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]

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

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Rubi [A]  time = 0.03, antiderivative size = 44, normalized size of antiderivative = 1.00, 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 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
]

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]

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*}

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Mathematica [A]  time = 0.01, 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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fricas [A]  time = 0.45, size = 47, normalized size = 1.07 \[ -\frac {{\left (b x - {\left (a n + {\left (b n + b\right )} x + a\right )} \log \left (b x + a\right ) + a\right )} {\left (b x + a\right )}^{n}}{b n^{2} + 2 \, b n + b} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b x + a\right )}^{n} \log \left (b x + a\right )\,{d x} \]

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)

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maple [B]  time = 0.07, size = 96, normalized size = 2.18 \[ \frac {x \,{\mathrm e}^{n \ln \left (b x +a \right )} \ln \left (b x +a \right )}{n +1}+\frac {a \,{\mathrm e}^{n \ln \left (b x +a \right )} \ln \left (b x +a \right )}{\left (n +1\right ) b}-\frac {x \,{\mathrm e}^{n \ln \left (b x +a \right )}}{n^{2}+2 n +1}-\frac {a \,{\mathrm e}^{n \ln \left (b x +a \right )}}{\left (n^{2}+2 n +1\right ) b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.53, size = 44, normalized size = 1.00 \[ \frac {{\left (b x + a\right )}^{n + 1} \log \left (b x + a\right )}{b {\left (n + 1\right )}} - \frac {{\left (b x + a\right )}^{n + 1}}{b {\left (n + 1\right )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.46, size = 52, normalized size = 1.18 \[ \left \{\begin {array}{cl} \frac {{\ln \left (a+b\,x\right )}^2}{2\,b} & \text {\ if\ \ }n=-1\\ \frac {\left (\ln \left (a+b\,x\right )-\frac {1}{n+1}\right )\,{\left (a+b\,x\right )}^{n+1}}{b\,\left (n+1\right )} & \text {\ if\ \ }n\neq -1 \end {array}\right . \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(a + b*x)*(a + b*x)^n,x)

[Out]

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

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sympy [A]  time = 1.47, size = 185, normalized size = 4.20 \[ \begin {cases} \frac {x \log {\relax (a )}}{a} & \text {for}\: b = 0 \wedge n = -1 \\a^{n} x \log {\relax (a )} & \text {for}\: b = 0 \\\frac {\log {\left (a + b x \right )}^{2}}{2 b} & \text {for}\: n = -1 \\\frac {a n \left (a + b x\right )^{n} \log {\left (a + b x \right )}}{b n^{2} + 2 b n + b} + \frac {a \left (a + b x\right )^{n} \log {\left (a + b x \right )}}{b n^{2} + 2 b n + b} - \frac {a \left (a + b x\right )^{n}}{b n^{2} + 2 b n + b} + \frac {b n x \left (a + b x\right )^{n} \log {\left (a + b x \right )}}{b n^{2} + 2 b n + b} + \frac {b x \left (a + b x\right )^{n} \log {\left (a + b x \right )}}{b n^{2} + 2 b n + b} - \frac {b x \left (a + b x\right )^{n}}{b n^{2} + 2 b n + b} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((x*log(a)/a, Eq(b, 0) & Eq(n, -1)), (a**n*x*log(a), Eq(b, 0)), (log(a + b*x)**2/(2*b), Eq(n, -1)), (
a*n*(a + b*x)**n*log(a + b*x)/(b*n**2 + 2*b*n + b) + a*(a + b*x)**n*log(a + b*x)/(b*n**2 + 2*b*n + b) - a*(a +
b*x)**n/(b*n**2 + 2*b*n + b) + b*n*x*(a + b*x)**n*log(a + b*x)/(b*n**2 + 2*b*n + b) + b*x*(a + b*x)**n*log(a
+ b*x)/(b*n**2 + 2*b*n + b) - b*x*(a + b*x)**n/(b*n**2 + 2*b*n + b), True))

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