Integrand size = 21, antiderivative size = 257 \[ \int \tan ^4(e+f x) \sqrt {1+\tan (e+f x)} \, dx=-\frac {\sqrt {\frac {1}{2} \left (1+\sqrt {2}\right )} \arctan \left (\frac {\sqrt {2 \left (1+\sqrt {2}\right )}-2 \sqrt {1+\tan (e+f x)}}{\sqrt {2 \left (-1+\sqrt {2}\right )}}\right )}{f}+\frac {\sqrt {\frac {1}{2} \left (1+\sqrt {2}\right )} \arctan \left (\frac {\sqrt {2 \left (1+\sqrt {2}\right )}+2 \sqrt {1+\tan (e+f x)}}{\sqrt {2 \left (-1+\sqrt {2}\right )}}\right )}{f}-\frac {\text {arctanh}\left (\frac {\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {1+\tan (e+f x)}}{1+\sqrt {2}+\tan (e+f x)}\right )}{\sqrt {2 \left (1+\sqrt {2}\right )} f}-\frac {18 (1+\tan (e+f x))^{3/2}}{35 f}-\frac {8 \tan (e+f x) (1+\tan (e+f x))^{3/2}}{35 f}+\frac {2 \tan ^2(e+f x) (1+\tan (e+f x))^{3/2}}{7 f} \] Output:
-1/2*(2+2*2^(1/2))^(1/2)*arctan(((2+2*2^(1/2))^(1/2)-2*(1+tan(f*x+e))^(1/2 ))/(-2+2*2^(1/2))^(1/2))/f+1/2*(2+2*2^(1/2))^(1/2)*arctan(((2+2*2^(1/2))^( 1/2)+2*(1+tan(f*x+e))^(1/2))/(-2+2*2^(1/2))^(1/2))/f-arctanh((2+2*2^(1/2)) ^(1/2)*(1+tan(f*x+e))^(1/2)/(1+2^(1/2)+tan(f*x+e)))/(2+2*2^(1/2))^(1/2)/f- 18/35*(1+tan(f*x+e))^(3/2)/f-8/35*tan(f*x+e)*(1+tan(f*x+e))^(3/2)/f+2/7*ta n(f*x+e)^2*(1+tan(f*x+e))^(3/2)/f
Result contains complex when optimal does not.
Time = 0.74 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.46 \[ \int \tan ^4(e+f x) \sqrt {1+\tan (e+f x)} \, dx=\frac {-35 i \sqrt {1-i} \text {arctanh}\left (\frac {\sqrt {1+\tan (e+f x)}}{\sqrt {1-i}}\right )+35 i \sqrt {1+i} \text {arctanh}\left (\frac {\sqrt {1+\tan (e+f x)}}{\sqrt {1+i}}\right )+2 \sqrt {1+\tan (e+f x)} \left (\sec ^2(e+f x) (1+5 \tan (e+f x))-2 (5+9 \tan (e+f x))\right )}{35 f} \] Input:
Integrate[Tan[e + f*x]^4*Sqrt[1 + Tan[e + f*x]],x]
Output:
((-35*I)*Sqrt[1 - I]*ArcTanh[Sqrt[1 + Tan[e + f*x]]/Sqrt[1 - I]] + (35*I)* Sqrt[1 + I]*ArcTanh[Sqrt[1 + Tan[e + f*x]]/Sqrt[1 + I]] + 2*Sqrt[1 + Tan[e + f*x]]*(Sec[e + f*x]^2*(1 + 5*Tan[e + f*x]) - 2*(5 + 9*Tan[e + f*x])))/( 35*f)
Time = 0.87 (sec) , antiderivative size = 327, normalized size of antiderivative = 1.27, number of steps used = 19, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.857, Rules used = {3042, 4049, 27, 3042, 4130, 27, 3042, 4114, 3042, 3966, 483, 1447, 1475, 1083, 217, 1478, 25, 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.
