\(\int x^2 \log (b+a x) \, dx\) [69]

Optimal result

Integrand size = 10, antiderivative size = 59 \[ \int x^2 \log (b+a x) \, dx=-\frac {b^2 x}{3 a^2}+\frac {b x^2}{6 a}-\frac {x^3}{9}+\frac {b^3 \log (b+a x)}{3 a^3}+\frac {1}{3} x^3 \log (b+a x) \]

[Out]

-1/3*b^2*x/a^2+1/6*b*x^2/a-1/9*x^3+1/3*b^3*ln(a*x+b)/a^3+1/3*x^3*ln(a*x+b)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2442, 45} \[ \int x^2 \log (b+a x) \, dx=\frac {b^3 \log (a x+b)}{3 a^3}-\frac {b^2 x}{3 a^2}+\frac {1}{3} x^3 \log (a x+b)+\frac {b x^2}{6 a}-\frac {x^3}{9} \]

[In]

Int[x^2*Log[b + a*x],x]

[Out]

-1/3*(b^2*x)/a^2 + (b*x^2)/(6*a) - x^3/9 + (b^3*Log[b + a*x])/(3*a^3) + (x^3*Log[b + a*x])/3

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*} \text {integral}& = \frac {1}{3} x^3 \log (b+a x)-\frac {1}{3} a \int \frac {x^3}{b+a x} \, dx \\ & = \frac {1}{3} x^3 \log (b+a x)-\frac {1}{3} a \int \left (\frac {b^2}{a^3}-\frac {b x}{a^2}+\frac {x^2}{a}-\frac {b^3}{a^3 (b+a x)}\right ) \, dx \\ & = -\frac {b^2 x}{3 a^2}+\frac {b x^2}{6 a}-\frac {x^3}{9}+\frac {b^3 \log (b+a x)}{3 a^3}+\frac {1}{3} x^3 \log (b+a x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.00 \[ \int x^2 \log (b+a x) \, dx=-\frac {b^2 x}{3 a^2}+\frac {b x^2}{6 a}-\frac {x^3}{9}+\frac {b^3 \log (b+a x)}{3 a^3}+\frac {1}{3} x^3 \log (b+a x) \]

[In]

Integrate[x^2*Log[b + a*x],x]

[Out]

-1/3*(b^2*x)/a^2 + (b*x^2)/(6*a) - x^3/9 + (b^3*Log[b + a*x])/(3*a^3) + (x^3*Log[b + a*x])/3

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.85

method	result	size
norman	\(-\frac {b^{2} x}{3 a^{2}}+\frac {b \,x^{2}}{6 a}-\frac {x^{3}}{9}+\frac {b^{3} \ln \left (a x +b \right )}{3 a^{3}}+\frac {x^{3} \ln \left (a x +b \right )}{3}\)	\(50\)
risch	\(-\frac {b^{2} x}{3 a^{2}}+\frac {b \,x^{2}}{6 a}-\frac {x^{3}}{9}+\frac {b^{3} \ln \left (a x +b \right )}{3 a^{3}}+\frac {x^{3} \ln \left (a x +b \right )}{3}\)	\(50\)
parts	\(\frac {x^{3} \ln \left (a x +b \right )}{3}-\frac {a \left (\frac {\frac {1}{3} a^{2} x^{3}-\frac {1}{2} a b \,x^{2}+b^{2} x}{a^{3}}-\frac {b^{3} \ln \left (a x +b \right )}{a^{4}}\right )}{3}\)	\(56\)
parallelrisch	\(\frac {6 x^{3} \ln \left (a x +b \right ) a^{3}-2 a^{3} x^{3}+3 a^{2} b \,x^{2}-6 a \,b^{2} x +6 b^{3} \ln \left (a x +b \right )+6 b^{3}}{18 a^{3}}\)	\(61\)
derivativedivides	\(\frac {b^{2} \left (\ln \left (a x +b \right ) \left (a x +b \right )-a x -b \right )-2 b \left (\frac {\left (a x +b \right )^{2} \ln \left (a x +b \right )}{2}-\frac {\left (a x +b \right )^{2}}{4}\right )+\frac {\left (a x +b \right )^{3} \ln \left (a x +b \right )}{3}-\frac {\left (a x +b \right )^{3}}{9}}{a^{3}}\)	\(82\)
default	\(\frac {b^{2} \left (\ln \left (a x +b \right ) \left (a x +b \right )-a x -b \right )-2 b \left (\frac {\left (a x +b \right )^{2} \ln \left (a x +b \right )}{2}-\frac {\left (a x +b \right )^{2}}{4}\right )+\frac {\left (a x +b \right )^{3} \ln \left (a x +b \right )}{3}-\frac {\left (a x +b \right )^{3}}{9}}{a^{3}}\)	\(82\)

