Integrand size = 35, antiderivative size = 255 \[ \int \frac {(a+i a \tan (e+f x))^{9/2}}{(c-i c \tan (e+f x))^{3/2}} \, dx=-\frac {35 i a^{9/2} \arctan \left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{c^{3/2} f}-\frac {2 i a (a+i a \tan (e+f x))^{7/2}}{3 f (c-i c \tan (e+f x))^{3/2}}+\frac {14 i a^2 (a+i a \tan (e+f x))^{5/2}}{3 c f \sqrt {c-i c \tan (e+f x)}}+\frac {35 i a^4 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 c^2 f}+\frac {35 i a^3 (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{6 c^2 f} \]
-35*I*a^(9/2)*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^(3/2)/f+35/2*I*a^4*(a+I*a*tan(f*x+e))^(1/2)*(c-I*c*tan(f*x +e))^(1/2)/c^2/f+35/6*I*a^3*(c-I*c*tan(f*x+e))^(1/2)*(a+I*a*tan(f*x+e))^(3 /2)/c^2/f+14/3*I*a^2*(a+I*a*tan(f*x+e))^(5/2)/c/f/(c-I*c*tan(f*x+e))^(1/2) -2/3*I*a*(a+I*a*tan(f*x+e))^(7/2)/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 = 9.65 (sec) , antiderivative size = 524, normalized size of antiderivative = 2.05 \[ \int \frac {(a+i a \tan (e+f x))^{9/2}}{(c-i c \tan (e+f x))^{3/2}} \, dx=\frac {i a \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {7}{2},\frac {9}{2},\frac {1}{2} (1+i \tan (e+f x))\right ) \sqrt {1-i \tan (e+f x)} (a+i a \tan (e+f x))^{7/2}}{7 \sqrt {2} c f \sqrt {c-i c \tan (e+f x)}}+\frac {2 a \left (\frac {i a \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {5}{2},\frac {7}{2},\frac {a+i a \tan (e+f x)}{2 a}\right ) (a+i a \tan (e+f x))^{5/2} \sqrt {\frac {c-i c \tan (e+f x)}{c}}}{5 \sqrt {2} c \sqrt {c-i c \tan (e+f x)}}+2 a \left (\frac {2 i a^2 (-i+\tan (e+f x)) \sqrt {i a (-i+\tan (e+f x))}}{3 c (i+\tan (e+f x)) \sqrt {-i c (i+\tan (e+f x))}}+\frac {2 i \sqrt {2} a^3 \sqrt {\frac {c-i c \tan (e+f x)}{c}} \left (-1+\frac {a+i a \tan (e+f x)}{2 a}\right ) \left (\frac {\arcsin \left (\frac {\sqrt {a+i a \tan (e+f x)}}{\sqrt {2} \sqrt {a}}\right ) \sqrt {a+i a \tan (e+f x)}}{\sqrt {2} \sqrt {a} \sqrt {1-\frac {a+i a \tan (e+f x)}{2 a}}}+\frac {a+i a \tan (e+f x)}{2 a \left (-1+\frac {a+i a \tan (e+f x)}{2 a}\right )}\right )}{c \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)} \sqrt {1-\frac {a+i a \tan (e+f x)}{2 a}}}\right )\right )}{f} \]
((I/7)*a*Hypergeometric2F1[3/2, 7/2, 9/2, (1 + I*Tan[e + f*x])/2]*Sqrt[1 - I*Tan[e + f*x]]*(a + I*a*Tan[e + f*x])^(7/2))/(Sqrt[2]*c*f*Sqrt[c - I*c*T an[e + f*x]]) + (2*a*(((I/5)*a*Hypergeometric2F1[3/2, 5/2, 7/2, (a + I*a*T an[e + f*x])/(2*a)]*(a + I*a*Tan[e + f*x])^(5/2)*Sqrt[(c - I*c*Tan[e + f*x ])/c])/(Sqrt[2]*c*Sqrt[c - I*c*Tan[e + f*x]]) + 2*a*((((2*I)/3)*a^2*(-I + Tan[e + f*x])*Sqrt[I*a*(-I + Tan[e + f*x])])/(c*(I + Tan[e + f*x])*Sqrt[(- I)*c*(I + Tan[e + f*x])]) + ((2*I)*Sqrt[2]*a^3*Sqrt[(c - I*c*Tan[e + f*x]) /c]*(-1 + (a + I*a*Tan[e + f*x])/(2*a))*((ArcSin[Sqrt[a + I*a*Tan[e + f*x] ]/(Sqrt[2]*Sqrt[a])]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[2]*Sqrt[a]*Sqrt[1 - (a + I*a*Tan[e + f*x])/(2*a)]) + (a + I*a*Tan[e + f*x])/(2*a*(-1 + (a + I *a*Tan[e + f*x])/(2*a)))))/(c*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[ e + f*x]]*Sqrt[1 - (a + I*a*Tan[e + f*x])/(2*a)]))))/f
Time = 0.37 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.01, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.229, Rules used = {3042, 4006, 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.
