3.1.26 \(\int \frac {(d+e x)^3 (a+b \log (c x^n))}{x^4} \, dx\) [26]

3.1.26.1 Optimal result
3.1.26.2 Mathematica [A] (verified)
3.1.26.3 Rubi [A] (verified)
3.1.26.4 Maple [A] (verified)
3.1.26.5 Fricas [A] (verification not implemented)
3.1.26.6 Sympy [A] (verification not implemented)
3.1.26.7 Maxima [A] (verification not implemented)
3.1.26.8 Giac [A] (verification not implemented)
3.1.26.9 Mupad [B] (verification not implemented)

3.1.26.1 Optimal result

Integrand size = 21, antiderivative size = 126 \[ \int \frac {(d+e x)^3 \left (a+b \log \left (c x^n\right )\right )}{x^4} \, dx=-\frac {b d^3 n}{9 x^3}-\frac {3 b d^2 e n}{4 x^2}-\frac {3 b d e^2 n}{x}-\frac {1}{2} b e^3 n \log ^2(x)-\frac {d^3 \left (a+b \log \left (c x^n\right )\right )}{3 x^3}-\frac {3 d^2 e \left (a+b \log \left (c x^n\right )\right )}{2 x^2}-\frac {3 d e^2 \left (a+b \log \left (c x^n\right )\right )}{x}+e^3 \log (x) \left (a+b \log \left (c x^n\right )\right ) \]

output
-1/9*b*d^3*n/x^3-3/4*b*d^2*e*n/x^2-3*b*d*e^2*n/x-1/2*b*e^3*n*ln(x)^2-1/3*d 
^3*(a+b*ln(c*x^n))/x^3-3/2*d^2*e*(a+b*ln(c*x^n))/x^2-3*d*e^2*(a+b*ln(c*x^n 
))/x+e^3*ln(x)*(a+b*ln(c*x^n))
 
3.1.26.2 Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.97 \[ \int \frac {(d+e x)^3 \left (a+b \log \left (c x^n\right )\right )}{x^4} \, dx=-\frac {b d^3 n}{9 x^3}-\frac {3 b d^2 e n}{4 x^2}-\frac {3 b d e^2 n}{x}-\frac {d^3 \left (a+b \log \left (c x^n\right )\right )}{3 x^3}-\frac {3 d^2 e \left (a+b \log \left (c x^n\right )\right )}{2 x^2}-\frac {3 d e^2 \left (a+b \log \left (c x^n\right )\right )}{x}+\frac {e^3 \left (a+b \log \left (c x^n\right )\right )^2}{2 b n} \]

input
Integrate[((d + e*x)^3*(a + b*Log[c*x^n]))/x^4,x]
 
output
-1/9*(b*d^3*n)/x^3 - (3*b*d^2*e*n)/(4*x^2) - (3*b*d*e^2*n)/x - (d^3*(a + b 
*Log[c*x^n]))/(3*x^3) - (3*d^2*e*(a + b*Log[c*x^n]))/(2*x^2) - (3*d*e^2*(a 
 + b*Log[c*x^n]))/x + (e^3*(a + b*Log[c*x^n])^2)/(2*b*n)
 
3.1.26.3 Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.98, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2772, 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.

\(\displaystyle \int \frac {(d+e x)^3 \left (a+b \log \left (c x^n\right )\right )}{x^4} \, dx\)

\(\Big \downarrow \) 2772

\(\displaystyle -b n \int \left (\frac {e^3 \log (x)}{x}-\frac {d \left (2 d^2+9 e x d+18 e^2 x^2\right )}{6 x^4}\right )dx-\frac {d^3 \left (a+b \log \left (c x^n\right )\right )}{3 x^3}-\frac {3 d^2 e \left (a+b \log \left (c x^n\right )\right )}{2 x^2}-\frac {3 d e^2 \left (a+b \log \left (c x^n\right )\right )}{x}+e^3 \log (x) \left (a+b \log \left (c x^n\right )\right )\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {d^3 \left (a+b \log \left (c x^n\right )\right )}{3 x^3}-\frac {3 d^2 e \left (a+b \log \left (c x^n\right )\right )}{2 x^2}-\frac {3 d e^2 \left (a+b \log \left (c x^n\right )\right )}{x}+e^3 \log (x) \left (a+b \log \left (c x^n\right )\right )-b n \left (\frac {d^3}{9 x^3}+\frac {3 d^2 e}{4 x^2}+\frac {3 d e^2}{x}+\frac {1}{2} e^3 \log ^2(x)\right )\)

input
Int[((d + e*x)^3*(a + b*Log[c*x^n]))/x^4,x]
 
output
-(b*n*(d^3/(9*x^3) + (3*d^2*e)/(4*x^2) + (3*d*e^2)/x + (e^3*Log[x]^2)/2)) 
- (d^3*(a + b*Log[c*x^n]))/(3*x^3) - (3*d^2*e*(a + b*Log[c*x^n]))/(2*x^2) 
- (3*d*e^2*(a + b*Log[c*x^n]))/x + e^3*Log[x]*(a + b*Log[c*x^n])
 

