Integrand size = 25, antiderivative size = 138 \[ \int \sqrt {d \tan (e+f x)} (a+a \tan (e+f x))^3 \, dx=\frac {2 \sqrt {2} a^3 \sqrt {d} \arctan \left (\frac {\sqrt {d}-\sqrt {d} \tan (e+f x)}{\sqrt {2} \sqrt {d \tan (e+f x)}}\right )}{f}+\frac {4 a^3 \sqrt {d \tan (e+f x)}}{f}+\frac {8 a^3 (d \tan (e+f x))^{3/2}}{5 d f}+\frac {2 (d \tan (e+f x))^{3/2} \left (a^3+a^3 \tan (e+f x)\right )}{5 d f} \]
2*a^3*arctan(1/2*(d^(1/2)-d^(1/2)*tan(f*x+e))*2^(1/2)/(d*tan(f*x+e))^(1/2) )*2^(1/2)*d^(1/2)/f+4*a^3*(d*tan(f*x+e))^(1/2)/f+8/5*a^3*(d*tan(f*x+e))^(3 /2)/d/f+2/5*(d*tan(f*x+e))^(3/2)*(a^3+a^3*tan(f*x+e))/d/f
Leaf count is larger than twice the leaf count of optimal. \(360\) vs. \(2(138)=276\).
Time = 3.56 (sec) , antiderivative size = 360, normalized size of antiderivative = 2.61 \[ \int \sqrt {d \tan (e+f x)} (a+a \tan (e+f x))^3 \, dx=\frac {a^3 \left (-20 \arctan \left (\sqrt [4]{-\tan ^2(e+f x)}\right ) \cos ^3(e+f x) \sqrt [4]{-\tan (e+f x)}+20 \text {arctanh}\left (\sqrt [4]{-\tan ^2(e+f x)}\right ) \cos ^3(e+f x) \sqrt [4]{-\tan (e+f x)}+\cos (e+f x) \left (10 \sqrt {2} \arctan \left (1-\sqrt {2} \sqrt {\tan (e+f x)}\right ) \cos ^2(e+f x)-10 \sqrt {2} \arctan \left (1+\sqrt {2} \sqrt {\tan (e+f x)}\right ) \cos ^2(e+f x)+5 \sqrt {2} \cos ^2(e+f x) \log \left (1-\sqrt {2} \sqrt {\tan (e+f x)}+\tan (e+f x)\right )-5 \sqrt {2} \cos ^2(e+f x) \log \left (1+\sqrt {2} \sqrt {\tan (e+f x)}+\tan (e+f x)\right )+40 \cos ^2(e+f x) \sqrt {\tan (e+f x)}+4 \sin ^2(e+f x) \sqrt {\tan (e+f x)}+10 \sin (2 (e+f x)) \sqrt {\tan (e+f x)}\right ) \sqrt [4]{\tan (e+f x)}\right ) \sqrt {d \tan (e+f x)} (1+\tan (e+f x))^3}{10 f (\cos (e+f x)+\sin (e+f x))^3 \tan ^{\frac {3}{4}}(e+f x)} \]
(a^3*(-20*ArcTan[(-Tan[e + f*x]^2)^(1/4)]*Cos[e + f*x]^3*(-Tan[e + f*x])^( 1/4) + 20*ArcTanh[(-Tan[e + f*x]^2)^(1/4)]*Cos[e + f*x]^3*(-Tan[e + f*x])^ (1/4) + Cos[e + f*x]*(10*Sqrt[2]*ArcTan[1 - Sqrt[2]*Sqrt[Tan[e + f*x]]]*Co s[e + f*x]^2 - 10*Sqrt[2]*ArcTan[1 + Sqrt[2]*Sqrt[Tan[e + f*x]]]*Cos[e + f *x]^2 + 5*Sqrt[2]*Cos[e + f*x]^2*Log[1 - Sqrt[2]*Sqrt[Tan[e + f*x]] + Tan[ e + f*x]] - 5*Sqrt[2]*Cos[e + f*x]^2*Log[1 + Sqrt[2]*Sqrt[Tan[e + f*x]] + Tan[e + f*x]] + 40*Cos[e + f*x]^2*Sqrt[Tan[e + f*x]] + 4*Sin[e + f*x]^2*Sq rt[Tan[e + f*x]] + 10*Sin[2*(e + f*x)]*Sqrt[Tan[e + f*x]])*Tan[e + f*x]^(1 /4))*Sqrt[d*Tan[e + f*x]]*(1 + Tan[e + f*x])^3)/(10*f*(Cos[e + f*x] + Sin[ e + f*x])^3*Tan[e + f*x]^(3/4))
Time = 0.66 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.11, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {3042, 4049, 3042, 4113, 3042, 4011, 3042, 4015, 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.
