Integrand size = 21, antiderivative size = 91 \[ \int x \left (d+e x^2\right )^3 \left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {1}{4} b d^3 n x^2-\frac {3}{16} b d^2 e n x^4-\frac {1}{12} b d e^2 n x^6-\frac {1}{64} b e^3 n x^8-\frac {b d^4 n \log (x)}{8 e}+\frac {\left (d+e x^2\right )^4 \left (a+b \log \left (c x^n\right )\right )}{8 e} \]
-1/4*b*d^3*n*x^2-3/16*b*d^2*e*n*x^4-1/12*b*d*e^2*n*x^6-1/64*b*e^3*n*x^8-1/ 8*b*d^4*n*ln(x)/e+1/8*(e*x^2+d)^4*(a+b*ln(c*x^n))/e
Time = 0.04 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.30 \[ \int x \left (d+e x^2\right )^3 \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {1}{192} x^2 \left (24 a \left (4 d^3+6 d^2 e x^2+4 d e^2 x^4+e^3 x^6\right )-b n \left (48 d^3+36 d^2 e x^2+16 d e^2 x^4+3 e^3 x^6\right )+24 b \left (4 d^3+6 d^2 e x^2+4 d e^2 x^4+e^3 x^6\right ) \log \left (c x^n\right )\right ) \]
(x^2*(24*a*(4*d^3 + 6*d^2*e*x^2 + 4*d*e^2*x^4 + e^3*x^6) - b*n*(48*d^3 + 3 6*d^2*e*x^2 + 16*d*e^2*x^4 + 3*e^3*x^6) + 24*b*(4*d^3 + 6*d^2*e*x^2 + 4*d* e^2*x^4 + e^3*x^6)*Log[c*x^n]))/192
Time = 0.26 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.95, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {2771, 27, 243, 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.
\(\displaystyle \int x \left (d+e x^2\right )^3 \left (a+b \log \left (c x^n\right )\right ) \, dx\) |
\(\Big \downarrow \) 2771 |
\(\displaystyle \frac {\left (d+e x^2\right )^4 \left (a+b \log \left (c x^n\right )\right )}{8 e}-b n \int \frac {\left (e x^2+d\right )^4}{8 e x}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\left (d+e x^2\right )^4 \left (a+b \log \left (c x^n\right )\right )}{8 e}-\frac {b n \int \frac {\left (e x^2+d\right )^4}{x}dx}{8 e}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {\left (d+e x^2\right )^4 \left (a+b \log \left (c x^n\right )\right )}{8 e}-\frac {b n \int \frac {\left (e x^2+d\right )^4}{x^2}dx^2}{16 e}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {\left (d+e x^2\right )^4 \left (a+b \log \left (c x^n\right )\right )}{8 e}-\frac {b n \int \left (e^4 x^6+4 d e^3 x^4+6 d^2 e^2 x^2+4 d^3 e+\frac {d^4}{x^2}\right )dx^2}{16 e}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\left (d+e x^2\right )^4 \left (a+b \log \left (c x^n\right )\right )}{8 e}-\frac {b n \left (d^4 \log \left (x^2\right )+4 d^3 e x^2+3 d^2 e^2 x^4+\frac {4}{3} d e^3 x^6+\frac {e^4 x^8}{4}\right )}{16 e}\) |
-1/16*(b*n*(4*d^3*e*x^2 + 3*d^2*e^2*x^4 + (4*d*e^3*x^6)/3 + (e^4*x^8)/4 + d^4*Log[x^2]))/e + ((d + e*x^2)^4*(a + b*Log[c*x^n]))/(8*e)
3.2.98.3.1 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[((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[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(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 = 0.73 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.58
method | result | size |
parallelrisch | \(\frac {x^{8} \ln \left (c \,x^{n}\right ) b \,e^{3}}{8}-\frac {b \,e^{3} n \,x^{8}}{64}+\frac {a \,e^{3} x^{8}}{8}+\frac {x^{6} \ln \left (c \,x^{n}\right ) b d \,e^{2}}{2}-\frac {b d \,e^{2} n \,x^{6}}{12}+\frac {a d \,e^{2} x^{6}}{2}+\frac {3 x^{4} b \ln \left (c \,x^{n}\right ) d^{2} e}{4}-\frac {3 b \,d^{2} e n \,x^{4}}{16}+\frac {3 x^{4} a \,d^{2} e}{4}+\frac {x^{2} b \ln \left (c \,x^{n}\right ) d^{3}}{2}-\frac {b \,d^{3} n \,x^{2}}{4}+\frac {a \,d^{3} x^{2}}{2}\) | \(144\) |
