\(\int (a+b x)^3 \log (x) \, dx\) [612]

Optimal result

Integrand size = 10, antiderivative size = 67 \[ \int (a+b x)^3 \log (x) \, dx=-a^3 x-\frac {3}{4} a^2 b x^2-\frac {1}{3} a b^2 x^3-\frac {b^3 x^4}{16}-\frac {a^4 \log (x)}{4 b}+\frac {(a+b x)^4 \log (x)}{4 b} \]

[Out]

-a^3*x-3/4*a^2*b*x^2-1/3*a*b^2*x^3-1/16*b^3*x^4-1/4*a^4*ln(x)/b+1/4*(b*x+a)^4*ln(x)/b

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {32, 2350, 12, 45} \[ \int (a+b x)^3 \log (x) \, dx=-\frac {a^4 \log (x)}{4 b}-a^3 x-\frac {3}{4} a^2 b x^2-\frac {1}{3} a b^2 x^3+\frac {\log (x) (a+b x)^4}{4 b}-\frac {b^3 x^4}{16} \]

[In]

Int[(a + b*x)^3*Log[x],x]

[Out]

-(a^3*x) - (3*a^2*b*x^2)/4 - (a*b^2*x^3)/3 - (b^3*x^4)/16 - (a^4*Log[x])/(4*b) + ((a + b*x)^4*Log[x])/(4*b)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2350

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> With[{u = IntHide[(d +
e*x^r)^q, x]}, Dist[a + b*Log[c*x^n], u, x] - Dist[b*n, Int[SimplifyIntegrand[u/x, x], x], x]] /; FreeQ[{a, b
, c, d, e, n, r}, x] && IGtQ[q, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {(a+b x)^4 \log (x)}{4 b}-\int \frac {(a+b x)^4}{4 b x} \, dx \\ & = \frac {(a+b x)^4 \log (x)}{4 b}-\frac {\int \frac {(a+b x)^4}{x} \, dx}{4 b} \\ & = \frac {(a+b x)^4 \log (x)}{4 b}-\frac {\int \left (4 a^3 b+\frac {a^4}{x}+6 a^2 b^2 x+4 a b^3 x^2+b^4 x^3\right ) \, dx}{4 b} \\ & = -a^3 x-\frac {3}{4} a^2 b x^2-\frac {1}{3} a b^2 x^3-\frac {b^3 x^4}{16}-\frac {a^4 \log (x)}{4 b}+\frac {(a+b x)^4 \log (x)}{4 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.21 \[ \int (a+b x)^3 \log (x) \, dx=-a^3 x-\frac {3}{4} a^2 b x^2-\frac {1}{3} a b^2 x^3-\frac {b^3 x^4}{16}+a^3 x \log (x)+\frac {3}{2} a^2 b x^2 \log (x)+a b^2 x^3 \log (x)+\frac {1}{4} b^3 x^4 \log (x) \]

[In]

Integrate[(a + b*x)^3*Log[x],x]

[Out]

-(a^3*x) - (3*a^2*b*x^2)/4 - (a*b^2*x^3)/3 - (b^3*x^4)/16 + a^3*x*Log[x] + (3*a^2*b*x^2*Log[x])/2 + a*b^2*x^3*
Log[x] + (b^3*x^4*Log[x])/4

Maple [A] (verified)

Time = 0.30 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.87

method	result	size
risch	\(-a^{3} x -\frac {3 a^{2} b \,x^{2}}{4}-\frac {a \,b^{2} x^{3}}{3}-\frac {b^{3} x^{4}}{16}-\frac {a^{4} \ln \left (x \right )}{4 b}+\frac {\left (b x +a \right )^{4} \ln \left (x \right )}{4 b}\)	\(58\)
default	\(b^{3} \left (-\frac {x^{4}}{16}+\frac {x^{4} \ln \left (x \right )}{4}\right )+3 b^{2} a \left (-\frac {x^{3}}{9}+\frac {x^{3} \ln \left (x \right )}{3}\right )+3 a^{2} b \left (-\frac {x^{2}}{4}+\frac {x^{2} \ln \left (x \right )}{2}\right )+a^{3} \left (-x +x \ln \left (x \right )\right )\)	\(69\)
norman	\(a^{3} x \ln \left (x \right )+a \,b^{2} x^{3} \ln \left (x \right )-a^{3} x -\frac {b^{3} x^{4}}{16}-\frac {a \,b^{2} x^{3}}{3}-\frac {3 a^{2} b \,x^{2}}{4}+\frac {b^{3} x^{4} \ln \left (x \right )}{4}+\frac {3 a^{2} b \,x^{2} \ln \left (x \right )}{2}\)	\(72\)
parallelrisch	\(a^{3} x \ln \left (x \right )+a \,b^{2} x^{3} \ln \left (x \right )-a^{3} x -\frac {b^{3} x^{4}}{16}-\frac {a \,b^{2} x^{3}}{3}-\frac {3 a^{2} b \,x^{2}}{4}+\frac {b^{3} x^{4} \ln \left (x \right )}{4}+\frac {3 a^{2} b \,x^{2} \ln \left (x \right )}{2}\)	\(72\)
parts	\(\frac {b^{3} x^{4} \ln \left (x \right )}{4}+a \,b^{2} x^{3} \ln \left (x \right )+\frac {3 a^{2} b \,x^{2} \ln \left (x \right )}{2}+a^{3} x \ln \left (x \right )+\frac {a^{4} \ln \left (x \right )}{4 b}-\frac {\frac {b^{4} x^{4}}{4}+\frac {4 a \,b^{3} x^{3}}{3}+3 a^{2} b^{2} x^{2}+4 a^{3} b x +a^{4} \ln \left (x \right )}{4 b}\)	\(97\)

