Integrand size = 21, antiderivative size = 173 \[ \int \tan ^2(e+f x) (1+\tan (e+f x))^{3/2} \, dx=\frac {\sqrt {-1+\sqrt {2}} \arctan \left (\frac {3-2 \sqrt {2}+\left (1-\sqrt {2}\right ) \tan (e+f x)}{\sqrt {2 \left (-7+5 \sqrt {2}\right )} \sqrt {1+\tan (e+f x)}}\right )}{f}+\frac {\sqrt {1+\sqrt {2}} \text {arctanh}\left (\frac {3+2 \sqrt {2}+\left (1+\sqrt {2}\right ) \tan (e+f x)}{\sqrt {2 \left (7+5 \sqrt {2}\right )} \sqrt {1+\tan (e+f x)}}\right )}{f}-\frac {2 \sqrt {1+\tan (e+f x)}}{f}+\frac {2 (1+\tan (e+f x))^{5/2}}{5 f} \] Output:
(2^(1/2)-1)^(1/2)*arctan((3-2*2^(1/2)+(1-2^(1/2))*tan(f*x+e))/(-14+10*2^(1 /2))^(1/2)/(1+tan(f*x+e))^(1/2))/f+(1+2^(1/2))^(1/2)*arctanh((3+2*2^(1/2)+ (1+2^(1/2))*tan(f*x+e))/(14+10*2^(1/2))^(1/2)/(1+tan(f*x+e))^(1/2))/f-2*(1 +tan(f*x+e))^(1/2)/f+2/5*(1+tan(f*x+e))^(5/2)/f
Result contains complex when optimal does not.
Time = 0.27 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.58 \[ \int \tan ^2(e+f x) (1+\tan (e+f x))^{3/2} \, dx=\frac {\frac {10 \text {arctanh}\left (\frac {\sqrt {1+\tan (e+f x)}}{\sqrt {1-i}}\right )}{\sqrt {1-i}}+\frac {10 \text {arctanh}\left (\frac {\sqrt {1+\tan (e+f x)}}{\sqrt {1+i}}\right )}{\sqrt {1+i}}+2 \sqrt {1+\tan (e+f x)} \left (-4+2 \tan (e+f x)+\tan ^2(e+f x)\right )}{5 f} \] Input:
Integrate[Tan[e + f*x]^2*(1 + Tan[e + f*x])^(3/2),x]
Output:
((10*ArcTanh[Sqrt[1 + Tan[e + f*x]]/Sqrt[1 - I]])/Sqrt[1 - I] + (10*ArcTan h[Sqrt[1 + Tan[e + f*x]]/Sqrt[1 + I]])/Sqrt[1 + I] + 2*Sqrt[1 + Tan[e + f* x]]*(-4 + 2*Tan[e + f*x] + Tan[e + f*x]^2))/(5*f)
Time = 0.60 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.18, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {3042, 4026, 25, 3042, 3963, 27, 3042, 4019, 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.
| \(\displaystyle \int \tan ^2(e+f x) (\tan (e+f x)+1)^{3/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \tan (e+f x)^2 (\tan (e+f x)+1)^{3/2}dx\) |
| \(\Big \downarrow \) 4026 |
| \(\displaystyle \int -(\tan (e+f x)+1)^{3/2}dx+\frac {2 (\tan (e+f x)+1)^{5/2}}{5 f}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {2 (\tan (e+f x)+1)^{5/2}}{5 f}-\int (\tan (e+f x)+1)^{3/2}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 (\tan (e+f x)+1)^{5/2}}{5 f}-\int (\tan (e+f x)+1)^{3/2}dx\) |
| \(\Big \downarrow \) 3963 |
| \(\displaystyle -\int \frac {2 \tan (e+f x)}{\sqrt {\tan (e+f x)+1}}dx+\frac {2 (\tan (e+f x)+1)^{5/2}}{5 f}-\frac {2 \sqrt {\tan (e+f x)+1}}{f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -2 \int \frac {\tan (e+f x)}{\sqrt {\tan (e+f x)+1}}dx+\frac {2 (\tan (e+f x)+1)^{5/2}}{5 f}-\frac {2 \sqrt {\tan (e+f x)+1}}{f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -2 \int \frac {\tan (e+f x)}{\sqrt {\tan (e+f x)+1}}dx+\frac {2 (\tan (e+f x)+1)^{5/2}}{5 f}-\frac {2 \sqrt {\tan (e+f x)+1}}{f}\) |
