Integrand size = 35, antiderivative size = 304 \[ \int \frac {(a+i a \tan (e+f x))^{11/2}}{(c-i c \tan (e+f x))^{5/2}} \, dx=\frac {63 i a^{11/2} \arctan \left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{c^{5/2} f}-\frac {2 i a (a+i a \tan (e+f x))^{9/2}}{5 f (c-i c \tan (e+f x))^{5/2}}+\frac {6 i a^2 (a+i a \tan (e+f x))^{7/2}}{5 c f (c-i c \tan (e+f x))^{3/2}}-\frac {42 i a^3 (a+i a \tan (e+f x))^{5/2}}{5 c^2 f \sqrt {c-i c \tan (e+f x)}}-\frac {63 i a^5 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 c^3 f}-\frac {21 i a^4 (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{2 c^3 f} \]
63*I*a^(11/2)*arctan(c^(1/2)*(a+I*a*tan(f*x+e))^(1/2)/a^(1/2)/(c-I*c*tan(f *x+e))^(1/2))/c^(5/2)/f-63/2*I*a^5*(a+I*a*tan(f*x+e))^(1/2)*(c-I*c*tan(f*x +e))^(1/2)/c^3/f-21/2*I*a^4*(c-I*c*tan(f*x+e))^(1/2)*(a+I*a*tan(f*x+e))^(3 /2)/c^3/f-42/5*I*a^3*(a+I*a*tan(f*x+e))^(5/2)/c^2/f/(c-I*c*tan(f*x+e))^(1/ 2)-2/5*I*a*(a+I*a*tan(f*x+e))^(9/2)/f/(c-I*c*tan(f*x+e))^(5/2)+6/5*I*a^2*( a+I*a*tan(f*x+e))^(7/2)/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 = 10.72 (sec) , antiderivative size = 366, normalized size of antiderivative = 1.20 \[ \int \frac {(a+i a \tan (e+f x))^{11/2}}{(c-i c \tan (e+f x))^{5/2}} \, dx=\frac {a \left (-90 i \sqrt {2} a^5 \operatorname {Hypergeometric2F1}\left (\frac {5}{2},\frac {7}{2},\frac {9}{2},\frac {1}{2} (1+i \tan (e+f x))\right ) \sec ^8(e+f x)+2016 a^5 (1-i \tan (e+f x))^{3/2} (-i+\tan (e+f x))^3+35 \sqrt {2} a^5 \operatorname {Hypergeometric2F1}\left (\frac {5}{2},\frac {9}{2},\frac {11}{2},\frac {1}{2} (1+i \tan (e+f x))\right ) (-i+\tan (e+f x))^5 (i+\tan (e+f x))^4+3360 i a^{5/2} (i+\tan (e+f x))^2 \left (-3 a^{5/2} \sec ^2(e+f x) \sqrt {1-i \tan (e+f x)}+3 \arcsin \left (\frac {\sqrt {a+i a \tan (e+f x)}}{\sqrt {2} \sqrt {a}}\right ) (a-i a \tan (e+f x))^2 \sqrt {a+i a \tan (e+f x)}+\sqrt {a} \sqrt {1-i \tan (e+f x)} (a+i a \tan (e+f x))^2\right )\right )}{1260 c^2 f (1-i \tan (e+f x))^{3/2} (i+\tan (e+f x))^2 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}} \]
(a*((-90*I)*Sqrt[2]*a^5*Hypergeometric2F1[5/2, 7/2, 9/2, (1 + I*Tan[e + f* x])/2]*Sec[e + f*x]^8 + 2016*a^5*(1 - I*Tan[e + f*x])^(3/2)*(-I + Tan[e + f*x])^3 + 35*Sqrt[2]*a^5*Hypergeometric2F1[5/2, 9/2, 11/2, (1 + I*Tan[e + f*x])/2]*(-I + Tan[e + f*x])^5*(I + Tan[e + f*x])^4 + (3360*I)*a^(5/2)*(I + Tan[e + f*x])^2*(-3*a^(5/2)*Sec[e + f*x]^2*Sqrt[1 - I*Tan[e + f*x]] + 3* ArcSin[Sqrt[a + I*a*Tan[e + f*x]]/(Sqrt[2]*Sqrt[a])]*(a - I*a*Tan[e + f*x] )^2*Sqrt[a + I*a*Tan[e + f*x]] + Sqrt[a]*Sqrt[1 - I*Tan[e + f*x]]*(a + I*a *Tan[e + f*x])^2)))/(1260*c^2*f*(1 - I*Tan[e + f*x])^(3/2)*(I + Tan[e + f* x])^2*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]])
Time = 0.37 (sec) , antiderivative size = 309, normalized size of antiderivative = 1.02, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.257, Rules used = {3042, 4006, 57, 57, 57, 60, 60, 45, 218}
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.
