Integrand size = 29, antiderivative size = 76 \[ \int \sec ^3\left (\frac {\pi }{4}+\frac {x}{2}\right ) \tan ^2\left (\frac {\pi }{4}+\frac {x}{2}\right ) \, dx=-\frac {1}{4} \text {arctanh}\left (\sin \left (\frac {\pi }{4}+\frac {x}{2}\right )\right )-\frac {1}{4} \sec \left (\frac {\pi }{4}+\frac {x}{2}\right ) \tan \left (\frac {\pi }{4}+\frac {x}{2}\right )+\frac {1}{2} \sec ^3\left (\frac {\pi }{4}+\frac {x}{2}\right ) \tan \left (\frac {\pi }{4}+\frac {x}{2}\right ) \] Output:
-1/4*arctanh(sin(1/4*Pi+1/2*x))-1/4*sec(1/4*Pi+1/2*x)*tan(1/4*Pi+1/2*x)+1/ 2*sec(1/4*Pi+1/2*x)^3*tan(1/4*Pi+1/2*x)
Time = 0.02 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.97 \[ \int \sec ^3\left (\frac {\pi }{4}+\frac {x}{2}\right ) \tan ^2\left (\frac {\pi }{4}+\frac {x}{2}\right ) \, dx=-\frac {1}{4} \text {arctanh}\left (\sin \left (\frac {\pi }{4}+\frac {x}{2}\right )\right )-\frac {1}{4} \sec ^2\left (\frac {1}{4} (\pi +2 x)\right ) \sin \left (\frac {\pi }{4}+\frac {x}{2}\right )+\frac {1}{2} \sec ^4\left (\frac {1}{4} (\pi +2 x)\right ) \sin \left (\frac {\pi }{4}+\frac {x}{2}\right ) \] Input:
Integrate[Sec[Pi/4 + x/2]^3*Tan[Pi/4 + x/2]^2,x]
Output:
-1/4*ArcTanh[Sin[Pi/4 + x/2]] - (Sec[(Pi + 2*x)/4]^2*Sin[Pi/4 + x/2])/4 + (Sec[(Pi + 2*x)/4]^4*Sin[Pi/4 + x/2])/2
Time = 0.33 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.01, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {3042, 3091, 3042, 4255, 3042, 4257}
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.
| \(\displaystyle \int \tan ^2\left (\frac {x}{2}+\frac {\pi }{4}\right ) \sec ^3\left (\frac {x}{2}+\frac {\pi }{4}\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \tan \left (\frac {x}{2}+\frac {\pi }{4}\right )^2 \sec \left (\frac {x}{2}+\frac {\pi }{4}\right )^3dx\) |
| \(\Big \downarrow \) 3091 |
| \(\displaystyle \frac {1}{2} \tan \left (\frac {x}{2}+\frac {\pi }{4}\right ) \sec ^3\left (\frac {x}{2}+\frac {\pi }{4}\right )-\frac {1}{4} \int \sec ^3\left (\frac {x}{2}+\frac {\pi }{4}\right )dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \tan \left (\frac {x}{2}+\frac {\pi }{4}\right ) \sec ^3\left (\frac {x}{2}+\frac {\pi }{4}\right )-\frac {1}{4} \int \csc \left (\frac {x}{2}+\frac {3 \pi }{4}\right )^3dx\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {1}{4} \left (\tan \left (\frac {x}{2}+\frac {\pi }{4}\right ) \left (-\sec \left (\frac {x}{2}+\frac {\pi }{4}\right )\right )-\frac {1}{2} \int \sec \left (\frac {x}{2}+\frac {\pi }{4}\right )dx\right )+\frac {1}{2} \tan \left (\frac {x}{2}+\frac {\pi }{4}\right ) \sec ^3\left (\frac {x}{2}+\frac {\pi }{4}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \left (\tan \left (\frac {x}{2}+\frac {\pi }{4}\right ) \left (-\sec \left (\frac {x}{2}+\frac {\pi }{4}\right )\right )-\frac {1}{2} \int \csc \left (\frac {x}{2}+\frac {3 \pi }{4}\right )dx\right )+\frac {1}{2} \tan \left (\frac {x}{2}+\frac {\pi }{4}\right ) \sec ^3\left (\frac {x}{2}+\frac {\pi }{4}\right )\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {1}{4} \left (-\text {arctanh}\left (\sin \left (\frac {x}{2}+\frac {\pi }{4}\right )\right )-\tan \left (\frac {x}{2}+\frac {\pi }{4}\right ) \sec \left (\frac {x}{2}+\frac {\pi }{4}\right )\right )+\frac {1}{2} \tan \left (\frac {x}{2}+\frac {\pi }{4}\right ) \sec ^3\left (\frac {x}{2}+\frac {\pi }{4}\right )\) |
