Integrand size = 28, antiderivative size = 329 \[ \int \frac {(d \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^3} \, dx=\frac {d^{5/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{8 \sqrt {2} a^3 f}-\frac {d^{5/2} \arctan \left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{8 \sqrt {2} a^3 f}-\frac {d^{5/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{16 \sqrt {2} a^3 f}+\frac {d^{5/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{16 \sqrt {2} a^3 f}-\frac {d (d \tan (e+f x))^{3/2}}{6 f (a+i a \tan (e+f x))^3}+\frac {i d^2 \sqrt {d \tan (e+f x)}}{4 a f (a+i a \tan (e+f x))^2}-\frac {i d^2 \sqrt {d \tan (e+f x)}}{4 f \left (a^3+i a^3 \tan (e+f x)\right )} \]
1/16*d^(5/2)*arctan(1-2^(1/2)*(d*tan(f*x+e))^(1/2)/d^(1/2))/a^3/f*2^(1/2)- 1/16*d^(5/2)*arctan(1+2^(1/2)*(d*tan(f*x+e))^(1/2)/d^(1/2))/a^3/f*2^(1/2)- 1/32*d^(5/2)*ln(d^(1/2)-2^(1/2)*(d*tan(f*x+e))^(1/2)+d^(1/2)*tan(f*x+e))/a ^3/f*2^(1/2)+1/32*d^(5/2)*ln(d^(1/2)+2^(1/2)*(d*tan(f*x+e))^(1/2)+d^(1/2)* tan(f*x+e))/a^3/f*2^(1/2)-1/6*d*(d*tan(f*x+e))^(3/2)/f/(a+I*a*tan(f*x+e))^ 3+1/4*I*d^2*(d*tan(f*x+e))^(1/2)/a/f/(a+I*a*tan(f*x+e))^2-1/4*I*d^2*(d*tan (f*x+e))^(1/2)/f/(a^3+I*a^3*tan(f*x+e))
Time = 1.56 (sec) , antiderivative size = 195, normalized size of antiderivative = 0.59 \[ \int \frac {(d \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^3} \, dx=\frac {d \sec ^3(e+f x) \left (6 i \left (-d^2\right )^{3/4} \arctan \left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt [4]{-d^2}}\right ) (\cos (3 (e+f x))+i \sin (3 (e+f x)))+6 \left (-d^2\right )^{3/4} \text {arctanh}\left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt [4]{-d^2}}\right ) (-i \cos (3 (e+f x))+\sin (3 (e+f x)))+d \left (-3 \cos (e+f x)+3 \cos (3 (e+f x))+4 i \cos ^2(e+f x) \sin (e+f x)\right ) \sqrt {d \tan (e+f x)}\right )}{48 a^3 f (-i+\tan (e+f x))^3} \]
(d*Sec[e + f*x]^3*((6*I)*(-d^2)^(3/4)*ArcTan[Sqrt[d*Tan[e + f*x]]/(-d^2)^( 1/4)]*(Cos[3*(e + f*x)] + I*Sin[3*(e + f*x)]) + 6*(-d^2)^(3/4)*ArcTanh[Sqr t[d*Tan[e + f*x]]/(-d^2)^(1/4)]*((-I)*Cos[3*(e + f*x)] + Sin[3*(e + f*x)]) + d*(-3*Cos[e + f*x] + 3*Cos[3*(e + f*x)] + (4*I)*Cos[e + f*x]^2*Sin[e + f*x])*Sqrt[d*Tan[e + f*x]]))/(48*a^3*f*(-I + Tan[e + f*x])^3)
Time = 1.10 (sec) , antiderivative size = 307, normalized size of antiderivative = 0.93, number of steps used = 22, number of rules used = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {3042, 4041, 27, 3042, 4078, 27, 3042, 4079, 27, 2030, 3042, 3957, 266, 826, 1476, 1082, 217, 1479, 25, 27, 1103}
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.
