Integrand size = 39, antiderivative size = 20 \[ \int \frac {a x^3+2 b n x^2 \log \left (c x^n\right )}{\left (a x^2+b x \log ^2\left (c x^n\right )\right )^3} \, dx=-\frac {1}{2 \left (a x+b \log ^2\left (c x^n\right )\right )^2} \] Output:
-1/2/(a*x+b*ln(c*x^n)^2)^2
Time = 0.02 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {a x^3+2 b n x^2 \log \left (c x^n\right )}{\left (a x^2+b x \log ^2\left (c x^n\right )\right )^3} \, dx=-\frac {1}{2 \left (a x+b \log ^2\left (c x^n\right )\right )^2} \] Input:
Integrate[(a*x^3 + 2*b*n*x^2*Log[c*x^n])/(a*x^2 + b*x*Log[c*x^n]^2)^3,x]
Output:
-1/2*1/(a*x + b*Log[c*x^n]^2)^2
Time = 0.37 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {3041, 3041, 3024}
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.
\(\displaystyle \int \frac {a x^3+2 b n x^2 \log \left (c x^n\right )}{\left (a x^2+b x \log ^2\left (c x^n\right )\right )^3} \, dx\) |
\(\Big \downarrow \) 3041 |
\(\displaystyle \int \frac {x^2 \left (a x+2 b n \log \left (c x^n\right )\right )}{\left (a x^2+b x \log ^2\left (c x^n\right )\right )^3}dx\) |
\(\Big \downarrow \) 3041 |
\(\displaystyle \int \frac {a x+2 b n \log \left (c x^n\right )}{x \left (a x+b \log ^2\left (c x^n\right )\right )^3}dx\) |
\(\Big \downarrow \) 3024 |
\(\displaystyle -\frac {1}{2 \left (a x+b \log ^2\left (c x^n\right )\right )^2}\) |
Input:
Int[(a*x^3 + 2*b*n*x^2*Log[c*x^n])/(a*x^2 + b*x*Log[c*x^n]^2)^3,x]
Output:
-1/2*1/(a*x + b*Log[c*x^n]^2)^2
Int[((Log[(c_.)*(x_)^(n_.)]^(q_)*(b_.) + (a_.)*(x_)^(m_.))^(p_.)*(Log[(c_.) *(x_)^(n_.)]^(r_.)*(e_.) + (d_.)*(x_)^(m_.)))/(x_), x_Symbol] :> Simp[e*((a *x^m + b*Log[c*x^n]^q)^(p + 1)/(b*n*q*(p + 1))), x] /; FreeQ[{a, b, c, d, e , m, n, p, q, r}, x] && EqQ[r, q - 1] && NeQ[p, -1] && EqQ[a*e*m - b*d*n*q, 0]
Int[(u_.)*((a_.)*(x_)^(m_.) + Log[(c_.)*(x_)^(n_.)]^(q_.)*(b_.)*(x_)^(r_.)) ^(p_.), x_Symbol] :> Int[u*x^(p*r)*(a*x^(m - r) + b*Log[c*x^n]^q)^p, x] /; FreeQ[{a, b, c, m, n, p, q, r}, x] && IntegerQ[p]
Time = 1.49 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95
method | result | size |
parallelrisch | \(-\frac {1}{2 {\left (a x +b \ln \left (c \,x^{n}\right )^{2}\right )}^{2}}\) | \(19\) |
risch | \(-\frac {8}{\left (-b \,\pi ^{2} \operatorname {csgn}\left (i x^{n}\right )^{2} \operatorname {csgn}\left (i c \,x^{n}\right )^{4}+2 b \,\pi ^{2} \operatorname {csgn}\left (i x^{n}\right )^{2} \operatorname {csgn}\left (i c \,x^{n}\right )^{3} \operatorname {csgn}\left (i c \right )-b \,\pi ^{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}+2 b \,\pi ^{2} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{5}-4 b \,\pi ^{2} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{4} \operatorname {csgn}\left (i c \right )+2 b \,\pi ^{2} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{3} \operatorname {csgn}\left (i c \right )^{2}-b \,\pi ^{2} \operatorname {csgn}\left (i c \,x^{n}\right )^{6}+2 b \,\pi ^{2} \operatorname {csgn}\left (i c \,x^{n}\right )^{5} \operatorname {csgn}\left (i c \right )-b \,\pi ^{2} \operatorname {csgn}\left (i c \,x^{n}\right )^{4} \operatorname {csgn}\left (i c \right )^{2}-4 i b \ln \left (c \right ) \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}+4 i b \ln \left (c \right ) \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}+4 i b \ln \left (c \right ) \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{2} \operatorname {csgn}\left (i c \right )+4 i b \ln \left (x^{n}\right ) \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{2} \operatorname {csgn}\left (i c \right )+4 i b \ln \left (x^{n}\right ) \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}-4 i b \ln \left (c \right ) \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) \operatorname {csgn}\left (i c \right )-4 i b \ln \left (x^{n}\right ) \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}-4 i b \ln \left (x^{n}\right ) \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) \operatorname {csgn}\left (i c \right )+4 b \ln \left (c \right )^{2}+8 b \ln \left (c \right ) \ln \left (x^{n}\right )+4 b \ln \left (x^{n}\right )^{2}+4 a x \right )^{2}}\) | \(451\) |
Input:
int((a*x^3+2*b*n*x^2*ln(c*x^n))/(a*x^2+b*x*ln(c*x^n)^2)^3,x,method=_RETURN VERBOSE)
Output:
-1/2/(a*x+b*ln(c*x^n)^2)^2
Leaf count of result is larger than twice the leaf count of optimal. 101 vs. \(2 (18) = 36\).
