3.602 \(\int \frac{x^4 \sec ^2(a x)}{(\cos (a x)+a x \sin (a x))^2} \, dx\)

Optimal. Leaf size=124 \[ -\frac{2 i \text{PolyLog}\left (2,-e^{2 i a x}\right )}{a^5}-\frac{2 i x^2}{a^3}+\frac{2 x^2 \tan (a x)}{a^3}+\frac{x^2 \tan (a x) \sec ^2(a x)}{a^3}-\frac{x^3 \sec ^3(a x)}{a^2 (a x \sin (a x)+\cos (a x))}+\frac{4 x \log \left (1+e^{2 i a x}\right )}{a^4}+\frac{\tan (a x)}{a^5}-\frac{x \sec ^2(a x)}{a^4} \]

[Out]

((-2*I)*x^2)/a^3 + (4*x*Log[1 + E^((2*I)*a*x)])/a^4 - ((2*I)*PolyLog[2, -E^((2*I)*a*x)])/a^5 - (x*Sec[a*x]^2)/
a^4 - (x^3*Sec[a*x]^3)/(a^2*(Cos[a*x] + a*x*Sin[a*x])) + Tan[a*x]/a^5 + (2*x^2*Tan[a*x])/a^3 + (x^2*Sec[a*x]^2
*Tan[a*x])/a^3

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Rubi [A]  time = 0.183291, antiderivative size = 124, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 9, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.375, Rules used = {4601, 4186, 3767, 8, 4184, 3719, 2190, 2279, 2391} \[ -\frac{2 i \text{PolyLog}\left (2,-e^{2 i a x}\right )}{a^5}-\frac{2 i x^2}{a^3}+\frac{2 x^2 \tan (a x)}{a^3}+\frac{x^2 \tan (a x) \sec ^2(a x)}{a^3}-\frac{x^3 \sec ^3(a x)}{a^2 (a x \sin (a x)+\cos (a x))}+\frac{4 x \log \left (1+e^{2 i a x}\right )}{a^4}+\frac{\tan (a x)}{a^5}-\frac{x \sec ^2(a x)}{a^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*I)*x^2)/a^3 + (4*x*Log[1 + E^((2*I)*a*x)])/a^4 - ((2*I)*PolyLog[2, -E^((2*I)*a*x)])/a^5 - (x*Sec[a*x]^2)/
a^4 - (x^3*Sec[a*x]^3)/(a^2*(Cos[a*x] + a*x*Sin[a*x])) + Tan[a*x]/a^5 + (2*x^2*Tan[a*x])/a^3 + (x^2*Sec[a*x]^2
*Tan[a*x])/a^3

Rule 4601

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

Rule 4186

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> -Simp[(b^2*(c + d*x)^m*Cot[e
+ f*x]*(b*Csc[e + f*x])^(n - 2))/(f*(n - 1)), x] + (Dist[(b^2*d^2*m*(m - 1))/(f^2*(n - 1)*(n - 2)), Int[(c + d
*x)^(m - 2)*(b*Csc[e + f*x])^(n - 2), x], x] + Dist[(b^2*(n - 2))/(n - 1), Int[(c + d*x)^m*(b*Csc[e + f*x])^(n
- 2), x], x] - Simp[(b^2*d*m*(c + d*x)^(m - 1)*(b*Csc[e + f*x])^(n - 2))/(f^2*(n - 1)*(n - 2)), x]) /; FreeQ[
{b, c, d, e, f}, x] && GtQ[n, 1] && NeQ[n, 2] && GtQ[m, 1]

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 8

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

Rule 4184

Int[csc[(e_.) + (f_.)*(x_)]^2*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> -Simp[((c + d*x)^m*Cot[e + f*x])/f, x]
+ Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cot[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 3719

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[(I*(c + d*x)^(m + 1))/(d*(m + 1)), x
] - Dist[2*I, Int[((c + d*x)^m*E^(2*I*(e + f*x)))/(1 + E^(2*I*(e + f*x))), x], x] /; FreeQ[{c, d, e, f}, x] &&
IGtQ[m, 0]

