3.7.76 \(\int \frac {(1+x^2) \tan ^{-1}(x)}{x^2} \, dx\) [676]

Optimal. Leaf size=22 \[ -\frac {\tan ^{-1}(x)}{x}+x \tan ^{-1}(x)+\log (x)-\log \left (1+x^2\right ) \]

[Out]

-arctan(x)/x+x*arctan(x)+ln(x)-ln(x^2+1)

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Rubi [A]
time = 0.02, antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.727, Rules used = {5070, 4946, 272, 36, 29, 31, 4930, 266} \begin {gather*} x \text {ArcTan}(x)-\frac {\text {ArcTan}(x)}{x}-\log \left (x^2+1\right )+\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((1 + x^2)*ArcTan[x])/x^2,x]

[Out]

-(ArcTan[x]/x) + x*ArcTan[x] + Log[x] - Log[1 + x^2]

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 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 266

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 272

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 4930

Int[((a_.) + ArcTan[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*ArcTan[c*x^n])^p, x] - Dist[b*c
*n*p, Int[x^n*((a + b*ArcTan[c*x^n])^(p - 1)/(1 + c^2*x^(2*n))), x], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[p, 0
] && (EqQ[n, 1] || EqQ[p, 1])

Rule 4946

Int[((a_.) + ArcTan[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)*((a + b*ArcTan[c*x^
n])^p/(m + 1)), x] - Dist[b*c*n*(p/(m + 1)), Int[x^(m + n)*((a + b*ArcTan[c*x^n])^(p - 1)/(1 + c^2*x^(2*n))),
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0] && (EqQ[p, 1] || (EqQ[n, 1] && IntegerQ[m])) && NeQ[m, -1]

Rule 5070

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]))

Rubi steps

\begin {align*} \int \frac {\left (1+x^2\right ) \tan ^{-1}(x)}{x^2} \, dx &=\int \tan ^{-1}(x) \, dx+\int \frac {\tan ^{-1}(x)}{x^2} \, dx\\ &=-\frac {\tan ^{-1}(x)}{x}+x \tan ^{-1}(x)+\int \frac {1}{x \left (1+x^2\right )} \, dx-\int \frac {x}{1+x^2} \, dx\\ &=-\frac {\tan ^{-1}(x)}{x}+x \tan ^{-1}(x)-\frac {1}{2} \log \left (1+x^2\right )+\frac {1}{2} \text {Subst}\left (\int \frac {1}{x (1+x)} \, dx,x,x^2\right )\\ &=-\frac {\tan ^{-1}(x)}{x}+x \tan ^{-1}(x)-\frac {1}{2} \log \left (1+x^2\right )+\frac {1}{2} \text {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )-\frac {1}{2} \text {Subst}\left (\int \frac {1}{1+x} \, dx,x,x^2\right )\\ &=-\frac {\tan ^{-1}(x)}{x}+x \tan ^{-1}(x)+\log (x)-\log \left (1+x^2\right )\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 22, normalized size = 1.00 \begin {gather*} -\frac {\tan ^{-1}(x)}{x}+x \tan ^{-1}(x)+\log (x)-\log \left (1+x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((1 + x^2)*ArcTan[x])/x^2,x]

[Out]

-(ArcTan[x]/x) + x*ArcTan[x] + Log[x] - Log[1 + x^2]

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Maple [A]
time = 0.05, size = 23, normalized size = 1.05

method	result	size
default	\(-\frac {\arctan \left (x \right )}{x}+x \arctan \left (x \right )+\ln \left (x \right )-\ln \left (x^{2}+1\right )\)	\(23\)
meijerg	\(\ln \left (x \right )-\frac {\arctan \left (\sqrt {x^{2}}\right )}{\sqrt {x^{2}}}-\ln \left (x^{2}+1\right )+\frac {x^{2} \arctan \left (\sqrt {x^{2}}\right )}{\sqrt {x^{2}}}\)	\(40\)
risch	\(-\frac {i \left (x^{2}-1\right ) \ln \left (i x +1\right )}{2 x}+\frac {i \left (-2 i \ln \left (x \right ) x +2 i \ln \left (x^{2}+1\right ) x +x^{2} \ln \left (-i x +1\right )-\ln \left (-i x +1\right )\right )}{2 x}\)	\(63\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2+1)*arctan(x)/x^2,x,method=_RETURNVERBOSE)

[Out]

-1/x*arctan(x)+x*arctan(x)+ln(x)-ln(x^2+1)

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Maxima [A]
time = 1.14, size = 21, normalized size = 0.95 \begin {gather*} {\left (x - \frac {1}{x}\right )} \arctan \left (x\right ) - \log \left (x^{2} + 1\right ) + \log \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2+1)*arctan(x)/x^2,x, algorithm="maxima")

[Out]

(x - 1/x)*arctan(x) - log(x^2 + 1) + log(x)

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Fricas [A]
time = 0.58, size = 26, normalized size = 1.18 \begin {gather*} \frac {{\left (x^{2} - 1\right )} \arctan \left (x\right ) - x \log \left (x^{2} + 1\right ) + x \log \left (x\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2+1)*arctan(x)/x^2,x, algorithm="fricas")

[Out]

((x^2 - 1)*arctan(x) - x*log(x^2 + 1) + x*log(x))/x

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Sympy [A]
time = 0.12, size = 19, normalized size = 0.86 \begin {gather*} x \operatorname {atan}{\left (x \right )} + \log {\left (x \right )} - \log {\left (x^{2} + 1 \right )} - \frac {\operatorname {atan}{\left (x \right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**2+1)*atan(x)/x**2,x)

[Out]

x*atan(x) + log(x) - log(x**2 + 1) - atan(x)/x

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Giac [A]
time = 1.28, size = 25, normalized size = 1.14 \begin {gather*} {\left (x - \frac {1}{x}\right )} \arctan \left (x\right ) - \log \left (x^{2} + 1\right ) + \frac {1}{2} \, \log \left (x^{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2+1)*arctan(x)/x^2,x, algorithm="giac")

[Out]

(x - 1/x)*arctan(x) - log(x^2 + 1) + 1/2*log(x^2)

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Mupad [B]
time = 0.07, size = 22, normalized size = 1.00 \begin {gather*} \ln \left (x\right )-\ln \left (x^2+1\right )-\frac {\mathrm {atan}\left (x\right )}{x}+x\,\mathrm {atan}\left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((atan(x)*(x^2 + 1))/x^2,x)

[Out]

log(x) - log(x^2 + 1) - atan(x)/x + x*atan(x)

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