Integrand size = 28, antiderivative size = 292 \[ \int \frac {\sqrt {d \tan (e+f x)}}{(a+i a \tan (e+f x))^3} \, dx=\frac {i \sqrt {d} \arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{8 \sqrt {2} a^3 f}-\frac {i \sqrt {d} \arctan \left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{8 \sqrt {2} a^3 f}+\frac {i \sqrt {d} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{16 \sqrt {2} a^3 f}-\frac {i \sqrt {d} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{16 \sqrt {2} a^3 f}+\frac {i \sqrt {d \tan (e+f x)}}{6 f (a+i a \tan (e+f x))^3}+\frac {i \sqrt {d \tan (e+f x)}}{12 a f (a+i a \tan (e+f x))^2} \]
1/16*I*arctan(1-2^(1/2)*(d*tan(f*x+e))^(1/2)/d^(1/2))*d^(1/2)/a^3/f*2^(1/2 )-1/16*I*arctan(1+2^(1/2)*(d*tan(f*x+e))^(1/2)/d^(1/2))*d^(1/2)/a^3/f*2^(1 /2)+1/32*I*ln(d^(1/2)-2^(1/2)*(d*tan(f*x+e))^(1/2)+d^(1/2)*tan(f*x+e))*d^( 1/2)/a^3/f*2^(1/2)-1/32*I*ln(d^(1/2)+2^(1/2)*(d*tan(f*x+e))^(1/2)+d^(1/2)* tan(f*x+e))*d^(1/2)/a^3/f*2^(1/2)+1/6*I*(d*tan(f*x+e))^(1/2)/f/(a+I*a*tan( f*x+e))^3+1/12*I*(d*tan(f*x+e))^(1/2)/a/f/(a+I*a*tan(f*x+e))^2
Time = 1.47 (sec) , antiderivative size = 267, normalized size of antiderivative = 0.91 \[ \int \frac {\sqrt {d \tan (e+f x)}}{(a+i a \tan (e+f x))^3} \, dx=\frac {\frac {6 i \sqrt {2} \sqrt {d} \arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{a^3}-\frac {6 i \sqrt {2} \sqrt {d} \arctan \left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{a^3}+\frac {3 i \sqrt {2} \sqrt {d} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{a^3}-\frac {3 i \sqrt {2} \sqrt {d} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{a^3}-\frac {24 i \sqrt {d \tan (e+f x)}}{a^3 (-i+\tan (e+f x))^2}+\frac {16 (d \tan (e+f x))^{3/2}}{d (a+i a \tan (e+f x))^3}}{96 f} \]
(((6*I)*Sqrt[2]*Sqrt[d]*ArcTan[1 - (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]] )/a^3 - ((6*I)*Sqrt[2]*Sqrt[d]*ArcTan[1 + (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/S qrt[d]])/a^3 + ((3*I)*Sqrt[2]*Sqrt[d]*Log[Sqrt[d] + Sqrt[d]*Tan[e + f*x] - Sqrt[2]*Sqrt[d*Tan[e + f*x]]])/a^3 - ((3*I)*Sqrt[2]*Sqrt[d]*Log[Sqrt[d] + Sqrt[d]*Tan[e + f*x] + Sqrt[2]*Sqrt[d*Tan[e + f*x]]])/a^3 - ((24*I)*Sqrt[ d*Tan[e + f*x]])/(a^3*(-I + Tan[e + f*x])^2) + (16*(d*Tan[e + f*x])^(3/2)) /(d*(a + I*a*Tan[e + f*x])^3))/(96*f)
Time = 0.84 (sec) , antiderivative size = 267, normalized size of antiderivative = 0.91, number of steps used = 18, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.607, Rules used = {3042, 4040, 3042, 4079, 27, 2011, 3042, 3957, 266, 755, 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.