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rubi steps

\begin{align*} \int \frac{x^4 \sec ^2(a x)}{(\cos (a x)+a x \sin (a x))^2} \, dx &=-\frac{x^3 \sec ^3(a x)}{a^2 (\cos (a x)+a x \sin (a x))}+\frac{3 \int x^2 \sec ^4(a x) \, dx}{a^2}\\ &=-\frac{x \sec ^2(a x)}{a^4}-\frac{x^3 \sec ^3(a x)}{a^2 (\cos (a x)+a x \sin (a x))}+\frac{x^2 \sec ^2(a x) \tan (a x)}{a^3}+\frac{\int \sec ^2(a x) \, dx}{a^4}+\frac{2 \int x^2 \sec ^2(a x) \, dx}{a^2}\\ &=-\frac{x \sec ^2(a x)}{a^4}-\frac{x^3 \sec ^3(a x)}{a^2 (\cos (a x)+a x \sin (a x))}+\frac{2 x^2 \tan (a x)}{a^3}+\frac{x^2 \sec ^2(a x) \tan (a x)}{a^3}-\frac{\operatorname{Subst}(\int 1 \, dx,x,-\tan (a x))}{a^5}-\frac{4 \int x \tan (a x) \, dx}{a^3}\\ &=-\frac{2 i x^2}{a^3}-\frac{x \sec ^2(a x)}{a^4}-\frac{x^3 \sec ^3(a x)}{a^2 (\cos (a x)+a x \sin (a x))}+\frac{\tan (a x)}{a^5}+\frac{2 x^2 \tan (a x)}{a^3}+\frac{x^2 \sec ^2(a x) \tan (a x)}{a^3}+\frac{(8 i) \int \frac{e^{2 i a x} x}{1+e^{2 i a x}} \, dx}{a^3}\\ &=-\frac{2 i x^2}{a^3}+\frac{4 x \log \left (1+e^{2 i a x}\right )}{a^4}-\frac{x \sec ^2(a x)}{a^4}-\frac{x^3 \sec ^3(a x)}{a^2 (\cos (a x)+a x \sin (a x))}+\frac{\tan (a x)}{a^5}+\frac{2 x^2 \tan (a x)}{a^3}+\frac{x^2 \sec ^2(a x) \tan (a x)}{a^3}-\frac{4 \int \log \left (1+e^{2 i a x}\right ) \, dx}{a^4}\\ &=-\frac{2 i x^2}{a^3}+\frac{4 x \log \left (1+e^{2 i a x}\right )}{a^4}-\frac{x \sec ^2(a x)}{a^4}-\frac{x^3 \sec ^3(a x)}{a^2 (\cos (a x)+a x \sin (a x))}+\frac{\tan (a x)}{a^5}+\frac{2 x^2 \tan (a x)}{a^3}+\frac{x^2 \sec ^2(a x) \tan (a x)}{a^3}+\frac{(2 i) \operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{2 i a x}\right )}{a^5}\\ &=-\frac{2 i x^2}{a^3}+\frac{4 x \log \left (1+e^{2 i a x}\right )}{a^4}-\frac{2 i \text{Li}_2\left (-e^{2 i a x}\right )}{a^5}-\frac{x \sec ^2(a x)}{a^4}-\frac{x^3 \sec ^3(a x)}{a^2 (\cos (a x)+a x \sin (a x))}+\frac{\tan (a x)}{a^5}+\frac{2 x^2 \tan (a x)}{a^3}+\frac{x^2 \sec ^2(a x) \tan (a x)}{a^3}\\ \end{align*}

Mathematica [A]  time = 1.08472, size = 130, normalized size = 1.05 \[ \frac{-2 i (a x \tan (a x)+1) \text{PolyLog}\left (2,-e^{2 i a x}\right )-a x \left (a^2 x^2+2 i a x-4 \log \left (1+e^{2 i a x}\right )+1\right )+a^3 x^3 \tan ^2(a x)+\left (-2 i a^3 x^3+2 a^2 x^2+4 a^2 x^2 \log \left (1+e^{2 i a x}\right )+1\right ) \tan (a x)}{a^5 (a x \tan (a x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(a*x*(1 + (2*I)*a*x + a^2*x^2 - 4*Log[1 + E^((2*I)*a*x)])) + (1 + 2*a^2*x^2 - (2*I)*a^3*x^3 + 4*a^2*x^2*Log[
1 + E^((2*I)*a*x)])*Tan[a*x] + a^3*x^3*Tan[a*x]^2 - (2*I)*PolyLog[2, -E^((2*I)*a*x)]*(1 + a*x*Tan[a*x]))/(a^5*
(1 + a*x*Tan[a*x]))

