### 3.249 $$\int (a+b x)^n \log (a+b x) \, dx$$

Optimal. Leaf size=44 $\frac{(a+b x)^{n+1} \log (a+b x)}{b (n+1)}-\frac{(a+b x)^{n+1}}{b (n+1)^2}$

[Out]

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

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Rubi [A]  time = 0.0302658, antiderivative size = 44, 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} $\frac{(a+b x)^{n+1} \log (a+b x)}{b (n+1)}-\frac{(a+b x)^{n+1}}{b (n+1)^2}$

Antiderivative was successfully veriﬁed.

[In]

Int[(a + b*x)^n*Log[a + b*x],x]

[Out]

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

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

Mathematica [A]  time = 0.0142109, size = 30, normalized size = 0.68 $\frac{(a+b x)^{n+1} ((n+1) \log (a+b x)-1)}{b (n+1)^2}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(a + b*x)^n*Log[a + b*x],x]

[Out]

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

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Maple [B]  time = 0.013, size = 96, normalized size = 2.2 \begin{align*}{\frac{\ln \left ( bx+a \right ) x{{\rm e}^{\ln \left ( bx+a \right ) n}}}{1+n}}+{\frac{a\ln \left ( bx+a \right ){{\rm e}^{\ln \left ( bx+a \right ) n}}}{b \left ( 1+n \right ) }}-{\frac{x{{\rm e}^{\ln \left ( bx+a \right ) n}}}{{n}^{2}+2\,n+1}}-{\frac{a{{\rm e}^{\ln \left ( bx+a \right ) n}}}{b \left ({n}^{2}+2\,n+1 \right ) }} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.83709, size = 112, normalized size = 2.55 \begin{align*} -\frac{{\left (b x -{\left (a n +{\left (b n + b\right )} x + a\right )} \log \left (b x + a\right ) + a\right )}{\left (b x + a\right )}^{n}}{b n^{2} + 2 \, b n + b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b x + a\right )}^{n} \log \left (b x + a\right )\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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