3.163 \(\int \frac{(c+a^2 c x^2)^2 \tan ^{-1}(a x)}{x^3} \, dx\)

Optimal. Leaf size=90 \[ i a^2 c^2 \text{PolyLog}(2,-i a x)-i a^2 c^2 \text{PolyLog}(2,i a x)+\frac{1}{2} a^4 c^2 x^2 \tan ^{-1}(a x)-\frac{1}{2} a^3 c^2 x-\frac{c^2 \tan ^{-1}(a x)}{2 x^2}-\frac{a c^2}{2 x} \]

[Out]

-(a*c^2)/(2*x) - (a^3*c^2*x)/2 - (c^2*ArcTan[a*x])/(2*x^2) + (a^4*c^2*x^2*ArcTan[a*x])/2 + I*a^2*c^2*PolyLog[2
, (-I)*a*x] - I*a^2*c^2*PolyLog[2, I*a*x]

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Rubi [A]  time = 0.123163, antiderivative size = 90, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 7, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.35, Rules used = {4948, 4852, 325, 203, 4848, 2391, 321} \[ i a^2 c^2 \text{PolyLog}(2,-i a x)-i a^2 c^2 \text{PolyLog}(2,i a x)+\frac{1}{2} a^4 c^2 x^2 \tan ^{-1}(a x)-\frac{1}{2} a^3 c^2 x-\frac{c^2 \tan ^{-1}(a x)}{2 x^2}-\frac{a c^2}{2 x} \]

Antiderivative was successfully verified.

[In]

Int[((c + a^2*c*x^2)^2*ArcTan[a*x])/x^3,x]

[Out]

-(a*c^2)/(2*x) - (a^3*c^2*x)/2 - (c^2*ArcTan[a*x])/(2*x^2) + (a^4*c^2*x^2*ArcTan[a*x])/2 + I*a^2*c^2*PolyLog[2
, (-I)*a*x] - I*a^2*c^2*PolyLog[2, I*a*x]

Rule 4948

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> Int[Ex
pandIntegrand[(f*x)^m*(d + e*x^2)^q*(a + b*ArcTan[c*x])^p, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[e,
c^2*d] && IGtQ[p, 0] && IGtQ[q, 1] && (EqQ[p, 1] || IntegerQ[m])

Rule 4852

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcTa
n[c*x])^p)/(d*(m + 1)), x] - Dist[(b*c*p)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcTan[c*x])^(p - 1))/(1 + c^
2*x^2), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[p, 0] && (EqQ[p, 1] || IntegerQ[m]) && NeQ[m, -1]

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 4848

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))/(x_), x_Symbol] :> Simp[a*Log[x], x] + (Dist[(I*b)/2, Int[Log[1 - I*c*x
]/x, x], x] - Dist[(I*b)/2, Int[Log[1 + I*c*x]/x, x], x]) /; FreeQ[{a, b, c}, x]

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]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rubi steps

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

Mathematica [C]  time = 0.044859, size = 103, normalized size = 1.14 \[ \frac{c^2 \left (-a x \text{Hypergeometric2F1}\left (-\frac{1}{2},1,\frac{1}{2},-a^2 x^2\right )+2 i a^2 x^2 \text{PolyLog}(2,-i a x)-2 i a^2 x^2 \text{PolyLog}(2,i a x)-a^3 x^3+a^4 x^4 \tan ^{-1}(a x)+a^2 x^2 \tan ^{-1}(a x)-\tan ^{-1}(a x)\right )}{2 x^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[((c + a^2*c*x^2)^2*ArcTan[a*x])/x^3,x]

[Out]

(c^2*(-(a^3*x^3) - ArcTan[a*x] + a^2*x^2*ArcTan[a*x] + a^4*x^4*ArcTan[a*x] - a*x*Hypergeometric2F1[-1/2, 1, 1/
2, -(a^2*x^2)] + (2*I)*a^2*x^2*PolyLog[2, (-I)*a*x] - (2*I)*a^2*x^2*PolyLog[2, I*a*x]))/(2*x^2)

