\(\int \frac {(d+e x^2)^3 (a+b \log (c x^n))}{x^5} \, dx\) [201]

Optimal result

Integrand size = 23, antiderivative size = 126 \[ \int \frac {\left (d+e x^2\right )^3 \left (a+b \log \left (c x^n\right )\right )}{x^5} \, dx=-\frac {1}{4} b e^3 n x^2-\frac {b d n \left (d+6 e x^2\right )^2}{16 x^4}-\frac {3}{2} b d e^2 n \log ^2(x)-\frac {d^3 \left (a+b \log \left (c x^n\right )\right )}{4 x^4}-\frac {3 d^2 e \left (a+b \log \left (c x^n\right )\right )}{2 x^2}+\frac {1}{2} e^3 x^2 \left (a+b \log \left (c x^n\right )\right )+3 d e^2 \log (x) \left (a+b \log \left (c x^n\right )\right ) \] Output:

-1/4*b*e^3*n*x^2-1/16*b*d*n*(6*e*x^2+d)^2/x^4-3/2*b*d*e^2*n*ln(x)^2-1/4*d^ 
3*(a+b*ln(c*x^n))/x^4-3/2*d^2*e*(a+b*ln(c*x^n))/x^2+1/2*e^3*x^2*(a+b*ln(c* 
x^n))+3*d*e^2*ln(x)*(a+b*ln(c*x^n))
 

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.91 \[ \int \frac {\left (d+e x^2\right )^3 \left (a+b \log \left (c x^n\right )\right )}{x^5} \, dx=\frac {1}{16} \left (-\frac {b d^3 n}{x^4}-\frac {12 b d^2 e n}{x^2}-4 b e^3 n x^2-\frac {4 d^3 \left (a+b \log \left (c x^n\right )\right )}{x^4}-\frac {24 d^2 e \left (a+b \log \left (c x^n\right )\right )}{x^2}+8 e^3 x^2 \left (a+b \log \left (c x^n\right )\right )+\frac {24 d e^2 \left (a+b \log \left (c x^n\right )\right )^2}{b n}\right ) \] Input:

Integrate[((d + e*x^2)^3*(a + b*Log[c*x^n]))/x^5,x]
 

Output:

(-((b*d^3*n)/x^4) - (12*b*d^2*e*n)/x^2 - 4*b*e^3*n*x^2 - (4*d^3*(a + b*Log 
[c*x^n]))/x^4 - (24*d^2*e*(a + b*Log[c*x^n]))/x^2 + 8*e^3*x^2*(a + b*Log[c 
*x^n]) + (24*d*e^2*(a + b*Log[c*x^n])^2)/(b*n))/16
 

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.98, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {2772, 27, 2010, 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.

\(\displaystyle \int \frac {\left (d+e x^2\right )^3 \left (a+b \log \left (c x^n\right )\right )}{x^5} \, dx\)

\(\Big \downarrow \) 2772

\(\displaystyle -b n \int -\frac {-2 e^3 x^6-12 d e^2 \log (x) x^4+6 d^2 e x^2+d^3}{4 x^5}dx-\frac {d^3 \left (a+b \log \left (c x^n\right )\right )}{4 x^4}-\frac {3 d^2 e \left (a+b \log \left (c x^n\right )\right )}{2 x^2}+3 d e^2 \log (x) \left (a+b \log \left (c x^n\right )\right )+\frac {1}{2} e^3 x^2 \left (a+b \log \left (c x^n\right )\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{4} b n \int \frac {-2 e^3 x^6-12 d e^2 \log (x) x^4+6 d^2 e x^2+d^3}{x^5}dx-\frac {d^3 \left (a+b \log \left (c x^n\right )\right )}{4 x^4}-\frac {3 d^2 e \left (a+b \log \left (c x^n\right )\right )}{2 x^2}+3 d e^2 \log (x) \left (a+b \log \left (c x^n\right )\right )+\frac {1}{2} e^3 x^2 \left (a+b \log \left (c x^n\right )\right )\)

