Integrand size = 12, antiderivative size = 212 \[ \int (d \tan (e+f x))^{5/2} \, dx=\frac {d^{5/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} f}-\frac {d^{5/2} \arctan \left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} f}-\frac {d^{5/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{2 \sqrt {2} f}+\frac {d^{5/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{2 \sqrt {2} f}+\frac {2 d (d \tan (e+f x))^{3/2}}{3 f} \]
1/2*d^(5/2)*arctan(1-2^(1/2)*(d*tan(f*x+e))^(1/2)/d^(1/2))/f*2^(1/2)-1/2*d ^(5/2)*arctan(1+2^(1/2)*(d*tan(f*x+e))^(1/2)/d^(1/2))/f*2^(1/2)-1/4*d^(5/2 )*ln(d^(1/2)-2^(1/2)*(d*tan(f*x+e))^(1/2)+d^(1/2)*tan(f*x+e))/f*2^(1/2)+1/ 4*d^(5/2)*ln(d^(1/2)+2^(1/2)*(d*tan(f*x+e))^(1/2)+d^(1/2)*tan(f*x+e))/f*2^ (1/2)+2/3*d*(d*tan(f*x+e))^(3/2)/f
Time = 0.16 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.48 \[ \int (d \tan (e+f x))^{5/2} \, dx=\frac {d (d \tan (e+f x))^{3/2} \left (-3 \arctan \left (\sqrt [4]{-\tan ^2(e+f x)}\right ) \sqrt [4]{-\tan (e+f x)}+3 \text {arctanh}\left (\sqrt [4]{-\tan ^2(e+f x)}\right ) \sqrt [4]{-\tan (e+f x)}+2 \tan ^{\frac {7}{4}}(e+f x)\right )}{3 f \tan ^{\frac {7}{4}}(e+f x)} \]
(d*(d*Tan[e + f*x])^(3/2)*(-3*ArcTan[(-Tan[e + f*x]^2)^(1/4)]*(-Tan[e + f* x])^(1/4) + 3*ArcTanh[(-Tan[e + f*x]^2)^(1/4)]*(-Tan[e + f*x])^(1/4) + 2*T an[e + f*x]^(7/4)))/(3*f*Tan[e + f*x]^(7/4))
Time = 0.44 (sec) , antiderivative size = 197, normalized size of antiderivative = 0.93, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.083, Rules used = {3042, 3954, 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.
| \(\displaystyle \int (d \tan (e+f x))^{5/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int (d \tan (e+f x))^{5/2}dx\) |
| \(\Big \downarrow \) 3954 |
| \(\displaystyle \frac {2 d (d \tan (e+f x))^{3/2}}{3 f}-d^2 \int \sqrt {d \tan (e+f x)}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 d (d \tan (e+f x))^{3/2}}{3 f}-d^2 \int \sqrt {d \tan (e+f x)}dx\) |
| \(\Big \downarrow \) 3957 |
| \(\displaystyle \frac {2 d (d \tan (e+f x))^{3/2}}{3 f}-\frac {d^3 \int \frac {\sqrt {d \tan (e+f x)}}{\tan ^2(e+f x) d^2+d^2}d(d \tan (e+f x))}{f}\) |
| \(\Big \downarrow \) 266 |
| \(\displaystyle \frac {2 d (d \tan (e+f x))^{3/2}}{3 f}-\frac {2 d^3 \int \frac {d^2 \tan ^2(e+f x)}{d^4 \tan ^4(e+f x)+d^2}d\sqrt {d \tan (e+f x)}}{f}\) |
| \(\Big \downarrow \) 826 |
| \(\displaystyle \frac {2 d (d \tan (e+f x))^{3/2}}{3 f}-\frac {2 d^3 \left (\frac {1}{2} \int \frac {d^2 \tan ^2(e+f x)+d}{d^4 \tan ^4(e+f x)+d^2}d\sqrt {d \tan (e+f x)}-\frac {1}{2} \int \frac {d-d^2 \tan ^2(e+f x)}{d^4 \tan ^4(e+f x)+d^2}d\sqrt {d \tan (e+f x)}\right )}{f}\) |
| \(\Big \downarrow \) 1476 |
| \(\displaystyle \frac {2 d (d \tan (e+f x))^{3/2}}{3 f}-\frac {2 d^3 \left (\frac {1}{2} \left (\frac {1}{2} \int \frac {1}{d^2 \tan ^2(e+f x)-\sqrt {2} d^{3/2} \tan (e+f x)+d}d\sqrt {d \tan (e+f x)}+\frac {1}{2} \int \frac {1}{d^2 \tan ^2(e+f x)+\sqrt {2} d^{3/2} \tan (e+f x)+d}d\sqrt {d \tan (e+f x)}\right )-\frac {1}{2} \int \frac {d-d^2 \tan ^2(e+f x)}{d^4 \tan ^4(e+f x)+d^2}d\sqrt {d \tan (e+f x)}\right )}{f}\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle \frac {2 d (d \tan (e+f x))^{3/2}}{3 f}-\frac {2 d^3 \left (\frac {1}{2} \left (\frac {\int \frac {1}{-d^2 \tan ^2(e+f x)-1}d\left (1-\sqrt {2} \sqrt {d} \tan (e+f x)\right )}{\sqrt {2} \sqrt {d}}-\frac {\int \frac {1}{-d^2 \tan ^2(e+f x)-1}d\left (\sqrt {2} \sqrt {d} \tan (e+f x)+1\right )}{\sqrt {2} \sqrt {d}}\right )-\frac {1}{2} \int \frac {d-d^2 \tan ^2(e+f x)}{d^4 \tan ^4(e+f x)+d^2}d\sqrt {d \tan (e+f x)}\right )}{f}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {2 d (d \tan (e+f x))^{3/2}}{3 f}-\frac {2 d^3 \left (\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {d} \tan (e+f x)+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {d} \tan (e+f x)\right )}{\sqrt {2} \sqrt {d}}\right )-\frac {1}{2} \int \frac {d-d^2 \tan ^2(e+f x)}{d^4 \tan ^4(e+f x)+d^2}d\sqrt {d \tan (e+f x)}\right )}{f}\) |
