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

Optimal. Leaf size=32 \[ \frac {1}{2} \log \left (\tan \left (2 x+\frac {\pi }{4}\right )\right )-\frac {1}{4} \cot ^2\left (2 x+\frac {\pi }{4}\right ) \]

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Rubi [A]  time = 0.02, antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {2620, 14} \[ \frac {1}{2} \log \left (\tan \left (2 x+\frac {\pi }{4}\right )\right )-\frac {1}{4} \cot ^2\left (2 x+\frac {\pi }{4}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-Cot[Pi/4 + 2*x]^2/4 + Log[Tan[Pi/4 + 2*x]]/2

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2620

Int[csc[(e_.) + (f_.)*(x_)]^(m_.)*sec[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(1 + x^2)^((
m + n)/2 - 1)/x^m, x], x, Tan[e + f*x]], x] /; FreeQ[{e, f}, x] && IntegersQ[m, n, (m + n)/2]

Rubi steps

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

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Mathematica [A]  time = 0.07, size = 45, normalized size = 1.41 \[ \frac {1}{4} \left (-\csc ^2\left (2 x+\frac {\pi }{4}\right )+2 \log \left (\sin \left (2 x+\frac {\pi }{4}\right )\right )-2 \log \left (\cos \left (\frac {1}{4} (8 x+\pi )\right )\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-Csc[Pi/4 + 2*x]^2 - 2*Log[Cos[(Pi + 8*x)/4]] + 2*Log[Sin[Pi/4 + 2*x]])/4

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

Verification is Not applicable to the result.

[In]

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

[Out]

Could not integrate

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fricas [B]  time = 0.86, size = 71, normalized size = 2.22 \[ -\frac {{\left (\cos \left (\frac {1}{4} \, \pi + 2 \, x\right )^{2} - 1\right )} \log \left (\cos \left (\frac {1}{4} \, \pi + 2 \, x\right )^{2}\right ) - {\left (\cos \left (\frac {1}{4} \, \pi + 2 \, x\right )^{2} - 1\right )} \log \left (-\frac {1}{4} \, \cos \left (\frac {1}{4} \, \pi + 2 \, x\right )^{2} + \frac {1}{4}\right ) - 1}{4 \, {\left (\cos \left (\frac {1}{4} \, \pi + 2 \, x\right )^{2} - 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*((cos(1/4*pi + 2*x)^2 - 1)*log(cos(1/4*pi + 2*x)^2) - (cos(1/4*pi + 2*x)^2 - 1)*log(-1/4*cos(1/4*pi + 2*x
)^2 + 1/4) - 1)/(cos(1/4*pi + 2*x)^2 - 1)

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giac [B]  time = 0.75, size = 132, normalized size = 4.12 \[ -\frac {{\left (\frac {4 \, {\left (\cos \left (\frac {1}{4} \, \pi + 2 \, x\right ) - 1\right )}}{\cos \left (\frac {1}{4} \, \pi + 2 \, x\right ) + 1} - 1\right )} {\left (\cos \left (\frac {1}{4} \, \pi + 2 \, x\right ) + 1\right )}}{16 \, {\left (\cos \left (\frac {1}{4} \, \pi + 2 \, x\right ) - 1\right )}} + \frac {\cos \left (\frac {1}{4} \, \pi + 2 \, x\right ) - 1}{16 \, {\left (\cos \left (\frac {1}{4} \, \pi + 2 \, x\right ) + 1\right )}} + \frac {1}{4} \, \log \left (-\frac {\cos \left (\frac {1}{4} \, \pi + 2 \, x\right ) - 1}{\cos \left (\frac {1}{4} \, \pi + 2 \, x\right ) + 1}\right ) - \frac {1}{2} \, \log \left ({\left | -\frac {\cos \left (\frac {1}{4} \, \pi + 2 \, x\right ) - 1}{\cos \left (\frac {1}{4} \, \pi + 2 \, x\right ) + 1} - 1 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(4*(cos(1/4*pi + 2*x) - 1)/(cos(1/4*pi + 2*x) + 1) - 1)*(cos(1/4*pi + 2*x) + 1)/(cos(1/4*pi + 2*x) - 1)
+ 1/16*(cos(1/4*pi + 2*x) - 1)/(cos(1/4*pi + 2*x) + 1) + 1/4*log(-(cos(1/4*pi + 2*x) - 1)/(cos(1/4*pi + 2*x) +
1)) - 1/2*log(abs(-(cos(1/4*pi + 2*x) - 1)/(cos(1/4*pi + 2*x) + 1) - 1))

