3.587 \(\int \frac{\sin ^4(a x)}{x^2 (a x \cos (a x)-\sin (a x))^2} \, dx\)
Optimal. Leaf size=80 \[ \frac{\sin ^2(a x)}{a^2 x^3}+\frac{\sin ^3(a x)}{a^2 x^3 (a x \cos (a x)-\sin (a x))}+2 a \text{Si}(2 a x)+\frac{\sin (a x) \cos (a x)}{a x^2}-\frac{2 \sin ^2(a x)}{x}+\frac{1}{x} \]
[Out]
x^(-1) + (Cos[a*x]*Sin[a*x])/(a*x^2) + Sin[a*x]^2/(a^2*x^3) - (2*Sin[a*x]^2)/x + Sin[a*x]^3/(a^2*x^3*(a*x*Cos[
a*x] - Sin[a*x])) + 2*a*SinIntegral[2*a*x]
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Rubi [A] time = 0.130805, antiderivative size = 80, normalized size of antiderivative = 1.,
number of steps used = 6, number of rules used = 6, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used =
{4598, 3314, 30, 3313, 12, 3299} \[ \frac{\sin ^2(a x)}{a^2 x^3}+\frac{\sin ^3(a x)}{a^2 x^3 (a x \cos (a x)-\sin (a x))}+2 a \text{Si}(2 a x)+\frac{\sin (a x) \cos (a x)}{a x^2}-\frac{2 \sin ^2(a x)}{x}+\frac{1}{x} \]
Antiderivative was successfully verified.
[In]
Int[Sin[a*x]^4/(x^2*(a*x*Cos[a*x] - Sin[a*x])^2),x]
[Out]
x^(-1) + (Cos[a*x]*Sin[a*x])/(a*x^2) + Sin[a*x]^2/(a^2*x^3) - (2*Sin[a*x]^2)/x + Sin[a*x]^3/(a^2*x^3*(a*x*Cos[
a*x] - Sin[a*x])) + 2*a*SinIntegral[2*a*x]
Rule 4598
Int[(((b_.)*(x_))^(m_)*Sin[(a_.)*(x_)]^(n_))/(Cos[(a_.)*(x_)]*(d_.)*(x_) + (c_.)*Sin[(a_.)*(x_)])^2, x_Symbol]
:> Simp[(b*(b*x)^(m - 1)*Sin[a*x]^(n - 1))/(a*d*(c*Sin[a*x] + d*x*Cos[a*x])), x] - Dist[(b^2*(n - 1))/d^2, In
t[(b*x)^(m - 2)*Sin[a*x]^(n - 2), x], x] /; FreeQ[{a, b, c, d, m, n}, x] && EqQ[a*c + d, 0] && EqQ[m, 2 - n]
Rule 3314
Int[((c_.) + (d_.)*(x_))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((c + d*x)^(m + 1)*(b*Si
n[e + f*x])^n)/(d*(m + 1)), x] + (Dist[(b^2*f^2*n*(n - 1))/(d^2*(m + 1)*(m + 2)), Int[(c + d*x)^(m + 2)*(b*Sin
[e + f*x])^(n - 2), x], x] - Dist[(f^2*n^2)/(d^2*(m + 1)*(m + 2)), Int[(c + d*x)^(m + 2)*(b*Sin[e + f*x])^n, x
], x] - Simp[(b*f*n*(c + d*x)^(m + 2)*Cos[e + f*x]*(b*Sin[e + f*x])^(n - 1))/(d^2*(m + 1)*(m + 2)), x]) /; Fre
eQ[{b, c, d, e, f}, x] && GtQ[n, 1] && LtQ[m, -2]
Rule 30
Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]
Rule 3313
Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Simp[((c + d*x)^(m + 1)*Sin[e + f*x]^
n)/(d*(m + 1)), x] - Dist[(f*n)/(d*(m + 1)), Int[ExpandTrigReduce[(c + d*x)^(m + 1), Cos[e + f*x]*Sin[e + f*x]
^(n - 1), x], x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && GeQ[m, -2] && LtQ[m, -1]
Rule 12
Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] && !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]
Rule 3299
Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[SinIntegral[e + f*x]/d, x] /; FreeQ[{c, d,
e, f}, x] && EqQ[d*e - c*f, 0]
Rubi steps
