3.601 \(\int \frac{x^3 \sec (a x)}{(\cos (a x)+a x \sin (a x))^2} \, dx\)

Optimal. Leaf size=110 \[ \frac{i \text{PolyLog}\left (2,-i e^{i a x}\right )}{a^4}-\frac{i \text{PolyLog}\left (2,i e^{i a x}\right )}{a^4}-\frac{x^2 \sec ^2(a x)}{a^2 (a x \sin (a x)+\cos (a x))}-\frac{2 i x \tan ^{-1}\left (e^{i a x}\right )}{a^3}-\frac{\sec (a x)}{a^4}+\frac{x \tan (a x) \sec (a x)}{a^3} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0934006, antiderivative size = 110, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.227, Rules used = {4601, 4185, 4181, 2279, 2391} \[ \frac{i \text{PolyLog}\left (2,-i e^{i a x}\right )}{a^4}-\frac{i \text{PolyLog}\left (2,i e^{i a x}\right )}{a^4}-\frac{x^2 \sec ^2(a x)}{a^2 (a x \sin (a x)+\cos (a x))}-\frac{2 i x \tan ^{-1}\left (e^{i a x}\right )}{a^3}-\frac{\sec (a x)}{a^4}+\frac{x \tan (a x) \sec (a x)}{a^3} \]

Antiderivative was successfully verified.

[In]

Int[(x^3*Sec[a*x])/(Cos[a*x] + a*x*Sin[a*x])^2,x]

[Out]

((-2*I)*x*ArcTan[E^(I*a*x)])/a^3 + (I*PolyLog[2, (-I)*E^(I*a*x)])/a^4 - (I*PolyLog[2, I*E^(I*a*x)])/a^4 - Sec[
a*x]/a^4 - (x^2*Sec[a*x]^2)/(a^2*(Cos[a*x] + a*x*Sin[a*x])) + (x*Sec[a*x]*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 4185

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

Rule 4181

Int[csc[(e_.) + Pi*(k_.) + (f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c + d*x)^m*ArcTanh[E
^(I*k*Pi)*E^(I*(e + f*x))])/f, x] + (-Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Log[1 - E^(I*k*Pi)*E^(I*(e + f*x))],
x], x] + Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Log[1 + E^(I*k*Pi)*E^(I*(e + f*x))], x], x]) /; FreeQ[{c, d, e,
f}, x] && IntegerQ[2*k] && 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^3 \sec (a x)}{(\cos (a x)+a x \sin (a x))^2} \, dx &=-\frac{x^2 \sec ^2(a x)}{a^2 (\cos (a x)+a x \sin (a x))}+\frac{2 \int x \sec ^3(a x) \, dx}{a^2}\\ &=-\frac{\sec (a x)}{a^4}-\frac{x^2 \sec ^2(a x)}{a^2 (\cos (a x)+a x \sin (a x))}+\frac{x \sec (a x) \tan (a x)}{a^3}+\frac{\int x \sec (a x) \, dx}{a^2}\\ &=-\frac{2 i x \tan ^{-1}\left (e^{i a x}\right )}{a^3}-\frac{\sec (a x)}{a^4}-\frac{x^2 \sec ^2(a x)}{a^2 (\cos (a x)+a x \sin (a x))}+\frac{x \sec (a x) \tan (a x)}{a^3}-\frac{\int \log \left (1-i e^{i a x}\right ) \, dx}{a^3}+\frac{\int \log \left (1+i e^{i a x}\right ) \, dx}{a^3}\\ &=-\frac{2 i x \tan ^{-1}\left (e^{i a x}\right )}{a^3}-\frac{\sec (a x)}{a^4}-\frac{x^2 \sec ^2(a x)}{a^2 (\cos (a x)+a x \sin (a x))}+\frac{x \sec (a x) \tan (a x)}{a^3}+\frac{i \operatorname{Subst}\left (\int \frac{\log (1-i x)}{x} \, dx,x,e^{i a x}\right )}{a^4}-\frac{i \operatorname{Subst}\left (\int \frac{\log (1+i x)}{x} \, dx,x,e^{i a x}\right )}{a^4}\\ &=-\frac{2 i x \tan ^{-1}\left (e^{i a x}\right )}{a^3}+\frac{i \text{Li}_2\left (-i e^{i a x}\right )}{a^4}-\frac{i \text{Li}_2\left (i e^{i a x}\right )}{a^4}-\frac{\sec (a x)}{a^4}-\frac{x^2 \sec ^2(a x)}{a^2 (\cos (a x)+a x \sin (a x))}+\frac{x \sec (a x) \tan (a x)}{a^3}\\ \end{align*}

