Integrand size = 23, antiderivative size = 74 \[ \int x^5 \left (d+e x^2\right )^2 \left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {1}{36} b d^2 n x^6-\frac {1}{32} b d e n x^8-\frac {1}{100} b e^2 n x^{10}+\frac {1}{60} \left (10 d^2 x^6+15 d e x^8+6 e^2 x^{10}\right ) \left (a+b \log \left (c x^n\right )\right ) \]
-1/36*b*d^2*n*x^6-1/32*b*d*e*n*x^8-1/100*b*e^2*n*x^10+1/60*(6*e^2*x^10+15* d*e*x^8+10*d^2*x^6)*(a+b*ln(c*x^n))
Time = 0.03 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.14 \[ \int x^5 \left (d+e x^2\right )^2 \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {x^6 \left (-200 b d^2 n-225 b d e n x^2-72 b e^2 n x^4+1200 d^2 \left (a+b \log \left (c x^n\right )\right )+1800 d e x^2 \left (a+b \log \left (c x^n\right )\right )+720 e^2 x^4 \left (a+b \log \left (c x^n\right )\right )\right )}{7200} \]
(x^6*(-200*b*d^2*n - 225*b*d*e*n*x^2 - 72*b*e^2*n*x^4 + 1200*d^2*(a + b*Lo g[c*x^n]) + 1800*d*e*x^2*(a + b*Log[c*x^n]) + 720*e^2*x^4*(a + b*Log[c*x^n ])))/7200
Time = 0.28 (sec) , antiderivative size = 75, 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 = {2771, 27, 1433, 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.
\(\displaystyle \int x^5 \left (d+e x^2\right )^2 \left (a+b \log \left (c x^n\right )\right ) \, dx\) |
\(\Big \downarrow \) 2771 |
\(\displaystyle \frac {1}{60} \left (10 d^2 x^6+15 d e x^8+6 e^2 x^{10}\right ) \left (a+b \log \left (c x^n\right )\right )-b n \int \frac {1}{60} x^5 \left (6 e^2 x^4+15 d e x^2+10 d^2\right )dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{60} \left (10 d^2 x^6+15 d e x^8+6 e^2 x^{10}\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{60} b n \int x^5 \left (6 e^2 x^4+15 d e x^2+10 d^2\right )dx\) |
\(\Big \downarrow \) 1433 |
\(\displaystyle \frac {1}{60} \left (10 d^2 x^6+15 d e x^8+6 e^2 x^{10}\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{60} b n \int \left (6 e^2 x^9+15 d e x^7+10 d^2 x^5\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{60} \left (10 d^2 x^6+15 d e x^8+6 e^2 x^{10}\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{60} b n \left (\frac {5 d^2 x^6}{3}+\frac {15}{8} d e x^8+\frac {3 e^2 x^{10}}{5}\right )\) |
-1/60*(b*n*((5*d^2*x^6)/3 + (15*d*e*x^8)/8 + (3*e^2*x^10)/5)) + ((10*d^2*x ^6 + 15*d*e*x^8 + 6*e^2*x^10)*(a + b*Log[c*x^n]))/60
3.2.83.3.1 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[((d_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d*x)^m*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[p, 0] && (EqQ[p, 1] || !IntegerQ[(m + 1)/2])
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[u*(a + b*Log[c*x^n]), x] - Simp[b*n Int[SimplifyIntegrand[u/x, x], x], x]] /; FreeQ[{a, b, c, d, e, n, r}, x] && IGtQ[q, 0] && IGtQ[m, 0]
Time = 1.37 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.36
method | result | size |