\(\Big \downarrow \) 2010

\(\displaystyle \frac {1}{4} b n \int \left (\frac {-2 e^3 x^6+6 d^2 e x^2+d^3}{x^5}-\frac {12 d e^2 \log (x)}{x}\right )dx-\frac {d^3 \left (a+b \log \left (c x^n\right )\right )}{4 x^4}-\frac {3 d^2 e \left (a+b \log \left (c x^n\right )\right )}{2 x^2}+3 d e^2 \log (x) \left (a+b \log \left (c x^n\right )\right )+\frac {1}{2} e^3 x^2 \left (a+b \log \left (c x^n\right )\right )\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {d^3 \left (a+b \log \left (c x^n\right )\right )}{4 x^4}-\frac {3 d^2 e \left (a+b \log \left (c x^n\right )\right )}{2 x^2}+3 d e^2 \log (x) \left (a+b \log \left (c x^n\right )\right )+\frac {1}{2} e^3 x^2 \left (a+b \log \left (c x^n\right )\right )+\frac {1}{4} b n \left (-\frac {d^3}{4 x^4}-\frac {3 d^2 e}{x^2}-6 d e^2 \log ^2(x)-e^3 x^2\right )\)

Input:

Int[((d + e*x^2)^3*(a + b*Log[c*x^n]))/x^5,x]
 

Output:

(b*n*(-1/4*d^3/x^4 - (3*d^2*e)/x^2 - e^3*x^2 - 6*d*e^2*Log[x]^2))/4 - (d^3 
*(a + b*Log[c*x^n]))/(4*x^4) - (3*d^2*e*(a + b*Log[c*x^n]))/(2*x^2) + (e^3 
*x^2*(a + b*Log[c*x^n]))/2 + 3*d*e^2*Log[x]*(a + b*Log[c*x^n])
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 2009

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

rule 2010

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x] 
, x] /; FreeQ[{c, m}, x] && SumQ[u] &&  !LinearQ[u, x] &&  !MatchQ[u, (a_) 
+ (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]
 

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])
 

Maple [A] (verified)

Time = 1.17 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.18

method	result	size
parallelrisch	\(\frac {8 x^{6} \ln \left (c \,x^{n}\right ) b \,e^{3} n -4 x^{6} b \,e^{3} n^{2}+8 x^{6} a \,e^{3} n +48 \ln \left (x \right ) x^{4} a d \,e^{2} n +24 e^{2} d b \ln \left (c \,x^{n}\right )^{2} x^{4}-24 x^{2} \ln \left (c \,x^{n}\right ) b \,d^{2} e n -12 x^{2} b \,d^{2} e \,n^{2}-24 x^{2} a \,d^{2} e n -4 \ln \left (c \,x^{n}\right ) b \,d^{3} n -b \,d^{3} n^{2}-4 a \,d^{3} n}{16 x^{4} n}\)	\(149\)
risch	\(-\frac {b \left (-2 e^{3} x^{6}-12 e^{2} d \ln \left (x \right ) x^{4}+6 d^{2} e \,x^{2}+d^{3}\right ) \ln \left (x^{n}\right )}{4 x^{4}}-\frac {4 a \,d^{3}-48 \ln \left (x \right ) e^{2} d a \,x^{4}+2 i \pi b \,d^{3} \operatorname {csgn}\left (i c \,x^{n}\right )^{2} \operatorname {csgn}\left (i c \right )-8 x^{6} a \,e^{3}+24 a \,d^{2} e \,x^{2}+4 d^{3} b \ln \left (c \right )+24 \ln \left (c \right ) b \,d^{2} x^{2} e +24 i \ln \left (x \right ) \pi b d \,e^{2} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) \operatorname {csgn}\left (i c \right ) x^{4}-2 i \pi b \,d^{3} \operatorname {csgn}\left (i c \,x^{n}\right )^{3}+12 i \pi b \,d^{2} e \,x^{2} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}+12 i \pi b \,d^{2} e \,x^{2} \operatorname {csgn}\left (i c \,x^{n}\right )^{2} \operatorname {csgn}\left (i c \right )+b \,d^{3} n -2 i \pi b \,d^{3} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) \operatorname {csgn}\left (i c \right )+4 i \pi b \,e^{3} x^{6} \operatorname {csgn}\left (i c \,x^{n}\right )^{3}+4 i \pi b \,e^{3} x^{6} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) \operatorname {csgn}\left (i c \right )-48 \ln \left (x \right ) \ln \left (c \right ) b d \,e^{2} x^{4}+24 e^{2} d b n \ln \left (x \right )^{2} x^{4}-24 i \ln \left (x \right ) \pi b d \,e^{2} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2} x^{4}-24 i \ln \left (x \right ) \pi b d \,e^{2} \operatorname {csgn}\left (i c \,x^{n}\right )^{2} \operatorname {csgn}\left (i c \right ) x^{4}-12 i \pi b \,d^{2} e \,x^{2} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) \operatorname {csgn}\left (i c \right )+24 i \ln \left (x \right ) \pi b d \,e^{2} \operatorname {csgn}\left (i c \,x^{n}\right )^{3} x^{4}+2 i \pi b \,d^{3} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}+12 b \,d^{2} e n \,x^{2}-8 \ln \left (c \right ) b \,e^{3} x^{6}-12 i \pi b \,d^{2} e \,x^{2} \operatorname {csgn}\left (i c \,x^{n}\right )^{3}+4 b \,e^{3} n \,x^{6}-4 i \pi b \,e^{3} x^{6} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}-4 i \pi b \,e^{3} x^{6} \operatorname {csgn}\left (i c \,x^{n}\right )^{2} \operatorname {csgn}\left (i c \right )}{16 x^{4}}\)	\(602\)

