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

Optimal. Leaf size=53 \[ -\frac{a+b \tan ^{-1}(c x)}{3 x^3}+\frac{1}{6} b c^3 \log \left (c^2 x^2+1\right )-\frac{1}{3} b c^3 \log (x)-\frac{b c}{6 x^2} \]

[Out]

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

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Rubi [A]  time = 0.033227, antiderivative size = 53, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {4852, 266, 44} \[ -\frac{a+b \tan ^{-1}(c x)}{3 x^3}+\frac{1}{6} b c^3 \log \left (c^2 x^2+1\right )-\frac{1}{3} b c^3 \log (x)-\frac{b c}{6 x^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTan[c*x])/x^4,x]

[Out]

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

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])

Rubi steps

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

Mathematica [A]  time = 0.015549, size = 54, normalized size = 1.02 \[ -\frac{a}{3 x^3}+\frac{1}{6} b c \left (c^2 \log \left (c^2 x^2+1\right )-2 c^2 \log (x)-\frac{1}{x^2}\right )-\frac{b \tan ^{-1}(c x)}{3 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcTan[c*x])/x^4,x]

[Out]

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

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Maple [A]  time = 0.007, size = 51, normalized size = 1. \begin{align*} -{\frac{a}{3\,{x}^{3}}}-{\frac{b\arctan \left ( cx \right ) }{3\,{x}^{3}}}+{\frac{b{c}^{3}\ln \left ({c}^{2}{x}^{2}+1 \right ) }{6}}-{\frac{bc}{6\,{x}^{2}}}-{\frac{{c}^{3}b\ln \left ( cx \right ) }{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.969048, size = 69, normalized size = 1.3 \begin{align*} \frac{1}{6} \,{\left ({\left (c^{2} \log \left (c^{2} x^{2} + 1\right ) - c^{2} \log \left (x^{2}\right ) - \frac{1}{x^{2}}\right )} c - \frac{2 \, \arctan \left (c x\right )}{x^{3}}\right )} b - \frac{a}{3 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.48142, size = 123, normalized size = 2.32 \begin{align*} \frac{b c^{3} x^{3} \log \left (c^{2} x^{2} + 1\right ) - 2 \, b c^{3} x^{3} \log \left (x\right ) - b c x - 2 \, b \arctan \left (c x\right ) - 2 \, a}{6 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 1.5361, size = 61, normalized size = 1.15 \begin{align*} \begin{cases} - \frac{a}{3 x^{3}} - \frac{b c^{3} \log{\left (x \right )}}{3} + \frac{b c^{3} \log{\left (x^{2} + \frac{1}{c^{2}} \right )}}{6} - \frac{b c}{6 x^{2}} - \frac{b \operatorname{atan}{\left (c x \right )}}{3 x^{3}} & \text{for}\: c \neq 0 \\- \frac{a}{3 x^{3}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.80703, size = 68, normalized size = 1.28 \begin{align*} \frac{b c^{3} x^{3} \log \left (c^{2} x^{2} + 1\right ) - 2 \, b c^{3} x^{3} \log \left (x\right ) - b c x - 2 \, b \arctan \left (c x\right ) - 2 \, a}{6 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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