\(\displaystyle \int \frac {(a+i a \tan (e+f x))^{11/2}}{(c-i c \tan (e+f x))^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a+i a \tan (e+f x))^{11/2}}{(c-i c \tan (e+f x))^{5/2}}dx\) |
\(\Big \downarrow \) 4006 |
\(\displaystyle \frac {a c \int \frac {(i \tan (e+f x) a+a)^{9/2}}{(c-i c \tan (e+f x))^{7/2}}d\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 57 |
\(\displaystyle \frac {a c \left (-\frac {9 a \int \frac {(i \tan (e+f x) a+a)^{7/2}}{(c-i c \tan (e+f x))^{5/2}}d\tan (e+f x)}{5 c}-\frac {2 i (a+i a \tan (e+f x))^{9/2}}{5 c (c-i c \tan (e+f x))^{5/2}}\right )}{f}\) |
\(\Big \downarrow \) 57 |
\(\displaystyle \frac {a c \left (-\frac {9 a \left (-\frac {7 a \int \frac {(i \tan (e+f x) a+a)^{5/2}}{(c-i c \tan (e+f x))^{3/2}}d\tan (e+f x)}{3 c}-\frac {2 i (a+i a \tan (e+f x))^{7/2}}{3 c (c-i c \tan (e+f x))^{3/2}}\right )}{5 c}-\frac {2 i (a+i a \tan (e+f x))^{9/2}}{5 c (c-i c \tan (e+f x))^{5/2}}\right )}{f}\) |
\(\Big \downarrow \) 57 |
\(\displaystyle \frac {a c \left (-\frac {9 a \left (-\frac {7 a \left (-\frac {5 a \int \frac {(i \tan (e+f x) a+a)^{3/2}}{\sqrt {c-i c \tan (e+f x)}}d\tan (e+f x)}{c}-\frac {2 i (a+i a \tan (e+f x))^{5/2}}{c \sqrt {c-i c \tan (e+f x)}}\right )}{3 c}-\frac {2 i (a+i a \tan (e+f x))^{7/2}}{3 c (c-i c \tan (e+f x))^{3/2}}\right )}{5 c}-\frac {2 i (a+i a \tan (e+f x))^{9/2}}{5 c (c-i c \tan (e+f x))^{5/2}}\right )}{f}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {a c \left (-\frac {9 a \left (-\frac {7 a \left (-\frac {5 a \left (\frac {3}{2} a \int \frac {\sqrt {i \tan (e+f x) a+a}}{\sqrt {c-i c \tan (e+f x)}}d\tan (e+f x)+\frac {i (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{2 c}\right )}{c}-\frac {2 i (a+i a \tan (e+f x))^{5/2}}{c \sqrt {c-i c \tan (e+f x)}}\right )}{3 c}-\frac {2 i (a+i a \tan (e+f x))^{7/2}}{3 c (c-i c \tan (e+f x))^{3/2}}\right )}{5 c}-\frac {2 i (a+i a \tan (e+f x))^{9/2}}{5 c (c-i c \tan (e+f x))^{5/2}}\right )}{f}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {a c \left (-\frac {9 a \left (-\frac {7 a \left (-\frac {5 a \left (\frac {3}{2} a \left (a \int \frac {1}{\sqrt {i \tan (e+f x) a+a} \sqrt {c-i c \tan (e+f x)}}d\tan (e+f x)+\frac {i \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{c}\right )+\frac {i (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{2 c}\right )}{c}-\frac {2 i (a+i a \tan (e+f x))^{5/2}}{c \sqrt {c-i c \tan (e+f x)}}\right )}{3 c}-\frac {2 i (a+i a \tan (e+f x))^{7/2}}{3 c (c-i c \tan (e+f x))^{3/2}}\right )}{5 c}-\frac {2 i (a+i a \tan (e+f x))^{9/2}}{5 c (c-i c \tan (e+f x))^{5/2}}\right )}{f}\) |
\(\Big \downarrow \) 45 |
