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

Optimal result

Integrand size = 10, antiderivative size = 61 \[ \int \log ^3(c (d+e x)) \, dx=-6 x+\frac {6 (d+e x) \log (c (d+e x))}{e}-\frac {3 (d+e x) \log ^2(c (d+e x))}{e}+\frac {(d+e x) \log ^3(c (d+e x))}{e} \]

[Out]

-6*x+6*(e*x+d)*ln(c*(e*x+d))/e-3*(e*x+d)*ln(c*(e*x+d))^2/e+(e*x+d)*ln(c*(e*x+d))^3/e

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {2436, 2333, 2332} \[ \int \log ^3(c (d+e x)) \, dx=\frac {(d+e x) \log ^3(c (d+e x))}{e}-\frac {3 (d+e x) \log ^2(c (d+e x))}{e}+\frac {6 (d+e x) \log (c (d+e x))}{e}-6 x \]

[In]

Int[Log[c*(d + e*x)]^3,x]

[Out]

-6*x + (6*(d + e*x)*Log[c*(d + e*x)])/e - (3*(d + e*x)*Log[c*(d + e*x)]^2)/e + ((d + e*x)*Log[c*(d + e*x)]^3)/
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*} \text {integral}& = \frac {\text {Subst}\left (\int \log ^3(c x) \, dx,x,d+e x\right )}{e} \\ & = \frac {(d+e x) \log ^3(c (d+e x))}{e}-\frac {3 \text {Subst}\left (\int \log ^2(c x) \, dx,x,d+e x\right )}{e} \\ & = -\frac {3 (d+e x) \log ^2(c (d+e x))}{e}+\frac {(d+e x) \log ^3(c (d+e x))}{e}+\frac {6 \text {Subst}(\int \log (c x) \, dx,x,d+e x)}{e} \\ & = -6 x+\frac {6 (d+e x) \log (c (d+e x))}{e}-\frac {3 (d+e x) \log ^2(c (d+e x))}{e}+\frac {(d+e x) \log ^3(c (d+e x))}{e} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.93 \[ \int \log ^3(c (d+e x)) \, dx=\frac {-6 e x+6 (d+e x) \log (c (d+e x))-3 (d+e x) \log ^2(c (d+e x))+(d+e x) \log ^3(c (d+e x))}{e} \]

[In]

Integrate[Log[c*(d + e*x)]^3,x]

[Out]

(-6*e*x + 6*(d + e*x)*Log[c*(d + e*x)] - 3*(d + e*x)*Log[c*(d + e*x)]^2 + (d + e*x)*Log[c*(d + e*x)]^3)/e

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.10

method	result	size
risch	\(\frac {\left (e x +d \right ) \ln \left (c \left (e x +d \right )\right )^{3}}{e}-\frac {3 \left (e x +d \right ) \ln \left (c \left (e x +d \right )\right )^{2}}{e}+6 x \ln \left (c \left (e x +d \right )\right )-6 x +\frac {6 d \ln \left (e x +d \right )}{e}\)	\(67\)
derivativedivides	\(\frac {\left (c e x +c d \right ) \ln \left (c e x +c d \right )^{3}-3 \left (c e x +c d \right ) \ln \left (c e x +c d \right )^{2}+6 \left (c e x +c d \right ) \ln \left (c e x +c d \right )-6 c e x -6 c d}{c e}\)	\(78\)
default	\(\frac {\left (c e x +c d \right ) \ln \left (c e x +c d \right )^{3}-3 \left (c e x +c d \right ) \ln \left (c e x +c d \right )^{2}+6 \left (c e x +c d \right ) \ln \left (c e x +c d \right )-6 c e x -6 c d}{c e}\)	\(78\)
norman	\(x \ln \left (c \left (e x +d \right )\right )^{3}+\frac {d \ln \left (c \left (e x +d \right )\right )^{3}}{e}-6 x +6 x \ln \left (c \left (e x +d \right )\right )-3 x \ln \left (c \left (e x +d \right )\right )^{2}+\frac {6 d \ln \left (c \left (e x +d \right )\right )}{e}-\frac {3 d \ln \left (c \left (e x +d \right )\right )^{2}}{e}\)	\(86\)
parallelrisch	\(\frac {x \ln \left (c \left (e x +d \right )\right )^{3} e -3 x \ln \left (c \left (e x +d \right )\right )^{2} e +\ln \left (c \left (e x +d \right )\right )^{3} d +6 \ln \left (c \left (e x +d \right )\right ) x e -3 \ln \left (c \left (e x +d \right )\right )^{2} d -6 e x +6 d \ln \left (c \left (e x +d \right )\right )+6 d}{e}\)	\(88\)

[In]

int(ln(c*(e*x+d))^3,x,method=_RETURNVERBOSE)

[Out]

(e*x+d)*ln(c*(e*x+d))^3/e-3*(e*x+d)*ln(c*(e*x+d))^2/e+6*x*ln(c*(e*x+d))-6*x+6*d/e*ln(e*x+d)

