3.6 \(\int \csc ^6(a+b x) \, dx\)

Optimal. Leaf size=42 \[ -\frac {\cot ^5(a+b x)}{5 b}-\frac {2 \cot ^3(a+b x)}{3 b}-\frac {\cot (a+b x)}{b} \]

[Out]

-cot(b*x+a)/b-2/3*cot(b*x+a)^3/b-1/5*cot(b*x+a)^5/b

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Rubi [A]  time = 0.01, antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {3767} \[ -\frac {\cot ^5(a+b x)}{5 b}-\frac {2 \cot ^3(a+b x)}{3 b}-\frac {\cot (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[Csc[a + b*x]^6,x]

[Out]

-(Cot[a + b*x]/b) - (2*Cot[a + b*x]^3)/(3*b) - Cot[a + b*x]^5/(5*b)

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 56, normalized size = 1.33 \[ -\frac {8 \cot (a+b x)}{15 b}-\frac {\cot (a+b x) \csc ^4(a+b x)}{5 b}-\frac {4 \cot (a+b x) \csc ^2(a+b x)}{15 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[a + b*x]^6,x]

[Out]

(-8*Cot[a + b*x])/(15*b) - (4*Cot[a + b*x]*Csc[a + b*x]^2)/(15*b) - (Cot[a + b*x]*Csc[a + b*x]^4)/(5*b)

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fricas [A]  time = 0.53, size = 64, normalized size = 1.52 \[ -\frac {8 \, \cos \left (b x + a\right )^{5} - 20 \, \cos \left (b x + a\right )^{3} + 15 \, \cos \left (b x + a\right )}{15 \, {\left (b \cos \left (b x + a\right )^{4} - 2 \, b \cos \left (b x + a\right )^{2} + b\right )} \sin \left (b x + a\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)^6,x, algorithm="fricas")

[Out]

-1/15*(8*cos(b*x + a)^5 - 20*cos(b*x + a)^3 + 15*cos(b*x + a))/((b*cos(b*x + a)^4 - 2*b*cos(b*x + a)^2 + b)*si
n(b*x + a))

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giac [A]  time = 0.19, size = 35, normalized size = 0.83 \[ -\frac {15 \, \tan \left (b x + a\right )^{4} + 10 \, \tan \left (b x + a\right )^{2} + 3}{15 \, b \tan \left (b x + a\right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)^6,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.79, size = 33, normalized size = 0.79 \[ \frac {\left (-\frac {8}{15}-\frac {\left (\csc ^{4}\left (b x +a \right )\right )}{5}-\frac {4 \left (\csc ^{2}\left (b x +a \right )\right )}{15}\right ) \cot \left (b x +a \right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(b*x+a)^6,x)

[Out]

1/b*(-8/15-1/5*csc(b*x+a)^4-4/15*csc(b*x+a)^2)*cot(b*x+a)

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maxima [A]  time = 0.35, size = 35, normalized size = 0.83 \[ -\frac {15 \, \tan \left (b x + a\right )^{4} + 10 \, \tan \left (b x + a\right )^{2} + 3}{15 \, b \tan \left (b x + a\right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)^6,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.10, size = 38, normalized size = 0.90 \[ -\frac {\mathrm {cot}\left (a+b\,x\right )}{b}-\frac {2\,{\mathrm {cot}\left (a+b\,x\right )}^3}{3\,b}-\frac {{\mathrm {cot}\left (a+b\,x\right )}^5}{5\,b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/sin(a + b*x)^6,x)

[Out]

- cot(a + b*x)/b - (2*cot(a + b*x)^3)/(3*b) - cot(a + b*x)^5/(5*b)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \csc ^{6}{\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)**6,x)

[Out]

Integral(csc(a + b*x)**6, x)

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