\(\displaystyle \int (a \tan (e+f x)+a)^3 \sqrt {d \tan (e+f x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (a \tan (e+f x)+a)^3 \sqrt {d \tan (e+f x)}dx\) |
\(\Big \downarrow \) 4049 |
\(\displaystyle \frac {2 \int \sqrt {d \tan (e+f x)} \left (6 d \tan ^2(e+f x) a^3+d a^3+5 d \tan (e+f x) a^3\right )dx}{5 d}+\frac {2 \left (a^3 \tan (e+f x)+a^3\right ) (d \tan (e+f x))^{3/2}}{5 d f}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 \int \sqrt {d \tan (e+f x)} \left (6 d \tan (e+f x)^2 a^3+d a^3+5 d \tan (e+f x) a^3\right )dx}{5 d}+\frac {2 \left (a^3 \tan (e+f x)+a^3\right ) (d \tan (e+f x))^{3/2}}{5 d f}\) |
\(\Big \downarrow \) 4113 |
\(\displaystyle \frac {2 \left (\int \sqrt {d \tan (e+f x)} \left (5 a^3 d \tan (e+f x)-5 a^3 d\right )dx+\frac {4 a^3 (d \tan (e+f x))^{3/2}}{f}\right )}{5 d}+\frac {2 \left (a^3 \tan (e+f x)+a^3\right ) (d \tan (e+f x))^{3/2}}{5 d f}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 \left (\int \sqrt {d \tan (e+f x)} \left (5 a^3 d \tan (e+f x)-5 a^3 d\right )dx+\frac {4 a^3 (d \tan (e+f x))^{3/2}}{f}\right )}{5 d}+\frac {2 \left (a^3 \tan (e+f x)+a^3\right ) (d \tan (e+f x))^{3/2}}{5 d f}\) |
\(\Big \downarrow \) 4011 |
\(\displaystyle \frac {2 \left (\int \frac {-5 d^2 a^3-5 d^2 \tan (e+f x) a^3}{\sqrt {d \tan (e+f x)}}dx+\frac {4 a^3 (d \tan (e+f x))^{3/2}}{f}+\frac {10 a^3 d \sqrt {d \tan (e+f x)}}{f}\right )}{5 d}+\frac {2 \left (a^3 \tan (e+f x)+a^3\right ) (d \tan (e+f x))^{3/2}}{5 d f}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 \left (\int \frac {-5 d^2 a^3-5 d^2 \tan (e+f x) a^3}{\sqrt {d \tan (e+f x)}}dx+\frac {4 a^3 (d \tan (e+f x))^{3/2}}{f}+\frac {10 a^3 d \sqrt {d \tan (e+f x)}}{f}\right )}{5 d}+\frac {2 \left (a^3 \tan (e+f x)+a^3\right ) (d \tan (e+f x))^{3/2}}{5 d f}\) |
\(\Big \downarrow \) 4015 |
\(\displaystyle \frac {2 \left (-\frac {50 a^6 d^4 \int \frac {1}{50 d^4 a^6+25 \cot (e+f x) \left (a^3 d^2-a^3 d^2 \tan (e+f x)\right )^2}d\left (-\frac {5 \left (a^3 d^2-a^3 d^2 \tan (e+f x)\right )}{\sqrt {d \tan (e+f x)}}\right )}{f}+\frac {4 a^3 (d \tan (e+f x))^{3/2}}{f}+\frac {10 a^3 d \sqrt {d \tan (e+f x)}}{f}\right )}{5 d}+\frac {2 \left (a^3 \tan (e+f x)+a^3\right ) (d \tan (e+f x))^{3/2}}{5 d f}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {2 \left (\frac {5 \sqrt {2} a^3 d^{3/2} \arctan \left (\frac {a^3 d^2-a^3 d^2 \tan (e+f x)}{\sqrt {2} a^3 d^{3/2} \sqrt {d \tan (e+f x)}}\right )}{f}+\frac {4 a^3 (d \tan (e+f x))^{3/2}}{f}+\frac {10 a^3 d \sqrt {d \tan (e+f x)}}{f}\right )}{5 d}+\frac {2 \left (a^3 \tan (e+f x)+a^3\right ) (d \tan (e+f x))^{3/2}}{5 d f}\) |
(2*(d*Tan[e + f*x])^(3/2)*(a^3 + a^3*Tan[e + f*x]))/(5*d*f) + (2*((5*Sqrt[ 2]*a^3*d^(3/2)*ArcTan[(a^3*d^2 - a^3*d^2*Tan[e + f*x])/(Sqrt[2]*a^3*d^(3/2 )*Sqrt[d*Tan[e + f*x]])])/f + (10*a^3*d*Sqrt[d*Tan[e + f*x]])/f + (4*a^3*( d*Tan[e + f*x])^(3/2))/f))/(5*d)
3.4.52.3.1 Defintions of rubi rules used
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_)]), 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[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_ )]], x_Symbol] :> Simp[-2*(d^2/f) Subst[Int[1/(2*c*d + b*x^2), x], x, (c - d*Tan[e + f*x])/Sqrt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && EqQ[c^2 - d^2, 0]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b^2*(a + b*Tan[e + f*x])^(m - 2)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(m + n - 1))), x] + Simp[1/(d*(m + n - 1)) Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^n*Simp[a^3*d*(m + n - 1) - b^2*(b*c*(m - 2) + a*d*(1 + n)) + b*d*(m + n - 1)*(3*a^2 - b^2)*Tan[ e + f*x] - b^2*(b*c*(m - 2) - a*d*(3*m + 2*n - 4))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2 , 0] && NeQ[c^2 + d^2, 0] && IntegerQ[2*m] && GtQ[m, 2] && (GeQ[n, -1] || I ntegerQ[m]) && !(IGtQ[n, 2] && ( !IntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])) )
