Integrand size = 23, antiderivative size = 74 \[ \int x^3 \left (d+e x^2\right )^2 \left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {1}{16} b d^2 n x^4-\frac {1}{18} b d e n x^6-\frac {1}{64} b e^2 n x^8+\frac {1}{24} \left (6 d^2 x^4+8 d e x^6+3 e^2 x^8\right ) \left (a+b \log \left (c x^n\right )\right ) \]
-1/16*b*d^2*n*x^4-1/18*b*d*e*n*x^6-1/64*b*e^2*n*x^8+1/24*(3*e^2*x^8+8*d*e* x^6+6*d^2*x^4)*(a+b*ln(c*x^n))
Time = 0.03 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.18 \[ \int x^3 \left (d+e x^2\right )^2 \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {1}{576} x^4 \left (24 a \left (6 d^2+8 d e x^2+3 e^2 x^4\right )-b n \left (36 d^2+32 d e x^2+9 e^2 x^4\right )+24 b \left (6 d^2+8 d e x^2+3 e^2 x^4\right ) \log \left (c x^n\right )\right ) \]
(x^4*(24*a*(6*d^2 + 8*d*e*x^2 + 3*e^2*x^4) - b*n*(36*d^2 + 32*d*e*x^2 + 9* e^2*x^4) + 24*b*(6*d^2 + 8*d*e*x^2 + 3*e^2*x^4)*Log[c*x^n]))/576
Time = 0.30 (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.
\(\displaystyle \int x^3 \left (d+e x^2\right )^2 \left (a+b \log \left (c x^n\right )\right ) \, dx\) |
\(\Big \downarrow \) 2771 |
\(\displaystyle \frac {1}{24} \left (6 d^2 x^4+8 d e x^6+3 e^2 x^8\right ) \left (a+b \log \left (c x^n\right )\right )-b n \int \frac {1}{24} x^3 \left (3 e^2 x^4+8 d e x^2+6 d^2\right )dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{24} \left (6 d^2 x^4+8 d e x^6+3 e^2 x^8\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{24} b n \int x^3 \left (3 e^2 x^4+8 d e x^2+6 d^2\right )dx\) |
\(\Big \downarrow \) 1433 |
\(\displaystyle \frac {1}{24} \left (6 d^2 x^4+8 d e x^6+3 e^2 x^8\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{24} b n \int \left (3 e^2 x^7+8 d e x^5+6 d^2 x^3\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{24} \left (6 d^2 x^4+8 d e x^6+3 e^2 x^8\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{24} b n \left (\frac {3 d^2 x^4}{2}+\frac {4}{3} d e x^6+\frac {3 e^2 x^8}{8}\right )\) |
-1/24*(b*n*((3*d^2*x^4)/2 + (4*d*e*x^6)/3 + (3*e^2*x^8)/8)) + ((6*d^2*x^4 + 8*d*e*x^6 + 3*e^2*x^8)*(a + b*Log[c*x^n]))/24
3.2.84.3.1 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 = 0.79 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.36
method | result | size |
parallelrisch | \(\frac {x^{8} \ln \left (c \,x^{n}\right ) b \,e^{2}}{8}-\frac {b \,e^{2} n \,x^{8}}{64}+\frac {a \,e^{2} x^{8}}{8}+\frac {x^{6} \ln \left (c \,x^{n}\right ) b d e}{3}-\frac {b d e n \,x^{6}}{18}+\frac {a d e \,x^{6}}{3}+\frac {x^{4} \ln \left (c \,x^{n}\right ) b \,d^{2}}{4}-\frac {b \,d^{2} n \,x^{4}}{16}+\frac {a \,d^{2} x^{4}}{4}\) | \(101\) |
risch | \(\frac {b \,x^{4} \left (3 e^{2} x^{4}+8 d e \,x^{2}+6 d^{2}\right ) \ln \left (x^{n}\right )}{24}+\frac {i \pi b \,d^{2} x^{4} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{8}-\frac {i \pi b d e \,x^{6} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )}{6}+\frac {i \pi b \,e^{2} x^{8} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{16}+\frac {i \pi b d e \,x^{6} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{6}+\frac {\ln \left (c \right ) b \,e^{2} x^{8}}{8}-\frac {b \,e^{2} n \,x^{8}}{64}+\frac {a \,e^{2} x^{8}}{8}-\frac {i \pi b \,e^{2} x^{8} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )}{16}+\frac {i \pi b \,d^{2} x^{4} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{8}-\frac {i \pi b \,d^{2} x^{4} \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{8}+\frac {i \pi b \,e^{2} x^{8} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{16}+\frac {\ln \left (c \right ) b d e \,x^{6}}{3}-\frac {b d e n \,x^{6}}{18}+\frac {a d e \,x^{6}}{3}+\frac {i \pi b d e \,x^{6} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{6}-\frac {i \pi b d e \,x^{6} \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{6}-\frac {i \pi b \,e^{2} x^{8} \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{16}-\frac {i \pi b \,d^{2} x^{4} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )}{8}+\frac {\ln \left (c \right ) b \,d^{2} x^{4}}{4}-\frac {b \,d^{2} n \,x^{4}}{16}+\frac {a \,d^{2} x^{4}}{4}\) | \(434\) |
