3.138 \(\int \frac{a+b \tan ^{-1}(\frac{c}{x})}{x^3} \, dx\)

Optimal. Leaf size=43 \[ -\frac{a+b \tan ^{-1}\left (\frac{c}{x}\right )}{2 x^2}+\frac{b \tan ^{-1}\left (\frac{x}{c}\right )}{2 c^2}+\frac{b}{2 c x} \]

[Out]

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

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Rubi [A]  time = 0.0238361, antiderivative size = 43, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {5033, 263, 325, 203} \[ -\frac{a+b \tan ^{-1}\left (\frac{c}{x}\right )}{2 x^2}+\frac{b \tan ^{-1}\left (\frac{x}{c}\right )}{2 c^2}+\frac{b}{2 c x} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTan[c/x])/x^3,x]

[Out]

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

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 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 0.0098911, size = 48, normalized size = 1.12 \[ -\frac{a}{2 x^2}+\frac{b \tan ^{-1}\left (\frac{x}{c}\right )}{2 c^2}-\frac{b \tan ^{-1}\left (\frac{c}{x}\right )}{2 x^2}+\frac{b}{2 c x} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcTan[c/x])/x^3,x]

[Out]

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

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Maple [A]  time = 0.026, size = 41, normalized size = 1. \begin{align*} -{\frac{a}{2\,{x}^{2}}}-{\frac{b}{2\,{x}^{2}}\arctan \left ({\frac{c}{x}} \right ) }+{\frac{b}{2\,cx}}+{\frac{b}{2\,{c}^{2}}\arctan \left ({\frac{x}{c}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctan(c/x))/x^3,x)

[Out]

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

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Maxima [A]  time = 1.52523, size = 57, normalized size = 1.33 \begin{align*} \frac{1}{2} \,{\left (c{\left (\frac{\arctan \left (\frac{x}{c}\right )}{c^{3}} + \frac{1}{c^{2} x}\right )} - \frac{\arctan \left (\frac{c}{x}\right )}{x^{2}}\right )} b - \frac{a}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(c/x))/x^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.06838, size = 84, normalized size = 1.95 \begin{align*} -\frac{a c^{2} - b c x +{\left (b c^{2} + b x^{2}\right )} \arctan \left (\frac{c}{x}\right )}{2 \, c^{2} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(c/x))/x^3,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 1.23747, size = 44, normalized size = 1.02 \begin{align*} \begin{cases} - \frac{a}{2 x^{2}} - \frac{b \operatorname{atan}{\left (\frac{c}{x} \right )}}{2 x^{2}} + \frac{b}{2 c x} - \frac{b \operatorname{atan}{\left (\frac{c}{x} \right )}}{2 c^{2}} & \text{for}\: c \neq 0 \\- \frac{a}{2 x^{2}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atan(c/x))/x**3,x)

[Out]

Piecewise((-a/(2*x**2) - b*atan(c/x)/(2*x**2) + b/(2*c*x) - b*atan(c/x)/(2*c**2), Ne(c, 0)), (-a/(2*x**2), Tru
e))

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Giac [A]  time = 1.14501, size = 84, normalized size = 1.95 \begin{align*} -\frac{2 \, b c^{2} i \arctan \left (\frac{c}{x}\right ) + 2 \, a c^{2} i - 2 \, b c i x - b x^{2} \log \left (i x + c\right ) + b x^{2} \log \left (-i x + c\right )}{4 \, c^{2} i x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(c/x))/x^3,x, algorithm="giac")

[Out]

-1/4*(2*b*c^2*i*arctan(c/x) + 2*a*c^2*i - 2*b*c*i*x - b*x^2*log(i*x + c) + b*x^2*log(-i*x + c))/(c^2*i*x^2)