Integrand size = 23, antiderivative size = 123 \[ \int \frac {\left (d+e x^r\right )^2 \left (a+b \log \left (c x^n\right )\right )}{x^2} \, dx=-\frac {b d^2 n}{x}-\frac {2 b d e n x^{-1+r}}{(1-r)^2}-\frac {b e^2 n x^{-1+2 r}}{(1-2 r)^2}-\frac {d^2 \left (a+b \log \left (c x^n\right )\right )}{x}-\frac {2 d e x^{-1+r} \left (a+b \log \left (c x^n\right )\right )}{1-r}-\frac {e^2 x^{-1+2 r} \left (a+b \log \left (c x^n\right )\right )}{1-2 r} \]
-b*d^2*n/x-2*b*d*e*n*x^(-1+r)/(1-r)^2-b*e^2*n*x^(-1+2*r)/(1-2*r)^2-d^2*(a+ b*ln(c*x^n))/x-2*d*e*x^(-1+r)*(a+b*ln(c*x^n))/(1-r)-e^2*x^(-1+2*r)*(a+b*ln (c*x^n))/(1-2*r)
Time = 0.20 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.98 \[ \int \frac {\left (d+e x^r\right )^2 \left (a+b \log \left (c x^n\right )\right )}{x^2} \, dx=\frac {b n \left (-d^2-\frac {2 d e x^r}{(-1+r)^2}-\frac {e^2 x^{2 r}}{(1-2 r)^2}\right )+a \left (-d^2+\frac {2 d e x^r}{-1+r}+\frac {e^2 x^{2 r}}{-1+2 r}\right )+b \left (-d^2+\frac {2 d e x^r}{-1+r}+\frac {e^2 x^{2 r}}{-1+2 r}\right ) \log \left (c x^n\right )}{x} \]
(b*n*(-d^2 - (2*d*e*x^r)/(-1 + r)^2 - (e^2*x^(2*r))/(1 - 2*r)^2) + a*(-d^2 + (2*d*e*x^r)/(-1 + r) + (e^2*x^(2*r))/(-1 + 2*r)) + b*(-d^2 + (2*d*e*x^r )/(-1 + r) + (e^2*x^(2*r))/(-1 + 2*r))*Log[c*x^n])/x
Time = 0.41 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.98, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {2772, 25, 1691, 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.
\(\displaystyle \int \frac {\left (d+e x^r\right )^2 \left (a+b \log \left (c x^n\right )\right )}{x^2} \, dx\) |
\(\Big \downarrow \) 2772 |
\(\displaystyle -b n \int -\frac {\frac {2 d e x^r}{1-r}+\frac {e^2 x^{2 r}}{1-2 r}+d^2}{x^2}dx-\frac {d^2 \left (a+b \log \left (c x^n\right )\right )}{x}-\frac {2 d e x^{r-1} \left (a+b \log \left (c x^n\right )\right )}{1-r}-\frac {e^2 x^{2 r-1} \left (a+b \log \left (c x^n\right )\right )}{1-2 r}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle b n \int \frac {\frac {2 d e x^r}{1-r}+\frac {e^2 x^{2 r}}{1-2 r}+d^2}{x^2}dx-\frac {d^2 \left (a+b \log \left (c x^n\right )\right )}{x}-\frac {2 d e x^{r-1} \left (a+b \log \left (c x^n\right )\right )}{1-r}-\frac {e^2 x^{2 r-1} \left (a+b \log \left (c x^n\right )\right )}{1-2 r}\) |
\(\Big \downarrow \) 1691 |
\(\displaystyle b n \int \left (-\frac {2 d e x^{r-2}}{r-1}+\frac {e^2 x^{2 (r-1)}}{1-2 r}+\frac {d^2}{x^2}\right )dx-\frac {d^2 \left (a+b \log \left (c x^n\right )\right )}{x}-\frac {2 d e x^{r-1} \left (a+b \log \left (c x^n\right )\right )}{1-r}-\frac {e^2 x^{2 r-1} \left (a+b \log \left (c x^n\right )\right )}{1-2 r}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {d^2 \left (a+b \log \left (c x^n\right )\right )}{x}-\frac {2 d e x^{r-1} \left (a+b \log \left (c x^n\right )\right )}{1-r}-\frac {e^2 x^{2 r-1} \left (a+b \log \left (c x^n\right )\right )}{1-2 r}+b n \left (-\frac {d^2}{x}-\frac {2 d e x^{r-1}}{(1-r)^2}-\frac {e^2 x^{2 r-1}}{(1-2 r)^2}\right )\) |
b*n*(-(d^2/x) - (2*d*e*x^(-1 + r))/(1 - r)^2 - (e^2*x^(-1 + 2*r))/(1 - 2*r )^2) - (d^2*(a + b*Log[c*x^n]))/x - (2*d*e*x^(-1 + r)*(a + b*Log[c*x^n]))/ (1 - r) - (e^2*x^(-1 + 2*r)*(a + b*Log[c*x^n]))/(1 - 2*r)
3.4.88.3.1 Defintions of rubi rules used
Int[((d_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d*x)^m*(a + b*x^n + c*x^(2*n))^p, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && EqQ[n2, 2*n] && IGtQ[p, 0] && !IntegerQ [Simplify[(m + 1)/n]]
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[(a + b*Log[c*x^n]) u, x] - Simp[b*n Int[SimplifyIntegrand[u/x, x], x], x]] /; FreeQ[{a, b, c, d, e, n, r}, x] && IGtQ[q, 0] && IntegerQ[m] && !(EqQ[q , 1] && EqQ[m, -1])
Leaf count of result is larger than twice the leaf count of optimal. \(470\) vs. \(2(123)=246\).
