Integrand size = 23, antiderivative size = 161 \[ \int \frac {x^2 \left (a+b \log \left (c x^n\right )\right )^2}{(d+e x)^4} \, dx=\frac {b n x^2 \left (a+b \log \left (c x^n\right )\right )}{3 d e (d+e x)^2}+\frac {x^3 \left (a+b \log \left (c x^n\right )\right )^2}{3 d (d+e x)^3}+\frac {b n x \left (2 a+b n+2 b \log \left (c x^n\right )\right )}{3 d e^2 (d+e x)}-\frac {b n \left (2 a+3 b n+2 b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )}{3 d e^3}-\frac {2 b^2 n^2 \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{3 d e^3} \]
1/3*b*n*x^2*(a+b*ln(c*x^n))/d/e/(e*x+d)^2+1/3*x^3*(a+b*ln(c*x^n))^2/d/(e*x +d)^3+1/3*b*n*x*(2*a+b*n+2*b*ln(c*x^n))/d/e^2/(e*x+d)-1/3*b*n*(2*a+3*b*n+2 *b*ln(c*x^n))*ln(1+e*x/d)/d/e^3-2/3*b^2*n^2*polylog(2,-e*x/d)/d/e^3
Leaf count is larger than twice the leaf count of optimal. \(371\) vs. \(2(161)=322\).
Time = 0.32 (sec) , antiderivative size = 371, normalized size of antiderivative = 2.30 \[ \int \frac {x^2 \left (a+b \log \left (c x^n\right )\right )^2}{(d+e x)^4} \, dx=-\frac {-\frac {a^2}{d}+\frac {a^2 d^2}{(d+e x)^3}-\frac {3 a^2 d}{(d+e x)^2}-\frac {a b d n}{(d+e x)^2}+\frac {3 a^2}{d+e x}+\frac {4 a b n}{d+e x}+\frac {b^2 n^2}{d+e x}-\frac {3 b^2 n^2 \log (x)}{d}-\frac {2 a b \log \left (c x^n\right )}{d}+\frac {2 a b d^2 \log \left (c x^n\right )}{(d+e x)^3}-\frac {6 a b d \log \left (c x^n\right )}{(d+e x)^2}-\frac {b^2 d n \log \left (c x^n\right )}{(d+e x)^2}+\frac {6 a b \log \left (c x^n\right )}{d+e x}+\frac {4 b^2 n \log \left (c x^n\right )}{d+e x}-\frac {b^2 \log ^2\left (c x^n\right )}{d}+\frac {b^2 d^2 \log ^2\left (c x^n\right )}{(d+e x)^3}-\frac {3 b^2 d \log ^2\left (c x^n\right )}{(d+e x)^2}+\frac {3 b^2 \log ^2\left (c x^n\right )}{d+e x}+\frac {3 b^2 n^2 \log (d+e x)}{d}+\frac {2 a b n \log \left (1+\frac {e x}{d}\right )}{d}+\frac {2 b^2 n \log \left (c x^n\right ) \log \left (1+\frac {e x}{d}\right )}{d}+\frac {2 b^2 n^2 \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{d}}{3 e^3} \]
-1/3*(-(a^2/d) + (a^2*d^2)/(d + e*x)^3 - (3*a^2*d)/(d + e*x)^2 - (a*b*d*n) /(d + e*x)^2 + (3*a^2)/(d + e*x) + (4*a*b*n)/(d + e*x) + (b^2*n^2)/(d + e* x) - (3*b^2*n^2*Log[x])/d - (2*a*b*Log[c*x^n])/d + (2*a*b*d^2*Log[c*x^n])/ (d + e*x)^3 - (6*a*b*d*Log[c*x^n])/(d + e*x)^2 - (b^2*d*n*Log[c*x^n])/(d + e*x)^2 + (6*a*b*Log[c*x^n])/(d + e*x) + (4*b^2*n*Log[c*x^n])/(d + e*x) - (b^2*Log[c*x^n]^2)/d + (b^2*d^2*Log[c*x^n]^2)/(d + e*x)^3 - (3*b^2*d*Log[c *x^n]^2)/(d + e*x)^2 + (3*b^2*Log[c*x^n]^2)/(d + e*x) + (3*b^2*n^2*Log[d + e*x])/d + (2*a*b*n*Log[1 + (e*x)/d])/d + (2*b^2*n*Log[c*x^n]*Log[1 + (e*x )/d])/d + (2*b^2*n^2*PolyLog[2, -((e*x)/d)])/d)/e^3
Time = 0.54 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.96, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {2781, 2784, 2784, 2754, 2838}
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.
