Integrand size = 27, antiderivative size = 142 \[ \int (f x)^{-1+m} \left (d+e x^m\right )^2 \left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {b d^2 n x (f x)^{-1+m}}{m^2}-\frac {b d e n x^{1+m} (f x)^{-1+m}}{2 m^2}-\frac {b e^2 n x^{1+2 m} (f x)^{-1+m}}{9 m^2}-\frac {b d^3 n x^{1-m} (f x)^{-1+m} \log (x)}{3 e m}+\frac {x^{1-m} (f x)^{-1+m} \left (d+e x^m\right )^3 \left (a+b \log \left (c x^n\right )\right )}{3 e m} \]
-b*d^2*n*x*(f*x)^(-1+m)/m^2-1/2*b*d*e*n*x^(1+m)*(f*x)^(-1+m)/m^2-1/9*b*e^2 *n*x^(1+2*m)*(f*x)^(-1+m)/m^2-1/3*b*d^3*n*x^(1-m)*(f*x)^(-1+m)*ln(x)/e/m+1 /3*x^(1-m)*(f*x)^(-1+m)*(d+e*x^m)^3*(a+b*ln(c*x^n))/e/m
Time = 0.08 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.71 \[ \int (f x)^{-1+m} \left (d+e x^m\right )^2 \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {(f x)^m \left (6 a m \left (3 d^2+3 d e x^m+e^2 x^{2 m}\right )-b n \left (18 d^2+9 d e x^m+2 e^2 x^{2 m}\right )+6 b m \left (3 d^2+3 d e x^m+e^2 x^{2 m}\right ) \log \left (c x^n\right )\right )}{18 f m^2} \]
((f*x)^m*(6*a*m*(3*d^2 + 3*d*e*x^m + e^2*x^(2*m)) - b*n*(18*d^2 + 9*d*e*x^ m + 2*e^2*x^(2*m)) + 6*b*m*(3*d^2 + 3*d*e*x^m + e^2*x^(2*m))*Log[c*x^n]))/ (18*f*m^2)
Time = 0.39 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.70, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {2777, 2776, 798, 49, 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.
| \(\displaystyle \int (f x)^{m-1} \left (d+e x^m\right )^2 \left (a+b \log \left (c x^n\right )\right ) \, dx\) |
| \(\Big \downarrow \) 2777 |
| \(\displaystyle x^{1-m} (f x)^{m-1} \int x^{m-1} \left (e x^m+d\right )^2 \left (a+b \log \left (c x^n\right )\right )dx\) |
| \(\Big \downarrow \) 2776 |
| \(\displaystyle x^{1-m} (f x)^{m-1} \left (\frac {\left (d+e x^m\right )^3 \left (a+b \log \left (c x^n\right )\right )}{3 e m}-\frac {b n \int \frac {\left (e x^m+d\right )^3}{x}dx}{3 e m}\right )\) |
| \(\Big \downarrow \) 798 |
| \(\displaystyle x^{1-m} (f x)^{m-1} \left (\frac {\left (d+e x^m\right )^3 \left (a+b \log \left (c x^n\right )\right )}{3 e m}-\frac {b n \int x^{-m} \left (e x^m+d\right )^3dx^m}{3 e m^2}\right )\) |
| \(\Big \downarrow \) 49 |
| \(\displaystyle x^{1-m} (f x)^{m-1} \left (\frac {\left (d+e x^m\right )^3 \left (a+b \log \left (c x^n\right )\right )}{3 e m}-\frac {b n \int \left (d^3 x^{-m}+3 d e^2 x^m+e^3 x^{2 m}+3 d^2 e\right )dx^m}{3 e m^2}\right )\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle x^{1-m} (f x)^{m-1} \left (\frac {\left (d+e x^m\right )^3 \left (a+b \log \left (c x^n\right )\right )}{3 e m}-\frac {b n \left (d^3 \log \left (x^m\right )+3 d^2 e x^m+\frac {3}{2} d e^2 x^{2 m}+\frac {1}{3} e^3 x^{3 m}\right )}{3 e m^2}\right )\) |
x^(1 - m)*(f*x)^(-1 + m)*(-1/3*(b*n*(3*d^2*e*x^m + (3*d*e^2*x^(2*m))/2 + ( e^3*x^(3*m))/3 + d^3*Log[x^m]))/(e*m^2) + ((d + e*x^m)^3*(a + b*Log[c*x^n] ))/(3*e*m))
