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

Optimal. Leaf size=81 \[ -\frac{1}{6} a^3 c^2 x^2-\frac{4}{3} a c^2 \log \left (a^2 x^2+1\right )+\frac{1}{3} a^4 c^2 x^3 \tan ^{-1}(a x)+2 a^2 c^2 x \tan ^{-1}(a x)+a c^2 \log (x)-\frac{c^2 \tan ^{-1}(a x)}{x} \]

[Out]

-(a^3*c^2*x^2)/6 - (c^2*ArcTan[a*x])/x + 2*a^2*c^2*x*ArcTan[a*x] + (a^4*c^2*x^3*ArcTan[a*x])/3 + a*c^2*Log[x]
- (4*a*c^2*Log[1 + a^2*x^2])/3

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Rubi [A]  time = 0.116758, antiderivative size = 81, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 9, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.45, Rules used = {4948, 4846, 260, 4852, 266, 36, 29, 31, 43} \[ -\frac{1}{6} a^3 c^2 x^2-\frac{4}{3} a c^2 \log \left (a^2 x^2+1\right )+\frac{1}{3} a^4 c^2 x^3 \tan ^{-1}(a x)+2 a^2 c^2 x \tan ^{-1}(a x)+a c^2 \log (x)-\frac{c^2 \tan ^{-1}(a x)}{x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^3*c^2*x^2)/6 - (c^2*ArcTan[a*x])/x + 2*a^2*c^2*x*ArcTan[a*x] + (a^4*c^2*x^3*ArcTan[a*x])/3 + a*c^2*Log[x]
- (4*a*c^2*Log[1 + a^2*x^2])/3

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 4846

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*ArcTan[c*x])^p, x] - Dist[b*c*p, Int[
(x*(a + b*ArcTan[c*x])^(p - 1))/(1 + c^2*x^2), x], x] /; FreeQ[{a, b, c}, x] && IGtQ[p, 0]

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

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 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{\left (c+a^2 c x^2\right )^2 \tan ^{-1}(a x)}{x^2} \, dx &=\int \left (2 a^2 c^2 \tan ^{-1}(a x)+\frac{c^2 \tan ^{-1}(a x)}{x^2}+a^4 c^2 x^2 \tan ^{-1}(a x)\right ) \, dx\\ &=c^2 \int \frac{\tan ^{-1}(a x)}{x^2} \, dx+\left (2 a^2 c^2\right ) \int \tan ^{-1}(a x) \, dx+\left (a^4 c^2\right ) \int x^2 \tan ^{-1}(a x) \, dx\\ &=-\frac{c^2 \tan ^{-1}(a x)}{x}+2 a^2 c^2 x \tan ^{-1}(a x)+\frac{1}{3} a^4 c^2 x^3 \tan ^{-1}(a x)+\left (a c^2\right ) \int \frac{1}{x \left (1+a^2 x^2\right )} \, dx-\left (2 a^3 c^2\right ) \int \frac{x}{1+a^2 x^2} \, dx-\frac{1}{3} \left (a^5 c^2\right ) \int \frac{x^3}{1+a^2 x^2} \, dx\\ &=-\frac{c^2 \tan ^{-1}(a x)}{x}+2 a^2 c^2 x \tan ^{-1}(a x)+\frac{1}{3} a^4 c^2 x^3 \tan ^{-1}(a x)-a c^2 \log \left (1+a^2 x^2\right )+\frac{1}{2} \left (a c^2\right ) \operatorname{Subst}\left (\int \frac{1}{x \left (1+a^2 x\right )} \, dx,x,x^2\right )-\frac{1}{6} \left (a^5 c^2\right ) \operatorname{Subst}\left (\int \frac{x}{1+a^2 x} \, dx,x,x^2\right )\\ &=-\frac{c^2 \tan ^{-1}(a x)}{x}+2 a^2 c^2 x \tan ^{-1}(a x)+\frac{1}{3} a^4 c^2 x^3 \tan ^{-1}(a x)-a c^2 \log \left (1+a^2 x^2\right )+\frac{1}{2} \left (a c^2\right ) \operatorname{Subst}\left (\int \frac{1}{x} \, dx,x,x^2\right )-\frac{1}{2} \left (a^3 c^2\right ) \operatorname{Subst}\left (\int \frac{1}{1+a^2 x} \, dx,x,x^2\right )-\frac{1}{6} \left (a^5 c^2\right ) \operatorname{Subst}\left (\int \left (\frac{1}{a^2}-\frac{1}{a^2 \left (1+a^2 x\right )}\right ) \, dx,x,x^2\right )\\ &=-\frac{1}{6} a^3 c^2 x^2-\frac{c^2 \tan ^{-1}(a x)}{x}+2 a^2 c^2 x \tan ^{-1}(a x)+\frac{1}{3} a^4 c^2 x^3 \tan ^{-1}(a x)+a c^2 \log (x)-\frac{4}{3} a c^2 \log \left (1+a^2 x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.0513391, size = 62, normalized size = 0.77 \[ \frac{c^2 \left (2 \left (a^4 x^4+6 a^2 x^2-3\right ) \tan ^{-1}(a x)-a x \left (a^2 x^2+8 \log \left (a^2 x^2+1\right )-6 \log (x)\right )\right )}{6 x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c^2*(2*(-3 + 6*a^2*x^2 + a^4*x^4)*ArcTan[a*x] - a*x*(a^2*x^2 - 6*Log[x] + 8*Log[1 + a^2*x^2])))/(6*x)