[In]

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

[Out]

-1/3*b^2*x/a^2+1/6*b*x^2/a-1/9*x^3+1/3*b^3*ln(a*x+b)/a^3+1/3*x^3*ln(a*x+b)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.83 \[ \int x^2 \log (b+a x) \, dx=-\frac {2 \, a^{3} x^{3} - 3 \, a^{2} b x^{2} + 6 \, a b^{2} x - 6 \, {\left (a^{3} x^{3} + b^{3}\right )} \log \left (a x + b\right )}{18 \, a^{3}} \]

[In]

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

[Out]

-1/18*(2*a^3*x^3 - 3*a^2*b*x^2 + 6*a*b^2*x - 6*(a^3*x^3 + b^3)*log(a*x + b))/a^3

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.92 \[ \int x^2 \log (b+a x) \, dx=- a \left (\frac {x^{3}}{9 a} - \frac {b x^{2}}{6 a^{2}} + \frac {b^{2} x}{3 a^{3}} - \frac {b^{3} \log {\left (a x + b \right )}}{3 a^{4}}\right ) + \frac {x^{3} \log {\left (a x + b \right )}}{3} \]

[In]

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

[Out]

-a*(x**3/(9*a) - b*x**2/(6*a**2) + b**2*x/(3*a**3) - b**3*log(a*x + b)/(3*a**4)) + x**3*log(a*x + b)/3

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.97 \[ \int x^2 \log (b+a x) \, dx=\frac {1}{3} \, x^{3} \log \left (a x + b\right ) + \frac {1}{18} \, a {\left (\frac {6 \, b^{3} \log \left (a x + b\right )}{a^{4}} - \frac {2 \, a^{2} x^{3} - 3 \, a b x^{2} + 6 \, b^{2} x}{a^{3}}\right )} \]

[In]

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

[Out]

1/3*x^3*log(a*x + b) + 1/18*a*(6*b^3*log(a*x + b)/a^4 - (2*a^2*x^3 - 3*a*b*x^2 + 6*b^2*x)/a^3)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.59 \[ \int x^2 \log (b+a x) \, dx=\frac {{\left (a x + b\right )}^{3} \log \left (a x + b\right )}{3 \, a^{3}} - \frac {{\left (a x + b\right )}^{2} b \log \left (a x + b\right )}{a^{3}} + \frac {{\left (a x + b\right )} b^{2} \log \left (a x + b\right )}{a^{3}} - \frac {{\left (a x + b\right )}^{3}}{9 \, a^{3}} + \frac {{\left (a x + b\right )}^{2} b}{2 \, a^{3}} - \frac {{\left (a x + b\right )} b^{2}}{a^{3}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.27 \[ \int x^2 \log (b+a x) \, dx=\left \{\begin {array}{cl} \frac {x^3\,\left (\ln \left (a\,x\right )-\frac {1}{3}\right )}{3} & \text {\ if\ \ }b=0\\ \frac {\ln \left (b+a\,x\right )\,\left (x^3+\frac {b^3}{a^3}\right )}{3}-\frac {b^3\,\left (\frac {a^3\,x^3}{3\,b^3}-\frac {a^2\,x^2}{2\,b^2}+\frac {a\,x}{b}\right )}{3\,a^3} & \text {\ if\ \ }b\neq 0 \end {array}\right . \]

[In]

int(x^2*log(b + a*x),x)

[Out]

piecewise(b == 0, (x^3*(log(a*x) - 1/3))/3, b ~= 0, (log(b + a*x)*(x^3 + b^3/a^3))/3 - (b^3*(- (a^2*x^2)/(2*b^
2) + (a^3*x^3)/(3*b^3) + (a*x)/b))/(3*a^3))