Input:

int((e*x^2+d)^3*(a+b*ln(c*x^n))/x^5,x,method=_RETURNVERBOSE)
 

Output:

1/16/x^4*(8*x^6*ln(c*x^n)*b*e^3*n-4*x^6*b*e^3*n^2+8*x^6*a*e^3*n+48*ln(x)*x 
^4*a*d*e^2*n+24*e^2*d*b*ln(c*x^n)^2*x^4-24*x^2*ln(c*x^n)*b*d^2*e*n-12*x^2* 
b*d^2*e*n^2-24*x^2*a*d^2*e*n-4*ln(c*x^n)*b*d^3*n-b*d^3*n^2-4*a*d^3*n)/n
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.25 \[ \int \frac {\left (d+e x^2\right )^3 \left (a+b \log \left (c x^n\right )\right )}{x^5} \, dx=\frac {24 \, b d e^{2} n x^{4} \log \left (x\right )^{2} - 4 \, {\left (b e^{3} n - 2 \, a e^{3}\right )} x^{6} - b d^{3} n - 4 \, a d^{3} - 12 \, {\left (b d^{2} e n + 2 \, a d^{2} e\right )} x^{2} + 4 \, {\left (2 \, b e^{3} x^{6} - 6 \, b d^{2} e x^{2} - b d^{3}\right )} \log \left (c\right ) + 4 \, {\left (2 \, b e^{3} n x^{6} + 12 \, b d e^{2} x^{4} \log \left (c\right ) + 12 \, a d e^{2} x^{4} - 6 \, b d^{2} e n x^{2} - b d^{3} n\right )} \log \left (x\right )}{16 \, x^{4}} \] Input:

integrate((e*x^2+d)^3*(a+b*log(c*x^n))/x^5,x, algorithm="fricas")
 

Output:

1/16*(24*b*d*e^2*n*x^4*log(x)^2 - 4*(b*e^3*n - 2*a*e^3)*x^6 - b*d^3*n - 4* 
a*d^3 - 12*(b*d^2*e*n + 2*a*d^2*e)*x^2 + 4*(2*b*e^3*x^6 - 6*b*d^2*e*x^2 - 
b*d^3)*log(c) + 4*(2*b*e^3*n*x^6 + 12*b*d*e^2*x^4*log(c) + 12*a*d*e^2*x^4 
- 6*b*d^2*e*n*x^2 - b*d^3*n)*log(x))/x^4
 

Sympy [A] (verification not implemented)

