### 3.246 $$\int (a+b x)^2 \log (a+b x) \, dx$$

Optimal. Leaf size=35 $\frac{(a+b x)^3 \log (a+b x)}{3 b}-\frac{(a+b x)^3}{9 b}$

[Out]

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

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Rubi [A]  time = 0.0246622, antiderivative size = 35, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 14, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.143, Rules used = {2390, 2304} $\frac{(a+b x)^3 \log (a+b x)}{3 b}-\frac{(a+b x)^3}{9 b}$

Antiderivative was successfully veriﬁed.

[In]

Int[(a + b*x)^2*Log[a + b*x],x]

[Out]

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

Rule 2390

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]

Rule 2304

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
]

Rubi steps

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

Mathematica [A]  time = 0.0167406, size = 44, normalized size = 1.26 $\frac{(a+b x)^3 \log (a+b x)}{3 b}-\frac{1}{9} x \left (3 a^2+3 a b x+b^2 x^2\right )$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(a + b*x)^2*Log[a + b*x],x]

[Out]

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

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Maple [B]  time = 0.001, size = 82, normalized size = 2.3 \begin{align*}{\frac{{b}^{2}\ln \left ( bx+a \right ){x}^{3}}{3}}+b\ln \left ( bx+a \right ){x}^{2}a+\ln \left ( bx+a \right ) x{a}^{2}+{\frac{\ln \left ( bx+a \right ){a}^{3}}{3\,b}}-{\frac{{b}^{2}{x}^{3}}{9}}-{\frac{b{x}^{2}a}{3}}-{\frac{x{a}^{2}}{3}}-{\frac{{a}^{3}}{9\,b}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.06081, size = 100, normalized size = 2.86 \begin{align*} \frac{1}{9} \,{\left (\frac{3 \, a^{3} \log \left (b x + a\right )}{b^{2}} - \frac{b^{2} x^{3} + 3 \, a b x^{2} + 3 \, a^{2} x}{b}\right )} b + \frac{1}{3} \,{\left (b^{2} x^{3} + 3 \, a b x^{2} + 3 \, a^{2} x\right )} \log \left (b x + a\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.7631, size = 139, normalized size = 3.97 \begin{align*} -\frac{b^{3} x^{3} + 3 \, a b^{2} x^{2} + 3 \, a^{2} b x - 3 \,{\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + 3 \, a^{2} b x + a^{3}\right )} \log \left (b x + a\right )}{9 \, b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B]  time = 0.365034, size = 63, normalized size = 1.8 \begin{align*} \frac{a^{3} \log{\left (a + b x \right )}}{3 b} - \frac{a^{2} x}{3} - \frac{a b x^{2}}{3} - \frac{b^{2} x^{3}}{9} + \left (a^{2} x + a b x^{2} + \frac{b^{2} x^{3}}{3}\right ) \log{\left (a + b x \right )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.32667, size = 42, normalized size = 1.2 \begin{align*} \frac{{\left (b x + a\right )}^{3} \log \left (b x + a\right )}{3 \, b} - \frac{{\left (b x + a\right )}^{3}}{9 \, b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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