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.98 \[ \int \log ^3(c (d+e x)) \, dx=\frac {{\left (e x + d\right )} \log \left (c e x + c d\right )^{3} - 3 \, {\left (e x + d\right )} \log \left (c e x + c d\right )^{2} - 6 \, e x + 6 \, {\left (e x + d\right )} \log \left (c e x + c d\right )}{e} \]

[In]

integrate(log(c*(e*x+d))^3,x, algorithm="fricas")

[Out]

((e*x + d)*log(c*e*x + c*d)^3 - 3*(e*x + d)*log(c*e*x + c*d)^2 - 6*e*x + 6*(e*x + d)*log(c*e*x + c*d))/e

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.11 \[ \int \log ^3(c (d+e x)) \, dx=- 6 e \left (- \frac {d \log {\left (d + e x \right )}}{e^{2}} + \frac {x}{e}\right ) + 6 x \log {\left (c \left (d + e x\right ) \right )} + \frac {\left (- 3 d - 3 e x\right ) \log {\left (c \left (d + e x\right ) \right )}^{2}}{e} + \frac {\left (d + e x\right ) \log {\left (c \left (d + e x\right ) \right )}^{3}}{e} \]

[In]

integrate(ln(c*(e*x+d))**3,x)

[Out]

-6*e*(-d*log(d + e*x)/e**2 + x/e) + 6*x*log(c*(d + e*x)) + (-3*d - 3*e*x)*log(c*(d + e*x))**2/e + (d + e*x)*lo
g(c*(d + e*x))**3/e

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 125 vs. \(2 (61) = 122\).

Time = 0.19 (sec) , antiderivative size = 125, normalized size of antiderivative = 2.05 \[ \int \log ^3(c (d+e x)) \, dx=-3 \, e {\left (\frac {x}{e} - \frac {d \log \left (e x + d\right )}{e^{2}}\right )} \log \left ({\left (e x + d\right )} c\right )^{2} + x \log \left ({\left (e x + d\right )} c\right )^{3} - e {\left (\frac {3 \, {\left (d \log \left (e x + d\right )^{2} - 2 \, e x + 2 \, d \log \left (e x + d\right )\right )} \log \left ({\left (e x + d\right )} c\right )}{e^{2}} - \frac {d \log \left (e x + d\right )^{3} + 3 \, d \log \left (e x + d\right )^{2} - 6 \, e x + 6 \, d \log \left (e x + d\right )}{e^{2}}\right )} \]

[In]

integrate(log(c*(e*x+d))^3,x, algorithm="maxima")

[Out]

-3*e*(x/e - d*log(e*x + d)/e^2)*log((e*x + d)*c)^2 + x*log((e*x + d)*c)^3 - e*(3*(d*log(e*x + d)^2 - 2*e*x + 2
*d*log(e*x + d))*log((e*x + d)*c)/e^2 - (d*log(e*x + d)^3 + 3*d*log(e*x + d)^2 - 6*e*x + 6*d*log(e*x + d))/e^2
)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.11 \[ \int \log ^3(c (d+e x)) \, dx=\frac {{\left (e x + d\right )} \log \left ({\left (e x + d\right )} c\right )^{3}}{e} - \frac {3 \, {\left (e x + d\right )} \log \left ({\left (e x + d\right )} c\right )^{2}}{e} + \frac {6 \, {\left (e x + d\right )} \log \left ({\left (e x + d\right )} c\right )}{e} - \frac {6 \, {\left (e x + d\right )}}{e} \]

[In]

integrate(log(c*(e*x+d))^3,x, algorithm="giac")

[Out]

(e*x + d)*log((e*x + d)*c)^3/e - 3*(e*x + d)*log((e*x + d)*c)^2/e + 6*(e*x + d)*log((e*x + d)*c)/e - 6*(e*x +
d)/e

Mupad [B] (verification not implemented)

Time = 1.45 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.44 \[ \int \log ^3(c (d+e x)) \, dx=6\,x\,\ln \left (c\,d+c\,e\,x\right )-6\,x-3\,x\,{\ln \left (c\,d+c\,e\,x\right )}^2+x\,{\ln \left (c\,d+c\,e\,x\right )}^3-\frac {3\,d\,{\ln \left (c\,d+c\,e\,x\right )}^2}{e}+\frac {d\,{\ln \left (c\,d+c\,e\,x\right )}^3}{e}+\frac {6\,d\,\ln \left (d+e\,x\right )}{e} \]

[In]

int(log(c*(d + e*x))^3,x)

[Out]

6*x*log(c*d + c*e*x) - 6*x - 3*x*log(c*d + c*e*x)^2 + x*log(c*d + c*e*x)^3 - (3*d*log(c*d + c*e*x)^2)/e + (d*l
og(c*d + c*e*x)^3)/e + (6*d*log(d + e*x))/e