3.362 \(\int \sec ^3(\frac {\pi }{4}+\frac {x}{2}) \tan ^2(\frac {\pi }{4}+\frac {x}{2}) \, dx\)

Optimal. Leaf size=76 \[ -\frac {1}{4} \tanh ^{-1}\left (\sin \left (\frac {x}{2}+\frac {\pi }{4}\right )\right )+\frac {1}{2} \tan \left (\frac {x}{2}+\frac {\pi }{4}\right ) \sec ^3\left (\frac {x}{2}+\frac {\pi }{4}\right )-\frac {1}{4} \tan \left (\frac {x}{2}+\frac {\pi }{4}\right ) \sec \left (\frac {x}{2}+\frac {\pi }{4}\right ) \]

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Rubi [A]  time = 0.04, antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {2611, 3768, 3770} \[ -\frac {1}{4} \tanh ^{-1}\left (\sin \left (\frac {x}{2}+\frac {\pi }{4}\right )\right )+\frac {1}{2} \tan \left (\frac {x}{2}+\frac {\pi }{4}\right ) \sec ^3\left (\frac {x}{2}+\frac {\pi }{4}\right )-\frac {1}{4} \tan \left (\frac {x}{2}+\frac {\pi }{4}\right ) \sec \left (\frac {x}{2}+\frac {\pi }{4}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Sec[Pi/4 + x/2]^3*Tan[Pi/4 + x/2]^2,x]

[Out]

-ArcTanh[Sin[Pi/4 + x/2]]/4 - (Sec[Pi/4 + x/2]*Tan[Pi/4 + x/2])/4 + (Sec[Pi/4 + x/2]^3*Tan[Pi/4 + x/2])/2

Rule 2611

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] - Dist[(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] && Integers
Q[2*m, 2*n]

Rule 3768

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] + Dist[(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]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin {align*} \int \sec ^3\left (\frac {\pi }{4}+\frac {x}{2}\right ) \tan ^2\left (\frac {\pi }{4}+\frac {x}{2}\right ) \, dx &=\frac {1}{2} \sec ^3\left (\frac {\pi }{4}+\frac {x}{2}\right ) \tan \left (\frac {\pi }{4}+\frac {x}{2}\right )-\frac {1}{4} \int \sec ^3\left (\frac {\pi }{4}+\frac {x}{2}\right ) \, dx\\ &=-\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 )-\frac {1}{8} \int \csc \left (\frac {\pi }{4}-\frac {x}{2}\right ) \, dx\\ &=-\frac {1}{4} \tanh ^{-1}\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 )\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 74, normalized size = 0.97 \[ -\frac {1}{4} \tanh ^{-1}\left (\sin \left (\frac {x}{2}+\frac {\pi }{4}\right )\right )+\frac {1}{2} \sin \left (\frac {x}{2}+\frac {\pi }{4}\right ) \sec ^4\left (\frac {1}{4} (2 x+\pi )\right )-\frac {1}{4} \sin \left (\frac {x}{2}+\frac {\pi }{4}\right ) \sec ^2\left (\frac {1}{4} (2 x+\pi )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[Pi/4 + x/2]^3*Tan[Pi/4 + x/2]^2,x]

[Out]

-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

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sec ^3\left (\frac {\pi }{4}+\frac {x}{2}\right ) \tan ^2\left (\frac {\pi }{4}+\frac {x}{2}\right ) \, dx \]

Verification is Not applicable to the result.

[In]

IntegrateAlgebraic[Sec[Pi/4 + x/2]^3*Tan[Pi/4 + x/2]^2,x]

[Out]

Could not integrate

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fricas [A]  time = 0.73, size = 82, normalized size = 1.08 \[ -\frac {\cos \left (\frac {1}{4} \, \pi + \frac {1}{2} \, x\right )^{4} \log \left (\sin \left (\frac {1}{4} \, \pi + \frac {1}{2} \, x\right ) + 1\right ) - \cos \left (\frac {1}{4} \, \pi + \frac {1}{2} \, x\right )^{4} \log \left (-\sin \left (\frac {1}{4} \, \pi + \frac {1}{2} \, x\right ) + 1\right ) + 2 \, {\left (\cos \left (\frac {1}{4} \, \pi + \frac {1}{2} \, x\right )^{2} - 2\right )} \sin \left (\frac {1}{4} \, \pi + \frac {1}{2} \, x\right )}{8 \, \cos \left (\frac {1}{4} \, \pi + \frac {1}{2} \, x\right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(1/4*pi+1/2*x)^3*tan(1/4*pi+1/2*x)^2,x, algorithm="fricas")

[Out]

-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

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giac [A]  time = 0.95, size = 95, normalized size = 1.25 \[ \frac {\frac {1}{\sin \left (\frac {1}{4} \, \pi + \frac {1}{2} \, x\right )} + \sin \left (\frac {1}{4} \, \pi + \frac {1}{2} \, x\right )}{4 \, {\left ({\left (\frac {1}{\sin \left (\frac {1}{4} \, \pi + \frac {1}{2} \, x\right )} + \sin \left (\frac {1}{4} \, \pi + \frac {1}{2} \, x\right )\right )}^{2} - 4\right )}} - \frac {1}{16} \, \log \left ({\left | \frac {1}{\sin \left (\frac {1}{4} \, \pi + \frac {1}{2} \, x\right )} + \sin \left (\frac {1}{4} \, \pi + \frac {1}{2} \, x\right ) + 2 \right |}\right ) + \frac {1}{16} \, \log \left ({\left | \frac {1}{\sin \left (\frac {1}{4} \, \pi + \frac {1}{2} \, x\right )} + \sin \left (\frac {1}{4} \, \pi + \frac {1}{2} \, x\right ) - 2 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(1/4*pi+1/2*x)^3*tan(1/4*pi+1/2*x)^2,x, algorithm="giac")

