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

Optimal. Leaf size=136 \[ \frac{a^3}{4 c^2 \left (a^2 x^2+1\right )}+\frac{7 a^3 \log \left (a^2 x^2+1\right )}{6 c^2}+\frac{a^4 x \tan ^{-1}(a x)}{2 c^2 \left (a^2 x^2+1\right )}-\frac{7 a^3 \log (x)}{3 c^2}+\frac{5 a^3 \tan ^{-1}(a x)^2}{4 c^2}+\frac{2 a^2 \tan ^{-1}(a x)}{c^2 x}-\frac{a}{6 c^2 x^2}-\frac{\tan ^{-1}(a x)}{3 c^2 x^3} \]

[Out]

-a/(6*c^2*x^2) + a^3/(4*c^2*(1 + a^2*x^2)) - ArcTan[a*x]/(3*c^2*x^3) + (2*a^2*ArcTan[a*x])/(c^2*x) + (a^4*x*Ar
cTan[a*x])/(2*c^2*(1 + a^2*x^2)) + (5*a^3*ArcTan[a*x]^2)/(4*c^2) - (7*a^3*Log[x])/(3*c^2) + (7*a^3*Log[1 + a^2
*x^2])/(6*c^2)

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Rubi [A]  time = 0.374903, antiderivative size = 136, normalized size of antiderivative = 1., number of steps used = 23, number of rules used = 11, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.55, Rules used = {4966, 4918, 4852, 266, 44, 36, 29, 31, 4884, 4892, 261} \[ \frac{a^3}{4 c^2 \left (a^2 x^2+1\right )}+\frac{7 a^3 \log \left (a^2 x^2+1\right )}{6 c^2}+\frac{a^4 x \tan ^{-1}(a x)}{2 c^2 \left (a^2 x^2+1\right )}-\frac{7 a^3 \log (x)}{3 c^2}+\frac{5 a^3 \tan ^{-1}(a x)^2}{4 c^2}+\frac{2 a^2 \tan ^{-1}(a x)}{c^2 x}-\frac{a}{6 c^2 x^2}-\frac{\tan ^{-1}(a x)}{3 c^2 x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-a/(6*c^2*x^2) + a^3/(4*c^2*(1 + a^2*x^2)) - ArcTan[a*x]/(3*c^2*x^3) + (2*a^2*ArcTan[a*x])/(c^2*x) + (a^4*x*Ar
cTan[a*x])/(2*c^2*(1 + a^2*x^2)) + (5*a^3*ArcTan[a*x]^2)/(4*c^2) - (7*a^3*Log[x])/(3*c^2) + (7*a^3*Log[1 + a^2
*x^2])/(6*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 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]

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

Mathematica [A]  time = 0.0968163, size = 124, normalized size = 0.91 \[ \frac{a^3}{4 c^2 \left (a^2 x^2+1\right )}+\frac{7 a^3 \log \left (a^2 x^2+1\right )}{6 c^2}+\frac{\left (15 a^4 x^4+10 a^2 x^2-2\right ) \tan ^{-1}(a x)}{6 c^2 x^3 \left (a^2 x^2+1\right )}-\frac{7 a^3 \log (x)}{3 c^2}+\frac{5 a^3 \tan ^{-1}(a x)^2}{4 c^2}-\frac{a}{6 c^2 x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-a/(6*c^2*x^2) + a^3/(4*c^2*(1 + a^2*x^2)) + ((-2 + 10*a^2*x^2 + 15*a^4*x^4)*ArcTan[a*x])/(6*c^2*x^3*(1 + a^2*
x^2)) + (5*a^3*ArcTan[a*x]^2)/(4*c^2) - (7*a^3*Log[x])/(3*c^2) + (7*a^3*Log[1 + a^2*x^2])/(6*c^2)

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Maple [A]  time = 0.05, size = 125, normalized size = 0.9 \begin{align*}{\frac{{a}^{4}x\arctan \left ( ax \right ) }{2\,{c}^{2} \left ({a}^{2}{x}^{2}+1 \right ) }}+{\frac{5\,{a}^{3} \left ( \arctan \left ( ax \right ) \right ) ^{2}}{4\,{c}^{2}}}-{\frac{\arctan \left ( ax \right ) }{3\,{c}^{2}{x}^{3}}}+2\,{\frac{{a}^{2}\arctan \left ( ax \right ) }{{c}^{2}x}}+{\frac{7\,{a}^{3}\ln \left ({a}^{2}{x}^{2}+1 \right ) }{6\,{c}^{2}}}+{\frac{{a}^{3}}{4\,{c}^{2} \left ({a}^{2}{x}^{2}+1 \right ) }}-{\frac{a}{6\,{c}^{2}{x}^{2}}}-{\frac{7\,{a}^{3}\ln \left ( ax \right ) }{3\,{c}^{2}}} \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)^2,x)

[Out]

1/2*a^4*x*arctan(a*x)/c^2/(a^2*x^2+1)+5/4*a^3*arctan(a*x)^2/c^2-1/3*arctan(a*x)/c^2/x^3+2*a^2*arctan(a*x)/c^2/
x+7/6*a^3*ln(a^2*x^2+1)/c^2+1/4*a^3/c^2/(a^2*x^2+1)-1/6*a/c^2/x^2-7/3*a^3/c^2*ln(a*x)

