c-I*c*tan(f*x+e))^(5/ 2)+9/16*I/a^2/f/(1+I*tan(f*x+e))/(c-I*c*tan(f*x+e))^(5/2)-21/64*I/a^2/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 = 2.33 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.23 \[ \int \frac {1}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}} \, dx=-\frac {i \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},3,-\frac {3}{2},-\frac {1}{2} i (i+\tan (e+f x))\right )}{20 a^2 f (c-i c \tan (e+f x))^{5/2}} \]
((-1/20*I)*Hypergeometric2F1[-5/2, 3, -3/2, (-1/2*I)*(I + Tan[e + f*x])])/ (a^2*f*(c - I*c*Tan[e + f*x])^(5/2))
Time = 0.46 (sec) , antiderivative size = 226, normalized size of antiderivative = 0.98, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 4005, 3042, 3968, 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.
| \(\displaystyle \int \frac {1}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}}dx\) |
| \(\Big \downarrow \) 4005 |
| \(\displaystyle \frac {\int \frac {\cos ^4(e+f x)}{\sqrt {c-i c \tan (e+f x)}}dx}{a^2 c^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {1}{\sec (e+f x)^4 \sqrt {c-i c \tan (e+f x)}}dx}{a^2 c^2}\) |
| \(\Big \downarrow \) 3968 |
| \(\displaystyle \frac {i c^3 \int \frac {1}{(c-i c \tan (e+f x))^{7/2} (i \tan (e+f x) c+c)^3}d(-i c \tan (e+f x))}{a^2 f}\) |
| \(\Big \downarrow \) 52 |
| \(\displaystyle \frac {i c^3 \left (\frac {9 \int \frac {1}{(c-i c \tan (e+f x))^{7/2} (i \tan (e+f x) c+c)^2}d(-i c \tan (e+f x))}{8 c}+\frac {1}{4 c (c-i c \tan (e+f x))^{5/2} (c+i c \tan (e+f x))^2}\right )}{a^2 f}\) |
| \(\Big \downarrow \) 52 |
| \(\displaystyle \frac {i c^3 \left (\frac {9 \left (\frac {7 \int \frac {1}{(c-i c \tan (e+f x))^{7/2} (i \tan (e+f x) c+c)}d(-i c \tan (e+f x))}{4 c}+\frac {1}{2 c (c-i c \tan (e+f x))^{5/2} (c+i c \tan (e+f x))}\right )}{8 c}+\frac {1}{4 c (c-i c \tan (e+f x))^{5/2} (c+i c \tan (e+f x))^2}\right )}{a^2 f}\) |
| \(\Big \downarrow \) 61 |
| \(\displaystyle \frac {i c^3 \left (\frac {9 \left (\frac {7 \left (\frac {\int \frac {1}{(c-i c \tan (e+f x))^{5/2} (i \tan (e+f x) c+c)}d(-i c \tan (e+f x))}{2 c}-\frac {1}{5 c (c-i c \tan (e+f x))^{5/2}}\right )}{4 c}+\frac {1}{2 c (c-i c \tan (e+f x))^{5/2} (c+i c \tan (e+f x))}\right )}{8 c}+\frac {1}{4 c (c-i c \tan (e+f x))^{5/2} (c+i c \tan (e+f x))^2}\right )}{a^2 f}\) |
| \(\Big \downarrow \) 61 |
| \(\displaystyle \frac {i c^3 \left (\frac {9 \left (\frac {7 \left (\frac {\frac {\int \frac {1}{(c-i c \tan (e+f x))^{3/2} (i \tan (e+f x) c+c)}d(-i c \tan (e+f x))}{2 c}-\frac {1}{3 c (c-i c \tan (e+f x))^{3/2}}}{2 c}-\frac {1}{5 c (c-i c \tan (e+f x))^{5/2}}\right )}{4 c}+\frac {1}{2 c (c-i c \tan (e+f x))^{5/2} (c+i c \tan (e+f x))}\right )}{8 c}+\frac {1}{4 c (c-i c \tan (e+f x))^{5/2} (c+i c \tan (e+f x))^2}\right )}{a^2 f}\) |
| \(\Big \downarrow \) 61 |
