Integrand size = 21, antiderivative size = 255 \[ \int \tan ^3(e+f x) (1+\tan (e+f x))^{3/2} \, dx=-\frac {\sqrt {1+\sqrt {2}} \arctan \left (\frac {\sqrt {2 \left (1+\sqrt {2}\right )}-2 \sqrt {1+\tan (e+f x)}}{\sqrt {2 \left (-1+\sqrt {2}\right )}}\right )}{f}+\frac {\sqrt {1+\sqrt {2}} \arctan \left (\frac {\sqrt {2 \left (1+\sqrt {2}\right )}+2 \sqrt {1+\tan (e+f x)}}{\sqrt {2 \left (-1+\sqrt {2}\right )}}\right )}{f}+\frac {\text {arctanh}\left (\frac {\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {1+\tan (e+f x)}}{1+\sqrt {2}+\tan (e+f x)}\right )}{\sqrt {1+\sqrt {2}} f}-\frac {2 \sqrt {1+\tan (e+f x)}}{f}-\frac {2 (1+\tan (e+f x))^{3/2}}{3 f}-\frac {4 (1+\tan (e+f x))^{5/2}}{35 f}+\frac {2 \tan (e+f x) (1+\tan (e+f x))^{5/2}}{7 f} \] Output:
-(1+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^(1/2))^(1/2)*arctan(((2+2*2^(1/2))^(1/2)+2*(1+ta n(f*x+e))^(1/2))/(-2+2*2^(1/2))^(1/2))/f+arctanh((2+2*2^(1/2))^(1/2)*(1+ta n(f*x+e))^(1/2)/(1+2^(1/2)+tan(f*x+e)))/(1+2^(1/2))^(1/2)/f-2*(1+tan(f*x+e ))^(1/2)/f-2/3*(1+tan(f*x+e))^(3/2)/f-4/35*(1+tan(f*x+e))^(5/2)/f+2/7*tan( f*x+e)*(1+tan(f*x+e))^(5/2)/f
Result contains complex when optimal does not.
Time = 0.39 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.44 \[ \int \tan ^3(e+f x) (1+\tan (e+f x))^{3/2} \, dx=\frac {105 (1-i)^{3/2} \text {arctanh}\left (\frac {\sqrt {1+\tan (e+f x)}}{\sqrt {1-i}}\right )+105 (1+i)^{3/2} \text {arctanh}\left (\frac {\sqrt {1+\tan (e+f x)}}{\sqrt {1+i}}\right )+2 \sqrt {1+\tan (e+f x)} \left (-146-32 \tan (e+f x)+24 \tan ^2(e+f x)+15 \tan ^3(e+f x)\right )}{105 f} \] Input:
Integrate[Tan[e + f*x]^3*(1 + Tan[e + f*x])^(3/2),x]
Output:
(105*(1 - I)^(3/2)*ArcTanh[Sqrt[1 + Tan[e + f*x]]/Sqrt[1 - I]] + 105*(1 + I)^(3/2)*ArcTanh[Sqrt[1 + Tan[e + f*x]]/Sqrt[1 + I]] + 2*Sqrt[1 + Tan[e + f*x]]*(-146 - 32*Tan[e + f*x] + 24*Tan[e + f*x]^2 + 15*Tan[e + f*x]^3))/(1 05*f)
Time = 0.89 (sec) , antiderivative size = 348, normalized size of antiderivative = 1.36, number of steps used = 21, number of rules used = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.952, Rules used = {3042, 4049, 27, 3042, 4113, 27, 3042, 4011, 3042, 4011, 27, 3042, 3966, 484, 1407, 1142, 25, 1083, 217, 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.