[In]

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

[Out]

-a^3*x-3/4*a^2*b*x^2-1/3*a*b^2*x^3-1/16*b^3*x^4-1/4*a^4*ln(x)/b+1/4*(b*x+a)^4*ln(x)/b

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.03 \[ \int (a+b x)^3 \log (x) \, dx=-\frac {1}{16} \, b^{3} x^{4} - \frac {1}{3} \, a b^{2} x^{3} - \frac {3}{4} \, a^{2} b x^{2} - a^{3} x + \frac {1}{4} \, {\left (b^{3} x^{4} + 4 \, a b^{2} x^{3} + 6 \, a^{2} b x^{2} + 4 \, a^{3} x\right )} \log \left (x\right ) \]

[In]

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

[Out]

-1/16*b^3*x^4 - 1/3*a*b^2*x^3 - 3/4*a^2*b*x^2 - a^3*x + 1/4*(b^3*x^4 + 4*a*b^2*x^3 + 6*a^2*b*x^2 + 4*a^3*x)*lo
g(x)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.06 \[ \int (a+b x)^3 \log (x) \, dx=- a^{3} x - \frac {3 a^{2} b x^{2}}{4} - \frac {a b^{2} x^{3}}{3} - \frac {b^{3} x^{4}}{16} + \left (a^{3} x + \frac {3 a^{2} b x^{2}}{2} + a b^{2} x^{3} + \frac {b^{3} x^{4}}{4}\right ) \log {\left (x \right )} \]

[In]

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

[Out]

-a**3*x - 3*a**2*b*x**2/4 - a*b**2*x**3/3 - b**3*x**4/16 + (a**3*x + 3*a**2*b*x**2/2 + a*b**2*x**3 + b**3*x**4
/4)*log(x)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.03 \[ \int (a+b x)^3 \log (x) \, dx=-\frac {1}{16} \, b^{3} x^{4} - \frac {1}{3} \, a b^{2} x^{3} - \frac {3}{4} \, a^{2} b x^{2} - a^{3} x + \frac {1}{4} \, {\left (b^{3} x^{4} + 4 \, a b^{2} x^{3} + 6 \, a^{2} b x^{2} + 4 \, a^{3} x\right )} \log \left (x\right ) \]

[In]

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

[Out]

-1/16*b^3*x^4 - 1/3*a*b^2*x^3 - 3/4*a^2*b*x^2 - a^3*x + 1/4*(b^3*x^4 + 4*a*b^2*x^3 + 6*a^2*b*x^2 + 4*a^3*x)*lo
g(x)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.06 \[ \int (a+b x)^3 \log (x) \, dx=\frac {1}{4} \, b^{3} x^{4} \log \left (x\right ) - \frac {1}{16} \, b^{3} x^{4} + a b^{2} x^{3} \log \left (x\right ) - \frac {1}{3} \, a b^{2} x^{3} + \frac {3}{2} \, a^{2} b x^{2} \log \left (x\right ) - \frac {3}{4} \, a^{2} b x^{2} + a^{3} x \log \left (x\right ) - a^{3} x \]

[In]

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

[Out]

1/4*b^3*x^4*log(x) - 1/16*b^3*x^4 + a*b^2*x^3*log(x) - 1/3*a*b^2*x^3 + 3/2*a^2*b*x^2*log(x) - 3/4*a^2*b*x^2 +
a^3*x*log(x) - a^3*x

Mupad [B] (verification not implemented)

Time = 0.44 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.06 \[ \int (a+b x)^3 \log (x) \, dx=a^3\,x\,\ln \left (x\right )-\frac {b^3\,x^4}{16}-\frac {3\,a^2\,b\,x^2}{4}-\frac {a\,b^2\,x^3}{3}-a^3\,x+\frac {b^3\,x^4\,\ln \left (x\right )}{4}+\frac {3\,a^2\,b\,x^2\,\ln \left (x\right )}{2}+a\,b^2\,x^3\,\ln \left (x\right ) \]

[In]

int(log(x)*(a + b*x)^3,x)

[Out]

a^3*x*log(x) - (b^3*x^4)/16 - (3*a^2*b*x^2)/4 - (a*b^2*x^3)/3 - a^3*x + (b^3*x^4*log(x))/4 + (3*a^2*b*x^2*log(
x))/2 + a*b^2*x^3*log(x)