Integrand size = 32, antiderivative size = 560 \[ \int \frac {x^3 \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{f-g x^2} \, dx=-\frac {a n x}{2 b g}+\frac {c n x}{2 d g}+\frac {a^2 n \log (a+b x)}{2 b^2 g}-\frac {n x^2 \log (a+b x)}{2 g}-\frac {c^2 n \log (c+d x)}{2 d^2 g}+\frac {n x^2 \log (c+d x)}{2 g}+\frac {x^2 \left (n \log (a+b x)-\log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )-n \log (c+d x)\right )}{2 g}-\frac {f n \log (a+b x) \log \left (\frac {b \left (\sqrt {f}-\sqrt {g} x\right )}{b \sqrt {f}+a \sqrt {g}}\right )}{2 g^2}+\frac {f n \log (c+d x) \log \left (\frac {d \left (\sqrt {f}-\sqrt {g} x\right )}{d \sqrt {f}+c \sqrt {g}}\right )}{2 g^2}-\frac {f n \log (a+b x) \log \left (\frac {b \left (\sqrt {f}+\sqrt {g} x\right )}{b \sqrt {f}-a \sqrt {g}}\right )}{2 g^2}+\frac {f n \log (c+d x) \log \left (\frac {d \left (\sqrt {f}+\sqrt {g} x\right )}{d \sqrt {f}-c \sqrt {g}}\right )}{2 g^2}+\frac {f \left (n \log (a+b x)-\log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )-n \log (c+d x)\right ) \log \left (f-g x^2\right )}{2 g^2}-\frac {f n \operatorname {PolyLog}\left (2,-\frac {\sqrt {g} (a+b x)}{b \sqrt {f}-a \sqrt {g}}\right )}{2 g^2}-\frac {f n \operatorname {PolyLog}\left (2,\frac {\sqrt {g} (a+b x)}{b \sqrt {f}+a \sqrt {g}}\right )}{2 g^2}+\frac {f n \operatorname {PolyLog}\left (2,-\frac {\sqrt {g} (c+d x)}{d \sqrt {f}-c \sqrt {g}}\right )}{2 g^2}+\frac {f n \operatorname {PolyLog}\left (2,\frac {\sqrt {g} (c+d x)}{d \sqrt {f}+c \sqrt {g}}\right )}{2 g^2} \]
-1/2*a*n*x/b/g+1/2*c*n*x/d/g+1/2*a^2*n*ln(b*x+a)/b^2/g-1/2*n*x^2*ln(b*x+a) /g-1/2*c^2*n*ln(d*x+c)/d^2/g+1/2*n*x^2*ln(d*x+c)/g+1/2*x^2*(n*ln(b*x+a)-ln (e*((b*x+a)/(d*x+c))^n)-n*ln(d*x+c))/g+1/2*f*(n*ln(b*x+a)-ln(e*((b*x+a)/(d *x+c))^n)-n*ln(d*x+c))*ln(-g*x^2+f)/g^2-1/2*f*n*ln(b*x+a)*ln(b*(f^(1/2)-x* g^(1/2))/(b*f^(1/2)+a*g^(1/2)))/g^2+1/2*f*n*ln(d*x+c)*ln(d*(f^(1/2)-x*g^(1 /2))/(d*f^(1/2)+c*g^(1/2)))/g^2-1/2*f*n*ln(b*x+a)*ln(b*(f^(1/2)+x*g^(1/2)) /(b*f^(1/2)-a*g^(1/2)))/g^2+1/2*f*n*ln(d*x+c)*ln(d*(f^(1/2)+x*g^(1/2))/(d* f^(1/2)-c*g^(1/2)))/g^2-1/2*f*n*polylog(2,-(b*x+a)*g^(1/2)/(b*f^(1/2)-a*g^ (1/2)))/g^2-1/2*f*n*polylog(2,(b*x+a)*g^(1/2)/(b*f^(1/2)+a*g^(1/2)))/g^2+1 /2*f*n*polylog(2,-(d*x+c)*g^(1/2)/(d*f^(1/2)-c*g^(1/2)))/g^2+1/2*f*n*polyl og(2,(d*x+c)*g^(1/2)/(d*f^(1/2)+c*g^(1/2)))/g^2