\(\displaystyle \int \tan ^3(e+f x) (\tan (e+f x)+1)^{3/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \tan (e+f x)^3 (\tan (e+f x)+1)^{3/2}dx\) |
\(\Big \downarrow \) 4049 |
\(\displaystyle \frac {2}{7} \int -\frac {1}{2} (\tan (e+f x)+1)^{3/2} \left (2 \tan ^2(e+f x)+7 \tan (e+f x)+2\right )dx+\frac {2 \tan (e+f x) (\tan (e+f x)+1)^{5/2}}{7 f}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2 \tan (e+f x) (\tan (e+f x)+1)^{5/2}}{7 f}-\frac {1}{7} \int (\tan (e+f x)+1)^{3/2} \left (2 \tan ^2(e+f x)+7 \tan (e+f x)+2\right )dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 \tan (e+f x) (\tan (e+f x)+1)^{5/2}}{7 f}-\frac {1}{7} \int (\tan (e+f x)+1)^{3/2} \left (2 \tan (e+f x)^2+7 \tan (e+f x)+2\right )dx\) |
\(\Big \downarrow \) 4113 |
\(\displaystyle \frac {1}{7} \left (-\int 7 \tan (e+f x) (\tan (e+f x)+1)^{3/2}dx-\frac {4 (\tan (e+f x)+1)^{5/2}}{5 f}\right )+\frac {2 \tan (e+f x) (\tan (e+f x)+1)^{5/2}}{7 f}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{7} \left (-7 \int \tan (e+f x) (\tan (e+f x)+1)^{3/2}dx-\frac {4 (\tan (e+f x)+1)^{5/2}}{5 f}\right )+\frac {2 \tan (e+f x) (\tan (e+f x)+1)^{5/2}}{7 f}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{7} \left (-7 \int \tan (e+f x) (\tan (e+f x)+1)^{3/2}dx-\frac {4 (\tan (e+f x)+1)^{5/2}}{5 f}\right )+\frac {2 \tan (e+f x) (\tan (e+f x)+1)^{5/2}}{7 f}\) |
\(\Big \downarrow \) 4011 |
\(\displaystyle \frac {1}{7} \left (-7 \left (\int (\tan (e+f x)-1) \sqrt {\tan (e+f x)+1}dx+\frac {2 (\tan (e+f x)+1)^{3/2}}{3 f}\right )-\frac {4 (\tan (e+f x)+1)^{5/2}}{5 f}\right )+\frac {2 \tan (e+f x) (\tan (e+f x)+1)^{5/2}}{7 f}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{7} \left (-7 \left (\int (\tan (e+f x)-1) \sqrt {\tan (e+f x)+1}dx+\frac {2 (\tan (e+f x)+1)^{3/2}}{3 f}\right )-\frac {4 (\tan (e+f x)+1)^{5/2}}{5 f}\right )+\frac {2 \tan (e+f x) (\tan (e+f x)+1)^{5/2}}{7 f}\) |
\(\Big \downarrow \) 4011 |
\(\displaystyle \frac {1}{7} \left (-7 \left (\int -\frac {2}{\sqrt {\tan (e+f x)+1}}dx+\frac {2 (\tan (e+f x)+1)^{3/2}}{3 f}+\frac {2 \sqrt {\tan (e+f x)+1}}{f}\right )-\frac {4 (\tan (e+f x)+1)^{5/2}}{5 f}\right )+\frac {2 \tan (e+f x) (\tan (e+f x)+1)^{5/2}}{7 f}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{7} \left (-7 \left (-2 \int \frac {1}{\sqrt {\tan (e+f x)+1}}dx+\frac {2 (\tan (e+f x)+1)^{3/2}}{3 f}+\frac {2 \sqrt {\tan (e+f x)+1}}{f}\right )-\frac {4 (\tan (e+f x)+1)^{5/2}}{5 f}\right )+\frac {2 \tan (e+f x) (\tan (e+f x)+1)^{5/2}}{7 f}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{7} \left (-7 \left (-2 \int \frac {1}{\sqrt {\tan (e+f x)+1}}dx+\frac {2 (\tan (e+f x)+1)^{3/2}}{3 f}+\frac {2 \sqrt {\tan (e+f x)+1}}{f}\right )-\frac {4 (\tan (e+f x)+1)^{5/2}}{5 f}\right )+\frac {2 \tan (e+f x) (\tan (e+f x)+1)^{5/2}}{7 f}\) |
