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] + (a*Log[1 + x^2])/2

________________________________________________________________________________________

Rubi [A]  time = 0.0063259, antiderivative size = 16, normalized size of antiderivative = 1., 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 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)]

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]

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*}

Mathematica [A]  time = 0.0059073, size = 16, normalized size = 1. \[ \frac{1}{2} a \log \left (x^2+1\right )+b \tan ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Integrate[(b + a*x)/(1 + x^2),x]

[Out]

b*ArcTan[x] + (a*Log[1 + x^2])/2

________________________________________________________________________________________

Maple [A]  time = 0.002, size = 15, normalized size = 0.9 \begin{align*} b\arctan \left ( x \right ) +{\frac{a\ln \left ({x}^{2}+1 \right ) }{2}} \end{align*}

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)

________________________________________________________________________________________

Maxima [A]  time = 1.41959, size = 19, normalized size = 1.19 \begin{align*} b \arctan \left (x\right ) + \frac{1}{2} \, a \log \left (x^{2} + 1\right ) \end{align*}

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)

________________________________________________________________________________________

Fricas [A]  time = 1.98883, size = 46, normalized size = 2.88 \begin{align*} b \arctan \left (x\right ) + \frac{1}{2} \, a \log \left (x^{2} + 1\right ) \end{align*}

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)

________________________________________________________________________________________

Sympy [C]  time = 0.153833, size = 26, normalized size = 1.62 \begin{align*} \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 )} \end{align*}

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)

________________________________________________________________________________________

Giac [A]  time = 1.07504, size = 19, normalized size = 1.19 \begin{align*} b \arctan \left (x\right ) + \frac{1}{2} \, a \log \left (x^{2} + 1\right ) \end{align*}

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)