3.1.26.3.1 Defintions of rubi rules used

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2772
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(x_)^(m_.)*((d_) + (e_.)*(x_)^(r_ 
.))^(q_.), x_Symbol] :> With[{u = IntHide[x^m*(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] && IntegerQ[m] &&  !(EqQ[q 
, 1] && EqQ[m, -1])
 
3.1.26.4 Maple [A] (verified)

Time = 0.50 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.14

method result size
parallelrisch \(\frac {36 \ln \left (x \right ) x^{3} a \,e^{3} n +18 e^{3} b \ln \left (c \,x^{n}\right )^{2} x^{3}-108 x^{2} \ln \left (c \,x^{n}\right ) b d \,e^{2} n -108 x^{2} b d \,e^{2} n^{2}-108 x^{2} a d \,e^{2} n -54 x \ln \left (c \,x^{n}\right ) b \,d^{2} e n -27 x b \,d^{2} e \,n^{2}-54 x a \,d^{2} e n -12 \ln \left (c \,x^{n}\right ) b \,d^{3} n -4 b \,d^{3} n^{2}-12 a \,d^{3} n}{36 x^{3} n}\) \(144\)
risch \(-\frac {b \left (-6 e^{3} \ln \left (x \right ) x^{3}+18 d \,e^{2} x^{2}+9 d^{2} e x +2 d^{3}\right ) \ln \left (x^{n}\right )}{6 x^{3}}-\frac {-54 i \pi b d \,e^{2} x^{2} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )-27 i \pi b \,d^{2} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) e x +54 i \pi b d \,e^{2} x^{2} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}+54 i \pi b d \,e^{2} x^{2} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}+27 i \pi b \,d^{2} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2} e x +27 i \pi b \,d^{2} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2} e x -6 i \pi b \,d^{3} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )+12 a \,d^{3}-36 \ln \left (x \right ) a \,e^{3} x^{3}+12 d^{3} b \ln \left (c \right )-18 i \ln \left (x \right ) \pi b \,e^{3} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2} x^{3}-18 i \ln \left (x \right ) \pi b \,e^{3} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2} x^{3}-27 i \pi b \,d^{2} \operatorname {csgn}\left (i c \,x^{n}\right )^{3} e x -54 i \pi b d \,e^{2} x^{2} \operatorname {csgn}\left (i c \,x^{n}\right )^{3}+4 b \,d^{3} n -6 i \pi b \,d^{3} \operatorname {csgn}\left (i c \,x^{n}\right )^{3}+18 i \ln \left (x \right ) \pi b \,e^{3} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) x^{3}+108 a d \,e^{2} x^{2}+54 a \,d^{2} e x +18 i \ln \left (x \right ) \pi b \,e^{3} \operatorname {csgn}\left (i c \,x^{n}\right )^{3} x^{3}+108 \ln \left (c \right ) b d \,e^{2} x^{2}+54 \ln \left (c \right ) b \,d^{2} e x -36 \ln \left (x \right ) \ln \left (c \right ) b \,e^{3} x^{3}+18 e^{3} b n \ln \left (x \right )^{2} x^{3}+27 b \,d^{2} e n x +108 b d \,e^{2} n \,x^{2}+6 i \pi b \,d^{3} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}+6 i \pi b \,d^{3} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{36 x^{3}}\) \(589\)

input
int((e*x+d)^3*(a+b*ln(c*x^n))/x^4,x,method=_RETURNVERBOSE)
 
output
1/36/x^3*(36*ln(x)*x^3*a*e^3*n+18*e^3*b*ln(c*x^n)^2*x^3-108*x^2*ln(c*x^n)* 
b*d*e^2*n-108*x^2*b*d*e^2*n^2-108*x^2*a*d*e^2*n-54*x*ln(c*x^n)*b*d^2*e*n-2 
7*x*b*d^2*e*n^2-54*x*a*d^2*e*n-12*ln(c*x^n)*b*d^3*n-4*b*d^3*n^2-12*a*d^3*n 
)/n
 