Input:
Int[Sec[Pi/4 + x/2]^3*Tan[Pi/4 + x/2]^2,x]
Output:
(Sec[Pi/4 + x/2]^3*Tan[Pi/4 + x/2])/2 + (-ArcTanh[Sin[Pi/4 + x/2]] - Sec[P i/4 + x/2]*Tan[Pi/4 + x/2])/4
Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( n_), x_Symbol] :> Simp[b*(a*Sec[e + f*x])^m*((b*Tan[e + f*x])^(n - 1)/(f*(m + n - 1))), x] - Simp[b^2*((n - 1)/(m + n - 1)) Int[(a*Sec[e + f*x])^m*( b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] & & NeQ[m + n - 1, 0] && IntegersQ[2*m, 2*n]
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[2*n]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 0.55 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.00
| method | result | size |
| derivativedivides | \(\frac {\sin \left (\frac {\pi }{4}+\frac {x}{2}\right )^{3}}{2 \cos \left (\frac {\pi }{4}+\frac {x}{2}\right )^{4}}+\frac {\sin \left (\frac {\pi }{4}+\frac {x}{2}\right )^{3}}{4 \cos \left (\frac {\pi }{4}+\frac {x}{2}\right )^{2}}+\frac {\sin \left (\frac {\pi }{4}+\frac {x}{2}\right )}{4}-\frac {\ln \left (\sec \left (\frac {\pi }{4}+\frac {x}{2}\right )+\tan \left (\frac {\pi }{4}+\frac {x}{2}\right )\right )}{4}\) | \(76\) |
| default | \(\frac {\sin \left (\frac {\pi }{4}+\frac {x}{2}\right )^{3}}{2 \cos \left (\frac {\pi }{4}+\frac {x}{2}\right )^{4}}+\frac {\sin \left (\frac {\pi }{4}+\frac {x}{2}\right )^{3}}{4 \cos \left (\frac {\pi }{4}+\frac {x}{2}\right )^{2}}+\frac {\sin \left (\frac {\pi }{4}+\frac {x}{2}\right )}{4}-\frac {\ln \left (\sec \left (\frac {\pi }{4}+\frac {x}{2}\right )+\tan \left (\frac {\pi }{4}+\frac {x}{2}\right )\right )}{4}\) | \(76\) |
| risch | \(\frac {i \left (-\left (-1\right )^{\frac {3}{4}} {\mathrm e}^{\frac {7 i x}{2}}+7 \left (-1\right )^{\frac {1}{4}} {\mathrm e}^{\frac {5 i x}{2}}+7 \left (-1\right )^{\frac {3}{4}} {\mathrm e}^{\frac {3 i x}{2}}-\left (-1\right )^{\frac {1}{4}} {\mathrm e}^{\frac {i x}{2}}\right )}{2 \left (i {\mathrm e}^{i x}+1\right )^{4}}+\frac {\ln \left ({\mathrm e}^{\frac {i \left (\pi +2 x \right )}{4}}-i\right )}{4}-\frac {\ln \left ({\mathrm e}^{\frac {i \left (\pi +2 x \right )}{4}}+i\right )}{4}\) | \(88\) |
Input:
int(sec(1/4*Pi+1/2*x)^3*tan(1/4*Pi+1/2*x)^2,x,method=_RETURNVERBOSE)
Output:
1/2*sin(1/4*Pi+1/2*x)^3/cos(1/4*Pi+1/2*x)^4+1/4*sin(1/4*Pi+1/2*x)^3/cos(1/ 4*Pi+1/2*x)^2+1/4*sin(1/4*Pi+1/2*x)-1/4*ln(sec(1/4*Pi+1/2*x)+tan(1/4*Pi+1/ 2*x))