\(\Big \downarrow \) 3966 |
\(\displaystyle \frac {1}{7} \left (-7 \left (-\frac {2 \int \frac {1}{\sqrt {\tan (e+f x)+1} \left (\tan ^2(e+f x)+1\right )}d\tan (e+f x)}{f}+\frac {2 (\tan (e+f x)+1)^{3/2}}{3 f}+\frac {2 \sqrt {\tan (e+f x)+1}}{f}\right )-\frac {4 (\tan (e+f x)+1)^{5/2}}{5 f}\right )+\frac {2 \tan (e+f x) (\tan (e+f x)+1)^{5/2}}{7 f}\) |
\(\Big \downarrow \) 484 |
\(\displaystyle \frac {1}{7} \left (-7 \left (-\frac {4 \int \frac {1}{(\tan (e+f x)+1)^2-2 (\tan (e+f x)+1)+2}d\sqrt {\tan (e+f x)+1}}{f}+\frac {2 (\tan (e+f x)+1)^{3/2}}{3 f}+\frac {2 \sqrt {\tan (e+f x)+1}}{f}\right )-\frac {4 (\tan (e+f x)+1)^{5/2}}{5 f}\right )+\frac {2 \tan (e+f x) (\tan (e+f x)+1)^{5/2}}{7 f}\) |
\(\Big \downarrow \) 1407 |
\(\displaystyle \frac {1}{7} \left (-7 \left (-\frac {4 \left (\frac {\int \frac {\sqrt {2 \left (1+\sqrt {2}\right )}-\sqrt {\tan (e+f x)+1}}{\tan (e+f x)-\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {\tan (e+f x)+1}+\sqrt {2}+1}d\sqrt {\tan (e+f x)+1}}{4 \sqrt {1+\sqrt {2}}}+\frac {\int \frac {\sqrt {\tan (e+f x)+1}+\sqrt {2 \left (1+\sqrt {2}\right )}}{\tan (e+f x)+\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {\tan (e+f x)+1}+\sqrt {2}+1}d\sqrt {\tan (e+f x)+1}}{4 \sqrt {1+\sqrt {2}}}\right )}{f}+\frac {2 (\tan (e+f x)+1)^{3/2}}{3 f}+\frac {2 \sqrt {\tan (e+f x)+1}}{f}\right )-\frac {4 (\tan (e+f x)+1)^{5/2}}{5 f}\right )+\frac {2 \tan (e+f x) (\tan (e+f x)+1)^{5/2}}{7 f}\) |
\(\Big \downarrow \) 1142 |
\(\displaystyle \frac {1}{7} \left (-7 \left (-\frac {4 \left (\frac {\sqrt {\frac {1}{2} \left (1+\sqrt {2}\right )} \int \frac {1}{\tan (e+f x)-\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {\tan (e+f x)+1}+\sqrt {2}+1}d\sqrt {\tan (e+f x)+1}-\frac {1}{2} \int -\frac {\sqrt {2 \left (1+\sqrt {2}\right )}-2 \sqrt {\tan (e+f x)+1}}{\tan (e+f x)-\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {\tan (e+f x)+1}+\sqrt {2}+1}d\sqrt {\tan (e+f x)+1}}{4 \sqrt {1+\sqrt {2}}}+\frac {\sqrt {\frac {1}{2} \left (1+\sqrt {2}\right )} \int \frac {1}{\tan (e+f x)+\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {\tan (e+f x)+1}+\sqrt {2}+1}d\sqrt {\tan (e+f x)+1}+\frac {1}{2} \int \frac {2 \sqrt {\tan (e+f x)+1}+\sqrt {2 \left (1+\sqrt {2}\right )}}{\tan (e+f x)+\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {\tan (e+f x)+1}+\sqrt {2}+1}d\sqrt {\tan (e+f x)+1}}{4 \sqrt {1+\sqrt {2}}}\right )}{f}+\frac {2 (\tan (e+f x)+1)^{3/2}}{3 f}+\frac {2 \sqrt {\tan (e+f x)+1}}{f}\right )-\frac {4 (\tan (e+f x)+1)^{5/2}}{5 f}\right )+\frac {2 \tan (e+f x) (\tan (e+f x)+1)^{5/2}}{7 f}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{7} \left (-7 \left (-\frac {4 \left (\frac {\sqrt {\frac {1}{2} \left (1+\sqrt {2}\right )} \int \frac {1}{\tan (e+f x)-\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {\tan (e+f x)+1}+\sqrt {2}+1}d\sqrt {\tan (e+f x)+1}+\frac {1}{2} \int \frac {\sqrt {2 \left (1+\sqrt {2}\right )}-2 \sqrt {\tan (e+f x)+1}}{\tan (e+f x)-\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {\tan (e+f x)+1}+\sqrt {2}+1}d\sqrt {\tan (e+f x)+1}}{4 \sqrt {1+\sqrt {2}}}+\frac {\sqrt {\frac {1}{2} \left (1+\sqrt {2}\right )} \int \frac {1}{\tan (e+f x)+\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {\tan (e+f x)+1}+\sqrt {2}+1}d\sqrt {\tan (e+f x)+1}+\frac {1}{2} \int \frac {2 \sqrt {\tan (e+f x)+1}+\sqrt {2 \left (1+\sqrt {2}\right )}}{\tan (e+f x)+\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {\tan (e+f x)+1}+\sqrt {2}+1}d\sqrt {\tan (e+f x)+1}}{4 \sqrt {1+\sqrt {2}}}\right )}{f}+\frac {2 (\tan (e+f x)+1)^{3/2}}{3 f}+\frac {2 \sqrt {\tan (e+f x)+1}}{f}\right )-\frac {4 (\tan (e+f x)+1)^{5/2}}{5 f}\right )+\frac {2 \tan (e+f x) (\tan (e+f x)+1)^{5/2}}{7 f}\) |
\(\Big \downarrow \) 1083 |
\(\displaystyle \frac {1}{7} \left (-7 \left (-\frac {4 \left (\frac {\frac {1}{2} \int \frac {\sqrt {2 \left (1+\sqrt {2}\right )}-2 \sqrt {\tan (e+f x)+1}}{\tan (e+f x)-\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {\tan (e+f x)+1}+\sqrt {2}+1}d\sqrt {\tan (e+f x)+1}-\sqrt {2 \left (1+\sqrt {2}\right )} \int \frac {1}{-\tan (e+f x)+2 \left (1-\sqrt {2}\right )-1}d\left (2 \sqrt {\tan (e+f x)+1}-\sqrt {2 \left (1+\sqrt {2}\right )}\right )}{4 \sqrt {1+\sqrt {2}}}+\frac {\frac {1}{2} \int \frac {2 \sqrt {\tan (e+f x)+1}+\sqrt {2 \left (1+\sqrt {2}\right )}}{\tan (e+f x)+\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {\tan (e+f x)+1}+\sqrt {2}+1}d\sqrt {\tan (e+f x)+1}-\sqrt {2 \left (1+\sqrt {2}\right )} \int \frac {1}{-\tan (e+f x)+2 \left (1-\sqrt {2}\right )-1}d\left (2 \sqrt {\tan (e+f x)+1}+\sqrt {2 \left (1+\sqrt {2}\right )}\right )}{4 \sqrt {1+\sqrt {2}}}\right )}{f}+\frac {2 (\tan (e+f x)+1)^{3/2}}{3 f}+\frac {2 \sqrt {\tan (e+f x)+1}}{f}\right )-\frac {4 (\tan (e+f x)+1)^{5/2}}{5 f}\right )+\frac {2 \tan (e+f x) (\tan (e+f x)+1)^{5/2}}{7 f}\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {1}{7} \left (-7 \left (-\frac {4 \left (\frac {\frac {1}{2} \int \frac {\sqrt {2 \left (1+\sqrt {2}\right )}-2 \sqrt {\tan (e+f x)+1}}{\tan (e+f x)-\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {\tan (e+f x)+1}+\sqrt {2}+1}d\sqrt {\tan (e+f x)+1}+\sqrt {\frac {1+\sqrt {2}}{\sqrt {2}-1}} \arctan \left (\frac {2 \sqrt {\tan (e+f x)+1}-\sqrt {2 \left (1+\sqrt {2}\right )}}{\sqrt {2 \left (\sqrt {2}-1\right )}}\right )}{4 \sqrt {1+\sqrt {2}}}+\frac {\frac {1}{2} \int \frac {2 \sqrt {\tan (e+f x)+1}+\sqrt {2 \left (1+\sqrt {2}\right )}}{\tan (e+f x)+\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {\tan (e+f x)+1}+\sqrt {2}+1}d\sqrt {\tan (e+f x)+1}+\sqrt {\frac {1+\sqrt {2}}{\sqrt {2}-1}} \arctan \left (\frac {2 \sqrt {\tan (e+f x)+1}+\sqrt {2 \left (1+\sqrt {2}\right )}}{\sqrt {2 \left (\sqrt {2}-1\right )}}\right )}{4 \sqrt {1+\sqrt {2}}}\right )}{f}+\frac {2 (\tan (e+f x)+1)^{3/2}}{3 f}+\frac {2 \sqrt {\tan (e+f x)+1}}{f}\right )-\frac {4 (\tan (e+f x)+1)^{5/2}}{5 f}\right )+\frac {2 \tan (e+f x) (\tan (e+f x)+1)^{5/2}}{7 f}\) |
\(\Big \downarrow \) 1103 |
\(\displaystyle \frac {1}{7} \left (-7 \left (-\frac {4 \left (\frac {\sqrt {\frac {1+\sqrt {2}}{\sqrt {2}-1}} \arctan \left (\frac {2 \sqrt {\tan (e+f x)+1}-\sqrt {2 \left (1+\sqrt {2}\right )}}{\sqrt {2 \left (\sqrt {2}-1\right )}}\right )-\frac {1}{2} \log \left (\tan (e+f x)-\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {\tan (e+f x)+1}+\sqrt {2}+1\right )}{4 \sqrt {1+\sqrt {2}}}+\frac {\sqrt {\frac {1+\sqrt {2}}{\sqrt {2}-1}} \arctan \left (\frac {2 \sqrt {\tan (e+f x)+1}+\sqrt {2 \left (1+\sqrt {2}\right )}}{\sqrt {2 \left (\sqrt {2}-1\right )}}\right )+\frac {1}{2} \log \left (\tan (e+f x)+\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {\tan (e+f x)+1}+\sqrt {2}+1\right )}{4 \sqrt {1+\sqrt {2}}}\right )}{f}+\frac {2 (\tan (e+f x)+1)^{3/2}}{3 f}+\frac {2 \sqrt {\tan (e+f x)+1}}{f}\right )-\frac {4 (\tan (e+f x)+1)^{5/2}}{5 f}\right )+\frac {2 \tan (e+f x) (\tan (e+f x)+1)^{5/2}}{7 f}\) |
Input:
Int[Tan[e + f*x]^3*(1 + Tan[e + f*x])^(3/2),x]
Output:
(2*Tan[e + f*x]*(1 + Tan[e + f*x])^(5/2))/(7*f) + ((-4*(1 + Tan[e + f*x])^ (5/2))/(5*f) - 7*((-4*((Sqrt[(1 + Sqrt[2])/(-1 + Sqrt[2])]*ArcTan[(-Sqrt[2 *(1 + Sqrt[2])] + 2*Sqrt[1 + Tan[e + f*x]])/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)/(4*Sqrt[1 + Sqrt[2]]) + (Sqrt[(1 + Sqrt[2])/(-1 + Sqrt[2])]*ArcTan[(Sq rt[2*(1 + Sqrt[2])] + 2*Sqrt[1 + Tan[e + f*x]])/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)/(4*Sqrt[1 + Sqrt[2]])))/f + (2*Sqrt[1 + Tan[e + f*x]])/f + (2*(1 + Tan[e + f*x])^(3/2))/(3*f)))/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[1/(Sqrt[(c_) + (d_.)*(x_)]*((a_) + (b_.)*(x_)^2)), x_Symbol] :> Simp[2* d Subst[Int[1/(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[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[(2*c*d - b*e)/(2*c) Int[1/(a + b*x + c*x^2), x], x] + Simp[e/(2*c) Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x]
Int[((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(-1), x_Symbol] :> With[{q = Rt[a/ c, 2]}, With[{r = Rt[2*q - b/c, 2]}, Simp[1/(2*c*q*r) Int[(r - x)/(q - r* x + x^2), x], x] + Simp[1/(2*c*q*r) Int[(r + x)/(q + r*x + x^2), x], x]]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && NegQ[b^2 - 4*a*c]