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Maple [A]  time = 0.353, size = 141, normalized size = 1.1 \begin{align*}{\frac{-2\,i \left ( -2\,i{a}^{2}{x}^{2}{{\rm e}^{2\,iax}}+2\,{x}^{3}{a}^{3}-2\,i{a}^{2}{x}^{2}+ax{{\rm e}^{2\,iax}}-i{{\rm e}^{2\,iax}}+ax-i \right ) }{ \left ( 1+{{\rm e}^{2\,iax}} \right ) \left ( ax{{\rm e}^{2\,iax}}-ax+i{{\rm e}^{2\,iax}}+i \right ){a}^{5}}}-{\frac{4\,i{x}^{2}}{{a}^{3}}}+4\,{\frac{x\ln \left ( 1+{{\rm e}^{2\,iax}} \right ) }{{a}^{4}}}-{\frac{2\,i{\it polylog} \left ( 2,-{{\rm e}^{2\,iax}} \right ) }{{a}^{5}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*I*(-2*I*a^2*x^2*exp(2*I*a*x)+2*x^3*a^3-2*I*a^2*x^2+a*x*exp(2*I*a*x)-I*exp(2*I*a*x)+a*x-I)/(1+exp(2*I*a*x))/
(a*x*exp(2*I*a*x)-a*x+I*exp(2*I*a*x)+I)/a^5-4*I/a^3*x^2+4*x*ln(1+exp(2*I*a*x))/a^4-2*I*polylog(2,-exp(2*I*a*x)
)/a^5

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Maxima [B]  time = 1.65258, size = 514, normalized size = 4.15 \begin{align*} -\frac{2 \, a x +{\left (4 \, a^{2} x^{2} - 8 i \, a x \cos \left (2 \, a x\right ) + 8 \, a x \sin \left (2 \, a x\right ) - 4 i \, a x -{\left (4 \, a^{2} x^{2} + 4 i \, a x\right )} \cos \left (4 \, a x\right ) + 4 \,{\left (-i \, a^{2} x^{2} + a x\right )} \sin \left (4 \, a x\right )\right )} \arctan \left (\sin \left (2 \, a x\right ), \cos \left (2 \, a x\right ) + 1\right ) + 4 \,{\left (a^{3} x^{3} + i \, a^{2} x^{2}\right )} \cos \left (4 \, a x\right ) -{\left (-4 i \, a^{2} x^{2} - 2 \, a x + 2 i\right )} \cos \left (2 \, a x\right ) -{\left (2 \, a x -{\left (2 \, a x + 2 i\right )} \cos \left (4 \, a x\right ) - 2 \,{\left (i \, a x - 1\right )} \sin \left (4 \, a x\right ) - 4 i \, \cos \left (2 \, a x\right ) + 4 \, \sin \left (2 \, a x\right ) - 2 i\right )}{\rm Li}_2\left (-e^{\left (2 i \, a x\right )}\right ) -{\left (2 i \, a^{2} x^{2} + 4 \, a x \cos \left (2 \, a x\right ) + 4 i \, a x \sin \left (2 \, a x\right ) + 2 \, a x - 2 \,{\left (i \, a^{2} x^{2} - a x\right )} \cos \left (4 \, a x\right ) +{\left (2 \, a^{2} x^{2} + 2 i \, a x\right )} \sin \left (4 \, a x\right )\right )} \log \left (\cos \left (2 \, a x\right )^{2} + \sin \left (2 \, a x\right )^{2} + 2 \, \cos \left (2 \, a x\right ) + 1\right ) -{\left (-4 i \, a^{3} x^{3} + 4 \, a^{2} x^{2}\right )} \sin \left (4 \, a x\right ) -{\left (4 \, a^{2} x^{2} - 2 i \, a x - 2\right )} \sin \left (2 \, a x\right ) - 2 i}{{\left (i \, a x +{\left (-i \, a x + 1\right )} \cos \left (4 \, a x\right ) +{\left (a x + i\right )} \sin \left (4 \, a x\right ) + 2 \, \cos \left (2 \, a x\right ) + 2 i \, \sin \left (2 \, a x\right ) + 1\right )} a^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*a*x + (4*a^2*x^2 - 8*I*a*x*cos(2*a*x) + 8*a*x*sin(2*a*x) - 4*I*a*x - (4*a^2*x^2 + 4*I*a*x)*cos(4*a*x) + 4*
(-I*a^2*x^2 + a*x)*sin(4*a*x))*arctan2(sin(2*a*x), cos(2*a*x) + 1) + 4*(a^3*x^3 + I*a^2*x^2)*cos(4*a*x) - (-4*
I*a^2*x^2 - 2*a*x + 2*I)*cos(2*a*x) - (2*a*x - (2*a*x + 2*I)*cos(4*a*x) - 2*(I*a*x - 1)*sin(4*a*x) - 4*I*cos(2
*a*x) + 4*sin(2*a*x) - 2*I)*dilog(-e^(2*I*a*x)) - (2*I*a^2*x^2 + 4*a*x*cos(2*a*x) + 4*I*a*x*sin(2*a*x) + 2*a*x
- 2*(I*a^2*x^2 - a*x)*cos(4*a*x) + (2*a^2*x^2 + 2*I*a*x)*sin(4*a*x))*log(cos(2*a*x)^2 + sin(2*a*x)^2 + 2*cos(
2*a*x) + 1) - (-4*I*a^3*x^3 + 4*a^2*x^2)*sin(4*a*x) - (4*a^2*x^2 - 2*I*a*x - 2)*sin(2*a*x) - 2*I)/((I*a*x + (-
I*a*x + 1)*cos(4*a*x) + (a*x + I)*sin(4*a*x) + 2*cos(2*a*x) + 2*I*sin(2*a*x) + 1)*a^5)