risch | \(\frac {\left (e \,x^{2}+d \right )^{4} b \ln \left (x^{n}\right )}{8 e}+\frac {a \,e^{3} x^{8}}{8}-\frac {i e^{3} \pi b \,x^{8} \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{16}+\frac {a d \,e^{2} x^{6}}{2}-\frac {i \pi b \,d^{3} x^{2} \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{4}-\frac {i \pi b d \,e^{2} x^{6} \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{4}+\frac {e^{3} \ln \left (c \right ) b \,x^{8}}{8}-\frac {3 i \pi b \,d^{2} e \,x^{4} \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{8}+\frac {i e^{3} \pi b \,x^{8} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{16}+\frac {i e^{3} \pi b \,x^{8} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{16}+\frac {3 x^{4} a \,d^{2} e}{4}-\frac {i e^{3} \pi b \,x^{8} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )}{16}+\frac {\ln \left (c \right ) b \,d^{3} x^{2}}{2}+\frac {i \pi b \,d^{3} x^{2} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{4}+\frac {i \pi b \,d^{3} x^{2} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{4}+\frac {a \,d^{3} x^{2}}{2}+\frac {3 i \pi b \,d^{2} e \,x^{4} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{8}+\frac {3 i \pi b \,d^{2} e \,x^{4} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{8}-\frac {3 i \pi b \,d^{2} e \,x^{4} \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^{3} x^{2} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )}{4}-\frac {i \pi b d \,e^{2} x^{6} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )}{4}-\frac {b d \,e^{2} n \,x^{6}}{12}-\frac {3 b \,d^{2} e n \,x^{4}}{16}-\frac {b \,d^{4} n \ln \left (x \right )}{8 e}-\frac {b \,d^{3} n \,x^{2}}{4}+\frac {3 \ln \left (c \right ) b \,d^{2} e \,x^{4}}{4}+\frac {\ln \left (c \right ) b d \,e^{2} x^{6}}{2}-\frac {b \,e^{3} n \,x^{8}}{64}+\frac {i \pi b d \,e^{2} x^{6} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{4}+\frac {i \pi b d \,e^{2} x^{6} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{4}\) | \(591\) |
1/8*x^8*ln(c*x^n)*b*e^3-1/64*b*e^3*n*x^8+1/8*a*e^3*x^8+1/2*x^6*ln(c*x^n)*b *d*e^2-1/12*b*d*e^2*n*x^6+1/2*a*d*e^2*x^6+3/4*x^4*b*ln(c*x^n)*d^2*e-3/16*b *d^2*e*n*x^4+3/4*x^4*a*d^2*e+1/2*x^2*b*ln(c*x^n)*d^3-1/4*b*d^3*n*x^2+1/2*a *d^3*x^2
Leaf count of result is larger than twice the leaf count of optimal. 165 vs. \(2 (79) = 158\).
Time = 0.32 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.81 \[ \int x \left (d+e x^2\right )^3 \left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {1}{64} \, {\left (b e^{3} n - 8 \, a e^{3}\right )} x^{8} - \frac {1}{12} \, {\left (b d e^{2} n - 6 \, a d e^{2}\right )} x^{6} - \frac {3}{16} \, {\left (b d^{2} e n - 4 \, a d^{2} e\right )} x^{4} - \frac {1}{4} \, {\left (b d^{3} n - 2 \, a d^{3}\right )} x^{2} + \frac {1}{8} \, {\left (b e^{3} x^{8} + 4 \, b d e^{2} x^{6} + 6 \, b d^{2} e x^{4} + 4 \, b d^{3} x^{2}\right )} \log \left (c\right ) + \frac {1}{8} \, {\left (b e^{3} n x^{8} + 4 \, b d e^{2} n x^{6} + 6 \, b d^{2} e n x^{4} + 4 \, b d^{3} n x^{2}\right )} \log \left (x\right ) \]
-1/64*(b*e^3*n - 8*a*e^3)*x^8 - 1/12*(b*d*e^2*n - 6*a*d*e^2)*x^6 - 3/16*(b *d^2*e*n - 4*a*d^2*e)*x^4 - 1/4*(b*d^3*n - 2*a*d^3)*x^2 + 1/8*(b*e^3*x^8 + 4*b*d*e^2*x^6 + 6*b*d^2*e*x^4 + 4*b*d^3*x^2)*log(c) + 1/8*(b*e^3*n*x^8 + 4*b*d*e^2*n*x^6 + 6*b*d^2*e*n*x^4 + 4*b*d^3*n*x^2)*log(x)