| \(\displaystyle \int \frac {(d \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(d \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^3}dx\) |
| \(\Big \downarrow \) 4041 |
| \(\displaystyle -\frac {\int -\frac {3 \sqrt {d \tan (e+f x)} \left (a d^2-3 i a d^2 \tan (e+f x)\right )}{2 (i \tan (e+f x) a+a)^2}dx}{6 a^2}-\frac {d (d \tan (e+f x))^{3/2}}{6 f (a+i a \tan (e+f x))^3}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {\sqrt {d \tan (e+f x)} \left (a d^2-3 i a d^2 \tan (e+f x)\right )}{(i \tan (e+f x) a+a)^2}dx}{4 a^2}-\frac {d (d \tan (e+f x))^{3/2}}{6 f (a+i a \tan (e+f x))^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\sqrt {d \tan (e+f x)} \left (a d^2-3 i a d^2 \tan (e+f x)\right )}{(i \tan (e+f x) a+a)^2}dx}{4 a^2}-\frac {d (d \tan (e+f x))^{3/2}}{6 f (a+i a \tan (e+f x))^3}\) |
| \(\Big \downarrow \) 4078 |
| \(\displaystyle \frac {\frac {i a d^2 \sqrt {d \tan (e+f x)}}{f (a+i a \tan (e+f x))^2}-\frac {\int \frac {2 \left (i a^2 d^3+3 a^2 \tan (e+f x) d^3\right )}{\sqrt {d \tan (e+f x)} (i \tan (e+f x) a+a)}dx}{4 a^2}}{4 a^2}-\frac {d (d \tan (e+f x))^{3/2}}{6 f (a+i a \tan (e+f x))^3}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {i a d^2 \sqrt {d \tan (e+f x)}}{f (a+i a \tan (e+f x))^2}-\frac {\int \frac {i a^2 d^3+3 a^2 \tan (e+f x) d^3}{\sqrt {d \tan (e+f x)} (i \tan (e+f x) a+a)}dx}{2 a^2}}{4 a^2}-\frac {d (d \tan (e+f x))^{3/2}}{6 f (a+i a \tan (e+f x))^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {i a d^2 \sqrt {d \tan (e+f x)}}{f (a+i a \tan (e+f x))^2}-\frac {\int \frac {i a^2 d^3+3 a^2 \tan (e+f x) d^3}{\sqrt {d \tan (e+f x)} (i \tan (e+f x) a+a)}dx}{2 a^2}}{4 a^2}-\frac {d (d \tan (e+f x))^{3/2}}{6 f (a+i a \tan (e+f x))^3}\) |
| \(\Big \downarrow \) 4079 |
| \(\displaystyle \frac {\frac {i a d^2 \sqrt {d \tan (e+f x)}}{f (a+i a \tan (e+f x))^2}-\frac {\frac {\int \frac {2 a^3 d^4 \tan (e+f x)}{\sqrt {d \tan (e+f x)}}dx}{2 a^2 d}+\frac {2 i a^2 d^2 \sqrt {d \tan (e+f x)}}{f (a+i a \tan (e+f x))}}{2 a^2}}{4 a^2}-\frac {d (d \tan (e+f x))^{3/2}}{6 f (a+i a \tan (e+f x))^3}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {i a d^2 \sqrt {d \tan (e+f x)}}{f (a+i a \tan (e+f x))^2}-\frac {a d^3 \int \frac {\tan (e+f x)}{\sqrt {d \tan (e+f x)}}dx+\frac {2 i a^2 d^2 \sqrt {d \tan (e+f x)}}{f (a+i a \tan (e+f x))}}{2 a^2}}{4 a^2}-\frac {d (d \tan (e+f x))^{3/2}}{6 f (a+i a \tan (e+f x))^3}\) |
| \(\Big \downarrow \) 2030 |
| \(\displaystyle \frac {\frac {i a d^2 \sqrt {d \tan (e+f x)}}{f (a+i a \tan (e+f x))^2}-\frac {a d^2 \int \sqrt {d \tan (e+f x)}dx+\frac {2 i a^2 d^2 \sqrt {d \tan (e+f x)}}{f (a+i a \tan (e+f x))}}{2 a^2}}{4 a^2}-\frac {d (d \tan (e+f x))^{3/2}}{6 f (a+i a \tan (e+f x))^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {i a d^2 \sqrt {d \tan (e+f x)}}{f (a+i a \tan (e+f x))^2}-\frac {a d^2 \int \sqrt {d \tan (e+f x)}dx+\frac {2 i a^2 d^2 \sqrt {d \tan (e+f x)}}{f (a+i a \tan (e+f x))}}{2 a^2}}{4 a^2}-\frac {d (d \tan (e+f x))^{3/2}}{6 f (a+i a \tan (e+f x))^3}\) |