\(\displaystyle \frac {a c \left (-\frac {9 a \left (-\frac {7 a \left (-\frac {5 a \left (\frac {3}{2} a \left (2 a \int \frac {1}{i a+\frac {i c (i \tan (e+f x) a+a)}{c-i c \tan (e+f x)}}d\frac {\sqrt {i \tan (e+f x) a+a}}{\sqrt {c-i c \tan (e+f x)}}+\frac {i \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{c}\right )+\frac {i (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{2 c}\right )}{c}-\frac {2 i (a+i a \tan (e+f x))^{5/2}}{c \sqrt {c-i c \tan (e+f x)}}\right )}{3 c}-\frac {2 i (a+i a \tan (e+f x))^{7/2}}{3 c (c-i c \tan (e+f x))^{3/2}}\right )}{5 c}-\frac {2 i (a+i a \tan (e+f x))^{9/2}}{5 c (c-i c \tan (e+f x))^{5/2}}\right )}{f}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {a c \left (-\frac {9 a \left (-\frac {7 a \left (-\frac {5 a \left (\frac {3}{2} a \left (\frac {i \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{c}-\frac {2 i \sqrt {a} \arctan \left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{\sqrt {c}}\right )+\frac {i (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{2 c}\right )}{c}-\frac {2 i (a+i a \tan (e+f x))^{5/2}}{c \sqrt {c-i c \tan (e+f x)}}\right )}{3 c}-\frac {2 i (a+i a \tan (e+f x))^{7/2}}{3 c (c-i c \tan (e+f x))^{3/2}}\right )}{5 c}-\frac {2 i (a+i a \tan (e+f x))^{9/2}}{5 c (c-i c \tan (e+f x))^{5/2}}\right )}{f}\) |
(a*c*((((-2*I)/5)*(a + I*a*Tan[e + f*x])^(9/2))/(c*(c - I*c*Tan[e + f*x])^ (5/2)) - (9*a*((((-2*I)/3)*(a + I*a*Tan[e + f*x])^(7/2))/(c*(c - I*c*Tan[e + f*x])^(3/2)) - (7*a*(((-2*I)*(a + I*a*Tan[e + f*x])^(5/2))/(c*Sqrt[c - I*c*Tan[e + f*x]]) - (5*a*(((I/2)*(a + I*a*Tan[e + f*x])^(3/2)*Sqrt[c - I* c*Tan[e + f*x]])/c + (3*a*(((-2*I)*Sqrt[a]*ArcTan[(Sqrt[c]*Sqrt[a + I*a*Ta n[e + f*x]])/(Sqrt[a]*Sqrt[c - I*c*Tan[e + f*x]])])/Sqrt[c] + (I*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]])/c))/2))/c))/(3*c)))/(5*c)))/ f
3.11.33.3.1 Defintions of rubi rules used
Int[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[ 2 Subst[Int[1/(b - d*x^2), x], x, Sqrt[a + b*x]/Sqrt[c + d*x]], x] /; Fre eQ[{a, b, c, d}, x] && EqQ[b*c + a*d, 0] && !GtQ[c, 0]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + 1))), x] - Simp[d*(n/(b*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] & & GtQ[n, 0] && LtQ[m, -1] && !(IntegerQ[n] && !IntegerQ[m]) && !(ILeQ[m + n + 2, 0] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c , d, m, n, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( b*(m + n + 1))) Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[a, b, c, d, m, n, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
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]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 489 vs. \(2 (244 ) = 488\).