Time = 0.08 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.08 \[ \int \sec ^3\left (\frac {\pi }{4}+\frac {x}{2}\right ) \tan ^2\left (\frac {\pi }{4}+\frac {x}{2}\right ) \, dx=-\frac {\cos \left (\frac {1}{4} \, \pi + \frac {1}{2} \, x\right )^{4} \log \left (\sin \left (\frac {1}{4} \, \pi + \frac {1}{2} \, x\right ) + 1\right ) - \cos \left (\frac {1}{4} \, \pi + \frac {1}{2} \, x\right )^{4} \log \left (-\sin \left (\frac {1}{4} \, \pi + \frac {1}{2} \, x\right ) + 1\right ) + 2 \, {\left (\cos \left (\frac {1}{4} \, \pi + \frac {1}{2} \, x\right )^{2} - 2\right )} \sin \left (\frac {1}{4} \, \pi + \frac {1}{2} \, x\right )}{8 \, \cos \left (\frac {1}{4} \, \pi + \frac {1}{2} \, x\right )^{4}} \] Input:
integrate(sec(1/4*pi+1/2*x)^3*tan(1/4*pi+1/2*x)^2,x, algorithm="fricas")
Output:
-1/8*(cos(1/4*pi + 1/2*x)^4*log(sin(1/4*pi + 1/2*x) + 1) - cos(1/4*pi + 1/ 2*x)^4*log(-sin(1/4*pi + 1/2*x) + 1) + 2*(cos(1/4*pi + 1/2*x)^2 - 2)*sin(1 /4*pi + 1/2*x))/cos(1/4*pi + 1/2*x)^4
\[ \int \sec ^3\left (\frac {\pi }{4}+\frac {x}{2}\right ) \tan ^2\left (\frac {\pi }{4}+\frac {x}{2}\right ) \, dx=\int \tan ^{2}{\left (\frac {x}{2} + \frac {\pi }{4} \right )} \sec ^{3}{\left (\frac {x}{2} + \frac {\pi }{4} \right )}\, dx \] Input:
integrate(sec(1/4*pi+1/2*x)**3*tan(1/4*pi+1/2*x)**2,x)
Output:
Integral(tan(x/2 + pi/4)**2*sec(x/2 + pi/4)**3, x)
Time = 0.03 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.97 \[ \int \sec ^3\left (\frac {\pi }{4}+\frac {x}{2}\right ) \tan ^2\left (\frac {\pi }{4}+\frac {x}{2}\right ) \, dx=\frac {\sin \left (\frac {1}{4} \, \pi + \frac {1}{2} \, x\right )^{3} + \sin \left (\frac {1}{4} \, \pi + \frac {1}{2} \, x\right )}{4 \, {\left (\sin \left (\frac {1}{4} \, \pi + \frac {1}{2} \, x\right )^{4} - 2 \, \sin \left (\frac {1}{4} \, \pi + \frac {1}{2} \, x\right )^{2} + 1\right )}} - \frac {1}{8} \, \log \left (\sin \left (\frac {1}{4} \, \pi + \frac {1}{2} \, x\right ) + 1\right ) + \frac {1}{8} \, \log \left (\sin \left (\frac {1}{4} \, \pi + \frac {1}{2} \, x\right ) - 1\right ) \] Input:
integrate(sec(1/4*pi+1/2*x)^3*tan(1/4*pi+1/2*x)^2,x, algorithm="maxima")
Output:
1/4*(sin(1/4*pi + 1/2*x)^3 + sin(1/4*pi + 1/2*x))/(sin(1/4*pi + 1/2*x)^4 - 2*sin(1/4*pi + 1/2*x)^2 + 1) - 1/8*log(sin(1/4*pi + 1/2*x) + 1) + 1/8*log (sin(1/4*pi + 1/2*x) - 1)
Time = 0.18 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.25 \[ \int \sec ^3\left (\frac {\pi }{4}+\frac {x}{2}\right ) \tan ^2\left (\frac {\pi }{4}+\frac {x}{2}\right ) \, dx=\frac {\frac {1}{\sin \left (\frac {1}{4} \, \pi + \frac {1}{2} \, x\right )} + \sin \left (\frac {1}{4} \, \pi + \frac {1}{2} \, x\right )}{4 \, {\left ({\left (\frac {1}{\sin \left (\frac {1}{4} \, \pi + \frac {1}{2} \, x\right )} + \sin \left (\frac {1}{4} \, \pi + \frac {1}{2} \, x\right )\right )}^{2} - 4\right )}} - \frac {1}{16} \, \log \left ({\left | \frac {1}{\sin \left (\frac {1}{4} \, \pi + \frac {1}{2} \, x\right )} + \sin \left (\frac {1}{4} \, \pi + \frac {1}{2} \, x\right ) + 2 \right |}\right ) + \frac {1}{16} \, \log \left ({\left | \frac {1}{\sin \left (\frac {1}{4} \, \pi + \frac {1}{2} \, x\right )} + \sin \left (\frac {1}{4} \, \pi + \frac {1}{2} \, x\right ) - 2 \right |}\right ) \] Input:
integrate(sec(1/4*pi+1/2*x)^3*tan(1/4*pi+1/2*x)^2,x, algorithm="giac")
Output:
1/4*(1/sin(1/4*pi + 1/2*x) + sin(1/4*pi + 1/2*x))/((1/sin(1/4*pi + 1/2*x) + sin(1/4*pi + 1/2*x))^2 - 4) - 1/16*log(abs(1/sin(1/4*pi + 1/2*x) + sin(1 /4*pi + 1/2*x) + 2)) + 1/16*log(abs(1/sin(1/4*pi + 1/2*x) + sin(1/4*pi + 1 /2*x) - 2))
Time = 3.91 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.99 \[ \int \sec ^3\left (\frac {\pi }{4}+\frac {x}{2}\right ) \tan ^2\left (\frac {\pi }{4}+\frac {x}{2}\right ) \, dx=\frac {2\,\left (\frac {{\mathrm {tan}\left (\frac {\Pi }{8}+\frac {x}{4}\right )}^7}{4}+\frac {7\,{\mathrm {tan}\left (\frac {\Pi }{8}+\frac {x}{4}\right )}^5}{4}+\frac {7\,{\mathrm {tan}\left (\frac {\Pi }{8}+\frac {x}{4}\right )}^3}{4}+\frac {\mathrm {tan}\left (\frac {\Pi }{8}+\frac {x}{4}\right )}{4}\right )}{{\left ({\mathrm {tan}\left (\frac {\Pi }{8}+\frac {x}{4}\right )}^2-1\right )}^4}-\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {\Pi }{8}+\frac {x}{4}\right )\right )}{2} \] Input:
int(tan(Pi/4 + x/2)^2/cos(Pi/4 + x/2)^3,x)
Output:
(2*(tan(Pi/8 + x/4)/4 + (7*tan(Pi/8 + x/4)^3)/4 + (7*tan(Pi/8 + x/4)^5)/4 + tan(Pi/8 + x/4)^7/4))/(tan(Pi/8 + x/4)^2 - 1)^4 - atanh(tan(Pi/8 + x/4)) /2
Time = 0.16 (sec) , antiderivative size = 163, normalized size of antiderivative = 2.14 \[ \int \sec ^3\left (\frac {\pi }{4}+\frac {x}{2}\right ) \tan ^2\left (\frac {\pi }{4}+\frac {x}{2}\right ) \, dx=\frac {\mathrm {log}\left (\tan \left (\frac {\pi }{8}+\frac {x}{4}\right )-1\right ) \sin \left (\frac {\pi }{4}+\frac {x}{2}\right )^{4}-2 \,\mathrm {log}\left (\tan \left (\frac {\pi }{8}+\frac {x}{4}\right )-1\right ) \sin \left (\frac {\pi }{4}+\frac {x}{2}\right )^{2}+\mathrm {log}\left (\tan \left (\frac {\pi }{8}+\frac {x}{4}\right )-1\right )-\mathrm {log}\left (\tan \left (\frac {\pi }{8}+\frac {x}{4}\right )+1\right ) \sin \left (\frac {\pi }{4}+\frac {x}{2}\right )^{4}+2 \,\mathrm {log}\left (\tan \left (\frac {\pi }{8}+\frac {x}{4}\right )+1\right ) \sin \left (\frac {\pi }{4}+\frac {x}{2}\right )^{2}-\mathrm {log}\left (\tan \left (\frac {\pi }{8}+\frac {x}{4}\right )+1\right )+\sin \left (\frac {\pi }{4}+\frac {x}{2}\right )^{3}+\sin \left (\frac {\pi }{4}+\frac {x}{2}\right )}{4 \sin \left (\frac {\pi }{4}+\frac {x}{2}\right )^{4}-8 \sin \left (\frac {\pi }{4}+\frac {x}{2}\right )^{2}+4} \] Input:
int(sec(1/4*Pi+1/2*x)^3*tan(1/4*Pi+1/2*x)^2,x)
Output:
(log(tan((pi + 2*x)/8) - 1)*sin((pi + 2*x)/4)**4 - 2*log(tan((pi + 2*x)/8) - 1)*sin((pi + 2*x)/4)**2 + log(tan((pi + 2*x)/8) - 1) - log(tan((pi + 2* x)/8) + 1)*sin((pi + 2*x)/4)**4 + 2*log(tan((pi + 2*x)/8) + 1)*sin((pi + 2 *x)/4)**2 - log(tan((pi + 2*x)/8) + 1) + sin((pi + 2*x)/4)**3 + sin((pi + 2*x)/4))/(4*(sin((pi + 2*x)/4)**4 - 2*sin((pi + 2*x)/4)**2 + 1))