3.591 \(\int \frac {x^2}{(a x \cos (a x)-\sin (a x))^2} \, dx\)

Optimal. Leaf size=35 \[ \frac {x \csc (a x)}{a^2 (a x \cos (a x)-\sin (a x))}-\frac {\cot (a x)}{a^3} \]

[Out]

-cot(a*x)/a^3+x*csc(a*x)/a^2/(a*x*cos(a*x)-sin(a*x))

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Rubi [A]  time = 0.04, antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {4594, 3767, 8} \[ \frac {x \csc (a x)}{a^2 (a x \cos (a x)-\sin (a x))}-\frac {\cot (a x)}{a^3} \]

Antiderivative was successfully verified.

[In]

Int[x^2/(a*x*Cos[a*x] - Sin[a*x])^2,x]

[Out]

-(Cot[a*x]/a^3) + (x*Csc[a*x])/(a^2*(a*x*Cos[a*x] - Sin[a*x]))

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

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]

Rule 4594

Int[(x_)^2/(Cos[(a_.)*(x_)]*(d_.)*(x_) + (c_.)*Sin[(a_.)*(x_)])^2, x_Symbol] :> Simp[x/(a*d*Sin[a*x]*(c*Sin[a*
x] + d*x*Cos[a*x])), x] + Dist[1/d^2, Int[1/Sin[a*x]^2, x], x] /; FreeQ[{a, c, d}, x] && EqQ[a*c + d, 0]

Rubi steps

\begin {align*} \int \frac {x^2}{(a x \cos (a x)-\sin (a x))^2} \, dx &=\frac {x \csc (a x)}{a^2 (a x \cos (a x)-\sin (a x))}+\frac {\int \csc ^2(a x) \, dx}{a^2}\\ &=\frac {x \csc (a x)}{a^2 (a x \cos (a x)-\sin (a x))}-\frac {\operatorname {Subst}(\int 1 \, dx,x,\cot (a x))}{a^3}\\ &=-\frac {\cot (a x)}{a^3}+\frac {x \csc (a x)}{a^2 (a x \cos (a x)-\sin (a x))}\\ \end {align*}

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Mathematica [A]  time = 0.46, size = 32, normalized size = 0.91 \[ \frac {a x \sin (a x)+\cos (a x)}{a^3 (a x \cos (a x)-\sin (a x))} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2/(a*x*Cos[a*x] - Sin[a*x])^2,x]

[Out]

(Cos[a*x] + a*x*Sin[a*x])/(a^3*(a*x*Cos[a*x] - Sin[a*x]))

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fricas [A]  time = 0.82, size = 34, normalized size = 0.97 \[ \frac {a x \sin \left (a x\right ) + \cos \left (a x\right )}{a^{4} x \cos \left (a x\right ) - a^{3} \sin \left (a x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(a*x*cos(a*x)-sin(a*x))^2,x, algorithm="fricas")

[Out]

(a*x*sin(a*x) + cos(a*x))/(a^4*x*cos(a*x) - a^3*sin(a*x))

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giac [A]  time = 0.15, size = 53, normalized size = 1.51 \[ -\frac {2 \, a x \tan \left (\frac {1}{2} \, a x\right ) - \tan \left (\frac {1}{2} \, a x\right )^{2} + 1}{a^{4} x \tan \left (\frac {1}{2} \, a x\right )^{2} - a^{4} x + 2 \, a^{3} \tan \left (\frac {1}{2} \, a x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(a*x*cos(a*x)-sin(a*x))^2,x, algorithm="giac")

[Out]

-(2*a*x*tan(1/2*a*x) - tan(1/2*a*x)^2 + 1)/(a^4*x*tan(1/2*a*x)^2 - a^4*x + 2*a^3*tan(1/2*a*x))

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maple [A]  time = 1.23, size = 54, normalized size = 1.54 \[ \frac {\frac {\tan ^{2}\left (\frac {a x}{2}\right )}{a^{3}}-\frac {1}{a^{3}}-\frac {2 x \tan \left (\frac {a x}{2}\right )}{a^{2}}}{a x \left (\tan ^{2}\left (\frac {a x}{2}\right )\right )-a x +2 \tan \left (\frac {a x}{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(a*x*cos(a*x)-sin(a*x))^2,x)

[Out]

(1/a^3*tan(1/2*a*x)^2-1/a^3-2*x/a^2*tan(1/2*a*x))/(a*x*tan(1/2*a*x)^2-a*x+2*tan(1/2*a*x))

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maxima [B]  time = 0.31, size = 100, normalized size = 2.86 \[ \frac {2 \, {\left (2 \, a x \cos \left (2 \, a x\right ) + {\left (a^{2} x^{2} - 1\right )} \sin \left (2 \, a x\right )\right )}}{{\left (a^{2} x^{2} + {\left (a^{2} x^{2} + 1\right )} \cos \left (2 \, a x\right )^{2} - 4 \, a x \sin \left (2 \, a x\right ) + {\left (a^{2} x^{2} + 1\right )} \sin \left (2 \, a x\right )^{2} + 2 \, {\left (a^{2} x^{2} - 1\right )} \cos \left (2 \, a x\right ) + 1\right )} a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(a*x*cos(a*x)-sin(a*x))^2,x, algorithm="maxima")

[Out]

2*(2*a*x*cos(2*a*x) + (a^2*x^2 - 1)*sin(2*a*x))/((a^2*x^2 + (a^2*x^2 + 1)*cos(2*a*x)^2 - 4*a*x*sin(2*a*x) + (a
^2*x^2 + 1)*sin(2*a*x)^2 + 2*(a^2*x^2 - 1)*cos(2*a*x) + 1)*a^3)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \int \frac {x^2}{{\left (\sin \left (a\,x\right )-a\,x\,\cos \left (a\,x\right )\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(sin(a*x) - a*x*cos(a*x))^2,x)

[Out]

int(x^2/(sin(a*x) - a*x*cos(a*x))^2, x)

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sympy [B]  time = 5.05, size = 112, normalized size = 3.20 \[ - \frac {2 a x \tan {\left (\frac {a x}{2} \right )}}{a^{4} x \tan ^{2}{\left (\frac {a x}{2} \right )} - a^{4} x + 2 a^{3} \tan {\left (\frac {a x}{2} \right )}} + \frac {\tan ^{2}{\left (\frac {a x}{2} \right )}}{a^{4} x \tan ^{2}{\left (\frac {a x}{2} \right )} - a^{4} x + 2 a^{3} \tan {\left (\frac {a x}{2} \right )}} - \frac {1}{a^{4} x \tan ^{2}{\left (\frac {a x}{2} \right )} - a^{4} x + 2 a^{3} \tan {\left (\frac {a x}{2} \right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2/(a*x*cos(a*x)-sin(a*x))**2,x)

[Out]

-2*a*x*tan(a*x/2)/(a**4*x*tan(a*x/2)**2 - a**4*x + 2*a**3*tan(a*x/2)) + tan(a*x/2)**2/(a**4*x*tan(a*x/2)**2 -
a**4*x + 2*a**3*tan(a*x/2)) - 1/(a**4*x*tan(a*x/2)**2 - a**4*x + 2*a**3*tan(a*x/2))

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