| \(\Big \downarrow \) 3957 |
| \(\displaystyle \frac {\frac {i a d^2 \sqrt {d \tan (e+f x)}}{f (a+i a \tan (e+f x))^2}-\frac {\frac {a d^3 \int \frac {\sqrt {d \tan (e+f x)}}{\tan ^2(e+f x) d^2+d^2}d(d \tan (e+f x))}{f}+\frac {2 i a^2 d^2 \sqrt {d \tan (e+f x)}}{f (a+i a \tan (e+f x))}}{2 a^2}}{4 a^2}-\frac {d (d \tan (e+f x))^{3/2}}{6 f (a+i a \tan (e+f x))^3}\) |
| \(\Big \downarrow \) 266 |
| \(\displaystyle \frac {\frac {i a d^2 \sqrt {d \tan (e+f x)}}{f (a+i a \tan (e+f x))^2}-\frac {\frac {2 a d^3 \int \frac {d^2 \tan ^2(e+f x)}{d^4 \tan ^4(e+f x)+d^2}d\sqrt {d \tan (e+f x)}}{f}+\frac {2 i a^2 d^2 \sqrt {d \tan (e+f x)}}{f (a+i a \tan (e+f x))}}{2 a^2}}{4 a^2}-\frac {d (d \tan (e+f x))^{3/2}}{6 f (a+i a \tan (e+f x))^3}\) |
| \(\Big \downarrow \) 826 |
| \(\displaystyle \frac {\frac {i a d^2 \sqrt {d \tan (e+f x)}}{f (a+i a \tan (e+f x))^2}-\frac {\frac {2 a d^3 \left (\frac {1}{2} \int \frac {d^2 \tan ^2(e+f x)+d}{d^4 \tan ^4(e+f x)+d^2}d\sqrt {d \tan (e+f x)}-\frac {1}{2} \int \frac {d-d^2 \tan ^2(e+f x)}{d^4 \tan ^4(e+f x)+d^2}d\sqrt {d \tan (e+f x)}\right )}{f}+\frac {2 i a^2 d^2 \sqrt {d \tan (e+f x)}}{f (a+i a \tan (e+f x))}}{2 a^2}}{4 a^2}-\frac {d (d \tan (e+f x))^{3/2}}{6 f (a+i a \tan (e+f x))^3}\) |
| \(\Big \downarrow \) 1476 |
| \(\displaystyle \frac {\frac {i a d^2 \sqrt {d \tan (e+f x)}}{f (a+i a \tan (e+f x))^2}-\frac {\frac {2 a d^3 \left (\frac {1}{2} \left (\frac {1}{2} \int \frac {1}{d^2 \tan ^2(e+f x)-\sqrt {2} d^{3/2} \tan (e+f x)+d}d\sqrt {d \tan (e+f x)}+\frac {1}{2} \int \frac {1}{d^2 \tan ^2(e+f x)+\sqrt {2} d^{3/2} \tan (e+f x)+d}d\sqrt {d \tan (e+f x)}\right )-\frac {1}{2} \int \frac {d-d^2 \tan ^2(e+f x)}{d^4 \tan ^4(e+f x)+d^2}d\sqrt {d \tan (e+f x)}\right )}{f}+\frac {2 i a^2 d^2 \sqrt {d \tan (e+f x)}}{f (a+i a \tan (e+f x))}}{2 a^2}}{4 a^2}-\frac {d (d \tan (e+f x))^{3/2}}{6 f (a+i a \tan (e+f x))^3}\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle \frac {\frac {i a d^2 \sqrt {d \tan (e+f x)}}{f (a+i a \tan (e+f x))^2}-\frac {\frac {2 a d^3 \left (\frac {1}{2} \left (\frac {\int \frac {1}{-d^2 \tan ^2(e+f x)-1}d\left (1-\sqrt {2} \sqrt {d} \tan (e+f x)\right )}{\sqrt {2} \sqrt {d}}-\frac {\int \frac {1}{-d^2 \tan ^2(e+f x)-1}d\left (\sqrt {2} \sqrt {d} \tan (e+f x)+1\right )}{\sqrt {2} \sqrt {d}}\right )-\frac {1}{2} \int \frac {d-d^2 \tan ^2(e+f x)}{d^4 \tan ^4(e+f x)+d^2}d\sqrt {d \tan (e+f x)}\right )}{f}+\frac {2 i a^2 d^2 \sqrt {d \tan (e+f x)}}{f (a+i a \tan (e+f x))}}{2 a^2}}{4 a^2}-\frac {d (d \tan (e+f x))^{3/2}}{6 f (a+i a \tan (e+f