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[C*((a + b*Tan[e + f*x])^(m + 1)/(b*f*(m + 1))), x] + Int[(a + b*Tan[e + f*x])^m*Si mp[A - C + B*Tan[e + f*x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && NeQ[A*b^2 - a*b*B + a^2*C, 0] && !LeQ[m, -1]
Leaf count of result is larger than twice the leaf count of optimal. \(322\) vs. \(2(117)=234\).
Time = 0.80 (sec) , antiderivative size = 323, normalized size of antiderivative = 2.34
method | result | size |
derivativedivides | \(\frac {2 a^{3} \left (\frac {\left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}+d \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}+2 d^{2} \sqrt {d \tan \left (f x +e \right )}-2 d^{3} \left (\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 d}+\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (d^{2}\right )^{\frac {1}{4}}}\right )\right )}{f \,d^{2}}\) | \(323\) |
default | \(\frac {2 a^{3} \left (\frac {\left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}+d \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}+2 d^{2} \sqrt {d \tan \left (f x +e \right )}-2 d^{3} \left (\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 d}+\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (d^{2}\right )^{\frac {1}{4}}}\right )\right )}{f \,d^{2}}\) | \(323\) |
parts | \(\frac {a^{3} d \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 )}{4 f \left (d^{2}\right )^{\frac {1}{4}}}+\frac {2 a^{3} \left (\frac {\left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}-d^{2} \sqrt {d \tan \left (f x +e \right )}+\frac {d^{2} \left (d^{2}\right )^{\frac {1}{4}} \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}\right )}{f \,d^{2}}+\frac {3 a^{3} \left (2 \sqrt {d \tan \left (f x +e \right )}-\frac {\left (d^{2}\right )^{\frac {1}{4}} \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 )}{4}\right )}{f}+\frac {6 a^{3} \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 d}\) | \(623\) |
2/f*a^3/d^2*(1/5*(d*tan(f*x+e))^(5/2)+d*(d*tan(f*x+e))^(3/2)+2*d^2*(d*tan( f*x+e))^(1/2)-2*d^3*(1/8/d*(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*ta n(f*x+e))^(1/2)+1)-2*arctan(-2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1))+ 1/8/(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*ar ctan(-2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1))))
Time = 0.25 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.57 \[ \int \sqrt {d \tan (e+f x)} (a+a \tan (e+f x))^3 \, dx=\left [\frac {5 \, \sqrt {2} a^{3} \sqrt {-d} \log \left (\frac {d \tan \left (f x + e\right )^{2} - 2 \, \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {-d} {\left (\tan \left (f x + e\right ) - 1\right )} - 4 \, d \tan \left (f x + e\right ) + d}{\tan \left (f x + e\right )^{2} + 1}\right ) + 2 \, {\left (a^{3} \tan \left (f x + e\right )^{2} + 5 \, a^{3} \tan \left (f x + e\right ) + 10 \, a^{3}\right )} \sqrt {d \tan \left (f x + e\right )}}{5 \, f}, -\frac {2 \, {\left (5 \, \sqrt {2} a^{3} \sqrt {d} \arctan \left (\frac {\sqrt {2} \sqrt {d \tan \left (f x + e\right )} {\left (\tan \left (f x + e\right ) - 1\right )}}{2 \, \sqrt {d} \tan \left (f x + e\right )}\right ) - {\left (a^{3} \tan \left (f x + e\right )^{2} + 5 \, a^{3} \tan \left (f x + e\right ) + 10 \, a^{3}\right )} \sqrt {d \tan \left (f x + e\right )}\right )}}{5 \, f}\right ] \]