| \(\Big \downarrow \) 4019 |
| \(\displaystyle -2 \left (\frac {\int \frac {1-\left (1-\sqrt {2}\right ) \tan (e+f x)}{\sqrt {\tan (e+f x)+1}}dx}{2 \sqrt {2}}-\frac {\int \frac {1-\left (1+\sqrt {2}\right ) \tan (e+f x)}{\sqrt {\tan (e+f x)+1}}dx}{2 \sqrt {2}}\right )+\frac {2 (\tan (e+f x)+1)^{5/2}}{5 f}-\frac {2 \sqrt {\tan (e+f x)+1}}{f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -2 \left (\frac {\int \frac {1-\left (1-\sqrt {2}\right ) \tan (e+f x)}{\sqrt {\tan (e+f x)+1}}dx}{2 \sqrt {2}}-\frac {\int \frac {1-\left (1+\sqrt {2}\right ) \tan (e+f x)}{\sqrt {\tan (e+f x)+1}}dx}{2 \sqrt {2}}\right )+\frac {2 (\tan (e+f x)+1)^{5/2}}{5 f}-\frac {2 \sqrt {\tan (e+f x)+1}}{f}\) |
| \(\Big \downarrow \) 4018 |
| \(\displaystyle -2 \left (\frac {\left (3+2 \sqrt {2}\right ) \int \frac {1}{\frac {\left (\left (1+\sqrt {2}\right ) \tan (e+f x)+2 \sqrt {2}+3\right )^2}{\tan (e+f x)+1}-2 \left (7+5 \sqrt {2}\right )}d\frac {\left (1+\sqrt {2}\right ) \tan (e+f x)+2 \sqrt {2}+3}{\sqrt {\tan (e+f x)+1}}}{\sqrt {2} f}-\frac {\left (3-2 \sqrt {2}\right ) \int \frac {1}{\frac {\left (\left (1-\sqrt {2}\right ) \tan (e+f x)-2 \sqrt {2}+3\right )^2}{\tan (e+f x)+1}-2 \left (7-5 \sqrt {2}\right )}d\frac {\left (1-\sqrt {2}\right ) \tan (e+f x)-2 \sqrt {2}+3}{\sqrt {\tan (e+f x)+1}}}{\sqrt {2} f}\right )+\frac {2 (\tan (e+f x)+1)^{5/2}}{5 f}-\frac {2 \sqrt {\tan (e+f x)+1}}{f}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle -2 \left (\frac {\left (3+2 \sqrt {2}\right ) \int \frac {1}{\frac {\left (\left (1+\sqrt {2}\right ) \tan (e+f x)+2 \sqrt {2}+3\right )^2}{\tan (e+f x)+1}-2 \left (7+5 \sqrt {2}\right )}d\frac {\left (1+\sqrt {2}\right ) \tan (e+f x)+2 \sqrt {2}+3}{\sqrt {\tan (e+f x)+1}}}{\sqrt {2} f}-\frac {\left (3-2 \sqrt {2}\right ) \arctan \left (\frac {\left (1-\sqrt {2}\right ) \tan (e+f x)-2 \sqrt {2}+3}{\sqrt {2 \left (5 \sqrt {2}-7\right )} \sqrt {\tan (e+f x)+1}}\right )}{2 \sqrt {5 \sqrt {2}-7} f}\right )+\frac {2 (\tan (e+f x)+1)^{5/2}}{5 f}-\frac {2 \sqrt {\tan (e+f x)+1}}{f}\) |
| \(\Big \downarrow \) 220 |
| \(\displaystyle -2 \left (-\frac {\left (3-2 \sqrt {2}\right ) \arctan \left (\frac {\left (1-\sqrt {2}\right ) \tan (e+f x)-2 \sqrt {2}+3}{\sqrt {2 \left (5 \sqrt {2}-7\right )} \sqrt {\tan (e+f x)+1}}\right )}{2 \sqrt {5 \sqrt {2}-7} f}-\frac {\left (3+2 \sqrt {2}\right ) \text {arctanh}\left (\frac {\left (1+\sqrt {2}\right ) \tan (e+f x)+2 \sqrt {2}+3}{\sqrt {2 \left (7+5 \sqrt {2}\right )} \sqrt {\tan (e+f x)+1}}\right )}{2 \sqrt {7+5 \sqrt {2}} f}\right )+\frac {2 (\tan (e+f x)+1)^{5/2}}{5 f}-\frac {2 \sqrt {\tan (e+f x)+1}}{f}\) |
