3.66 \(\int (b+a x)^2 \log (x) \, dx\)

Optimal. Leaf size=54 \[ -\frac {a^2 x^3}{9}-\frac {b^3 \log (x)}{3 a}-\frac {1}{2} a b x^2+\frac {\log (x) (a x+b)^3}{3 a}-b^2 x \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {32, 2313, 12, 43} \[ -\frac {a^2 x^3}{9}-\frac {b^3 \log (x)}{3 a}-\frac {1}{2} a b x^2+\frac {\log (x) (a x+b)^3}{3 a}-b^2 x \]

Antiderivative was successfully verified.

[In]

Int[(b + a*x)^2*Log[x],x]

[Out]

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

Rule 12

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

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 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])

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]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 53, normalized size = 0.98 \[ -\frac {1}{9} a^2 x^3+\frac {1}{3} a^2 x^3 \log (x)-\frac {1}{2} a b x^2+a b x^2 \log (x)-b^2 x+b^2 x \log (x) \]

Antiderivative was successfully verified.

[In]

Integrate[(b + a*x)^2*Log[x],x]

[Out]

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

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fricas [A]  time = 0.40, size = 47, normalized size = 0.87 \[ -\frac {1}{9} \, a^{2} x^{3} - \frac {1}{2} \, a b x^{2} - b^{2} x + \frac {1}{3} \, {\left (a^{2} x^{3} + 3 \, a b x^{2} + 3 \, b^{2} x\right )} \log \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 1.10, size = 47, normalized size = 0.87 \[ \frac {1}{3} \, a^{2} x^{3} \log \relax (x) - \frac {1}{9} \, a^{2} x^{3} + a b x^{2} \log \relax (x) - \frac {1}{2} \, a b x^{2} + b^{2} x \log \relax (x) - b^{2} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.00, size = 48, normalized size = 0.89 \[ \frac {a^{2} x^{3} \ln \relax (x )}{3}-\frac {a^{2} x^{3}}{9}+a b \,x^{2} \ln \relax (x )-\frac {a b \,x^{2}}{2}+b^{2} x \ln \relax (x )-b^{2} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.42, size = 47, normalized size = 0.87 \[ -\frac {1}{9} \, a^{2} x^{3} - \frac {1}{2} \, a b x^{2} - b^{2} x + \frac {1}{3} \, {\left (a^{2} x^{3} + 3 \, a b x^{2} + 3 \, b^{2} x\right )} \log \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.17, size = 47, normalized size = 0.87 \[ b^2\,x\,\ln \relax (x)-\frac {a^2\,x^3}{9}-b^2\,x+\frac {a^2\,x^3\,\ln \relax (x)}{3}-\frac {a\,b\,x^2}{2}+a\,b\,x^2\,\ln \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(x)*(b + a*x)^2,x)

[Out]

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

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sympy [A]  time = 0.14, size = 44, normalized size = 0.81 \[ - \frac {a^{2} x^{3}}{9} - \frac {a b x^{2}}{2} - b^{2} x + \left (\frac {a^{2} x^{3}}{3} + a b x^{2} + b^{2} x\right ) \log {\relax (x )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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