| \(\displaystyle \int \frac {\sqrt {d \tan (e+f x)}}{(a+i a \tan (e+f x))^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sqrt {d \tan (e+f x)}}{(a+i a \tan (e+f x))^3}dx\) |
| \(\Big \downarrow \) 4040 |
| \(\displaystyle \frac {i \sqrt {d \tan (e+f x)}}{6 f (a+i a \tan (e+f x))^3}-\frac {\int \frac {i a d-5 a d \tan (e+f x)}{\sqrt {d \tan (e+f x)} (i \tan (e+f x) a+a)^2}dx}{12 a^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {i \sqrt {d \tan (e+f x)}}{6 f (a+i a \tan (e+f x))^3}-\frac {\int \frac {i a d-5 a d \tan (e+f x)}{\sqrt {d \tan (e+f x)} (i \tan (e+f x) a+a)^2}dx}{12 a^2}\) |
| \(\Big \downarrow \) 4079 |
| \(\displaystyle \frac {i \sqrt {d \tan (e+f x)}}{6 f (a+i a \tan (e+f x))^3}-\frac {\frac {\int \frac {6 \left (i a^2 d^2-a^2 d^2 \tan (e+f x)\right )}{\sqrt {d \tan (e+f x)} (i \tan (e+f x) a+a)}dx}{4 a^2 d}-\frac {i a \sqrt {d \tan (e+f x)}}{f (a+i a \tan (e+f x))^2}}{12 a^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {i \sqrt {d \tan (e+f x)}}{6 f (a+i a \tan (e+f x))^3}-\frac {\frac {3 \int \frac {i a^2 d^2-a^2 d^2 \tan (e+f x)}{\sqrt {d \tan (e+f x)} (i \tan (e+f x) a+a)}dx}{2 a^2 d}-\frac {i a \sqrt {d \tan (e+f x)}}{f (a+i a \tan (e+f x))^2}}{12 a^2}\) |
| \(\Big \downarrow \) 2011 |
| \(\displaystyle \frac {i \sqrt {d \tan (e+f x)}}{6 f (a+i a \tan (e+f x))^3}-\frac {\frac {3 i d \int \frac {1}{\sqrt {d \tan (e+f x)}}dx}{2 a}-\frac {i a \sqrt {d \tan (e+f x)}}{f (a+i a \tan (e+f x))^2}}{12 a^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {i \sqrt {d \tan (e+f x)}}{6 f (a+i a \tan (e+f x))^3}-\frac {\frac {3 i d \int \frac {1}{\sqrt {d \tan (e+f x)}}dx}{2 a}-\frac {i a \sqrt {d \tan (e+f x)}}{f (a+i a \tan (e+f x))^2}}{12 a^2}\) |
| \(\Big \downarrow \) 3957 |
| \(\displaystyle \frac {i \sqrt {d \tan (e+f x)}}{6 f (a+i a \tan (e+f x))^3}-\frac {\frac {3 i d^2 \int \frac {1}{\sqrt {d \tan (e+f x)} \left (\tan ^2(e+f x) d^2+d^2\right )}d(d \tan (e+f x))}{2 a f}-\frac {i a \sqrt {d \tan (e+f x)}}{f (a+i a \tan (e+f x))^2}}{12 a^2}\) |
| \(\Big \downarrow \) 266 |
| \(\displaystyle \frac {i \sqrt {d \tan (e+f x)}}{6 f (a+i a \tan (e+f x))^3}-\frac {\frac {3 i d^2 \int \frac {1}{d^4 \tan ^4(e+f x)+d^2}d\sqrt {d \tan (e+f x)}}{a f}-\frac {i a \sqrt {d \tan (e+f x)}}{f (a+i a \tan (e+f x))^2}}{12 a^2}\) |
| \(\Big \downarrow \) 755 |
| \(\displaystyle \frac {i \sqrt {d \tan (e+f x)}}{6 f (a+i a \tan (e+f x))^3}-\frac {\frac {3 i d^2 \left (\frac {\int \frac {d-d^2 \tan ^2(e+f x)}{d^4 \tan ^4(e+f x)+d^2}d\sqrt {d \tan (e+f x)}}{2 d}+\frac {\int \frac {d^2 \tan ^2(e+f x)+d}{d^4 \tan ^4(e+f x)+d^2}d\sqrt {d \tan (e+f x)}}{2 d}\right )}{a f}-\frac {i a \sqrt {d \tan (e+f x)}}{f (a+i a \tan (e+f x))^2}}{12 a^2}\) |
| \(\Big \downarrow \) 1476 |