1/8*x^8*ln(c*x^n)*b*e^2-1/64*b*e^2*n*x^8+1/8*a*e^2*x^8+1/3*x^6*ln(c*x^n)*b *d*e-1/18*b*d*e*n*x^6+1/3*a*d*e*x^6+1/4*x^4*ln(c*x^n)*b*d^2-1/16*b*d^2*n*x ^4+1/4*a*d^2*x^4
Time = 0.28 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.59 \[ \int x^3 \left (d+e x^2\right )^2 \left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {1}{64} \, {\left (b e^{2} n - 8 \, a e^{2}\right )} x^{8} - \frac {1}{18} \, {\left (b d e n - 6 \, a d e\right )} x^{6} - \frac {1}{16} \, {\left (b d^{2} n - 4 \, a d^{2}\right )} x^{4} + \frac {1}{24} \, {\left (3 \, b e^{2} x^{8} + 8 \, b d e x^{6} + 6 \, b d^{2} x^{4}\right )} \log \left (c\right ) + \frac {1}{24} \, {\left (3 \, b e^{2} n x^{8} + 8 \, b d e n x^{6} + 6 \, b d^{2} n x^{4}\right )} \log \left (x\right ) \]
-1/64*(b*e^2*n - 8*a*e^2)*x^8 - 1/18*(b*d*e*n - 6*a*d*e)*x^6 - 1/16*(b*d^2 *n - 4*a*d^2)*x^4 + 1/24*(3*b*e^2*x^8 + 8*b*d*e*x^6 + 6*b*d^2*x^4)*log(c) + 1/24*(3*b*e^2*n*x^8 + 8*b*d*e*n*x^6 + 6*b*d^2*n*x^4)*log(x)
Time = 0.87 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.57 \[ \int x^3 \left (d+e x^2\right )^2 \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {a d^{2} x^{4}}{4} + \frac {a d e x^{6}}{3} + \frac {a e^{2} x^{8}}{8} - \frac {b d^{2} n x^{4}}{16} + \frac {b d^{2} x^{4} \log {\left (c x^{n} \right )}}{4} - \frac {b d e n x^{6}}{18} + \frac {b d e x^{6} \log {\left (c x^{n} \right )}}{3} - \frac {b e^{2} n x^{8}}{64} + \frac {b e^{2} x^{8} \log {\left (c x^{n} \right )}}{8} \]
a*d**2*x**4/4 + a*d*e*x**6/3 + a*e**2*x**8/8 - b*d**2*n*x**4/16 + b*d**2*x **4*log(c*x**n)/4 - b*d*e*n*x**6/18 + b*d*e*x**6*log(c*x**n)/3 - b*e**2*n* x**8/64 + b*e**2*x**8*log(c*x**n)/8
Time = 0.20 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.35 \[ \int x^3 \left (d+e x^2\right )^2 \left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {1}{64} \, b e^{2} n x^{8} + \frac {1}{8} \, b e^{2} x^{8} \log \left (c x^{n}\right ) + \frac {1}{8} \, a e^{2} x^{8} - \frac {1}{18} \, b d e n x^{6} + \frac {1}{3} \, b d e x^{6} \log \left (c x^{n}\right ) + \frac {1}{3} \, a d e x^{6} - \frac {1}{16} \, b d^{2} n x^{4} + \frac {1}{4} \, b d^{2} x^{4} \log \left (c x^{n}\right ) + \frac {1}{4} \, a d^{2} x^{4} \]
-1/64*b*e^2*n*x^8 + 1/8*b*e^2*x^8*log(c*x^n) + 1/8*a*e^2*x^8 - 1/18*b*d*e* n*x^6 + 1/3*b*d*e*x^6*log(c*x^n) + 1/3*a*d*e*x^6 - 1/16*b*d^2*n*x^4 + 1/4* b*d^2*x^4*log(c*x^n) + 1/4*a*d^2*x^4
Time = 0.33 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.66 \[ \int x^3 \left (d+e x^2\right )^2 \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {1}{8} \, b e^{2} n x^{8} \log \left (x\right ) - \frac {1}{64} \, b e^{2} n x^{8} + \frac {1}{8} \, b e^{2} x^{8} \log \left (c\right ) + \frac {1}{8} \, a e^{2} x^{8} + \frac {1}{3} \, b d e n x^{6} \log \left (x\right ) - \frac {1}{18} \, b d e n x^{6} + \frac {1}{3} \, b d e x^{6} \log \left (c\right ) + \frac {1}{3} \, a d e x^{6} + \frac {1}{4} \, b d^{2} n x^{4} \log \left (x\right ) - \frac {1}{16} \, b d^{2} n x^{4} + \frac {1}{4} \, b d^{2} x^{4} \log \left (c\right ) + \frac {1}{4} \, a d^{2} x^{4} \]
1/8*b*e^2*n*x^8*log(x) - 1/64*b*e^2*n*x^8 + 1/8*b*e^2*x^8*log(c) + 1/8*a*e ^2*x^8 + 1/3*b*d*e*n*x^6*log(x) - 1/18*b*d*e*n*x^6 + 1/3*b*d*e*x^6*log(c) + 1/3*a*d*e*x^6 + 1/4*b*d^2*n*x^4*log(x) - 1/16*b*d^2*n*x^4 + 1/4*b*d^2*x^ 4*log(c) + 1/4*a*d^2*x^4
Time = 0.37 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.11 \[ \int x^3 \left (d+e x^2\right )^2 \left (a+b \log \left (c x^n\right )\right ) \, dx=\ln \left (c\,x^n\right )\,\left (\frac {b\,d^2\,x^4}{4}+\frac {b\,d\,e\,x^6}{3}+\frac {b\,e^2\,x^8}{8}\right )+\frac {d^2\,x^4\,\left (4\,a-b\,n\right )}{16}+\frac {e^2\,x^8\,\left (8\,a-b\,n\right )}{64}+\frac {d\,e\,x^6\,\left (6\,a-b\,n\right )}{18} \]