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Maxima [A]  time = 1.63609, size = 216, normalized size = 1.59 \begin{align*} \frac{1}{6} \,{\left (\frac{15 \, a^{3} \arctan \left (a x\right )}{c^{2}} + \frac{15 \, a^{4} x^{4} + 10 \, a^{2} x^{2} - 2}{a^{2} c^{2} x^{5} + c^{2} x^{3}}\right )} \arctan \left (a x\right ) + \frac{{\left (a^{2} x^{2} - 15 \,{\left (a^{4} x^{4} + a^{2} x^{2}\right )} \arctan \left (a x\right )^{2} + 14 \,{\left (a^{4} x^{4} + a^{2} x^{2}\right )} \log \left (a^{2} x^{2} + 1\right ) - 28 \,{\left (a^{4} x^{4} + a^{2} x^{2}\right )} \log \left (x\right ) - 2\right )} a}{12 \,{\left (a^{2} c^{2} x^{4} + c^{2} x^{2}\right )}} \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)^2,x, algorithm="maxima")

[Out]

1/6*(15*a^3*arctan(a*x)/c^2 + (15*a^4*x^4 + 10*a^2*x^2 - 2)/(a^2*c^2*x^5 + c^2*x^3))*arctan(a*x) + 1/12*(a^2*x
^2 - 15*(a^4*x^4 + a^2*x^2)*arctan(a*x)^2 + 14*(a^4*x^4 + a^2*x^2)*log(a^2*x^2 + 1) - 28*(a^4*x^4 + a^2*x^2)*l
og(x) - 2)*a/(a^2*c^2*x^4 + c^2*x^2)

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Fricas [A]  time = 1.75338, size = 279, normalized size = 2.05 \begin{align*} \frac{a^{3} x^{3} + 15 \,{\left (a^{5} x^{5} + a^{3} x^{3}\right )} \arctan \left (a x\right )^{2} - 2 \, a x + 2 \,{\left (15 \, a^{4} x^{4} + 10 \, a^{2} x^{2} - 2\right )} \arctan \left (a x\right ) + 14 \,{\left (a^{5} x^{5} + a^{3} x^{3}\right )} \log \left (a^{2} x^{2} + 1\right ) - 28 \,{\left (a^{5} x^{5} + a^{3} x^{3}\right )} \log \left (x\right )}{12 \,{\left (a^{2} c^{2} x^{5} + c^{2} x^{3}\right )}} \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)^2,x, algorithm="fricas")

[Out]

1/12*(a^3*x^3 + 15*(a^5*x^5 + a^3*x^3)*arctan(a*x)^2 - 2*a*x + 2*(15*a^4*x^4 + 10*a^2*x^2 - 2)*arctan(a*x) + 1
4*(a^5*x^5 + a^3*x^3)*log(a^2*x^2 + 1) - 28*(a^5*x^5 + a^3*x^3)*log(x))/(a^2*c^2*x^5 + c^2*x^3)

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Sympy [B]  time = 3.4458, size = 360, normalized size = 2.65 \begin{align*} - \frac{28 a^{5} x^{5} \log{\left (x \right )}}{12 a^{2} c^{2} x^{5} + 12 c^{2} x^{3}} + \frac{14 a^{5} x^{5} \log{\left (x^{2} + \frac{1}{a^{2}} \right )}}{12 a^{2} c^{2} x^{5} + 12 c^{2} x^{3}} + \frac{15 a^{5} x^{5} \operatorname{atan}^{2}{\left (a x \right )}}{12 a^{2} c^{2} x^{5} + 12 c^{2} x^{3}} + \frac{30 a^{4} x^{4} \operatorname{atan}{\left (a x \right )}}{12 a^{2} c^{2} x^{5} + 12 c^{2} x^{3}} - \frac{28 a^{3} x^{3} \log{\left (x \right )}}{12 a^{2} c^{2} x^{5} + 12 c^{2} x^{3}} + \frac{14 a^{3} x^{3} \log{\left (x^{2} + \frac{1}{a^{2}} \right )}}{12 a^{2} c^{2} x^{5} + 12 c^{2} x^{3}} + \frac{15 a^{3} x^{3} \operatorname{atan}^{2}{\left (a x \right )}}{12 a^{2} c^{2} x^{5} + 12 c^{2} x^{3}} + \frac{a^{3} x^{3}}{12 a^{2} c^{2} x^{5} + 12 c^{2} x^{3}} + \frac{20 a^{2} x^{2} \operatorname{atan}{\left (a x \right )}}{12 a^{2} c^{2} x^{5} + 12 c^{2} x^{3}} - \frac{2 a x}{12 a^{2} c^{2} x^{5} + 12 c^{2} x^{3}} - \frac{4 \operatorname{atan}{\left (a x \right )}}{12 a^{2} c^{2} x^{5} + 12 c^{2} x^{3}} \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)**2,x)

[Out]

-28*a**5*x**5*log(x)/(12*a**2*c**2*x**5 + 12*c**2*x**3) + 14*a**5*x**5*log(x**2 + a**(-2))/(12*a**2*c**2*x**5
+ 12*c**2*x**3) + 15*a**5*x**5*atan(a*x)**2/(12*a**2*c**2*x**5 + 12*c**2*x**3) + 30*a**4*x**4*atan(a*x)/(12*a*
*2*c**2*x**5 + 12*c**2*x**3) - 28*a**3*x**3*log(x)/(12*a**2*c**2*x**5 + 12*c**2*x**3) + 14*a**3*x**3*log(x**2
+ a**(-2))/(12*a**2*c**2*x**5 + 12*c**2*x**3) + 15*a**3*x**3*atan(a*x)**2/(12*a**2*c**2*x**5 + 12*c**2*x**3) +
a**3*x**3/(12*a**2*c**2*x**5 + 12*c**2*x**3) + 20*a**2*x**2*atan(a*x)/(12*a**2*c**2*x**5 + 12*c**2*x**3) - 2*
a*x/(12*a**2*c**2*x**5 + 12*c**2*x**3) - 4*atan(a*x)/(12*a**2*c**2*x**5 + 12*c**2*x**3)

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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^{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)^2,x, algorithm="giac")

[Out]

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