[Out]

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))

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maple [A]  time = 0.56, size = 76, normalized size = 1.00

method	result	size
derivativedivides	\(\frac {\sin ^{3}\left (\frac {\pi }{4}+\frac {x}{2}\right )}{2 \cos \left (\frac {\pi }{4}+\frac {x}{2}\right )^{4}}+\frac {\sin ^{3}\left (\frac {\pi }{4}+\frac {x}{2}\right )}{4 \cos \left (\frac {\pi }{4}+\frac {x}{2}\right )^{2}}+\frac {\sin \left (\frac {\pi }{4}+\frac {x}{2}\right )}{4}-\frac {\ln \left (\sec \left (\frac {\pi }{4}+\frac {x}{2}\right )+\tan \left (\frac {\pi }{4}+\frac {x}{2}\right )\right )}{4}\)	\(76\)
default	\(\frac {\sin ^{3}\left (\frac {\pi }{4}+\frac {x}{2}\right )}{2 \cos \left (\frac {\pi }{4}+\frac {x}{2}\right )^{4}}+\frac {\sin ^{3}\left (\frac {\pi }{4}+\frac {x}{2}\right )}{4 \cos \left (\frac {\pi }{4}+\frac {x}{2}\right )^{2}}+\frac {\sin \left (\frac {\pi }{4}+\frac {x}{2}\right )}{4}-\frac {\ln \left (\sec \left (\frac {\pi }{4}+\frac {x}{2}\right )+\tan \left (\frac {\pi }{4}+\frac {x}{2}\right )\right )}{4}\)	\(76\)
risch	\(\frac {i \left (-\left (-1\right )^{\frac {3}{4}} {\mathrm e}^{\frac {7 i x}{2}}+7 \left (-1\right )^{\frac {1}{4}} {\mathrm e}^{\frac {5 i x}{2}}+7 \left (-1\right )^{\frac {3}{4}} {\mathrm e}^{\frac {3 i x}{2}}-\left (-1\right )^{\frac {1}{4}} {\mathrm e}^{\frac {i x}{2}}\right )}{2 \left (i {\mathrm e}^{i x}+1\right )^{4}}+\frac {\ln \left ({\mathrm e}^{\frac {i \left (\pi +2 x \right )}{4}}-i\right )}{4}-\frac {\ln \left ({\mathrm e}^{\frac {i \left (\pi +2 x \right )}{4}}+i\right )}{4}\)	\(88\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(1/4*Pi+1/2*x)^3*tan(1/4*Pi+1/2*x)^2,x,method=_RETURNVERBOSE)

[Out]

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))

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maxima [A]  time = 0.67, size = 74, normalized size = 0.97 \[ \frac {\sin \left (\frac {1}{4} \, \pi + \frac {1}{2} \, x\right )^{3} + \sin \left (\frac {1}{4} \, \pi + \frac {1}{2} \, x\right )}{4 \, {\left (\sin \left (\frac {1}{4} \, \pi + \frac {1}{2} \, x\right )^{4} - 2 \, \sin \left (\frac {1}{4} \, \pi + \frac {1}{2} \, x\right )^{2} + 1\right )}} - \frac {1}{8} \, \log \left (\sin \left (\frac {1}{4} \, \pi + \frac {1}{2} \, x\right ) + 1\right ) + \frac {1}{8} \, \log \left (\sin \left (\frac {1}{4} \, \pi + \frac {1}{2} \, x\right ) - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(1/4*pi+1/2*x)^3*tan(1/4*pi+1/2*x)^2,x, algorithm="maxima")

[Out]

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)

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mupad [B]  time = 6.38, size = 75, normalized size = 0.99 \[ \frac {2\,\left (\frac {{\mathrm {tan}\left (\frac {\Pi }{8}+\frac {x}{4}\right )}^7}{4}+\frac {7\,{\mathrm {tan}\left (\frac {\Pi }{8}+\frac {x}{4}\right )}^5}{4}+\frac {7\,{\mathrm {tan}\left (\frac {\Pi }{8}+\frac {x}{4}\right )}^3}{4}+\frac {\mathrm {tan}\left (\frac {\Pi }{8}+\frac {x}{4}\right )}{4}\right )}{{\left ({\mathrm {tan}\left (\frac {\Pi }{8}+\frac {x}{4}\right )}^2-1\right )}^4}-\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {\Pi }{8}+\frac {x}{4}\right )\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(Pi/4 + x/2)^2/cos(Pi/4 + x/2)^3,x)

[Out]

(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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \tan ^{2}{\left (\frac {x}{2} + \frac {\pi }{4} \right )} \sec ^{3}{\left (\frac {x}{2} + \frac {\pi }{4} \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(1/4*pi+1/2*x)**3*tan(1/4*pi+1/2*x)**2,x)

[Out]

Integral(tan(x/2 + pi/4)**2*sec(x/2 + pi/4)**3, x)

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