Integrand size = 25, antiderivative size = 281 \[ \int \frac {x^3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{f+g x} \, dx=\frac {a f^2 x}{g^3}-\frac {b f^2 n x}{g^3}-\frac {b d f n x}{2 e g^2}-\frac {b d^2 n x}{3 e^2 g}+\frac {b f n x^2}{4 g^2}+\frac {b d n x^2}{6 e g}-\frac {b n x^3}{9 g}+\frac {b d^2 f n \log (d+e x)}{2 e^2 g^2}+\frac {b d^3 n \log (d+e x)}{3 e^3 g}+\frac {b f^2 (d+e x) \log \left (c (d+e x)^n\right )}{e g^3}-\frac {f x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g^2}+\frac {x^3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{3 g}-\frac {f^3 \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e (f+g x)}{e f-d g}\right )}{g^4}-\frac {b f^3 n \operatorname {PolyLog}\left (2,-\frac {g (d+e x)}{e f-d g}\right )}{g^4} \] Output:
a*f^2*x/g^3-b*f^2*n*x/g^3-1/2*b*d*f*n*x/e/g^2-1/3*b*d^2*n*x/e^2/g+1/4*b*f* n*x^2/g^2+1/6*b*d*n*x^2/e/g-1/9*b*n*x^3/g+1/2*b*d^2*f*n*ln(e*x+d)/e^2/g^2+ 1/3*b*d^3*n*ln(e*x+d)/e^3/g+b*f^2*(e*x+d)*ln(c*(e*x+d)^n)/e/g^3-1/2*f*x^2* (a+b*ln(c*(e*x+d)^n))/g^2+1/3*x^3*(a+b*ln(c*(e*x+d)^n))/g-f^3*(a+b*ln(c*(e *x+d)^n))*ln(e*(g*x+f)/(-d*g+e*f))/g^4-b*f^3*n*polylog(2,-g*(e*x+d)/(-d*g+ e*f))/g^4
Time = 0.27 (sec) , antiderivative size = 241, normalized size of antiderivative = 0.86 \[ \int \frac {x^3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{f+g x} \, dx=\frac {6 b d^2 g^2 (3 e f+2 d g) n \log (d+e x)+e \left (g x \left (6 a e^2 \left (6 f^2-3 f g x+2 g^2 x^2\right )-b n \left (12 d^2 g^2-6 d e g (-3 f+g x)+e^2 \left (36 f^2-9 f g x+4 g^2 x^2\right )\right )\right )-36 a e^2 f^3 \log \left (\frac {e (f+g x)}{e f-d g}\right )+6 b e \log \left (c (d+e x)^n\right ) \left (6 d f^2 g+e g x \left (6 f^2-3 f g x+2 g^2 x^2\right )-6 e f^3 \log \left (\frac {e (f+g x)}{e f-d g}\right )\right )\right )-36 b e^3 f^3 n \operatorname {PolyLog}\left (2,\frac {g (d+e x)}{-e f+d g}\right )}{36 e^3 g^4} \] Input:
Integrate[(x^3*(a + b*Log[c*(d + e*x)^n]))/(f + g*x),x]
Output:
(6*b*d^2*g^2*(3*e*f + 2*d*g)*n*Log[d + e*x] + e*(g*x*(6*a*e^2*(6*f^2 - 3*f *g*x + 2*g^2*x^2) - b*n*(12*d^2*g^2 - 6*d*e*g*(-3*f + g*x) + e^2*(36*f^2 - 9*f*g*x + 4*g^2*x^2))) - 36*a*e^2*f^3*Log[(e*(f + g*x))/(e*f - d*g)] + 6* b*e*Log[c*(d + e*x)^n]*(6*d*f^2*g + e*g*x*(6*f^2 - 3*f*g*x + 2*g^2*x^2) - 6*e*f^3*Log[(e*(f + g*x))/(e*f - d*g)])) - 36*b*e^3*f^3*n*PolyLog[2, (g*(d + e*x))/(-(e*f) + d*g)])/(36*e^3*g^4)
Time = 0.93 (sec) , antiderivative size = 281, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {2863, 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.