| \(\displaystyle \frac {i c^3 \left (\frac {9 \left (\frac {7 \left (\frac {\frac {\frac {\int \frac {1}{\sqrt {c-i c \tan (e+f x)} (i \tan (e+f x) c+c)}d(-i c \tan (e+f x))}{2 c}-\frac {1}{c \sqrt {c-i c \tan (e+f x)}}}{2 c}-\frac {1}{3 c (c-i c \tan (e+f x))^{3/2}}}{2 c}-\frac {1}{5 c (c-i c \tan (e+f x))^{5/2}}\right )}{4 c}+\frac {1}{2 c (c-i c \tan (e+f x))^{5/2} (c+i c \tan (e+f x))}\right )}{8 c}+\frac {1}{4 c (c-i c \tan (e+f x))^{5/2} (c+i c \tan (e+f x))^2}\right )}{a^2 f}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {i c^3 \left (\frac {9 \left (\frac {7 \left (\frac {\frac {\frac {\int \frac {1}{c^2 \tan ^2(e+f x)+2 c}d\sqrt {c-i c \tan (e+f x)}}{c}-\frac {1}{c \sqrt {c-i c \tan (e+f x)}}}{2 c}-\frac {1}{3 c (c-i c \tan (e+f x))^{3/2}}}{2 c}-\frac {1}{5 c (c-i c \tan (e+f x))^{5/2}}\right )}{4 c}+\frac {1}{2 c (c-i c \tan (e+f x))^{5/2} (c+i c \tan (e+f x))}\right )}{8 c}+\frac {1}{4 c (c-i c \tan (e+f x))^{5/2} (c+i c \tan (e+f x))^2}\right )}{a^2 f}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {i c^3 \left (\frac {9 \left (\frac {7 \left (\frac {\frac {-\frac {i \arctan \left (\frac {\sqrt {c} \tan (e+f x)}{\sqrt {2}}\right )}{\sqrt {2} c^{3/2}}-\frac {1}{c \sqrt {c-i c \tan (e+f x)}}}{2 c}-\frac {1}{3 c (c-i c \tan (e+f x))^{3/2}}}{2 c}-\frac {1}{5 c (c-i c \tan (e+f x))^{5/2}}\right )}{4 c}+\frac {1}{2 c (c-i c \tan (e+f x))^{5/2} (c+i c \tan (e+f x))}\right )}{8 c}+\frac {1}{4 c (c-i c \tan (e+f x))^{5/2} (c+i c \tan (e+f x))^2}\right )}{a^2 f}\) |
(I*c^3*(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*Sqrt[c - I*c*Tan[e + f*x]]))/(2*c))/(2*c)))/(4*c)))/(8*c)))/(a^2*f)
3.10.90.3.1 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 = 0.59 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.68
| method | result | size |
| derivativedivides | \(\frac {2 i c^{3} \left (\frac {\frac {-\frac {15 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{16}+\frac {17 c \sqrt {c -i c \tan \left (f x +e \right )}}{8}}{\left (c +i c \tan \left (f x +e \right )\right )^{2}}+\frac {63 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{32 \sqrt {c}}}{16 c^{5}}-\frac {3}{16 c^{5} \sqrt {c -i c \tan \left (f x +e \right )}}-\frac {1}{16 c^{4} \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}-\frac {1}{40 c^{3} \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}\right )}{f \,a^{2}}\) | \(158\) |
| default | \(\frac {2 i c^{3} \left (\frac {\frac {-\frac {15 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{16}+\frac {17 c \sqrt {c -i c \tan \left (f x +e \right )}}{8}}{\left (c +i c \tan \left (f x +e \right )\right )^{2}}+\frac {63 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{32 \sqrt {c}}}{16 c^{5}}-\frac {3}{16 c^{5} \sqrt {c -i c \tan \left (f x +e \right )}}-\frac {1}{16 c^{4} \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}-\frac {1}{40 c^{3} \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}\right )}{f \,a^{2}}\) | \(158\) |