3.1.26.5 Fricas [A] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.20 \[ \int \frac {(d+e x)^3 \left (a+b \log \left (c x^n\right )\right )}{x^4} \, dx=\frac {18 \, b e^{3} n x^{3} \log \left (x\right )^{2} - 4 \, b d^{3} n - 12 \, a d^{3} - 108 \, {\left (b d e^{2} n + a d e^{2}\right )} x^{2} - 27 \, {\left (b d^{2} e n + 2 \, a d^{2} e\right )} x - 6 \, {\left (18 \, b d e^{2} x^{2} + 9 \, b d^{2} e x + 2 \, b d^{3}\right )} \log \left (c\right ) + 6 \, {\left (6 \, b e^{3} x^{3} \log \left (c\right ) - 18 \, b d e^{2} n x^{2} + 6 \, a e^{3} x^{3} - 9 \, b d^{2} e n x - 2 \, b d^{3} n\right )} \log \left (x\right )}{36 \, x^{3}} \]

input
integrate((e*x+d)^3*(a+b*log(c*x^n))/x^4,x, algorithm="fricas")
 
output
1/36*(18*b*e^3*n*x^3*log(x)^2 - 4*b*d^3*n - 12*a*d^3 - 108*(b*d*e^2*n + a* 
d*e^2)*x^2 - 27*(b*d^2*e*n + 2*a*d^2*e)*x - 6*(18*b*d*e^2*x^2 + 9*b*d^2*e* 
x + 2*b*d^3)*log(c) + 6*(6*b*e^3*x^3*log(c) - 18*b*d*e^2*n*x^2 + 6*a*e^3*x 
^3 - 9*b*d^2*e*n*x - 2*b*d^3*n)*log(x))/x^3
 
3.1.26.6 Sympy [A] (verification not implemented)

Time = 2.46 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.14 \[ \int \frac {(d+e x)^3 \left (a+b \log \left (c x^n\right )\right )}{x^4} \, dx=- \frac {a d^{3}}{3 x^{3}} - \frac {3 a d^{2} e}{2 x^{2}} - \frac {3 a d e^{2}}{x} + a e^{3} \log {\left (x \right )} + b d^{3} \left (- \frac {n}{9 x^{3}} - \frac {\log {\left (c x^{n} \right )}}{3 x^{3}}\right ) + 3 b d^{2} e \left (- \frac {n}{4 x^{2}} - \frac {\log {\left (c x^{n} \right )}}{2 x^{2}}\right ) + 3 b d e^{2} \left (- \frac {n}{x} - \frac {\log {\left (c x^{n} \right )}}{x}\right ) - b e^{3} \left (\begin {cases} - \log {\left (c \right )} \log {\left (x \right )} & \text {for}\: n = 0 \\- \frac {\log {\left (c x^{n} \right )}^{2}}{2 n} & \text {otherwise} \end {cases}\right ) \]

input
integrate((e*x+d)**3*(a+b*ln(c*x**n))/x**4,x)
 
output
-a*d**3/(3*x**3) - 3*a*d**2*e/(2*x**2) - 3*a*d*e**2/x + a*e**3*log(x) + b* 
d**3*(-n/(9*x**3) - log(c*x**n)/(3*x**3)) + 3*b*d**2*e*(-n/(4*x**2) - log( 
c*x**n)/(2*x**2)) + 3*b*d*e**2*(-n/x - log(c*x**n)/x) - b*e**3*Piecewise(( 
-log(c)*log(x), Eq(n, 0)), (-log(c*x**n)**2/(2*n), True))
 
3.1.26.7 Maxima [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.06 \[ \int \frac {(d+e x)^3 \left (a+b \log \left (c x^n\right )\right )}{x^4} \, dx=\frac {b e^{3} \log \left (c x^{n}\right )^{2}}{2 \, n} + a e^{3} \log \left (x\right ) - \frac {3 \, b d e^{2} n}{x} - \frac {3 \, b d e^{2} \log \left (c x^{n}\right )}{x} - \frac {3 \, b d^{2} e n}{4 \, x^{2}} - \frac {3 \, a d e^{2}}{x} - \frac {3 \, b d^{2} e \log \left (c x^{n}\right )}{2 \, x^{2}} - \frac {b d^{3} n}{9 \, x^{3}} - \frac {3 \, a d^{2} e}{2 \, x^{2}} - \frac {b d^{3} \log \left (c x^{n}\right )}{3 \, x^{3}} - \frac {a d^{3}}{3 \, x^{3}} \]

input
integrate((e*x+d)^3*(a+b*log(c*x^n))/x^4,x, algorithm="maxima")
 
output
1/2*b*e^3*log(c*x^n)^2/n + a*e^3*log(x) - 3*b*d*e^2*n/x - 3*b*d*e^2*log(c* 
x^n)/x - 3/4*b*d^2*e*n/x^2 - 3*a*d*e^2/x - 3/2*b*d^2*e*log(c*x^n)/x^2 - 1/ 
9*b*d^3*n/x^3 - 3/2*a*d^2*e/x^2 - 1/3*b*d^3*log(c*x^n)/x^3 - 1/3*a*d^3/x^3
 
