Integrand size = 10, antiderivative size = 67 \[ \int (a+b x)^3 \log (x) \, dx=-a^3 x-\frac {3}{4} a^2 b x^2-\frac {1}{3} a b^2 x^3-\frac {b^3 x^4}{16}-\frac {a^4 \log (x)}{4 b}+\frac {(a+b x)^4 \log (x)}{4 b} \] Output:
-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
Time = 0.01 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.21 \[ \int (a+b x)^3 \log (x) \, dx=-a^3 x-\frac {3}{4} a^2 b x^2-\frac {1}{3} a b^2 x^3-\frac {b^3 x^4}{16}+a^3 x \log (x)+\frac {3}{2} a^2 b x^2 \log (x)+a b^2 x^3 \log (x)+\frac {1}{4} b^3 x^4 \log (x) \] Input:
Integrate[(a + b*x)^3*Log[x],x]
Output:
-(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
Time = 0.26 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.04, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {2750, 27, 49, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \log (x) (a+b x)^3 \, dx\) |
\(\Big \downarrow \) 2750 |
\(\displaystyle \frac {\log (x) (a+b x)^4}{4 b}-\int \frac {(a+b x)^4}{4 b x}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\log (x) (a+b x)^4}{4 b}-\frac {\int \frac {(a+b x)^4}{x}dx}{4 b}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {\log (x) (a+b x)^4}{4 b}-\frac {\int \left (\frac {a^4}{x}+4 b a^3+6 b^2 x a^2+4 b^3 x^2 a+b^4 x^3\right )dx}{4 b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\log (x) (a+b x)^4}{4 b}-\frac {a^4 \log (x)+4 a^3 b x+3 a^2 b^2 x^2+\frac {4}{3} a b^3 x^3+\frac {b^4 x^4}{4}}{4 b}\) |
Input:
Int[(a + b*x)^3*Log[x],x]
Output:
((a + b*x)^4*Log[x])/(4*b) - (4*a^3*b*x + 3*a^2*b^2*x^2 + (4*a*b^3*x^3)/3 + (b^4*x^4)/4 + a^4*Log[x])/(4*b)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> With[{u = IntHide[(d + e*x^r)^q, x]}, Simp[(a + b*Log[c*x^n]) u, x] - Simp[b*n Int[SimplifyIntegrand[u/x, x], x], x]] /; FreeQ[{a, b, c, d, e, n, r}, x] && IGtQ[q, 0]
Time = 0.14 (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 | \(b^{3} \left (-\frac {x^{4}}{16}+\frac {x^{4} \ln \left (x \right )}{4}\right )+3 b^{2} a \left (\frac {x^{3} \ln \left (x \right )}{3}-\frac {x^{3}}{9}\right )+3 a^{2} b \left (-\frac {x^{2}}{4}+\frac {x^{2} \ln \left (x \right )}{2}\right )+a^{3} \left (-x +x \ln \left (x \right )\right )\) | \(69\) |
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 | \(\frac {b^{3} x^{4} \ln \left (x \right )}{4}+a \,b^{2} x^{3} \ln \left (x \right )+\frac {3 a^{2} b \,x^{2} \ln \left (x \right )}{2}+a^{3} x \ln \left (x \right )+\frac {a^{4} \ln \left (x \right )}{4 b}-\frac {\frac {b^{4} x^{4}}{4}+\frac {4 a \,b^{3} x^{3}}{3}+3 a^{2} b^{2} x^{2}+4 a^{3} b x +a^{4} \ln \left (x \right )}{4 b}\) | \(97\) |
orering | \(\frac {x \left (7 b^{4} x^{4}+36 a \,b^{3} x^{3}+76 a^{2} b^{2} x^{2}+88 a^{3} b x +16 a^{4}\right ) \ln \left (x \right )}{16 b x +16 a}-\frac {x^{2} \left (3 x^{3} b^{3}+16 a \,b^{2} x^{2}+36 a^{2} b x +48 a^{3}\right ) \left (3 \left (b x +a \right )^{2} \ln \left (x \right ) b +\frac {\left (b x +a \right )^{3}}{x}\right )}{48 \left (b x +a \right )^{3}}\) | \(121\) |