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maple [A]  time = 0.09, size = 25, normalized size = 0.78

method	result	size
derivativedivides	\(-\frac {1}{4 \sin \left (\frac {\pi }{4}+2 x \right )^{2}}+\frac {\ln \left (\tan \left (\frac {\pi }{4}+2 x \right )\right )}{2}\)	\(25\)
default	\(-\frac {1}{4 \sin \left (\frac {\pi }{4}+2 x \right )^{2}}+\frac {\ln \left (\tan \left (\frac {\pi }{4}+2 x \right )\right )}{2}\)	\(25\)
risch	\(\frac {i {\mathrm e}^{4 i x}}{\left (i {\mathrm e}^{4 i x}-1\right )^{2}}+\frac {\ln \left (i {\mathrm e}^{4 i x}-1\right )}{2}-\frac {\ln \left (i {\mathrm e}^{4 i x}+1\right )}{2}\)	\(48\)
norman	\(\frac {-\frac {1}{16}-\frac {\left (\tan ^{4}\left (\frac {\pi }{8}+x \right )\right )}{16}}{\tan \left (\frac {\pi }{8}+x \right )^{2}}+\frac {\ln \left (\tan \left (\frac {\pi }{8}+x \right )\right )}{2}-\frac {\ln \left (\tan \left (\frac {\pi }{8}+x \right )-1\right )}{2}-\frac {\ln \left (\tan \left (\frac {\pi }{8}+x \right )+1\right )}{2}\)	\(54\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4/sin(1/4*Pi+2*x)^2+1/2*ln(tan(1/4*Pi+2*x))

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maxima [A]  time = 0.47, size = 41, normalized size = 1.28 \[ -\frac {1}{4 \, \sin \left (\frac {1}{4} \, \pi + 2 \, x\right )^{2}} - \frac {1}{4} \, \log \left (\sin \left (\frac {1}{4} \, \pi + 2 \, x\right )^{2} - 1\right ) + \frac {1}{4} \, \log \left (\sin \left (\frac {1}{4} \, \pi + 2 \, x\right )^{2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4/sin(1/4*pi + 2*x)^2 - 1/4*log(sin(1/4*pi + 2*x)^2 - 1) + 1/4*log(sin(1/4*pi + 2*x)^2)

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mupad [B]  time = 0.28, size = 24, normalized size = 0.75 \[ \frac {\ln \left (\mathrm {tan}\left (\frac {\Pi }{4}+2\,x\right )\right )}{2}-\frac {1}{4\,{\sin \left (\frac {\Pi }{4}+2\,x\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(Pi/4 + 2*x)*sin(Pi/4 + 2*x)^3),x)

[Out]

log(tan(Pi/4 + 2*x))/2 - 1/(4*sin(Pi/4 + 2*x)^2)

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sympy [B]  time = 1.66, size = 54, normalized size = 1.69 \[ - \frac {\log {\left (\tan {\left (x + \frac {\pi }{8} \right )} - 1 \right )}}{2} - \frac {\log {\left (\tan {\left (x + \frac {\pi }{8} \right )} + 1 \right )}}{2} + \frac {\log {\left (\tan {\left (x + \frac {\pi }{8} \right )} \right )}}{2} - \frac {\tan ^{2}{\left (x + \frac {\pi }{8} \right )}}{16} - \frac {1}{16 \tan ^{2}{\left (x + \frac {\pi }{8} \right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/cos(1/4*pi+2*x)/sin(1/4*pi+2*x)**3,x)

[Out]

-log(tan(x + pi/8) - 1)/2 - log(tan(x + pi/8) + 1)/2 + log(tan(x + pi/8))/2 - tan(x + pi/8)**2/16 - 1/(16*tan(
x + pi/8)**2)

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