Integrand size = 30, antiderivative size = 388 \[ \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{(f+g x)^5} \, dx=-\frac {B (b c-a d) n}{12 (b f-a g) (d f-c g) (f+g x)^3}-\frac {B (b c-a d) (2 b d f-b c g-a d g) n}{8 (b f-a g)^2 (d f-c g)^2 (f+g x)^2}-\frac {B (b c-a d) \left (a^2 d^2 g^2-a b d g (3 d f-c g)+b^2 \left (3 d^2 f^2-3 c d f g+c^2 g^2\right )\right ) n}{4 (b f-a g)^3 (d f-c g)^3 (f+g x)}+\frac {b^4 B n \log (a+b x)}{4 g (b f-a g)^4}-\frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{4 g (f+g x)^4}-\frac {B d^4 n \log (c+d x)}{4 g (d f-c g)^4}-\frac {B (b c-a d) (2 b d f-b c g-a d g) \left (2 a b d^2 f g-a^2 d^2 g^2-b^2 \left (2 d^2 f^2-2 c d f g+c^2 g^2\right )\right ) n \log (f+g x)}{4 (b f-a g)^4 (d f-c g)^4} \]
-1/12*B*(-a*d+b*c)*n/(-a*g+b*f)/(-c*g+d*f)/(g*x+f)^3-1/8*B*(-a*d+b*c)*(-a* d*g-b*c*g+2*b*d*f)*n/(-a*g+b*f)^2/(-c*g+d*f)^2/(g*x+f)^2-1/4*B*(-a*d+b*c)* (a^2*d^2*g^2-a*b*d*g*(-c*g+3*d*f)+b^2*(c^2*g^2-3*c*d*f*g+3*d^2*f^2))*n/(-a *g+b*f)^3/(-c*g+d*f)^3/(g*x+f)+1/4*b^4*B*n*ln(b*x+a)/g/(-a*g+b*f)^4+1/4*(- A-B*ln(e*((b*x+a)/(d*x+c))^n))/g/(g*x+f)^4-1/4*B*d^4*n*ln(d*x+c)/g/(-c*g+d *f)^4-1/4*B*(-a*d+b*c)*(-a*d*g-b*c*g+2*b*d*f)*(2*a*b*d^2*f*g-a^2*d^2*g^2-b ^2*(c^2*g^2-2*c*d*f*g+2*d^2*f^2))*n*ln(g*x+f)/(-a*g+b*f)^4/(-c*g+d*f)^4
Time = 0.61 (sec) , antiderivative size = 359, normalized size of antiderivative = 0.93 \[ \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{(f+g x)^5} \, dx=\frac {-\frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{(f+g x)^4}+B (b c-a d) n \left (-\frac {g}{3 (b f-a g) (d f-c g) (f+g x)^3}+\frac {g (-2 b d f+b c g+a d g)}{2 (b f-a g)^2 (d f-c g)^2 (f+g x)^2}-\frac {g \left (a^2 d^2 g^2+a b d g (-3 d f+c g)+b^2 \left (3 d^2 f^2-3 c d f g+c^2 g^2\right )\right )}{(b f-a g)^3 (d f-c g)^3 (f+g x)}+\frac {b^4 \log (a+b x)}{(b c-a d) (b f-a g)^4}-\frac {d^4 \log (c+d x)}{(b c-a d) (d f-c g)^4}-\frac {g (-2 b d f+b c g+a d g) \left (-2 a b d^2 f g+a^2 d^2 g^2+b^2 \left (2 d^2 f^2-2 c d f g+c^2 g^2\right )\right ) \log (f+g x)}{(b f-a g)^4 (d f-c g)^4}\right )}{4 g} \]
(-((A + B*Log[e*((a + b*x)/(c + d*x))^n])/(f + g*x)^4) + B*(b*c - a*d)*n*( -1/3*g/((b*f - a*g)*(d*f - c*g)*(f + g*x)^3) + (g*(-2*b*d*f + b*c*g + a*d* g))/(2*(b*f - a*g)^2*(d*f - c*g)^2*(f + g*x)^2) - (g*(a^2*d^2*g^2 + a*b*d* g*(-3*d*f + c*g) + b^2*(3*d^2*f^2 - 3*c*d*f*g + c^2*g^2)))/((b*f - a*g)^3* (d*f - c*g)^3*(f + g*x)) + (b^4*Log[a + b*x])/((b*c - a*d)*(b*f - a*g)^4) - (d^4*Log[c + d*x])/((b*c - a*d)*(d*f - c*g)^4) - (g*(-2*b*d*f + b*c*g + a*d*g)*(-2*a*b*d^2*f*g + a^2*d^2*g^2 + b^2*(2*d^2*f^2 - 2*c*d*f*g + c^2*g^ 2))*Log[f + g*x])/((b*f - a*g)^4*(d*f - c*g)^4)))/(4*g)
Time = 0.76 (sec) , antiderivative size = 371, normalized size of antiderivative = 0.96, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {2947, 93, 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.