| \(\displaystyle \frac {i \sqrt {d \tan (e+f x)}}{6 f (a+i a \tan (e+f x))^3}-\frac {\frac {3 i d^2 \left (\frac {\frac {1}{2} \int \frac {1}{d^2 \tan ^2(e+f x)-\sqrt {2} d^{3/2} \tan (e+f x)+d}d\sqrt {d \tan (e+f x)}+\frac {1}{2} \int \frac {1}{d^2 \tan ^2(e+f x)+\sqrt {2} d^{3/2} \tan (e+f x)+d}d\sqrt {d \tan (e+f x)}}{2 d}+\frac {\int \frac {d-d^2 \tan ^2(e+f x)}{d^4 \tan ^4(e+f x)+d^2}d\sqrt {d \tan (e+f x)}}{2 d}\right )}{a f}-\frac {i a \sqrt {d \tan (e+f x)}}{f (a+i a \tan (e+f x))^2}}{12 a^2}\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle \frac {i \sqrt {d \tan (e+f x)}}{6 f (a+i a \tan (e+f x))^3}-\frac {\frac {3 i d^2 \left (\frac {\frac {\int \frac {1}{-d^2 \tan ^2(e+f x)-1}d\left (1-\sqrt {2} \sqrt {d} \tan (e+f x)\right )}{\sqrt {2} \sqrt {d}}-\frac {\int \frac {1}{-d^2 \tan ^2(e+f x)-1}d\left (\sqrt {2} \sqrt {d} \tan (e+f x)+1\right )}{\sqrt {2} \sqrt {d}}}{2 d}+\frac {\int \frac {d-d^2 \tan ^2(e+f x)}{d^4 \tan ^4(e+f x)+d^2}d\sqrt {d \tan (e+f x)}}{2 d}\right )}{a f}-\frac {i a \sqrt {d \tan (e+f x)}}{f (a+i a \tan (e+f x))^2}}{12 a^2}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {i \sqrt {d \tan (e+f x)}}{6 f (a+i a \tan (e+f x))^3}-\frac {\frac {3 i d^2 \left (\frac {\int \frac {d-d^2 \tan ^2(e+f x)}{d^4 \tan ^4(e+f x)+d^2}d\sqrt {d \tan (e+f x)}}{2 d}+\frac {\frac {\arctan \left (\sqrt {2} \sqrt {d} \tan (e+f x)+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {d} \tan (e+f x)\right )}{\sqrt {2} \sqrt {d}}}{2 d}\right )}{a f}-\frac {i a \sqrt {d \tan (e+f x)}}{f (a+i a \tan (e+f x))^2}}{12 a^2}\) |
| \(\Big \downarrow \) 1479 |
| \(\displaystyle \frac {i \sqrt {d \tan (e+f x)}}{6 f (a+i a \tan (e+f x))^3}-\frac {\frac {3 i d^2 \left (\frac {-\frac {\int -\frac {\sqrt {2} \sqrt {d}-2 \sqrt {d \tan (e+f x)}}{d^2 \tan ^2(e+f x)-\sqrt {2} d^{3/2} \tan (e+f x)+d}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {d}+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{d^2 \tan ^2(e+f x)+\sqrt {2} d^{3/2} \tan (e+f x)+d}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}}{2 d}+\frac {\frac {\arctan \left (\sqrt {2} \sqrt {d} \tan (e+f x)+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {d} \tan (e+f x)\right )}{\sqrt {2} \sqrt {d}}}{2 d}\right )}{a f}-\frac {i a \sqrt {d \tan (e+f x)}}{f (a+i a \tan (e+f x))^2}}{12 a^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {i \sqrt {d \tan (e+f x)}}{6 f (a+i a \tan (e+f x))^3}-\frac {\frac {3 i d^2 \left (\frac {\frac {\int \frac {\sqrt {2} \sqrt {d}-2 \sqrt {d \tan (e+f x)}}{d^2 \tan ^2(e+f x)-\sqrt {2} d^{3/2} \tan (e+f x)+d}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {d}+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{d^2 \tan ^2(e+f x)+\sqrt {2} d^{3/2} \tan (e+f x)+d}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}}{2 d}+\frac {\frac {\arctan \left (\sqrt {2} \sqrt {d} \tan (e+f x)+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {d} \tan (e+f x)\right )}{\sqrt {2} \sqrt {d}}}{2 d}\right )}{a f}-\frac {i a \sqrt {d \tan (e+f x)}}{f (a+i a \tan (e+f x))^2}}{12 a^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {i \sqrt {d \tan (e+f x)}}{6 f (a+i a \tan (e+f x))^3}-\frac {\frac {3 i d^2 \left (\frac {\frac {\int \frac {\sqrt {2} \sqrt {d}-2 \sqrt {d \tan (e+f x)}}{d^2 \tan ^2(e+f x)-\sqrt {2} d^{3/2} \tan (e+f x)+d}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}+\frac {\int \frac {\sqrt {d}+\sqrt {2} \sqrt {d \tan (e+f x)}}{d^2 \tan ^2(e+f x)+\sqrt {2} d^{3/2} \tan (e+f x)+d}d\sqrt {d \tan (e+f x)}}{2 \sqrt {d}}}{2 d}+\frac {\frac {\arctan \left (\sqrt {2} \sqrt {d} \tan (e+f x)+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {d} \tan (e+f x)\right )}{\sqrt {2} \sqrt {d}}}{2 d}\right )}{a f}-\frac {i a \sqrt {d \tan (e+f x)}}{f (a+i a \tan (e+f x))^2}}{12 a^2}\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle \frac {i \sqrt {d \tan (e+f x)}}{6 f (a+i a \tan (e+f x))^3}-\frac {\frac {3 i d^2 \left (\frac {\frac {\arctan \left (\sqrt {2} \sqrt {d} \tan (e+f x)+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {d} \tan (e+f x)\right )}{\sqrt {2} \sqrt {d}}}{2 d}+\frac {\frac {\log \left (\sqrt {2} d^{3/2} \tan (e+f x)+d^2 \tan ^2(e+f x)+d\right )}{2 \sqrt {2} \sqrt {d}}-\frac {\log \left (-\sqrt {2} d^{3/2} \tan (e+f x)+d^2 \tan ^2(e+f x)+d\right )}{2 \sqrt {2} \sqrt {d}}}{2 d}\right )}{a f}-\frac {i a \sqrt {d \tan (e+f x)}}{f (a+i a \tan (e+f x))^2}}{12 a^2}\) |
((I/6)*Sqrt[d*Tan[e + f*x]])/(f*(a + I*a*Tan[e + f*x])^3) - (((3*I)*d^2*(( -(ArcTan[1 - Sqrt[2]*Sqrt[d]*Tan[e + f*x]]/(Sqrt[2]*Sqrt[d])) + ArcTan[1 + Sqrt[2]*Sqrt[d]*Tan[e + f*x]]/(Sqrt[2]*Sqrt[d]))/(2*d) + (-1/2*Log[d - Sq rt[2]*d^(3/2)*Tan[e + f*x] + d^2*Tan[e + f*x]^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*d)))/(a*f) - (I*a*Sqrt[d*Tan[e + f*x]])/(f*(a + I*a*Tan[e + f*x])^2))/ (12*a^2)
3.2.83.3.1 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[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2] ], s = Denominator[Rt[a/b, 2]]}, Simp[1/(2*r) Int[(r - s*x^2)/(a + b*x^4) , x], x] + Simp[1/(2*r) 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[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Simp[(b/d)^m Int[u*(c + d*v)^(m + n), x], x] /; FreeQ[{a, b, c, d, n}, x ] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c + d*x , a + b*x])
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_)*Sqrt[(c_.) + (d_.)*tan[(e_. ) + (f_.)*(x_)]], x_Symbol] :> Simp[(-b)*(a + b*Tan[e + f*x])^m*(Sqrt[c + d *Tan[e + f*x]]/(2*a*f*m)), x] + Simp[1/(4*a^2*m) Int[(a + b*Tan[e + f*x]) ^(m + 1)*(Simp[2*a*c*m + b*d + a*d*(2*m + 1)*Tan[e + f*x], x]/Sqrt[c + d*Ta n[e + f*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] && IntegersQ[2*m]
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 = 0.84 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.41
| method | result | size |
| derivativedivides | \(\frac {2 d^{4} \left (\frac {\frac {-\frac {2 i d \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-2 d^{2} \sqrt {d \tan \left (f x +e \right )}}{\left (d \tan \left (f x +e \right )-i d \right )^{3}}-\frac {\arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {-i d}}\right )}{\sqrt {-i d}}}{16 d^{3}}+\frac {\arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {i d}}\right )}{16 d^{3} \sqrt {i d}}\right )}{f \,a^{3}}\) | \(120\) |
