Integrand size = 16, antiderivative size = 52 \[ \int x^3 \left (a+b \log \left (c x^n\right )\right )^2 \, dx=\frac {1}{32} b^2 n^2 x^4-\frac {1}{8} b n x^4 \left (a+b \log \left (c x^n\right )\right )+\frac {1}{4} x^4 \left (a+b \log \left (c x^n\right )\right )^2 \] Output:
1/32*b^2*n^2*x^4-1/8*b*n*x^4*(a+b*ln(c*x^n))+1/4*x^4*(a+b*ln(c*x^n))^2
Time = 0.02 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.83 \[ \int x^3 \left (a+b \log \left (c x^n\right )\right )^2 \, dx=\frac {1}{32} x^4 \left (b^2 n^2-4 b n \left (a+b \log \left (c x^n\right )\right )+8 \left (a+b \log \left (c x^n\right )\right )^2\right ) \] Input:
Integrate[x^3*(a + b*Log[c*x^n])^2,x]
Output:
(x^4*(b^2*n^2 - 4*b*n*(a + b*Log[c*x^n]) + 8*(a + b*Log[c*x^n])^2))/32
Time = 0.23 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.02, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {2742, 2741}
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.
\(\displaystyle \int x^3 \left (a+b \log \left (c x^n\right )\right )^2 \, dx\) |
\(\Big \downarrow \) 2742 |
\(\displaystyle \frac {1}{4} x^4 \left (a+b \log \left (c x^n\right )\right )^2-\frac {1}{2} b n \int x^3 \left (a+b \log \left (c x^n\right )\right )dx\) |
\(\Big \downarrow \) 2741 |
\(\displaystyle \frac {1}{4} x^4 \left (a+b \log \left (c x^n\right )\right )^2-\frac {1}{2} b n \left (\frac {1}{4} x^4 \left (a+b \log \left (c x^n\right )\right )-\frac {1}{16} b n x^4\right )\) |
Input:
Int[x^3*(a + b*Log[c*x^n])^2,x]
Output:
(x^4*(a + b*Log[c*x^n])^2)/4 - (b*n*(-1/16*(b*n*x^4) + (x^4*(a + b*Log[c*x ^n]))/4))/2
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*Log[c*x^n])/(d*(m + 1))), x] - Simp[b*n*((d*x)^( m + 1)/(d*(m + 1)^2)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1]
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbo l] :> Simp[(d*x)^(m + 1)*((a + b*Log[c*x^n])^p/(d*(m + 1))), x] - Simp[b*n* (p/(m + 1)) Int[(d*x)^m*(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{a, b , c, d, m, n}, x] && NeQ[m, -1] && GtQ[p, 0]
Time = 0.31 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.40
method | result | size |
parallelrisch | \(\frac {x^{4} \ln \left (c \,x^{n}\right )^{2} b^{2}}{4}-\frac {\ln \left (c \,x^{n}\right ) x^{4} n \,b^{2}}{8}+\frac {b^{2} n^{2} x^{4}}{32}+\frac {\ln \left (c \,x^{n}\right ) x^{4} a b}{2}-\frac {a b n \,x^{4}}{8}+\frac {a^{2} x^{4}}{4}\) | \(73\) |
risch | \(\frac {b^{2} x^{4} \ln \left (x^{n}\right )^{2}}{4}+\frac {b \,x^{4} \left (2 i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}-2 i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) \operatorname {csgn}\left (i c \right )-2 i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}+2 i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{2} \operatorname {csgn}\left (i c \right )+4 b \ln \left (c \right )-n b +4 a \right ) \ln \left (x^{n}\right )}{8}+\frac {x^{4} \left (-8 i \pi a b \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) \operatorname {csgn}\left (i c \right )-8 i \pi \ln \left (c \right ) b^{2} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) \operatorname {csgn}\left (i c \right )+2 i \pi \,b^{2} n \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) \operatorname {csgn}\left (i c \right )+8 i \pi \ln \left (c \right ) b^{2} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}+8 i \pi a b \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}+8 i \pi \ln \left (c \right ) b^{2} \operatorname {csgn}\left (i c \,x^{n}\right )^{2} \operatorname {csgn}\left (i c \right )-4 b^{2} \ln \left (c \right ) n +16 a b \ln \left (c \right )+8 a^{2}+8 b^{2} \ln \left (c \right )^{2}+b^{2} n^{2}+4 \pi ^{2} b^{2} \operatorname {csgn}\left (i c \,x^{n}\right )^{5} \operatorname {csgn}\left (i c \right )-2 \pi ^{2} b^{2} \operatorname {csgn}\left (i c \,x^{n}\right )^{4} \operatorname {csgn}\left (i c \right )^{2}-2 \pi ^{2} b^{2} \operatorname {csgn}\left (i x^{n}\right )^{2} \operatorname {csgn}\left (i c \,x^{n}\right )^{4}+4 \pi ^{2} b^{2} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{5}-2 i \pi \,b^{2} n \operatorname {csgn}\left (i c \,x^{n}\right )^{2} \operatorname {csgn}\left (i c \right )-2 i \pi \,b^{2} n \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}+8 i \pi a b \operatorname {csgn}\left (i c \,x^{n}\right )^{2} \operatorname {csgn}\left (i c \right )-4 a n b -2 \pi ^{2} b^{2} \operatorname {csgn}\left (i c \,x^{n}\right )^{6}-8 i \pi a b \operatorname {csgn}\left (i c \,x^{n}\right )^{3}+4 \pi ^{2} b^{2} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{3} \operatorname {csgn}\left (i c \right )^{2}+4 \pi ^{2} b^{2} \operatorname {csgn}\left (i x^{n}\right )^{2} \operatorname {csgn}\left (i c \,x^{n}\right )^{3} \operatorname {csgn}\left (i c \right )-8 \pi ^{2} b^{2} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{4} \operatorname {csgn}\left (i c \right )+2 i \pi \,b^{2} n \operatorname {csgn}\left (i c \,x^{n}\right )^{3}-8 i \pi \ln \left (c \right ) b^{2} \operatorname {csgn}\left (i c \,x^{n}\right )^{3}-2 \pi ^{2} b^{2} \operatorname {csgn}\left (i x^{n}\right )^{2} \operatorname {csgn}\left (i c \,x^{n}\right )^{2} \operatorname {csgn}\left (i c \right )^{2}\right )}{32}\) | \(691\) |
Input:
int(x^3*(a+b*ln(c*x^n))^2,x,method=_RETURNVERBOSE)
Output:
1/4*x^4*ln(c*x^n)^2*b^2-1/8*ln(c*x^n)*x^4*n*b^2+1/32*b^2*n^2*x^4+1/2*ln(c* x^n)*x^4*a*b-1/8*a*b*n*x^4+1/4*a^2*x^4
Leaf count of result is larger than twice the leaf count of optimal. 102 vs. \(2 (46) = 92\).
Time = 0.07 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.96 \[ \int x^3 \left (a+b \log \left (c x^n\right )\right )^2 \, dx=\frac {1}{4} \, b^{2} n^{2} x^{4} \log \left (x\right )^{2} + \frac {1}{4} \, b^{2} x^{4} \log \left (c\right )^{2} - \frac {1}{8} \, {\left (b^{2} n - 4 \, a b\right )} x^{4} \log \left (c\right ) + \frac {1}{32} \, {\left (b^{2} n^{2} - 4 \, a b n + 8 \, a^{2}\right )} x^{4} + \frac {1}{8} \, {\left (4 \, b^{2} n x^{4} \log \left (c\right ) - {\left (b^{2} n^{2} - 4 \, a b n\right )} x^{4}\right )} \log \left (x\right ) \] Input:
integrate(x^3*(a+b*log(c*x^n))^2,x, algorithm="fricas")
Output:
1/4*b^2*n^2*x^4*log(x)^2 + 1/4*b^2*x^4*log(c)^2 - 1/8*(b^2*n - 4*a*b)*x^4* log(c) + 1/32*(b^2*n^2 - 4*a*b*n + 8*a^2)*x^4 + 1/8*(4*b^2*n*x^4*log(c) - (b^2*n^2 - 4*a*b*n)*x^4)*log(x)
Time = 0.49 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.50 \[ \int x^3 \left (a+b \log \left (c x^n\right )\right )^2 \, dx=\frac {a^{2} x^{4}}{4} - \frac {a b n x^{4}}{8} + \frac {a b x^{4} \log {\left (c x^{n} \right )}}{2} + \frac {b^{2} n^{2} x^{4}}{32} - \frac {b^{2} n x^{4} \log {\left (c x^{n} \right )}}{8} + \frac {b^{2} x^{4} \log {\left (c x^{n} \right )}^{2}}{4} \] Input:
integrate(x**3*(a+b*ln(c*x**n))**2,x)
Output:
a**2*x**4/4 - a*b*n*x**4/8 + a*b*x**4*log(c*x**n)/2 + b**2*n**2*x**4/32 - b**2*n*x**4*log(c*x**n)/8 + b**2*x**4*log(c*x**n)**2/4
Time = 0.04 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.37 \[ \int x^3 \left (a+b \log \left (c x^n\right )\right )^2 \, dx=\frac {1}{4} \, b^{2} x^{4} \log \left (c x^{n}\right )^{2} - \frac {1}{8} \, a b n x^{4} + \frac {1}{2} \, a b x^{4} \log \left (c x^{n}\right ) + \frac {1}{4} \, a^{2} x^{4} + \frac {1}{32} \, {\left (n^{2} x^{4} - 4 \, n x^{4} \log \left (c x^{n}\right )\right )} b^{2} \] Input:
integrate(x^3*(a+b*log(c*x^n))^2,x, algorithm="maxima")
Output:
1/4*b^2*x^4*log(c*x^n)^2 - 1/8*a*b*n*x^4 + 1/2*a*b*x^4*log(c*x^n) + 1/4*a^ 2*x^4 + 1/32*(n^2*x^4 - 4*n*x^4*log(c*x^n))*b^2
Leaf count of result is larger than twice the leaf count of optimal. 111 vs. \(2 (46) = 92\).