\begin{align*} \int \frac{\sin ^4(a x)}{x^2 (a x \cos (a x)-\sin (a x))^2} \, dx &=\frac{\sin ^3(a x)}{a^2 x^3 (a x \cos (a x)-\sin (a x))}-\frac{3 \int \frac{\sin ^2(a x)}{x^4} \, dx}{a^2}\\ &=\frac{\cos (a x) \sin (a x)}{a x^2}+\frac{\sin ^2(a x)}{a^2 x^3}+\frac{\sin ^3(a x)}{a^2 x^3 (a x \cos (a x)-\sin (a x))}+2 \int \frac{\sin ^2(a x)}{x^2} \, dx-\int \frac{1}{x^2} \, dx\\ &=\frac{1}{x}+\frac{\cos (a x) \sin (a x)}{a x^2}+\frac{\sin ^2(a x)}{a^2 x^3}-\frac{2 \sin ^2(a x)}{x}+\frac{\sin ^3(a x)}{a^2 x^3 (a x \cos (a x)-\sin (a x))}+(4 a) \int \frac{\sin (2 a x)}{2 x} \, dx\\ &=\frac{1}{x}+\frac{\cos (a x) \sin (a x)}{a x^2}+\frac{\sin ^2(a x)}{a^2 x^3}-\frac{2 \sin ^2(a x)}{x}+\frac{\sin ^3(a x)}{a^2 x^3 (a x \cos (a x)-\sin (a x))}+(2 a) \int \frac{\sin (2 a x)}{x} \, dx\\ &=\frac{1}{x}+\frac{\cos (a x) \sin (a x)}{a x^2}+\frac{\sin ^2(a x)}{a^2 x^3}-\frac{2 \sin ^2(a x)}{x}+\frac{\sin ^3(a x)}{a^2 x^3 (a x \cos (a x)-\sin (a x))}+2 a \text{Si}(2 a x)\\ \end{align*}
Mathematica [A] time = 0.85417, size = 77, normalized size = 0.96 \[ \frac{8 a x \text{Si}(2 a x) (a x \cos (a x)-\sin (a x))+3 \sin (a x)-\sin (3 a x)+2 a x \cos (a x)+2 a x \cos (3 a x)}{4 x (a x \cos (a x)-\sin (a x))} \]
Antiderivative was successfully verified.
[In]
Integrate[Sin[a*x]^4/(x^2*(a*x*Cos[a*x] - Sin[a*x])^2),x]
[Out]
(2*a*x*Cos[a*x] + 2*a*x*Cos[3*a*x] + 3*Sin[a*x] - Sin[3*a*x] + 8*a*x*(a*x*Cos[a*x] - Sin[a*x])*SinIntegral[2*a
*x])/(4*x*(a*x*Cos[a*x] - Sin[a*x]))
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Maple [F] time = 180., size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( \sin \left ( ax \right ) \right ) ^{4}}{{x}^{2} \left ( ax\cos \left ( ax \right ) -\sin \left ( ax \right ) \right ) ^{2}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(sin(a*x)^4/x^2/(a*x*cos(a*x)-sin(a*x))^2,x)
[Out]
int(sin(a*x)^4/x^2/(a*x*cos(a*x)-sin(a*x))^2,x)
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sin(a*x)^4/x^2/(a*x*cos(a*x)-sin(a*x))^2,x, algorithm="maxima")
[Out]
Exception raised: RuntimeError
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Fricas [A] time = 2.22975, size = 209, normalized size = 2.61 \begin{align*} \frac{2 \, a x \cos \left (a x\right )^{3} +{\left (2 \, a^{2} x^{2} \operatorname{Si}\left (2 \, a x\right ) - a x\right )} \cos \left (a x\right ) -{\left (2 \, a x \operatorname{Si}\left (2 \, a x\right ) + \cos \left (a x\right )^{2} - 1\right )} \sin \left (a x\right )}{a x^{2} \cos \left (a x\right ) - x \sin \left (a x\right )} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sin(a*x)^4/x^2/(a*x*cos(a*x)-sin(a*x))^2,x, algorithm="fricas")
[Out]
(2*a*x*cos(a*x)^3 + (2*a^2*x^2*sin_integral(2*a*x) - a*x)*cos(a*x) - (2*a*x*sin_integral(2*a*x) + cos(a*x)^2 -
1)*sin(a*x))/(a*x^2*cos(a*x) - x*sin(a*x))
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin ^{4}{\left (a x \right )}}{x^{2} \left (a x \cos{\left (a x \right )} - \sin{\left (a x \right )}\right )^{2}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sin(a*x)**4/x**2/(a*x*cos(a*x)-sin(a*x))**2,x)