| default | \(\frac {2 d^{4} \left (\frac {\frac {-\frac {2 i d \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-2 d^{2} \sqrt {d \tan \left (f x +e \right )}}{\left (d \tan \left (f x +e \right )-i d \right )^{3}}-\frac {\arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {-i d}}\right )}{\sqrt {-i d}}}{16 d^{3}}+\frac {\arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {i d}}\right )}{16 d^{3} \sqrt {i d}}\right )}{f \,a^{3}}\) | \(120\) |
2/f/a^3*d^4*(1/16/d^3*((-2/3*I*d*(d*tan(f*x+e))^(3/2)-2*d^2*(d*tan(f*x+e)) ^(1/2))/(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)))+1/16/d^3/(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. 547 vs. \(2 (216) = 432\).
Time = 0.25 (sec) , antiderivative size = 547, normalized size of antiderivative = 1.87 \[ \int \frac {\sqrt {d \tan (e+f x)}}{(a+i a \tan (e+f x))^3} \, dx=-\frac {{\left (12 \, a^{3} f \sqrt {\frac {i \, d}{64 \, a^{6} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (-2 \, {\left (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}{64 \, a^{6} f^{2}}} + i \, d e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}\right ) - 12 \, a^{3} f \sqrt {\frac {i \, d}{64 \, a^{6} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (-2 \, {\left (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}{64 \, a^{6} f^{2}}} + i \, d e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}\right ) - 12 \, a^{3} f \sqrt {-\frac {i \, d}{64 \, a^{6} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (\frac {{\left (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}{64 \, a^{6} f^{2}}} + d\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{8 \, a^{3} f}\right ) + 12 \, a^{3} f \sqrt {-\frac {i \, d}{64 \, a^{6} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (-\frac {{\left (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}{64 \, a^{6} f^{2}}} - d\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{8 \, 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}} {\left (2 i \, e^{\left (6 i \, f x + 6 i \, e\right )} + 5 i \, e^{\left (4 i \, f x + 4 i \, e\right )} + 4 i \, e^{\left (2 i \, f x + 2 i \, e\right )} + i\right )}\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/(a^6*f^2))*e^(6*I*f*x + 6*I*e)*log(-2*(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/(a^6*f^2)) + I*d*e^(2*I*f*x + 2* I*e))*e^(-2*I*f*x - 2*I*e)) - 12*a^3*f*sqrt(1/64*I*d/(a^6*f^2))*e^(6*I*f*x + 6*I*e)*log(-2*(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/(a^6*f^2 )) + I*d*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)) - 12*a^3*f*sqrt(-1/64* I*d/(a^6*f^2))*e^(6*I*f*x + 6*I*e)*log(1/8*(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))*s qrt(-1/64*I*d/(a^6*f^2)) + d)*e^(-2*I*f*x - 2*I*e)/(a^3*f)) + 12*a^3*f*sqr t(-1/64*I*d/(a^6*f^2))*e^(6*I*f*x + 6*I*e)*log(-1/8*(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/(a^6*f^2)) - d)*e^(-2*I*f*x - 2*I*e)/(a^3*f)) - sqr t((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*(2*I*e^(6*I* f*x + 6*I*e) + 5*I*e^(4*I*f*x + 4*I*e) + 4*I*e^(2*I*f*x + 2*I*e) + I))*e^( -6*I*f*x - 6*I*e)/(a^3*f)