| \(\displaystyle \int \tan ^4(e+f x) \sqrt {\tan (e+f x)+1} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \tan (e+f x)^4 \sqrt {\tan (e+f x)+1}dx\) |
| \(\Big \downarrow \) 4049 |
| \(\displaystyle \frac {2}{7} \int -\frac {1}{2} \tan (e+f x) \sqrt {\tan (e+f x)+1} \left (4 \tan ^2(e+f x)+7 \tan (e+f x)+4\right )dx+\frac {2 (\tan (e+f x)+1)^{3/2} \tan ^2(e+f x)}{7 f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2 \tan ^2(e+f x) (\tan (e+f x)+1)^{3/2}}{7 f}-\frac {1}{7} \int \tan (e+f x) \sqrt {\tan (e+f x)+1} \left (4 \tan ^2(e+f x)+7 \tan (e+f x)+4\right )dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 \tan ^2(e+f x) (\tan (e+f x)+1)^{3/2}}{7 f}-\frac {1}{7} \int \tan (e+f x) \sqrt {\tan (e+f x)+1} \left (4 \tan (e+f x)^2+7 \tan (e+f x)+4\right )dx\) |
| \(\Big \downarrow \) 4130 |
| \(\displaystyle \frac {1}{7} \left (-\frac {2}{5} \int -\frac {1}{2} \sqrt {\tan (e+f x)+1} \left (8-27 \tan ^2(e+f x)\right )dx-\frac {8 \tan (e+f x) (\tan (e+f x)+1)^{3/2}}{5 f}\right )+\frac {2 (\tan (e+f x)+1)^{3/2} \tan ^2(e+f x)}{7 f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{7} \left (\frac {1}{5} \int \sqrt {\tan (e+f x)+1} \left (8-27 \tan ^2(e+f x)\right )dx-\frac {8 \tan (e+f x) (\tan (e+f x)+1)^{3/2}}{5 f}\right )+\frac {2 (\tan (e+f x)+1)^{3/2} \tan ^2(e+f x)}{7 f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{7} \left (\frac {1}{5} \int \sqrt {\tan (e+f x)+1} \left (8-27 \tan (e+f x)^2\right )dx-\frac {8 \tan (e+f x) (\tan (e+f x)+1)^{3/2}}{5 f}\right )+\frac {2 (\tan (e+f x)+1)^{3/2} \tan ^2(e+f x)}{7 f}\) |
| \(\Big \downarrow \) 4114 |
| \(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (35 \int \sqrt {\tan (e+f x)+1}dx-\frac {18 (\tan (e+f x)+1)^{3/2}}{f}\right )-\frac {8 \tan (e+f x) (\tan (e+f x)+1)^{3/2}}{5 f}\right )+\frac {2 (\tan (e+f x)+1)^{3/2} \tan ^2(e+f x)}{7 f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (35 \int \sqrt {\tan (e+f x)+1}dx-\frac {18 (\tan (e+f x)+1)^{3/2}}{f}\right )-\frac {8 \tan (e+f x) (\tan (e+f x)+1)^{3/2}}{5 f}\right )+\frac {2 (\tan (e+f x)+1)^{3/2} \tan ^2(e+f x)}{7 f}\) |
| \(\Big \downarrow \) 3966 |
| \(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (\frac {35 \int \frac {\sqrt {\tan (e+f x)+1}}{\tan ^2(e+f x)+1}d\tan (e+f x)}{f}-\frac {18 (\tan (e+f x)+1)^{3/2}}{f}\right )-\frac {8 \tan (e+f x) (\tan (e+f x)+1)^{3/2}}{5 f}\right )+\frac {2 (\tan (e+f x)+1)^{3/2} \tan ^2(e+f x)}{7 f}\) |
| \(\Big \downarrow \) 483 |
| \(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (\frac {70 \int \frac {\tan (e+f x)+1}{(\tan (e+f x)+1)^2-2 (\tan (e+f x)+1)+2}d\sqrt {\tan (e+f x)+1}}{f}-\frac {18 (\tan (e+f x)+1)^{3/2}}{f}\right )-\frac {8 \tan (e+f x) (\tan (e+f x)+1)^{3/2}}{5 f}\right )+\frac {2 (\tan (e+f x)+1)^{3/2} \tan ^2(e+f x)}{7 f}\) |
