\(\int \frac {\cos ^4(a x)}{x^2 (\cos (a x)+a x \sin (a x))^2} \, dx\) [596]

Optimal result

Integrand size = 24, antiderivative size = 80 \[ \int \frac {\cos ^4(a x)}{x^2 (\cos (a x)+a x \sin (a x))^2} \, dx=\frac {1}{x}+\frac {\cos ^2(a x)}{a^2 x^3}-\frac {2 \cos ^2(a x)}{x}-\frac {\cos (a x) \sin (a x)}{a x^2}-\frac {\cos ^3(a x)}{a^2 x^3 (\cos (a x)+a x \sin (a x))}-2 a \text {Si}(2 a x) \]

[Out]

1/x+cos(a*x)^2/a^2/x^3-2*cos(a*x)^2/x-2*a*Si(2*a*x)-cos(a*x)*sin(a*x)/a/x^2-cos(a*x)^3/a^2/x^3/(cos(a*x)+a*x*s
in(a*x))

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {4695, 3395, 30, 3394, 12, 3380} \[ \int \frac {\cos ^4(a x)}{x^2 (\cos (a x)+a x \sin (a x))^2} \, dx=\frac {\cos ^2(a x)}{a^2 x^3}-\frac {\cos ^3(a x)}{a^2 x^3 (a x \sin (a x)+\cos (a x))}-2 a \text {Si}(2 a x)-\frac {\sin (a x) \cos (a x)}{a x^2}-\frac {2 \cos ^2(a x)}{x}+\frac {1}{x} \]

[In]

Int[Cos[a*x]^4/(x^2*(Cos[a*x] + a*x*Sin[a*x])^2),x]

[Out]

x^(-1) + Cos[a*x]^2/(a^2*x^3) - (2*Cos[a*x]^2)/x - (Cos[a*x]*Sin[a*x])/(a*x^2) - Cos[a*x]^3/(a^2*x^3*(Cos[a*x]
+ a*x*Sin[a*x])) - 2*a*SinIntegral[2*a*x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 3380

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]

Rule 3394

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 3395

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 4695

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

Rubi steps \begin{align*} \text {integral}& = -\frac {\cos ^3(a x)}{a^2 x^3 (\cos (a x)+a x \sin (a x))}-\frac {3 \int \frac {\cos ^2(a x)}{x^4} \, dx}{a^2} \\ & = \frac {\cos ^2(a x)}{a^2 x^3}-\frac {\cos (a x) \sin (a x)}{a x^2}-\frac {\cos ^3(a x)}{a^2 x^3 (\cos (a x)+a x \sin (a x))}+2 \int \frac {\cos ^2(a x)}{x^2} \, dx-\int \frac {1}{x^2} \, dx \\ & = \frac {1}{x}+\frac {\cos ^2(a x)}{a^2 x^3}-\frac {2 \cos ^2(a x)}{x}-\frac {\cos (a x) \sin (a x)}{a x^2}-\frac {\cos ^3(a x)}{a^2 x^3 (\cos (a x)+a x \sin (a x))}+(4 a) \int -\frac {\sin (2 a x)}{2 x} \, dx \\ & = \frac {1}{x}+\frac {\cos ^2(a x)}{a^2 x^3}-\frac {2 \cos ^2(a x)}{x}-\frac {\cos (a x) \sin (a x)}{a x^2}-\frac {\cos ^3(a x)}{a^2 x^3 (\cos (a x)+a x \sin (a x))}-(2 a) \int \frac {\sin (2 a x)}{x} \, dx \\ & = \frac {1}{x}+\frac {\cos ^2(a x)}{a^2 x^3}-\frac {2 \cos ^2(a x)}{x}-\frac {\cos (a x) \sin (a x)}{a x^2}-\frac {\cos ^3(a x)}{a^2 x^3 (\cos (a x)+a x \sin (a x))}-2 a \text {Si}(2 a x) \\ \end{align*}

Mathematica [A] (verified)

Time = 1.01 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.89 \[ \int \frac {\cos ^4(a x)}{x^2 (\cos (a x)+a x \sin (a x))^2} \, dx=-\frac {3 \cos (a x)+\cos (3 a x)-2 a x \sin (a x)+2 a x \sin (3 a x)+8 a x (\cos (a x)+a x \sin (a x)) \text {Si}(2 a x)}{4 x (\cos (a x)+a x \sin (a x))} \]

[In]

Integrate[Cos[a*x]^4/(x^2*(Cos[a*x] + a*x*Sin[a*x])^2),x]

[Out]

-1/4*(3*Cos[a*x] + Cos[3*a*x] - 2*a*x*Sin[a*x] + 2*a*x*Sin[3*a*x] + 8*a*x*(Cos[a*x] + a*x*Sin[a*x])*SinIntegra
l[2*a*x])/(x*(Cos[a*x] + a*x*Sin[a*x]))

Maple [F(-1)]

Timed out.

\[\int \frac {\cos \left (a x \right )^{4}}{x^{2} \left (\cos \left (a x \right )+a x \sin \left (a x \right )\right )^{2}}d x\]

[In]

int(cos(a*x)^4/x^2/(cos(a*x)+a*x*sin(a*x))^2,x)

[Out]

int(cos(a*x)^4/x^2/(cos(a*x)+a*x*sin(a*x))^2,x)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.91 \[ \int \frac {\cos ^4(a x)}{x^2 (\cos (a x)+a x \sin (a x))^2} \, dx=-\frac {2 \, a x \cos \left (a x\right ) \operatorname {Si}\left (2 \, a x\right ) + \cos \left (a x\right )^{3} + {\left (2 \, a^{2} x^{2} \operatorname {Si}\left (2 \, a x\right ) + 2 \, a x \cos \left (a x\right )^{2} - a x\right )} \sin \left (a x\right )}{a x^{2} \sin \left (a x\right ) + x \cos \left (a x\right )} \]

