Integrand size = 21, antiderivative size = 208 \[ \int \tan ^3(e+f x) \sqrt {1+\tan (e+f x)} \, dx=\frac {\sqrt {\frac {1}{2} \left (-1+\sqrt {2}\right )} \arctan \left (\frac {4-3 \sqrt {2}+\left (2-\sqrt {2}\right ) \tan (e+f x)}{2 \sqrt {-7+5 \sqrt {2}} \sqrt {1+\tan (e+f x)}}\right )}{f}+\frac {\sqrt {\frac {1}{2} \left (1+\sqrt {2}\right )} \text {arctanh}\left (\frac {4+3 \sqrt {2}+\left (2+\sqrt {2}\right ) \tan (e+f x)}{2 \sqrt {7+5 \sqrt {2}} \sqrt {1+\tan (e+f x)}}\right )}{f}-\frac {2 \sqrt {1+\tan (e+f x)}}{f}-\frac {4 (1+\tan (e+f x))^{3/2}}{15 f}+\frac {2 \tan (e+f x) (1+\tan (e+f x))^{3/2}}{5 f} \] Output:
1/2*(-2+2*2^(1/2))^(1/2)*arctan(1/2*(4-3*2^(1/2)+(2-2^(1/2))*tan(f*x+e))/( -7+5*2^(1/2))^(1/2)/(1+tan(f*x+e))^(1/2))/f+1/2*(2+2*2^(1/2))^(1/2)*arctan h(1/2*(4+3*2^(1/2)+(2+2^(1/2))*tan(f*x+e))/(7+5*2^(1/2))^(1/2)/(1+tan(f*x+ e))^(1/2))/f-2*(1+tan(f*x+e))^(1/2)/f-4/15*(1+tan(f*x+e))^(3/2)/f+2/5*tan( f*x+e)*(1+tan(f*x+e))^(3/2)/f
Result contains complex when optimal does not.
Time = 0.23 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.48 \[ \int \tan ^3(e+f x) \sqrt {1+\tan (e+f x)} \, dx=\frac {15 \sqrt {1-i} \text {arctanh}\left (\frac {\sqrt {1+\tan (e+f x)}}{\sqrt {1-i}}\right )+15 \sqrt {1+i} \text {arctanh}\left (\frac {\sqrt {1+\tan (e+f x)}}{\sqrt {1+i}}\right )+2 \sqrt {1+\tan (e+f x)} \left (-17+\tan (e+f x)+3 \tan ^2(e+f x)\right )}{15 f} \] Input:
Integrate[Tan[e + f*x]^3*Sqrt[1 + Tan[e + f*x]],x]
Output:
(15*Sqrt[1 - I]*ArcTanh[Sqrt[1 + Tan[e + f*x]]/Sqrt[1 - I]] + 15*Sqrt[1 + I]*ArcTanh[Sqrt[1 + Tan[e + f*x]]/Sqrt[1 + I]] + 2*Sqrt[1 + Tan[e + f*x]]* (-17 + Tan[e + f*x] + 3*Tan[e + f*x]^2))/(15*f)
Time = 0.79 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.13, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.714, Rules used = {3042, 4049, 27, 3042, 4113, 27, 3042, 4011, 3042, 4019, 25, 3042, 4018, 216, 220}
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.