2*I/f/a^2*c^3*(1/16/c^5*(4*(-15/64*(c-I*c*tan(f*x+e))^(3/2)+17/32*c*(c-I*c *tan(f*x+e))^(1/2))/(c+I*c*tan(f*x+e))^2+63/32*2^(1/2)/c^(1/2)*arctanh(1/2 *(c-I*c*tan(f*x+e))^(1/2)*2^(1/2)/c^(1/2)))-3/16/c^5/(c-I*c*tan(f*x+e))^(1 /2)-1/16/c^4/(c-I*c*tan(f*x+e))^(3/2)-1/40/c^3/(c-I*c*tan(f*x+e))^(5/2))
Time = 0.27 (sec) , antiderivative size = 339, normalized size of antiderivative = 1.47 \[ \int \frac {1}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}} \, dx=\frac {{\left (-315 i \, \sqrt {\frac {1}{2}} a^{2} c^{3} f \sqrt {\frac {1}{a^{4} c^{5} f^{2}}} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (-\frac {63 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (i \, a^{2} c^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + i \, a^{2} c^{2} f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {1}{a^{4} c^{5} f^{2}}} - i\right )} e^{\left (-i \, f x - i \, e\right )}}{64 \, a^{2} c^{2} f}\right ) + 315 i \, \sqrt {\frac {1}{2}} a^{2} c^{3} f \sqrt {\frac {1}{a^{4} c^{5} f^{2}}} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (-\frac {63 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (-i \, a^{2} c^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} - i \, a^{2} c^{2} f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {1}{a^{4} c^{5} f^{2}}} - i\right )} e^{\left (-i \, f x - i \, e\right )}}{64 \, a^{2} c^{2} f}\right ) + \sqrt {2} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} {\left (-8 i \, e^{\left (10 i \, f x + 10 i \, e\right )} - 64 i \, e^{\left (8 i \, f x + 8 i \, e\right )} - 344 i \, e^{\left (6 i \, f x + 6 i \, e\right )} - 203 i \, e^{\left (4 i \, f x + 4 i \, e\right )} + 95 i \, e^{\left (2 i \, f x + 2 i \, e\right )} + 10 i\right )}\right )} e^{\left (-4 i \, f x - 4 i \, e\right )}}{1280 \, a^{2} c^{3} f} \]
1/1280*(-315*I*sqrt(1/2)*a^2*c^3*f*sqrt(1/(a^4*c^5*f^2))*e^(4*I*f*x + 4*I* e)*log(-63/64*(sqrt(2)*sqrt(1/2)*(I*a^2*c^2*f*e^(2*I*f*x + 2*I*e) + I*a^2* c^2*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(1/(a^4*c^5*f^2)) - I)*e^(-I* f*x - I*e)/(a^2*c^2*f)) + 315*I*sqrt(1/2)*a^2*c^3*f*sqrt(1/(a^4*c^5*f^2))* e^(4*I*f*x + 4*I*e)*log(-63/64*(sqrt(2)*sqrt(1/2)*(-I*a^2*c^2*f*e^(2*I*f*x + 2*I*e) - I*a^2*c^2*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(1/(a^4*c^5 *f^2)) - I)*e^(-I*f*x - I*e)/(a^2*c^2*f)) + sqrt(2)*sqrt(c/(e^(2*I*f*x + 2 *I*e) + 1))*(-8*I*e^(10*I*f*x + 10*I*e) - 64*I*e^(8*I*f*x + 8*I*e) - 344*I *e^(6*I*f*x + 6*I*e) - 203*I*e^(4*I*f*x + 4*I*e) + 95*I*e^(2*I*f*x + 2*I*e ) + 10*I))*e^(-4*I*f*x - 4*I*e)/(a^2*c^3*f)
\[ \int \frac {1}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}} \, dx=- \frac {\int \frac {1}{- c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{4}{\left (e + f x \right )} - 2 c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )} - c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c}}\, dx}{a^{2}} \]
-Integral(1/(-c**2*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**4 - 2*c**2*sq rt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**2 - c**2*sqrt(-I*c*tan(e + f*x) + c)), x)/a**2
