\(\int (a+b x) \log (a+b x) \, dx\) [245]

Optimal result

Integrand size = 12, antiderivative size = 35 \[ \int (a+b x) \log (a+b x) \, dx=-\frac {(a+b x)^2}{4 b}+\frac {(a+b x)^2 \log (a+b x)}{2 b} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2437, 2341} \[ \int (a+b x) \log (a+b x) \, dx=\frac {(a+b x)^2 \log (a+b x)}{2 b}-\frac {(a+b x)^2}{4 b} \]

[In]

Int[(a + b*x)*Log[a + b*x],x]

[Out]

-1/4*(a + b*x)^2/b + ((a + b*x)^2*Log[a + b*x])/(2*b)

Rule 2341

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*Log[c*x^
n])/(d*(m + 1))), x] - Simp[b*n*((d*x)^(m + 1)/(d*(m + 1)^2)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

Rule 2437

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

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}(\int x \log (x) \, dx,x,a+b x)}{b} \\ & = -\frac {(a+b x)^2}{4 b}+\frac {(a+b x)^2 \log (a+b x)}{2 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.94 \[ \int (a+b x) \log (a+b x) \, dx=-\frac {1}{4} x (2 a+b x)+\frac {(a+b x)^2 \log (a+b x)}{2 b} \]

[In]

Integrate[(a + b*x)*Log[a + b*x],x]

[Out]

-1/4*(x*(2*a + b*x)) + ((a + b*x)^2*Log[a + b*x])/(2*b)

Maple [A] (verified)

Time = 0.20 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.86

method	result	size
derivativedivides	\(\frac {\frac {\left (b x +a \right )^{2} \ln \left (b x +a \right )}{2}-\frac {\left (b x +a \right )^{2}}{4}}{b}\)	\(30\)
default	\(\frac {\frac {\left (b x +a \right )^{2} \ln \left (b x +a \right )}{2}-\frac {\left (b x +a \right )^{2}}{4}}{b}\)	\(30\)
risch	\(\left (\frac {1}{2} b \,x^{2}+a x \right ) \ln \left (b x +a \right )-\frac {b \,x^{2}}{4}-\frac {a x}{2}+\frac {a^{2} \ln \left (b x +a \right )}{2 b}\)	\(43\)
norman	\(a x \ln \left (b x +a \right )-\frac {a x}{2}-\frac {b \,x^{2}}{4}+\frac {a^{2} \ln \left (b x +a \right )}{2 b}+\frac {b \,x^{2} \ln \left (b x +a \right )}{2}\)	\(47\)
parts	\(\frac {b \,x^{2} \ln \left (b x +a \right )}{2}+a x \ln \left (b x +a \right )-\frac {b \left (\frac {\frac {1}{2} b \,x^{2}+a x}{b}-\frac {a^{2} \ln \left (b x +a \right )}{b^{2}}\right )}{2}\)	\(55\)
parallelrisch	\(\frac {2 x^{2} \ln \left (b x +a \right ) b^{2}-b^{2} x^{2}+4 x \ln \left (b x +a \right ) a b -2 a b x +2 a^{2} \ln \left (b x +a \right )+2 a^{2}}{4 b}\)	\(61\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.20 \[ \int (a+b x) \log (a+b x) \, dx=-\frac {b^{2} x^{2} + 2 \, a b x - 2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \log \left (b x + a\right )}{4 \, b} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.17 \[ \int (a+b x) \log (a+b x) \, dx=\frac {a^{2} \log {\left (a + b x \right )}}{2 b} - \frac {a x}{2} - \frac {b x^{2}}{4} + \left (a x + \frac {b x^{2}}{2}\right ) \log {\left (a + b x \right )} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.49 \[ \int (a+b x) \log (a+b x) \, dx=\frac {1}{4} \, b {\left (\frac {2 \, a^{2} \log \left (b x + a\right )}{b^{2}} - \frac {b x^{2} + 2 \, a x}{b}\right )} + \frac {1}{2} \, {\left (b x^{2} + 2 \, a x\right )} \log \left (b x + a\right ) \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.89 \[ \int (a+b x) \log (a+b x) \, dx=\frac {{\left (b x + a\right )}^{2} \log \left (b x + a\right )}{2 \, b} - \frac {{\left (b x + a\right )}^{2}}{4 \, b} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 1.61 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.31 \[ \int (a+b x) \log (a+b x) \, dx=\frac {a^2\,\ln \left (a+b\,x\right )}{2\,b}-\frac {b\,x^2}{4}-\frac {a\,x}{2}+a\,x\,\ln \left (a+b\,x\right )+\frac {b\,x^2\,\ln \left (a+b\,x\right )}{2} \]

[In]

int(log(a + b*x)*(a + b*x),x)

[Out]

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