Input:
Int[Tan[e + f*x]^2*(1 + Tan[e + f*x])^(3/2),x]
Output:
-2*(-1/2*((3 - 2*Sqrt[2])*ArcTan[(3 - 2*Sqrt[2] + (1 - Sqrt[2])*Tan[e + f* x])/(Sqrt[2*(-7 + 5*Sqrt[2])]*Sqrt[1 + Tan[e + f*x]])])/(Sqrt[-7 + 5*Sqrt[ 2]]*f) - ((3 + 2*Sqrt[2])*ArcTanh[(3 + 2*Sqrt[2] + (1 + Sqrt[2])*Tan[e + f *x])/(Sqrt[2*(7 + 5*Sqrt[2])]*Sqrt[1 + Tan[e + f*x]])])/(2*Sqrt[7 + 5*Sqrt [2]]*f)) - (2*Sqrt[1 + Tan[e + f*x]])/f + (2*(1 + Tan[e + f*x])^(5/2))/(5* f)
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[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*((a + b*Tan[c + d*x])^(n - 1)/(d*(n - 1))), x] + Int[(a^2 - b^2 + 2*a*b*Tan[c + d *x])*(a + b*Tan[c + d*x])^(n - 2), x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 + b^2, 0] && GtQ[n, 1]
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_)])^2, x_Symbol] :> Simp[d^2*((a + b*Tan[e + f*x])^(m + 1)/(b*f*( m + 1))), x] + Int[(a + b*Tan[e + f*x])^m*Simp[c^2 - d^2 + 2*c*d*Tan[e + f* x], x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && !LeQ [m, -1] && !(EqQ[m, 2] && EqQ[a, 0])
Time = 1.02 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.35
| method | result | size |
| derivativedivides | \(\frac {\frac {2 \left (\tan \left (f x +e \right )+1\right )^{\frac {5}{2}}}{5}-2 \sqrt {\tan \left (f x +e \right )+1}+\frac {\sqrt {2}\, \left (\frac {\sqrt {2 \sqrt {2}+2}\, \ln \left (\tan \left (f x +e \right )+1+\sqrt {\tan \left (f x +e \right )+1}\, \sqrt {2 \sqrt {2}+2}+\sqrt {2}\right )}{2}+\frac {2 \left (1-\sqrt {2}\right ) \arctan \left (\frac {\sqrt {2 \sqrt {2}+2}+2 \sqrt {\tan \left (f x +e \right )+1}}{\sqrt {-2+2 \sqrt {2}}}\right )}{\sqrt {-2+2 \sqrt {2}}}\right )}{2}+\frac {\sqrt {2}\, \left (-\frac {\sqrt {2 \sqrt {2}+2}\, \ln \left (\tan \left (f x +e \right )+1-\sqrt {\tan \left (f x +e \right )+1}\, \sqrt {2 \sqrt {2}+2}+\sqrt {2}\right )}{2}+\frac {2 \left (1-\sqrt {2}\right ) \arctan \left (\frac {2 \sqrt {\tan \left (f x +e \right )+1}-\sqrt {2 \sqrt {2}+2}}{\sqrt {-2+2 \sqrt {2}}}\right )}{\sqrt {-2+2 \sqrt {2}}}\right )}{2}}{f}\) | \(233\) |
| default | \(\frac {\frac {2 \left (\tan \left (f x +e \right )+1\right )^{\frac {5}{2}}}{5}-2 \sqrt {\tan \left (f x +e \right )+1}+\frac {\sqrt {2}\, \left (\frac {\sqrt {2 \sqrt {2}+2}\, \ln \left (\tan \left (f x +e \right )+1+\sqrt {\tan \left (f x +e \right )+1}\, \sqrt {2 \sqrt {2}+2}+\sqrt {2}\right )}{2}+\frac {2 \left (1-\sqrt {2}\right ) \arctan \left (\frac {\sqrt {2 \sqrt {2}+2}+2 \sqrt {\tan \left (f x +e \right )+1}}{\sqrt {-2+2 \sqrt {2}}}\right )}{\sqrt {-2+2 \sqrt {2}}}\right )}{2}+\frac {\sqrt {2}\, \left (-\frac {\sqrt {2 \sqrt {2}+2}\, \ln \left (\tan \left (f x +e \right )+1-\sqrt {\tan \left (f x +e \right )+1}\, \sqrt {2 \sqrt {2}+2}+\sqrt {2}\right )}{2}+\frac {2 \left (1-\sqrt {2}\right ) \arctan \left (\frac {2 \sqrt {\tan \left (f x +e \right )+1}-\sqrt {2 \sqrt {2}+2}}{\sqrt {-2+2 \sqrt {2}}}\right )}{\sqrt {-2+2 \sqrt {2}}}\right )}{2}}{f}\) | \(233\) |