x))^3}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {\frac {i a d^2 \sqrt {d \tan (e+f x)}}{f (a+i a \tan (e+f x))^2}-\frac {\frac {2 a d^3 \left (\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {d} \tan (e+f x)+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {d} \tan (e+f x)\right )}{\sqrt {2} \sqrt {d}}\right )-\frac {1}{2} \int \frac {d-d^2 \tan ^2(e+f x)}{d^4 \tan ^4(e+f x)+d^2}d\sqrt {d \tan (e+f x)}\right )}{f}+\frac {2 i a^2 d^2 \sqrt {d \tan (e+f x)}}{f (a+i a \tan (e+f x))}}{2 a^2}}{4 a^2}-\frac {d (d \tan (e+f x))^{3/2}}{6 f (a+i a \tan (e+f x))^3}\) |
| \(\Big \downarrow \) 1479 |
| \(\displaystyle \frac {\frac {i a d^2 \sqrt {d \tan (e+f x)}}{f (a+i a \tan (e+f x))^2}-\frac {\frac {2 a d^3 \left (\frac {1}{2} \left (\frac {\int -\frac {\sqrt {2} \sqrt {d}-2 \sqrt {d \tan (e+f x)}}{d^2 \tan ^2(e+f x)-\sqrt {2} d^{3/2} \tan (e+f x)+d}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}+\frac {\int -\frac {\sqrt {2} \left (\sqrt {d}+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{d^2 \tan ^2(e+f x)+\sqrt {2} d^{3/2} \tan (e+f x)+d}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {d} \tan (e+f x)+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {d} \tan (e+f x)\right )}{\sqrt {2} \sqrt {d}}\right )\right )}{f}+\frac {2 i a^2 d^2 \sqrt {d \tan (e+f x)}}{f (a+i a \tan (e+f x))}}{2 a^2}}{4 a^2}-\frac {d (d \tan (e+f x))^{3/2}}{6 f (a+i a \tan (e+f x))^3}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {i a d^2 \sqrt {d \tan (e+f x)}}{f (a+i a \tan (e+f x))^2}-\frac {\frac {2 a d^3 \left (\frac {1}{2} \left (-\frac {\int \frac {\sqrt {2} \sqrt {d}-2 \sqrt {d \tan (e+f x)}}{d^2 \tan ^2(e+f x)-\sqrt {2} d^{3/2} \tan (e+f x)+d}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}-\frac {\int \frac {\sqrt {2} \left (\sqrt {d}+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{d^2 \tan ^2(e+f x)+\sqrt {2} d^{3/2} \tan (e+f x)+d}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {d} \tan (e+f x)+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {d} \tan (e+f x)\right )}{\sqrt {2} \sqrt {d}}\right )\right )}{f}+\frac {2 i a^2 d^2 \sqrt {d \tan (e+f x)}}{f (a+i a \tan (e+f x))}}{2 a^2}}{4 a^2}-\frac {d (d \tan (e+f x))^{3/2}}{6 f (a+i a \tan (e+f x))^3}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {i a d^2 \sqrt {d \tan (e+f x)}}{f (a+i a \tan (e+f x))^2}-\frac {\frac {2 a d^3 \left (\frac {1}{2} \left (-\frac {\int \frac {\sqrt {2} \sqrt {d}-2 \sqrt {d \tan (e+f x)}}{d^2 \tan ^2(e+f x)-\sqrt {2} d^{3/2} \tan (e+f x)+d}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}-\frac {\int \frac {\sqrt {d}+\sqrt {2} \sqrt {d \tan (e+f x)}}{d^2 \tan ^2(e+f x)+\sqrt {2} d^{3/2} \tan (e+f x)+d}d\sqrt {d \tan (e+f x)}}{2 \sqrt {d}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {d} \tan (e+f x)+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {d} \tan (e+f x)\right )}{\sqrt {2} \sqrt {d}}\right )\right )}{f}+\frac {2 i a^2 d^2 \sqrt {d \tan (e+f x)}}{f (a+i a \tan (e+f x))}}{2 a^2}}{4 a^2}-\frac {d (d \tan (e+f x))^{3/2}}{6 f (a+i a \tan (e+f x))^3}\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle \frac {\frac {i a d^2 \sqrt {d \tan (e+f x)}}{f (a+i a \tan (e+f x))^2}-\frac {\frac {2 a d^3 \left (\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {d} \tan (e+f x)+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {d} \tan (e+f x)\right )}{\sqrt {2} \sqrt {d}}\right )+\frac {1}{2} \left (\frac {\log \left (-\sqrt {2} d^{3/2} \tan (e+f x)+d^2 \tan ^2(e+f x)+d\right )}{2 \sqrt {2} \sqrt {d}}-\frac {\log \left (\sqrt {2} d^{3/2} \tan (e+f x)+d^2 \tan ^2(e+f x)+d\right )}{2 \sqrt {2} \sqrt {d}}\right )\right )}{f}+\frac {2 i a^2 d^2 \sqrt {d \tan (e+f x)}}{f (a+i a \tan (e+f x))}}{2 a^2}}{4 a^2}-\frac {d (d \tan (e+f x))^{3/2}}{6 f (a+i a \tan (e+f x))^3}\) |
-1/6*(d*(d*Tan[e + f*x])^(3/2))/(f*(a + I*a*Tan[e + f*x])^3) + ((I*a*d^2*S qrt[d*Tan[e + f*x]])/(f*(a + I*a*Tan[e + f*x])^2) - ((2*a*d^3*((-(ArcTan[1 - Sqrt[2]*Sqrt[d]*Tan[e + f*x]]/(Sqrt[2]*Sqrt[d])) + ArcTan[1 + Sqrt[2]*S qrt[d]*Tan[e + f*x]]/(Sqrt[2]*Sqrt[d]))/2 + (Log[d - Sqrt[2]*d^(3/2)*Tan[e + f*x] + d^2*Tan[e + f*x]^2]/(2*Sqrt[2]*Sqrt[d]) - Log[d + Sqrt[2]*d^(3/2 )*Tan[e + f*x] + d^2*Tan[e + f*x]^2]/(2*Sqrt[2]*Sqrt[d]))/2))/f + ((2*I)*a ^2*d^2*Sqrt[d*Tan[e + f*x]])/(f*(a + I*a*Tan[e + f*x])))/(2*a^2))/(4*a^2)
3.2.81.3.1 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_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De nominator[m]}, Simp[k/c Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) ^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I ntBinomialQ[a, b, c, 2, m, p, x]
Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Simp[1/(2*s) Int[(r + s*x^2)/(a + b*x^ 4), x], x] - Simp[1/(2*s) Int[(r - s*x^2)/(a + b*x^4), x], x]] /; FreeQ[{ a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ, b]]))
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S implify[a*(c/b^2)]}, Simp[-2/b Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b )], x] /; RationalQ[q] && (EqQ[q^2, 1] || !RationalQ[b^2 - 4*a*c])] /; Fre eQ[{a, b, c}, x]
Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 2*(d/e), 2]}, Simp[e/(2*c) Int[1/Simp[d/e + q*x + x^2, x], x], x] + Simp[ e/(2*c) Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ -2*(d/e), 2]}, Simp[e/(2*c*q) Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Simp[e/(2*c*q) Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /; F reeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]
Int[(Fx_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Simp[1/b^m Int[(b*v) ^(m + n)*Fx, x], x] /; FreeQ[{b, n}, x] && IntegerQ[m]
Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b/d Subst[Int [x^n/(b^2 + x^2), x], x, b*Tan[c + d*x]], x] /; FreeQ[{b, c, d, n}, x] && !IntegerQ[n]
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-(b*c - a*d))*(a + b*Tan[e + f*x])^m* ((c + d*Tan[e + f*x])^(n - 1)/(2*a*f*m)), x] + Simp[1/(2*a^2*m) Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n - 2)*Simp[c*(a*c*m + b*d*(n - 1)) - d*(b*c*m + a*d*(n - 1)) - d*(b*d*(m - n + 1) - a*c*(m + n - 1))*Ta n[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && LtQ[m, 0] && GtQ[n, 1] && (In tegerQ[m] || IntegersQ[2*m, 2*n])
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[(-(A*b - a*B))*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^n/(2*a*f*m)), x] + Simp[1/(2*a^2*m) Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f* x])^(n - 1)*Simp[A*(a*c*m + b*d*n) - B*(b*c*m + a*d*n) - d*(b*B*(m - n) - a *A*(m + n))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[m, 0] && GtQ[n, 0]
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[(a*A + b*B)*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(2*f*m*( b*c - a*d))), x] + Simp[1/(2*a*m*(b*c - a*d)) Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[A*(b*c*m - a*d*(2*m + n + 1)) + B*(a*c*m - b*d*(n + 1)) + d*(A*b - a*B)*(m + n + 1)*Tan[e + f*x], x], x], x] /; Free Q[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[m, 0] && !GtQ[n, 0]
Time = 0.78 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.35
| method | result | size |
| derivativedivides | \(\frac {2 d^{4} \left (-\frac {\arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {i d}}\right )}{16 d \sqrt {i d}}-\frac {\frac {2 \left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}-\frac {2 i d \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}}{\left (d \tan \left (f x +e \right )-i d \right )^{3}}+\frac {\arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {-i d}}\right )}{\sqrt {-i d}}}{16 d}\right )}{f \,a^{3}}\) | \(116\) |
| default | \(\frac {2 d^{4} \left (-\frac {\arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {i d}}\right )}{16 d \sqrt {i d}}-\frac {\frac {2 \left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}-\frac {2 i d \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}}{\left (d \tan \left (f x +e \right )-i d \right )^{3}}+\frac {\arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {-i d}}\right )}{\sqrt {-i d}}}{16 d}\right )}{f \,a^{3}}\) | \(116\) |
2/f/a^3*d^4*(-1/16/d/(I*d)^(1/2)*arctan((d*tan(f*x+e))^(1/2)/(I*d)^(1/2))- 1/16/d*((2*(d*tan(f*x+e))^(5/2)-2/3*I*d*(d*tan(f*x+e))^(3/2))/(d*tan(f*x+e )-I*d)^3+1/(-I*d)^(1/2)*arctan((d*tan(f*x+e))^(1/2)/(-I*d)^(1/2))))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 587 vs. \(2 (255) = 510\).
