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

Optimal. Leaf size=49 \[ \frac{(a+x (b+c)) \log ^2(a+x (b+c))}{b+c}-\frac{2 (a+x (b+c)) \log (a+x (b+c))}{b+c}+2 x \]

[Out]

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

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Rubi [A]  time = 0.0252992, antiderivative size = 49, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.364, Rules used = {2444, 2389, 2296, 2295} \[ \frac{(a+x (b+c)) \log ^2(a+x (b+c))}{b+c}-\frac{2 (a+x (b+c)) \log (a+x (b+c))}{b+c}+2 x \]

Antiderivative was successfully verified.

[In]

Int[Log[a + b*x + c*x]^2,x]

[Out]

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

Rule 2444

Int[((a_.) + Log[(c_.)*(v_)^(n_.)]*(b_.))^(p_.)*(u_.), x_Symbol] :> Int[u*(a + b*Log[c*ExpandToSum[v, x]^n])^p
, x] /; FreeQ[{a, b, c, n, p}, x] && LinearQ[v, x] &&  !LinearMatchQ[v, x] &&  !(EqQ[n, 1] && MatchQ[c*v, (e_.
)*((f_) + (g_.)*x) /; FreeQ[{e, f, g}, x]])

Rule 2389

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.), x_Symbol] :> Dist[1/e, Subst[Int[(a + b*Log[c*
x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, n, p}, x]

Rule 2296

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*Log[c*x^n])^p, x] - Dist[b*n*p, In
t[(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{a, b, c, n}, x] && GtQ[p, 0] && IntegerQ[2*p]

Rule 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rubi steps

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

Mathematica [A]  time = 0.0062025, size = 48, normalized size = 0.98 \[ \frac{(a+x (b+c)) \log ^2(a+x (b+c))-2 (a+x (b+c)) \log (a+x (b+c))+2 x (b+c)}{b+c} \]

Antiderivative was successfully verified.

[In]

Integrate[Log[a + b*x + c*x]^2,x]

[Out]

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

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Maple [B]  time = 0.057, size = 131, normalized size = 2.7 \begin{align*}{\frac{ \left ( \ln \left ( a+ \left ( b+c \right ) x \right ) \right ) ^{2}xb}{b+c}}+{\frac{ \left ( \ln \left ( a+ \left ( b+c \right ) x \right ) \right ) ^{2}xc}{b+c}}+{\frac{ \left ( \ln \left ( a+ \left ( b+c \right ) x \right ) \right ) ^{2}a}{b+c}}-2\,{\frac{\ln \left ( a+ \left ( b+c \right ) x \right ) xb}{b+c}}-2\,{\frac{\ln \left ( a+ \left ( b+c \right ) x \right ) xc}{b+c}}-2\,{\frac{\ln \left ( a+ \left ( b+c \right ) x \right ) a}{b+c}}+2\,{\frac{bx}{b+c}}+2\,{\frac{cx}{b+c}}+2\,{\frac{a}{b+c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.18191, size = 51, normalized size = 1.04 \begin{align*} \frac{{\left (b x + c x + a\right )}{\left (\log \left (b x + c x + a\right )^{2} - 2 \, \log \left (b x + c x + a\right ) + 2\right )}}{b + c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.91871, size = 136, normalized size = 2.78 \begin{align*} \frac{{\left ({\left (b + c\right )} x + a\right )} \log \left ({\left (b + c\right )} x + a\right )^{2} + 2 \,{\left (b + c\right )} x - 2 \,{\left ({\left (b + c\right )} x + a\right )} \log \left ({\left (b + c\right )} x + a\right )}{b + c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.501055, size = 63, normalized size = 1.29 \begin{align*} - 2 x \log{\left (a + b x + c x \right )} + \left (2 b + 2 c\right ) \left (- \frac{a \log{\left (a + x \left (b + c\right ) \right )}}{\left (b + c\right )^{2}} + \frac{x}{b + c}\right ) + \frac{\left (a + b x + c x\right ) \log{\left (a + b x + c x \right )}^{2}}{b + c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*x*log(a + b*x + c*x) + (2*b + 2*c)*(-a*log(a + x*(b + c))/(b + c)**2 + x/(b + c)) + (a + b*x + c*x)*log(a +
b*x + c*x)**2/(b + c)

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Giac [A]  time = 1.21706, size = 88, normalized size = 1.8 \begin{align*} \frac{{\left (b x + c x + a\right )} \log \left (b x + c x + a\right )^{2}}{b + c} - \frac{2 \,{\left (b x + c x + a\right )} \log \left (b x + c x + a\right )}{b + c} + \frac{2 \,{\left (b x + c x + a\right )}}{b + c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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