\[ \int \frac {\sqrt {d \tan (e+f x)}}{(a+i a \tan (e+f x))^3} \, dx=\frac {i \int \frac {\sqrt {d \tan {\left (e + f x \right )}}}{\tan ^{3}{\left (e + f x \right )} - 3 i \tan ^{2}{\left (e + f x \right )} - 3 \tan {\left (e + f x \right )} + i}\, dx}{a^{3}} \]
I*Integral(sqrt(d*tan(e + f*x))/(tan(e + f*x)**3 - 3*I*tan(e + f*x)**2 - 3 *tan(e + f*x) + I), x)/a**3
Exception generated. \[ \int \frac {\sqrt {d \tan (e+f x)}}{(a+i a \tan (e+f x))^3} \, dx=\text {Exception raised: RuntimeError} \]
Time = 0.79 (sec) , antiderivative size = 200, normalized size of antiderivative = 0.68 \[ \int \frac {\sqrt {d \tan (e+f x)}}{(a+i a \tan (e+f x))^3} \, dx=\frac {1}{24} \, d^{4} {\left (\frac {3 \, \sqrt {2} \arctan \left (\frac {8 \, \sqrt {d^{2}} \sqrt {d \tan \left (f x + e\right )}}{4 i \, \sqrt {2} d^{\frac {3}{2}} + 4 \, \sqrt {2} \sqrt {d^{2}} \sqrt {d}}\right )}{a^{3} d^{\frac {7}{2}} f {\left (\frac {i \, d}{\sqrt {d^{2}}} + 1\right )}} - \frac {3 \, \sqrt {2} \arctan \left (\frac {8 \, \sqrt {d^{2}} \sqrt {d \tan \left (f x + e\right )}}{-4 i \, \sqrt {2} d^{\frac {3}{2}} + 4 \, \sqrt {2} \sqrt {d^{2}} \sqrt {d}}\right )}{a^{3} d^{\frac {7}{2}} f {\left (-\frac {i \, d}{\sqrt {d^{2}}} + 1\right )}} - \frac {2 \, {\left (i \, \sqrt {d \tan \left (f x + e\right )} d \tan \left (f x + e\right ) + 3 \, \sqrt {d \tan \left (f x + e\right )} d\right )}}{{\left (d \tan \left (f x + e\right ) - i \, d\right )}^{3} a^{3} d^{2} f}\right )} \]
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^(7/2)*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^(7/2)*f*(-I*d/sqrt(d^2) + 1)) - 2* (I*sqrt(d*tan(f*x + e))*d*tan(f*x + e) + 3*sqrt(d*tan(f*x + e))*d)/((d*tan (f*x + e) - I*d)^3*a^3*d^2*f))
Time = 5.05 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.54 \[ \int \frac {\sqrt {d \tan (e+f x)}}{(a+i a \tan (e+f x))^3} \, dx=\frac {\frac {d^3\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{4\,a^3\,f}+\frac {d^2\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}\,1{}\mathrm {i}}{12\,a^3\,f}}{-d^3\,{\mathrm {tan}\left (e+f\,x\right )}^3+d^3\,{\mathrm {tan}\left (e+f\,x\right )}^2\,3{}\mathrm {i}+3\,d^3\,\mathrm {tan}\left (e+f\,x\right )-d^3\,1{}\mathrm {i}}-\frac {{\left (-1\right )}^{1/4}\,\sqrt {d}\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )}{8\,a^3\,f}-\frac {{\left (-1\right )}^{1/4}\,\sqrt {d}\,\mathrm {atanh}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )}{8\,a^3\,f} \]
((d^3*(d*tan(e + f*x))^(1/2))/(4*a^3*f) + (d^2*(d*tan(e + f*x))^(3/2)*1i)/ (12*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) - ((-1)^(1/4)*d^(1/2)*atan(((-1)^(1/4)*(d*tan(e + f*x))^(1/2) )/d^(1/2)))/(8*a^3*f) - ((-1)^(1/4)*d^(1/2)*atanh(((-1)^(1/4)*(d*tan(e + f *x))^(1/2))/d^(1/2)))/(8*a^3*f)