| \(\displaystyle \int \frac {(a+i a \tan (e+f x))^{9/2}}{(c-i c \tan (e+f x))^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a+i a \tan (e+f x))^{9/2}}{(c-i c \tan (e+f x))^{3/2}}dx\) |
| \(\Big \downarrow \) 4006 |
| \(\displaystyle \frac {a c \int \frac {(i \tan (e+f x) a+a)^{7/2}}{(c-i c \tan (e+f x))^{5/2}}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 57 |
| \(\displaystyle \frac {a c \left (-\frac {7 a \int \frac {(i \tan (e+f x) a+a)^{5/2}}{(c-i c \tan (e+f x))^{3/2}}d\tan (e+f x)}{3 c}-\frac {2 i (a+i a \tan (e+f x))^{7/2}}{3 c (c-i c \tan (e+f x))^{3/2}}\right )}{f}\) |
| \(\Big \downarrow \) 57 |
| \(\displaystyle \frac {a c \left (-\frac {7 a \left (-\frac {5 a \int \frac {(i \tan (e+f x) a+a)^{3/2}}{\sqrt {c-i c \tan (e+f x)}}d\tan (e+f x)}{c}-\frac {2 i (a+i a \tan (e+f x))^{5/2}}{c \sqrt {c-i c \tan (e+f x)}}\right )}{3 c}-\frac {2 i (a+i a \tan (e+f x))^{7/2}}{3 c (c-i c \tan (e+f x))^{3/2}}\right )}{f}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {a c \left (-\frac {7 a \left (-\frac {5 a \left (\frac {3}{2} a \int \frac {\sqrt {i \tan (e+f x) a+a}}{\sqrt {c-i c \tan (e+f x)}}d\tan (e+f x)+\frac {i (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{2 c}\right )}{c}-\frac {2 i (a+i a \tan (e+f x))^{5/2}}{c \sqrt {c-i c \tan (e+f x)}}\right )}{3 c}-\frac {2 i (a+i a \tan (e+f x))^{7/2}}{3 c (c-i c \tan (e+f x))^{3/2}}\right )}{f}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {a c \left (-\frac {7 a \left (-\frac {5 a \left (\frac {3}{2} a \left (a \int \frac {1}{\sqrt {i \tan (e+f x) a+a} \sqrt {c-i c \tan (e+f x)}}d\tan (e+f x)+\frac {i \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{c}\right )+\frac {i (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{2 c}\right )}{c}-\frac {2 i (a+i a \tan (e+f x))^{5/2}}{c \sqrt {c-i c \tan (e+f x)}}\right )}{3 c}-\frac {2 i (a+i a \tan (e+f x))^{7/2}}{3 c (c-i c \tan (e+f x))^{3/2}}\right )}{f}\) |
| \(\Big \downarrow \) 45 |
| \(\displaystyle \frac {a c \left (-\frac {7 a \left (-\frac {5 a \left (\frac {3}{2} a \left (2 a \int \frac {1}{i a+\frac {i c (i \tan (e+f x) a+a)}{c-i c \tan (e+f x)}}d\frac {\sqrt {i \tan (e+f x) a+a}}{\sqrt {c-i c \tan (e+f x)}}+\frac {i \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{c}\right )+\frac {i (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{2 c}\right )}{c}-\frac {2 i (a+i a \tan (e+f x))^{5/2}}{c \sqrt {c-i c \tan (e+f x)}}\right )}{3 c}-\frac {2 i (a+i a \tan (e+f x))^{7/2}}{3 c (c-i c \tan (e+f x))^{3/2}}\right )}{f}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {a c \left (-\frac {7 a \left (-\frac {5 a \left (\frac {3}{2} a \left (\frac {i \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{c}-\frac {2 i \sqrt {a} \arctan \left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{\sqrt {c}}\right )+\frac {i (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{2 c}\right )}{c}-\frac {2 i (a+i a \tan (e+f x))^{5/2}}{c \sqrt {c-i c \tan (e+f x)}}\right )}{3 c}-\frac {2 i (a+i a \tan (e+f x))^{7/2}}{3 c (c-i c \tan (e+f x))^{3/2}}\right )}{f}\) |
(a*c*((((-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*Tan[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)))/f
3.11.24.3.1 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]
Time = 0.81 (sec) , antiderivative size = 409, normalized size of antiderivative = 1.60
| method | result | size |