| \(\displaystyle \int \frac {x^2 \left (a+b \log \left (c x^n\right )\right )^2}{(d+e x)^4} \, dx\) |
| \(\Big \downarrow \) 2781 |
| \(\displaystyle \frac {x^3 \left (a+b \log \left (c x^n\right )\right )^2}{3 d (d+e x)^3}-\frac {2 b n \int \frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^3}dx}{3 d}\) |
| \(\Big \downarrow \) 2784 |
| \(\displaystyle \frac {x^3 \left (a+b \log \left (c x^n\right )\right )^2}{3 d (d+e x)^3}-\frac {2 b n \left (\frac {\int \frac {x \left (2 a+b n+2 b \log \left (c x^n\right )\right )}{(d+e x)^2}dx}{2 e}-\frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{2 e (d+e x)^2}\right )}{3 d}\) |
| \(\Big \downarrow \) 2784 |
| \(\displaystyle \frac {x^3 \left (a+b \log \left (c x^n\right )\right )^2}{3 d (d+e x)^3}-\frac {2 b n \left (\frac {\frac {\int \frac {2 a+3 b n+2 b \log \left (c x^n\right )}{d+e x}dx}{e}-\frac {x \left (2 a+2 b \log \left (c x^n\right )+b n\right )}{e (d+e x)}}{2 e}-\frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{2 e (d+e x)^2}\right )}{3 d}\) |
| \(\Big \downarrow \) 2754 |
| \(\displaystyle \frac {x^3 \left (a+b \log \left (c x^n\right )\right )^2}{3 d (d+e x)^3}-\frac {2 b n \left (\frac {\frac {\frac {\log \left (\frac {e x}{d}+1\right ) \left (2 a+2 b \log \left (c x^n\right )+3 b n\right )}{e}-\frac {2 b n \int \frac {\log \left (\frac {e x}{d}+1\right )}{x}dx}{e}}{e}-\frac {x \left (2 a+2 b \log \left (c x^n\right )+b n\right )}{e (d+e x)}}{2 e}-\frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{2 e (d+e x)^2}\right )}{3 d}\) |
| \(\Big \downarrow \) 2838 |
| \(\displaystyle \frac {x^3 \left (a+b \log \left (c x^n\right )\right )^2}{3 d (d+e x)^3}-\frac {2 b n \left (\frac {\frac {\frac {\log \left (\frac {e x}{d}+1\right ) \left (2 a+2 b \log \left (c x^n\right )+3 b n\right )}{e}+\frac {2 b n \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{e}}{e}-\frac {x \left (2 a+2 b \log \left (c x^n\right )+b n\right )}{e (d+e x)}}{2 e}-\frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{2 e (d+e x)^2}\right )}{3 d}\) |
(x^3*(a + b*Log[c*x^n])^2)/(3*d*(d + e*x)^3) - (2*b*n*(-1/2*(x^2*(a + b*Lo g[c*x^n]))/(e*(d + e*x)^2) + (-((x*(2*a + b*n + 2*b*Log[c*x^n]))/(e*(d + e *x))) + (((2*a + 3*b*n + 2*b*Log[c*x^n])*Log[1 + (e*x)/d])/e + (2*b*n*Poly Log[2, -((e*x)/d)])/e)/e)/(2*e)))/(3*d)
3.2.15.3.1 Defintions of rubi rules used
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symb ol] :> Simp[Log[1 + e*(x/d)]*((a + b*Log[c*x^n])^p/e), x] - Simp[b*n*(p/e) Int[Log[1 + e*(x/d)]*((a + b*Log[c*x^n])^(p - 1)/x), x], x] /; FreeQ[{a, b, c, d, e, n}, x] && IGtQ[p, 0]