\(\displaystyle \int \frac {B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A}{(f+g x)^5} \, dx\) |
\(\Big \downarrow \) 2947 |
\(\displaystyle \frac {B n (b c-a d) \int \frac {1}{(a+b x) (c+d x) (f+g x)^4}dx}{4 g}-\frac {B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A}{4 g (f+g x)^4}\) |
\(\Big \downarrow \) 93 |
\(\displaystyle \frac {B n (b c-a d) \int \left (\frac {b^5}{(b c-a d) (b f-a g)^4 (a+b x)}-\frac {d^5}{(b c-a d) (c g-d f)^4 (c+d x)}+\frac {g^2 (2 b d f-b c g-a d g) \left (2 d^2 f^2 b^2+c^2 g^2 b^2-2 c d f g b^2-2 a d^2 f g b+a^2 d^2 g^2\right )}{(b f-a g)^4 (d f-c g)^4 (f+g x)}+\frac {g^2 \left (\left (3 d^2 f^2-3 c d g f+c^2 g^2\right ) b^2-a d g (3 d f-c g) b+a^2 d^2 g^2\right )}{(b f-a g)^3 (d f-c g)^3 (f+g x)^2}-\frac {g^2 (-2 b d f+b c g+a d g)}{(b f-a g)^2 (d f-c g)^2 (f+g x)^3}+\frac {g^2}{(b f-a g) (d f-c g) (f+g x)^4}\right )dx}{4 g}-\frac {B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A}{4 g (f+g x)^4}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {B n (b c-a d) \left (-\frac {g \left (a^2 d^2 g^2-a b d g (3 d f-c g)+b^2 \left (c^2 g^2-3 c d f g+3 d^2 f^2\right )\right )}{(f+g x) (b f-a g)^3 (d f-c g)^3}-\frac {g \log (f+g x) (-a d g-b c g+2 b d f) \left (-a^2 d^2 g^2+2 a b d^2 f g-\left (b^2 \left (c^2 g^2-2 c d f g+2 d^2 f^2\right )\right )\right )}{(b f-a g)^4 (d f-c g)^4}+\frac {b^4 \log (a+b x)}{(b c-a d) (b f-a g)^4}-\frac {d^4 \log (c+d x)}{(b c-a d) (d f-c g)^4}-\frac {g (-a d g-b c g+2 b d f)}{2 (f+g x)^2 (b f-a g)^2 (d f-c g)^2}-\frac {g}{3 (f+g x)^3 (b f-a g) (d f-c g)}\right )}{4 g}-\frac {B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A}{4 g (f+g x)^4}\) |
-1/4*(A + B*Log[e*((a + b*x)/(c + d*x))^n])/(g*(f + g*x)^4) + (B*(b*c - a* d)*n*(-1/3*g/((b*f - a*g)*(d*f - c*g)*(f + g*x)^3) - (g*(2*b*d*f - b*c*g - a*d*g))/(2*(b*f - a*g)^2*(d*f - c*g)^2*(f + g*x)^2) - (g*(a^2*d^2*g^2 - a *b*d*g*(3*d*f - c*g) + b^2*(3*d^2*f^2 - 3*c*d*f*g + c^2*g^2)))/((b*f - a*g )^3*(d*f - c*g)^3*(f + g*x)) + (b^4*Log[a + b*x])/((b*c - a*d)*(b*f - a*g) ^4) - (d^4*Log[c + d*x])/((b*c - a*d)*(d*f - c*g)^4) - (g*(2*b*d*f - b*c*g - a*d*g)*(2*a*b*d^2*f*g - a^2*d^2*g^2 - b^2*(2*d^2*f^2 - 2*c*d*f*g + c^2* g^2))*Log[f + g*x])/((b*f - a*g)^4*(d*f - c*g)^4)))/(4*g)