Time = 0.25 (sec) , antiderivative size = 587, normalized size of antiderivative = 1.78 \[ \int \frac {(d \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^3} \, dx=\frac {{\left (12 \, a^{3} f \sqrt {\frac {i \, d^{5}}{64 \, a^{6} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (-\frac {2 \, {\left (i \, d^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + 8 \, {\left (i \, a^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + i \, a^{3} f\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {i \, d^{5}}{64 \, a^{6} f^{2}}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{d^{2}}\right ) - 12 \, a^{3} f \sqrt {\frac {i \, d^{5}}{64 \, a^{6} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (-\frac {2 \, {\left (i \, d^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + 8 \, {\left (-i \, a^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} - i \, a^{3} f\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {i \, d^{5}}{64 \, a^{6} f^{2}}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{d^{2}}\right ) + 12 \, a^{3} f \sqrt {-\frac {i \, d^{5}}{64 \, a^{6} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (\frac {{\left (d^{3} + 8 \, {\left (a^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{3} f\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {i \, d^{5}}{64 \, a^{6} f^{2}}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{8 \, a^{3} f}\right ) - 12 \, a^{3} f \sqrt {-\frac {i \, d^{5}}{64 \, a^{6} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (\frac {{\left (d^{3} - 8 \, {\left (a^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{3} f\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {i \, d^{5}}{64 \, a^{6} f^{2}}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{8 \, a^{3} f}\right ) + {\left (-2 i \, d^{2} e^{\left (6 i \, f x + 6 i \, e\right )} + i \, d^{2} e^{\left (4 i \, f x + 4 i \, e\right )} + 2 i \, d^{2} e^{\left (2 i \, f x + 2 i \, e\right )} - i \, d^{2}\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-6 i \, f x - 6 i \, e\right )}}{48 \, a^{3} f} \]
1/48*(12*a^3*f*sqrt(1/64*I*d^5/(a^6*f^2))*e^(6*I*f*x + 6*I*e)*log(-2*(I*d^ 3*e^(2*I*f*x + 2*I*e) + 8*(I*a^3*f*e^(2*I*f*x + 2*I*e) + I*a^3*f)*sqrt((-I *d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(1/64*I*d^5/( a^6*f^2)))*e^(-2*I*f*x - 2*I*e)/d^2) - 12*a^3*f*sqrt(1/64*I*d^5/(a^6*f^2)) *e^(6*I*f*x + 6*I*e)*log(-2*(I*d^3*e^(2*I*f*x + 2*I*e) + 8*(-I*a^3*f*e^(2* I*f*x + 2*I*e) - I*a^3*f)*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f* x + 2*I*e) + 1))*sqrt(1/64*I*d^5/(a^6*f^2)))*e^(-2*I*f*x - 2*I*e)/d^2) + 1 2*a^3*f*sqrt(-1/64*I*d^5/(a^6*f^2))*e^(6*I*f*x + 6*I*e)*log(1/8*(d^3 + 8*( a^3*f*e^(2*I*f*x + 2*I*e) + a^3*f)*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/( e^(2*I*f*x + 2*I*e) + 1))*sqrt(-1/64*I*d^5/(a^6*f^2)))*e^(-2*I*f*x - 2*I*e )/(a^3*f)) - 12*a^3*f*sqrt(-1/64*I*d^5/(a^6*f^2))*e^(6*I*f*x + 6*I*e)*log( 1/8*(d^3 - 8*(a^3*f*e^(2*I*f*x + 2*I*e) + a^3*f)*sqrt((-I*d*e^(2*I*f*x + 2 *I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-1/64*I*d^5/(a^6*f^2)))*e^(-2 *I*f*x - 2*I*e)/(a^3*f)) + (-2*I*d^2*e^(6*I*f*x + 6*I*e) + I*d^2*e^(4*I*f* x + 4*I*e) + 2*I*d^2*e^(2*I*f*x + 2*I*e) - I*d^2)*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-6*I*f*x - 6*I*e)/(a^3*f)