| \(\displaystyle \int \tan ^3(e+f x) \sqrt {\tan (e+f x)+1} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \tan (e+f x)^3 \sqrt {\tan (e+f x)+1}dx\) |
| \(\Big \downarrow \) 4049 |
| \(\displaystyle \frac {2}{5} \int -\frac {1}{2} \sqrt {\tan (e+f x)+1} \left (2 \tan ^2(e+f x)+5 \tan (e+f x)+2\right )dx+\frac {2 \tan (e+f x) (\tan (e+f x)+1)^{3/2}}{5 f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2 \tan (e+f x) (\tan (e+f x)+1)^{3/2}}{5 f}-\frac {1}{5} \int \sqrt {\tan (e+f x)+1} \left (2 \tan ^2(e+f x)+5 \tan (e+f x)+2\right )dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 \tan (e+f x) (\tan (e+f x)+1)^{3/2}}{5 f}-\frac {1}{5} \int \sqrt {\tan (e+f x)+1} \left (2 \tan (e+f x)^2+5 \tan (e+f x)+2\right )dx\) |
| \(\Big \downarrow \) 4113 |
| \(\displaystyle \frac {1}{5} \left (-\int 5 \tan (e+f x) \sqrt {\tan (e+f x)+1}dx-\frac {4 (\tan (e+f x)+1)^{3/2}}{3 f}\right )+\frac {2 \tan (e+f x) (\tan (e+f x)+1)^{3/2}}{5 f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \left (-5 \int \tan (e+f x) \sqrt {\tan (e+f x)+1}dx-\frac {4 (\tan (e+f x)+1)^{3/2}}{3 f}\right )+\frac {2 \tan (e+f x) (\tan (e+f x)+1)^{3/2}}{5 f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (-5 \int \tan (e+f x) \sqrt {\tan (e+f x)+1}dx-\frac {4 (\tan (e+f x)+1)^{3/2}}{3 f}\right )+\frac {2 \tan (e+f x) (\tan (e+f x)+1)^{3/2}}{5 f}\) |
| \(\Big \downarrow \) 4011 |
| \(\displaystyle \frac {1}{5} \left (-5 \left (\int \frac {\tan (e+f x)-1}{\sqrt {\tan (e+f x)+1}}dx+\frac {2 \sqrt {\tan (e+f x)+1}}{f}\right )-\frac {4 (\tan (e+f x)+1)^{3/2}}{3 f}\right )+\frac {2 \tan (e+f x) (\tan (e+f x)+1)^{3/2}}{5 f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (-5 \left (\int \frac {\tan (e+f x)-1}{\sqrt {\tan (e+f x)+1}}dx+\frac {2 \sqrt {\tan (e+f x)+1}}{f}\right )-\frac {4 (\tan (e+f x)+1)^{3/2}}{3 f}\right )+\frac {2 \tan (e+f x) (\tan (e+f x)+1)^{3/2}}{5 f}\) |
| \(\Big \downarrow \) 4019 |
| \(\displaystyle \frac {1}{5} \left (-5 \left (\frac {\int -\frac {\left (2-\sqrt {2}\right ) \tan (e+f x)+\sqrt {2}}{\sqrt {\tan (e+f x)+1}}dx}{2 \sqrt {2}}-\frac {\int \frac {\sqrt {2}-\left (2+\sqrt {2}\right ) \tan (e+f x)}{\sqrt {\tan (e+f x)+1}}dx}{2 \sqrt {2}}+\frac {2 \sqrt {\tan (e+f x)+1}}{f}\right )-\frac {4 (\tan (e+f x)+1)^{3/2}}{3 f}\right )+\frac {2 \tan (e+f x) (\tan (e+f x)+1)^{3/2}}{5 f}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{5} \left (-5 \left (-\frac {\int \frac {\left (2-\sqrt {2}\right ) \tan (e+f x)+\sqrt {2}}{\sqrt {\tan (e+f x)+1}}dx}{2 \sqrt {2}}-\frac {\int \frac {\sqrt {2}-\left (2+\sqrt {2}\right ) \tan (e+f x)}{\sqrt {\tan (e+f x)+1}}dx}{2 \sqrt {2}}+\frac {2 \sqrt {\tan (e+f x)+1}}{f}\right )-\frac {4 (\tan (e+f x)+1)^{3/2}}{3 