3.1.26.8 Giac [A] (verification not implemented)

Time = 0.39 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.18 \[ \int \frac {(d+e x)^3 \left (a+b \log \left (c x^n\right )\right )}{x^4} \, dx=\frac {1}{2} \, b e^{3} n \log \left (x\right )^{2} - 3 \, b d e^{2} n {\left (\frac {\log \left (x\right )}{x} + \frac {1}{x}\right )} - \frac {3}{4} \, b d^{2} e n {\left (\frac {2 \, \log \left (x\right )}{x^{2}} + \frac {1}{x^{2}}\right )} - \frac {1}{9} \, b d^{3} n {\left (\frac {3 \, \log \left (x\right )}{x^{3}} + \frac {1}{x^{3}}\right )} + b e^{3} \log \left (c\right ) \log \left ({\left | x \right |}\right ) + a e^{3} \log \left ({\left | x \right |}\right ) - \frac {3 \, b d e^{2} \log \left (c\right )}{x} - \frac {3 \, a d e^{2}}{x} - \frac {3 \, b d^{2} e \log \left (c\right )}{2 \, x^{2}} - \frac {3 \, a d^{2} e}{2 \, x^{2}} - \frac {b d^{3} \log \left (c\right )}{3 \, x^{3}} - \frac {a d^{3}}{3 \, x^{3}} \]

input
integrate((e*x+d)^3*(a+b*log(c*x^n))/x^4,x, algorithm="giac")
 
output
1/2*b*e^3*n*log(x)^2 - 3*b*d*e^2*n*(log(x)/x + 1/x) - 3/4*b*d^2*e*n*(2*log 
(x)/x^2 + 1/x^2) - 1/9*b*d^3*n*(3*log(x)/x^3 + 1/x^3) + b*e^3*log(c)*log(a 
bs(x)) + a*e^3*log(abs(x)) - 3*b*d*e^2*log(c)/x - 3*a*d*e^2/x - 3/2*b*d^2* 
e*log(c)/x^2 - 3/2*a*d^2*e/x^2 - 1/3*b*d^3*log(c)/x^3 - 1/3*a*d^3/x^3
 
3.1.26.9 Mupad [B] (verification not implemented)

Time = 0.44 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.08 \[ \int \frac {(d+e x)^3 \left (a+b \log \left (c x^n\right )\right )}{x^4} \, dx=\ln \left (x\right )\,\left (a\,e^3+\frac {11\,b\,e^3\,n}{6}\right )-\frac {x\,\left (9\,a\,d^2\,e+\frac {9\,b\,d^2\,e\,n}{2}\right )+2\,a\,d^3+x^2\,\left (18\,a\,d\,e^2+18\,b\,d\,e^2\,n\right )+\frac {2\,b\,d^3\,n}{3}}{6\,x^3}-\frac {\ln \left (c\,x^n\right )\,\left (\frac {b\,d^3}{3}+\frac {3\,b\,d^2\,e\,x}{2}+3\,b\,d\,e^2\,x^2+\frac {11\,b\,e^3\,x^3}{6}\right )}{x^3}+\frac {b\,e^3\,{\ln \left (c\,x^n\right )}^2}{2\,n} \]

input
int(((a + b*log(c*x^n))*(d + e*x)^3)/x^4,x)
 
output
log(x)*(a*e^3 + (11*b*e^3*n)/6) - (x*(9*a*d^2*e + (9*b*d^2*e*n)/2) + 2*a*d 
^3 + x^2*(18*a*d*e^2 + 18*b*d*e^2*n) + (2*b*d^3*n)/3)/(6*x^3) - (log(c*x^n 
)*((b*d^3)/3 + (11*b*e^3*x^3)/6 + (3*b*d^2*e*x)/2 + 3*b*d*e^2*x^2))/x^3 + 
(b*e^3*log(c*x^n)^2)/(2*n)