Input:
int((b*x+a)^3*ln(x),x,method=_RETURNVERBOSE)
Output:
-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
Time = 0.06 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.03 \[ \int (a+b x)^3 \log (x) \, dx=-\frac {1}{16} \, b^{3} x^{4} - \frac {1}{3} \, a b^{2} x^{3} - \frac {3}{4} \, a^{2} b x^{2} - a^{3} x + \frac {1}{4} \, {\left (b^{3} x^{4} + 4 \, a b^{2} x^{3} + 6 \, a^{2} b x^{2} + 4 \, a^{3} x\right )} \log \left (x\right ) \] Input:
integrate((b*x+a)^3*log(x),x, algorithm="fricas")
Output:
-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)*log(x)
Time = 0.06 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.06 \[ \int (a+b x)^3 \log (x) \, dx=- a^{3} x - \frac {3 a^{2} b x^{2}}{4} - \frac {a b^{2} x^{3}}{3} - \frac {b^{3} x^{4}}{16} + \left (a^{3} x + \frac {3 a^{2} b x^{2}}{2} + a b^{2} x^{3} + \frac {b^{3} x^{4}}{4}\right ) \log {\left (x \right )} \] Input:
integrate((b*x+a)**3*ln(x),x)
Output:
-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)
Time = 0.03 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.03 \[ \int (a+b x)^3 \log (x) \, dx=-\frac {1}{16} \, b^{3} x^{4} - \frac {1}{3} \, a b^{2} x^{3} - \frac {3}{4} \, a^{2} b x^{2} - a^{3} x + \frac {1}{4} \, {\left (b^{3} x^{4} + 4 \, a b^{2} x^{3} + 6 \, a^{2} b x^{2} + 4 \, a^{3} x\right )} \log \left (x\right ) \] Input:
integrate((b*x+a)^3*log(x),x, algorithm="maxima")
Output:
-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)*log(x)
Time = 0.12 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.06 \[ \int (a+b x)^3 \log (x) \, dx=\frac {1}{4} \, b^{3} x^{4} \log \left (x\right ) - \frac {1}{16} \, b^{3} x^{4} + a b^{2} x^{3} \log \left (x\right ) - \frac {1}{3} \, a b^{2} x^{3} + \frac {3}{2} \, a^{2} b x^{2} \log \left (x\right ) - \frac {3}{4} \, a^{2} b x^{2} + a^{3} x \log \left (x\right ) - a^{3} x \] Input:
integrate((b*x+a)^3*log(x),x, algorithm="giac")
Output:
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
Time = 0.22 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.06 \[ \int (a+b x)^3 \log (x) \, dx=a^3\,x\,\ln \left (x\right )-\frac {b^3\,x^4}{16}-\frac {3\,a^2\,b\,x^2}{4}-\frac {a\,b^2\,x^3}{3}-a^3\,x+\frac {b^3\,x^4\,\ln \left (x\right )}{4}+\frac {3\,a^2\,b\,x^2\,\ln \left (x\right )}{2}+a\,b^2\,x^3\,\ln \left (x\right ) \] Input:
int(log(x)*(a + b*x)^3,x)
Output:
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)
Time = 0.15 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.04 \[ \int (a+b x)^3 \log (x) \, dx=\frac {x \left (48 \,\mathrm {log}\left (x \right ) a^{3}+72 \,\mathrm {log}\left (x \right ) a^{2} b x +48 \,\mathrm {log}\left (x \right ) a \,b^{2} x^{2}+12 \,\mathrm {log}\left (x \right ) b^{3} x^{3}-48 a^{3}-36 a^{2} b x -16 a \,b^{2} x^{2}-3 b^{3} x^{3}\right )}{48} \] Input:
int((b*x+a)^3*log(x),x)
Output:
(x*(48*log(x)*a**3 + 72*log(x)*a**2*b*x + 48*log(x)*a*b**2*x**2 + 12*log(x )*b**3*x**3 - 48*a**3 - 36*a**2*b*x - 16*a*b**2*x**2 - 3*b**3*x**3))/48