f}\right )+\frac {2 \tan (e+f x) (\tan (e+f x)+1)^{3/2}}{5 f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (-5 \left (-\frac {\int \frac {\left (2-\sqrt {2}\right ) \tan (e+f x)+\sqrt {2}}{\sqrt {\tan (e+f x)+1}}dx}{2 \sqrt {2}}-\frac {\int \frac {\sqrt {2}-\left (2+\sqrt {2}\right ) \tan (e+f x)}{\sqrt {\tan (e+f x)+1}}dx}{2 \sqrt {2}}+\frac {2 \sqrt {\tan (e+f x)+1}}{f}\right )-\frac {4 (\tan (e+f x)+1)^{3/2}}{3 f}\right )+\frac {2 \tan (e+f x) (\tan (e+f x)+1)^{3/2}}{5 f}\) |
| \(\Big \downarrow \) 4018 |
| \(\displaystyle \frac {1}{5} \left (-5 \left (\frac {\sqrt {2} \left (3-2 \sqrt {2}\right ) \int \frac {1}{\frac {\left (\left (2-\sqrt {2}\right ) \tan (e+f x)-3 \sqrt {2}+4\right )^2}{\tan (e+f x)+1}-4 \left (7-5 \sqrt {2}\right )}d\left (-\frac {\left (2-\sqrt {2}\right ) \tan (e+f x)-3 \sqrt {2}+4}{\sqrt {\tan (e+f x)+1}}\right )}{f}+\frac {\sqrt {2} \left (3+2 \sqrt {2}\right ) \int \frac {1}{\frac {\left (\left (2+\sqrt {2}\right ) \tan (e+f x)+3 \sqrt {2}+4\right )^2}{\tan (e+f x)+1}-4 \left (7+5 \sqrt {2}\right )}d\frac {\left (2+\sqrt {2}\right ) \tan (e+f x)+3 \sqrt {2}+4}{\sqrt {\tan (e+f x)+1}}}{f}+\frac {2 \sqrt {\tan (e+f x)+1}}{f}\right )-\frac {4 (\tan (e+f x)+1)^{3/2}}{3 f}\right )+\frac {2 \tan (e+f x) (\tan (e+f x)+1)^{3/2}}{5 f}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {1}{5} \left (-5 \left (\frac {\sqrt {2} \left (3+2 \sqrt {2}\right ) \int \frac {1}{\frac {\left (\left (2+\sqrt {2}\right ) \tan (e+f x)+3 \sqrt {2}+4\right )^2}{\tan (e+f x)+1}-4 \left (7+5 \sqrt {2}\right )}d\frac {\left (2+\sqrt {2}\right ) \tan (e+f x)+3 \sqrt {2}+4}{\sqrt {\tan (e+f x)+1}}}{f}-\frac {\left (3-2 \sqrt {2}\right ) \arctan \left (\frac {\left (2-\sqrt {2}\right ) \tan (e+f x)-3 \sqrt {2}+4}{2 \sqrt {5 \sqrt {2}-7} \sqrt {\tan (e+f x)+1}}\right )}{\sqrt {2 \left (5 \sqrt {2}-7\right )} f}+\frac {2 \sqrt {\tan (e+f x)+1}}{f}\right )-\frac {4 (\tan (e+f x)+1)^{3/2}}{3 f}\right )+\frac {2 \tan (e+f x) (\tan (e+f x)+1)^{3/2}}{5 f}\) |
| \(\Big \downarrow \) 220 |
| \(\displaystyle \frac {1}{5} \left (-5 \left (-\frac {\left (3-2 \sqrt {2}\right ) \arctan \left (\frac {\left (2-\sqrt {2}\right ) \tan (e+f x)-3 \sqrt {2}+4}{2 \sqrt {5 \sqrt {2}-7} \sqrt {\tan (e+f x)+1}}\right )}{\sqrt {2 \left (5 \sqrt {2}-7\right )} f}-\frac {\left (3+2 \sqrt {2}\right ) \text {arctanh}\left (\frac {\left (2+\sqrt {2}\right ) \tan (e+f x)+3 \sqrt {2}+4}{2 \sqrt {7+5 \sqrt {2}} \sqrt {\tan (e+f x)+1}}\right )}{\sqrt {2 \left (7+5 \sqrt {2}\right )} f}+\frac {2 \sqrt {\tan (e+f x)+1}}{f}\right )-\frac {4 (\tan (e+f x)+1)^{3/2}}{3 f}\right )+\frac {2 \tan (e+f x) (\tan (e+f x)+1)^{3/2}}{5 f}\) |
Input:
Int[Tan[e + f*x]^3*Sqrt[1 + Tan[e + f*x]],x]
Output:
(2*Tan[e + f*x]*(1 + Tan[e + f*x])^(3/2))/(5*f) + ((-4*(1 + Tan[e + f*x])^ (3/2))/(3*f) - 5*(-(((3 - 2*Sqrt[2])*ArcTan[(4 - 3*Sqrt[2] + (2 - Sqrt[2]) *Tan[e + f*x])/(2*Sqrt[-7 + 5*Sqrt[2]]*Sqrt[1 + Tan[e + f*x]])])/(Sqrt[2*( -7 + 5*Sqrt[2])]*f)) - ((3 + 2*Sqrt[2])*ArcTanh[(4 + 3*Sqrt[2] + (2 + Sqrt [2])*Tan[e + f*x])/(2*Sqrt[7 + 5*Sqrt[2]]*Sqrt[1 + Tan[e + f*x]])])/(Sqrt[ 2*(7 + 5*Sqrt[2])]*f) + (2*Sqrt[1 + Tan[e + f*x]])/f))/5
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[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(- 1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])
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[(a_) + (b_.)*tan[(e_.) + ( f_.)*(x_)]], x_Symbol] :> Simp[-2*(d^2/f) Subst[Int[1/(2*b*c*d - 4*a*d^2 + x^2), x], x, (b*c - 2*a*d - b*d*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*x]]], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0 ] && NeQ[c^2 + d^2, 0] && EqQ[2*a*c*d - b*(c^2 - d^2), 0]
Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(a_) + (b_.)*tan[(e_.) + ( f_.)*(x_)]], x_Symbol] :> With[{q = Rt[a^2 + b^2, 2]}, Simp[1/(2*q) Int[( a*c + b*d + c*q + (b*c - a*d + d*q)*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*x]], x], x] - Simp[1/(2*q) Int[(a*c + b*d - c*q + (b*c - a*d - d*q)*Tan[e + f *x])/Sqrt[a + b*Tan[e + f*x]], x], x]] /; FreeQ[{a, b, c, d, e, f}, x] && N eQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && NeQ[2*a*c*d - b*(c^2 - d^2), 0] && NiceSqrtQ[a^2 + b^2]
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]
Time = 1.06 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.11
| method | result | size |
| derivativedivides | \(\frac {\frac {2 \left (\tan \left (f x +e \right )+1\right )^{\frac {5}{2}}}{5}-\frac {2 \left (\tan \left (f x +e \right )+1\right )^{\frac {3}{2}}}{3}-2 \sqrt {\tan \left (f x +e \right )+1}+\frac {\sqrt {2 \sqrt {2}+2}\, \ln \left (\tan \left (f x +e \right )+1+\sqrt {\tan \left (f x +e \right )+1}\, \sqrt {2 \sqrt {2}+2}+\sqrt {2}\right )}{4}+\frac {\left (\sqrt {2}-1\right ) \arctan \left (\frac {\sqrt {2 \sqrt {2}+2}+2 \sqrt {\tan \left (f x +e \right )+1}}{\sqrt {-2+2 \sqrt {2}}}\right )}{\sqrt {-2+2 \sqrt {2}}}-\frac {\sqrt {2 \sqrt {2}+2}\, \ln \left (\tan \left (f x +e \right )+1-\sqrt {\tan \left (f x +e \right )+1}\, \sqrt {2 \sqrt {2}+2}+\sqrt {2}\right )}{4}-\frac {\left (1-\sqrt {2}\right ) \arctan \left (\frac {2 \sqrt {\tan \left (f x +e \right )+1}-\sqrt {2 \sqrt {2}+2}}{\sqrt {-2+2 \sqrt {2}}}\right )}{\sqrt {-2+2 \sqrt {2}}}}{f}\) | \(230\) |