\[ \int \frac {(d \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^3} \, dx=\frac {i \int \frac {\left (d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}}}{\tan ^{3}{\left (e + f x \right )} - 3 i \tan ^{2}{\left (e + f x \right )} - 3 \tan {\left (e + f x \right )} + i}\, dx}{a^{3}} \]
I*Integral((d*tan(e + f*x))**(5/2)/(tan(e + f*x)**3 - 3*I*tan(e + f*x)**2 - 3*tan(e + f*x) + I), x)/a**3
Exception generated. \[ \int \frac {(d \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^3} \, dx=\text {Exception raised: RuntimeError} \]
Time = 0.83 (sec) , antiderivative size = 212, normalized size of antiderivative = 0.64 \[ \int \frac {(d \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^3} \, dx=-\frac {1}{24} \, d^{4} {\left (\frac {3 \, \sqrt {2} \arctan \left (\frac {8 \, \sqrt {d^{2}} \sqrt {d \tan \left (f x + e\right )}}{4 i \, \sqrt {2} d^{\frac {3}{2}} + 4 \, \sqrt {2} \sqrt {d^{2}} \sqrt {d}}\right )}{a^{3} d^{\frac {3}{2}} f {\left (\frac {i \, d}{\sqrt {d^{2}}} + 1\right )}} + \frac {3 \, \sqrt {2} \arctan \left (\frac {8 \, \sqrt {d^{2}} \sqrt {d \tan \left (f x + e\right )}}{-4 i \, \sqrt {2} d^{\frac {3}{2}} + 4 \, \sqrt {2} \sqrt {d^{2}} \sqrt {d}}\right )}{a^{3} d^{\frac {3}{2}} f {\left (-\frac {i \, d}{\sqrt {d^{2}}} + 1\right )}} + \frac {2 \, {\left (3 \, \sqrt {d \tan \left (f x + e\right )} d^{2} \tan \left (f x + e\right )^{2} - i \, \sqrt {d \tan \left (f x + e\right )} d^{2} \tan \left (f x + e\right )\right )}}{{\left (d \tan \left (f x + e\right ) - i \, d\right )}^{3} a^{3} d f}\right )} \]
-1/24*d^4*(3*sqrt(2)*arctan(8*sqrt(d^2)*sqrt(d*tan(f*x + e))/(4*I*sqrt(2)* d^(3/2) + 4*sqrt(2)*sqrt(d^2)*sqrt(d)))/(a^3*d^(3/2)*f*(I*d/sqrt(d^2) + 1) ) + 3*sqrt(2)*arctan(8*sqrt(d^2)*sqrt(d*tan(f*x + e))/(-4*I*sqrt(2)*d^(3/2 ) + 4*sqrt(2)*sqrt(d^2)*sqrt(d)))/(a^3*d^(3/2)*f*(-I*d/sqrt(d^2) + 1)) + 2 *(3*sqrt(d*tan(f*x + e))*d^2*tan(f*x + e)^2 - I*sqrt(d*tan(f*x + e))*d^2*t an(f*x + e))/((d*tan(f*x + e) - I*d)^3*a^3*d*f))
Time = 5.63 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.48 \[ \int \frac {(d \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^3} \, dx=\frac {{\left (-1\right )}^{1/4}\,d^{5/2}\,\mathrm {atanh}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )}{8\,a^3\,f}-\frac {{\left (-1\right )}^{1/4}\,d^{5/2}\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )}{8\,a^3\,f}-\frac {-\frac {d^3\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}}{4\,a^3\,f}+\frac {d^4\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}\,1{}\mathrm {i}}{12\,a^3\,f}}{-d^3\,{\mathrm {tan}\left (e+f\,x\right )}^3+d^3\,{\mathrm {tan}\left (e+f\,x\right )}^2\,3{}\mathrm {i}+3\,d^3\,\mathrm {tan}\left (e+f\,x\right )-d^3\,1{}\mathrm {i}} \]
((-1)^(1/4)*d^(5/2)*atanh(((-1)^(1/4)*(d*tan(e + f*x))^(1/2))/d^(1/2)))/(8 *a^3*f) - ((-1)^(1/4)*d^(5/2)*atan(((-1)^(1/4)*(d*tan(e + f*x))^(1/2))/d^( 1/2)))/(8*a^3*f) - ((d^4*(d*tan(e + f*x))^(3/2)*1i)/(12*a^3*f) - (d^3*(d*t an(e + f*x))^(5/2))/(4*a^3*f))/(3*d^3*tan(e + f*x) - d^3*1i + d^3*tan(e + f*x)^2*3i - d^3*tan(e + f*x)^3)