3.4.52.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[1/n Subst [Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_))^(q_.), x_Symbol] :> Simp[f^m*(d + e*x^r)^(q + 1)*((a + b*L og[c*x^n])^p/(e*r*(q + 1))), x] - Simp[b*f^m*n*(p/(e*r*(q + 1))) Int[(d + e*x^r)^(q + 1)*((a + b*Log[c*x^n])^(p - 1)/x), x], x] /; FreeQ[{a, b, c, d , e, f, m, n, q, r}, x] && EqQ[m, r - 1] && IGtQ[p, 0] && (IntegerQ[m] || G tQ[f, 0]) && NeQ[r, n] && NeQ[q, -1]
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((f_)*(x_))^(m_.)*((d_) + ( e_.)*(x_)^(r_))^(q_.), x_Symbol] :> Simp[(f*x)^m/x^m Int[x^m*(d + e*x^r)^ q*(a + b*Log[c*x^n])^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, q, r}, x] && EqQ[m, r - 1] && IGtQ[p, 0] && !(IntegerQ[m] || GtQ[f, 0])
Time = 12.18 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.27
| method | result | size |
| parallelrisch | \(-\frac {-6 e^{2} b \left (f x \right )^{m -1} \ln \left (c \,x^{n}\right ) x^{2 m} x m -6 x \,x^{2 m} \left (f x \right )^{m -1} a \,e^{2} m +2 x \,x^{2 m} \left (f x \right )^{m -1} b \,e^{2} n -18 b d e \left (f x \right )^{m -1} \ln \left (c \,x^{n}\right ) x^{m} x m -18 x \,x^{m} \left (f x \right )^{m -1} a d e m +9 x \,x^{m} \left (f x \right )^{m -1} b d e n -18 b \,d^{2} \left (f x \right )^{m -1} \ln \left (c \,x^{n}\right ) x m -18 x \left (f x \right )^{m -1} a \,d^{2} m +18 x \left (f x \right )^{m -1} b \,d^{2} n}{18 m^{2}}\) | \(181\) |
| risch | \(\frac {b \left (e^{2} x^{2 m}+3 d e \,x^{m}+3 d^{2}\right ) x \,{\mathrm e}^{\frac {\left (m -1\right ) \left (-i \pi \operatorname {csgn}\left (i f x \right )^{3}+i \pi \operatorname {csgn}\left (i f x \right )^{2} \operatorname {csgn}\left (i f \right )+i \pi \operatorname {csgn}\left (i f x \right )^{2} \operatorname {csgn}\left (i x \right )-i \pi \,\operatorname {csgn}\left (i f x \right ) \operatorname {csgn}\left (i f \right ) \operatorname {csgn}\left (i x \right )+2 \ln \left (x \right )+2 \ln \left (f \right )\right )}{2}} \ln \left (x^{n}\right )}{3 m}+\frac {\left (-9 i \pi b d e \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) x^{m} m +9 i \pi b d e \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2} x^{m} m -3 i \pi b \,e^{2} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) x^{2 m} m -9 i \pi b d e \operatorname {csgn}\left (i c \,x^{n}\right )^{3} x^{m} m -3 i \pi b \,e^{2} \operatorname {csgn}\left (i c \,x^{n}\right )^{3} x^{2 m} m -9 i \pi b \,d^{2} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) m +9 i \pi b \,d^{2} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2} m +3 i \pi b \,e^{2} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2} x^{2 m} m +3 i \pi b \,e^{2} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2} x^{2 m} m -9 i \pi b \,d^{2} \operatorname {csgn}\left (i c \,x^{n}\right )^{3} m +9 i \pi b d e \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2} x^{m} m +9 i \pi b \,d^{2} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2} m +6 \ln \left (c \right ) b \,e^{2} x^{2 m} m +18 \ln \left (c \right ) b d e \,x^{m} m +6 a \,e^{2} x^{2 m} m -2 b \,e^{2} n \,x^{2 m}+18 \ln \left (c \right ) b \,d^{2} m +18 a d e \,x^{m} m -9 b d e n \,x^{m}+18 a \,d^{2} m -18 b \,d^{2} n \right ) x \,{\mathrm e}^{\frac {\left (m -1\right ) \left (-i \pi \operatorname {csgn}\left (i f x \right )^{3}+i \pi \operatorname {csgn}\left (i f x \right )^{2} \operatorname {csgn}\left (i f \right )+i \pi \operatorname {csgn}\left (i f x \right )^{2} \operatorname {csgn}\left (i x \right )-i \pi \,\operatorname {csgn}\left (i f x \right ) \operatorname {csgn}\left (i f \right ) \operatorname {csgn}\left (i x \right )+2 \ln \left (x \right )+2 \ln \left (f \right )\right )}{2}}}{18 m^{2}}\) | \(616\) |
-1/18*(-6*e^2*b*(f*x)^(m-1)*ln(c*x^n)*(x^m)^2*x*m-6*x*(x^m)^2*(f*x)^(m-1)* a*e^2*m+2*x*(x^m)^2*(f*x)^(m-1)*b*e^2*n-18*b*d*e*(f*x)^(m-1)*ln(c*x^n)*x^m *x*m-18*x*x^m*(f*x)^(m-1)*a*d*e*m+9*x*x^m*(f*x)^(m-1)*b*d*e*n-18*b*d^2*(f* x)^(m-1)*ln(c*x^n)*x*m-18*x*(f*x)^(m-1)*a*d^2*m+18*x*(f*x)^(m-1)*b*d^2*n)/ m^2
Time = 0.29 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.95 \[ \int (f x)^{-1+m} \left (d+e x^m\right )^2 \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {2 \, {\left (3 \, b e^{2} m n \log \left (x\right ) + 3 \, b e^{2} m \log \left (c\right ) + 3 \, a e^{2} m - b e^{2} n\right )} f^{m - 1} x^{3 \, m} + 9 \, {\left (2 \, b d e m n \log \left (x\right ) + 2 \, b d e m \log \left (c\right ) + 2 \, a d e m - b d e n\right )} f^{m - 1} x^{2 \, m} + 18 \, {\left (b d^{2} m n \log \left (x\right ) + b d^{2} m \log \left (c\right ) + a d^{2} m - b d^{2} n\right )} f^{m - 1} x^{m}}{18 \, m^{2}} \]
1/18*(2*(3*b*e^2*m*n*log(x) + 3*b*e^2*m*log(c) + 3*a*e^2*m - b*e^2*n)*f^(m - 1)*x^(3*m) + 9*(2*b*d*e*m*n*log(x) + 2*b*d*e*m*log(c) + 2*a*d*e*m - b*d *e*n)*f^(m - 1)*x^(2*m) + 18*(b*d^2*m*n*log(x) + b*d^2*m*log(c) + a*d^2*m - b*d^2*n)*f^(m - 1)*x^m)/m^2
Time = 4.90 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.70 \[ \int (f x)^{-1+m} \left (d+e x^m\right )^2 \left (a+b \log \left (c x^n\right )\right ) \, dx=\begin {cases} \frac {a d^{2} x \left (f x\right )^{m - 1}}{m} + \frac {a d e x x^{m} \left (f x\right )^{m - 1}}{m} + \frac {a e^{2} x x^{2 m} \left (f x\right )^{m - 1}}{3 m} + \frac {b d^{2} x \left (f x\right )^{m - 1} \log {\left (c x^{n} \right )}}{m} - \frac {b d^{2} n x \left (f x\right )^{m - 1}}{m^{2}} + \frac {b d e x x^{m} \left (f x\right )^{m - 1} \log {\left (c x^{n} \right )}}{m} - \frac {b d e n x x^{m} \left (f x\right )^{m - 1}}{2 m^{2}} + \frac {b e^{2} x x^{2 m} \left (f x\right )^{m - 1} \log {\left (c x^{n} \right )}}{3 m} - \frac {b e^{2} n x x^{2 m} \left (f x\right )^{m - 1}}{9 m^{2}} & \text {for}\: m \neq 0 \\\frac {\left (d + e\right )^{2} \left (\begin {cases} a \log {\left (x \right )} & \text {for}\: b = 0 \\- \left (- a - b \log {\left (c \right )}\right ) \log {\left (x \right )} & \text {for}\: n = 0 \\\frac {\left (- a - b \log {\left (c x^{n} \right )}\right )^{2}}{2 b n} & \text {otherwise} \end {cases}\right )}{f} & \text {otherwise} \end {cases} \]