| default | \(\frac {\frac {2 \left (\tan \left (f x +e \right )+1\right )^{\frac {5}{2}}}{5}-\frac {2 \left (\tan \left (f x +e \right )+1\right )^{\frac {3}{2}}}{3}-2 \sqrt {\tan \left (f x +e \right )+1}+\frac {\sqrt {2 \sqrt {2}+2}\, \ln \left (\tan \left (f x +e \right )+1+\sqrt {\tan \left (f x +e \right )+1}\, \sqrt {2 \sqrt {2}+2}+\sqrt {2}\right )}{4}+\frac {\left (\sqrt {2}-1\right ) \arctan \left (\frac {\sqrt {2 \sqrt {2}+2}+2 \sqrt {\tan \left (f x +e \right )+1}}{\sqrt {-2+2 \sqrt {2}}}\right )}{\sqrt {-2+2 \sqrt {2}}}-\frac {\sqrt {2 \sqrt {2}+2}\, \ln \left (\tan \left (f x +e \right )+1-\sqrt {\tan \left (f x +e \right )+1}\, \sqrt {2 \sqrt {2}+2}+\sqrt {2}\right )}{4}-\frac {\left (1-\sqrt {2}\right ) \arctan \left (\frac {2 \sqrt {\tan \left (f x +e \right )+1}-\sqrt {2 \sqrt {2}+2}}{\sqrt {-2+2 \sqrt {2}}}\right )}{\sqrt {-2+2 \sqrt {2}}}}{f}\) | \(230\) |
Input:
int(tan(f*x+e)^3*(tan(f*x+e)+1)^(1/2),x,method=_RETURNVERBOSE)
Output:
1/f*(2/5*(tan(f*x+e)+1)^(5/2)-2/3*(tan(f*x+e)+1)^(3/2)-2*(tan(f*x+e)+1)^(1 /2)+1/4*(2*2^(1/2)+2)^(1/2)*ln(tan(f*x+e)+1+(tan(f*x+e)+1)^(1/2)*(2*2^(1/2 )+2)^(1/2)+2^(1/2))+(2^(1/2)-1)/(-2+2*2^(1/2))^(1/2)*arctan(((2*2^(1/2)+2) ^(1/2)+2*(tan(f*x+e)+1)^(1/2))/(-2+2*2^(1/2))^(1/2))-1/4*(2*2^(1/2)+2)^(1/ 2)*ln(tan(f*x+e)+1-(tan(f*x+e)+1)^(1/2)*(2*2^(1/2)+2)^(1/2)+2^(1/2))-(1-2^ (1/2))/(-2+2*2^(1/2))^(1/2)*arctan((2*(tan(f*x+e)+1)^(1/2)-(2*2^(1/2)+2)^( 1/2))/(-2+2*2^(1/2))^(1/2)))
Time = 0.09 (sec) , antiderivative size = 262, normalized size of antiderivative = 1.26 \[ \int \tan ^3(e+f x) \sqrt {1+\tan (e+f x)} \, dx=\frac {15 \, f \sqrt {\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} + 1}{f^{2}}} \log \left (f \sqrt {\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} + 1}{f^{2}}} + \sqrt {\tan \left (f x + e\right ) + 1}\right ) - 15 \, f \sqrt {\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} + 1}{f^{2}}} \log \left (-f \sqrt {\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} + 1}{f^{2}}} + \sqrt {\tan \left (f x + e\right ) + 1}\right ) + 15 \, f \sqrt {-\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} - 1}{f^{2}}} \log \left (f \sqrt {-\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} - 1}{f^{2}}} + \sqrt {\tan \left (f x + e\right ) + 1}\right ) - 15 \, f \sqrt {-\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} - 1}{f^{2}}} \log \left (-f \sqrt {-\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} - 1}{f^{2}}} + \sqrt {\tan \left (f x + e\right ) + 1}\right ) + 4 \, {\left (3 \, \tan \left (f x + e\right )^{2} + \tan \left (f x + e\right ) - 17\right )} \sqrt {\tan \left (f x + e\right ) + 1}}{30 \, f} \] Input:
integrate(tan(f*x+e)^3*(1+tan(f*x+e))^(1/2),x, algorithm="fricas")
Output:
1/30*(15*f*sqrt((f^2*sqrt(-1/f^4) + 1)/f^2)*log(f*sqrt((f^2*sqrt(-1/f^4) + 1)/f^2) + sqrt(tan(f*x + e) + 1)) - 15*f*sqrt((f^2*sqrt(-1/f^4) + 1)/f^2) *log(-f*sqrt((f^2*sqrt(-1/f^4) + 1)/f^2) + sqrt(tan(f*x + e) + 1)) + 15*f* sqrt(-(f^2*sqrt(-1/f^4) - 1)/f^2)*log(f*sqrt(-(f^2*sqrt(-1/f^4) - 1)/f^2) + sqrt(tan(f*x + e) + 1)) - 15*f*sqrt(-(f^2*sqrt(-1/f^4) - 1)/f^2)*log(-f* sqrt(-(f^2*sqrt(-1/f^4) - 1)/f^2) + sqrt(tan(f*x + e) + 1)) + 4*(3*tan(f*x + e)^2 + tan(f*x + e) - 17)*sqrt(tan(f*x + e) + 1))/f
\[ \int \tan ^3(e+f x) \sqrt {1+\tan (e+f x)} \, dx=\int \sqrt {\tan {\left (e + f x \right )} + 1} \tan ^{3}{\left (e + f x \right )}\, dx \] Input:
integrate(tan(f*x+e)**3*(1+tan(f*x+e))**(1/2),x)
Output:
Integral(sqrt(tan(e + f*x) + 1)*tan(e + f*x)**3, x)
\[ \int \tan ^3(e+f x) \sqrt {1+\tan (e+f x)} \, dx=\int { \sqrt {\tan \left (f x + e\right ) + 1} \tan \left (f x + e\right )^{3} \,d x } \] Input:
integrate(tan(f*x+e)^3*(1+tan(f*x+e))^(1/2),x, algorithm="maxima")
Output:
integrate(sqrt(tan(f*x + e) + 1)*tan(f*x + e)^3, x)
Exception generated. \[ \int \tan ^3(e+f x) \sqrt {1+\tan (e+f x)} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(tan(f*x+e)^3*(1+tan(f*x+e))^(1/2),x, algorithm="giac")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Unable to divide, perhaps due to ro unding error%%%{%%%{1,[19]%%%}+%%%{8,[17]%%%}+%%%{28,[15]%%%}+%%%{56,[13]% %%}+%%%{7
Time = 1.83 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.58 \[ \int \tan ^3(e+f x) \sqrt {1+\tan (e+f x)} \, dx=\frac {2\,{\left (\mathrm {tan}\left (e+f\,x\right )+1\right )}^{5/2}}{5\,f}-\frac {2\,{\left (\mathrm {tan}\left (e+f\,x\right )+1\right )}^{3/2}}{3\,f}-\frac {2\,\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}}{f}+\mathrm {atan}\left (f\,\sqrt {\frac {\frac {1}{4}-\frac {1}{4}{}\mathrm {i}}{f^2}}\,\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}\,\left (1-\mathrm {i}\right )\right )\,\sqrt {\frac {\frac {1}{4}-\frac {1}{4}{}\mathrm {i}}{f^2}}\,2{}\mathrm {i}-\mathrm {atan}\left (f\,\sqrt {\frac {\frac {1}{4}+\frac {1}{4}{}\mathrm {i}}{f^2}}\,\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}\,\left (1+1{}\mathrm {i}\right )\right )\,\sqrt {\frac {\frac {1}{4}+\frac {1}{4}{}\mathrm {i}}{f^2}}\,2{}\mathrm {i} \] Input:
int(tan(e + f*x)^3*(tan(e + f*x) + 1)^(1/2),x)
Output:
(2*(tan(e + f*x) + 1)^(5/2))/(5*f) - (2*(tan(e + f*x) + 1)^(3/2))/(3*f) - (2*(tan(e + f*x) + 1)^(1/2))/f + atan(f*((1/4 - 1i/4)/f^2)^(1/2)*(tan(e + f*x) + 1)^(1/2)*(1 - 1i))*((1/4 - 1i/4)/f^2)^(1/2)*2i - atan(f*((1/4 + 1i/ 4)/f^2)^(1/2)*(tan(e + f*x) + 1)^(1/2)*(1 + 1i))*((1/4 + 1i/4)/f^2)^(1/2)* 2i
\[ \int \tan ^3(e+f x) \sqrt {1+\tan (e+f x)} \, dx=\int \sqrt {\tan \left (f x +e \right )+1}\, \tan \left (f x +e \right )^{3}d x \] Input:
int(tan(f*x+e)^3*(1+tan(f*x+e))^(1/2),x)
Output:
int(sqrt(tan(e + f*x) + 1)*tan(e + f*x)**3,x)