Time = 0.12 (sec) , antiderivative size = 111, normalized size of antiderivative = 2.13 \[ \int x^3 \left (a+b \log \left (c x^n\right )\right )^2 \, dx=\frac {1}{4} \, b^{2} n^{2} x^{4} \log \left (x\right )^{2} - \frac {1}{8} \, b^{2} n^{2} x^{4} \log \left (x\right ) + \frac {1}{2} \, b^{2} n x^{4} \log \left (c\right ) \log \left (x\right ) + \frac {1}{32} \, b^{2} n^{2} x^{4} - \frac {1}{8} \, b^{2} n x^{4} \log \left (c\right ) + \frac {1}{4} \, b^{2} x^{4} \log \left (c\right )^{2} + \frac {1}{2} \, a b n x^{4} \log \left (x\right ) - \frac {1}{8} \, a b n x^{4} + \frac {1}{2} \, a b x^{4} \log \left (c\right ) + \frac {1}{4} \, a^{2} x^{4} \] Input:
integrate(x^3*(a+b*log(c*x^n))^2,x, algorithm="giac")
Output:
1/4*b^2*n^2*x^4*log(x)^2 - 1/8*b^2*n^2*x^4*log(x) + 1/2*b^2*n*x^4*log(c)*l og(x) + 1/32*b^2*n^2*x^4 - 1/8*b^2*n*x^4*log(c) + 1/4*b^2*x^4*log(c)^2 + 1 /2*a*b*n*x^4*log(x) - 1/8*a*b*n*x^4 + 1/2*a*b*x^4*log(c) + 1/4*a^2*x^4
Time = 27.08 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.17 \[ \int x^3 \left (a+b \log \left (c x^n\right )\right )^2 \, dx=x^4\,\left (\frac {a^2}{4}-\frac {a\,b\,n}{8}+\frac {b^2\,n^2}{32}\right )+\frac {x^4\,\ln \left (c\,x^n\right )\,\left (a\,b-\frac {b^2\,n}{4}\right )}{2}+\frac {b^2\,x^4\,{\ln \left (c\,x^n\right )}^2}{4} \] Input:
int(x^3*(a + b*log(c*x^n))^2,x)
Output:
x^4*(a^2/4 + (b^2*n^2)/32 - (a*b*n)/8) + (x^4*log(c*x^n)*(a*b - (b^2*n)/4) )/2 + (b^2*x^4*log(c*x^n)^2)/4
Time = 0.15 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.12 \[ \int x^3 \left (a+b \log \left (c x^n\right )\right )^2 \, dx=\frac {x^{4} \left (8 \mathrm {log}\left (x^{n} c \right )^{2} b^{2}+16 \,\mathrm {log}\left (x^{n} c \right ) a b -4 \,\mathrm {log}\left (x^{n} c \right ) b^{2} n +8 a^{2}-4 a b n +b^{2} n^{2}\right )}{32} \] Input:
int(x^3*(a+b*log(c*x^n))^2,x)
Output:
(x**4*(8*log(x**n*c)**2*b**2 + 16*log(x**n*c)*a*b - 4*log(x**n*c)*b**2*n + 8*a**2 - 4*a*b*n + b**2*n**2))/32