3.347 \(\int \csc ^6(x) \sec ^6(x) \, dx\)

Optimal. Leaf size=41 \[ \frac {\tan ^5(x)}{5}+\frac {5 \tan ^3(x)}{3}+10 \tan (x)-\frac {1}{5} \cot ^5(x)-\frac {5 \cot ^3(x)}{3}-10 \cot (x) \]

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Rubi [A]  time = 0.03, antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2620, 270} \[ \frac {\tan ^5(x)}{5}+\frac {5 \tan ^3(x)}{3}+10 \tan (x)-\frac {1}{5} \cot ^5(x)-\frac {5 \cot ^3(x)}{3}-10 \cot (x) \]

Antiderivative was successfully verified.

[In]

Int[Csc[x]^6*Sec[x]^6,x]

[Out]

-10*Cot[x] - (5*Cot[x]^3)/3 - Cot[x]^5/5 + 10*Tan[x] + (5*Tan[x]^3)/3 + Tan[x]^5/5

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

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 ^6(x) \sec ^6(x) \, dx &=\operatorname {Subst}\left (\int \frac {\left (1+x^2\right )^5}{x^6} \, dx,x,\tan (x)\right )\\ &=\operatorname {Subst}\left (\int \left (10+\frac {1}{x^6}+\frac {5}{x^4}+\frac {10}{x^2}+5 x^2+x^4\right ) \, dx,x,\tan (x)\right )\\ &=-10 \cot (x)-\frac {5 \cot ^3(x)}{3}-\frac {\cot ^5(x)}{5}+10 \tan (x)+\frac {5 \tan ^3(x)}{3}+\frac {\tan ^5(x)}{5}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 53, normalized size = 1.29 \[ \frac {128 \tan (x)}{15}-\frac {128 \cot (x)}{15}-\frac {1}{5} \cot (x) \csc ^4(x)-\frac {19}{15} \cot (x) \csc ^2(x)+\frac {1}{5} \tan (x) \sec ^4(x)+\frac {19}{15} \tan (x) \sec ^2(x) \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[x]^6*Sec[x]^6,x]

[Out]

(-128*Cot[x])/15 - (19*Cot[x]*Csc[x]^2)/15 - (Cot[x]*Csc[x]^4)/5 + (128*Tan[x])/15 + (19*Sec[x]^2*Tan[x])/15 +
(Sec[x]^4*Tan[x])/5

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \csc ^6(x) \sec ^6(x) \, dx \]

Verification is Not applicable to the result.

[In]

IntegrateAlgebraic[Csc[x]^6*Sec[x]^6,x]

[Out]

Could not integrate

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fricas [A]  time = 0.63, size = 55, normalized size = 1.34 \[ -\frac {256 \, \cos \relax (x)^{10} - 640 \, \cos \relax (x)^{8} + 480 \, \cos \relax (x)^{6} - 80 \, \cos \relax (x)^{4} - 10 \, \cos \relax (x)^{2} - 3}{15 \, {\left (\cos \relax (x)^{9} - 2 \, \cos \relax (x)^{7} + \cos \relax (x)^{5}\right )} \sin \relax (x)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/cos(x)^6/sin(x)^6,x, algorithm="fricas")

[Out]

-1/15*(256*cos(x)^10 - 640*cos(x)^8 + 480*cos(x)^6 - 80*cos(x)^4 - 10*cos(x)^2 - 3)/((cos(x)^9 - 2*cos(x)^7 +
cos(x)^5)*sin(x))

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giac [A]  time = 0.83, size = 26, normalized size = 0.63 \[ -\frac {32 \, {\left (15 \, \tan \left (2 \, x\right )^{4} + 10 \, \tan \left (2 \, x\right )^{2} + 3\right )}}{15 \, \tan \left (2 \, x\right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/cos(x)^6/sin(x)^6,x, algorithm="giac")

[Out]

-32/15*(15*tan(2*x)^4 + 10*tan(2*x)^2 + 3)/tan(2*x)^5

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maple [C]  time = 0.33, size = 38, normalized size = 0.93

method	result	size
risch	\(-\frac {512 i \left (10 \,{\mathrm e}^{8 i x}-5 \,{\mathrm e}^{4 i x}+1\right )}{15 \left ({\mathrm e}^{2 i x}-1\right )^{5} \left (1+{\mathrm e}^{2 i x}\right )^{5}}\)	\(38\)
default	\(\frac {1}{5 \sin \relax (x )^{5} \cos \relax (x )^{5}}-\frac {2}{5 \sin \relax (x )^{5} \cos \relax (x )^{3}}+\frac {16}{15 \sin \relax (x )^{3} \cos \relax (x )^{3}}-\frac {32}{15 \sin \relax (x )^{3} \cos \relax (x )}+\frac {128}{15 \sin \relax (x ) \cos \relax (x )}-\frac {256 \cot \relax (x )}{15}\)	\(56\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/cos(x)^6/sin(x)^6,x,method=_RETURNVERBOSE)

[Out]

-512/15*I*(10*exp(8*I*x)-5*exp(4*I*x)+1)/(exp(2*I*x)-1)^5/(1+exp(2*I*x))^5

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maxima [A]  time = 0.49, size = 37, normalized size = 0.90 \[ \frac {1}{5} \, \tan \relax (x)^{5} + \frac {5}{3} \, \tan \relax (x)^{3} - \frac {150 \, \tan \relax (x)^{4} + 25 \, \tan \relax (x)^{2} + 3}{15 \, \tan \relax (x)^{5}} + 10 \, \tan \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/cos(x)^6/sin(x)^6,x, algorithm="maxima")

[Out]

1/5*tan(x)^5 + 5/3*tan(x)^3 - 1/15*(150*tan(x)^4 + 25*tan(x)^2 + 3)/tan(x)^5 + 10*tan(x)

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mupad [B]  time = 0.12, size = 27, normalized size = 0.66 \[ -\frac {32\,\left (\frac {\cos \left (2\,x\right )}{3}-\frac {\cos \left (6\,x\right )}{6}+\frac {\cos \left (10\,x\right )}{30}\right )}{{\sin \left (2\,x\right )}^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(x)^6*sin(x)^6),x)

[Out]

-(32*(cos(2*x)/3 - cos(6*x)/6 + cos(10*x)/30))/sin(2*x)^5

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sympy [A]  time = 0.08, size = 44, normalized size = 1.07 \[ - \frac {256 \cos {\left (2 x \right )}}{15 \sin {\left (2 x \right )}} - \frac {128 \cos {\left (2 x \right )}}{15 \sin ^{3}{\left (2 x \right )}} - \frac {32 \cos {\left (2 x \right )}}{5 \sin ^{5}{\left (2 x \right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/cos(x)**6/sin(x)**6,x)

[Out]

-256*cos(2*x)/(15*sin(2*x)) - 128*cos(2*x)/(15*sin(2*x)**3) - 32*cos(2*x)/(5*sin(2*x)**5)

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