3.1.68 \(\int x \log (b+a x) \, dx\) [68]

Optimal. Leaf size=46 \[ \frac {b x}{2 a}-\frac {x^2}{4}-\frac {b^2 \log (b+a x)}{2 a^2}+\frac {1}{2} x^2 \log (b+a x) \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2442, 45} \begin {gather*} -\frac {b^2 \log (a x+b)}{2 a^2}+\frac {1}{2} x^2 \log (a x+b)+\frac {b x}{2 a}-\frac {x^2}{4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x*Log[b + a*x],x]

[Out]

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

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2442

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))*((f_.) + (g_.)*(x_))^(q_.), x_Symbol] :> Simp[(f + g*
x)^(q + 1)*((a + b*Log[c*(d + e*x)^n])/(g*(q + 1))), x] - Dist[b*e*(n/(g*(q + 1))), Int[(f + g*x)^(q + 1)/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, q}, x] && NeQ[e*f - d*g, 0] && NeQ[q, -1]

Rubi steps

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

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Mathematica [A]
time = 0.01, size = 46, normalized size = 1.00 \begin {gather*} \frac {b x}{2 a}-\frac {x^2}{4}-\frac {b^2 \log (b+a x)}{2 a^2}+\frac {1}{2} x^2 \log (b+a x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x*Log[b + a*x],x]

[Out]

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

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Mathics [A]
time = 2.19, size = 38, normalized size = 0.83 \begin {gather*} \frac {b x}{2 a}-\frac {b^2 \text {Log}\left [a x+b\right ]}{2 a^2}-\frac {x^2}{4}+\frac {x^2 \text {Log}\left [a x+b\right ]}{2} \end {gather*}

Antiderivative was successfully verified.

[In]

mathics('Integrate[x*Log[a*x+b],x]')

[Out]

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

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Maple [A]
time = 0.02, size = 53, normalized size = 1.15

method	result	size
norman	\(\frac {b x}{2 a}-\frac {x^{2}}{4}-\frac {b^{2} \ln \left (a x +b \right )}{2 a^{2}}+\frac {x^{2} \ln \left (a x +b \right )}{2}\)	\(39\)
risch	\(\frac {b x}{2 a}-\frac {x^{2}}{4}-\frac {b^{2} \ln \left (a x +b \right )}{2 a^{2}}+\frac {x^{2} \ln \left (a x +b \right )}{2}\)	\(39\)
derivativedivides	\(\frac {-b \left (\ln \left (a x +b \right ) \left (a x +b \right )-a x -b \right )+\frac {\left (a x +b \right )^{2} \ln \left (a x +b \right )}{2}-\frac {\left (a x +b \right )^{2}}{4}}{a^{2}}\)	\(53\)
default	\(\frac {-b \left (\ln \left (a x +b \right ) \left (a x +b \right )-a x -b \right )+\frac {\left (a x +b \right )^{2} \ln \left (a x +b \right )}{2}-\frac {\left (a x +b \right )^{2}}{4}}{a^{2}}\)	\(53\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.26, size = 44, normalized size = 0.96 \begin {gather*} \frac {1}{2} \, x^{2} \log \left (a x + b\right ) - \frac {1}{4} \, a {\left (\frac {2 \, b^{2} \log \left (a x + b\right )}{a^{3}} + \frac {a x^{2} - 2 \, b x}{a^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^2*log(a*x + b) - 1/4*a*(2*b^2*log(a*x + b)/a^3 + (a*x^2 - 2*b*x)/a^2)

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Fricas [A]
time = 0.34, size = 39, normalized size = 0.85 \begin {gather*} -\frac {a^{2} x^{2} - 2 \, a b x - 2 \, {\left (a^{2} x^{2} - b^{2}\right )} \log \left (a x + b\right )}{4 \, a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.09, size = 42, normalized size = 0.91 \begin {gather*} - a \left (\frac {x^{2}}{4 a} - \frac {b x}{2 a^{2}} + \frac {b^{2} \log {\left (a x + b \right )}}{2 a^{3}}\right ) + \frac {x^{2} \log {\left (a x + b \right )}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a*(x**2/(4*a) - b*x/(2*a**2) + b**2*log(a*x + b)/(2*a**3)) + x**2*log(a*x + b)/2

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Giac [A]
time = 0.00, size = 65, normalized size = 1.41 \begin {gather*} \frac {b \left (a x+b\right )}{a^{2}}-\frac {\left (a x+b\right )^{2}}{4 a^{2}}+\frac {\left (a x+b\right )^{2} \ln \left (a x+b\right )}{2 a^{2}}-\frac {b \left (a x+b\right ) \ln \left (a x+b\right )}{a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.19, size = 66, normalized size = 1.43 \begin {gather*} \left \{\begin {array}{cl} \frac {x^2\,\left (\ln \left (a\,x\right )-\frac {1}{2}\right )}{2} & \text {\ if\ \ }b=0\\ \frac {\ln \left (b+a\,x\right )\,\left (x^2-\frac {b^2}{a^2}\right )}{2}-\frac {b^2\,\left (\frac {a^2\,x^2}{2\,b^2}-\frac {a\,x}{b}\right )}{2\,a^2} & \text {\ if\ \ }b\neq 0 \end {array}\right . \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*log(b + a*x),x)

[Out]

piecewise(b == 0, (x^2*(log(a*x) - 1/2))/2, b ~= 0, (log(b + a*x)*(x^2 - b^2/a^2))/2 - (b^2*((a^2*x^2)/(2*b^2)
- (a*x)/b))/(2*a^2))

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