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_)]), 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[((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 = 1.03 (sec) , antiderivative size = 354, normalized size of antiderivative = 1.39
method | result | size |
derivativedivides | \(\frac {2 \left (\tan \left (f x +e \right )+1\right )^{\frac {7}{2}}}{7 f}-\frac {2 \left (\tan \left (f x +e \right )+1\right )^{\frac {5}{2}}}{5 f}-\frac {2 \left (\tan \left (f x +e \right )+1\right )^{\frac {3}{2}}}{3 f}-\frac {2 \sqrt {\tan \left (f x +e \right )+1}}{f}+\frac {\sqrt {2 \sqrt {2}+2}\, \sqrt {2}\, \ln \left (\tan \left (f x +e \right )+1-\sqrt {\tan \left (f x +e \right )+1}\, \sqrt {2 \sqrt {2}+2}+\sqrt {2}\right )}{4 f}-\frac {\sqrt {2 \sqrt {2}+2}\, \ln \left (\tan \left (f x +e \right )+1-\sqrt {\tan \left (f x +e \right )+1}\, \sqrt {2 \sqrt {2}+2}+\sqrt {2}\right )}{2 f}+\frac {\arctan \left (\frac {2 \sqrt {\tan \left (f x +e \right )+1}-\sqrt {2 \sqrt {2}+2}}{\sqrt {-2+2 \sqrt {2}}}\right ) \sqrt {2}}{f \sqrt {-2+2 \sqrt {2}}}-\frac {\sqrt {2 \sqrt {2}+2}\, \sqrt {2}\, \ln \left (\tan \left (f x +e \right )+1+\sqrt {\tan \left (f x +e \right )+1}\, \sqrt {2 \sqrt {2}+2}+\sqrt {2}\right )}{4 f}+\frac {\sqrt {2 \sqrt {2}+2}\, \ln \left (\tan \left (f x +e \right )+1+\sqrt {\tan \left (f x +e \right )+1}\, \sqrt {2 \sqrt {2}+2}+\sqrt {2}\right )}{2 f}+\frac {\arctan \left (\frac {\sqrt {2 \sqrt {2}+2}+2 \sqrt {\tan \left (f x +e \right )+1}}{\sqrt {-2+2 \sqrt {2}}}\right ) \sqrt {2}}{f \sqrt {-2+2 \sqrt {2}}}\) | \(354\) |
default | \(\frac {2 \left (\tan \left (f x +e \right )+1\right )^{\frac {7}{2}}}{7 f}-\frac {2 \left (\tan \left (f x +e \right )+1\right )^{\frac {5}{2}}}{5 f}-\frac {2 \left (\tan \left (f x +e \right )+1\right )^{\frac {3}{2}}}{3 f}-\frac {2 \sqrt {\tan \left (f x +e \right )+1}}{f}+\frac {\sqrt {2 \sqrt {2}+2}\, \sqrt {2}\, \ln \left (\tan \left (f x +e \right )+1-\sqrt {\tan \left (f x +e \right )+1}\, \sqrt {2 \sqrt {2}+2}+\sqrt {2}\right )}{4 f}-\frac {\sqrt {2 \sqrt {2}+2}\, \ln \left (\tan \left (f x +e \right )+1-\sqrt {\tan \left (f x +e \right )+1}\, \sqrt {2 \sqrt {2}+2}+\sqrt {2}\right )}{2 f}+\frac {\arctan \left (\frac {2 \sqrt {\tan \left (f x +e \right )+1}-\sqrt {2 \sqrt {2}+2}}{\sqrt {-2+2 \sqrt {2}}}\right ) \sqrt {2}}{f \sqrt {-2+2 \sqrt {2}}}-\frac {\sqrt {2 \sqrt {2}+2}\, \sqrt {2}\, \ln \left (\tan \left (f x +e \right )+1+\sqrt {\tan \left (f x +e \right )+1}\, \sqrt {2 \sqrt {2}+2}+\sqrt {2}\right )}{4 f}+\frac {\sqrt {2 \sqrt {2}+2}\, \ln \left (\tan \left (f x +e \right )+1+\sqrt {\tan \left (f x +e \right )+1}\, \sqrt {2 \sqrt {2}+2}+\sqrt {2}\right )}{2 f}+\frac {\arctan \left (\frac {\sqrt {2 \sqrt {2}+2}+2 \sqrt {\tan \left (f x +e \right )+1}}{\sqrt {-2+2 \sqrt {2}}}\right ) \sqrt {2}}{f \sqrt {-2+2 \sqrt {2}}}\) | \(354\) |