Mathematica [A]  time = 1.14046, size = 176, normalized size = 1.6 \[ -\frac{-i (a x \tan (a x)+1) \text{PolyLog}\left (2,-i e^{i a x}\right )+i (a x \tan (a x)+1) \text{PolyLog}\left (2,i e^{i a x}\right )+a^2 x^2 \sec (a x)-a^2 x^2 \log \left (1-i e^{i a x}\right ) \tan (a x)+a^2 x^2 \log \left (1+i e^{i a x}\right ) \tan (a x)-a x \log \left (1-i e^{i a x}\right )+a x \log \left (1+i e^{i a x}\right )+\sec (a x)}{a^4 (a x \tan (a x)+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^3*Sec[a*x])/(Cos[a*x] + a*x*Sin[a*x])^2,x]

[Out]

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

________________________________________________________________________________________

Maple [F]  time = 1.057, size = 0, normalized size = 0. \begin{align*} \int{\frac{{x}^{3}\sec \left ( ax \right ) }{ \left ( \cos \left ( ax \right ) +ax\sin \left ( ax \right ) \right ) ^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*sec(a*x)/(cos(a*x)+a*x*sin(a*x))^2,x)

[Out]

int(x^3*sec(a*x)/(cos(a*x)+a*x*sin(a*x))^2,x)

________________________________________________________________________________________

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(x^3*sec(a*x)/(cos(a*x)+a*x*sin(a*x))^2,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

________________________________________________________________________________________

Fricas [B]  time = 2.32017, size = 772, normalized size = 7.02 \begin{align*} -\frac{2 \, a^{2} x^{2} -{\left (-i \, a x \sin \left (a x\right ) - i \, \cos \left (a x\right )\right )}{\rm Li}_2\left (i \, \cos \left (a x\right ) + \sin \left (a x\right )\right ) -{\left (-i \, a x \sin \left (a x\right ) - i \, \cos \left (a x\right )\right )}{\rm Li}_2\left (i \, \cos \left (a x\right ) - \sin \left (a x\right )\right ) -{\left (i \, a x \sin \left (a x\right ) + i \, \cos \left (a x\right )\right )}{\rm Li}_2\left (-i \, \cos \left (a x\right ) + \sin \left (a x\right )\right ) -{\left (i \, a x \sin \left (a x\right ) + i \, \cos \left (a x\right )\right )}{\rm Li}_2\left (-i \, \cos \left (a x\right ) - \sin \left (a x\right )\right ) -{\left (a^{2} x^{2} \sin \left (a x\right ) + a x \cos \left (a x\right )\right )} \log \left (i \, \cos \left (a x\right ) + \sin \left (a x\right ) + 1\right ) +{\left (a^{2} x^{2} \sin \left (a x\right ) + a x \cos \left (a x\right )\right )} \log \left (i \, \cos \left (a x\right ) - \sin \left (a x\right ) + 1\right ) -{\left (a^{2} x^{2} \sin \left (a x\right ) + a x \cos \left (a x\right )\right )} \log \left (-i \, \cos \left (a x\right ) + \sin \left (a x\right ) + 1\right ) +{\left (a^{2} x^{2} \sin \left (a x\right ) + a x \cos \left (a x\right )\right )} \log \left (-i \, \cos \left (a x\right ) - \sin \left (a x\right ) + 1\right ) + 2}{2 \,{\left (a^{5} x \sin \left (a x\right ) + a^{4} \cos \left (a x\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*sec(a*x)/(cos(a*x)+a*x*sin(a*x))^2,x, algorithm="fricas")

[Out]

-1/2*(2*a^2*x^2 - (-I*a*x*sin(a*x) - I*cos(a*x))*dilog(I*cos(a*x) + sin(a*x)) - (-I*a*x*sin(a*x) - I*cos(a*x))
*dilog(I*cos(a*x) - sin(a*x)) - (I*a*x*sin(a*x) + I*cos(a*x))*dilog(-I*cos(a*x) + sin(a*x)) - (I*a*x*sin(a*x)
+ I*cos(a*x))*dilog(-I*cos(a*x) - sin(a*x)) - (a^2*x^2*sin(a*x) + a*x*cos(a*x))*log(I*cos(a*x) + sin(a*x) + 1)
+ (a^2*x^2*sin(a*x) + a*x*cos(a*x))*log(I*cos(a*x) - sin(a*x) + 1) - (a^2*x^2*sin(a*x) + a*x*cos(a*x))*log(-I
*cos(a*x) + sin(a*x) + 1) + (a^2*x^2*sin(a*x) + a*x*cos(a*x))*log(-I*cos(a*x) - sin(a*x) + 1) + 2)/(a^5*x*sin(
a*x) + a^4*cos(a*x))

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3} \sec{\left (a x \right )}}{\left (a x \sin{\left (a x \right )} + \cos{\left (a x \right )}\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*sec(a*x)/(cos(a*x)+a*x*sin(a*x))**2,x)

[Out]

Integral(x**3*sec(a*x)/(a*x*sin(a*x) + cos(a*x))**2, x)

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3} \sec \left (a x\right )}{{\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^3*sec(a*x)/(cos(a*x)+a*x*sin(a*x))^2,x, algorithm="giac")

[Out]

integrate(x^3*sec(a*x)/(a*x*sin(a*x) + cos(a*x))^2, x)