| \(\displaystyle \int \frac {x^3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{f+g x} \, dx\) |
| \(\Big \downarrow \) 2863 |
| \(\displaystyle \int \left (-\frac {f^3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{g^3 (f+g x)}+\frac {f^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{g^3}-\frac {f x \left (a+b \log \left (c (d+e x)^n\right )\right )}{g^2}+\frac {x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{g}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {f^3 \log \left (\frac {e (f+g x)}{e f-d g}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{g^4}-\frac {f x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g^2}+\frac {x^3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{3 g}+\frac {a f^2 x}{g^3}+\frac {b f^2 (d+e x) \log \left (c (d+e x)^n\right )}{e g^3}+\frac {b d^3 n \log (d+e x)}{3 e^3 g}+\frac {b d^2 f n \log (d+e x)}{2 e^2 g^2}-\frac {b d^2 n x}{3 e^2 g}-\frac {b f^3 n \operatorname {PolyLog}\left (2,-\frac {g (d+e x)}{e f-d g}\right )}{g^4}-\frac {b d f n x}{2 e g^2}+\frac {b d n x^2}{6 e g}-\frac {b f^2 n x}{g^3}+\frac {b f n x^2}{4 g^2}-\frac {b n x^3}{9 g}\) |
Input:
Int[(x^3*(a + b*Log[c*(d + e*x)^n]))/(f + g*x),x]
Output:
(a*f^2*x)/g^3 - (b*f^2*n*x)/g^3 - (b*d*f*n*x)/(2*e*g^2) - (b*d^2*n*x)/(3*e ^2*g) + (b*f*n*x^2)/(4*g^2) + (b*d*n*x^2)/(6*e*g) - (b*n*x^3)/(9*g) + (b*d ^2*f*n*Log[d + e*x])/(2*e^2*g^2) + (b*d^3*n*Log[d + e*x])/(3*e^3*g) + (b*f ^2*(d + e*x)*Log[c*(d + e*x)^n])/(e*g^3) - (f*x^2*(a + b*Log[c*(d + e*x)^n ]))/(2*g^2) + (x^3*(a + b*Log[c*(d + e*x)^n]))/(3*g) - (f^3*(a + b*Log[c*( d + e*x)^n])*Log[(e*(f + g*x))/(e*f - d*g)])/g^4 - (b*f^3*n*PolyLog[2, -(( g*(d + e*x))/(e*f - d*g))])/g^4
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((h_.)*(x_)) ^(m_.)*((f_) + (g_.)*(x_)^(r_.))^(q_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*Log[c*(d + e*x)^n])^p, (h*x)^m*(f + g*x^r)^q, x], x] /; FreeQ[{a, b, c , d, e, f, g, h, m, n, p, q, r}, x] && IntegerQ[m] && IntegerQ[q]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 1.93 (sec) , antiderivative size = 507, normalized size of antiderivative = 1.80
| method | result | size |
| risch | \(\frac {b \ln \left (\left (e x +d \right )^{n}\right ) x^{3}}{3 g}-\frac {b \ln \left (\left (e x +d \right )^{n}\right ) f \,x^{2}}{2 g^{2}}+\frac {b \ln \left (\left (e x +d \right )^{n}\right ) x \,f^{2}}{g^{3}}-\frac {b \ln \left (\left (e x +d \right )^{n}\right ) f^{3} \ln \left (g x +f \right )}{g^{4}}-\frac {b n \,x^{3}}{9 g}+\frac {b f n \,x^{2}}{4 g^{2}}-\frac {b \,f^{2} n x}{g^{3}}-\frac {49 b n \,f^{3}}{36 g^{4}}+\frac {b d n \,x^{2}}{6 e g}-\frac {b d f n x}{2 e \,g^{2}}-\frac {2 b n d \,f^{2}}{3 e \,g^{3}}-\frac {b \,d^{2} n x}{3 e^{2} g}-\frac {b n \,d^{2} f}{3 e^{2} g^{2}}+\frac {b n \,d^{3} \ln \left (\left (g x +f \right ) e +d g -e f \right )}{3 e^{3} g}+\frac {b n \,d^{2} \ln \left (\left (g x +f \right ) e +d g -e f \right ) f}{2 e^{2} g^{2}}+\frac {b n d \ln \left (\left (g x +f \right ) e +d g -e f \right ) f^{2}}{e \,g^{3}}+\frac {b n \,f^{3} \operatorname {dilog}\left (\frac {\left (g x +f \right ) e +d g -e f}{d g -e f}\right )}{g^{4}}+\frac {b n \,f^{3} \ln \left (g x +f \right ) \ln \left (\frac {\left (g x +f \right ) e +d g -e f}{d g -e f}\right )}{g^{4}}+\left (\frac {i b \pi \,\operatorname {csgn}\left (i \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2}}{2}-\frac {i b \pi \,\operatorname {csgn}\left (i \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \right )}{2}-\frac {i b \pi \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{3}}{2}+\frac {i b \pi \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2} \operatorname {csgn}\left (i c \right )}{2}+b \ln \left (c \right )+a \right ) \left (\frac {\frac {1}{3} g^{2} x^{3}-\frac {1}{2} f g \,x^{2}+f^{2} x}{g^{3}}-\frac {f^{3} \ln \left (g x +f \right )}{g^{4}}\right )\) | \(507\) |
Input:
int(x^3*(a+b*ln(c*(e*x+d)^n))/(g*x+f),x,method=_RETURNVERBOSE)
Output:
1/3*b*ln((e*x+d)^n)/g*x^3-1/2*b*ln((e*x+d)^n)/g^2*f*x^2+b*ln((e*x+d)^n)/g^ 3*x*f^2-b*ln((e*x+d)^n)*f^3/g^4*ln(g*x+f)-1/9*b*n*x^3/g+1/4*b*f*n*x^2/g^2- b*f^2*n*x/g^3-49/36*b*n/g^4*f^3+1/6*b*d*n*x^2/e/g-1/2*b*d*f*n*x/e/g^2-2/3* b/e*n/g^3*d*f^2-1/3*b*d^2*n*x/e^2/g-1/3*b/e^2*n/g^2*d^2*f+1/3*b/e^3*n/g*d^ 3*ln((g*x+f)*e+d*g-e*f)+1/2*b/e^2*n/g^2*d^2*ln((g*x+f)*e+d*g-e*f)*f+b/e*n/ g^3*d*ln((g*x+f)*e+d*g-e*f)*f^2+b*n/g^4*f^3*dilog(((g*x+f)*e+d*g-e*f)/(d*g -e*f))+b*n/g^4*f^3*ln(g*x+f)*ln(((g*x+f)*e+d*g-e*f)/(d*g-e*f))+(1/2*I*b*Pi *csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2-1/2*I*b*Pi*csgn(I*(e*x+d)^n)*csgn (I*c*(e*x+d)^n)*csgn(I*c)-1/2*I*b*Pi*csgn(I*c*(e*x+d)^n)^3+1/2*I*b*Pi*csgn (I*c*(e*x+d)^n)^2*csgn(I*c)+b*ln(c)+a)*(1/g^3*(1/3*g^2*x^3-1/2*f*g*x^2+f^2 *x)-f^3/g^4*ln(g*x+f))
\[ \int \frac {x^3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{f+g x} \, dx=\int { \frac {{\left (b \log \left ({\left (e x + d\right )}^{n} c\right ) + a\right )} x^{3}}{g x + f} \,d x } \] Input:
integrate(x^3*(a+b*log(c*(e*x+d)^n))/(g*x+f),x, algorithm="fricas")
Output:
integral((b*x^3*log((e*x + d)^n*c) + a*x^3)/(g*x + f), x)
\[ \int \frac {x^3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{f+g x} \, dx=\int \frac {x^{3} \left (a + b \log {\left (c \left (d + e x\right )^{n} \right )}\right )}{f + g x}\, dx \] Input:
integrate(x**3*(a+b*ln(c*(e*x+d)**n))/(g*x+f),x)
Output:
Integral(x**3*(a + b*log(c*(d + e*x)**n))/(f + g*x), x)
\[ \int \frac {x^3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{f+g x} \, dx=\int { \frac {{\left (b \log \left ({\left (e x + d\right )}^{n} c\right ) + a\right )} x^{3}}{g x + f} \,d x } \] Input:
integrate(x^3*(a+b*log(c*(e*x+d)^n))/(g*x+f),x, algorithm="maxima")
Output:
-1/6*a*(6*f^3*log(g*x + f)/g^4 - (2*g^2*x^3 - 3*f*g*x^2 + 6*f^2*x)/g^3) + b*integrate((x^3*log((e*x + d)^n) + x^3*log(c))/(g*x + f), x)
