3.1.66 \(\int \log ^3(a+b x) \, dx\) [66]

Optimal. Leaf size=55 \[ -6 x+\frac {6 (a+b x) \log (a+b x)}{b}-\frac {3 (a+b x) \log ^2(a+b x)}{b}+\frac {(a+b x) \log ^3(a+b x)}{b} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {2436, 2333, 2332} \begin {gather*} \frac {(a+b x) \log ^3(a+b x)}{b}-\frac {3 (a+b x) \log ^2(a+b x)}{b}+\frac {6 (a+b x) \log (a+b x)}{b}-6 x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Log[a + b*x]^3,x]

[Out]

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

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

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Mathematica [A]
time = 0.01, size = 51, normalized size = 0.93 \begin {gather*} \frac {-6 b x+6 (a+b x) \log (a+b x)-3 (a+b x) \log ^2(a+b x)+(a+b x) \log ^3(a+b x)}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Log[a + b*x]^3,x]

[Out]

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

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Maple [A]
time = 0.13, size = 55, normalized size = 1.00

method	result	size
derivativedivides	\(\frac {\ln \left (b x +a \right )^{3} \left (b x +a \right )-3 \ln \left (b x +a \right )^{2} \left (b x +a \right )+6 \left (b x +a \right ) \ln \left (b x +a \right )-6 b x -6 a}{b}\)	\(55\)
default	\(\frac {\ln \left (b x +a \right )^{3} \left (b x +a \right )-3 \ln \left (b x +a \right )^{2} \left (b x +a \right )+6 \left (b x +a \right ) \ln \left (b x +a \right )-6 b x -6 a}{b}\)	\(55\)
risch	\(\frac {\left (b x +a \right ) \ln \left (b x +a \right )^{3}}{b}-\frac {3 \left (b x +a \right ) \ln \left (b x +a \right )^{2}}{b}+6 x \ln \left (b x +a \right )-6 x +\frac {6 a \ln \left (b x +a \right )}{b}\)	\(61\)
norman	\(x \ln \left (b x +a \right )^{3}+\frac {a \ln \left (b x +a \right )^{3}}{b}-6 x +6 x \ln \left (b x +a \right )-3 x \ln \left (b x +a \right )^{2}+\frac {6 a \ln \left (b x +a \right )}{b}-\frac {3 a \ln \left (b x +a \right )^{2}}{b}\)	\(74\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(ln(b*x+a)^3,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.28, size = 37, normalized size = 0.67 \begin {gather*} \frac {{\left (\log \left (b x + a\right )^{3} - 3 \, \log \left (b x + a\right )^{2} + 6 \, \log \left (b x + a\right ) - 6\right )} {\left (b x + a\right )}}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.36, size = 51, normalized size = 0.93 \begin {gather*} \frac {{\left (b x + a\right )} \log \left (b x + a\right )^{3} - 3 \, {\left (b x + a\right )} \log \left (b x + a\right )^{2} - 6 \, b x + 6 \, {\left (b x + a\right )} \log \left (b x + a\right )}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.10, size = 63, normalized size = 1.15 \begin {gather*} - 6 b \left (- \frac {a \log {\left (a + b x \right )}}{b^{2}} + \frac {x}{b}\right ) + 6 x \log {\left (a + b x \right )} + \frac {\left (- 3 a - 3 b x\right ) \log {\left (a + b x \right )}^{2}}{b} + \frac {\left (a + b x\right ) \log {\left (a + b x \right )}^{3}}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 5.49, size = 62, normalized size = 1.13 \begin {gather*} \frac {{\left (b x + a\right )} \log \left (b x + a\right )^{3}}{b} - \frac {3 \, {\left (b x + a\right )} \log \left (b x + a\right )^{2}}{b} + \frac {6 \, {\left (b x + a\right )} \log \left (b x + a\right )}{b} - \frac {6 \, {\left (b x + a\right )}}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.25, size = 73, normalized size = 1.33 \begin {gather*} 6\,x\,\ln \left (a+b\,x\right )-6\,x-3\,x\,{\ln \left (a+b\,x\right )}^2+x\,{\ln \left (a+b\,x\right )}^3-\frac {3\,a\,{\ln \left (a+b\,x\right )}^2}{b}+\frac {a\,{\ln \left (a+b\,x\right )}^3}{b}+\frac {6\,a\,\ln \left (a+b\,x\right )}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(a + b*x)^3,x)

[Out]

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

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