Time = 1.12 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.66 \[ \int \frac {\left (d+e x^2\right )^3 \left (a+b \log \left (c x^n\right )\right )}{x^5} \, dx=\begin {cases} - \frac {a d^{3}}{4 x^{4}} - \frac {3 a d^{2} e}{2 x^{2}} + \frac {3 a d e^{2} \log {\left (c x^{n} \right )}}{n} + \frac {a e^{3} x^{2}}{2} - \frac {b d^{3} n}{16 x^{4}} - \frac {b d^{3} \log {\left (c x^{n} \right )}}{4 x^{4}} - \frac {3 b d^{2} e n}{4 x^{2}} - \frac {3 b d^{2} e \log {\left (c x^{n} \right )}}{2 x^{2}} + \frac {3 b d e^{2} \log {\left (c x^{n} \right )}^{2}}{2 n} - \frac {b e^{3} n x^{2}}{4} + \frac {b e^{3} x^{2} \log {\left (c x^{n} \right )}}{2} & \text {for}\: n \neq 0 \\\left (a + b \log {\left (c \right )}\right ) \left (- \frac {d^{3}}{4 x^{4}} - \frac {3 d^{2} e}{2 x^{2}} + 3 d e^{2} \log {\left (x \right )} + \frac {e^{3} x^{2}}{2}\right ) & \text {otherwise} \end {cases} \] Input:

integrate((e*x**2+d)**3*(a+b*ln(c*x**n))/x**5,x)
 

Output:

Piecewise((-a*d**3/(4*x**4) - 3*a*d**2*e/(2*x**2) + 3*a*d*e**2*log(c*x**n) 
/n + a*e**3*x**2/2 - b*d**3*n/(16*x**4) - b*d**3*log(c*x**n)/(4*x**4) - 3* 
b*d**2*e*n/(4*x**2) - 3*b*d**2*e*log(c*x**n)/(2*x**2) + 3*b*d*e**2*log(c*x 
**n)**2/(2*n) - b*e**3*n*x**2/4 + b*e**3*x**2*log(c*x**n)/2, Ne(n, 0)), (( 
a + b*log(c))*(-d**3/(4*x**4) - 3*d**2*e/(2*x**2) + 3*d*e**2*log(x) + e**3 
*x**2/2), True))
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.06 \[ \int \frac {\left (d+e x^2\right )^3 \left (a+b \log \left (c x^n\right )\right )}{x^5} \, dx=-\frac {1}{4} \, b e^{3} n x^{2} + \frac {1}{2} \, b e^{3} x^{2} \log \left (c x^{n}\right ) + \frac {1}{2} \, a e^{3} x^{2} + \frac {3 \, b d e^{2} \log \left (c x^{n}\right )^{2}}{2 \, n} + 3 \, a d e^{2} \log \left (x\right ) - \frac {3 \, b d^{2} e n}{4 \, x^{2}} - \frac {3 \, b d^{2} e \log \left (c x^{n}\right )}{2 \, x^{2}} - \frac {3 \, a d^{2} e}{2 \, x^{2}} - \frac {b d^{3} n}{16 \, x^{4}} - \frac {b d^{3} \log \left (c x^{n}\right )}{4 \, x^{4}} - \frac {a d^{3}}{4 \, x^{4}} \] Input:

integrate((e*x^2+d)^3*(a+b*log(c*x^n))/x^5,x, algorithm="maxima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.22 \[ \int \frac {\left (d+e x^2\right )^3 \left (a+b \log \left (c x^n\right )\right )}{x^5} \, dx=\frac {1}{2} \, b e^{3} x^{2} \log \left (c\right ) + \frac {3}{2} \, b d e^{2} n \log \left (x\right )^{2} + \frac {1}{4} \, {\left (2 \, x^{2} \log \left (x\right ) - x^{2}\right )} b e^{3} n + \frac {1}{2} \, a e^{3} x^{2} - \frac {3}{4} \, b d^{2} e n {\left (\frac {2 \, \log \left (x\right )}{x^{2}} + \frac {1}{x^{2}}\right )} - \frac {1}{16} \, b d^{3} n {\left (\frac {4 \, \log \left (x\right )}{x^{4}} + \frac {1}{x^{4}}\right )} + 3 \, b d e^{2} \log \left (c\right ) \log \left ({\left | x \right |}\right ) + 3 \, a d e^{2} \log \left ({\left | x \right |}\right ) - \frac {3 \, b d^{2} e \log \left (c\right )}{2 \, x^{2}} - \frac {3 \, a d^{2} e}{2 \, x^{2}} - \frac {b d^{3} \log \left (c\right )}{4 \, x^{4}} - \frac {a d^{3}}{4 \, x^{4}} \] Input:

integrate((e*x^2+d)^3*(a+b*log(c*x^n))/x^5,x, algorithm="giac")
 