3.1.66.3.1 Defintions of rubi rules used
Int[((e_.) + (f_.)*(x_))^(p_)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_] :> Int[ExpandIntegrand[(e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; Fre eQ[{a, b, c, d, e, f}, x] && IntegerQ[p]
Int[((A_.) + Log[(e_.)*(((a_.) + (b_.)*(x_))/((c_.) + (d_.)*(x_)))^(n_.)]*( B_.))*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> Simp[(f + g*x)^(m + 1)*((A + B*Log[e*((a + b*x)/(c + d*x))^n])/(g*(m + 1))), x] - Simp[B*n*((b*c - a*d) /(g*(m + 1))) Int[(f + g*x)^(m + 1)/((a + b*x)*(c + d*x)), x], x] /; Free Q[{a, b, c, d, e, f, g, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, -2]
Leaf count of result is larger than twice the leaf count of optimal. \(5205\) vs. \(2(377)=754\).
Time = 256.76 (sec) , antiderivative size = 5206, normalized size of antiderivative = 13.42
Timed out. \[ \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{(f+g x)^5} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{(f+g x)^5} \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 1761 vs. \(2 (374) = 748\).
Time = 0.31 (sec) , antiderivative size = 1761, normalized size of antiderivative = 4.54 \[ \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{(f+g x)^5} \, dx=\text {Too large to display} \]
1/24*(6*b^4*log(b*x + a)/(b^4*f^4*g - 4*a*b^3*f^3*g^2 + 6*a^2*b^2*f^2*g^3 - 4*a^3*b*f*g^4 + a^4*g^5) - 6*d^4*log(d*x + c)/(d^4*f^4*g - 4*c*d^3*f^3*g ^2 + 6*c^2*d^2*f^2*g^3 - 4*c^3*d*f*g^4 + c^4*g^5) + 6*(4*(b^4*c*d^3 - a*b^ 3*d^4)*f^3 - 6*(b^4*c^2*d^2 - a^2*b^2*d^4)*f^2*g + 4*(b^4*c^3*d - a^3*b*d^ 4)*f*g^2 - (b^4*c^4 - a^4*d^4)*g^3)*log(g*x + f)/(b^4*d^4*f^8 + a^4*c^4*g^ 8 - 4*(b^4*c*d^3 + a*b^3*d^4)*f^7*g + 