Time = 0.26 (sec) , antiderivative size = 461, normalized size of antiderivative = 0.82 \[ \int \frac {x^3 \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{f-g x^2} \, dx=\frac {-g x^2 \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+\frac {g n \left (a^2 d^2 \log (a+b x)-b \left (d (-b c+a d) x+b c^2 \log (c+d x)\right )\right )}{b^2 d^2}-f \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right ) \log \left (\sqrt {f}-\sqrt {g} x\right )-f \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right ) \log \left (\sqrt {f}+\sqrt {g} x\right )+f n \left (\left (\log \left (\frac {\sqrt {g} (a+b x)}{b \sqrt {f}+a \sqrt {g}}\right )-\log \left (\frac {\sqrt {g} (c+d x)}{d \sqrt {f}+c \sqrt {g}}\right )\right ) \log \left (\sqrt {f}-\sqrt {g} x\right )+\operatorname {PolyLog}\left (2,\frac {b \left (\sqrt {f}-\sqrt {g} x\right )}{b \sqrt {f}+a \sqrt {g}}\right )-\operatorname {PolyLog}\left (2,\frac {d \left (\sqrt {f}-\sqrt {g} x\right )}{d \sqrt {f}+c \sqrt {g}}\right )\right )+f n \left (\left (\log \left (-\frac {\sqrt {g} (a+b x)}{b \sqrt {f}-a \sqrt {g}}\right )-\log \left (-\frac {\sqrt {g} (c+d x)}{d \sqrt {f}-c \sqrt {g}}\right )\right ) \log \left (\sqrt {f}+\sqrt {g} x\right )+\operatorname {PolyLog}\left (2,\frac {b \left (\sqrt {f}+\sqrt {g} x\right )}{b \sqrt {f}-a \sqrt {g}}\right )-\operatorname {PolyLog}\left (2,\frac {d \left (\sqrt {f}+\sqrt {g} x\right )}{d \sqrt {f}-c \sqrt {g}}\right )\right )}{2 g^2} \]
(-(g*x^2*Log[e*((a + b*x)/(c + d*x))^n]) + (g*n*(a^2*d^2*Log[a + b*x] - b* (d*(-(b*c) + a*d)*x + b*c^2*Log[c + d*x])))/(b^2*d^2) - f*Log[e*((a + b*x) /(c + d*x))^n]*Log[Sqrt[f] - Sqrt[g]*x] - f*Log[e*((a + b*x)/(c + d*x))^n] *Log[Sqrt[f] + Sqrt[g]*x] + f*n*((Log[(Sqrt[g]*(a + b*x))/(b*Sqrt[f] + a*S qrt[g])] - Log[(Sqrt[g]*(c + d*x))/(d*Sqrt[f] + c*Sqrt[g])])*Log[Sqrt[f] - Sqrt[g]*x] + PolyLog[2, (b*(Sqrt[f] - Sqrt[g]*x))/(b*Sqrt[f] + a*Sqrt[g]) ] - PolyLog[2, (d*(Sqrt[f] - Sqrt[g]*x))/(d*Sqrt[f] + c*Sqrt[g])]) + f*n*( (Log[-((Sqrt[g]*(a + b*x))/(b*Sqrt[f] - a*Sqrt[g]))] - Log[-((Sqrt[g]*(c + d*x))/(d*Sqrt[f] - c*Sqrt[g]))])*Log[Sqrt[f] + Sqrt[g]*x] + PolyLog[2, (b *(Sqrt[f] + Sqrt[g]*x))/(b*Sqrt[f] - a*Sqrt[g])] - PolyLog[2, (d*(Sqrt[f] + Sqrt[g]*x))/(d*Sqrt[f] - c*Sqrt[g])]))/(2*g^2)
Time = 1.09 (sec) , antiderivative size = 536, normalized size of antiderivative = 0.96, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {2993, 243, 49, 2009, 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.
