Integrand size = 23, antiderivative size = 91 \[ \int \frac {\left (d+e x^2\right )^2 \left (a+b \log \left (c x^n\right )\right )}{x^3} \, dx=-\frac {b d^2 n}{4 x^2}-\frac {1}{4} b e^2 n x^2-b d e n \log ^2(x)-\frac {d^2 \left (a+b \log \left (c x^n\right )\right )}{2 x^2}+\frac {1}{2} e^2 x^2 \left (a+b \log \left (c x^n\right )\right )+2 d e \log (x) \left (a+b \log \left (c x^n\right )\right ) \]
-1/4*b*d^2*n/x^2-1/4*b*e^2*n*x^2-b*d*e*n*ln(x)^2-1/2*d^2*(a+b*ln(c*x^n))/x ^2+1/2*e^2*x^2*(a+b*ln(c*x^n))+2*d*e*ln(x)*(a+b*ln(c*x^n))
Time = 0.04 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.91 \[ \int \frac {\left (d+e x^2\right )^2 \left (a+b \log \left (c x^n\right )\right )}{x^3} \, dx=\frac {1}{4} \left (-\frac {b d^2 n}{x^2}-b e^2 n x^2-\frac {2 d^2 \left (a+b \log \left (c x^n\right )\right )}{x^2}+2 e^2 x^2 \left (a+b \log \left (c x^n\right )\right )+\frac {4 d e \left (a+b \log \left (c x^n\right )\right )^2}{b n}\right ) \]
(-((b*d^2*n)/x^2) - b*e^2*n*x^2 - (2*d^2*(a + b*Log[c*x^n]))/x^2 + 2*e^2*x ^2*(a + b*Log[c*x^n]) + (4*d*e*(a + b*Log[c*x^n])^2)/(b*n))/4
Time = 0.32 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.01, 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.
\(\displaystyle \int \frac {\left (d+e x^2\right )^2 \left (a+b \log \left (c x^n\right )\right )}{x^3} \, dx\) |
\(\Big \downarrow \) 2772 |
\(\displaystyle -b n \int -\frac {-e^2 x^4-4 d e \log (x) x^2+d^2}{2 x^3}dx-\frac {d^2 \left (a+b \log \left (c x^n\right )\right )}{2 x^2}+2 d e \log (x) \left (a+b \log \left (c x^n\right )\right )+\frac {1}{2} e^2 x^2 \left (a+b \log \left (c x^n\right )\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} b n \int \frac {-e^2 x^4-4 d e \log (x) x^2+d^2}{x^3}dx-\frac {d^2 \left (a+b \log \left (c x^n\right )\right )}{2 x^2}+2 d e \log (x) \left (a+b \log \left (c x^n\right )\right )+\frac {1}{2} e^2 x^2 \left (a+b \log \left (c x^n\right )\right )\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle \frac {1}{2} b n \int \left (\frac {d^2-e^2 x^4}{x^3}-\frac {4 d e \log (x)}{x}\right )dx-\frac {d^2 \left (a+b \log \left (c x^n\right )\right )}{2 x^2}+2 d e \log (x) \left (a+b \log \left (c x^n\right )\right )+\frac {1}{2} e^2 x^2 \left (a+b \log \left (c x^n\right )\right )\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {d^2 \left (a+b \log \left (c x^n\right )\right )}{2 x^2}+2 d e \log (x) \left (a+b \log \left (c x^n\right )\right )+\frac {1}{2} e^2 x^2 \left (a+b \log \left (c x^n\right )\right )+\frac {1}{2} b n \left (-\frac {d^2}{2 x^2}-2 d e \log ^2(x)-\frac {e^2 x^2}{2}\right )\) |
(b*n*(-1/2*d^2/x^2 - (e^2*x^2)/2 - 2*d*e*Log[x]^2))/2 - (d^2*(a + b*Log[c* x^n]))/(2*x^2) + (e^2*x^2*(a + b*Log[c*x^n]))/2 + 2*d*e*Log[x]*(a + b*Log[ c*x^n])
3.2.87.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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]]