Input:
int(tan(f*x+e)^2*(tan(f*x+e)+1)^(3/2),x,method=_RETURNVERBOSE)
Output:
1/f*(2/5*(tan(f*x+e)+1)^(5/2)-2*(tan(f*x+e)+1)^(1/2)+1/2*2^(1/2)*(1/2*(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*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)))+1/2*2^(1/2)*(-1/2*(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*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))))
Leaf count of result is larger than twice the leaf count of optimal. 346 vs. \(2 (134) = 268\).
Time = 0.09 (sec) , antiderivative size = 346, normalized size of antiderivative = 2.00 \[ \int \tan ^2(e+f x) (1+\tan (e+f x))^{3/2} \, dx=-\frac {5 \, \sqrt {2} f \sqrt {\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} + 1}{f^{2}}} \log \left (\sqrt {2} {\left (f^{3} \sqrt {-\frac {1}{f^{4}}} - f\right )} \sqrt {\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} + 1}{f^{2}}} + 2 \, \sqrt {\tan \left (f x + e\right ) + 1}\right ) - 5 \, \sqrt {2} f \sqrt {\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} + 1}{f^{2}}} \log \left (-\sqrt {2} {\left (f^{3} \sqrt {-\frac {1}{f^{4}}} - f\right )} \sqrt {\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} + 1}{f^{2}}} + 2 \, \sqrt {\tan \left (f x + e\right ) + 1}\right ) - 5 \, \sqrt {2} f \sqrt {-\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} - 1}{f^{2}}} \log \left (\sqrt {2} {\left (f^{3} \sqrt {-\frac {1}{f^{4}}} + f\right )} \sqrt {-\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} - 1}{f^{2}}} + 2 \, \sqrt {\tan \left (f x + e\right ) + 1}\right ) + 5 \, \sqrt {2} f \sqrt {-\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} - 1}{f^{2}}} \log \left (-\sqrt {2} {\left (f^{3} \sqrt {-\frac {1}{f^{4}}} + f\right )} \sqrt {-\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} - 1}{f^{2}}} + 2 \, \sqrt {\tan \left (f x + e\right ) + 1}\right ) - 4 \, {\left (\tan \left (f x + e\right )^{2} + 2 \, \tan \left (f x + e\right ) - 4\right )} \sqrt {\tan \left (f x + e\right ) + 1}}{10 \, f} \] Input:
integrate(tan(f*x+e)^2*(1+tan(f*x+e))^(3/2),x, algorithm="fricas")
Output:
-1/10*(5*sqrt(2)*f*sqrt((f^2*sqrt(-1/f^4) + 1)/f^2)*log(sqrt(2)*(f^3*sqrt( -1/f^4) - f)*sqrt((f^2*sqrt(-1/f^4) + 1)/f^2) + 2*sqrt(tan(f*x + e) + 1)) - 5*sqrt(2)*f*sqrt((f^2*sqrt(-1/f^4) + 1)/f^2)*log(-sqrt(2)*(f^3*sqrt(-1/f ^4) - f)*sqrt((f^2*sqrt(-1/f^4) + 1)/f^2) + 2*sqrt(tan(f*x + e) + 1)) - 5* sqrt(2)*f*sqrt(-(f^2*sqrt(-1/f^4) - 1)/f^2)*log(sqrt(2)*(f^3*sqrt(-1/f^4) + f)*sqrt(-(f^2*sqrt(-1/f^4) - 1)/f^2) + 2*sqrt(tan(f*x + e) + 1)) + 5*sqr t(2)*f*sqrt(-(f^2*sqrt(-1/f^4) - 1)/f^2)*log(-sqrt(2)*(f^3*sqrt(-1/f^4) + f)*sqrt(-(f^2*sqrt(-1/f^4) - 1)/f^2) + 2*sqrt(tan(f*x + e) + 1)) - 4*(tan( f*x + e)^2 + 2*tan(f*x + e) - 4)*sqrt(tan(f*x + e) + 1))/f
\[ \int \tan ^2(e+f x) (1+\tan (e+f x))^{3/2} \, dx=\int \left (\tan {\left (e + f x \right )} + 1\right )^{\frac {3}{2}} \tan ^{2}{\left (e + f x \right )}\, dx \] Input:
integrate(tan(f*x+e)**2*(1+tan(f*x+e))**(3/2),x)
Output:
Integral((tan(e + f*x) + 1)**(3/2)*tan(e + f*x)**2, x)
\[ \int \tan ^2(e+f x) (1+\tan (e+f x))^{3/2} \, dx=\int { {\left (\tan \left (f x + e\right ) + 1\right )}^{\frac {3}{2}} \tan \left (f x + e\right )^{2} \,d x } \] Input:
integrate(tan(f*x+e)^2*(1+tan(f*x+e))^(3/2),x, algorithm="maxima")
Output:
integrate((tan(f*x + e) + 1)^(3/2)*tan(f*x + e)^2, x)
Exception generated. \[ \int \tan ^2(e+f x) (1+\tan (e+f x))^{3/2} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(tan(f*x+e)^2*(1+tan(f*x+e))^(3/2),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 = 0.96 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.60 \[ \int \tan ^2(e+f x) (1+\tan (e+f x))^{3/2} \, dx=\frac {2\,{\left (\mathrm {tan}\left (e+f\,x\right )+1\right )}^{5/2}}{5\,f}-\frac {2\,\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}}{f}-\mathrm {atan}\left (f\,\sqrt {\frac {\frac {1}{2}-\frac {1}{2}{}\mathrm {i}}{f^2}}\,\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}\,1{}\mathrm {i}\right )\,\sqrt {\frac {\frac {1}{2}-\frac {1}{2}{}\mathrm {i}}{f^2}}\,2{}\mathrm {i}-\mathrm {atan}\left (f\,\sqrt {\frac {\frac {1}{2}+\frac {1}{2}{}\mathrm {i}}{f^2}}\,\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}\,1{}\mathrm {i}\right )\,\sqrt {\frac {\frac {1}{2}+\frac {1}{2}{}\mathrm {i}}{f^2}}\,2{}\mathrm {i} \] Input:
int(tan(e + f*x)^2*(tan(e + f*x) + 1)^(3/2),x)
Output:
(2*(tan(e + f*x) + 1)^(5/2))/(5*f) - (2*(tan(e + f*x) + 1)^(1/2))/f - atan (f*((1/2 - 1i/2)/f^2)^(1/2)*(tan(e + f*x) + 1)^(1/2)*1i)*((1/2 - 1i/2)/f^2 )^(1/2)*2i - atan(f*((1/2 + 1i/2)/f^2)^(1/2)*(tan(e + f*x) + 1)^(1/2)*1i)* ((1/2 + 1i/2)/f^2)^(1/2)*2i
\[ \int \tan ^2(e+f x) (1+\tan (e+f x))^{3/2} \, dx=\int \sqrt {\tan \left (f x +e \right )+1}\, \tan \left (f x +e \right )^{3}d x +\int \sqrt {\tan \left (f x +e \right )+1}\, \tan \left (f x +e \right )^{2}d x \] Input:
int(tan(f*x+e)^2*(1+tan(f*x+e))^(3/2),x)
Output:
int(sqrt(tan(e + f*x) + 1)*tan(e + f*x)**3,x) + int(sqrt(tan(e + f*x) + 1) *tan(e + f*x)**2,x)