Time = 0.08 (sec) , antiderivative size = 101, normalized size of antiderivative = 5.05 \[ \int \frac {a x^3+2 b n x^2 \log \left (c x^n\right )}{\left (a x^2+b x \log ^2\left (c x^n\right )\right )^3} \, dx=-\frac {1}{2 \, {\left (b^{2} n^{4} \log \left (x\right )^{4} + 4 \, b^{2} n^{3} \log \left (c\right ) \log \left (x\right )^{3} + b^{2} \log \left (c\right )^{4} + 2 \, a b x \log \left (c\right )^{2} + a^{2} x^{2} + 2 \, {\left (3 \, b^{2} n^{2} \log \left (c\right )^{2} + a b n^{2} x\right )} \log \left (x\right )^{2} + 4 \, {\left (b^{2} n \log \left (c\right )^{3} + a b n x \log \left (c\right )\right )} \log \left (x\right )\right )}} \] Input:
integrate((a*x^3+2*b*n*x^2*log(c*x^n))/(a*x^2+b*x*log(c*x^n)^2)^3,x, algor ithm="fricas")
Output:
-1/2/(b^2*n^4*log(x)^4 + 4*b^2*n^3*log(c)*log(x)^3 + b^2*log(c)^4 + 2*a*b* x*log(c)^2 + a^2*x^2 + 2*(3*b^2*n^2*log(c)^2 + a*b*n^2*x)*log(x)^2 + 4*(b^ 2*n*log(c)^3 + a*b*n*x*log(c))*log(x))
\[ \int \frac {a x^3+2 b n x^2 \log \left (c x^n\right )}{\left (a x^2+b x \log ^2\left (c x^n\right )\right )^3} \, dx=\int \frac {a x + 2 b n \log {\left (c x^{n} \right )}}{x \left (a x + b \log {\left (c x^{n} \right )}^{2}\right )^{3}}\, dx \] Input:
integrate((a*x**3+2*b*n*x**2*ln(c*x**n))/(a*x**2+b*x*ln(c*x**n)**2)**3,x)
Output:
Integral((a*x + 2*b*n*log(c*x**n))/(x*(a*x + b*log(c*x**n)**2)**3), x)
Leaf count of result is larger than twice the leaf count of optimal. 95 vs. \(2 (18) = 36\).
Time = 0.11 (sec) , antiderivative size = 95, normalized size of antiderivative = 4.75 \[ \int \frac {a x^3+2 b n x^2 \log \left (c x^n\right )}{\left (a x^2+b x \log ^2\left (c x^n\right )\right )^3} \, dx=-\frac {1}{2 \, {\left (b^{2} \log \left (c\right )^{4} + 4 \, b^{2} \log \left (c\right ) \log \left (x^{n}\right )^{3} + b^{2} \log \left (x^{n}\right )^{4} + 2 \, a b x \log \left (c\right )^{2} + a^{2} x^{2} + 2 \, {\left (3 \, b^{2} \log \left (c\right )^{2} + a b x\right )} \log \left (x^{n}\right )^{2} + 4 \, {\left (b^{2} \log \left (c\right )^{3} + a b x \log \left (c\right )\right )} \log \left (x^{n}\right )\right )}} \] Input:
integrate((a*x^3+2*b*n*x^2*log(c*x^n))/(a*x^2+b*x*log(c*x^n)^2)^3,x, algor ithm="maxima")
Output:
-1/2/(b^2*log(c)^4 + 4*b^2*log(c)*log(x^n)^3 + b^2*log(x^n)^4 + 2*a*b*x*lo g(c)^2 + a^2*x^2 + 2*(3*b^2*log(c)^2 + a*b*x)*log(x^n)^2 + 4*(b^2*log(c)^3 + a*b*x*log(c))*log(x^n))
Leaf count of result is larger than twice the leaf count of optimal. 306 vs. \(2 (18) = 36\).
