3.69 \(\int \log ^3(a+b x+c x) \, dx\)

Optimal. Leaf size=73 \[ \frac {(a+x (b+c)) \log ^3(a+x (b+c))}{b+c}-\frac {3 (a+x (b+c)) \log ^2(a+x (b+c))}{b+c}+\frac {6 (a+x (b+c)) \log (a+x (b+c))}{b+c}-6 x \]

[Out]

-6*x+6*(a+(b+c)*x)*ln(a+(b+c)*x)/(b+c)-3*(a+(b+c)*x)*ln(a+(b+c)*x)^2/(b+c)+(a+(b+c)*x)*ln(a+(b+c)*x)^3/(b+c)

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Rubi [A]  time = 0.03, antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {2444, 2389, 2296, 2295} \[ \frac {(a+x (b+c)) \log ^3(a+x (b+c))}{b+c}-\frac {3 (a+x (b+c)) \log ^2(a+x (b+c))}{b+c}+\frac {6 (a+x (b+c)) \log (a+x (b+c))}{b+c}-6 x \]

Antiderivative was successfully verified.

[In]

Int[Log[a + b*x + c*x]^3,x]

[Out]

-6*x + (6*(a + (b + c)*x)*Log[a + (b + c)*x])/(b + c) - (3*(a + (b + c)*x)*Log[a + (b + c)*x]^2)/(b + c) + ((a
+ (b + c)*x)*Log[a + (b + c)*x]^3)/(b + c)

Rule 2295

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

Rubi steps

\begin {align*} \int \log ^3(a+b x+c x) \, dx &=\int \log ^3(a+(b+c) x) \, dx\\ &=\frac {\operatorname {Subst}\left (\int \log ^3(x) \, dx,x,a+(b+c) x\right )}{b+c}\\ &=\frac {(a+(b+c) x) \log ^3(a+(b+c) x)}{b+c}-\frac {3 \operatorname {Subst}\left (\int \log ^2(x) \, dx,x,a+(b+c) x\right )}{b+c}\\ &=-\frac {3 (a+(b+c) x) \log ^2(a+(b+c) x)}{b+c}+\frac {(a+(b+c) x) \log ^3(a+(b+c) x)}{b+c}+\frac {6 \operatorname {Subst}(\int \log (x) \, dx,x,a+(b+c) x)}{b+c}\\ &=-6 x+\frac {6 (a+(b+c) x) \log (a+(b+c) x)}{b+c}-\frac {3 (a+(b+c) x) \log ^2(a+(b+c) x)}{b+c}+\frac {(a+(b+c) x) \log ^3(a+(b+c) x)}{b+c}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 67, normalized size = 0.92 \[ \frac {(a+x (b+c)) \log ^3(a+x (b+c))-3 (a+x (b+c)) \log ^2(a+x (b+c))+6 (a+x (b+c)) \log (a+x (b+c))-6 x (b+c)}{b+c} \]

Antiderivative was successfully verified.

[In]

Integrate[Log[a + b*x + c*x]^3,x]

[Out]

(-6*(b + c)*x + 6*(a + (b + c)*x)*Log[a + (b + c)*x] - 3*(a + (b + c)*x)*Log[a + (b + c)*x]^2 + (a + (b + c)*x
)*Log[a + (b + c)*x]^3)/(b + c)

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fricas [A]  time = 0.59, size = 67, normalized size = 0.92 \[ \frac {{\left ({\left (b + c\right )} x + a\right )} \log \left ({\left (b + c\right )} x + a\right )^{3} - 3 \, {\left ({\left (b + c\right )} x + a\right )} \log \left ({\left (b + c\right )} x + a\right )^{2} - 6 \, {\left (b + c\right )} x + 6 \, {\left ({\left (b + c\right )} x + a\right )} \log \left ({\left (b + c\right )} x + a\right )}{b + c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(b*x+c*x+a)^3,x, algorithm="fricas")

[Out]

(((b + c)*x + a)*log((b + c)*x + a)^3 - 3*((b + c)*x + a)*log((b + c)*x + a)^2 - 6*(b + c)*x + 6*((b + c)*x +
a)*log((b + c)*x + a))/(b + c)

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giac [A]  time = 0.17, size = 91, normalized size = 1.25 \[ \frac {{\left (b x + c x + a\right )} \log \left (b x + c x + a\right )^{3}}{b + c} - \frac {3 \, {\left (b x + c x + a\right )} \log \left (b x + c x + a\right )^{2}}{b + c} + \frac {6 \, {\left (b x + c x + a\right )} \log \left (b x + c x + a\right )}{b + c} - \frac {6 \, {\left (b x + c x + a\right )}}{b + c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(b*x+c*x+a)^3,x, algorithm="giac")

[Out]

(b*x + c*x + a)*log(b*x + c*x + a)^3/(b + c) - 3*(b*x + c*x + a)*log(b*x + c*x + a)^2/(b + c) + 6*(b*x + c*x +
a)*log(b*x + c*x + a)/(b + c) - 6*(b*x + c*x + a)/(b + c)

