Integrand size = 28, antiderivative size = 131 \[ \int (a+i a \tan (e+f x)) (c+d \tan (e+f x))^{5/2} \, dx=-\frac {2 i a (c-i d)^{5/2} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f}+\frac {2 i a (c-i d)^2 \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 a (i c+d) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 i a (c+d \tan (e+f x))^{5/2}}{5 f} \]
-2*I*a*(c-I*d)^(5/2)*arctanh((c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2))/f+2*I*a *(c-I*d)^2*(c+d*tan(f*x+e))^(1/2)/f+2/3*a*(I*c+d)*(c+d*tan(f*x+e))^(3/2)/f +2/5*I*a*(c+d*tan(f*x+e))^(5/2)/f
Time = 1.18 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.89 \[ \int (a+i a \tan (e+f x)) (c+d \tan (e+f x))^{5/2} \, dx=\frac {2 a \left (-15 i (c-i d)^{5/2} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )+\sqrt {c+d \tan (e+f x)} \left (23 i c^2+35 c d-15 i d^2+d (11 i c+5 d) \tan (e+f x)+3 i d^2 \tan ^2(e+f x)\right )\right )}{15 f} \]
(2*a*((-15*I)*(c - I*d)^(5/2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I* d]] + Sqrt[c + d*Tan[e + f*x]]*((23*I)*c^2 + 35*c*d - (15*I)*d^2 + d*((11* I)*c + 5*d)*Tan[e + f*x] + (3*I)*d^2*Tan[e + f*x]^2)))/(15*f)
Time = 0.77 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.07, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.393, Rules used = {3042, 4011, 3042, 4011, 3042, 4011, 3042, 4020, 27, 73, 221}
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.
| \(\displaystyle \int (a+i a \tan (e+f x)) (c+d \tan (e+f x))^{5/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int (a+i a \tan (e+f x)) (c+d \tan (e+f x))^{5/2}dx\) |
| \(\Big \downarrow \) 4011 |
| \(\displaystyle \int (c+d \tan (e+f x))^{3/2} (a (c-i d)+a (i c+d) \tan (e+f x))dx+\frac {2 i a (c+d \tan (e+f x))^{5/2}}{5 f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int (c+d \tan (e+f x))^{3/2} (a (c-i d)+a (i c+d) \tan (e+f x))dx+\frac {2 i a (c+d \tan (e+f x))^{5/2}}{5 f}\) |
| \(\Big \downarrow \) 4011 |
| \(\displaystyle \int \left (a (c-i d)^2+i a \tan (e+f x) (c-i d)^2\right ) \sqrt {c+d \tan (e+f x)}dx+\frac {2 i a (c+d \tan (e+f x))^{5/2}}{5 f}+\frac {2 a (d+i c) (c+d \tan (e+f x))^{3/2}}{3 f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \left (a (c-i d)^2+i a \tan (e+f x) (c-i d)^2\right ) \sqrt {c+d \tan (e+f x)}dx+\frac {2 i a (c+d \tan (e+f x))^{5/2}}{5 f}+\frac {2 a (d+i c) (c+d \tan (e+f x))^{3/2}}{3 f}\) |
| \(\Big \downarrow \) 4011 |
| \(\displaystyle \int \frac {a (c-i d)^3-a (i c+d)^3 \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx+\frac {2 i a (c+d \tan (e+f x))^{5/2}}{5 f}+\frac {2 a (d+i c) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 i a (c-i d)^2 \sqrt {c+d \tan (e+f x)}}{f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {a (c-i d)^3-a (i c+d)^3 \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx+\frac {2 i a (c+d \tan (e+f x))^{5/2}}{5 f}+\frac {2 a (d+i c) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 i a (c-i d)^2 \sqrt {c+d \tan (e+f x)}}{f}\) |
| \(\Big \downarrow \) 4020 |