| \(\Big \downarrow \) 1479 |
| \(\displaystyle \frac {2 d (d \tan (e+f x))^{3/2}}{3 f}-\frac {2 d^3 \left (\frac {1}{2} \left (\frac {\int -\frac {\sqrt {2} \sqrt {d}-2 \sqrt {d \tan (e+f x)}}{d^2 \tan ^2(e+f x)-\sqrt {2} d^{3/2} \tan (e+f x)+d}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}+\frac {\int -\frac {\sqrt {2} \left (\sqrt {d}+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{d^2 \tan ^2(e+f x)+\sqrt {2} d^{3/2} \tan (e+f x)+d}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {d} \tan (e+f x)+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {d} \tan (e+f x)\right )}{\sqrt {2} \sqrt {d}}\right )\right )}{f}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {2 d (d \tan (e+f x))^{3/2}}{3 f}-\frac {2 d^3 \left (\frac {1}{2} \left (-\frac {\int \frac {\sqrt {2} \sqrt {d}-2 \sqrt {d \tan (e+f x)}}{d^2 \tan ^2(e+f x)-\sqrt {2} d^{3/2} \tan (e+f x)+d}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}-\frac {\int \frac {\sqrt {2} \left (\sqrt {d}+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{d^2 \tan ^2(e+f x)+\sqrt {2} d^{3/2} \tan (e+f x)+d}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {d} \tan (e+f x)+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {d} \tan (e+f x)\right )}{\sqrt {2} \sqrt {d}}\right )\right )}{f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2 d (d \tan (e+f x))^{3/2}}{3 f}-\frac {2 d^3 \left (\frac {1}{2} \left (-\frac {\int \frac {\sqrt {2} \sqrt {d}-2 \sqrt {d \tan (e+f x)}}{d^2 \tan ^2(e+f x)-\sqrt {2} d^{3/2} \tan (e+f x)+d}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}-\frac {\int \frac {\sqrt {d}+\sqrt {2} \sqrt {d \tan (e+f x)}}{d^2 \tan ^2(e+f x)+\sqrt {2} d^{3/2} \tan (e+f x)+d}d\sqrt {d \tan (e+f x)}}{2 \sqrt {d}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {d} \tan (e+f x)+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {d} \tan (e+f x)\right )}{\sqrt {2} \sqrt {d}}\right )\right )}{f}\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle \frac {2 d (d \tan (e+f x))^{3/2}}{3 f}-\frac {2 d^3 \left (\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {d} \tan (e+f x)+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {d} \tan (e+f x)\right )}{\sqrt {2} \sqrt {d}}\right )+\frac {1}{2} \left (\frac {\log \left (-\sqrt {2} d^{3/2} \tan (e+f x)+d^2 \tan ^2(e+f x)+d\right )}{2 \sqrt {2} \sqrt {d}}-\frac {\log \left (\sqrt {2} d^{3/2} \tan (e+f x)+d^2 \tan ^2(e+f x)+d\right )}{2 \sqrt {2} \sqrt {d}}\right )\right )}{f}\) |
(-2*d^3*((-(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 + (Log[d - S qrt[2]*d^(3/2)*Tan[e + f*x] + d^2*Tan[e + f*x]^2]/(2*Sqrt[2]*Sqrt[d]) - Lo g[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*d*(d*Tan[e + f*x])^(3/2))/(3*f)
3.3.50.3.1 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[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*((b*Tan[c + d *x])^(n - 1)/(d*(n - 1))), x] - Simp[b^2 Int[(b*Tan[c + d*x])^(n - 2), x] , x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]
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]
Time = 0.16 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.73
| method | result | size |
| derivativedivides | \(\frac {2 d \left (\frac {\left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-\frac {d^{2} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (d^{2}\right )^{\frac {1}{4}}}\right )}{f}\) | \(154\) |
| default | \(\frac {2 d \left (\frac {\left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-\frac {d^{2} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (d^{2}\right )^{\frac {1}{4}}}\right )}{f}\) | \(154\) |
2/f*d*(1/3*(d*tan(f*x+e))^(3/2)-1/8*d^2/(d^2)^(1/4)*2^(1/2)*(ln((d*tan(f*x +e)-(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)*2^(1/2)+(d^2)^(1/2))/(d*tan(f*x+e)+(d ^2)^(1/4)*(d*tan(f*x+e))^(1/2)*2^(1/2)+(d^2)^(1/2)))+2*arctan(2^(1/2)/(d^2 )^(1/4)*(d*tan(f*x+e))^(1/2)+1)-2*arctan(-2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e ))^(1/2)+1)))
Result contains complex when optimal does not.