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Fricas [B]  time = 2.5673, size = 1014, normalized size = 8.18 \begin{align*} \frac{a^{3} x^{3} -{\left (2 \, a^{3} x^{3} + a x\right )} \cos \left (a x\right )^{2} +{\left (2 \, a^{2} x^{2} + 1\right )} \cos \left (a x\right ) \sin \left (a x\right ) +{\left (2 i \, a x \cos \left (a x\right ) \sin \left (a x\right ) + 2 i \, \cos \left (a x\right )^{2}\right )}{\rm Li}_2\left (i \, \cos \left (a x\right ) + \sin \left (a x\right )\right ) +{\left (-2 i \, a x \cos \left (a x\right ) \sin \left (a x\right ) - 2 i \, \cos \left (a x\right )^{2}\right )}{\rm Li}_2\left (i \, \cos \left (a x\right ) - \sin \left (a x\right )\right ) +{\left (-2 i \, a x \cos \left (a x\right ) \sin \left (a x\right ) - 2 i \, \cos \left (a x\right )^{2}\right )}{\rm Li}_2\left (-i \, \cos \left (a x\right ) + \sin \left (a x\right )\right ) +{\left (2 i \, a x \cos \left (a x\right ) \sin \left (a x\right ) + 2 i \, \cos \left (a x\right )^{2}\right )}{\rm Li}_2\left (-i \, \cos \left (a x\right ) - \sin \left (a x\right )\right ) + 2 \,{\left (a^{2} x^{2} \cos \left (a x\right ) \sin \left (a x\right ) + a x \cos \left (a x\right )^{2}\right )} \log \left (i \, \cos \left (a x\right ) + \sin \left (a x\right ) + 1\right ) + 2 \,{\left (a^{2} x^{2} \cos \left (a x\right ) \sin \left (a x\right ) + a x \cos \left (a x\right )^{2}\right )} \log \left (i \, \cos \left (a x\right ) - \sin \left (a x\right ) + 1\right ) + 2 \,{\left (a^{2} x^{2} \cos \left (a x\right ) \sin \left (a x\right ) + a x \cos \left (a x\right )^{2}\right )} \log \left (-i \, \cos \left (a x\right ) + \sin \left (a x\right ) + 1\right ) + 2 \,{\left (a^{2} x^{2} \cos \left (a x\right ) \sin \left (a x\right ) + a x \cos \left (a x\right )^{2}\right )} \log \left (-i \, \cos \left (a x\right ) - \sin \left (a x\right ) + 1\right )}{a^{6} x \cos \left (a x\right ) \sin \left (a x\right ) + a^{5} \cos \left (a x\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a^3*x^3 - (2*a^3*x^3 + a*x)*cos(a*x)^2 + (2*a^2*x^2 + 1)*cos(a*x)*sin(a*x) + (2*I*a*x*cos(a*x)*sin(a*x) + 2*I
*cos(a*x)^2)*dilog(I*cos(a*x) + sin(a*x)) + (-2*I*a*x*cos(a*x)*sin(a*x) - 2*I*cos(a*x)^2)*dilog(I*cos(a*x) - s
in(a*x)) + (-2*I*a*x*cos(a*x)*sin(a*x) - 2*I*cos(a*x)^2)*dilog(-I*cos(a*x) + sin(a*x)) + (2*I*a*x*cos(a*x)*sin
(a*x) + 2*I*cos(a*x)^2)*dilog(-I*cos(a*x) - sin(a*x)) + 2*(a^2*x^2*cos(a*x)*sin(a*x) + a*x*cos(a*x)^2)*log(I*c
os(a*x) + sin(a*x) + 1) + 2*(a^2*x^2*cos(a*x)*sin(a*x) + a*x*cos(a*x)^2)*log(I*cos(a*x) - sin(a*x) + 1) + 2*(a
^2*x^2*cos(a*x)*sin(a*x) + a*x*cos(a*x)^2)*log(-I*cos(a*x) + sin(a*x) + 1) + 2*(a^2*x^2*cos(a*x)*sin(a*x) + a*
x*cos(a*x)^2)*log(-I*cos(a*x) - sin(a*x) + 1))/(a^6*x*cos(a*x)*sin(a*x) + a^5*cos(a*x)^2)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{4} \sec \left (a x\right )^{2}}{{\left (a x \sin \left (a x\right ) + \cos \left (a x\right )\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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