Time = 1.11 (sec) , antiderivative size = 471, normalized size of antiderivative = 3.83
method | result | size |
parallelrisch | \(-\frac {b \ln \left (c \,x^{n}\right ) d^{2}+2 b d e n \,x^{r}+13 b \,d^{2} n \,r^{2}+2 d e \,x^{r} a -6 b \,d^{2} n r +2 d e \,x^{r} b \ln \left (c \,x^{n}\right )+4 a \,d^{2} r^{4}-12 a \,d^{2} r^{3}-8 a d e \,r^{3} x^{r}+b \,d^{2} n +a \,d^{2}+4 \ln \left (c \,x^{n}\right ) b \,d^{2} r^{4}-12 \ln \left (c \,x^{n}\right ) b \,d^{2} r^{3}+13 \ln \left (c \,x^{n}\right ) b \,d^{2} r^{2}-6 \ln \left (c \,x^{n}\right ) b \,d^{2} r +4 b \,d^{2} n \,r^{4}-12 b \,d^{2} n \,r^{3}-8 b d e n r \,x^{r}+13 a \,d^{2} r^{2}-6 a \,d^{2} r +5 a \,e^{2} r^{2} x^{2 r}-4 a \,e^{2} r \,x^{2 r}+b \,e^{2} n \,x^{2 r}-2 a \,e^{2} r^{3} x^{2 r}+e^{2} x^{2 r} b \ln \left (c \,x^{n}\right )+8 b d e n \,r^{2} x^{r}+16 a d e \,r^{2} x^{r}-10 a d e r \,x^{r}-2 b \,e^{2} n r \,x^{2 r}-2 x^{2 r} \ln \left (c \,x^{n}\right ) b \,e^{2} r^{3}+5 x^{2 r} \ln \left (c \,x^{n}\right ) b \,e^{2} r^{2}-4 x^{2 r} \ln \left (c \,x^{n}\right ) b \,e^{2} r +b \,e^{2} n \,r^{2} x^{2 r}+e^{2} x^{2 r} a -8 x^{r} \ln \left (c \,x^{n}\right ) b d e \,r^{3}+16 x^{r} \ln \left (c \,x^{n}\right ) b d e \,r^{2}-10 x^{r} \ln \left (c \,x^{n}\right ) b d e r}{x \left (-1+2 r \right )^{2} \left (r^{2}-2 r +1\right )}\) | \(471\) |
risch | \(\text {Expression too large to display}\) | \(1927\) |
-1/x*(b*ln(c*x^n)*d^2-2*b*e^2*n*r*(x^r)^2+2*b*d*e*n*x^r+5*a*e^2*r^2*(x^r)^ 2-4*a*e^2*r*(x^r)^2+b*e^2*n*(x^r)^2-2*a*e^2*r^3*(x^r)^2+e^2*(x^r)^2*a+13*b *d^2*n*r^2+2*d*e*x^r*a+e^2*(x^r)^2*b*ln(c*x^n)-6*b*d^2*n*r+2*d*e*x^r*b*ln( c*x^n)+4*a*d^2*r^4-12*a*d^2*r^3-8*a*d*e*r^3*x^r+b*d^2*n+a*d^2+4*ln(c*x^n)* b*d^2*r^4-12*ln(c*x^n)*b*d^2*r^3+13*ln(c*x^n)*b*d^2*r^2-6*ln(c*x^n)*b*d^2* r+4*b*d^2*n*r^4-12*b*d^2*n*r^3-8*b*d*e*n*r*x^r+13*a*d^2*r^2-6*a*d^2*r-2*(x ^r)^2*ln(c*x^n)*b*e^2*r^3+5*(x^r)^2*ln(c*x^n)*b*e^2*r^2-4*(x^r)^2*ln(c*x^n )*b*e^2*r+8*b*d*e*n*r^2*x^r+16*a*d*e*r^2*x^r-10*a*d*e*r*x^r+b*e^2*n*r^2*(x ^r)^2-8*x^r*ln(c*x^n)*b*d*e*r^3+16*x^r*ln(c*x^n)*b*d*e*r^2-10*x^r*ln(c*x^n )*b*d*e*r)/(-1+2*r)^2/(r^2-2*r+1)
Leaf count of result is larger than twice the leaf count of optimal. 455 vs. \(2 (118) = 236\).