| derivativedivides | \(\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a^{4} \left (315 i \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c \left (\tan ^{2}\left (f x +e \right )\right )+27 i \sqrt {a c}\, \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \left (\tan ^{3}\left (f x +e \right )\right )+105 \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c \left (\tan ^{3}\left (f x +e \right )\right )-3 \sqrt {a c}\, \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \left (\tan ^{4}\left (f x +e \right )\right )-105 i \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c -393 i \sqrt {a c}\, \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \tan \left (f x +e \right )-315 \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c \tan \left (f x +e \right )-259 \sqrt {a c}\, \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \left (\tan ^{2}\left (f x +e \right )\right )+164 \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\right )}{6 f \,c^{2} \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\, \left (\tan \left (f x +e \right )+i\right )^{3}}\) | \(409\) |
| default | \(\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a^{4} \left (315 i \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c \left (\tan ^{2}\left (f x +e \right )\right )+27 i \sqrt {a c}\, \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \left (\tan ^{3}\left (f x +e \right )\right )+105 \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c \left (\tan ^{3}\left (f x +e \right )\right )-3 \sqrt {a c}\, \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \left (\tan ^{4}\left (f x +e \right )\right )-105 i \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c -393 i \sqrt {a c}\, \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \tan \left (f x +e \right )-315 \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c \tan \left (f x +e \right )-259 \sqrt {a c}\, \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \left (\tan ^{2}\left (f x +e \right )\right )+164 \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\right )}{6 f \,c^{2} \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}\, \left (\tan \left (f x +e \right )+i\right )^{3}}\) | \(409\) |
1/6/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(I*tan(f*x+e)-1))^(1/2)*a^4/c^2*(315* 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)^2+27*I*(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2)*tan(f*x+e )^3+105*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)^3-3*(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2)*tan(f* x+e)^4-105*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-393*I*(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2)*tan(f*x+e) -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)-259*(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2)*tan(f*x+e )^2+164*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2))/(a*c*(1+tan(f*x+e)^2))^( 1/2)/(a*c)^(1/2)/(tan(f*x+e)+I)^3
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 430 vs. \(2 (189) = 378\).
Time = 0.26 (sec) , antiderivative size = 430, normalized size of antiderivative = 1.69 \[ \int \frac {(a+i a \tan (e+f x))^{9/2}}{(c-i c \tan (e+f x))^{3/2}} \, dx=\frac {105 \, \sqrt {\frac {a^{9}}{c^{3} f^{2}}} {\left (c^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + c^{2} f\right )} \log \left (\frac {4 \, {\left (2 \, {\left (a^{4} e^{\left (3 i \, f x + 3 i \, e\right )} + a^{4} e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} - \sqrt {\frac {a^{9}}{c^{3} f^{2}}} {\left (i \, c^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} - i \, c^{2} f\right )}\right )}}{a^{4} e^{\left (2 i \, f x + 2 i \, e\right )} + a^{4}}\right ) - 105 \, \sqrt {\frac {a^{9}}{c^{3} f^{2}}} {\left (c^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + c^{2} f\right )} \log \left (\frac {4 \, {\left (2 \, {\left (a^{4} e^{\left (3 i \, f x + 3 i \, e\right )} + a^{4} e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} - \sqrt {\frac {a^{9}}{c^{3} f^{2}}} {\left (-i \, c^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + i \, c^{2} f\right )}\right )}}{a^{4} e^{\left (2 i \, f x + 2 i \, e\right )} + a^{4}}\right ) - 4 \, {\left (8 i \, a^{4} e^{\left (7 i \, f x + 7 i \, e\right )} - 56 i \, a^{4} e^{\left (5 i \, f x + 5 i \, e\right )} - 175 i \, a^{4} e^{\left (3 i \, f x + 3 i \, e\right )} - 105 i \, a^{4} e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{12 \, {\left (c^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + c^{2} f\right )}} \]
1/12*(105*sqrt(a^9/(c^3*f^2))*(c^2*f*e^(2*I*f*x + 2*I*e) + c^2*f)*log(4*(2 *(a^4*e^(3*I*f*x + 3*I*e) + a^4*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)) - sqrt(a^9/(c^3*f^2))*(I*c^2*f* e^(2*I*f*x + 2*I*e) - I*c^2*f))/(a^4*e^(2*I*f*x + 2*I*e) + a^4)) - 105*sqr t(a^9/(c^3*f^2))*(c^2*f*e^(2*I*f*x + 2*I*e) + c^2*f)*log(4*(2*(a^4*e^(3*I* f*x + 3*I*e) + a^4*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)) - sqrt(a^9/(c^3*f^2))*(-I*c^2*f*e^(2*I*f*x + 2*I*e) + I*c^2*f))/(a^4*e^(2*I*f*x + 2*I*e) + a^4)) - 4*(8*I*a^4*e^(7*I*f *x + 7*I*e) - 56*I*a^4*e^(5*I*f*x + 5*I*e) - 175*I*a^4*e^(3*I*f*x + 3*I*e) - 105*I*a^4*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^2*f*e^(2*I*f*x + 2*I*e) + c^2*f)
Timed out. \[ \int \frac {(a+i a \tan (e+f x))^{9/2}}{(c-i c \tan (e+f x))^{3/2}} \, dx=\text {Timed out} \]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 915 vs. \(2 (189) = 378\).
Time = 0.44 (sec) , antiderivative size = 915, normalized size of antiderivative = 3.59 \[ \int \frac {(a+i a \tan (e+f x))^{9/2}}{(c-i c \tan (e+f x))^{3/2}} \, dx=\text {Too large to display} \]
-6*(210*(a^4*cos(4*f*x + 4*e) + 2*a^4*cos(2*f*x + 2*e) + I*a^4*sin(4*f*x + 4*e) + 2*I*a^4*sin(2*f*x + 2*e) + a^4)*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) + 210*(a^4*cos(4*f*x + 4*e) + 2*a^4*cos(2*f*x + 2*e) + I*a^4*s in(4*f*x + 4*e) + 2*I*a^4*sin(2*f*x + 2*e) + a^4)*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) + 4*(8*a^4*cos(4*f*x + 4*e) + 16*a^4*cos(2*f*x + 2* e) + 8*I*a^4*sin(4*f*x + 4*e) + 16*I*a^4*sin(2*f*x + 2*e) - 31*a^4)*cos(3/ 2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 12*(24*a^4*cos(4*f*x + 4* e) + 48*a^4*cos(2*f*x + 2*e) + 24*I*a^4*sin(4*f*x + 4*e) + 48*I*a^4*sin(2* f*x + 2*e) + 35*a^4)*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 105*(I*a^4*cos(4*f*x + 4*e) + 2*I*a^4*cos(2*f*x + 2*e) - a^4*sin(4*f*x + 4*e) - 2*a^4*sin(2*f*x + 2*e) + I*a^4)*log(cos(1/2*arctan2(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) + 10 5*(-I*a^4*cos(4*f*x + 4*e) - 2*I*a^4*cos(2*f*x + 2*e) + a^4*sin(4*f*x + 4* e) + 2*a^4*sin(2*f*x + 2*e) - I*a^4)*log(cos(1/2*arctan2(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) + 4*(8* I*a^4*cos(4*f*x + 4*e) + 16*I*a^4*cos(2*f*x + 2*e) - 8*a^4*sin(4*f*x + ...
\[ \int \frac {(a+i a \tan (e+f x))^{9/2}}{(c-i c \tan (e+f x))^{3/2}} \, dx=\int { \frac {{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac {9}{2}}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {(a+i a \tan (e+f x))^{9/2}}{(c-i c \tan (e+f x))^{3/2}} \, dx=\int \frac {{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{9/2}}{{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}} \,d x \]