Input:
int(tan(f*x+e)^3*(tan(f*x+e)+1)^(3/2),x,method=_RETURNVERBOSE)
Output:
2/7/f*(tan(f*x+e)+1)^(7/2)-2/5*(tan(f*x+e)+1)^(5/2)/f-2/3*(tan(f*x+e)+1)^( 3/2)/f-2*(tan(f*x+e)+1)^(1/2)/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))-1/2/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/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))*2^(1/2)-1/4/f*(2*2^(1/2)+2)^(1/2)*2^(1/2)*ln(ta n(f*x+e)+1+(tan(f*x+e)+1)^(1/2)*(2*2^(1/2)+2)^(1/2)+2^(1/2))+1/2/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/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))*2^(1/2)
Time = 0.09 (sec) , antiderivative size = 358, normalized size of antiderivative = 1.40 \[ \int \tan ^3(e+f x) (1+\tan (e+f x))^{3/2} \, dx=\frac {105 \, \sqrt {2} f \sqrt {-\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} + 1}{f^{2}}} \log \left (\sqrt {2} {\left (f^{3} \sqrt {-\frac {1}{f^{4}}} + f\right )} \sqrt {-\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} + 1}{f^{2}}} + 2 \, \sqrt {\tan \left (f x + e\right ) + 1}\right ) - 105 \, \sqrt {2} f \sqrt {-\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} + 1}{f^{2}}} \log \left (-\sqrt {2} {\left (f^{3} \sqrt {-\frac {1}{f^{4}}} + f\right )} \sqrt {-\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} + 1}{f^{2}}} + 2 \, \sqrt {\tan \left (f x + e\right ) + 1}\right ) - 105 \, \sqrt {2} f \sqrt {\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} - 1}{f^{2}}} \log \left (\sqrt {2} {\left (f^{3} \sqrt {-\frac {1}{f^{4}}} - f\right )} \sqrt {\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} - 1}{f^{2}}} + 2 \, \sqrt {\tan \left (f x + e\right ) + 1}\right ) + 105 \, \sqrt {2} f \sqrt {\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} - 1}{f^{2}}} \log \left (-\sqrt {2} {\left (f^{3} \sqrt {-\frac {1}{f^{4}}} - f\right )} \sqrt {\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} - 1}{f^{2}}} + 2 \, \sqrt {\tan \left (f x + e\right ) + 1}\right ) + 4 \, {\left (15 \, \tan \left (f x + e\right )^{3} + 24 \, \tan \left (f x + e\right )^{2} - 32 \, \tan \left (f x + e\right ) - 146\right )} \sqrt {\tan \left (f x + e\right ) + 1}}{210 \, f} \] Input:
integrate(tan(f*x+e)^3*(1+tan(f*x+e))^(3/2),x, algorithm="fricas")
Output:
1/210*(105*sqrt(2)*f*sqrt(-(f^2*sqrt(-1/f^4) + 1)/f^2)*log(sqrt(2)*(f^3*sq rt(-1/f^4) + f)*sqrt(-(f^2*sqrt(-1/f^4) + 1)/f^2) + 2*sqrt(tan(f*x + e) + 1)) - 105*sqrt(2)*f*sqrt(-(f^2*sqrt(-1/f^4) + 1)/f^2)*log(-sqrt(2)*(f^3*sq rt(-1/f^4) + f)*sqrt(-(f^2*sqrt(-1/f^4) + 1)/f^2) + 2*sqrt(tan(f*x + e) + 1)) - 105*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)) + 105*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)) + 4*(15*tan(f*x + e)^3 + 24*tan(f*x + e)^2 - 32*tan(f*x + e) - 146)*sqrt(ta n(f*x + e) + 1))/f
\[ \int \tan ^3(e+f x) (1+\tan (e+f x))^{3/2} \, dx=\int \left (\tan {\left (e + f x \right )} + 1\right )^{\frac {3}{2}} \tan ^{3}{\left (e + f x \right )}\, dx \] Input:
integrate(tan(f*x+e)**3*(1+tan(f*x+e))**(3/2),x)
Output:
Integral((tan(e + f*x) + 1)**(3/2)*tan(e + f*x)**3, x)
\[ \int \tan ^3(e+f x) (1+\tan (e+f x))^{3/2} \, dx=\int { {\left (\tan \left (f x + e\right ) + 1\right )}^{\frac {3}{2}} \tan \left (f x + e\right )^{3} \,d x } \] Input:
integrate(tan(f*x+e)^3*(1+tan(f*x+e))^(3/2),x, algorithm="maxima")
Output:
integrate((tan(f*x + e) + 1)^(3/2)*tan(f*x + e)^3, x)
Exception generated. \[ \int \tan ^3(e+f x) (1+\tan (e+f x))^{3/2} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(tan(f*x+e)^3*(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,[24]%%%}+%%%{10,[22]%%%}+%%%{45,[20]%%%}+%%%{120,[18 ]%%%}+%%%
Time = 2.39 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.51 \[ \int \tan ^3(e+f x) (1+\tan (e+f x))^{3/2} \, dx=\frac {2\,{\left (\mathrm {tan}\left (e+f\,x\right )+1\right )}^{7/2}}{7\,f}-\frac {2\,{\left (\mathrm {tan}\left (e+f\,x\right )+1\right )}^{3/2}}{3\,f}-\frac {2\,{\left (\mathrm {tan}\left (e+f\,x\right )+1\right )}^{5/2}}{5\,f}-\frac {2\,\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}}{f}+\mathrm {atan}\left (f\,\sqrt {\frac {-\frac {1}{2}-\frac {1}{2}{}\mathrm {i}}{f^2}}\,\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}\right )\,\sqrt {\frac {-\frac {1}{2}-\frac {1}{2}{}\mathrm {i}}{f^2}}\,2{}\mathrm {i}-\mathrm {atan}\left (f\,\sqrt {\frac {-\frac {1}{2}+\frac {1}{2}{}\mathrm {i}}{f^2}}\,\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}\right )\,\sqrt {\frac {-\frac {1}{2}+\frac {1}{2}{}\mathrm {i}}{f^2}}\,2{}\mathrm {i} \] Input:
int(tan(e + f*x)^3*(tan(e + f*x) + 1)^(3/2),x)
Output:
(2*(tan(e + f*x) + 1)^(7/2))/(7*f) - (2*(tan(e + f*x) + 1)^(3/2))/(3*f) - (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))*((- 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))* ((- 1/2 + 1i/2)/f^2)^(1/2)*2i
\[ \int \tan ^3(e+f x) (1+\tan (e+f x))^{3/2} \, dx=\int \sqrt {\tan \left (f x +e \right )+1}\, \tan \left (f x +e \right )^{4}d x +\int \sqrt {\tan \left (f x +e \right )+1}\, \tan \left (f x +e \right )^{3}d x \] Input:
int(tan(f*x+e)^3*(1+tan(f*x+e))^(3/2),x)
Output:
int(sqrt(tan(e + f*x) + 1)*tan(e + f*x)**4,x) + int(sqrt(tan(e + f*x) + 1) *tan(e + f*x)**3,x)