parallelrisch | \(\frac {x^{10} \ln \left (c \,x^{n}\right ) b \,e^{2}}{10}-\frac {b \,e^{2} n \,x^{10}}{100}+\frac {a \,e^{2} x^{10}}{10}+\frac {x^{8} \ln \left (c \,x^{n}\right ) b d e}{4}-\frac {b d e n \,x^{8}}{32}+\frac {a d e \,x^{8}}{4}+\frac {x^{6} \ln \left (c \,x^{n}\right ) b \,d^{2}}{6}-\frac {b \,d^{2} n \,x^{6}}{36}+\frac {a \,d^{2} x^{6}}{6}\) | \(101\) |
risch | \(\frac {b \,x^{6} \left (6 e^{2} x^{4}+15 d e \,x^{2}+10 d^{2}\right ) \ln \left (x^{n}\right )}{60}-\frac {i \pi b d e \,x^{8} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )}{8}+\frac {i \pi b d e \,x^{8} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{8}+\frac {i \pi b \,d^{2} x^{6} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{12}-\frac {i \pi b \,d^{2} x^{6} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )}{12}+\frac {\ln \left (c \right ) b \,e^{2} x^{10}}{10}-\frac {b \,e^{2} n \,x^{10}}{100}+\frac {a \,e^{2} x^{10}}{10}-\frac {i \pi b d e \,x^{8} \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{8}+\frac {i \pi b \,e^{2} x^{10} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{20}+\frac {i \pi b \,e^{2} x^{10} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{20}+\frac {i \pi b d e \,x^{8} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{8}+\frac {\ln \left (c \right ) b d e \,x^{8}}{4}-\frac {b d e n \,x^{8}}{32}+\frac {a d e \,x^{8}}{4}+\frac {i \pi b \,d^{2} x^{6} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{12}-\frac {i \pi b \,e^{2} x^{10} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )}{20}-\frac {i \pi b \,d^{2} x^{6} \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{12}-\frac {i \pi b \,e^{2} x^{10} \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{20}+\frac {\ln \left (c \right ) b \,d^{2} x^{6}}{6}-\frac {b \,d^{2} n \,x^{6}}{36}+\frac {a \,d^{2} x^{6}}{6}\) | \(434\) |
1/10*x^10*ln(c*x^n)*b*e^2-1/100*b*e^2*n*x^10+1/10*a*e^2*x^10+1/4*x^8*ln(c* x^n)*b*d*e-1/32*b*d*e*n*x^8+1/4*a*d*e*x^8+1/6*x^6*ln(c*x^n)*b*d^2-1/36*b*d ^2*n*x^6+1/6*a*d^2*x^6
Time = 0.29 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.59 \[ \int x^5 \left (d+e x^2\right )^2 \left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {1}{100} \, {\left (b e^{2} n - 10 \, a e^{2}\right )} x^{10} - \frac {1}{32} \, {\left (b d e n - 8 \, a d e\right )} x^{8} - \frac {1}{36} \, {\left (b d^{2} n - 6 \, a d^{2}\right )} x^{6} + \frac {1}{60} \, {\left (6 \, b e^{2} x^{10} + 15 \, b d e x^{8} + 10 \, b d^{2} x^{6}\right )} \log \left (c\right ) + \frac {1}{60} \, {\left (6 \, b e^{2} n x^{10} + 15 \, b d e n x^{8} + 10 \, b d^{2} n x^{6}\right )} \log \left (x\right ) \]
-1/100*(b*e^2*n - 10*a*e^2)*x^10 - 1/32*(b*d*e*n - 8*a*d*e)*x^8 - 1/36*(b* d^2*n - 6*a*d^2)*x^6 + 1/60*(6*b*e^2*x^10 + 15*b*d*e*x^8 + 10*b*d^2*x^6)*l og(c) + 1/60*(6*b*e^2*n*x^10 + 15*b*d*e*n*x^8 + 10*b*d^2*n*x^6)*log(x)