Piecewise((a*d**2*x*(f*x)**(m - 1)/m + a*d*e*x*x**m*(f*x)**(m - 1)/m + a*e **2*x*x**(2*m)*(f*x)**(m - 1)/(3*m) + b*d**2*x*(f*x)**(m - 1)*log(c*x**n)/ m - b*d**2*n*x*(f*x)**(m - 1)/m**2 + b*d*e*x*x**m*(f*x)**(m - 1)*log(c*x** n)/m - b*d*e*n*x*x**m*(f*x)**(m - 1)/(2*m**2) + b*e**2*x*x**(2*m)*(f*x)**( m - 1)*log(c*x**n)/(3*m) - b*e**2*n*x*x**(2*m)*(f*x)**(m - 1)/(9*m**2), Ne (m, 0)), ((d + e)**2*Piecewise((a*log(x), Eq(b, 0)), (-(-a - b*log(c))*log (x), Eq(n, 0)), ((-a - b*log(c*x**n))**2/(2*b*n), True))/f, True))
Time = 0.20 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.27 \[ \int (f x)^{-1+m} \left (d+e x^m\right )^2 \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {b e^{2} f^{m - 1} x^{3 \, m} \log \left (c x^{n}\right )}{3 \, m} + \frac {b d e f^{m - 1} x^{2 \, m} \log \left (c x^{n}\right )}{m} + \frac {a e^{2} f^{m - 1} x^{3 \, m}}{3 \, m} - \frac {b e^{2} f^{m - 1} n x^{3 \, m}}{9 \, m^{2}} + \frac {a d e f^{m - 1} x^{2 \, m}}{m} - \frac {b d e f^{m - 1} n x^{2 \, m}}{2 \, m^{2}} - \frac {b d^{2} f^{m - 1} n x^{m}}{m^{2}} + \frac {\left (f x\right )^{m} b d^{2} \log \left (c x^{n}\right )}{f m} + \frac {\left (f x\right )^{m} a d^{2}}{f m} \]
1/3*b*e^2*f^(m - 1)*x^(3*m)*log(c*x^n)/m + b*d*e*f^(m - 1)*x^(2*m)*log(c*x ^n)/m + 1/3*a*e^2*f^(m - 1)*x^(3*m)/m - 1/9*b*e^2*f^(m - 1)*n*x^(3*m)/m^2 + a*d*e*f^(m - 1)*x^(2*m)/m - 1/2*b*d*e*f^(m - 1)*n*x^(2*m)/m^2 - b*d^2*f^ (m - 1)*n*x^m/m^2 + (f*x)^m*b*d^2*log(c*x^n)/(f*m) + (f*x)^m*a*d^2/(f*m)
Time = 0.42 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.70 \[ \int (f x)^{-1+m} \left (d+e x^m\right )^2 \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {b e^{2} f^{m} n x^{3 \, m} \log \left (x\right )}{3 \, f m} + \frac {b d e f^{m} n x^{2 \, m} \log \left (x\right )}{f m} + \frac {b d^{2} f^{m} n x^{m} \log \left (x\right )}{f m} + \frac {b e^{2} f^{m} x^{3 \, m} \log \left (c\right )}{3 \, f m} + \frac {b d e f^{m} x^{2 \, m} \log \left (c\right )}{f m} + \frac {b d^{2} f^{m} x^{m} \log \left (c\right )}{f m} + \frac {a e^{2} f^{m} x^{3 \, m}}{3 \, f m} - \frac {b e^{2} f^{m} n x^{3 \, m}}{9 \, f m^{2}} + \frac {a d e f^{m} x^{2 \, m}}{f m} - \frac {b d e f^{m} n x^{2 \, m}}{2 \, f m^{2}} + \frac {a d^{2} f^{m} x^{m}}{f m} - \frac {b d^{2} f^{m} n x^{m}}{f m^{2}} \]
1/3*b*e^2*f^m*n*x^(3*m)*log(x)/(f*m) + b*d*e*f^m*n*x^(2*m)*log(x)/(f*m) + b*d^2*f^m*n*x^m*log(x)/(f*m) + 1/3*b*e^2*f^m*x^(3*m)*log(c)/(f*m) + b*d*e* f^m*x^(2*m)*log(c)/(f*m) + b*d^2*f^m*x^m*log(c)/(f*m) + 1/3*a*e^2*f^m*x^(3 *m)/(f*m) - 1/9*b*e^2*f^m*n*x^(3*m)/(f*m^2) + a*d*e*f^m*x^(2*m)/(f*m) - 1/ 2*b*d*e*f^m*n*x^(2*m)/(f*m^2) + a*d^2*f^m*x^m/(f*m) - b*d^2*f^m*n*x^m/(f*m ^2)
Timed out. \[ \int (f x)^{-1+m} \left (d+e x^m\right )^2 \left (a+b \log \left (c x^n\right )\right ) \, dx=\int {\left (f\,x\right )}^{m-1}\,{\left (d+e\,x^m\right )}^2\,\left (a+b\,\ln \left (c\,x^n\right )\right ) \,d x \]