Time = 0.29 (sec) , antiderivative size = 205, normalized size of antiderivative = 0.89 \[ \int \frac {1}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}} \, dx=-\frac {i \, {\left (\frac {4 \, {\left (315 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{4} - 1050 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{3} c + 672 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{2} c^{2} + 192 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )} c^{3} + 128 \, c^{4}\right )}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {9}{2}} a^{2} c - 4 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {7}{2}} a^{2} c^{2} + 4 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} a^{2} c^{3}} + \frac {315 \, \sqrt {2} \log \left (-\frac {\sqrt {2} \sqrt {c} - \sqrt {-i \, c \tan \left (f x + e\right ) + c}}{\sqrt {2} \sqrt {c} + \sqrt {-i \, c \tan \left (f x + e\right ) + c}}\right )}{a^{2} c^{\frac {3}{2}}}\right )}}{2560 \, c f} \]
-1/2560*I*(4*(315*(-I*c*tan(f*x + e) + c)^4 - 1050*(-I*c*tan(f*x + e) + c) ^3*c + 672*(-I*c*tan(f*x + e) + c)^2*c^2 + 192*(-I*c*tan(f*x + e) + c)*c^3 + 128*c^4)/((-I*c*tan(f*x + e) + c)^(9/2)*a^2*c - 4*(-I*c*tan(f*x + e) + c)^(7/2)*a^2*c^2 + 4*(-I*c*tan(f*x + e) + c)^(5/2)*a^2*c^3) + 315*sqrt(2)* log(-(sqrt(2)*sqrt(c) - sqrt(-I*c*tan(f*x + e) + c))/(sqrt(2)*sqrt(c) + sq rt(-I*c*tan(f*x + e) + c)))/(a^2*c^(3/2)))/(c*f)
\[ \int \frac {1}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}} \, dx=\int { \frac {1}{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{2} {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}}} \,d x } \]
Time = 6.54 (sec) , antiderivative size = 208, normalized size of antiderivative = 0.90 \[ \int \frac {1}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}} \, dx=-\frac {\frac {{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2\,21{}\mathrm {i}}{20\,a^2\,f}+\frac {c^2\,1{}\mathrm {i}}{5\,a^2\,f}-\frac {{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^3\,105{}\mathrm {i}}{64\,a^2\,c\,f}+\frac {{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^4\,63{}\mathrm {i}}{128\,a^2\,c^2\,f}+\frac {c\,\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )\,3{}\mathrm {i}}{10\,a^2\,f}}{-4\,c\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{7/2}+{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{9/2}+4\,c^2\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}-\frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-c}}\right )\,63{}\mathrm {i}}{256\,a^2\,{\left (-c\right )}^{5/2}\,f} \]
- (((c - c*tan(e + f*x)*1i)^2*21i)/(20*a^2*f) + (c^2*1i)/(5*a^2*f) - ((c - c*tan(e + f*x)*1i)^3*105i)/(64*a^2*c*f) + ((c - c*tan(e + f*x)*1i)^4*63i) /(128*a^2*c^2*f) + (c*(c - c*tan(e + f*x)*1i)*3i)/(10*a^2*f))/((c - c*tan( e + f*x)*1i)^(9/2) - 4*c*(c - c*tan(e + f*x)*1i)^(7/2) + 4*c^2*(c - c*tan( e + f*x)*1i)^(5/2)) - (2^(1/2)*atan((2^(1/2)*(c - c*tan(e + f*x)*1i)^(1/2) )/(2*(-c)^(1/2)))*63i)/(256*a^2*(-c)^(5/2)*f)