\[ \int \frac {x^3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{f+g x} \, dx=\int { \frac {{\left (b \log \left ({\left (e x + d\right )}^{n} c\right ) + a\right )} x^{3}}{g x + f} \,d x } \] Input:
integrate(x^3*(a+b*log(c*(e*x+d)^n))/(g*x+f),x, algorithm="giac")
Output:
integrate((b*log((e*x + d)^n*c) + a)*x^3/(g*x + f), x)
Timed out. \[ \int \frac {x^3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{f+g x} \, dx=\int \frac {x^3\,\left (a+b\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )\right )}{f+g\,x} \,d x \] Input:
int((x^3*(a + b*log(c*(d + e*x)^n)))/(f + g*x),x)
Output:
int((x^3*(a + b*log(c*(d + e*x)^n)))/(f + g*x), x)
\[ \int \frac {x^3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{f+g x} \, dx=\frac {-36 \left (\int \frac {\mathrm {log}\left (\left (e x +d \right )^{n} c \right )}{e g \,x^{2}+d g x +e f x +d f}d x \right ) b d \,e^{3} f^{3} g n +36 \left (\int \frac {\mathrm {log}\left (\left (e x +d \right )^{n} c \right )}{e g \,x^{2}+d g x +e f x +d f}d x \right ) b \,e^{4} f^{4} n -36 \,\mathrm {log}\left (g x +f \right ) a \,e^{3} f^{3} n -18 \mathrm {log}\left (\left (e x +d \right )^{n} c \right )^{2} b \,e^{3} f^{3}+12 \,\mathrm {log}\left (\left (e x +d \right )^{n} c \right ) b \,d^{3} g^{3} n +18 \,\mathrm {log}\left (\left (e x +d \right )^{n} c \right ) b \,d^{2} e f \,g^{2} n +36 \,\mathrm {log}\left (\left (e x +d \right )^{n} c \right ) b d \,e^{2} f^{2} g n +36 \,\mathrm {log}\left (\left (e x +d \right )^{n} c \right ) b \,e^{3} f^{2} g n x -18 \,\mathrm {log}\left (\left (e x +d \right )^{n} c \right ) b \,e^{3} f \,g^{2} n \,x^{2}+12 \,\mathrm {log}\left (\left (e x +d \right )^{n} c \right ) b \,e^{3} g^{3} n \,x^{3}+36 a \,e^{3} f^{2} g n x -18 a \,e^{3} f \,g^{2} n \,x^{2}+12 a \,e^{3} g^{3} n \,x^{3}-12 b \,d^{2} e \,g^{3} n^{2} x -18 b d \,e^{2} f \,g^{2} n^{2} x +6 b d \,e^{2} g^{3} n^{2} x^{2}-36 b \,e^{3} f^{2} g \,n^{2} x +9 b \,e^{3} f \,g^{2} n^{2} x^{2}-4 b \,e^{3} g^{3} n^{2} x^{3}}{36 e^{3} g^{4} n} \] Input:
int(x^3*(a+b*log(c*(e*x+d)^n))/(g*x+f),x)
Output:
( - 36*int(log((d + e*x)**n*c)/(d*f + d*g*x + e*f*x + e*g*x**2),x)*b*d*e** 3*f**3*g*n + 36*int(log((d + e*x)**n*c)/(d*f + d*g*x + e*f*x + e*g*x**2),x )*b*e**4*f**4*n - 36*log(f + g*x)*a*e**3*f**3*n - 18*log((d + e*x)**n*c)** 2*b*e**3*f**3 + 12*log((d + e*x)**n*c)*b*d**3*g**3*n + 18*log((d + e*x)**n *c)*b*d**2*e*f*g**2*n + 36*log((d + e*x)**n*c)*b*d*e**2*f**2*g*n + 36*log( (d + e*x)**n*c)*b*e**3*f**2*g*n*x - 18*log((d + e*x)**n*c)*b*e**3*f*g**2*n *x**2 + 12*log((d + e*x)**n*c)*b*e**3*g**3*n*x**3 + 36*a*e**3*f**2*g*n*x - 18*a*e**3*f*g**2*n*x**2 + 12*a*e**3*g**3*n*x**3 - 12*b*d**2*e*g**3*n**2*x - 18*b*d*e**2*f*g**2*n**2*x + 6*b*d*e**2*g**3*n**2*x**2 - 36*b*e**3*f**2* g*n**2*x + 9*b*e**3*f*g**2*n**2*x**2 - 4*b*e**3*g**3*n**2*x**3)/(36*e**3*g **4*n)