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])
Time = 0.51 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.14
method | result | size |
parallelrisch | \(\frac {2 x^{4} \ln \left (c \,x^{n}\right ) b \,e^{2} n -x^{4} b \,e^{2} n^{2}+2 x^{4} a \,e^{2} n +8 \ln \left (x \right ) x^{2} a d e n +4 b d e \ln \left (c \,x^{n}\right )^{2} x^{2}-2 \ln \left (c \,x^{n}\right ) b \,d^{2} n -b \,d^{2} n^{2}-2 a \,d^{2} n}{4 x^{2} n}\) | \(104\) |
risch | \(-\frac {b \left (-e^{2} x^{4}-4 d e \ln \left (x \right ) x^{2}+d^{2}\right ) \ln \left (x^{n}\right )}{2 x^{2}}-\frac {-4 i \ln \left (x \right ) \pi b d e \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2} x^{2}+i \pi b \,d^{2} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}-i \pi b \,e^{2} x^{4} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}+4 i \ln \left (x \right ) \pi b d e \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) x^{2}+i \pi b \,e^{2} x^{4} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )-i \pi b \,d^{2} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )+i \pi b \,e^{2} x^{4} \operatorname {csgn}\left (i c \,x^{n}\right )^{3}+i \pi b \,d^{2} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}-i \pi b \,e^{2} x^{4} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}-i \pi b \,d^{2} \operatorname {csgn}\left (i c \,x^{n}\right )^{3}-4 i \ln \left (x \right ) \pi b d e \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2} x^{2}+4 i \ln \left (x \right ) \pi b d e \operatorname {csgn}\left (i c \,x^{n}\right )^{3} x^{2}-2 \ln \left (c \right ) b \,e^{2} x^{4}+4 b d e n \ln \left (x \right )^{2} x^{2}+b \,e^{2} n \,x^{4}-8 \ln \left (x \right ) \ln \left (c \right ) b d e \,x^{2}-2 x^{4} a \,e^{2}-8 \ln \left (x \right ) a d e \,x^{2}+2 d^{2} b \ln \left (c \right )+b \,d^{2} n +2 a \,d^{2}}{4 x^{2}}\) | \(433\) |
1/4/x^2*(2*x^4*ln(c*x^n)*b*e^2*n-x^4*b*e^2*n^2+2*x^4*a*e^2*n+8*ln(x)*x^2*a *d*e*n+4*b*d*e*ln(c*x^n)^2*x^2-2*ln(c*x^n)*b*d^2*n-b*d^2*n^2-2*a*d^2*n)/n
Time = 0.30 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.19 \[ \int \frac {\left (d+e x^2\right )^2 \left (a+b \log \left (c x^n\right )\right )}{x^3} \, dx=\frac {4 \, b d e n x^{2} \log \left (x\right )^{2} - {\left (b e^{2} n - 2 \, a e^{2}\right )} x^{4} - b d^{2} n - 2 \, a d^{2} + 2 \, {\left (b e^{2} x^{4} - b d^{2}\right )} \log \left (c\right ) + 2 \, {\left (b e^{2} n x^{4} + 4 \, b d e x^{2} \log \left (c\right ) + 4 \, a d e x^{2} - b d^{2} n\right )} \log \left (x\right )}{4 \, x^{2}} \]
1/4*(4*b*d*e*n*x^2*log(x)^2 - (b*e^2*n - 2*a*e^2)*x^4 - b*d^2*n - 2*a*d^2 + 2*(b*e^2*x^4 - b*d^2)*log(c) + 2*(b*e^2*n*x^4 + 4*b*d*e*x^2*log(c) + 4*a *d*e*x^2 - b*d^2*n)*log(x))/x^2