| \(\Big \downarrow \) 1447 |
| \(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (\frac {70 \left (\frac {1}{2} \int \frac {\tan (e+f x)+\sqrt {2}+1}{(\tan (e+f x)+1)^2-2 (\tan (e+f x)+1)+2}d\sqrt {\tan (e+f x)+1}-\frac {1}{2} \int \frac {-\tan (e+f x)+\sqrt {2}-1}{(\tan (e+f x)+1)^2-2 (\tan (e+f x)+1)+2}d\sqrt {\tan (e+f x)+1}\right )}{f}-\frac {18 (\tan (e+f x)+1)^{3/2}}{f}\right )-\frac {8 \tan (e+f x) (\tan (e+f x)+1)^{3/2}}{5 f}\right )+\frac {2 (\tan (e+f x)+1)^{3/2} \tan ^2(e+f x)}{7 f}\) |
| \(\Big \downarrow \) 1475 |
| \(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (\frac {70 \left (\frac {1}{2} \left (\frac {1}{2} \int \frac {1}{\tan (e+f x)-\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {\tan (e+f x)+1}+\sqrt {2}+1}d\sqrt {\tan (e+f x)+1}+\frac {1}{2} \int \frac {1}{\tan (e+f x)+\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {\tan (e+f x)+1}+\sqrt {2}+1}d\sqrt {\tan (e+f x)+1}\right )-\frac {1}{2} \int \frac {-\tan (e+f x)+\sqrt {2}-1}{(\tan (e+f x)+1)^2-2 (\tan (e+f x)+1)+2}d\sqrt {\tan (e+f x)+1}\right )}{f}-\frac {18 (\tan (e+f x)+1)^{3/2}}{f}\right )-\frac {8 \tan (e+f x) (\tan (e+f x)+1)^{3/2}}{5 f}\right )+\frac {2 (\tan (e+f x)+1)^{3/2} \tan ^2(e+f x)}{7 f}\) |
| \(\Big \downarrow \) 1083 |
| \(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (\frac {70 \left (\frac {1}{2} \left (-\int \frac {1}{-\tan (e+f x)+2 \left (1-\sqrt {2}\right )-1}d\left (2 \sqrt {\tan (e+f x)+1}-\sqrt {2 \left (1+\sqrt {2}\right )}\right )-\int \frac {1}{-\tan (e+f x)+2 \left (1-\sqrt {2}\right )-1}d\left (2 \sqrt {\tan (e+f x)+1}+\sqrt {2 \left (1+\sqrt {2}\right )}\right )\right )-\frac {1}{2} \int \frac {-\tan (e+f x)+\sqrt {2}-1}{(\tan (e+f x)+1)^2-2 (\tan (e+f x)+1)+2}d\sqrt {\tan (e+f x)+1}\right )}{f}-\frac {18 (\tan (e+f x)+1)^{3/2}}{f}\right )-\frac {8 \tan (e+f x) (\tan (e+f x)+1)^{3/2}}{5 f}\right )+\frac {2 (\tan (e+f x)+1)^{3/2} \tan ^2(e+f x)}{7 f}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (\frac {70 \left (\frac {1}{2} \left (\frac {\arctan \left (\frac {2 \sqrt {\tan (e+f x)+1}-\sqrt {2 \left (1+\sqrt {2}\right )}}{\sqrt {2 \left (\sqrt {2}-1\right )}}\right )}{\sqrt {2 \left (\sqrt {2}-1\right )}}+\frac {\arctan \left (\frac {2 \sqrt {\tan (e+f x)+1}+\sqrt {2 \left (1+\sqrt {2}\right )}}{\sqrt {2 \left (\sqrt {2}-1\right )}}\right )}{\sqrt {2 \left (\sqrt {2}-1\right )}}\right )-\frac {1}{2} \int \frac {-\tan (e+f x)+\sqrt {2}-1}{(\tan (e+f x)+1)^2-2 (\tan (e+f x)+1)+2}d\sqrt {\tan (e+f x)+1}\right )}{f}-\frac {18 (\tan (e+f x)+1)^{3/2}}{f}\right )-\frac {8 \tan (e+f x) (\tan (e+f x)+1)^{3/2}}{5 f}\right )+\frac {2 (\tan (e+f x)+1)^{3/2} \tan ^2(e+f x)}{7 f}\) |