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Maple [A]  time = 0.04, size = 139, normalized size = 1.5 \begin{align*}{\frac{{a}^{4}{c}^{2}{x}^{2}\arctan \left ( ax \right ) }{2}}-{\frac{{c}^{2}\arctan \left ( ax \right ) }{2\,{x}^{2}}}+2\,{a}^{2}{c}^{2}\arctan \left ( ax \right ) \ln \left ( ax \right ) -{\frac{{a}^{3}{c}^{2}x}{2}}-{\frac{a{c}^{2}}{2\,x}}+i{a}^{2}{c}^{2}\ln \left ( ax \right ) \ln \left ( 1+iax \right ) -i{a}^{2}{c}^{2}\ln \left ( ax \right ) \ln \left ( 1-iax \right ) +i{a}^{2}{c}^{2}{\it dilog} \left ( 1+iax \right ) -i{a}^{2}{c}^{2}{\it dilog} \left ( 1-iax \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a^2*c*x^2+c)^2*arctan(a*x)/x^3,x)

[Out]

1/2*a^4*c^2*x^2*arctan(a*x)-1/2*c^2*arctan(a*x)/x^2+2*a^2*c^2*arctan(a*x)*ln(a*x)-1/2*a^3*c^2*x-1/2*a*c^2/x+I*
a^2*c^2*ln(a*x)*ln(1+I*a*x)-I*a^2*c^2*ln(a*x)*ln(1-I*a*x)+I*a^2*c^2*dilog(1+I*a*x)-I*a^2*c^2*dilog(1-I*a*x)

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Maxima [A]  time = 1.65607, size = 182, normalized size = 2.02 \begin{align*} -\frac{a^{3} c^{2} x^{3} + \pi a^{2} c^{2} x^{2} \log \left (a^{2} x^{2} + 1\right ) - 4 \, a^{2} c^{2} x^{2} \arctan \left (a x\right ) \log \left (x{\left | a \right |}\right ) + 2 i \, a^{2} c^{2} x^{2}{\rm Li}_2\left (i \, a x + 1\right ) - 2 i \, a^{2} c^{2} x^{2}{\rm Li}_2\left (-i \, a x + 1\right ) + a c^{2} x -{\left (a^{4} c^{2} x^{4} + 4 i \, a^{2} c^{2} x^{2} \arctan \left (0, a\right ) - c^{2}\right )} \arctan \left (a x\right )}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a^2*c*x^2+c)^2*arctan(a*x)/x^3,x, algorithm="maxima")

[Out]

-1/2*(a^3*c^2*x^3 + pi*a^2*c^2*x^2*log(a^2*x^2 + 1) - 4*a^2*c^2*x^2*arctan(a*x)*log(x*abs(a)) + 2*I*a^2*c^2*x^
2*dilog(I*a*x + 1) - 2*I*a^2*c^2*x^2*dilog(-I*a*x + 1) + a*c^2*x - (a^4*c^2*x^4 + 4*I*a^2*c^2*x^2*arctan2(0, a
) - c^2)*arctan(a*x))/x^2

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (a^{4} c^{2} x^{4} + 2 \, a^{2} c^{2} x^{2} + c^{2}\right )} \arctan \left (a x\right )}{x^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a^2*c*x^2+c)^2*arctan(a*x)/x^3,x, algorithm="fricas")

[Out]

integral((a^4*c^2*x^4 + 2*a^2*c^2*x^2 + c^2)*arctan(a*x)/x^3, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} c^{2} \left (\int \frac{\operatorname{atan}{\left (a x \right )}}{x^{3}}\, dx + \int \frac{2 a^{2} \operatorname{atan}{\left (a x \right )}}{x}\, dx + \int a^{4} x \operatorname{atan}{\left (a x \right )}\, dx\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a**2*c*x**2+c)**2*atan(a*x)/x**3,x)

[Out]

c**2*(Integral(atan(a*x)/x**3, x) + Integral(2*a**2*atan(a*x)/x, x) + Integral(a**4*x*atan(a*x), x))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (a^{2} c x^{2} + c\right )}^{2} \arctan \left (a x\right )}{x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a^2*c*x^2+c)^2*arctan(a*x)/x^3,x, algorithm="giac")

[Out]

integrate((a^2*c*x^2 + c)^2*arctan(a*x)/x^3, x)