| \(\displaystyle \frac {i a^2 (c-i d)^6 \int \frac {1}{a \sqrt {c+d \tan (e+f x)} \left (a (i c+d)^6-a (c-i d)^3 (i c+d)^3 \tan (e+f x)\right )}d\left (-a (i c+d)^3 \tan (e+f x)\right )}{f}+\frac {2 i a (c-i d)^2 \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 i a (c+d \tan (e+f x))^{5/2}}{5 f}+\frac {2 a (d+i c) (c+d \tan (e+f x))^{3/2}}{3 f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {i a (c-i d)^6 \int \frac {1}{\sqrt {c+d \tan (e+f x)} \left (a (i c+d)^6-a (c-i d)^3 (i c+d)^3 \tan (e+f x)\right )}d\left (-a (i c+d)^3 \tan (e+f x)\right )}{f}+\frac {2 i a (c-i d)^2 \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 i a (c+d \tan (e+f x))^{5/2}}{5 f}+\frac {2 a (d+i c) (c+d \tan (e+f x))^{3/2}}{3 f}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle -\frac {2 i a^2 (d+i c)^3 (c-i d)^6 \int \frac {1}{\frac {a (i c+d)^7}{d}+\frac {i a^3 (c-i d)^6 \tan ^2(e+f x) (i c+d)^6}{d}}d\sqrt {c+d \tan (e+f x)}}{d f}+\frac {2 i a (c-i d)^2 \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 i a (c+d \tan (e+f x))^{5/2}}{5 f}+\frac {2 a (d+i c) (c+d \tan (e+f x))^{3/2}}{3 f}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle -\frac {2 a (d+i c)^3 \text {arctanh}\left (\frac {a (d+i c)^3 \tan (e+f x)}{\sqrt {c-i d}}\right )}{f \sqrt {c-i d}}+\frac {2 a (d+i c) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 i a (c+d \tan (e+f x))^{5/2}}{5 f}+\frac {2 i a (c-i d)^2 \sqrt {c+d \tan (e+f x)}}{f}\) |
(-2*a*(I*c + d)^3*ArcTanh[(a*(I*c + d)^3*Tan[e + f*x])/Sqrt[c - I*d]])/(Sq rt[c - I*d]*f) + ((2*I)*a*(c - I*d)^2*Sqrt[c + d*Tan[e + f*x]])/f + (2*a*( I*c + d)*(c + d*Tan[e + f*x])^(3/2))/(3*f) + (((2*I)/5)*a*(c + d*Tan[e + f *x])^(5/2))/f
3.12.15.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, 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[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[d*((a + b*Tan[e + f*x])^m/(f*m)), x] + Int [(a + b*Tan[e + f*x])^(m - 1)*Simp[a*c - b*d + (b*c + a*d)*Tan[e + f*x], x] , x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && GtQ[m, 0]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[c*(d/f) Subst[Int[(a + (b/d)*x)^m/(d^2 + c*x), x], x, d*Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[ b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && EqQ[c^2 + d^2, 0]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 969 vs. \(2 (108 ) = 216\).
Time = 0.65 (sec) , antiderivative size = 970, normalized size of antiderivative = 7.40
| method | result | size |
| derivativedivides | \(\frac {a \left (\frac {2 i \left (c +d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}+\frac {2 i c \left (c +d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}+2 i c^{2} \sqrt {c +d \tan \left (f x +e \right )}-2 i d^{2} \sqrt {c +d \tan \left (f x +e \right )}+\frac {2 d \left (c +d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}+4 c d \sqrt {c +d \tan \left (f x +e \right )}+\frac {\frac {\left (-i c^{3} \sqrt {c^{2}+d^{2}}+3 i c \,d^{2} \sqrt {c^{2}+d^{2}}-i c^{4}+i d^{4}-3 c^{2} d \sqrt {c^{2}+d^{2}}+d^{3} \sqrt {c^{2}+d^{2}}-2 c^{3} d -2 c \,d^{3}\right ) \ln \left (d \tan \left (f x +e \right )+c +\sqrt {c +d \tan \left (f x +e \right )}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}+\sqrt {c^{2}+d^{2}}\right )}{2}+\frac {2 \left (-i \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c^{4}+i \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, d^{4}-2 \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c^{3} d -2 \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c \,d^{3}-\frac {\left (-i c^{3} \sqrt {c^{2}+d^{2}}+3 i c \,d^{2} \sqrt {c^{2}+d^{2}}-i c^{4}+i d^{4}-3 c^{2} d \sqrt {c^{2}+d^{2}}+d^{3} \sqrt {c^{2}+d^{2}}-2 c^{3} d -2 c \,d^{3}\right ) \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}}{2}\right ) \arctan \left (\frac {2 \sqrt {c +d \tan \left (f x +e \right )}+\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}\right )}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}}{\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, \sqrt {c^{2}+d^{2}}}+\frac {-\frac {\left (-i c^{3} \sqrt {c^{2}+d^{2}}+3 i c \,d^{2} \sqrt {c^{2}+d^{2}}-i c^{4}+i d^{4}-3 c^{2} d \sqrt {c^{2}+d^{2}}+d^{3} \sqrt {c^{2}+d^{2}}-2 c^{3} d -2 c \,d^{3}\right ) \ln \left (\sqrt {c +d \tan \left (f x +e \right )}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}-d \tan \left (f x +e \right )-c -\sqrt {c^{2}+d^{2}}\right )}{2}+\frac {2 \left (i \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c^{4}-i \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, d^{4}+2 \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c^{3} d +2 \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c \,d^{3}+\frac {\left (-i c^{3} \sqrt {c^{2}+d^{2}}+3 i c \,d^{2} \sqrt {c^{2}+d^{2}}-i c^{4}+i d^{4}-3 c^{2} d \sqrt {c^{2}+d^{2}}+d^{3} \sqrt {c^{2}+d^{2}}-2 c^{3} d -2 c \,d^{3}\right ) \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}}{2}\right ) \arctan \left (\frac {\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}-2 \sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}\right )}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}}{\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, \sqrt {c^{2}+d^{2}}}\right )}{f}\) | \(970\) |
| default | \(\frac {a \left (\frac {2 i \left (c +d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}+\frac {2 i c \left (c +d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}+2 i c^{2} \sqrt {c +d \tan \left (f x +e \right )}-2 i d^{2} \sqrt {c +d \tan \left (f x +e \right )}+\frac {2 d \left (c +d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}+4 c d \sqrt {c +d \tan \left (f x +e \right )}+\frac {\frac {\left (-i c^{3} \sqrt {c^{2}+d^{2}}+3 i c \,d^{2} \sqrt {c^{2}+d^{2}}-i c^{4}+i d^{4}-3 c^{2} d \sqrt {c^{2}+d^{2}}+d^{3} \sqrt {c^{2}+d^{2}}-2 c^{3} d -2 c \,d^{3}\right ) \ln \left (d \tan \left (f x +e \right )+c +\sqrt {c +d \tan \left (f x +e \right )}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}+\sqrt {c^{2}+d^{2}}\right )}{2}+\frac {2 \left (-i \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c^{4}+i \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, d^{4}-2 \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c^{3} d -2 \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c \,d^{3}-\frac {\left (-i c^{3} \sqrt {c^{2}+d^{2}}+3 i c \,d^{2} \sqrt {c^{2}+d^{2}}-i c^{4}+i d^{4}-3 c^{2} d \sqrt {c^{2}+d^{2}}+d^{3} \sqrt {c^{2}+d^{2}}-2 c^{3} d -2 c \,d^{3}\right ) \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}}{2}\right ) \arctan \left (\frac {2 \sqrt {c +d \tan \left (f x +e \right )}+\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}\right )}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}}{\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, \sqrt {c^{2}+d^{2}}}+\frac {-\frac {\left (-i c^{3} \sqrt {c^{2}+d^{2}}+3 i c \,d^{2} \sqrt {c^{2}+d^{2}}-i c^{4}+i d^{4}-3 c^{2} d \sqrt {c^{2}+d^{2}}+d^{3} \sqrt {c^{2}+d^{2}}-2 c^{3} d -2 c \,d^{3}\right ) \ln \left (\sqrt {c +d \tan \left (f x +e \right )}\, \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}-d \tan \left (f x +e \right )-c -\sqrt {c^{2}+d^{2}}\right )}{2}+\frac {2 \left (i \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c^{4}-i \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, d^{4}+2 \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c^{3} d +2 \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, c \,d^{3}+\frac {\left (-i c^{3} \sqrt {c^{2}+d^{2}}+3 i c \,d^{2} \sqrt {c^{2}+d^{2}}-i c^{4}+i d^{4}-3 c^{2} d \sqrt {c^{2}+d^{2}}+d^{3} \sqrt {c^{2}+d^{2}}-2 c^{3} d -2 c \,d^{3}\right ) \sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}}{2}\right ) \arctan \left (\frac {\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}-2 \sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}\right )}{\sqrt {2 \sqrt {c^{2}+d^{2}}-2 c}}}{\sqrt {2 \sqrt {c^{2}+d^{2}}+2 c}\, \sqrt {c^{2}+d^{2}}}\right )}{f}\) | \(970\) |
| parts | \(\text {Expression too large to display}\) | \(2413\) |
1/f*a*(2/5*I*(c+d*tan(f*x+e))^(5/2)+2/3*I*c*(c+d*tan(f*x+e))^(3/2)+2*I*c^2 *(c+d*tan(f*x+e))^(1/2)-2*I*d^2*(c+d*tan(f*x+e))^(1/2)+2/3*d*(c+d*tan(f*x+ e))^(3/2)+4*c*d*(c+d*tan(f*x+e))^(1/2)+1/(2*(c^2+d^2)^(1/2)+2*c)^(1/2)/(c^ 2+d^2)^(1/2)*(1/2*(-I*c^3*(c^2+d^2)^(1/2)+3*I*c*d^2*(c^2+d^2)^(1/2)-I*c^4+ I*d^4-3*c^2*d*(c^2+d^2)^(1/2)+d^3*(c^2+d^2)^(1/2)-2*c^3*d-2*c*d^3)*ln(d*ta n(f*x+e)+c+(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+d^2)^ (1/2))+2*(-I*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c^4+I*(2*(c^2+d^2)^(1/2)+2*c)^( 1/2)*d^4-2*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c^3*d-2*(2*(c^2+d^2)^(1/2)+2*c)^( 1/2)*c*d^3-1/2*(-I*c^3*(c^2+d^2)^(1/2)+3*I*c*d^2*(c^2+d^2)^(1/2)-I*c^4+I*d ^4-3*c^2*d*(c^2+d^2)^(1/2)+d^3*(c^2+d^2)^(1/2)-2*c^3*d-2*c*d^3)*(2*(c^2+d^ 2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+ e))^(1/2)+(2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)))+1 /(2*(c^2+d^2)^(1/2)+2*c)^(1/2)/(c^2+d^2)^(1/2)*(-1/2*(-I*c^3*(c^2+d^2)^(1/ 2)+3*I*c*d^2*(c^2+d^2)^(1/2)-I*c^4+I*d^4-3*c^2*d*(c^2+d^2)^(1/2)+d^3*(c^2+ d^2)^(1/2)-2*c^3*d-2*c*d^3)*ln((c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2 *c)^(1/2)-d*tan(f*x+e)-c-(c^2+d^2)^(1/2))+2*(I*(2*(c^2+d^2)^(1/2)+2*c)^(1/ 2)*c^4-I*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*d^4+2*(2*(c^2+d^2)^(1/2)+2*c)^(1/2) *c^3*d+2*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c*d^3+1/2*(-I*c^3*(c^2+d^2)^(1/2)+3 *I*c*d^2*(c^2+d^2)^(1/2)-I*c^4+I*d^4-3*c^2*d*(c^2+d^2)^(1/2)+d^3*(c^2+d^2) ^(1/2)-2*c^3*d-2*c*d^3)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/...
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 732 vs. \(2 (101) = 202\).
