3.1.65 \(\int (b+a x) \log (x) \, dx\) [65]

Optimal. Leaf size=28 \[ -b x-\frac {a x^2}{4}+b x \log (x)+\frac {1}{2} a x^2 \log (x) \]

[Out]

-b*x-1/4*a*x^2+b*x*ln(x)+1/2*a*x^2*ln(x)

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Rubi [A]
time = 0.01, antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {2350} \begin {gather*} -\frac {a x^2}{4}+\frac {1}{2} a x^2 \log (x)-b x+b x \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b + a*x)*Log[x],x]

[Out]

-(b*x) - (a*x^2)/4 + b*x*Log[x] + (a*x^2*Log[x])/2

Rule 2350

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> With[{u = IntHide[(d +
e*x^r)^q, x]}, Dist[a + b*Log[c*x^n], u, x] - Dist[b*n, Int[SimplifyIntegrand[u/x, x], x], x]] /; FreeQ[{a, b
, c, d, e, n, r}, x] && IGtQ[q, 0]

Rubi steps

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

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Mathematica [A]
time = 0.00, size = 28, normalized size = 1.00 \begin {gather*} -b x-\frac {a x^2}{4}+b x \log (x)+\frac {1}{2} a x^2 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(b + a*x)*Log[x],x]

[Out]

-(b*x) - (a*x^2)/4 + b*x*Log[x] + (a*x^2*Log[x])/2

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Maple [A]
time = 0.01, size = 27, normalized size = 0.96

method	result	size
norman	\(-b x -\frac {a \,x^{2}}{4}+b x \ln \left (x \right )+\frac {a \,x^{2} \ln \left (x \right )}{2}\)	\(25\)
risch	\(\left (\frac {1}{2} a \,x^{2}+b x \right ) \ln \left (x \right )-\frac {a \,x^{2}}{4}-b x\)	\(25\)
default	\(a \left (-\frac {x^{2}}{4}+\frac {x^{2} \ln \left (x \right )}{2}\right )+b \left (-x +x \ln \left (x \right )\right )\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x+b)*ln(x),x,method=_RETURNVERBOSE)

[Out]

a*(-1/4*x^2+1/2*x^2*ln(x))+b*(-x+x*ln(x))

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Maxima [A]
time = 1.47, size = 25, normalized size = 0.89 \begin {gather*} -\frac {1}{4} \, a x^{2} - b x + \frac {1}{2} \, {\left (a x^{2} + 2 \, b x\right )} \log \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+b)*log(x),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.83, size = 25, normalized size = 0.89 \begin {gather*} -\frac {1}{4} \, a x^{2} - b x + \frac {1}{2} \, {\left (a x^{2} + 2 \, b x\right )} \log \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+b)*log(x),x, algorithm="fricas")

[Out]

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

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Sympy [A]
time = 0.04, size = 22, normalized size = 0.79 \begin {gather*} - \frac {a x^{2}}{4} - b x + \left (\frac {a x^{2}}{2} + b x\right ) \log {\left (x \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+b)*ln(x),x)

[Out]

-a*x**2/4 - b*x + (a*x**2/2 + b*x)*log(x)

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Giac [A]
time = 1.06, size = 24, normalized size = 0.86 \begin {gather*} \frac {1}{2} \, a x^{2} \log \left (x\right ) - \frac {1}{4} \, a x^{2} + b x \log \left (x\right ) - b x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+b)*log(x),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 0.14, size = 21, normalized size = 0.75 \begin {gather*} -\frac {x\,\left (4\,b+a\,x-4\,b\,\ln \left (x\right )-2\,a\,x\,\ln \left (x\right )\right )}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(x)*(b + a*x),x)

[Out]

-(x*(4*b + a*x - 4*b*log(x) - 2*a*x*log(x)))/4

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