[Out]
Integral(sin(a*x)**4/(x**2*(a*x*cos(a*x) - sin(a*x))**2), x)
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Giac [C] time = 1.50151, size = 1395, normalized size = 17.44 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sin(a*x)^4/x^2/(a*x*cos(a*x)-sin(a*x))^2,x, algorithm="giac")
[Out]
(a^4*x^4*imag_part(cos_integral(2*a*x))*tan(a*x)^2*tan(1/2*a*x)^2 - a^4*x^4*imag_part(cos_integral(-2*a*x))*ta
n(a*x)^2*tan(1/2*a*x)^2 + 2*a^4*x^4*sin_integral(2*a*x)*tan(a*x)^2*tan(1/2*a*x)^2 - a^4*x^4*imag_part(cos_inte
gral(2*a*x))*tan(a*x)^2 + a^4*x^4*imag_part(cos_integral(-2*a*x))*tan(a*x)^2 - 2*a^4*x^4*sin_integral(2*a*x)*t
an(a*x)^2 + a^4*x^4*imag_part(cos_integral(2*a*x))*tan(1/2*a*x)^2 - a^4*x^4*imag_part(cos_integral(-2*a*x))*ta
n(1/2*a*x)^2 + 2*a^4*x^4*sin_integral(2*a*x)*tan(1/2*a*x)^2 + 2*a^3*x^3*imag_part(cos_integral(2*a*x))*tan(a*x
)^2*tan(1/2*a*x) - 2*a^3*x^3*imag_part(cos_integral(-2*a*x))*tan(a*x)^2*tan(1/2*a*x) + 4*a^3*x^3*sin_integral(
2*a*x)*tan(a*x)^2*tan(1/2*a*x) - a^3*x^3*tan(a*x)^2*tan(1/2*a*x)^2 - a^4*x^4*imag_part(cos_integral(2*a*x)) +
a^4*x^4*imag_part(cos_integral(-2*a*x)) - 2*a^4*x^4*sin_integral(2*a*x) + a^2*x^2*imag_part(cos_integral(2*a*x
))*tan(a*x)^2*tan(1/2*a*x)^2 - a^2*x^2*imag_part(cos_integral(-2*a*x))*tan(a*x)^2*tan(1/2*a*x)^2 + 2*a^2*x^2*s
in_integral(2*a*x)*tan(a*x)^2*tan(1/2*a*x)^2 + a^3*x^3*tan(a*x)^2 + 2*a^3*x^3*imag_part(cos_integral(2*a*x))*t
an(1/2*a*x) - 2*a^3*x^3*imag_part(cos_integral(-2*a*x))*tan(1/2*a*x) + 4*a^3*x^3*sin_integral(2*a*x)*tan(1/2*a
*x) + a^3*x^3*tan(1/2*a*x)^2 - a^2*x^2*imag_part(cos_integral(2*a*x))*tan(a*x)^2 + a^2*x^2*imag_part(cos_integ
ral(-2*a*x))*tan(a*x)^2 - 2*a^2*x^2*sin_integral(2*a*x)*tan(a*x)^2 - 2*a^2*x^2*tan(a*x)^2*tan(1/2*a*x) + a^2*x
^2*imag_part(cos_integral(2*a*x))*tan(1/2*a*x)^2 - a^2*x^2*imag_part(cos_integral(-2*a*x))*tan(1/2*a*x)^2 + 2*
a^2*x^2*sin_integral(2*a*x)*tan(1/2*a*x)^2 + a^2*x^2*tan(a*x)*tan(1/2*a*x)^2 - a^3*x^3 + 2*a*x*imag_part(cos_i
ntegral(2*a*x))*tan(a*x)^2*tan(1/2*a*x) - 2*a*x*imag_part(cos_integral(-2*a*x))*tan(a*x)^2*tan(1/2*a*x) + 4*a*
x*sin_integral(2*a*x)*tan(a*x)^2*tan(1/2*a*x) - a^2*x^2*imag_part(cos_integral(2*a*x)) + a^2*x^2*imag_part(cos
_integral(-2*a*x)) - 2*a^2*x^2*sin_integral(2*a*x) - a^2*x^2*tan(a*x) + 2*a^2*x^2*tan(1/2*a*x) + 2*a*x*imag_pa
rt(cos_integral(2*a*x))*tan(1/2*a*x) - 2*a*x*imag_part(cos_integral(-2*a*x))*tan(1/2*a*x) + 4*a*x*sin_integral
(2*a*x)*tan(1/2*a*x) + 2*a*x*tan(a*x)*tan(1/2*a*x) + a*x*tan(1/2*a*x)^2 - 2*tan(a*x)^2*tan(1/2*a*x) - a*x)/(a^
3*x^4*tan(a*x)^2*tan(1/2*a*x)^2 - a^3*x^4*tan(a*x)^2 + a^3*x^4*tan(1/2*a*x)^2 + 2*a^2*x^3*tan(a*x)^2*tan(1/2*a
*x) - a^3*x^4 + a*x^2*tan(a*x)^2*tan(1/2*a*x)^2 + 2*a^2*x^3*tan(1/2*a*x) - a*x^2*tan(a*x)^2 + a*x^2*tan(1/2*a*
x)^2 + 2*x*tan(a*x)^2*tan(1/2*a*x) - a*x^2 + 2*x*tan(1/2*a*x))