Time = 0.32 (sec) , antiderivative size = 455, normalized size of antiderivative = 3.70 \[ \int \frac {\left (d+e x^r\right )^2 \left (a+b \log \left (c x^n\right )\right )}{x^2} \, dx=-\frac {4 \, {\left (b d^{2} n + a d^{2}\right )} r^{4} + b d^{2} n - 12 \, {\left (b d^{2} n + a d^{2}\right )} r^{3} + a d^{2} + 13 \, {\left (b d^{2} n + a d^{2}\right )} r^{2} - 6 \, {\left (b d^{2} n + a d^{2}\right )} r - {\left (2 \, a e^{2} r^{3} - b e^{2} n - a e^{2} - {\left (b e^{2} n + 5 \, a e^{2}\right )} r^{2} + 2 \, {\left (b e^{2} n + 2 \, a e^{2}\right )} r + {\left (2 \, b e^{2} r^{3} - 5 \, b e^{2} r^{2} + 4 \, b e^{2} r - b e^{2}\right )} \log \left (c\right ) + {\left (2 \, b e^{2} n r^{3} - 5 \, b e^{2} n r^{2} + 4 \, b e^{2} n r - b e^{2} n\right )} \log \left (x\right )\right )} x^{2 \, r} - 2 \, {\left (4 \, a d e r^{3} - b d e n - a d e - 4 \, {\left (b d e n + 2 \, a d e\right )} r^{2} + {\left (4 \, b d e n + 5 \, a d e\right )} r + {\left (4 \, b d e r^{3} - 8 \, b d e r^{2} + 5 \, b d e r - b d e\right )} \log \left (c\right ) + {\left (4 \, b d e n r^{3} - 8 \, b d e n r^{2} + 5 \, b d e n r - b d e n\right )} \log \left (x\right )\right )} x^{r} + {\left (4 \, b d^{2} r^{4} - 12 \, b d^{2} r^{3} + 13 \, b d^{2} r^{2} - 6 \, b d^{2} r + b d^{2}\right )} \log \left (c\right ) + {\left (4 \, b d^{2} n r^{4} - 12 \, b d^{2} n r^{3} + 13 \, b d^{2} n r^{2} - 6 \, b d^{2} n r + b d^{2} n\right )} \log \left (x\right )}{{\left (4 \, r^{4} - 12 \, r^{3} + 13 \, r^{2} - 6 \, r + 1\right )} x} \]
-(4*(b*d^2*n + a*d^2)*r^4 + b*d^2*n - 12*(b*d^2*n + a*d^2)*r^3 + a*d^2 + 1 3*(b*d^2*n + a*d^2)*r^2 - 6*(b*d^2*n + a*d^2)*r - (2*a*e^2*r^3 - b*e^2*n - a*e^2 - (b*e^2*n + 5*a*e^2)*r^2 + 2*(b*e^2*n + 2*a*e^2)*r + (2*b*e^2*r^3 - 5*b*e^2*r^2 + 4*b*e^2*r - b*e^2)*log(c) + (2*b*e^2*n*r^3 - 5*b*e^2*n*r^2 + 4*b*e^2*n*r - b*e^2*n)*log(x))*x^(2*r) - 2*(4*a*d*e*r^3 - b*d*e*n - a*d *e - 4*(b*d*e*n + 2*a*d*e)*r^2 + (4*b*d*e*n + 5*a*d*e)*r + (4*b*d*e*r^3 - 8*b*d*e*r^2 + 5*b*d*e*r - b*d*e)*log(c) + (4*b*d*e*n*r^3 - 8*b*d*e*n*r^2 + 5*b*d*e*n*r - b*d*e*n)*log(x))*x^r + (4*b*d^2*r^4 - 12*b*d^2*r^3 + 13*b*d ^2*r^2 - 6*b*d^2*r + b*d^2)*log(c) + (4*b*d^2*n*r^4 - 12*b*d^2*n*r^3 + 13* b*d^2*n*r^2 - 6*b*d^2*n*r + b*d^2*n)*log(x))/((4*r^4 - 12*r^3 + 13*r^2 - 6 *r + 1)*x)