Time = 1.69 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.57 \[ \int x^5 \left (d+e x^2\right )^2 \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {a d^{2} x^{6}}{6} + \frac {a d e x^{8}}{4} + \frac {a e^{2} x^{10}}{10} - \frac {b d^{2} n x^{6}}{36} + \frac {b d^{2} x^{6} \log {\left (c x^{n} \right )}}{6} - \frac {b d e n x^{8}}{32} + \frac {b d e x^{8} \log {\left (c x^{n} \right )}}{4} - \frac {b e^{2} n x^{10}}{100} + \frac {b e^{2} x^{10} \log {\left (c x^{n} \right )}}{10} \]
a*d**2*x**6/6 + a*d*e*x**8/4 + a*e**2*x**10/10 - b*d**2*n*x**6/36 + b*d**2 *x**6*log(c*x**n)/6 - b*d*e*n*x**8/32 + b*d*e*x**8*log(c*x**n)/4 - b*e**2* n*x**10/100 + b*e**2*x**10*log(c*x**n)/10
Time = 0.20 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.35 \[ \int x^5 \left (d+e x^2\right )^2 \left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {1}{100} \, b e^{2} n x^{10} + \frac {1}{10} \, b e^{2} x^{10} \log \left (c x^{n}\right ) + \frac {1}{10} \, a e^{2} x^{10} - \frac {1}{32} \, b d e n x^{8} + \frac {1}{4} \, b d e x^{8} \log \left (c x^{n}\right ) + \frac {1}{4} \, a d e x^{8} - \frac {1}{36} \, b d^{2} n x^{6} + \frac {1}{6} \, b d^{2} x^{6} \log \left (c x^{n}\right ) + \frac {1}{6} \, a d^{2} x^{6} \]
-1/100*b*e^2*n*x^10 + 1/10*b*e^2*x^10*log(c*x^n) + 1/10*a*e^2*x^10 - 1/32* b*d*e*n*x^8 + 1/4*b*d*e*x^8*log(c*x^n) + 1/4*a*d*e*x^8 - 1/36*b*d^2*n*x^6 + 1/6*b*d^2*x^6*log(c*x^n) + 1/6*a*d^2*x^6
Time = 0.37 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.66 \[ \int x^5 \left (d+e x^2\right )^2 \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {1}{10} \, b e^{2} n x^{10} \log \left (x\right ) - \frac {1}{100} \, b e^{2} n x^{10} + \frac {1}{10} \, b e^{2} x^{10} \log \left (c\right ) + \frac {1}{10} \, a e^{2} x^{10} + \frac {1}{4} \, b d e n x^{8} \log \left (x\right ) - \frac {1}{32} \, b d e n x^{8} + \frac {1}{4} \, b d e x^{8} \log \left (c\right ) + \frac {1}{4} \, a d e x^{8} + \frac {1}{6} \, b d^{2} n x^{6} \log \left (x\right ) - \frac {1}{36} \, b d^{2} n x^{6} + \frac {1}{6} \, b d^{2} x^{6} \log \left (c\right ) + \frac {1}{6} \, a d^{2} x^{6} \]
1/10*b*e^2*n*x^10*log(x) - 1/100*b*e^2*n*x^10 + 1/10*b*e^2*x^10*log(c) + 1 /10*a*e^2*x^10 + 1/4*b*d*e*n*x^8*log(x) - 1/32*b*d*e*n*x^8 + 1/4*b*d*e*x^8 *log(c) + 1/4*a*d*e*x^8 + 1/6*b*d^2*n*x^6*log(x) - 1/36*b*d^2*n*x^6 + 1/6* b*d^2*x^6*log(c) + 1/6*a*d^2*x^6
Time = 0.39 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.11 \[ \int x^5 \left (d+e x^2\right )^2 \left (a+b \log \left (c x^n\right )\right ) \, dx=\ln \left (c\,x^n\right )\,\left (\frac {b\,d^2\,x^6}{6}+\frac {b\,d\,e\,x^8}{4}+\frac {b\,e^2\,x^{10}}{10}\right )+\frac {d^2\,x^6\,\left (6\,a-b\,n\right )}{36}+\frac {e^2\,x^{10}\,\left (10\,a-b\,n\right )}{100}+\frac {d\,e\,x^8\,\left (8\,a-b\,n\right )}{32} \]