Time = 0.91 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.87 \[ \int x \left (d+e x^2\right )^3 \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {a d^{3} x^{2}}{2} + \frac {3 a d^{2} e x^{4}}{4} + \frac {a d e^{2} x^{6}}{2} + \frac {a e^{3} x^{8}}{8} - \frac {b d^{3} n x^{2}}{4} + \frac {b d^{3} x^{2} \log {\left (c x^{n} \right )}}{2} - \frac {3 b d^{2} e n x^{4}}{16} + \frac {3 b d^{2} e x^{4} \log {\left (c x^{n} \right )}}{4} - \frac {b d e^{2} n x^{6}}{12} + \frac {b d e^{2} x^{6} \log {\left (c x^{n} \right )}}{2} - \frac {b e^{3} n x^{8}}{64} + \frac {b e^{3} x^{8} \log {\left (c x^{n} \right )}}{8} \]
a*d**3*x**2/2 + 3*a*d**2*e*x**4/4 + a*d*e**2*x**6/2 + a*e**3*x**8/8 - b*d* *3*n*x**2/4 + b*d**3*x**2*log(c*x**n)/2 - 3*b*d**2*e*n*x**4/16 + 3*b*d**2* e*x**4*log(c*x**n)/4 - b*d*e**2*n*x**6/12 + b*d*e**2*x**6*log(c*x**n)/2 - b*e**3*n*x**8/64 + b*e**3*x**8*log(c*x**n)/8
Time = 0.19 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.57 \[ \int x \left (d+e x^2\right )^3 \left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {1}{64} \, b e^{3} n x^{8} + \frac {1}{8} \, b e^{3} x^{8} \log \left (c x^{n}\right ) + \frac {1}{8} \, a e^{3} x^{8} - \frac {1}{12} \, b d e^{2} n x^{6} + \frac {1}{2} \, b d e^{2} x^{6} \log \left (c x^{n}\right ) + \frac {1}{2} \, a d e^{2} x^{6} - \frac {3}{16} \, b d^{2} e n x^{4} + \frac {3}{4} \, b d^{2} e x^{4} \log \left (c x^{n}\right ) + \frac {3}{4} \, a d^{2} e x^{4} - \frac {1}{4} \, b d^{3} n x^{2} + \frac {1}{2} \, b d^{3} x^{2} \log \left (c x^{n}\right ) + \frac {1}{2} \, a d^{3} x^{2} \]
-1/64*b*e^3*n*x^8 + 1/8*b*e^3*x^8*log(c*x^n) + 1/8*a*e^3*x^8 - 1/12*b*d*e^ 2*n*x^6 + 1/2*b*d*e^2*x^6*log(c*x^n) + 1/2*a*d*e^2*x^6 - 3/16*b*d^2*e*n*x^ 4 + 3/4*b*d^2*e*x^4*log(c*x^n) + 3/4*a*d^2*e*x^4 - 1/4*b*d^3*n*x^2 + 1/2*b *d^3*x^2*log(c*x^n) + 1/2*a*d^3*x^2
Leaf count of result is larger than twice the leaf count of optimal. 177 vs. \(2 (79) = 158\).
Time = 0.38 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.95 \[ \int x \left (d+e x^2\right )^3 \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {1}{8} \, b e^{3} n x^{8} \log \left (x\right ) - \frac {1}{64} \, b e^{3} n x^{8} + \frac {1}{8} \, b e^{3} x^{8} \log \left (c\right ) + \frac {1}{8} \, a e^{3} x^{8} + \frac {1}{2} \, b d e^{2} n x^{6} \log \left (x\right ) - \frac {1}{12} \, b d e^{2} n x^{6} + \frac {1}{2} \, b d e^{2} x^{6} \log \left (c\right ) + \frac {1}{2} \, a d e^{2} x^{6} + \frac {3}{4} \, b d^{2} e n x^{4} \log \left (x\right ) - \frac {3}{16} \, b d^{2} e n x^{4} + \frac {3}{4} \, b d^{2} e x^{4} \log \left (c\right ) + \frac {3}{4} \, a d^{2} e x^{4} + \frac {1}{2} \, b d^{3} n x^{2} \log \left (x\right ) - \frac {1}{4} \, b d^{3} n x^{2} + \frac {1}{2} \, b d^{3} x^{2} \log \left (c\right ) + \frac {1}{2} \, a d^{3} x^{2} \]
1/8*b*e^3*n*x^8*log(x) - 1/64*b*e^3*n*x^8 + 1/8*b*e^3*x^8*log(c) + 1/8*a*e ^3*x^8 + 1/2*b*d*e^2*n*x^6*log(x) - 1/12*b*d*e^2*n*x^6 + 1/2*b*d*e^2*x^6*l og(c) + 1/2*a*d*e^2*x^6 + 3/4*b*d^2*e*n*x^4*log(x) - 3/16*b*d^2*e*n*x^4 + 3/4*b*d^2*e*x^4*log(c) + 3/4*a*d^2*e*x^4 + 1/2*b*d^3*n*x^2*log(x) - 1/4*b* d^3*n*x^2 + 1/2*b*d^3*x^2*log(c) + 1/2*a*d^3*x^2
Time = 0.36 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.24 \[ \int x \left (d+e x^2\right )^3 \left (a+b \log \left (c x^n\right )\right ) \, dx=\ln \left (c\,x^n\right )\,\left (\frac {b\,d^3\,x^2}{2}+\frac {3\,b\,d^2\,e\,x^4}{4}+\frac {b\,d\,e^2\,x^6}{2}+\frac {b\,e^3\,x^8}{8}\right )+\frac {d^3\,x^2\,\left (2\,a-b\,n\right )}{4}+\frac {e^3\,x^8\,\left (8\,a-b\,n\right )}{64}+\frac {3\,d^2\,e\,x^4\,\left (4\,a-b\,n\right )}{16}+\frac {d\,e^2\,x^6\,\left (6\,a-b\,n\right )}{12} \]