| \(\displaystyle \int \frac {x^3 \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{f-g x^2} \, dx\) |
| \(\Big \downarrow \) 2993 |
| \(\displaystyle -\left (\left (-\log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+n \log (a+b x)-n \log (c+d x)\right ) \int \frac {x^3}{f-g x^2}dx\right )+n \int \frac {x^3 \log (a+b x)}{f-g x^2}dx-n \int \frac {x^3 \log (c+d x)}{f-g x^2}dx\) |
| \(\Big \downarrow \) 243 |
| \(\displaystyle -\frac {1}{2} \left (-\log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+n \log (a+b x)-n \log (c+d x)\right ) \int \frac {x^2}{f-g x^2}dx^2+n \int \frac {x^3 \log (a+b x)}{f-g x^2}dx-n \int \frac {x^3 \log (c+d x)}{f-g x^2}dx\) |
| \(\Big \downarrow \) 49 |
| \(\displaystyle -\frac {1}{2} \left (-\log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+n \log (a+b x)-n \log (c+d x)\right ) \int \left (-\frac {f}{g \left (g x^2-f\right )}-\frac {1}{g}\right )dx^2+n \int \frac {x^3 \log (a+b x)}{f-g x^2}dx-n \int \frac {x^3 \log (c+d x)}{f-g x^2}dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle n \int \frac {x^3 \log (a+b x)}{f-g x^2}dx-n \int \frac {x^3 \log (c+d x)}{f-g x^2}dx-\frac {1}{2} \left (-\frac {f \log \left (f-g x^2\right )}{g^2}-\frac {x^2}{g}\right ) \left (-\log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+n \log (a+b x)-n \log (c+d x)\right )\) |
| \(\Big \downarrow \) 2863 |
| \(\displaystyle n \int \left (\frac {f x \log (a+b x)}{g \left (f-g x^2\right )}-\frac {x \log (a+b x)}{g}\right )dx-n \int \left (\frac {f x \log (c+d x)}{g \left (f-g x^2\right )}-\frac {x \log (c+d x)}{g}\right )dx-\frac {1}{2} \left (-\frac {f \log \left (f-g x^2\right )}{g^2}-\frac {x^2}{g}\right ) \left (-\log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+n \log (a+b x)-n \log (c+d x)\right )\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle n \left (\frac {a^2 \log (a+b x)}{2 b^2 g}-\frac {f \operatorname {PolyLog}\left (2,-\frac {\sqrt {g} (a+b x)}{b \sqrt {f}-a \sqrt {g}}\right )}{2 g^2}-\frac {f \operatorname {PolyLog}\left (2,\frac {\sqrt {g} (a+b x)}{\sqrt {g} a+b \sqrt {f}}\right )}{2 g^2}-\frac {f \log (a+b x) \log \left (\frac {b \left (\sqrt {f}-\sqrt {g} x\right )}{a \sqrt {g}+b \sqrt {f}}\right )}{2 g^2}-\frac {f \log (a+b x) \log \left (\frac {b \left (\sqrt {f}+\sqrt {g} x\right )}{b \sqrt {f}-a \sqrt {g}}\right )}{2 g^2}-\frac {x^2 \log (a+b x)}{2 g}-\frac {a x}{2 b g}+\frac {x^2}{4 g}\right )-\frac {1}{2} \left (-\frac {f \log \left (f-g x^2\right )}{g^2}-\frac {x^2}{g}\right ) \left (-\log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+n \log (a+b x)-n \log (c+d x)\right )-n \left (\frac {c^2 \log (c+d x)}{2 d^2 g}-\frac {f \operatorname {PolyLog}\left (2,-\frac {\sqrt {g} (c+d x)}{d \sqrt {f}-c \sqrt {g}}\right )}{2 g^2}-\frac {f \operatorname {PolyLog}\left (2,\frac {\sqrt {g} (c+d x)}{\sqrt {g} c+d \sqrt {f}}\right )}{2 g^2}-\frac {f \log (c+d x) \log \left (\frac {d \left (\sqrt {f}-\sqrt {g} x\right )}{c \sqrt {g}+d \sqrt {f}}\right )}{2 g^2}-\frac {f \log (c+d x) \log \left (\frac {d \left (\sqrt {f}+\sqrt {g} x\right )}{d \sqrt {f}-c \sqrt {g}}\right )}{2 g^2}-\frac {x^2 \log (c+d x)}{2 g}-\frac {c x}{2 d g}+\frac {x^2}{4 g}\right )\) |
-1/2*((n*Log[a + b*x] - Log[e*((a + b*x)/(c + d*x))^n] - n*Log[c + d*x])*( -(x^2/g) - (f*Log[f - g*x^2])/g^2)) + n*(-1/2*(a*x)/(b*g) + x^2/(4*g) + (a ^2*Log[a + b*x])/(2*b^2*g) - (x^2*Log[a + b*x])/(2*g) - (f*Log[a + b*x]*Lo g[(b*(Sqrt[f] - Sqrt[g]*x))/(b*Sqrt[f] + a*Sqrt[g])])/(2*g^2) - (f*Log[a + b*x]*Log[(b*(Sqrt[f] + Sqrt[g]*x))/(b*Sqrt[f] - a*Sqrt[g])])/(2*g^2) - (f *PolyLog[2, -((Sqrt[g]*(a + b*x))/(b*Sqrt[f] - a*Sqrt[g]))])/(2*g^2) - (f* PolyLog[2, (Sqrt[g]*(a + b*x))/(b*Sqrt[f] + a*Sqrt[g])])/(2*g^2)) - n*(-1/ 2*(c*x)/(d*g) + x^2/(4*g) + (c^2*Log[c + d*x])/(2*d^2*g) - (x^2*Log[c + d* x])/(2*g) - (f*Log[c + d*x]*Log[(d*(Sqrt[f] - Sqrt[g]*x))/(d*Sqrt[f] + c*S qrt[g])])/(2*g^2) - (f*Log[c + d*x]*Log[(d*(Sqrt[f] + Sqrt[g]*x))/(d*Sqrt[ f] - c*Sqrt[g])])/(2*g^2) - (f*PolyLog[2, -((Sqrt[g]*(c + d*x))/(d*Sqrt[f] - c*Sqrt[g]))])/(2*g^2) - (f*PolyLog[2, (Sqrt[g]*(c + d*x))/(d*Sqrt[f] + c*Sqrt[g])])/(2*g^2))
3.1.76.3.1 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_)^2)^(p_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
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]
Int[Log[(e_.)*((f_.)*((a_.) + (b_.)*(x_))^(p_.)*((c_.) + (d_.)*(x_))^(q_.)) ^(r_.)]*(RFx_.), x_Symbol] :> Simp[p*r Int[RFx*Log[a + b*x], x], x] + (Si mp[q*r Int[RFx*Log[c + d*x], x], x] - Simp[(p*r*Log[a + b*x] + q*r*Log[c + d*x] - Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r]) Int[RFx, x], x]) /; FreeQ[ {a, b, c, d, e, f, p, q, r}, x] && RationalFunctionQ[RFx, x] && NeQ[b*c - a *d, 0] && !MatchQ[RFx, (u_.)*(a + b*x)^(m_.)*(c + d*x)^(n_.) /; IntegersQ[ m, n]]
Time = 1.52 (sec) , antiderivative size = 538, normalized size of antiderivative = 0.96
| method | result | size |
| parts | \(-\frac {\ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right ) x^{2}}{2 g}-\frac {\ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right ) f \ln \left (-g \,x^{2}+f \right )}{2 g^{2}}-\frac {n \left (\frac {\left (a d -c b \right ) \left (\frac {x}{b d}+\frac {c^{2} \ln \left (d x +c \right )}{d^{2} \left (a d -c b \right )}-\frac {a^{2} \ln \left (b x +a \right )}{b^{2} \left (a d -c b \right )}\right )}{g}+\frac {f \left (a d -c b \right ) \left (\frac {\left (\frac {\ln \left (d x +c \right ) \ln \left (-g \,x^{2}+f \right )}{d}+\frac {2 g \left (-\frac {\ln \left (d x +c \right ) \left (\ln \left (\frac {d \sqrt {f g}-\left (d