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_))^(q_), x_Symbol] :> Simp[(-(f*x)^(m + 1))*(d + e*x)^(q + 1)*((a + b*Log[c*x^n])^p/(d*f*(q + 1))), x] + Simp[b*n*(p/(d*(q + 1))) Int[(f*x) ^m*(d + e*x)^(q + 1)*(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, f, m, n, q}, x] && EqQ[m + q + 2, 0] && IGtQ[p, 0] && LtQ[q, -1]
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((f_.)*(x_))^(m_.)*((d_) + (e_.)* (x_))^(q_.), x_Symbol] :> Simp[(f*x)^m*(d + e*x)^(q + 1)*((a + b*Log[c*x^n] )/(e*(q + 1))), x] - Simp[f/(e*(q + 1)) Int[(f*x)^(m - 1)*(d + e*x)^(q + 1)*(a*m + b*n + b*m*Log[c*x^n]), x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && ILtQ[q, -1] && GtQ[m, 0]
Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2 , (-c)*e*x^n]/n, x] /; FreeQ[{c, d, e, n}, x] && EqQ[c*d, 1]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.52 (sec) , antiderivative size = 593, normalized size of antiderivative = 3.68
| method | result | size |
| risch | \(-\frac {b^{2} \ln \left (x^{n}\right )^{2} d^{2}}{3 e^{3} \left (e x +d \right )^{3}}-\frac {b^{2} \ln \left (x^{n}\right )^{2}}{e^{3} \left (e x +d \right )}+\frac {b^{2} \ln \left (x^{n}\right )^{2} d}{e^{3} \left (e x +d \right )^{2}}-\frac {4 b^{2} n \ln \left (x^{n}\right )}{3 e^{3} \left (e x +d \right )}+\frac {b^{2} n \ln \left (x^{n}\right ) d}{3 e^{3} \left (e x +d \right )^{2}}-\frac {2 b^{2} n \ln \left (x^{n}\right ) \ln \left (e x +d \right )}{3 e^{3} d}+\frac {2 b^{2} n \ln \left (x^{n}\right ) \ln \left (x \right )}{3 e^{3} d}-\frac {b^{2} n^{2} \ln \left (x \right )^{2}}{3 e^{3} d}+\frac {2 b^{2} n^{2} \ln \left (e x +d \right ) \ln \left (-\frac {e x}{d}\right )}{3 e^{3} d}+\frac {2 b^{2} n^{2} \operatorname {dilog}\left (-\frac {e x}{d}\right )}{3 e^{3} d}-\frac {b^{2} n^{2}}{3 e^{3} \left (e x +d \right )}-\frac {b^{2} n^{2} \ln \left (e x +d \right )}{e^{3} d}+\frac {b^{2} n^{2} \ln \left (x \right )}{e^{3} d}+\left (-i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )+i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}+i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}-i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}+2 b \ln \left (c \right )+2 a \right ) b \left (-\frac {\ln \left (x^{n}\right ) d^{2}}{3 e^{3} \left (e x +d \right )^{3}}-\frac {\ln \left (x^{n}\right )}{e^{3} \left (e x +d \right )}+\frac {\ln \left (x^{n}\right ) d}{e^{3} \left (e x +d \right )^{2}}-\frac {n \left (-\frac {d}{2 \left (e x +d \right )^{2}}+\frac {\ln \left (e x +d \right )}{d}+\frac {2}{e x +d}-\frac {\ln \left (x \right )}{d}\right )}{3 e^{3}}\right )+\frac {{\left (-i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )+i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}+i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}-i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}+2 b \ln \left (c \right )+2 a \right )}^{2} \left (-\frac {d^{2}}{3 e^{3} \left (e x +d \right )^{3}}-\frac {1}{e^{3} \left (e x +d \right )}+\frac {d}{e^{3} \left (e x +d \right )^{2}}\right )}{4}\) | \(593\) |