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Maple [A]  time = 0.03, size = 78, normalized size = 1. \begin{align*}{\frac{{a}^{4}{c}^{2}{x}^{3}\arctan \left ( ax \right ) }{3}}+2\,{a}^{2}{c}^{2}x\arctan \left ( ax \right ) -{\frac{{c}^{2}\arctan \left ( ax \right ) }{x}}-{\frac{{c}^{2}{x}^{2}{a}^{3}}{6}}-{\frac{4\,a{c}^{2}\ln \left ({a}^{2}{x}^{2}+1 \right ) }{3}}+a{c}^{2}\ln \left ( ax \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^2,x)

[Out]

1/3*a^4*c^2*x^3*arctan(a*x)+2*a^2*c^2*x*arctan(a*x)-c^2*arctan(a*x)/x-1/6*c^2*x^2*a^3-4/3*a*c^2*ln(a^2*x^2+1)+
a*c^2*ln(a*x)

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Maxima [A]  time = 0.968203, size = 96, normalized size = 1.19 \begin{align*} -\frac{1}{6} \,{\left (a^{2} c^{2} x^{2} + 8 \, c^{2} \log \left (a^{2} x^{2} + 1\right ) - 6 \, c^{2} \log \left (x\right )\right )} a + \frac{1}{3} \,{\left (a^{4} c^{2} x^{3} + 6 \, a^{2} c^{2} x - \frac{3 \, c^{2}}{x}\right )} \arctan \left (a 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^2,x, algorithm="maxima")

[Out]

-1/6*(a^2*c^2*x^2 + 8*c^2*log(a^2*x^2 + 1) - 6*c^2*log(x))*a + 1/3*(a^4*c^2*x^3 + 6*a^2*c^2*x - 3*c^2/x)*arcta
n(a*x)

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Fricas [A]  time = 1.70552, size = 167, normalized size = 2.06 \begin{align*} -\frac{a^{3} c^{2} x^{3} + 8 \, a c^{2} x \log \left (a^{2} x^{2} + 1\right ) - 6 \, a c^{2} x \log \left (x\right ) - 2 \,{\left (a^{4} c^{2} x^{4} + 6 \, a^{2} c^{2} x^{2} - 3 \, c^{2}\right )} \arctan \left (a x\right )}{6 \, 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^2,x, algorithm="fricas")

[Out]

-1/6*(a^3*c^2*x^3 + 8*a*c^2*x*log(a^2*x^2 + 1) - 6*a*c^2*x*log(x) - 2*(a^4*c^2*x^4 + 6*a^2*c^2*x^2 - 3*c^2)*ar
ctan(a*x))/x

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Sympy [A]  time = 1.85078, size = 82, normalized size = 1.01 \begin{align*} \begin{cases} \frac{a^{4} c^{2} x^{3} \operatorname{atan}{\left (a x \right )}}{3} - \frac{a^{3} c^{2} x^{2}}{6} + 2 a^{2} c^{2} x \operatorname{atan}{\left (a x \right )} + a c^{2} \log{\left (x \right )} - \frac{4 a c^{2} \log{\left (x^{2} + \frac{1}{a^{2}} \right )}}{3} - \frac{c^{2} \operatorname{atan}{\left (a x \right )}}{x} & \text{for}\: a \neq 0 \\0 & \text{otherwise} \end{cases} \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**2,x)

[Out]

Piecewise((a**4*c**2*x**3*atan(a*x)/3 - a**3*c**2*x**2/6 + 2*a**2*c**2*x*atan(a*x) + a*c**2*log(x) - 4*a*c**2*
log(x**2 + a**(-2))/3 - c**2*atan(a*x)/x, Ne(a, 0)), (0, True))

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Giac [A]  time = 1.10871, size = 97, normalized size = 1.2 \begin{align*} -\frac{1}{6} \, a^{3} c^{2} x^{2} - \frac{4}{3} \, a c^{2} \log \left (a^{2} x^{2} + 1\right ) + \frac{1}{2} \, a c^{2} \log \left (x^{2}\right ) + \frac{1}{3} \,{\left (a^{4} c^{2} x^{3} + 6 \, a^{2} c^{2} x - \frac{3 \, c^{2}}{x}\right )} \arctan \left (a 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^2,x, algorithm="giac")

[Out]

-1/6*a^3*c^2*x^2 - 4/3*a*c^2*log(a^2*x^2 + 1) + 1/2*a*c^2*log(x^2) + 1/3*(a^4*c^2*x^3 + 6*a^2*c^2*x - 3*c^2/x)
*arctan(a*x)