Time = 3.29 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.70 \[ \int \frac {\left (d+e x^r\right )^2 \left (a+b \log \left (c x^n\right )\right )}{x^2} \, dx=- \frac {a d^{2}}{x} + 2 a d e \left (\begin {cases} \frac {x^{r}}{r x - x} & \text {for}\: r \neq 1 \\\frac {x^{r} \log {\left (x \right )}}{x} & \text {otherwise} \end {cases}\right ) + a e^{2} \left (\begin {cases} \frac {x^{2 r}}{2 r x - x} & \text {for}\: r \neq \frac {1}{2} \\\frac {x^{2 r} \log {\left (x \right )}}{x} & \text {otherwise} \end {cases}\right ) - \frac {b d^{2} n}{x} - \frac {b d^{2} \log {\left (c x^{n} \right )}}{x} - 2 b d e n \left (\begin {cases} \frac {\begin {cases} \frac {x^{r - 1}}{r - 1} & \text {for}\: r \neq 1 \\\log {\left (x \right )} & \text {otherwise} \end {cases}}{r - 1} & \text {for}\: r > -\infty \wedge r < \infty \wedge r \neq 1 \\\frac {\log {\left (x \right )}^{2}}{2} & \text {otherwise} \end {cases}\right ) + 2 b d e \left (\begin {cases} \frac {x^{r - 1}}{r - 1} & \text {for}\: r \neq 1 \\\log {\left (x \right )} & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )} - b e^{2} n \left (\begin {cases} \frac {\begin {cases} \frac {x^{2 r - 1}}{2 r - 1} & \text {for}\: r \neq \frac {1}{2} \\\log {\left (x \right )} & \text {otherwise} \end {cases}}{2 r - 1} & \text {for}\: r > -\infty \wedge r < \infty \wedge r \neq \frac {1}{2} \\\frac {\log {\left (x \right )}^{2}}{2} & \text {otherwise} \end {cases}\right ) + b e^{2} \left (\begin {cases} \frac {x^{2 r - 1}}{2 r - 1} & \text {for}\: r \neq \frac {1}{2} \\\log {\left (x \right )} & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )} \]
-a*d**2/x + 2*a*d*e*Piecewise((x**r/(r*x - x), Ne(r, 1)), (x**r*log(x)/x, True)) + a*e**2*Piecewise((x**(2*r)/(2*r*x - x), Ne(r, 1/2)), (x**(2*r)*lo g(x)/x, True)) - b*d**2*n/x - b*d**2*log(c*x**n)/x - 2*b*d*e*n*Piecewise(( Piecewise((x**(r - 1)/(r - 1), Ne(r, 1)), (log(x), True))/(r - 1), (r > -o o) & (r < oo) & Ne(r, 1)), (log(x)**2/2, True)) + 2*b*d*e*Piecewise((x**(r - 1)/(r - 1), Ne(r, 1)), (log(x), True))*log(c*x**n) - b*e**2*n*Piecewise ((Piecewise((x**(2*r - 1)/(2*r - 1), Ne(r, 1/2)), (log(x), True))/(2*r - 1 ), (r > -oo) & (r < oo) & Ne(r, 1/2)), (log(x)**2/2, True)) + b*e**2*Piece wise((x**(2*r - 1)/(2*r - 1), Ne(r, 1/2)), (log(x), True))*log(c*x**n)
Exception generated. \[ \int \frac {\left (d+e x^r\right )^2 \left (a+b \log \left (c x^n\right )\right )}{x^2} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(r-2>0)', see `assume?` for more details)Is
\[ \int \frac {\left (d+e x^r\right )^2 \left (a+b \log \left (c x^n\right )\right )}{x^2} \, dx=\int { \frac {{\left (e x^{r} + d\right )}^{2} {\left (b \log \left (c x^{n}\right ) + a\right )}}{x^{2}} \,d x } \]
Timed out. \[ \int \frac {\left (d+e x^r\right )^2 \left (a+b \log \left (c x^n\right )\right )}{x^2} \, dx=\int \frac {{\left (d+e\,x^r\right )}^2\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{x^2} \,d x \]