-1/3*b^2*ln(x^n)^2/e^3*d^2/(e*x+d)^3-b^2*ln(x^n)^2/e^3/(e*x+d)+b^2*ln(x^n) ^2/e^3*d/(e*x+d)^2-4/3*b^2*n*ln(x^n)/e^3/(e*x+d)+1/3*b^2*n*ln(x^n)/e^3*d/( e*x+d)^2-2/3*b^2*n*ln(x^n)/e^3/d*ln(e*x+d)+2/3*b^2*n*ln(x^n)/e^3/d*ln(x)-1 /3*b^2*n^2/e^3/d*ln(x)^2+2/3*b^2*n^2/e^3/d*ln(e*x+d)*ln(-e*x/d)+2/3*b^2*n^ 2/e^3/d*dilog(-e*x/d)-1/3*b^2*n^2/e^3/(e*x+d)-b^2*n^2/e^3/d*ln(e*x+d)+b^2* n^2/e^3/d*ln(x)+(-I*b*Pi*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)+I*b*Pi*csgn(I *c)*csgn(I*c*x^n)^2+I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2-I*b*Pi*csgn(I*c*x^n )^3+2*b*ln(c)+2*a)*b*(-1/3*ln(x^n)/e^3*d^2/(e*x+d)^3-ln(x^n)/e^3/(e*x+d)+l n(x^n)/e^3*d/(e*x+d)^2-1/3*n/e^3*(-1/2*d/(e*x+d)^2+1/d*ln(e*x+d)+2/(e*x+d) -1/d*ln(x)))+1/4*(-I*b*Pi*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)+I*b*Pi*csgn( I*c)*csgn(I*c*x^n)^2+I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2-I*b*Pi*csgn(I*c*x^ n)^3+2*b*ln(c)+2*a)^2*(-1/3/e^3*d^2/(e*x+d)^3-1/e^3/(e*x+d)+1/e^3*d/(e*x+d )^2)
\[ \int \frac {x^2 \left (a+b \log \left (c x^n\right )\right )^2}{(d+e x)^4} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )}^{2} x^{2}}{{\left (e x + d\right )}^{4}} \,d x } \]
integral((b^2*x^2*log(c*x^n)^2 + 2*a*b*x^2*log(c*x^n) + a^2*x^2)/(e^4*x^4 + 4*d*e^3*x^3 + 6*d^2*e^2*x^2 + 4*d^3*e*x + d^4), x)
\[ \int \frac {x^2 \left (a+b \log \left (c x^n\right )\right )^2}{(d+e x)^4} \, dx=\int \frac {x^{2} \left (a + b \log {\left (c x^{n} \right )}\right )^{2}}{\left (d + e x\right )^{4}}\, dx \]
\[ \int \frac {x^2 \left (a+b \log \left (c x^n\right )\right )^2}{(d+e x)^4} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )}^{2} x^{2}}{{\left (e x + d\right )}^{4}} \,d x } \]
-1/3*a*b*n*((4*e*x + 3*d)/(e^5*x^2 + 2*d*e^4*x + d^2*e^3) + 2*log(e*x + d) /(d*e^3) - 2*log(x)/(d*e^3)) - 1/3*((3*e^2*x^2 + 3*d*e*x + d^2)*log(x^n)^2 /(e^6*x^3 + 3*d*e^5*x^2 + 3*d^2*e^4*x + d^3*e^3) - 3*integrate(1/3*(3*e^3* x^3*log(c)^2 + 2*(6*d*e^2*n*x^2 + 4*d^2*e*n*x + d^3*n + 3*(e^3*n + e^3*log (c))*x^3)*log(x^n))/(e^7*x^5 + 4*d*e^6*x^4 + 6*d^2*e^5*x^3 + 4*d^3*e^4*x^2 + d^4*e^3*x), x))*b^2 - 2/3*(3*e^2*x^2 + 3*d*e*x + d^2)*a*b*log(c*x^n)/(e ^6*x^3 + 3*d*e^5*x^2 + 3*d^2*e^4*x + d^3*e^3) - 1/3*(3*e^2*x^2 + 3*d*e*x + d^2)*a^2/(e^6*x^3 + 3*d*e^5*x^2 + 3*d^2*e^4*x + d^3*e^3)
\[ \int \frac {x^2 \left (a+b \log \left (c x^n\right )\right )^2}{(d+e x)^4} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )}^{2} x^{2}}{{\left (e x + d\right )}^{4}} \,d x } \]
Timed out. \[ \int \frac {x^2 \left (a+b \log \left (c x^n\right )\right )^2}{(d+e x)^4} \, dx=\int \frac {x^2\,{\left (a+b\,\ln \left (c\,x^n\right )\right )}^2}{{\left (d+e\,x\right )}^4} \,d x \]