x +c \right ) g +c g}{d \sqrt {f g}+c g}\right )+\ln \left (\frac {d \sqrt {f g}+\left (d x +c \right ) g -c g}{d \sqrt {f g}-c g}\right )\right )}{2 g}-\frac {\operatorname {dilog}\left (\frac {d \sqrt {f g}-\left (d x +c \right ) g +c g}{d \sqrt {f g}+c g}\right )+\operatorname {dilog}\left (\frac {d \sqrt {f g}+\left (d x +c \right ) g -c g}{d \sqrt {f g}-c g}\right )}{2 g}\right )}{d}\right ) d}{a d -c b}-\frac {\left (\frac {\ln \left (b x +a \right ) \ln \left (-g \,x^{2}+f \right )}{b}+\frac {2 g \left (-\frac {\ln \left (b x +a \right ) \left (\ln \left (\frac {b \sqrt {f g}-g \left (b x +a \right )+a g}{b \sqrt {f g}+a g}\right )+\ln \left (\frac {b \sqrt {f g}+g \left (b x +a \right )-a g}{b \sqrt {f g}-a g}\right )\right )}{2 g}-\frac {\operatorname {dilog}\left (\frac {b \sqrt {f g}-g \left (b x +a \right )+a g}{b \sqrt {f g}+a g}\right )+\operatorname {dilog}\left (\frac {b \sqrt {f g}+g \left (b x +a \right )-a g}{b \sqrt {f g}-a g}\right )}{2 g}\right )}{b}\right ) b}{a d -c b}\right )}{g^{2}}\right )}{2}\) | \(538\) |
-1/2*ln(e*((b*x+a)/(d*x+c))^n)*x^2/g-1/2*ln(e*((b*x+a)/(d*x+c))^n)*f/g^2*l n(-g*x^2+f)-1/2*n*((a*d-b*c)/g*(x/b/d+1/d^2*c^2/(a*d-b*c)*ln(d*x+c)-1/b^2* a^2/(a*d-b*c)*ln(b*x+a))+f*(a*d-b*c)/g^2*((ln(d*x+c)/d*ln(-g*x^2+f)+2/d*g* (-1/2*ln(d*x+c)*(ln((d*(f*g)^(1/2)-(d*x+c)*g+c*g)/(d*(f*g)^(1/2)+c*g))+ln( (d*(f*g)^(1/2)+(d*x+c)*g-c*g)/(d*(f*g)^(1/2)-c*g)))/g-1/2*(dilog((d*(f*g)^ (1/2)-(d*x+c)*g+c*g)/(d*(f*g)^(1/2)+c*g))+dilog((d*(f*g)^(1/2)+(d*x+c)*g-c *g)/(d*(f*g)^(1/2)-c*g)))/g))*d/(a*d-b*c)-(ln(b*x+a)/b*ln(-g*x^2+f)+2/b*g* (-1/2*ln(b*x+a)*(ln((b*(f*g)^(1/2)-g*(b*x+a)+a*g)/(b*(f*g)^(1/2)+a*g))+ln( (b*(f*g)^(1/2)+g*(b*x+a)-a*g)/(b*(f*g)^(1/2)-a*g)))/g-1/2*(dilog((b*(f*g)^ (1/2)-g*(b*x+a)+a*g)/(b*(f*g)^(1/2)+a*g))+dilog((b*(f*g)^(1/2)+g*(b*x+a)-a *g)/(b*(f*g)^(1/2)-a*g)))/g))*b/(a*d-b*c)))
\[ \int \frac {x^3 \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{f-g x^2} \, dx=\int { -\frac {x^{3} \log \left (e \left (\frac {b x + a}{d x + c}\right )^{n}\right )}{g x^{2} - f} \,d x } \]
Timed out. \[ \int \frac {x^3 \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{f-g x^2} \, dx=\text {Timed out} \]
\[ \int \frac {x^3 \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{f-g x^2} \, dx=\int { -\frac {x^{3} \log \left (e \left (\frac {b x + a}{d x + c}\right )^{n}\right )}{g x^{2} - f} \,d x } \]
\[ \int \frac {x^3 \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{f-g x^2} \, dx=\int { -\frac {x^{3} \log \left (e \left (\frac {b x + a}{d x + c}\right )^{n}\right )}{g x^{2} - f} \,d x } \]
Timed out. \[ \int \frac {x^3 \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{f-g x^2} \, dx=\int \frac {x^3\,\ln \left (e\,{\left (\frac {a+b\,x}{c+d\,x}\right )}^n\right )}{f-g\,x^2} \,d x \]