3.3 \(\int \log ^2(c (d+e x)) \, dx\)

Optimal. Leaf size=41 \[ \frac{(d+e x) \log ^2(c (d+e x))}{e}-\frac{2 (d+e x) \log (c (d+e x))}{e}+2 x \]

[Out]

2*x - (2*(d + e*x)*Log[c*(d + e*x)])/e + ((d + e*x)*Log[c*(d + e*x)]^2)/e

________________________________________________________________________________________

Rubi [A]  time = 0.0184991, antiderivative size = 41, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {2389, 2296, 2295} \[ \frac{(d+e x) \log ^2(c (d+e x))}{e}-\frac{2 (d+e x) \log (c (d+e x))}{e}+2 x \]

Antiderivative was successfully verified.

[In]

Int[Log[c*(d + e*x)]^2,x]

[Out]

2*x - (2*(d + e*x)*Log[c*(d + e*x)])/e + ((d + e*x)*Log[c*(d + e*x)]^2)/e

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

Mathematica [A]  time = 0.0044105, size = 40, normalized size = 0.98 \[ \frac{(d+e x) \log ^2(c (d+e x))-2 (d+e x) \log (c (d+e x))+2 e x}{e} \]

Antiderivative was successfully verified.

[In]

Integrate[Log[c*(d + e*x)]^2,x]

[Out]

(2*e*x - 2*(d + e*x)*Log[c*(d + e*x)] + (d + e*x)*Log[c*(d + e*x)]^2)/e

________________________________________________________________________________________

Maple [A]  time = 0.056, size = 67, normalized size = 1.6 \begin{align*} \left ( \ln \left ( cex+cd \right ) \right ) ^{2}x+{\frac{ \left ( \ln \left ( cex+cd \right ) \right ) ^{2}d}{e}}-2\,\ln \left ( cex+cd \right ) x-2\,{\frac{\ln \left ( cex+cd \right ) d}{e}}+2\,x+2\,{\frac{d}{e}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(ln(c*(e*x+d))^2,x)

[Out]

ln(c*e*x+c*d)^2*x+1/e*ln(c*e*x+c*d)^2*d-2*ln(c*e*x+c*d)*x-2/e*ln(c*e*x+c*d)*d+2*x+2*d/e

________________________________________________________________________________________

Maxima [A]  time = 1.08999, size = 96, normalized size = 2.34 \begin{align*} -2 \, e{\left (\frac{x}{e} - \frac{d \log \left (e x + d\right )}{e^{2}}\right )} \log \left ({\left (e x + d\right )} c\right ) + x \log \left ({\left (e x + d\right )} c\right )^{2} - \frac{d \log \left (e x + d\right )^{2} - 2 \, e x + 2 \, d \log \left (e x + d\right )}{e} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(c*(e*x+d))^2,x, algorithm="maxima")

[Out]

-2*e*(x/e - d*log(e*x + d)/e^2)*log((e*x + d)*c) + x*log((e*x + d)*c)^2 - (d*log(e*x + d)^2 - 2*e*x + 2*d*log(
e*x + d))/e

________________________________________________________________________________________

Fricas [A]  time = 1.8574, size = 99, normalized size = 2.41 \begin{align*} \frac{{\left (e x + d\right )} \log \left (c e x + c d\right )^{2} + 2 \, e x - 2 \,{\left (e x + d\right )} \log \left (c e x + c d\right )}{e} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(c*(e*x+d))^2,x, algorithm="fricas")

[Out]

((e*x + d)*log(c*e*x + c*d)^2 + 2*e*x - 2*(e*x + d)*log(c*e*x + c*d))/e

________________________________________________________________________________________

Sympy [A]  time = 0.433221, size = 46, normalized size = 1.12 \begin{align*} 2 e \left (- \frac{d \log{\left (d + e x \right )}}{e^{2}} + \frac{x}{e}\right ) - 2 x \log{\left (c \left (d + e x\right ) \right )} + \frac{\left (d + e x\right ) \log{\left (c \left (d + e x\right ) \right )}^{2}}{e} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(ln(c*(e*x+d))**2,x)

[Out]

2*e*(-d*log(d + e*x)/e**2 + x/e) - 2*x*log(c*(d + e*x)) + (d + e*x)*log(c*(d + e*x))**2/e

________________________________________________________________________________________

Giac [A]  time = 1.13679, size = 68, normalized size = 1.66 \begin{align*}{\left (x e + d\right )} e^{\left (-1\right )} \log \left ({\left (x e + d\right )} c\right )^{2} - 2 \,{\left (x e + d\right )} e^{\left (-1\right )} \log \left ({\left (x e + d\right )} c\right ) + 2 \,{\left (x e + d\right )} e^{\left (-1\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(c*(e*x+d))^2,x, algorithm="giac")

[Out]

(x*e + d)*e^(-1)*log((x*e + d)*c)^2 - 2*(x*e + d)*e^(-1)*log((x*e + d)*c) + 2*(x*e + d)*e^(-1)