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

Optimal. Leaf size=55 \[ -\frac{a+b \tan ^{-1}\left (\frac{c}{x}\right )}{3 x^3}-\frac{b \log \left (c^2+x^2\right )}{6 c^3}+\frac{b \log (x)}{3 c^3}+\frac{b}{6 c x^2} \]

[Out]

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

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Rubi [A]  time = 0.0362819, antiderivative size = 55, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {5033, 263, 266, 44} \[ -\frac{a+b \tan ^{-1}\left (\frac{c}{x}\right )}{3 x^3}-\frac{b \log \left (c^2+x^2\right )}{6 c^3}+\frac{b \log (x)}{3 c^3}+\frac{b}{6 c x^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTan[c/x])/x^4,x]

[Out]

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

Rule 5033

Int[((a_.) + ArcTan[(c_.)*(x_)^(n_)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcTan
[c*x^n]))/(d*(m + 1)), x] - Dist[(b*c*n)/(d*(m + 1)), Int[(x^(n - 1)*(d*x)^(m + 1))/(1 + c^2*x^(2*n)), x], x]
/; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1]

Rule 263

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

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

Mathematica [A]  time = 0.0099862, size = 60, normalized size = 1.09 \[ -\frac{a}{3 x^3}-\frac{b \log \left (c^2+x^2\right )}{6 c^3}+\frac{b \log (x)}{3 c^3}+\frac{b}{6 c x^2}-\frac{b \tan ^{-1}\left (\frac{c}{x}\right )}{3 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcTan[c/x])/x^4,x]

[Out]

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

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Maple [A]  time = 0.026, size = 45, normalized size = 0.8 \begin{align*} -{\frac{a}{3\,{x}^{3}}}-{\frac{b}{3\,{x}^{3}}\arctan \left ({\frac{c}{x}} \right ) }+{\frac{b}{6\,c{x}^{2}}}-{\frac{b}{6\,{c}^{3}}\ln \left ( 1+{\frac{{c}^{2}}{{x}^{2}}} \right ) } \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/x^2-1/6/c^3*b*ln(1+c^2/x^2)

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Maxima [A]  time = 1.02616, size = 73, normalized size = 1.33 \begin{align*} -\frac{1}{6} \,{\left (c{\left (\frac{\log \left (c^{2} + x^{2}\right )}{c^{4}} - \frac{\log \left (x^{2}\right )}{c^{4}} - \frac{1}{c^{2} x^{2}}\right )} + \frac{2 \, \arctan \left (\frac{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*(log(c^2 + x^2)/c^4 - log(x^2)/c^4 - 1/(c^2*x^2)) + 2*arctan(c/x)/x^3)*b - 1/3*a/x^3

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Fricas [A]  time = 2.23219, size = 132, normalized size = 2.4 \begin{align*} -\frac{2 \, b c^{3} \arctan \left (\frac{c}{x}\right ) + b x^{3} \log \left (c^{2} + x^{2}\right ) - 2 \, b x^{3} \log \left (x\right ) + 2 \, a c^{3} - b c^{2} x}{6 \, c^{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="fricas")

[Out]

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

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Sympy [A]  time = 1.63721, size = 60, normalized size = 1.09 \begin{align*} \begin{cases} - \frac{a}{3 x^{3}} - \frac{b \operatorname{atan}{\left (\frac{c}{x} \right )}}{3 x^{3}} + \frac{b}{6 c x^{2}} + \frac{b \log{\left (x \right )}}{3 c^{3}} - \frac{b \log{\left (c^{2} + x^{2} \right )}}{6 c^{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*atan(c/x)/(3*x**3) + b/(6*c*x**2) + b*log(x)/(3*c**3) - b*log(c**2 + x**2)/(6*c**3)
, Ne(c, 0)), (-a/(3*x**3), True))

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Giac [A]  time = 1.13462, size = 74, normalized size = 1.35 \begin{align*} -\frac{2 \, b c^{3} \arctan \left (\frac{c}{x}\right ) + b x^{3} \log \left (c^{2} + x^{2}\right ) - 2 \, b x^{3} \log \left (x\right ) + 2 \, a c^{3} - b c^{2} x}{6 \, c^{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="giac")

[Out]

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