Time = 0.84 (sec) , antiderivative size = 490, normalized size of antiderivative = 1.61
method | result | size |
derivativedivides | \(-\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a^{5} \left (1260 i \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c \left (\tan ^{3}\left (f x +e \right )\right )+60 i \sqrt {a c}\, \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \left (\tan ^{4}\left (f x +e \right )\right )+315 \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c \left (\tan ^{4}\left (f x +e \right )\right )-5 \sqrt {a c}\, \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \left (\tan ^{5}\left (f x +e \right )\right )-1260 i \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c \tan \left (f x +e \right )-1964 i \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\, \left (\tan ^{2}\left (f x +e \right )\right )-1890 \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c \left (\tan ^{2}\left (f x +e \right )\right )-866 \left (\tan ^{3}\left (f x +e \right )\right ) \sqrt {a c}\, \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}+496 i \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}+315 a c \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}{\sqrt {a c}}\right )+1659 \tan \left (f x +e \right ) \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\right )}{10 f \,c^{3} \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\, \left (\tan \left (f x +e \right )+i\right )^{4}}\) | \(490\) |
default | \(-\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a^{5} \left (1260 i \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c \left (\tan ^{3}\left (f x +e \right )\right )+60 i \sqrt {a c}\, \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \left (\tan ^{4}\left (f x +e \right )\right )+315 \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c \left (\tan ^{4}\left (f x +e \right )\right )-5 \sqrt {a c}\, \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \left (\tan ^{5}\left (f x +e \right )\right )-1260 i \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c \tan \left (f x +e \right )-1964 i \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\, \left (\tan ^{2}\left (f x +e \right )\right )-1890 \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c \left (\tan ^{2}\left (f x +e \right )\right )-866 \left (\tan ^{3}\left (f x +e \right )\right ) \sqrt {a c}\, \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}+496 i \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}+315 a c \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}{\sqrt {a c}}\right )+1659 \tan \left (f x +e \right ) \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\right )}{10 f \,c^{3} \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\, \left (\tan \left (f x +e \right )+i\right )^{4}}\) | \(490\) |
-1/10/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(I*tan(f*x+e)-1))^(1/2)*a^5/c^3*(12 60*I*ln((c*a*tan(f*x+e)+(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2))/(a*c)^(1 /2))*a*c*tan(f*x+e)^3+60*I*(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2)*tan(f* x+e)^4+315*ln((c*a*tan(f*x+e)+(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2))/(a *c)^(1/2))*a*c*tan(f*x+e)^4-5*(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2)*tan (f*x+e)^5-1260*I*ln((c*a*tan(f*x+e)+(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/ 2))/(a*c)^(1/2))*a*c*tan(f*x+e)-1964*I*(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^ (1/2)*tan(f*x+e)^2-1890*ln((c*a*tan(f*x+e)+(a*c*(1+tan(f*x+e)^2))^(1/2)*(a *c)^(1/2))/(a*c)^(1/2))*a*c*tan(f*x+e)^2-866*tan(f*x+e)^3*(a*c)^(1/2)*(a*c *(1+tan(f*x+e)^2))^(1/2)+496*I*(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2)+31 5*a*c*ln((c*a*tan(f*x+e)+(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2))/(a*c)^( 1/2))+1659*tan(f*x+e)*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2))/(a*c*(1+ta n(f*x+e)^2))^(1/2)/(a*c)^(1/2)/(tan(f*x+e)+I)^4