Output:

1/2*b*e^3*x^2*log(c) + 3/2*b*d*e^2*n*log(x)^2 + 1/4*(2*x^2*log(x) - x^2)*b 
*e^3*n + 1/2*a*e^3*x^2 - 3/4*b*d^2*e*n*(2*log(x)/x^2 + 1/x^2) - 1/16*b*d^3 
*n*(4*log(x)/x^4 + 1/x^4) + 3*b*d*e^2*log(c)*log(abs(x)) + 3*a*d*e^2*log(a 
bs(x)) - 3/2*b*d^2*e*log(c)/x^2 - 3/2*a*d^2*e/x^2 - 1/4*b*d^3*log(c)/x^4 - 
 1/4*a*d^3/x^4
 

Mupad [B] (verification not implemented)

Time = 26.42 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.18 \[ \int \frac {\left (d+e x^2\right )^3 \left (a+b \log \left (c x^n\right )\right )}{x^5} \, dx=\ln \left (x\right )\,\left (3\,a\,d\,e^2+\frac {9\,b\,d\,e^2\,n}{4}\right )-\ln \left (c\,x^n\right )\,\left (\frac {\frac {b\,d^3}{4}+\frac {3\,b\,d^2\,e\,x^2}{2}+\frac {9\,b\,d\,e^2\,x^4}{4}+b\,e^3\,x^6}{x^4}-\frac {3\,b\,e^3\,x^2}{2}\right )-\frac {a\,d^3+x^2\,\left (6\,a\,d^2\,e+3\,b\,d^2\,e\,n\right )+\frac {b\,d^3\,n}{4}}{4\,x^4}+\frac {e^3\,x^2\,\left (2\,a-b\,n\right )}{4}+\frac {3\,b\,d\,e^2\,{\ln \left (c\,x^n\right )}^2}{2\,n} \] Input:

int(((d + e*x^2)^3*(a + b*log(c*x^n)))/x^5,x)
 

Output:

log(x)*(3*a*d*e^2 + (9*b*d*e^2*n)/4) - log(c*x^n)*(((b*d^3)/4 + b*e^3*x^6 
+ (3*b*d^2*e*x^2)/2 + (9*b*d*e^2*x^4)/4)/x^4 - (3*b*e^3*x^2)/2) - (a*d^3 + 
 x^2*(6*a*d^2*e + 3*b*d^2*e*n) + (b*d^3*n)/4)/(4*x^4) + (e^3*x^2*(2*a - b* 
n))/4 + (3*b*d*e^2*log(c*x^n)^2)/(2*n)
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.17 \[ \int \frac {\left (d+e x^2\right )^3 \left (a+b \log \left (c x^n\right )\right )}{x^5} \, dx=\frac {24 \mathrm {log}\left (x^{n} c \right )^{2} b d \,e^{2} x^{4}-4 \,\mathrm {log}\left (x^{n} c \right ) b \,d^{3} n -24 \,\mathrm {log}\left (x^{n} c \right ) b \,d^{2} e n \,x^{2}+8 \,\mathrm {log}\left (x^{n} c \right ) b \,e^{3} n \,x^{6}+48 \,\mathrm {log}\left (x \right ) a d \,e^{2} n \,x^{4}-4 a \,d^{3} n -24 a \,d^{2} e n \,x^{2}+8 a \,e^{3} n \,x^{6}-b \,d^{3} n^{2}-12 b \,d^{2} e \,n^{2} x^{2}-4 b \,e^{3} n^{2} x^{6}}{16 n \,x^{4}} \] Input:

int((e*x^2+d)^3*(a+b*log(c*x^n))/x^5,x)
 

Output:

(24*log(x**n*c)**2*b*d*e**2*x**4 - 4*log(x**n*c)*b*d**3*n - 24*log(x**n*c) 
*b*d**2*e*n*x**2 + 8*log(x**n*c)*b*e**3*n*x**6 + 48*log(x)*a*d*e**2*n*x**4 
 - 4*a*d**3*n - 24*a*d**2*e*n*x**2 + 8*a*e**3*n*x**6 - b*d**3*n**2 - 12*b* 
d**2*e*n**2*x**2 - 4*b*e**3*n**2*x**6)/(16*n*x**4)