| \(\Big \downarrow \) 1478 |
| \(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (\frac {70 \left (\frac {1}{2} \left (\frac {\int -\frac {\sqrt {2 \left (1+\sqrt {2}\right )}-2 \sqrt {\tan (e+f x)+1}}{\tan (e+f x)-\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {\tan (e+f x)+1}+\sqrt {2}+1}d\sqrt {\tan (e+f x)+1}}{2 \sqrt {2 \left (1+\sqrt {2}\right )}}+\frac {\int -\frac {2 \sqrt {\tan (e+f x)+1}+\sqrt {2 \left (1+\sqrt {2}\right )}}{\tan (e+f x)+\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {\tan (e+f x)+1}+\sqrt {2}+1}d\sqrt {\tan (e+f x)+1}}{2 \sqrt {2 \left (1+\sqrt {2}\right )}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\frac {2 \sqrt {\tan (e+f x)+1}-\sqrt {2 \left (1+\sqrt {2}\right )}}{\sqrt {2 \left (\sqrt {2}-1\right )}}\right )}{\sqrt {2 \left (\sqrt {2}-1\right )}}+\frac {\arctan \left (\frac {2 \sqrt {\tan (e+f x)+1}+\sqrt {2 \left (1+\sqrt {2}\right )}}{\sqrt {2 \left (\sqrt {2}-1\right )}}\right )}{\sqrt {2 \left (\sqrt {2}-1\right )}}\right )\right )}{f}-\frac {18 (\tan (e+f x)+1)^{3/2}}{f}\right )-\frac {8 \tan (e+f x) (\tan (e+f x)+1)^{3/2}}{5 f}\right )+\frac {2 (\tan (e+f x)+1)^{3/2} \tan ^2(e+f x)}{7 f}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (\frac {70 \left (\frac {1}{2} \left (-\frac {\int \frac {\sqrt {2 \left (1+\sqrt {2}\right )}-2 \sqrt {\tan (e+f x)+1}}{\tan (e+f x)-\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {\tan (e+f x)+1}+\sqrt {2}+1}d\sqrt {\tan (e+f x)+1}}{2 \sqrt {2 \left (1+\sqrt {2}\right )}}-\frac {\int \frac {2 \sqrt {\tan (e+f x)+1}+\sqrt {2 \left (1+\sqrt {2}\right )}}{\tan (e+f x)+\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {\tan (e+f x)+1}+\sqrt {2}+1}d\sqrt {\tan (e+f x)+1}}{2 \sqrt {2 \left (1+\sqrt {2}\right )}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\frac {2 \sqrt {\tan (e+f x)+1}-\sqrt {2 \left (1+\sqrt {2}\right )}}{\sqrt {2 \left (\sqrt {2}-1\right )}}\right )}{\sqrt {2 \left (\sqrt {2}-1\right )}}+\frac {\arctan \left (\frac {2 \sqrt {\tan (e+f x)+1}+\sqrt {2 \left (1+\sqrt {2}\right )}}{\sqrt {2 \left (\sqrt {2}-1\right )}}\right )}{\sqrt {2 \left (\sqrt {2}-1\right )}}\right )\right )}{f}-\frac {18 (\tan (e+f x)+1)^{3/2}}{f}\right )-\frac {8 \tan (e+f x) (\tan (e+f x)+1)^{3/2}}{5 f}\right )+\frac {2 (\tan (e+f x)+1)^{3/2} \tan ^2(e+f x)}{7 f}\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (\frac {70 \left (\frac {1}{2} \left (\frac {\arctan \left (\frac {2 \sqrt {\tan (e+f x)+1}-\sqrt {2 \left (1+\sqrt {2}\right )}}{\sqrt {2 \left (\sqrt {2}-1\right )}}\right )}{\sqrt {2 \left (\sqrt {2}-1\right )}}+\frac {\arctan \left (\frac {2 \sqrt {\tan (e+f x)+1}+\sqrt {2 \left (1+\sqrt {2}\right )}}{\sqrt {2 \left (\sqrt {2}-1\right )}}\right )}{\sqrt {2 \left (\sqrt {2}-1\right )}}\right )+\frac {1}{2} \left (\frac {\log \left (\tan (e+f x)-\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {\tan (e+f x)+1}+\sqrt {2}+1\right )}{2 \sqrt {2 \left (1+\sqrt {2}\right )}}-\frac {\log \left (\tan (e+f x)+\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {\tan (e+f x)+1}+\sqrt {2}+1\right )}{2 \sqrt {2 \left (1+\sqrt {2}\right )}}\right )\right )}{f}-\frac {18 (\tan (e+f x)+1)^{3/2}}{f}\right )-\frac {8 \tan (e+f x) (\tan (e+f x)+1)^{3/2}}{5 f}\right )+\frac {2 (\tan (e+f x)+1)^{3/2} \tan ^2(e+f x)}{7 f}\) |
Input:
Int[Tan[e + f*x]^4*Sqrt[1 + Tan[e + f*x]],x]
Output:
(2*Tan[e + f*x]^2*(1 + Tan[e + f*x])^(3/2))/(7*f) + ((-8*Tan[e + f*x]*(1 + Tan[e + f*x])^(3/2))/(5*f) + ((70*((ArcTan[(-Sqrt[2*(1 + Sqrt[2])] + 2*Sq rt[1 + Tan[e + f*x]])/Sqrt[2*(-1 + Sqrt[2])]]/Sqrt[2*(-1 + Sqrt[2])] + Arc Tan[(Sqrt[2*(1 + Sqrt[2])] + 2*Sqrt[1 + Tan[e + f*x]])/Sqrt[2*(-1 + Sqrt[2 ])]]/Sqrt[2*(-1 + Sqrt[2])])/2 + (Log[1 + Sqrt[2] + Tan[e + f*x] - Sqrt[2* (1 + Sqrt[2])]*Sqrt[1 + Tan[e + f*x]]]/(2*Sqrt[2*(1 + Sqrt[2])]) - Log[1 + Sqrt[2] + Tan[e + f*x] + Sqrt[2*(1 + Sqrt[2])]*Sqrt[1 + Tan[e + f*x]]]/(2 *Sqrt[2*(1 + Sqrt[2])]))/2))/f - (18*(1 + Tan[e + f*x])^(3/2))/f)/5)/7
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[Sqrt[(c_) + (d_.)*(x_)]/((a_) + (b_.)*(x_)^2), x_Symbol] :> Simp[2*d Subst[Int[x^2/(b*c^2 + a*d^2 - 2*b*c*x^2 + b*x^4), x], x, Sqrt[c + d*x]], x ] /; FreeQ[{a, b, c, d}, x]
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2 Subst[I nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{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[(x_)^2/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ a/c, 2]}, Simp[1/2 Int[(q + x^2)/(a + b*x^2 + c*x^4), x], x] - Simp[1/2 Int[(q - x^2)/(a + b*x^2 + c*x^4), x], x]] /; FreeQ[{a, b, c}, x] && LtQ[b ^2 - 4*a*c, 0] && PosQ[a*c]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] : > With[{q = Rt[2*(d/e) - b/c, 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]] /; F reeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - a*e^2, 0] && (GtQ[2*(d/e) - b/c, 0] || ( !LtQ[2*(d/e) - b/c, 0] && EqQ[d - e*Rt[a/c, 2] , 0]))
Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] : > With[{q = Rt[-2*(d/e) - b/c, 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]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ [c*d^2 - a*e^2, 0] && !GtQ[b^2 - 4*a*c, 0]
Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b/d Su bst[Int[(a + x)^n/(b^2 + x^2), x], x, b*Tan[c + d*x]], x] /; FreeQ[{a, b, c , d, n}, x] && NeQ[a^2 + b^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_.) + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[C*((a + b*Tan[e + f*x])^(m + 1)/(b*f*(m + 1))), x] + Simp[(A - C) Int[(a + b*Tan[e + f*x])^m, x], x] /; FreeQ[{a , b, e, f, A, C, m}, x] && NeQ[A*b^2 + a^2*C, 0] && !LeQ[m, -1]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_. ) + (f_.)*(x_)]^2), x_Symbol] :> Simp[C*(a + b*Tan[e + f*x])^m*((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 - 1)*(c + d*Tan[e + f*x])^n*Simp[a*A*d*(m + n + 1) - C *(b*c*m + a*d*(n + 1)) + d*(A*b + a*B - b*C)*(m + n + 1)*Tan[e + f*x] - (C* m*(b*c - a*d) - b*B*d*(m + n + 1))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 0] && !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[ c, 0] && NeQ[a, 0])))
Time = 1.13 (sec) , antiderivative size = 318, normalized size of antiderivative = 1.24
| method | result | size |
| derivativedivides | \(\frac {2 \left (\tan \left (f x +e \right )+1\right )^{\frac {7}{2}}}{7 f}-\frac {4 \left (\tan \left (f x +e \right )+1\right )^{\frac {5}{2}}}{5 f}+\frac {\sqrt {2 \sqrt {2}+2}\, \sqrt {2}\, \ln \left (\tan \left (f x +e \right )+1-\sqrt {\tan \left (f x +e \right )+1}\, \sqrt {2 \sqrt {2}+2}+\sqrt {2}\right )}{4 f}+\frac {\arctan \left (\frac {2 \sqrt {\tan \left (f x +e \right )+1}-\sqrt {2 \sqrt {2}+2}}{\sqrt {-2+2 \sqrt {2}}}\right )}{f \sqrt {-2+2 \sqrt {2}}}-\frac {\sqrt {2 \sqrt {2}+2}\, \ln \left (\tan \left (f x +e \right )+1-\sqrt {\tan \left (f x +e \right )+1}\, \sqrt {2 \sqrt {2}+2}+\sqrt {2}\right )}{4 f}-\frac {\sqrt {2 \sqrt {2}+2}\, \sqrt {2}\, \ln \left (\tan \left (f x +e \right )+1+\sqrt {\tan \left (f x +e \right )+1}\, \sqrt {2 \sqrt {2}+2}+\sqrt {2}\right )}{4 f}+\frac {\arctan \left (\frac {\sqrt {2 \sqrt {2}+2}+2 \sqrt {\tan \left (f x +e \right )+1}}{\sqrt {-2+2 \sqrt {2}}}\right )}{f \sqrt {-2+2 \sqrt {2}}}+\frac {\sqrt {2 \sqrt {2}+2}\, \ln \left (\tan \left (f x +e \right )+1+\sqrt {\tan \left (f x +e \right )+1}\, \sqrt {2 \sqrt {2}+2}+\sqrt {2}\right )}{4 f}\) | \(318\) |
| default | \(\frac {2 \left (\tan \left (f x +e \right )+1\right )^{\frac {7}{2}}}{7 f}-\frac {4 \left (\tan \left (f x +e \right )+1\right )^{\frac {5}{2}}}{5 f}+\frac {\sqrt {2 \sqrt {2}+2}\, \sqrt {2}\, \ln \left (\tan \left (f x +e \right )+1-\sqrt {\tan \left (f x +e \right )+1}\, \sqrt {2 \sqrt {2}+2}+\sqrt {2}\right )}{4 f}+\frac {\arctan \left (\frac {2 \sqrt {\tan \left (f x +e \right )+1}-\sqrt {2 \sqrt {2}+2}}{\sqrt {-2+2 \sqrt {2}}}\right )}{f \sqrt {-2+2 \sqrt {2}}}-\frac {\sqrt {2 \sqrt {2}+2}\, \ln \left (\tan \left (f x +e \right )+1-\sqrt {\tan \left (f x +e \right )+1}\, \sqrt {2 \sqrt {2}+2}+\sqrt {2}\right )}{4 f}-\frac {\sqrt {2 \sqrt {2}+2}\, \sqrt {2}\, \ln \left (\tan \left (f x +e \right )+1+\sqrt {\tan \left (f x +e \right )+1}\, \sqrt {2 \sqrt {2}+2}+\sqrt {2}\right )}{4 f}+\frac {\arctan \left (\frac {\sqrt {2 \sqrt {2}+2}+2 \sqrt {\tan \left (f x +e \right )+1}}{\sqrt {-2+2 \sqrt {2}}}\right )}{f \sqrt {-2+2 \sqrt {2}}}+\frac {\sqrt {2 \sqrt {2}+2}\, \ln \left (\tan \left (f x +e \right )+1+\sqrt {\tan \left (f x +e \right )+1}\, \sqrt {2 \sqrt {2}+2}+\sqrt {2}\right )}{4 f}\) | \(318\) |
Input:
int(tan(f*x+e)^4*(tan(f*x+e)+1)^(1/2),x,method=_RETURNVERBOSE)
Output:
2/7/f*(tan(f*x+e)+1)^(7/2)-4/5*(tan(f*x+e)+1)^(5/2)/f+1/4/f*(2*2^(1/2)+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/f/(-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))-1/4/f*(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))-1/4/f*(2*2^(1/2)+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/f/(-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/f*(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))
Time = 0.09 (sec) , antiderivative size = 308, normalized size of antiderivative = 1.20 \[ \int \tan ^4(e+f x) \sqrt {1+\tan (e+f x)} \, dx=-\frac {35 \, f \sqrt {-\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} + 1}{f^{2}}} \log \left (f^{3} \sqrt {-\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} + 1}{f^{2}}} \sqrt {-\frac {1}{f^{4}}} + \sqrt {\tan \left (f x + e\right ) + 1}\right ) - 35 \, f \sqrt {-\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} + 1}{f^{2}}} \log \left (-f^{3} \sqrt {-\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} + 1}{f^{2}}} \sqrt {-\frac {1}{f^{4}}} + \sqrt {\tan \left (f x + e\right ) + 1}\right ) - 35 \, f \sqrt {\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} - 1}{f^{2}}} \log \left (f^{3} \sqrt {\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} - 1}{f^{2}}} \sqrt {-\frac {1}{f^{4}}} + \sqrt {\tan \left (f x + e\right ) + 1}\right ) + 35 \, f \sqrt {\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} - 1}{f^{2}}} \log \left (-f^{3} \sqrt {\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} - 1}{f^{2}}} \sqrt {-\frac {1}{f^{4}}} + \sqrt {\tan \left (f x + e\right ) + 1}\right ) - 4 \, {\left (5 \, \tan \left (f x + e\right )^{3} + \tan \left (f x + e\right )^{2} - 13 \, \tan \left (f x + e\right ) - 9\right )} \sqrt {\tan \left (f x + e\right ) + 1}}{70 \, f} \] Input:
integrate(tan(f*x+e)^4*(1+tan(f*x+e))^(1/2),x, algorithm="fricas")
Output:
-1/70*(35*f*sqrt(-(f^2*sqrt(-1/f^4) + 1)/f^2)*log(f^3*sqrt(-(f^2*sqrt(-1/f ^4) + 1)/f^2)*sqrt(-1/f^4) + sqrt(tan(f*x + e) + 1)) - 35*f*sqrt(-(f^2*sqr t(-1/f^4) + 1)/f^2)*log(-f^3*sqrt(-(f^2*sqrt(-1/f^4) + 1)/f^2)*sqrt(-1/f^4 ) + sqrt(tan(f*x + e) + 1)) - 35*f*sqrt((f^2*sqrt(-1/f^4) - 1)/f^2)*log(f^ 3*sqrt((f^2*sqrt(-1/f^4) - 1)/f^2)*sqrt(-1/f^4) + sqrt(tan(f*x + e) + 1)) + 35*f*sqrt((f^2*sqrt(-1/f^4) - 1)/f^2)*log(-f^3*sqrt((f^2*sqrt(-1/f^4) - 1)/f^2)*sqrt(-1/f^4) + sqrt(tan(f*x + e) + 1)) - 4*(5*tan(f*x + e)^3 + tan (f*x + e)^2 - 13*tan(f*x + e) - 9)*sqrt(tan(f*x + e) + 1))/f
\[ \int \tan ^4(e+f x) \sqrt {1+\tan (e+f x)} \, dx=\int \sqrt {\tan {\left (e + f x \right )} + 1} \tan ^{4}{\left (e + f x \right )}\, dx \] Input:
integrate(tan(f*x+e)**4*(1+tan(f*x+e))**(1/2),x)
Output:
Integral(sqrt(tan(e + f*x) + 1)*tan(e + f*x)**4, x)
\[ \int \tan ^4(e+f x) \sqrt {1+\tan (e+f x)} \, dx=\int { \sqrt {\tan \left (f x + e\right ) + 1} \tan \left (f x + e\right )^{4} \,d x } \] Input:
integrate(tan(f*x+e)^4*(1+tan(f*x+e))^(1/2),x, algorithm="maxima")
Output:
integrate(sqrt(tan(f*x + e) + 1)*tan(f*x + e)^4, x)
Exception generated. \[ \int \tan ^4(e+f x) \sqrt {1+\tan (e+f x)} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(tan(f*x+e)^4*(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,[24]%%%}+%%%{10,[22]%%%}+%%%{45,[20]%%%}+%%%{120,[18 ]%%%}+%%%
Time = 2.40 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.42 \[ \int \tan ^4(e+f x) \sqrt {1+\tan (e+f x)} \, dx=\frac {2\,{\left (\mathrm {tan}\left (e+f\,x\right )+1\right )}^{7/2}}{7\,f}-\frac {4\,{\left (\mathrm {tan}\left (e+f\,x\right )+1\right )}^{5/2}}{5\,f}-\mathrm {atan}\left (f^3\,{\left (\frac {-\frac {1}{4}-\frac {1}{4}{}\mathrm {i}}{f^2}\right )}^{3/2}\,\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}\,4{}\mathrm {i}\right )\,\sqrt {\frac {-\frac {1}{4}-\frac {1}{4}{}\mathrm {i}}{f^2}}\,2{}\mathrm {i}-\mathrm {atan}\left (f^3\,{\left (\frac {-\frac {1}{4}+\frac {1}{4}{}\mathrm {i}}{f^2}\right )}^{3/2}\,\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}\,4{}\mathrm {i}\right )\,\sqrt {\frac {-\frac {1}{4}+\frac {1}{4}{}\mathrm {i}}{f^2}}\,2{}\mathrm {i} \] Input:
int(tan(e + f*x)^4*(tan(e + f*x) + 1)^(1/2),x)
Output:
(2*(tan(e + f*x) + 1)^(7/2))/(7*f) - atan(f^3*((- 1/4 + 1i/4)/f^2)^(3/2)*( tan(e + f*x) + 1)^(1/2)*4i)*((- 1/4 + 1i/4)/f^2)^(1/2)*2i - (4*(tan(e + f* x) + 1)^(5/2))/(5*f) - atan(f^3*((- 1/4 - 1i/4)/f^2)^(3/2)*(tan(e + f*x) + 1)^(1/2)*4i)*((- 1/4 - 1i/4)/f^2)^(1/2)*2i
\[ \int \tan ^4(e+f x) \sqrt {1+\tan (e+f x)} \, dx=\int \sqrt {\tan \left (f x +e \right )+1}\, \tan \left (f x +e \right )^{4}d x \] Input:
int(tan(f*x+e)^4*(1+tan(f*x+e))^(1/2),x)
Output:
int(sqrt(tan(e + f*x) + 1)*tan(e + f*x)**4,x)