Time = 0.27 (sec) , antiderivative size = 444, normalized size of antiderivative = 1.46 \[ \int \frac {(a+i a \tan (e+f x))^{11/2}}{(c-i c \tan (e+f x))^{5/2}} \, dx=-\frac {315 \, {\left (c^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + c^{3} f\right )} \sqrt {\frac {a^{11}}{c^{5} f^{2}}} \log \left (\frac {4 \, {\left (2 \, {\left (a^{5} e^{\left (3 i \, f x + 3 i \, e\right )} + a^{5} e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} - {\left (i \, c^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} - i \, c^{3} f\right )} \sqrt {\frac {a^{11}}{c^{5} f^{2}}}\right )}}{a^{5} e^{\left (2 i \, f x + 2 i \, e\right )} + a^{5}}\right ) - 315 \, {\left (c^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + c^{3} f\right )} \sqrt {\frac {a^{11}}{c^{5} f^{2}}} \log \left (\frac {4 \, {\left (2 \, {\left (a^{5} e^{\left (3 i \, f x + 3 i \, e\right )} + a^{5} e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} - {\left (-i \, c^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + i \, c^{3} f\right )} \sqrt {\frac {a^{11}}{c^{5} f^{2}}}\right )}}{a^{5} e^{\left (2 i \, f x + 2 i \, e\right )} + a^{5}}\right ) + 4 \, {\left (8 i \, a^{5} e^{\left (9 i \, f x + 9 i \, e\right )} - 24 i \, a^{5} e^{\left (7 i \, f x + 7 i \, e\right )} + 168 i \, a^{5} e^{\left (5 i \, f x + 5 i \, e\right )} + 525 i \, a^{5} e^{\left (3 i \, f x + 3 i \, e\right )} + 315 i \, a^{5} e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{20 \, {\left (c^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + c^{3} f\right )}} \]
-1/20*(315*(c^3*f*e^(2*I*f*x + 2*I*e) + c^3*f)*sqrt(a^11/(c^5*f^2))*log(4* (2*(a^5*e^(3*I*f*x + 3*I*e) + a^5*e^(I*f*x + I*e))*sqrt(a/(e^(2*I*f*x + 2* I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)) - (I*c^3*f*e^(2*I*f*x + 2*I*e ) - I*c^3*f)*sqrt(a^11/(c^5*f^2)))/(a^5*e^(2*I*f*x + 2*I*e) + a^5)) - 315* (c^3*f*e^(2*I*f*x + 2*I*e) + c^3*f)*sqrt(a^11/(c^5*f^2))*log(4*(2*(a^5*e^( 3*I*f*x + 3*I*e) + a^5*e^(I*f*x + I*e))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))* sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)) - (-I*c^3*f*e^(2*I*f*x + 2*I*e) + I*c^3* f)*sqrt(a^11/(c^5*f^2)))/(a^5*e^(2*I*f*x + 2*I*e) + a^5)) + 4*(8*I*a^5*e^( 9*I*f*x + 9*I*e) - 24*I*a^5*e^(7*I*f*x + 7*I*e) + 168*I*a^5*e^(5*I*f*x + 5 *I*e) + 525*I*a^5*e^(3*I*f*x + 3*I*e) + 315*I*a^5*e^(I*f*x + I*e))*sqrt(a/ (e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)))/(c^3*f*e^(2* I*f*x + 2*I*e) + c^3*f)
Timed out. \[ \int \frac {(a+i a \tan (e+f x))^{11/2}}{(c-i c \tan (e+f x))^{5/2}} \, dx=\text {Timed out} \]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1083 vs. \(2 (226) = 452\).
Time = 0.46 (sec) , antiderivative size = 1083, normalized size of antiderivative = 3.56 \[ \int \frac {(a+i a \tan (e+f x))^{11/2}}{(c-i c \tan (e+f x))^{5/2}} \, dx=\text {Too large to display} \]
10*(630*(a^5*cos(4*f*x + 4*e) + 2*a^5*cos(2*f*x + 2*e) + I*a^5*sin(4*f*x + 4*e) + 2*I*a^5*sin(2*f*x + 2*e) + a^5)*arctan2(cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))), sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 1) + 630*(a^5*cos(4*f*x + 4*e) + 2*a^5*cos(2*f*x + 2*e) + I*a^5*s in(4*f*x + 4*e) + 2*I*a^5*sin(2*f*x + 2*e) + a^5)*arctan2(cos(1/2*arctan2( sin(2*f*x + 2*e), cos(2*f*x + 2*e))), -sin(1/2*arctan2(sin(2*f*x + 2*e), c os(2*f*x + 2*e))) + 1) - 32*(a^5*cos(4*f*x + 4*e) + 2*a^5*cos(2*f*x + 2*e) + I*a^5*sin(4*f*x + 4*e) + 2*I*a^5*sin(2*f*x + 2*e) + a^5)*cos(5/2*arctan 2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 20*(8*a^5*cos(4*f*x + 4*e) + 16*a ^5*cos(2*f*x + 2*e) + 8*I*a^5*sin(4*f*x + 4*e) + 16*I*a^5*sin(2*f*x + 2*e) - 9*a^5)*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 60*(16*a^ 5*cos(4*f*x + 4*e) + 32*a^5*cos(2*f*x + 2*e) + 16*I*a^5*sin(4*f*x + 4*e) + 32*I*a^5*sin(2*f*x + 2*e) + 21*a^5)*cos(1/2*arctan2(sin(2*f*x + 2*e), cos (2*f*x + 2*e))) + 315*(I*a^5*cos(4*f*x + 4*e) + 2*I*a^5*cos(2*f*x + 2*e) - a^5*sin(4*f*x + 4*e) - 2*a^5*sin(2*f*x + 2*e) + I*a^5)*log(cos(1/2*arctan 2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + sin(1/2*arctan2(sin(2*f*x + 2*e ), cos(2*f*x + 2*e)))^2 + 2*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 1) + 315*(-I*a^5*cos(4*f*x + 4*e) - 2*I*a^5*cos(2*f*x + 2*e) + a^ 5*sin(4*f*x + 4*e) + 2*a^5*sin(2*f*x + 2*e) - I*a^5)*log(cos(1/2*arctan2(s in(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + sin(1/2*arctan2(sin(2*f*x + 2*e...
\[ \int \frac {(a+i a \tan (e+f x))^{11/2}}{(c-i c \tan (e+f x))^{5/2}} \, dx=\int { \frac {{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac {11}{2}}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {(a+i a \tan (e+f x))^{11/2}}{(c-i c \tan (e+f x))^{5/2}} \, dx=\int \frac {{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{11/2}}{{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}} \,d x \]