3.612 \(\int (a+b x)^3 \log (x) \, dx\)

Optimal. Leaf size=67 \[ -\frac{3}{4} a^2 b x^2-\frac{a^4 \log (x)}{4 b}-a^3 x-\frac{1}{3} a b^2 x^3+\frac{\log (x) (a+b x)^4}{4 b}-\frac{b^3 x^4}{16} \]

[Out]

-(a^3*x) - (3*a^2*b*x^2)/4 - (a*b^2*x^3)/3 - (b^3*x^4)/16 - (a^4*Log[x])/(4*b) + ((a + b*x)^4*Log[x])/(4*b)

________________________________________________________________________________________

Rubi [A]  time = 0.0333638, antiderivative size = 67, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {32, 2313, 12, 43} \[ -\frac{3}{4} a^2 b x^2-\frac{a^4 \log (x)}{4 b}-a^3 x-\frac{1}{3} a b^2 x^3+\frac{\log (x) (a+b x)^4}{4 b}-\frac{b^3 x^4}{16} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^3*Log[x],x]

[Out]

-(a^3*x) - (3*a^2*b*x^2)/4 - (a*b^2*x^3)/3 - (b^3*x^4)/16 - (a^4*Log[x])/(4*b) + ((a + b*x)^4*Log[x])/(4*b)

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 2313

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> With[{u = IntHide[(d +
e*x^r)^q, x]}, Simp[u*(a + b*Log[c*x^n]), x] - Dist[b*n, Int[SimplifyIntegrand[u/x, x], x], x]] /; FreeQ[{a,
b, c, d, e, n, r}, x] && IGtQ[q, 0]

Rule 12

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

Rule 43

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])

Rubi steps

\begin{align*} \int (a+b x)^3 \log (x) \, dx &=\frac{(a+b x)^4 \log (x)}{4 b}-\int \frac{(a+b x)^4}{4 b x} \, dx\\ &=\frac{(a+b x)^4 \log (x)}{4 b}-\frac{\int \frac{(a+b x)^4}{x} \, dx}{4 b}\\ &=\frac{(a+b x)^4 \log (x)}{4 b}-\frac{\int \left (4 a^3 b+\frac{a^4}{x}+6 a^2 b^2 x+4 a b^3 x^2+b^4 x^3\right ) \, dx}{4 b}\\ &=-a^3 x-\frac{3}{4} a^2 b x^2-\frac{1}{3} a b^2 x^3-\frac{b^3 x^4}{16}-\frac{a^4 \log (x)}{4 b}+\frac{(a+b x)^4 \log (x)}{4 b}\\ \end{align*}

Mathematica [A]  time = 0.0219597, size = 81, normalized size = 1.21 \[ -\frac{3}{4} a^2 b x^2+\frac{3}{2} a^2 b x^2 \log (x)-a^3 x+a^3 x \log (x)-\frac{1}{3} a b^2 x^3+a b^2 x^3 \log (x)-\frac{1}{16} b^3 x^4+\frac{1}{4} b^3 x^4 \log (x) \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^3*Log[x],x]

[Out]

-(a^3*x) - (3*a^2*b*x^2)/4 - (a*b^2*x^3)/3 - (b^3*x^4)/16 + a^3*x*Log[x] + (3*a^2*b*x^2*Log[x])/2 + a*b^2*x^3*
Log[x] + (b^3*x^4*Log[x])/4

________________________________________________________________________________________

Maple [A]  time = 0.001, size = 72, normalized size = 1.1 \begin{align*}{\frac{{b}^{3}{x}^{4}\ln \left ( x \right ) }{4}}-{\frac{{b}^{3}{x}^{4}}{16}}+{b}^{2}a{x}^{3}\ln \left ( x \right ) -{\frac{a{b}^{2}{x}^{3}}{3}}+{\frac{3\,{a}^{2}b{x}^{2}\ln \left ( x \right ) }{2}}-{\frac{3\,{a}^{2}b{x}^{2}}{4}}+\ln \left ( x \right ) x{a}^{3}-{a}^{3}x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^3*ln(x),x)

