\(\int (b+a x)^2 \log (x) \, dx\) [66]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 10, antiderivative size = 54 \[ \int (b+a x)^2 \log (x) \, dx=-b^2 x-\frac {1}{2} a b x^2-\frac {a^2 x^3}{9}-\frac {b^3 \log (x)}{3 a}+\frac {(b+a x)^3 \log (x)}{3 a} \]

[Out]

-b^2*x-1/2*a*b*x^2-1/9*a^2*x^3-1/3*b^3*ln(x)/a+1/3*(a*x+b)^3*ln(x)/a

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {32, 2350, 12, 45} \[ \int (b+a x)^2 \log (x) \, dx=-\frac {a^2 x^3}{9}-\frac {b^3 \log (x)}{3 a}-\frac {1}{2} a b x^2+\frac {\log (x) (a x+b)^3}{3 a}-b^2 x \]

[In]

Int[(b + a*x)^2*Log[x],x]

[Out]

-(b^2*x) - (a*b*x^2)/2 - (a^2*x^3)/9 - (b^3*Log[x])/(3*a) + ((b + a*x)^3*Log[x])/(3*a)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

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 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*} \text {integral}& = \frac {(b+a x)^3 \log (x)}{3 a}-\int \frac {(b+a x)^3}{3 a x} \, dx \\ & = \frac {(b+a x)^3 \log (x)}{3 a}-\frac {\int \frac {(b+a x)^3}{x} \, dx}{3 a} \\ & = \frac {(b+a x)^3 \log (x)}{3 a}-\frac {\int \left (3 a b^2+\frac {b^3}{x}+3 a^2 b x+a^3 x^2\right ) \, dx}{3 a} \\ & = -b^2 x-\frac {1}{2} a b x^2-\frac {a^2 x^3}{9}-\frac {b^3 \log (x)}{3 a}+\frac {(b+a x)^3 \log (x)}{3 a} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.98 \[ \int (b+a x)^2 \log (x) \, dx=-b^2 x-\frac {1}{2} a b x^2-\frac {a^2 x^3}{9}+b^2 x \log (x)+a b x^2 \log (x)+\frac {1}{3} a^2 x^3 \log (x) \]

[In]

Integrate[(b + a*x)^2*Log[x],x]

[Out]

-(b^2*x) - (a*b*x^2)/2 - (a^2*x^3)/9 + b^2*x*Log[x] + a*b*x^2*Log[x] + (a^2*x^3*Log[x])/3

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.87

method result size
risch \(-b^{2} x -\frac {a b \,x^{2}}{2}-\frac {a^{2} x^{3}}{9}-\frac {b^{3} \ln \left (x \right )}{3 a}+\frac {\left (a x +b \right )^{3} \ln \left (x \right )}{3 a}\) \(47\)
default \(a^{2} \left (-\frac {x^{3}}{9}+\frac {x^{3} \ln \left (x \right )}{3}\right )+2 a b \left (-\frac {x^{2}}{4}+\frac {x^{2} \ln \left (x \right )}{2}\right )+b^{2} \left (-x +x \ln \left (x \right )\right )\) \(48\)
norman \(b^{2} x \ln \left (x \right )+a b \,x^{2} \ln \left (x \right )-\frac {a^{2} x^{3}}{9}-b^{2} x -\frac {a b \,x^{2}}{2}+\frac {a^{2} x^{3} \ln \left (x \right )}{3}\) \(48\)
parallelrisch \(b^{2} x \ln \left (x \right )+a b \,x^{2} \ln \left (x \right )-\frac {a^{2} x^{3}}{9}-b^{2} x -\frac {a b \,x^{2}}{2}+\frac {a^{2} x^{3} \ln \left (x \right )}{3}\) \(48\)
parts \(\frac {a^{2} x^{3} \ln \left (x \right )}{3}+a b \,x^{2} \ln \left (x \right )+b^{2} x \ln \left (x \right )+\frac {b^{3} \ln \left (x \right )}{3 a}-\frac {\frac {a^{3} x^{3}}{3}+\frac {3 a^{2} b \,x^{2}}{2}+3 a \,b^{2} x +b^{3} \ln \left (x \right )}{3 a}\) \(73\)

[In]

int((a*x+b)^2*ln(x),x,method=_RETURNVERBOSE)

[Out]

-b^2*x-1/2*a*b*x^2-1/9*a^2*x^3-1/3*b^3*ln(x)/a+1/3*(a*x+b)^3*ln(x)/a

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.87 \[ \int (b+a x)^2 \log (x) \, dx=-\frac {1}{9} \, a^{2} x^{3} - \frac {1}{2} \, a b x^{2} - b^{2} x + \frac {1}{3} \, {\left (a^{2} x^{3} + 3 \, a b x^{2} + 3 \, b^{2} x\right )} \log \left (x\right ) \]

[In]

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

[Out]

-1/9*a^2*x^3 - 1/2*a*b*x^2 - b^2*x + 1/3*(a^2*x^3 + 3*a*b*x^2 + 3*b^2*x)*log(x)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.81 \[ \int (b+a x)^2 \log (x) \, dx=- \frac {a^{2} x^{3}}{9} - \frac {a b x^{2}}{2} - b^{2} x + \left (\frac {a^{2} x^{3}}{3} + a b x^{2} + b^{2} x\right ) \log {\left (x \right )} \]

[In]

integrate((a*x+b)**2*ln(x),x)

[Out]

-a**2*x**3/9 - a*b*x**2/2 - b**2*x + (a**2*x**3/3 + a*b*x**2 + b**2*x)*log(x)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.87 \[ \int (b+a x)^2 \log (x) \, dx=-\frac {1}{9} \, a^{2} x^{3} - \frac {1}{2} \, a b x^{2} - b^{2} x + \frac {1}{3} \, {\left (a^{2} x^{3} + 3 \, a b x^{2} + 3 \, b^{2} x\right )} \log \left (x\right ) \]

[In]

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

[Out]

-1/9*a^2*x^3 - 1/2*a*b*x^2 - b^2*x + 1/3*(a^2*x^3 + 3*a*b*x^2 + 3*b^2*x)*log(x)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.87 \[ \int (b+a x)^2 \log (x) \, dx=\frac {1}{3} \, a^{2} x^{3} \log \left (x\right ) - \frac {1}{9} \, a^{2} x^{3} + a b x^{2} \log \left (x\right ) - \frac {1}{2} \, a b x^{2} + b^{2} x \log \left (x\right ) - b^{2} x \]

[In]

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

[Out]

1/3*a^2*x^3*log(x) - 1/9*a^2*x^3 + a*b*x^2*log(x) - 1/2*a*b*x^2 + b^2*x*log(x) - b^2*x

Mupad [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.87 \[ \int (b+a x)^2 \log (x) \, dx=b^2\,x\,\ln \left (x\right )-\frac {a^2\,x^3}{9}-b^2\,x+\frac {a^2\,x^3\,\ln \left (x\right )}{3}-\frac {a\,b\,x^2}{2}+a\,b\,x^2\,\ln \left (x\right ) \]

[In]

int(log(x)*(b + a*x)^2,x)

[Out]

b^2*x*log(x) - (a^2*x^3)/9 - b^2*x + (a^2*x^3*log(x))/3 - (a*b*x^2)/2 + a*b*x^2*log(x)