3.181 \(\int \frac{\tan ^{-1}(a x)}{x^4 (c+a^2 c x^2)} \, dx\)

Optimal. Leaf size=88 \[ \frac{2 a^3 \log \left (a^2 x^2+1\right )}{3 c}-\frac{4 a^3 \log (x)}{3 c}+\frac{a^3 \tan ^{-1}(a x)^2}{2 c}+\frac{a^2 \tan ^{-1}(a x)}{c x}-\frac{a}{6 c x^2}-\frac{\tan ^{-1}(a x)}{3 c x^3} \]

[Out]

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

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Rubi [A]  time = 0.164881, antiderivative size = 88, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 8, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {4918, 4852, 266, 44, 36, 29, 31, 4884} \[ \frac{2 a^3 \log \left (a^2 x^2+1\right )}{3 c}-\frac{4 a^3 \log (x)}{3 c}+\frac{a^3 \tan ^{-1}(a x)^2}{2 c}+\frac{a^2 \tan ^{-1}(a x)}{c x}-\frac{a}{6 c x^2}-\frac{\tan ^{-1}(a x)}{3 c x^3} \]

Antiderivative was successfully verified.

[In]

Int[ArcTan[a*x]/(x^4*(c + a^2*c*x^2)),x]

[Out]

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

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 44

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

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]

Rubi steps

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

Mathematica [A]  time = 0.0178617, size = 88, normalized size = 1. \[ \frac{2 a^3 \log \left (a^2 x^2+1\right )}{3 c}-\frac{4 a^3 \log (x)}{3 c}+\frac{a^3 \tan ^{-1}(a x)^2}{2 c}+\frac{a^2 \tan ^{-1}(a x)}{c x}-\frac{a}{6 c x^2}-\frac{\tan ^{-1}(a x)}{3 c x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTan[a*x]/(x^4*(c + a^2*c*x^2)),x]

[Out]

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

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Maple [A]  time = 0.041, size = 81, normalized size = 0.9 \begin{align*}{\frac{{a}^{3} \left ( \arctan \left ( ax \right ) \right ) ^{2}}{2\,c}}-{\frac{\arctan \left ( ax \right ) }{3\,c{x}^{3}}}+{\frac{{a}^{2}\arctan \left ( ax \right ) }{cx}}+{\frac{2\,{a}^{3}\ln \left ({a}^{2}{x}^{2}+1 \right ) }{3\,c}}-{\frac{a}{6\,c{x}^{2}}}-{\frac{4\,{a}^{3}\ln \left ( ax \right ) }{3\,c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.65361, size = 122, normalized size = 1.39 \begin{align*} \frac{1}{3} \,{\left (\frac{3 \, a^{3} \arctan \left (a x\right )}{c} + \frac{3 \, a^{2} x^{2} - 1}{c x^{3}}\right )} \arctan \left (a x\right ) - \frac{{\left (3 \, a^{2} x^{2} \arctan \left (a x\right )^{2} - 4 \, a^{2} x^{2} \log \left (a^{2} x^{2} + 1\right ) + 8 \, a^{2} x^{2} \log \left (x\right ) + 1\right )} a}{6 \, c x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(a*x)/x^4/(a^2*c*x^2+c),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.63022, size = 169, normalized size = 1.92 \begin{align*} \frac{3 \, a^{3} x^{3} \arctan \left (a x\right )^{2} + 4 \, a^{3} x^{3} \log \left (a^{2} x^{2} + 1\right ) - 8 \, a^{3} x^{3} \log \left (x\right ) - a x + 2 \,{\left (3 \, a^{2} x^{2} - 1\right )} \arctan \left (a x\right )}{6 \, c x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(a*x)/x^4/(a^2*c*x^2+c),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 3.48911, size = 117, normalized size = 1.33 \begin{align*} \begin{cases} - \frac{4 a^{3} \log{\left (x \right )}}{3 c} + \frac{2 a^{3} \log{\left (x^{2} + \frac{1}{a^{2}} \right )}}{3 c} + \frac{a^{3} \operatorname{atan}^{2}{\left (a x \right )}}{2 c} + \frac{a^{2} \operatorname{atan}{\left (a x \right )}}{c x} - \frac{a}{6 c x^{2}} - \frac{\operatorname{atan}{\left (a x \right )}}{3 c x^{3}} & \text{for}\: c \neq 0 \\\tilde{\infty } \left (- \frac{a^{3} \log{\left (x \right )}}{3} + \frac{a^{3} \log{\left (a^{2} x^{2} + 1 \right )}}{6} - \frac{a}{6 x^{2}} - \frac{\operatorname{atan}{\left (a x \right )}}{3 x^{3}}\right ) & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-4*a**3*log(x)/(3*c) + 2*a**3*log(x**2 + a**(-2))/(3*c) + a**3*atan(a*x)**2/(2*c) + a**2*atan(a*x)/
(c*x) - a/(6*c*x**2) - atan(a*x)/(3*c*x**3), Ne(c, 0)), (zoo*(-a**3*log(x)/3 + a**3*log(a**2*x**2 + 1)/6 - a/(
6*x**2) - atan(a*x)/(3*x**3)), True))

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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 )} x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(a*x)/x^4/(a^2*c*x^2+c),x, algorithm="giac")

[Out]

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