Time = 0.33 (sec) , antiderivative size = 732, normalized size of antiderivative = 5.59 \[ \int (a+i a \tan (e+f x)) (c+d \tan (e+f x))^{5/2} \, dx=-\frac {15 \, {\left (f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \sqrt {-\frac {a^{2} c^{5} - 5 i \, a^{2} c^{4} d - 10 \, a^{2} c^{3} d^{2} + 10 i \, a^{2} c^{2} d^{3} + 5 \, a^{2} c d^{4} - i \, a^{2} d^{5}}{f^{2}}} \log \left (\frac {2 \, {\left (a c^{3} - 2 i \, a c^{2} d - a c d^{2} - {\left (i \, f e^{\left (2 i \, f x + 2 i \, e\right )} + i \, f\right )} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {a^{2} c^{5} - 5 i \, a^{2} c^{4} d - 10 \, a^{2} c^{3} d^{2} + 10 i \, a^{2} c^{2} d^{3} + 5 \, a^{2} c d^{4} - i \, a^{2} d^{5}}{f^{2}}} + {\left (a c^{3} - 3 i \, a c^{2} d - 3 \, a c d^{2} + i \, a d^{3}\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{a c^{2} - 2 i \, a c d - a d^{2}}\right ) - 15 \, {\left (f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \sqrt {-\frac {a^{2} c^{5} - 5 i \, a^{2} c^{4} d - 10 \, a^{2} c^{3} d^{2} + 10 i \, a^{2} c^{2} d^{3} + 5 \, a^{2} c d^{4} - i \, a^{2} d^{5}}{f^{2}}} \log \left (\frac {2 \, {\left (a c^{3} - 2 i \, a c^{2} d - a c d^{2} - {\left (-i \, f e^{\left (2 i \, f x + 2 i \, e\right )} - i \, f\right )} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {a^{2} c^{5} - 5 i \, a^{2} c^{4} d - 10 \, a^{2} c^{3} d^{2} + 10 i \, a^{2} c^{2} d^{3} + 5 \, a^{2} c d^{4} - i \, a^{2} d^{5}}{f^{2}}} + {\left (a c^{3} - 3 i \, a c^{2} d - 3 \, a c d^{2} + i \, a d^{3}\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{a c^{2} - 2 i \, a c d - a d^{2}}\right ) + 4 \, {\left (-23 i \, a c^{2} - 24 \, a c d + 13 i \, a d^{2} + 23 \, {\left (-i \, a c^{2} - 2 \, a c d + i \, a d^{2}\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, {\left (-23 i \, a c^{2} - 35 \, a c d + 12 i \, a d^{2}\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{30 \, {\left (f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \]
-1/30*(15*(f*e^(4*I*f*x + 4*I*e) + 2*f*e^(2*I*f*x + 2*I*e) + f)*sqrt(-(a^2 *c^5 - 5*I*a^2*c^4*d - 10*a^2*c^3*d^2 + 10*I*a^2*c^2*d^3 + 5*a^2*c*d^4 - I *a^2*d^5)/f^2)*log(2*(a*c^3 - 2*I*a*c^2*d - a*c*d^2 - (I*f*e^(2*I*f*x + 2* I*e) + I*f)*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2 *I*e) + 1))*sqrt(-(a^2*c^5 - 5*I*a^2*c^4*d - 10*a^2*c^3*d^2 + 10*I*a^2*c^2 *d^3 + 5*a^2*c*d^4 - I*a^2*d^5)/f^2) + (a*c^3 - 3*I*a*c^2*d - 3*a*c*d^2 + I*a*d^3)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/(a*c^2 - 2*I*a*c*d - a* d^2)) - 15*(f*e^(4*I*f*x + 4*I*e) + 2*f*e^(2*I*f*x + 2*I*e) + f)*sqrt(-(a^ 2*c^5 - 5*I*a^2*c^4*d - 10*a^2*c^3*d^2 + 10*I*a^2*c^2*d^3 + 5*a^2*c*d^4 - I*a^2*d^5)/f^2)*log(2*(a*c^3 - 2*I*a*c^2*d - a*c*d^2 - (-I*f*e^(2*I*f*x + 2*I*e) - I*f)*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-(a^2*c^5 - 5*I*a^2*c^4*d - 10*a^2*c^3*d^2 + 10*I*a^2*c ^2*d^3 + 5*a^2*c*d^4 - I*a^2*d^5)/f^2) + (a*c^3 - 3*I*a*c^2*d - 3*a*c*d^2 + I*a*d^3)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/(a*c^2 - 2*I*a*c*d - a*d^2)) + 4*(-23*I*a*c^2 - 24*a*c*d + 13*I*a*d^2 + 23*(-I*a*c^2 - 2*a*c*d + I*a*d^2)*e^(4*I*f*x + 4*I*e) + 2*(-23*I*a*c^2 - 35*a*c*d + 12*I*a*d^2)*e ^(2*I*f*x + 2*I*e))*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I *f*x + 2*I*e) + 1)))/(f*e^(4*I*f*x + 4*I*e) + 2*f*e^(2*I*f*x + 2*I*e) + f)
\[ \int (a+i a \tan (e+f x)) (c+d \tan (e+f x))^{5/2} \, dx=i a \left (\int \left (- i c^{2} \sqrt {c + d \tan {\left (e + f x \right )}}\right )\, dx + \int c^{2} \sqrt {c + d \tan {\left (e + f x \right )}} \tan {\left (e + f x \right )}\, dx + \int d^{2} \sqrt {c + d \tan {\left (e + f x \right )}} \tan ^{3}{\left (e + f x \right )}\, dx + \int \left (- i d^{2} \sqrt {c + d \tan {\left (e + f x \right )}} \tan ^{2}{\left (e + f x \right )}\right )\, dx + \int 2 c d \sqrt {c + d \tan {\left (e + f x \right )}} \tan ^{2}{\left (e + f x \right )}\, dx + \int \left (- 2 i c d \sqrt {c + d \tan {\left (e + f x \right )}} \tan {\left (e + f x \right )}\right )\, dx\right ) \]
I*a*(Integral(-I*c**2*sqrt(c + d*tan(e + f*x)), x) + Integral(c**2*sqrt(c + d*tan(e + f*x))*tan(e + f*x), x) + Integral(d**2*sqrt(c + d*tan(e + f*x) )*tan(e + f*x)**3, x) + Integral(-I*d**2*sqrt(c + d*tan(e + f*x))*tan(e + f*x)**2, x) + Integral(2*c*d*sqrt(c + d*tan(e + f*x))*tan(e + f*x)**2, x) + Integral(-2*I*c*d*sqrt(c + d*tan(e + f*x))*tan(e + f*x), x))
Timed out. \[ \int (a+i a \tan (e+f x)) (c+d \tan (e+f x))^{5/2} \, dx=\text {Timed out} \]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 297 vs. \(2 (101) = 202\).
Time = 0.71 (sec) , antiderivative size = 297, normalized size of antiderivative = 2.27 \[ \int (a+i a \tan (e+f x)) (c+d \tan (e+f x))^{5/2} \, dx=\frac {2}{15} i \, a {\left (\frac {30 \, {\left (c^{3} - 3 i \, c^{2} d - 3 \, c d^{2} + i \, d^{3}\right )} \arctan \left (\frac {2 \, {\left (\sqrt {d \tan \left (f x + e\right ) + c} c - \sqrt {c^{2} + d^{2}} \sqrt {d \tan \left (f x + e\right ) + c}\right )}}{c \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}} - i \, \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}} d - \sqrt {c^{2} + d^{2}} \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}}}\right )}{\sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}} f {\left (-\frac {i \, d}{c - \sqrt {c^{2} + d^{2}}} + 1\right )}} + \frac {3 \, {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} f^{4} + 5 \, {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} c f^{4} + 15 \, \sqrt {d \tan \left (f x + e\right ) + c} c^{2} f^{4} - 5 i \, {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} d f^{4} - 30 i \, \sqrt {d \tan \left (f x + e\right ) + c} c d f^{4} - 15 \, \sqrt {d \tan \left (f x + e\right ) + c} d^{2} f^{4}}{f^{5}}\right )} \]
2/15*I*a*(30*(c^3 - 3*I*c^2*d - 3*c*d^2 + I*d^3)*arctan(2*(sqrt(d*tan(f*x + e) + c)*c - sqrt(c^2 + d^2)*sqrt(d*tan(f*x + e) + c))/(c*sqrt(-2*c + 2*s qrt(c^2 + d^2)) - I*sqrt(-2*c + 2*sqrt(c^2 + d^2))*d - sqrt(c^2 + d^2)*sqr t(-2*c + 2*sqrt(c^2 + d^2))))/(sqrt(-2*c + 2*sqrt(c^2 + d^2))*f*(-I*d/(c - sqrt(c^2 + d^2)) + 1)) + (3*(d*tan(f*x + e) + c)^(5/2)*f^4 + 5*(d*tan(f*x + e) + c)^(3/2)*c*f^4 + 15*sqrt(d*tan(f*x + e) + c)*c^2*f^4 - 5*I*(d*tan( f*x + e) + c)^(3/2)*d*f^4 - 30*I*sqrt(d*tan(f*x + e) + c)*c*d*f^4 - 15*sqr t(d*tan(f*x + e) + c)*d^2*f^4)/f^5)
Time = 31.31 (sec) , antiderivative size = 3899, normalized size of antiderivative = 29.76 \[ \int (a+i a \tan (e+f x)) (c+d \tan (e+f x))^{5/2} \, dx=\text {Too large to display} \]
log(((((-a^4*d^2*f^4*(5*c^4 + d^4 - 10*c^2*d^2)^2)^(1/2) - a^2*c^5*f^2 - 5 *a^2*c*d^4*f^2 + 10*a^2*c^3*d^2*f^2)/f^4)^(1/2)*(((((-a^4*d^2*f^4*(5*c^4 + d^4 - 10*c^2*d^2)^2)^(1/2) - a^2*c^5*f^2 - 5*a^2*c*d^4*f^2 + 10*a^2*c^3*d ^2*f^2)/f^4)^(1/2)*(64*a*c^3*d^3 + 64*a*c*d^5 - 32*c*d^2*f*(((-a^4*d^2*f^4 *(5*c^4 + d^4 - 10*c^2*d^2)^2)^(1/2) - a^2*c^5*f^2 - 5*a^2*c*d^4*f^2 + 10* a^2*c^3*d^2*f^2)/f^4)^(1/2)*(c + d*tan(e + f*x))^(1/2)))/(2*f) - (16*a^2*d ^2*(c + d*tan(e + f*x))^(1/2)*(c^6 - d^6 + 15*c^2*d^4 - 15*c^4*d^2))/f^2)) /2 - (8*a^3*d^3*(3*c^2 - d^2)*(c^2 + d^2)^3)/f^3)*((20*a^4*c^2*d^8*f^4 - a ^4*d^10*f^4 - 110*a^4*c^4*d^6*f^4 + 100*a^4*c^6*d^4*f^4 - 25*a^4*c^8*d^2*f ^4)^(1/2)/(4*f^4) - (a^2*c^5)/(4*f^2) - (5*a^2*c*d^4)/(4*f^2) + (5*a^2*c^3 *d^2)/(2*f^2))^(1/2) - log(((((-a^4*d^2*f^4*(5*c^4 + d^4 - 10*c^2*d^2)^2)^ (1/2) - a^2*c^5*f^2 - 5*a^2*c*d^4*f^2 + 10*a^2*c^3*d^2*f^2)/f^4)^(1/2)*((( ((-a^4*d^2*f^4*(5*c^4 + d^4 - 10*c^2*d^2)^2)^(1/2) - a^2*c^5*f^2 - 5*a^2*c *d^4*f^2 + 10*a^2*c^3*d^2*f^2)/f^4)^(1/2)*(64*a*c^3*d^3 + 64*a*c*d^5 + 32* c*d^2*f*(((-a^4*d^2*f^4*(5*c^4 + d^4 - 10*c^2*d^2)^2)^(1/2) - a^2*c^5*f^2 - 5*a^2*c*d^4*f^2 + 10*a^2*c^3*d^2*f^2)/f^4)^(1/2)*(c + d*tan(e + f*x))^(1 /2)))/(2*f) + (16*a^2*d^2*(c + d*tan(e + f*x))^(1/2)*(c^6 - d^6 + 15*c^2*d ^4 - 15*c^4*d^2))/f^2))/2 - (8*a^3*d^3*(3*c^2 - d^2)*(c^2 + d^2)^3)/f^3)*( ((20*a^4*c^2*d^8*f^4 - a^4*d^10*f^4 - 110*a^4*c^4*d^6*f^4 + 100*a^4*c^6*d^ 4*f^4 - 25*a^4*c^8*d^2*f^4)^(1/2) - a^2*c^5*f^2 - 5*a^2*c*d^4*f^2 + 10*...