3.9 \(\int \frac {b+a x}{1+x^2} \, dx\)

Optimal. Leaf size=16 \[ \frac {1}{2} a \log \left (x^2+1\right )+b \tan ^{-1}(x) \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {635, 203, 260} \[ \frac {1}{2} a \log \left (x^2+1\right )+b \tan ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[(b + a*x)/(1 + x^2),x]

[Out]

b*ArcTan[x] + (a*Log[1 + x^2])/2

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])

Rule 260

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 635

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[-(a*c)]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 16, normalized size = 1.00 \[ \frac {1}{2} a \log \left (x^2+1\right )+b \tan ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Integrate[(b + a*x)/(1 + x^2),x]

[Out]

b*ArcTan[x] + (a*Log[1 + x^2])/2

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fricas [A]  time = 0.40, size = 14, normalized size = 0.88 \[ b \arctan \relax (x) + \frac {1}{2} \, a \log \left (x^{2} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*arctan(x) + 1/2*a*log(x^2 + 1)

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giac [A]  time = 1.19, size = 14, normalized size = 0.88 \[ b \arctan \relax (x) + \frac {1}{2} \, a \log \left (x^{2} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*arctan(x) + 1/2*a*log(x^2 + 1)

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maple [A]  time = 0.00, size = 15, normalized size = 0.94 \[ \frac {a \ln \left (x^{2}+1\right )}{2}+b \arctan \relax (x ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x+b)/(x^2+1),x)

[Out]

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

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maxima [A]  time = 0.97, size = 14, normalized size = 0.88 \[ b \arctan \relax (x) + \frac {1}{2} \, a \log \left (x^{2} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*arctan(x) + 1/2*a*log(x^2 + 1)

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mupad [B]  time = 0.04, size = 14, normalized size = 0.88 \[ \frac {a\,\ln \left (x^2+1\right )}{2}+b\,\mathrm {atan}\relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b + a*x)/(x^2 + 1),x)

[Out]

(a*log(x^2 + 1))/2 + b*atan(x)

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sympy [C]  time = 0.16, size = 26, normalized size = 1.62 \[ \left (\frac {a}{2} - \frac {i b}{2}\right ) \log {\left (x - i \right )} + \left (\frac {a}{2} + \frac {i b}{2}\right ) \log {\left (x + i \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+b)/(x**2+1),x)

[Out]

(a/2 - I*b/2)*log(x - I) + (a/2 + I*b/2)*log(x + I)

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