[In]

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

[Out]

-(2*a*x*cos(a*x)*sin_integral(2*a*x) + cos(a*x)^3 + (2*a^2*x^2*sin_integral(2*a*x) + 2*a*x*cos(a*x)^2 - a*x)*s
in(a*x))/(a*x^2*sin(a*x) + x*cos(a*x))

Sympy [F]

\[ \int \frac {\cos ^4(a x)}{x^2 (\cos (a x)+a x \sin (a x))^2} \, dx=\int \frac {\cos ^{4}{\left (a x \right )}}{x^{2} \left (a x \sin {\left (a x \right )} + \cos {\left (a x \right )}\right )^{2}}\, dx \]

[In]

integrate(cos(a*x)**4/x**2/(cos(a*x)+a*x*sin(a*x))**2,x)

[Out]

Integral(cos(a*x)**4/(x**2*(a*x*sin(a*x) + cos(a*x))**2), x)

Maxima [F(-2)]

Exception generated. \[ \int \frac {\cos ^4(a x)}{x^2 (\cos (a x)+a x \sin (a x))^2} \, dx=\text {Exception raised: RuntimeError} \]

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is undefined.

Giac [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.40 (sec) , antiderivative size = 997, normalized size of antiderivative = 12.46 \[ \int \frac {\cos ^4(a x)}{x^2 (\cos (a x)+a x \sin (a x))^2} \, dx=\text {Too large to display} \]

[In]

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

[Out]

-(2*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))
*tan(a*x)^2*tan(1/2*a*x) + 4*a^4*x^4*sin_integral(2*a*x)*tan(a*x)^2*tan(1/2*a*x) - a^3*x^3*imag_part(cos_integ
ral(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)^2 - 2*
a^3*x^3*sin_integral(2*a*x)*tan(a*x)^2*tan(1/2*a*x)^2 + 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))*tan(1/2*a*x) + 4*a^4*x^4*sin_integral(2*a*x)*tan(1/2*a*x) + a^3*x^
3*imag_part(cos_integral(2*a*x))*tan(a*x)^2 - a^3*x^3*imag_part(cos_integral(-2*a*x))*tan(a*x)^2 + 2*a^3*x^3*s
in_integral(2*a*x)*tan(a*x)^2 - 2*a^3*x^3*tan(a*x)^2*tan(1/2*a*x) - a^3*x^3*imag_part(cos_integral(2*a*x))*tan
(1/2*a*x)^2 + a^3*x^3*imag_part(cos_integral(-2*a*x))*tan(1/2*a*x)^2 - 2*a^3*x^3*sin_integral(2*a*x)*tan(1/2*a
*x)^2 + 2*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) + 4*a^2*x^2*sin_integral(2*a*x)*tan(a*x)^2*tan(1/2*a*x) + a^2*x^2*tan(a*x)^2*ta
n(1/2*a*x)^2 + a^3*x^3*imag_part(cos_integral(2*a*x)) - a^3*x^3*imag_part(cos_integral(-2*a*x)) + 2*a^3*x^3*si
n_integral(2*a*x) + 2*a^3*x^3*tan(1/2*a*x) - a*x*imag_part(cos_integral(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)^2 - 2*a*x*sin_integral(2*a*x)*tan(a*x)^2*tan(1/2*a*x
)^2 - a^2*x^2*tan(a*x)^2 + 2*a^2*x^2*imag_part(cos_integral(2*a*x))*tan(1/2*a*x) - 2*a^2*x^2*imag_part(cos_int
egral(-2*a*x))*tan(1/2*a*x) + 4*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) - a
^2*x^2*tan(1/2*a*x)^2 + a*x*imag_part(cos_integral(2*a*x))*tan(a*x)^2 - a*x*imag_part(cos_integral(-2*a*x))*ta
n(a*x)^2 + 2*a*x*sin_integral(2*a*x)*tan(a*x)^2 - 2*a*x*tan(a*x)^2*tan(1/2*a*x) - a*x*imag_part(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)^2 - 2*a*x*sin_integral(2*a*x)*tan(1/2
*a*x)^2 - a*x*tan(a*x)*tan(1/2*a*x)^2 + a^2*x^2 + a*x*imag_part(cos_integral(2*a*x)) - a*x*imag_part(cos_integ
ral(-2*a*x)) + 2*a*x*sin_integral(2*a*x) + a*x*tan(a*x) - tan(1/2*a*x)^2 + 1)/(2*a^3*x^4*tan(a*x)^2*tan(1/2*a*
x) - a^2*x^3*tan(a*x)^2*tan(1/2*a*x)^2 + 2*a^3*x^4*tan(1/2*a*x) + a^2*x^3*tan(a*x)^2 - a^2*x^3*tan(1/2*a*x)^2
+ 2*a*x^2*tan(a*x)^2*tan(1/2*a*x) + a^2*x^3 - x*tan(a*x)^2*tan(1/2*a*x)^2 + 2*a*x^2*tan(1/2*a*x) + x*tan(a*x)^
2 - x*tan(1/2*a*x)^2 + x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\cos ^4(a x)}{x^2 (\cos (a x)+a x \sin (a x))^2} \, dx=\int \frac {{\cos \left (a\,x\right )}^4}{x^2\,{\left (\cos \left (a\,x\right )+a\,x\,\sin \left (a\,x\right )\right )}^2} \,d x \]

[In]

int(cos(a*x)^4/(x^2*(cos(a*x) + a*x*sin(a*x))^2),x)

[Out]

int(cos(a*x)^4/(x^2*(cos(a*x) + a*x*sin(a*x))^2), x)