[1/5*(5*sqrt(2)*a^3*sqrt(-d)*log((d*tan(f*x + e)^2 - 2*sqrt(2)*sqrt(d*tan( f*x + e))*sqrt(-d)*(tan(f*x + e) - 1) - 4*d*tan(f*x + e) + d)/(tan(f*x + e )^2 + 1)) + 2*(a^3*tan(f*x + e)^2 + 5*a^3*tan(f*x + e) + 10*a^3)*sqrt(d*ta n(f*x + e)))/f, -2/5*(5*sqrt(2)*a^3*sqrt(d)*arctan(1/2*sqrt(2)*sqrt(d*tan( f*x + e))*(tan(f*x + e) - 1)/(sqrt(d)*tan(f*x + e))) - (a^3*tan(f*x + e)^2 + 5*a^3*tan(f*x + e) + 10*a^3)*sqrt(d*tan(f*x + e)))/f]
\[ \int \sqrt {d \tan (e+f x)} (a+a \tan (e+f x))^3 \, dx=a^{3} \left (\int \sqrt {d \tan {\left (e + f x \right )}}\, dx + \int 3 \sqrt {d \tan {\left (e + f x \right )}} \tan {\left (e + f x \right )}\, dx + \int 3 \sqrt {d \tan {\left (e + f x \right )}} \tan ^{2}{\left (e + f x \right )}\, dx + \int \sqrt {d \tan {\left (e + f x \right )}} \tan ^{3}{\left (e + f x \right )}\, dx\right ) \]
a**3*(Integral(sqrt(d*tan(e + f*x)), x) + Integral(3*sqrt(d*tan(e + f*x))* tan(e + f*x), x) + Integral(3*sqrt(d*tan(e + f*x))*tan(e + f*x)**2, x) + I ntegral(sqrt(d*tan(e + f*x))*tan(e + f*x)**3, x))
Time = 0.32 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.04 \[ \int \sqrt {d \tan (e+f x)} (a+a \tan (e+f x))^3 \, dx=-\frac {2 \, {\left (5 \, a^{3} d^{2} {\left (\frac {\sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} + 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} + \frac {\sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} - 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}}\right )} - \frac {\left (d \tan \left (f x + e\right )\right )^{\frac {5}{2}} a^{3} + 5 \, \left (d \tan \left (f x + e\right )\right )^{\frac {3}{2}} a^{3} d + 10 \, \sqrt {d \tan \left (f x + e\right )} a^{3} d^{2}}{d}\right )}}{5 \, d f} \]
-2/5*(5*a^3*d^2*(sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(d) + 2*sqrt(d*ta n(f*x + e)))/sqrt(d))/sqrt(d) + sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt( d) - 2*sqrt(d*tan(f*x + e)))/sqrt(d))/sqrt(d)) - ((d*tan(f*x + e))^(5/2)*a ^3 + 5*(d*tan(f*x + e))^(3/2)*a^3*d + 10*sqrt(d*tan(f*x + e))*a^3*d^2)/d)/ (d*f)
Timed out. \[ \int \sqrt {d \tan (e+f x)} (a+a \tan (e+f x))^3 \, dx=\text {Timed out} \]
Time = 6.10 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.99 \[ \int \sqrt {d \tan (e+f x)} (a+a \tan (e+f x))^3 \, dx=\frac {4\,a^3\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{f}+\frac {2\,a^3\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}{d\,f}+\frac {2\,a^3\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}}{5\,d^2\,f}-\frac {\sqrt {2}\,a^3\,\sqrt {d}\,\left (2\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{2\,\sqrt {d}}\right )+2\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{2\,\sqrt {d}}+\frac {\sqrt {2}\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}{2\,d^{3/2}}\right )\right )}{f} \]
(4*a^3*(d*tan(e + f*x))^(1/2))/f + (2*a^3*(d*tan(e + f*x))^(3/2))/(d*f) + (2*a^3*(d*tan(e + f*x))^(5/2))/(5*d^2*f) - (2^(1/2)*a^3*d^(1/2)*(2*atan((2 ^(1/2)*(d*tan(e + f*x))^(1/2))/(2*d^(1/2))) + 2*atan((2^(1/2)*(d*tan(e + f *x))^(1/2))/(2*d^(1/2)) + (2^(1/2)*(d*tan(e + f*x))^(3/2))/(2*d^(3/2)))))/ f