3.1.66 \(\int (b+a x)^2 \log (x) \, dx\) [66]

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

[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 = 0.02, 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, 2350, 12, 45} \begin {gather*} -\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 \end {gather*}

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 45

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 2350

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> With[{u = IntHide[(d +
e*x^r)^q, x]}, Dist[a + b*Log[c*x^n], u, 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 \begin {gather*} -b^2 x-\frac {1}{2} a b x^2-\frac {a^2 x^3}{9}+b^2 x \log (x)+a b x^2 \log (x)+\frac {1}{3} a^2 x^3 \log (x) \end {gather*}

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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Mathics [A]
time = 2.08, size = 44, normalized size = 0.81 \begin {gather*} \frac {x \left (-2 a^2 x^2-9 a b x-18 b^2+6 \text {Log}\left [x\right ] \left (a^2 x^2+3 a b x+3 b^2\right )\right )}{18} \end {gather*}

Antiderivative was successfully verified.

[In]

mathics('Integrate[(a*x+b)^2*Log[x],x]')

[Out]

x (-2 a ^ 2 x ^ 2 - 9 a b x - 18 b ^ 2 + 6 Log[x] (a ^ 2 x ^ 2 + 3 a b x + 3 b ^ 2)) / 18

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Maple [A]
time = 0.02, size = 48, normalized size = 0.89

method	result	size
risch	\(-b^{2} x -\frac {a b \,x^{2}}{2}-\frac {a^{2} x^{3}}{9}-\frac {b^{3} \ln \left (x \right )}{3 a}+\frac {\left (a x +b \right )^{3} \ln \left (x \right )}{3 a}\)	\(47\)
default	\(a^{2} \left (-\frac {x^{3}}{9}+\frac {x^{3} \ln \left (x \right )}{3}\right )+2 a b \left (-\frac {x^{2}}{4}+\frac {x^{2} \ln \left (x \right )}{2}\right )+b^{2} \left (-x +x \ln \left (x \right )\right )\)	\(48\)
norman	\(b^{2} x \ln \left (x \right )+a b \,x^{2} \ln \left (x \right )-\frac {a^{2} x^{3}}{9}-b^{2} x -\frac {a b \,x^{2}}{2}+\frac {a^{2} x^{3} \ln \left (x \right )}{3}\)	\(48\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^2*(-1/9*x^3+1/3*x^3*ln(x))+2*a*b*(-1/4*x^2+1/2*x^2*ln(x))+b^2*(-x+x*ln(x))

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Maxima [A]
time = 0.30, size = 47, normalized size = 0.87 \begin {gather*} -\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 \left (x\right ) \end {gather*}

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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Fricas [A]
time = 0.34, size = 47, normalized size = 0.87 \begin {gather*} -\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 \left (x\right ) \end {gather*}

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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Sympy [A]
time = 0.07, size = 44, normalized size = 0.81 \begin {gather*} - \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 {\left (x \right )} \end {gather*}

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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Giac [A]
time = 0.00, size = 52, normalized size = 0.96 \begin {gather*} -\frac {1}{9} a^{2} x^{3}-\frac {1}{2} a b x^{2}-b^{2} x+\frac {1}{3} a^{2} x^{3} \ln x+b^{2} x \ln x+a b x^{2} \ln x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+b)^2*log(x),x)

[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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Mupad [B]
time = 0.17, size = 47, normalized size = 0.87 \begin {gather*} b^2\,x\,\ln \left (x\right )-\frac {a^2\,x^3}{9}-b^2\,x+\frac {a^2\,x^3\,\ln \left (x\right )}{3}-\frac {a\,b\,x^2}{2}+a\,b\,x^2\,\ln \left (x\right ) \end {gather*}

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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