Time = 0.27 (sec) , antiderivative size = 202, normalized size of antiderivative = 0.95 \[ \int (d \tan (e+f x))^{5/2} \, dx=\frac {4 \, \sqrt {d \tan \left (f x + e\right )} d^{2} \tan \left (f x + e\right ) - 3 \, \left (-\frac {d^{10}}{f^{4}}\right )^{\frac {1}{4}} f \log \left (\sqrt {d \tan \left (f x + e\right )} d^{7} + \left (-\frac {d^{10}}{f^{4}}\right )^{\frac {3}{4}} f^{3}\right ) + 3 i \, \left (-\frac {d^{10}}{f^{4}}\right )^{\frac {1}{4}} f \log \left (\sqrt {d \tan \left (f x + e\right )} d^{7} + i \, \left (-\frac {d^{10}}{f^{4}}\right )^{\frac {3}{4}} f^{3}\right ) - 3 i \, \left (-\frac {d^{10}}{f^{4}}\right )^{\frac {1}{4}} f \log \left (\sqrt {d \tan \left (f x + e\right )} d^{7} - i \, \left (-\frac {d^{10}}{f^{4}}\right )^{\frac {3}{4}} f^{3}\right ) + 3 \, \left (-\frac {d^{10}}{f^{4}}\right )^{\frac {1}{4}} f \log \left (\sqrt {d \tan \left (f x + e\right )} d^{7} - \left (-\frac {d^{10}}{f^{4}}\right )^{\frac {3}{4}} f^{3}\right )}{6 \, f} \]
1/6*(4*sqrt(d*tan(f*x + e))*d^2*tan(f*x + e) - 3*(-d^10/f^4)^(1/4)*f*log(s qrt(d*tan(f*x + e))*d^7 + (-d^10/f^4)^(3/4)*f^3) + 3*I*(-d^10/f^4)^(1/4)*f *log(sqrt(d*tan(f*x + e))*d^7 + I*(-d^10/f^4)^(3/4)*f^3) - 3*I*(-d^10/f^4) ^(1/4)*f*log(sqrt(d*tan(f*x + e))*d^7 - I*(-d^10/f^4)^(3/4)*f^3) + 3*(-d^1 0/f^4)^(1/4)*f*log(sqrt(d*tan(f*x + e))*d^7 - (-d^10/f^4)^(3/4)*f^3))/f
\[ \int (d \tan (e+f x))^{5/2} \, dx=\int \left (d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}}\, dx \]
Time = 0.30 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.83 \[ \int (d \tan (e+f x))^{5/2} \, dx=-\frac {3 \, d^{4} {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} + 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} - 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} - \frac {\sqrt {2} \log \left (d \tan \left (f x + e\right ) + \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}} + \frac {\sqrt {2} \log \left (d \tan \left (f x + e\right ) - \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}}\right )} - 8 \, \left (d \tan \left (f x + e\right )\right )^{\frac {3}{2}} d^{2}}{12 \, d f} \]
-1/12*(3*d^4*(2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(d) + 2*sqrt(d*tan (f*x + e)))/sqrt(d))/sqrt(d) + 2*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt (d) - 2*sqrt(d*tan(f*x + e)))/sqrt(d))/sqrt(d) - sqrt(2)*log(d*tan(f*x + e ) + sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(d) + d)/sqrt(d) + sqrt(2)*log(d*tan( f*x + e) - sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(d) + d)/sqrt(d)) - 8*(d*tan(f *x + e))^(3/2)*d^2)/(d*f)
Timed out. \[ \int (d \tan (e+f x))^{5/2} \, dx=\text {Timed out} \]
Time = 3.42 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.35 \[ \int (d \tan (e+f x))^{5/2} \, dx=\frac {2\,d\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}{3\,f}-\frac {{\left (-1\right )}^{1/4}\,d^{5/2}\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )}{f}+\frac {{\left (-1\right )}^{1/4}\,d^{5/2}\,\mathrm {atanh}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )}{f} \]