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

Optimal. Leaf size=63 \[ -\frac{1}{3} a^3 c \log \left (a^2 x^2+1\right )+\frac{2}{3} a^3 c \log (x)-\frac{a^2 c \tan ^{-1}(a x)}{x}-\frac{a c}{6 x^2}-\frac{c \tan ^{-1}(a x)}{3 x^3} \]

[Out]

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

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Rubi [A]  time = 0.0822095, antiderivative size = 63, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 7, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.389, Rules used = {4950, 4852, 266, 44, 36, 29, 31} \[ -\frac{1}{3} a^3 c \log \left (a^2 x^2+1\right )+\frac{2}{3} a^3 c \log (x)-\frac{a^2 c \tan ^{-1}(a x)}{x}-\frac{a c}{6 x^2}-\frac{c \tan ^{-1}(a x)}{3 x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4950

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

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]

Rubi steps

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

Mathematica [A]  time = 0.0200303, size = 58, normalized size = 0.92 \[ \frac{c \left (a x \left (4 a^2 x^2 \log (x)-2 a^2 x^2 \log \left (a^2 x^2+1\right )-1\right )-2 \left (3 a^2 x^2+1\right ) \tan ^{-1}(a x)\right )}{6 x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.033, size = 58, normalized size = 0.9 \begin{align*} -{\frac{{a}^{2}c\arctan \left ( ax \right ) }{x}}-{\frac{c\arctan \left ( ax \right ) }{3\,{x}^{3}}}-{\frac{{a}^{3}c\ln \left ({a}^{2}{x}^{2}+1 \right ) }{3}}-{\frac{ac}{6\,{x}^{2}}}+{\frac{2\,{a}^{3}c\ln \left ( ax \right ) }{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.990587, size = 76, normalized size = 1.21 \begin{align*} -\frac{1}{6} \,{\left (2 \, a^{2} c \log \left (a^{2} x^{2} + 1\right ) - 2 \, a^{2} c \log \left (x^{2}\right ) + \frac{c}{x^{2}}\right )} a - \frac{{\left (3 \, a^{2} c x^{2} + c\right )} \arctan \left (a x\right )}{3 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.59627, size = 140, normalized size = 2.22 \begin{align*} -\frac{2 \, a^{3} c x^{3} \log \left (a^{2} x^{2} + 1\right ) - 4 \, a^{3} c x^{3} \log \left (x\right ) + a c x + 2 \,{\left (3 \, a^{2} c x^{2} + c\right )} \arctan \left (a x\right )}{6 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 1.52307, size = 61, normalized size = 0.97 \begin{align*} \begin{cases} \frac{2 a^{3} c \log{\left (x \right )}}{3} - \frac{a^{3} c \log{\left (x^{2} + \frac{1}{a^{2}} \right )}}{3} - \frac{a^{2} c \operatorname{atan}{\left (a x \right )}}{x} - \frac{a c}{6 x^{2}} - \frac{c \operatorname{atan}{\left (a x \right )}}{3 x^{3}} & \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)*atan(a*x)/x**4,x)

[Out]

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

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Giac [A]  time = 1.12177, size = 88, normalized size = 1.4 \begin{align*} -\frac{1}{3} \, a^{3} c \log \left (a^{2} x^{2} + 1\right ) + \frac{1}{3} \, a^{3} c \log \left (x^{2}\right ) - \frac{2 \, a^{3} c x^{2} + a c}{6 \, x^{2}} - \frac{{\left (3 \, a^{2} c x^{2} + c\right )} \arctan \left (a x\right )}{3 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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