3.1.3 \(\int \log ^2(c (d+e x)) \, dx\) [3]

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

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {2436, 2333, 2332} \begin {gather*} \frac {(d+e x) \log ^2(c (d+e x))}{e}-\frac {2 (d+e x) \log (c (d+e x))}{e}+2 x \end {gather*}

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 2332

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

Rule 2333

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 2436

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]

Rubi steps

\begin {align*} \int \log ^2(c (d+e x)) \, dx &=\frac {\text {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 \text {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*}

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Mathematica [A]
time = 0.00, size = 40, normalized size = 0.98 \begin {gather*} \frac {2 e x-2 (d+e x) \log (c (d+e x))+(d+e x) \log ^2(c (d+e x))}{e} \end {gather*}

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

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Maple [A]
time = 0.13, size = 57, normalized size = 1.39

method	result	size
risch	\(\frac {\left (e x +d \right ) \ln \left (c \left (e x +d \right )\right )^{2}}{e}-2 x \ln \left (c \left (e x +d \right )\right )+2 x -\frac {2 d \ln \left (e x +d \right )}{e}\)	\(47\)
derivativedivides	\(\frac {\left (c e x +c d \right ) \ln \left (c e x +c d \right )^{2}-2 \left (c e x +c d \right ) \ln \left (c e x +c d \right )+2 c e x +2 c d}{c e}\)	\(57\)
default	\(\frac {\left (c e x +c d \right ) \ln \left (c e x +c d \right )^{2}-2 \left (c e x +c d \right ) \ln \left (c e x +c d \right )+2 c e x +2 c d}{c e}\)	\(57\)
norman	\(x \ln \left (c \left (e x +d \right )\right )^{2}+\frac {d \ln \left (c \left (e x +d \right )\right )^{2}}{e}+2 x -2 x \ln \left (c \left (e x +d \right )\right )-\frac {2 d \ln \left (c \left (e x +d \right )\right )}{e}\)	\(57\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.27, size = 75, normalized size = 1.83 \begin {gather*} 2 \, {\left (d e^{\left (-2\right )} \log \left (x e + d\right ) - x e^{\left (-1\right )}\right )} e \log \left ({\left (x e + d\right )} c\right ) + x \log \left ({\left (x e + d\right )} c\right )^{2} - {\left (d \log \left (x e + d\right )^{2} - 2 \, x e + 2 \, d \log \left (x e + d\right )\right )} e^{\left (-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.33, size = 46, normalized size = 1.12 \begin {gather*} {\left ({\left (x e + d\right )} \log \left (c x e + c d\right )^{2} + 2 \, x e - 2 \, {\left (x e + d\right )} \log \left (c x e + c d\right )\right )} e^{\left (-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.08, size = 46, normalized size = 1.12 \begin {gather*} 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 {gather*}

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

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Giac [A]
time = 4.19, size = 50, normalized size = 1.22 \begin {gather*} {\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 {gather*}

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)

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Mupad [B]
time = 0.22, size = 57, normalized size = 1.39 \begin {gather*} 2\,x-2\,x\,\ln \left (c\,d+c\,e\,x\right )+x\,{\ln \left (c\,d+c\,e\,x\right )}^2+\frac {d\,{\ln \left (c\,d+c\,e\,x\right )}^2}{e}-\frac {2\,d\,\ln \left (d+e\,x\right )}{e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(c*(d + e*x))^2,x)

[Out]

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

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