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

Optimal. Leaf size=97 \[ -\frac{a}{4 c^2 \left (a^2 x^2+1\right )}-\frac{a \log \left (a^2 x^2+1\right )}{2 c^2}-\frac{a^2 x \tan ^{-1}(a x)}{2 c^2 \left (a^2 x^2+1\right )}+\frac{a \log (x)}{c^2}-\frac{3 a \tan ^{-1}(a x)^2}{4 c^2}-\frac{\tan ^{-1}(a x)}{c^2 x} \]

[Out]

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

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Rubi [A]  time = 0.161934, antiderivative size = 97, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 10, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {4966, 4918, 4852, 266, 36, 29, 31, 4884, 4892, 261} \[ -\frac{a}{4 c^2 \left (a^2 x^2+1\right )}-\frac{a \log \left (a^2 x^2+1\right )}{2 c^2}-\frac{a^2 x \tan ^{-1}(a x)}{2 c^2 \left (a^2 x^2+1\right )}+\frac{a \log (x)}{c^2}-\frac{3 a \tan ^{-1}(a x)^2}{4 c^2}-\frac{\tan ^{-1}(a x)}{c^2 x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4966

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> Dist[1/d, Int[
x^m*(d + e*x^2)^(q + 1)*(a + b*ArcTan[c*x])^p, x], x] - Dist[e/d, Int[x^(m + 2)*(d + e*x^2)^q*(a + b*ArcTan[c*
x])^p, x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && IntegersQ[p, 2*q] && LtQ[q, -1] && ILtQ[m, 0] &
& NeQ[p, -1]

Rule 4918

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[1/d,
Int[(f*x)^m*(a + b*ArcTan[c*x])^p, x], x] - Dist[e/(d*f^2), Int[((f*x)^(m + 2)*(a + b*ArcTan[c*x])^p)/(d + e*
x^2), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && GtQ[p, 0] && LtQ[m, -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 4884

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(a + b*ArcTan[c*x])^(p +
1)/(b*c*d*(p + 1)), x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[e, c^2*d] && NeQ[p, -1]

Rule 4892

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

Rule 261

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

Rubi steps

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

Mathematica [A]  time = 0.0732947, size = 94, normalized size = 0.97 \[ -\frac{a}{4 c^2 \left (a^2 x^2+1\right )}-\frac{a \log \left (a^2 x^2+1\right )}{2 c^2}-\frac{\left (3 a^2 x^2+2\right ) \tan ^{-1}(a x)}{2 c^2 x \left (a^2 x^2+1\right )}+\frac{a \log (x)}{c^2}-\frac{3 a \tan ^{-1}(a x)^2}{4 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.05, size = 92, normalized size = 1. \begin{align*} -{\frac{{a}^{2}x\arctan \left ( ax \right ) }{2\,{c}^{2} \left ({a}^{2}{x}^{2}+1 \right ) }}-{\frac{3\,a \left ( \arctan \left ( ax \right ) \right ) ^{2}}{4\,{c}^{2}}}-{\frac{\arctan \left ( ax \right ) }{{c}^{2}x}}-{\frac{a\ln \left ({a}^{2}{x}^{2}+1 \right ) }{2\,{c}^{2}}}-{\frac{a}{4\,{c}^{2} \left ({a}^{2}{x}^{2}+1 \right ) }}+{\frac{a\ln \left ( ax \right ) }{{c}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.59835, size = 161, normalized size = 1.66 \begin{align*} -\frac{1}{2} \,{\left (\frac{3 \, a^{2} x^{2} + 2}{a^{2} c^{2} x^{3} + c^{2} x} + \frac{3 \, a \arctan \left (a x\right )}{c^{2}}\right )} \arctan \left (a x\right ) + \frac{{\left (3 \,{\left (a^{2} x^{2} + 1\right )} \arctan \left (a x\right )^{2} - 2 \,{\left (a^{2} x^{2} + 1\right )} \log \left (a^{2} x^{2} + 1\right ) + 4 \,{\left (a^{2} x^{2} + 1\right )} \log \left (x\right ) - 1\right )} a}{4 \,{\left (a^{2} c^{2} x^{2} + c^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.68617, size = 221, normalized size = 2.28 \begin{align*} -\frac{3 \,{\left (a^{3} x^{3} + a x\right )} \arctan \left (a x\right )^{2} + a x + 2 \,{\left (3 \, a^{2} x^{2} + 2\right )} \arctan \left (a x\right ) + 2 \,{\left (a^{3} x^{3} + a x\right )} \log \left (a^{2} x^{2} + 1\right ) - 4 \,{\left (a^{3} x^{3} + a x\right )} \log \left (x\right )}{4 \,{\left (a^{2} c^{2} x^{3} + c^{2} x\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B]  time = 2.13428, size = 299, normalized size = 3.08 \begin{align*} \frac{12 a^{3} x^{3} \log{\left (x \right )}}{12 a^{2} c^{2} x^{3} + 12 c^{2} x} - \frac{6 a^{3} x^{3} \log{\left (x^{2} + \frac{1}{a^{2}} \right )}}{12 a^{2} c^{2} x^{3} + 12 c^{2} x} - \frac{9 a^{3} x^{3} \operatorname{atan}^{2}{\left (a x \right )}}{12 a^{2} c^{2} x^{3} + 12 c^{2} x} + \frac{a^{3} x^{3}}{12 a^{2} c^{2} x^{3} + 12 c^{2} x} - \frac{18 a^{2} x^{2} \operatorname{atan}{\left (a x \right )}}{12 a^{2} c^{2} x^{3} + 12 c^{2} x} + \frac{12 a x \log{\left (x \right )}}{12 a^{2} c^{2} x^{3} + 12 c^{2} x} - \frac{6 a x \log{\left (x^{2} + \frac{1}{a^{2}} \right )}}{12 a^{2} c^{2} x^{3} + 12 c^{2} x} - \frac{9 a x \operatorname{atan}^{2}{\left (a x \right )}}{12 a^{2} c^{2} x^{3} + 12 c^{2} x} - \frac{2 a x}{12 a^{2} c^{2} x^{3} + 12 c^{2} x} - \frac{12 \operatorname{atan}{\left (a x \right )}}{12 a^{2} c^{2} x^{3} + 12 c^{2} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

12*a**3*x**3*log(x)/(12*a**2*c**2*x**3 + 12*c**2*x) - 6*a**3*x**3*log(x**2 + a**(-2))/(12*a**2*c**2*x**3 + 12*
c**2*x) - 9*a**3*x**3*atan(a*x)**2/(12*a**2*c**2*x**3 + 12*c**2*x) + a**3*x**3/(12*a**2*c**2*x**3 + 12*c**2*x)
- 18*a**2*x**2*atan(a*x)/(12*a**2*c**2*x**3 + 12*c**2*x) + 12*a*x*log(x)/(12*a**2*c**2*x**3 + 12*c**2*x) - 6*
a*x*log(x**2 + a**(-2))/(12*a**2*c**2*x**3 + 12*c**2*x) - 9*a*x*atan(a*x)**2/(12*a**2*c**2*x**3 + 12*c**2*x) -
2*a*x/(12*a**2*c**2*x**3 + 12*c**2*x) - 12*atan(a*x)/(12*a**2*c**2*x**3 + 12*c**2*x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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