[Out]

1/4*b^3*x^4*ln(x)-1/16*b^3*x^4+b^2*a*x^3*ln(x)-1/3*a*b^2*x^3+3/2*a^2*b*x^2*ln(x)-3/4*a^2*b*x^2+ln(x)*x*a^3-a^3
*x

________________________________________________________________________________________

Maxima [A]  time = 0.938009, size = 93, normalized size = 1.39 \begin{align*} -\frac{1}{16} \, b^{3} x^{4} - \frac{1}{3} \, a b^{2} x^{3} - \frac{3}{4} \, a^{2} b x^{2} - a^{3} x + \frac{1}{4} \,{\left (b^{3} x^{4} + 4 \, a b^{2} x^{3} + 6 \, a^{2} b x^{2} + 4 \, a^{3} x\right )} \log \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^3*log(x),x, algorithm="maxima")

[Out]

-1/16*b^3*x^4 - 1/3*a*b^2*x^3 - 3/4*a^2*b*x^2 - a^3*x + 1/4*(b^3*x^4 + 4*a*b^2*x^3 + 6*a^2*b*x^2 + 4*a^3*x)*lo
g(x)

________________________________________________________________________________________

Fricas [A]  time = 2.0462, size = 157, normalized size = 2.34 \begin{align*} -\frac{1}{16} \, b^{3} x^{4} - \frac{1}{3} \, a b^{2} x^{3} - \frac{3}{4} \, a^{2} b x^{2} - a^{3} x + \frac{1}{4} \,{\left (b^{3} x^{4} + 4 \, a b^{2} x^{3} + 6 \, a^{2} b x^{2} + 4 \, a^{3} x\right )} \log \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^3*log(x),x, algorithm="fricas")

[Out]

-1/16*b^3*x^4 - 1/3*a*b^2*x^3 - 3/4*a^2*b*x^2 - a^3*x + 1/4*(b^3*x^4 + 4*a*b^2*x^3 + 6*a^2*b*x^2 + 4*a^3*x)*lo
g(x)

________________________________________________________________________________________

Sympy [A]  time = 0.133159, size = 71, normalized size = 1.06 \begin{align*} - a^{3} x - \frac{3 a^{2} b x^{2}}{4} - \frac{a b^{2} x^{3}}{3} - \frac{b^{3} x^{4}}{16} + \left (a^{3} x + \frac{3 a^{2} b x^{2}}{2} + a b^{2} x^{3} + \frac{b^{3} x^{4}}{4}\right ) \log{\left (x \right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**3*ln(x),x)

[Out]

-a**3*x - 3*a**2*b*x**2/4 - a*b**2*x**3/3 - b**3*x**4/16 + (a**3*x + 3*a**2*b*x**2/2 + a*b**2*x**3 + b**3*x**4
/4)*log(x)

________________________________________________________________________________________

Giac [A]  time = 1.1419, size = 96, normalized size = 1.43 \begin{align*} \frac{1}{4} \, b^{3} x^{4} \log \left (x\right ) - \frac{1}{16} \, b^{3} x^{4} + a b^{2} x^{3} \log \left (x\right ) - \frac{1}{3} \, a b^{2} x^{3} + \frac{3}{2} \, a^{2} b x^{2} \log \left (x\right ) - \frac{3}{4} \, a^{2} b x^{2} + a^{3} x \log \left (x\right ) - a^{3} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^3*log(x),x, algorithm="giac")

[Out]

1/4*b^3*x^4*log(x) - 1/16*b^3*x^4 + a*b^2*x^3*log(x) - 1/3*a*b^2*x^3 + 3/2*a^2*b*x^2*log(x) - 3/4*a^2*b*x^2 +
a^3*x*log(x) - a^3*x