Time = 0.11 (sec) , antiderivative size = 306, normalized size of antiderivative = 15.30 \[ \int \frac {a x^3+2 b n x^2 \log \left (c x^n\right )}{\left (a x^2+b x \log ^2\left (c x^n\right )\right )^3} \, dx=-\frac {4 \, a b n^{2} x + a^{2} x^{2}}{2 \, {\left (4 \, a b^{3} n^{6} x \log \left (x\right )^{4} + 16 \, a b^{3} n^{5} x \log \left (c\right ) \log \left (x\right )^{3} + a^{2} b^{2} n^{4} x^{2} \log \left (x\right )^{4} + 24 \, a b^{3} n^{4} x \log \left (c\right )^{2} \log \left (x\right )^{2} + 4 \, a^{2} b^{2} n^{3} x^{2} \log \left (c\right ) \log \left (x\right )^{3} + 16 \, a b^{3} n^{3} x \log \left (c\right )^{3} \log \left (x\right ) + 8 \, a^{2} b^{2} n^{4} x^{2} \log \left (x\right )^{2} + 6 \, a^{2} b^{2} n^{2} x^{2} \log \left (c\right )^{2} \log \left (x\right )^{2} + 4 \, a b^{3} n^{2} x \log \left (c\right )^{4} + 16 \, a^{2} b^{2} n^{3} x^{2} \log \left (c\right ) \log \left (x\right ) + 4 \, a^{2} b^{2} n x^{2} \log \left (c\right )^{3} \log \left (x\right ) + 2 \, a^{3} b n^{2} x^{3} \log \left (x\right )^{2} + 8 \, a^{2} b^{2} n^{2} x^{2} \log \left (c\right )^{2} + a^{2} b^{2} x^{2} \log \left (c\right )^{4} + 4 \, a^{3} b n x^{3} \log \left (c\right ) \log \left (x\right ) + 4 \, a^{3} b n^{2} x^{3} + 2 \, a^{3} b x^{3} \log \left (c\right )^{2} + a^{4} x^{4}\right )}} \] Input:
integrate((a*x^3+2*b*n*x^2*log(c*x^n))/(a*x^2+b*x*log(c*x^n)^2)^3,x, algor ithm="giac")
Output:
-1/2*(4*a*b*n^2*x + a^2*x^2)/(4*a*b^3*n^6*x*log(x)^4 + 16*a*b^3*n^5*x*log( c)*log(x)^3 + a^2*b^2*n^4*x^2*log(x)^4 + 24*a*b^3*n^4*x*log(c)^2*log(x)^2 + 4*a^2*b^2*n^3*x^2*log(c)*log(x)^3 + 16*a*b^3*n^3*x*log(c)^3*log(x) + 8*a ^2*b^2*n^4*x^2*log(x)^2 + 6*a^2*b^2*n^2*x^2*log(c)^2*log(x)^2 + 4*a*b^3*n^ 2*x*log(c)^4 + 16*a^2*b^2*n^3*x^2*log(c)*log(x) + 4*a^2*b^2*n*x^2*log(c)^3 *log(x) + 2*a^3*b*n^2*x^3*log(x)^2 + 8*a^2*b^2*n^2*x^2*log(c)^2 + a^2*b^2* x^2*log(c)^4 + 4*a^3*b*n*x^3*log(c)*log(x) + 4*a^3*b*n^2*x^3 + 2*a^3*b*x^3 *log(c)^2 + a^4*x^4)
Time = 25.78 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.95 \[ \int \frac {a x^3+2 b n x^2 \log \left (c x^n\right )}{\left (a x^2+b x \log ^2\left (c x^n\right )\right )^3} \, dx=-\frac {1}{2\,a^2\,x^2+4\,a\,b\,x\,{\ln \left (c\,x^n\right )}^2+2\,b^2\,{\ln \left (c\,x^n\right )}^4} \] Input:
int((a*x^3 + 2*b*n*x^2*log(c*x^n))/(a*x^2 + b*x*log(c*x^n)^2)^3,x)
Output:
-1/(2*b^2*log(c*x^n)^4 + 2*a^2*x^2 + 4*a*b*x*log(c*x^n)^2)
Time = 0.14 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.95 \[ \int \frac {a x^3+2 b n x^2 \log \left (c x^n\right )}{\left (a x^2+b x \log ^2\left (c x^n\right )\right )^3} \, dx=-\frac {1}{2 \mathrm {log}\left (x^{n} c \right )^{4} b^{2}+4 \mathrm {log}\left (x^{n} c \right )^{2} a b x +2 a^{2} x^{2}} \] Input:
int((a*x^3+2*b*n*x^2*log(c*x^n))/(a*x^2+b*x*log(c*x^n)^2)^3,x)
Output:
( - 1)/(2*(log(x**n*c)**4*b**2 + 2*log(x**n*c)**2*a*b*x + a**2*x**2))