Time = 0.55 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.53 \[ \int \frac {\left (d+e x^2\right )^2 \left (a+b \log \left (c x^n\right )\right )}{x^3} \, dx=\begin {cases} - \frac {a d^{2}}{2 x^{2}} + \frac {2 a d e \log {\left (c x^{n} \right )}}{n} + \frac {a e^{2} x^{2}}{2} - \frac {b d^{2} n}{4 x^{2}} - \frac {b d^{2} \log {\left (c x^{n} \right )}}{2 x^{2}} + \frac {b d e \log {\left (c x^{n} \right )}^{2}}{n} - \frac {b e^{2} n x^{2}}{4} + \frac {b e^{2} x^{2} \log {\left (c x^{n} \right )}}{2} & \text {for}\: n \neq 0 \\\left (a + b \log {\left (c \right )}\right ) \left (- \frac {d^{2}}{2 x^{2}} + 2 d e \log {\left (x \right )} + \frac {e^{2} x^{2}}{2}\right ) & \text {otherwise} \end {cases} \]
Piecewise((-a*d**2/(2*x**2) + 2*a*d*e*log(c*x**n)/n + a*e**2*x**2/2 - b*d* *2*n/(4*x**2) - b*d**2*log(c*x**n)/(2*x**2) + b*d*e*log(c*x**n)**2/n - b*e **2*n*x**2/4 + b*e**2*x**2*log(c*x**n)/2, Ne(n, 0)), ((a + b*log(c))*(-d** 2/(2*x**2) + 2*d*e*log(x) + e**2*x**2/2), True))
Time = 0.20 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.00 \[ \int \frac {\left (d+e x^2\right )^2 \left (a+b \log \left (c x^n\right )\right )}{x^3} \, dx=-\frac {1}{4} \, b e^{2} n x^{2} + \frac {1}{2} \, b e^{2} x^{2} \log \left (c x^{n}\right ) + \frac {1}{2} \, a e^{2} x^{2} + \frac {b d e \log \left (c x^{n}\right )^{2}}{n} + 2 \, a d e \log \left (x\right ) - \frac {b d^{2} n}{4 \, x^{2}} - \frac {b d^{2} \log \left (c x^{n}\right )}{2 \, x^{2}} - \frac {a d^{2}}{2 \, x^{2}} \]
-1/4*b*e^2*n*x^2 + 1/2*b*e^2*x^2*log(c*x^n) + 1/2*a*e^2*x^2 + b*d*e*log(c* x^n)^2/n + 2*a*d*e*log(x) - 1/4*b*d^2*n/x^2 - 1/2*b*d^2*log(c*x^n)/x^2 - 1 /2*a*d^2/x^2
Time = 0.30 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.16 \[ \int \frac {\left (d+e x^2\right )^2 \left (a+b \log \left (c x^n\right )\right )}{x^3} \, dx=\frac {1}{2} \, b e^{2} x^{2} \log \left (c\right ) + b d e n \log \left (x\right )^{2} + \frac {1}{4} \, {\left (2 \, x^{2} \log \left (x\right ) - x^{2}\right )} b e^{2} n + \frac {1}{2} \, a e^{2} x^{2} - \frac {1}{4} \, b d^{2} n {\left (\frac {2 \, \log \left (x\right )}{x^{2}} + \frac {1}{x^{2}}\right )} + 2 \, b d e \log \left (c\right ) \log \left ({\left | x \right |}\right ) + 2 \, a d e \log \left ({\left | x \right |}\right ) - \frac {b d^{2} \log \left (c\right )}{2 \, x^{2}} - \frac {a d^{2}}{2 \, x^{2}} \]
1/2*b*e^2*x^2*log(c) + b*d*e*n*log(x)^2 + 1/4*(2*x^2*log(x) - x^2)*b*e^2*n + 1/2*a*e^2*x^2 - 1/4*b*d^2*n*(2*log(x)/x^2 + 1/x^2) + 2*b*d*e*log(c)*log (abs(x)) + 2*a*d*e*log(abs(x)) - 1/2*b*d^2*log(c)/x^2 - 1/2*a*d^2/x^2
Time = 0.37 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.21 \[ \int \frac {\left (d+e x^2\right )^2 \left (a+b \log \left (c x^n\right )\right )}{x^3} \, dx=\ln \left (x\right )\,\left (2\,a\,d\,e+b\,d\,e\,n\right )-\frac {\frac {a\,d^2}{2}+\frac {b\,d^2\,n}{4}}{x^2}-\ln \left (c\,x^n\right )\,\left (\frac {\frac {b\,d^2}{2}+b\,d\,e\,x^2+\frac {b\,e^2\,x^4}{2}}{x^2}-b\,e^2\,x^2\right )+\frac {e^2\,x^2\,\left (2\,a-b\,n\right )}{4}+\frac {b\,d\,e\,{\ln \left (c\,x^n\right )}^2}{n} \]