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maple [B]  time = 0.04, size = 187, normalized size = 2.56 \[ \frac {b x \ln \left (a +\left (b +c \right ) x \right )^{3}}{b +c}+\frac {c x \ln \left (a +\left (b +c \right ) x \right )^{3}}{b +c}+\frac {a \ln \left (a +\left (b +c \right ) x \right )^{3}}{b +c}-\frac {3 b x \ln \left (a +\left (b +c \right ) x \right )^{2}}{b +c}-\frac {3 c x \ln \left (a +\left (b +c \right ) x \right )^{2}}{b +c}-\frac {3 a \ln \left (a +\left (b +c \right ) x \right )^{2}}{b +c}+\frac {6 b x \ln \left (a +\left (b +c \right ) x \right )}{b +c}+\frac {6 c x \ln \left (a +\left (b +c \right ) x \right )}{b +c}+\frac {6 a \ln \left (a +\left (b +c \right ) x \right )}{b +c}-\frac {6 b x}{b +c}-\frac {6 c x}{b +c}-\frac {6 a}{b +c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(ln(b*x+c*x+a)^3,x)

[Out]

1/(b+c)*ln(a+(b+c)*x)^3*x*b+1/(b+c)*ln(a+(b+c)*x)^3*x*c+1/(b+c)*ln(a+(b+c)*x)^3*a-3/(b+c)*b*x*ln(a+(b+c)*x)^2-
3/(b+c)*c*x*ln(a+(b+c)*x)^2-3/(b+c)*a*ln(a+(b+c)*x)^2+6/(b+c)*b*x*ln(a+(b+c)*x)+6/(b+c)*c*x*ln(a+(b+c)*x)+6/(b
+c)*a*ln(a+(b+c)*x)-6/(b+c)*b*x-6/(b+c)*c*x-6/(b+c)*a

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maxima [A]  time = 0.61, size = 51, normalized size = 0.70 \[ \frac {{\left (\log \left (b x + c x + a\right )^{3} - 3 \, \log \left (b x + c x + a\right )^{2} + 6 \, \log \left (b x + c x + a\right ) - 6\right )} {\left (b x + c x + a\right )}}{b + c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(b*x+c*x+a)^3,x, algorithm="maxima")

[Out]

(log(b*x + c*x + a)^3 - 3*log(b*x + c*x + a)^2 + 6*log(b*x + c*x + a) - 6)*(b*x + c*x + a)/(b + c)

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mupad [B]  time = 0.37, size = 138, normalized size = 1.89 \[ \frac {6\,a\,\ln \left (a+b\,x+c\,x\right )-6\,c\,x-6\,b\,x-3\,a\,{\ln \left (a+b\,x+c\,x\right )}^2+a\,{\ln \left (a+b\,x+c\,x\right )}^3-3\,b\,x\,{\ln \left (a+b\,x+c\,x\right )}^2+b\,x\,{\ln \left (a+b\,x+c\,x\right )}^3-3\,c\,x\,{\ln \left (a+b\,x+c\,x\right )}^2+c\,x\,{\ln \left (a+b\,x+c\,x\right )}^3+6\,b\,x\,\ln \left (a+b\,x+c\,x\right )+6\,c\,x\,\ln \left (a+b\,x+c\,x\right )}{b+c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(a + b*x + c*x)^3,x)

[Out]

(6*a*log(a + b*x + c*x) - 6*c*x - 6*b*x - 3*a*log(a + b*x + c*x)^2 + a*log(a + b*x + c*x)^3 - 3*b*x*log(a + b*
x + c*x)^2 + b*x*log(a + b*x + c*x)^3 - 3*c*x*log(a + b*x + c*x)^2 + c*x*log(a + b*x + c*x)^3 + 6*b*x*log(a +
b*x + c*x) + 6*c*x*log(a + b*x + c*x))/(b + c)

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sympy [A]  time = 0.65, size = 95, normalized size = 1.30 \[ 6 x \log {\left (a + b x + c x \right )} + \left (- 6 b - 6 c\right ) \left (- \frac {a \log {\left (a + x \left (b + c\right ) \right )}}{\left (b + c\right )^{2}} + \frac {x}{b + c}\right ) + \frac {\left (- 3 a - 3 b x - 3 c x\right ) \log {\left (a + b x + c x \right )}^{2}}{b + c} + \frac {\left (a + b x + c x\right ) \log {\left (a + b x + c x \right )}^{3}}{b + c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(ln(b*x+c*x+a)**3,x)

[Out]

6*x*log(a + b*x + c*x) + (-6*b - 6*c)*(-a*log(a + x*(b + c))/(b + c)**2 + x/(b + c)) + (-3*a - 3*b*x - 3*c*x)*
log(a + b*x + c*x)**2/(b + c) + (a + b*x + c*x)*log(a + b*x + c*x)**3/(b + c)

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