2*(3*b^4*c^2*d^2 + 8*a*b^3*c*d^3 + 3 *a^2*b^2*d^4)*f^6*g^2 - 4*(b^4*c^3*d + 6*a*b^3*c^2*d^2 + 6*a^2*b^2*c*d^3 + a^3*b*d^4)*f^5*g^3 + (b^4*c^4 + 16*a*b^3*c^3*d + 36*a^2*b^2*c^2*d^2 + 16* a^3*b*c*d^3 + a^4*d^4)*f^4*g^4 - 4*(a*b^3*c^4 + 6*a^2*b^2*c^3*d + 6*a^3*b* c^2*d^2 + a^4*c*d^3)*f^3*g^5 + 2*(3*a^2*b^2*c^4 + 8*a^3*b*c^3*d + 3*a^4*c^ 2*d^2)*f^2*g^6 - 4*(a^3*b*c^4 + a^4*c^3*d)*f*g^7) - (26*(b^3*c*d^2 - a*b^2 *d^3)*f^4 - 31*(b^3*c^2*d - a^2*b*d^3)*f^3*g + (11*b^3*c^3 + 15*a*b^2*c^2* d - 15*a^2*b*c*d^2 - 11*a^3*d^3)*f^2*g^2 - 7*(a*b^2*c^3 - a^3*c*d^2)*f*g^3 + 2*(a^2*b*c^3 - a^3*c^2*d)*g^4 + 6*(3*(b^3*c*d^2 - a*b^2*d^3)*f^2*g^2 - 3*(b^3*c^2*d - a^2*b*d^3)*f*g^3 + (b^3*c^3 - a^3*d^3)*g^4)*x^2 + 3*(14*(b^ 3*c*d^2 - a*b^2*d^3)*f^3*g - 15*(b^3*c^2*d - a^2*b*d^3)*f^2*g^2 + (5*b^3*c ^3 + 3*a*b^2*c^2*d - 3*a^2*b*c*d^2 - 5*a^3*d^3)*f*g^3 - (a*b^2*c^3 - a^3*c *d^2)*g^4)*x)/(b^3*d^3*f^9 + a^3*c^3*f^3*g^6 - 3*(b^3*c*d^2 + a*b^2*d^3)*f ^8*g + 3*(b^3*c^2*d + 3*a*b^2*c*d^2 + a^2*b*d^3)*f^7*g^2 - (b^3*c^3 + 9*a* b^2*c^2*d + 9*a^2*b*c*d^2 + a^3*d^3)*f^6*g^3 + 3*(a*b^2*c^3 + 3*a^2*b*c...
Leaf count of result is larger than twice the leaf count of optimal. 21743 vs. \(2 (374) = 748\).
Time = 1.21 (sec) , antiderivative size = 21743, normalized size of antiderivative = 56.04 \[ \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{(f+g x)^5} \, dx=\text {Too large to display} \]
1/24*(6*(4*B*b^5*c^2*d^3*f^3*n - 8*B*a*b^4*c*d^4*f^3*n + 4*B*a^2*b^3*d^5*f ^3*n - 6*B*b^5*c^3*d^2*f^2*g*n + 6*B*a*b^4*c^2*d^3*f^2*g*n + 6*B*a^2*b^3*c *d^4*f^2*g*n - 6*B*a^3*b^2*d^5*f^2*g*n + 4*B*b^5*c^4*d*f*g^2*n - 4*B*a*b^4 *c^3*d^2*f*g^2*n - 4*B*a^3*b^2*c*d^4*f*g^2*n + 4*B*a^4*b*d^5*f*g^2*n - B*b ^5*c^5*g^3*n + B*a*b^4*c^4*d*g^3*n + B*a^4*b*c*d^4*g^3*n - B*a^5*d^5*g^3*n )*log(-b*f + (b*x + a)*d*f/(d*x + c) + a*g - (b*x + a)*c*g/(d*x + c))/(b^4 *d^4*f^8 - 4*b^4*c*d^3*f^7*g - 4*a*b^3*d^4*f^7*g + 6*b^4*c^2*d^2*f^6*g^2 + 16*a*b^3*c*d^3*f^6*g^2 + 6*a^2*b^2*d^4*f^6*g^2 - 4*b^4*c^3*d*f^5*g^3 - 24 *a*b^3*c^2*d^2*f^5*g^3 - 24*a^2*b^2*c*d^3*f^5*g^3 - 4*a^3*b*d^4*f^5*g^3 + b^4*c^4*f^4*g^4 + 16*a*b^3*c^3*d*f^4*g^4 + 36*a^2*b^2*c^2*d^2*f^4*g^4 + 16 *a^3*b*c*d^3*f^4*g^4 + a^4*d^4*f^4*g^4 - 4*a*b^3*c^4*f^3*g^5 - 24*a^2*b^2* c^3*d*f^3*g^5 - 24*a^3*b*c^2*d^2*f^3*g^5 - 4*a^4*c*d^3*f^3*g^5 + 6*a^2*b^2 *c^4*f^2*g^6 + 16*a^3*b*c^3*d*f^2*g^6 + 6*a^4*c^2*d^2*f^2*g^6 - 4*a^3*b*c^ 4*f*g^7 - 4*a^4*c^3*d*f*g^7 + a^4*c^4*g^8) + 6*(4*B*b^5*c^2*d^3*f^3*n - 8* B*a*b^4*c*d^4*f^3*n - 12*(b*x + a)*B*b^4*c^2*d^4*f^3*n/(d*x + c) + 4*B*a^2 *b^3*d^5*f^3*n + 24*(b*x + a)*B*a*b^3*c*d^5*f^3*n/(d*x + c) + 12*(b*x + a) ^2*B*b^3*c^2*d^5*f^3*n/(d*x + c)^2 - 12*(b*x + a)*B*a^2*b^2*d^6*f^3*n/(d*x + c) - 24*(b*x + a)^2*B*a*b^2*c*d^6*f^3*n/(d*x + c)^2 - 4*(b*x + a)^3*B*b ^2*c^2*d^6*f^3*n/(d*x + c)^3 + 12*(b*x + a)^2*B*a^2*b*d^7*f^3*n/(d*x + c)^ 2 + 8*(b*x + a)^3*B*a*b*c*d^7*f^3*n/(d*x + c)^3 - 4*(b*x + a)^3*B*a^2*d...
Time = 10.34 (sec) , antiderivative size = 2569, normalized size of antiderivative = 6.62 \[ \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{(f+g x)^5} \, dx=\text {Too large to display} \]
((x^3*(B*a^3*d^3*g^6*n - B*b^3*c^3*g^6*n - 3*B*a^2*b*d^3*f*g^5*n + 3*B*b^3 *c^2*d*f*g^5*n + 3*B*a*b^2*d^3*f^2*g^4*n - 3*B*b^3*c*d^2*f^2*g^4*n))/(a^3* c^3*g^6 + b^3*d^3*f^6 - a^3*d^3*f^3*g^3 - b^3*c^3*f^3*g^3 - 3*a^2*b*c^3*f* g^5 - 3*a*b^2*d^3*f^5*g - 3*a^3*c^2*d*f*g^5 - 3*b^3*c*d^2*f^5*g + 3*a*b^2* c^3*f^2*g^4 + 3*a^2*b*d^3*f^4*g^2 + 3*a^3*c*d^2*f^2*g^4 + 3*b^3*c^2*d*f^4* g^2 + 9*a*b^2*c*d^2*f^4*g^2 - 9*a*b^2*c^2*d*f^3*g^3 - 9*a^2*b*c*d^2*f^3*g^ 3 + 9*a^2*b*c^2*d*f^2*g^4) - (6*A*a^3*c^3*g^6 + 6*A*b^3*d^3*f^6 - 6*A*a^3* d^3*f^3*g^3 - 6*A*b^3*c^3*f^3*g^3 + 18*A*a*b^2*c^3*f^2*g^4 + 18*A*a^2*b*d^ 3*f^4*g^2 + 18*A*a^3*c*d^2*f^2*g^4 + 18*A*b^3*c^2*d*f^4*g^2 - 11*B*a^3*d^3 *f^3*g^3*n + 11*B*b^3*c^3*f^3*g^3*n - 18*A*a^2*b*c^3*f*g^5 - 18*A*a*b^2*d^ 3*f^5*g - 18*A*a^3*c^2*d*f*g^5 - 18*A*b^3*c*d^2*f^5*g + 2*B*a^2*b*c^3*f*g^ 5*n - 26*B*a*b^2*d^3*f^5*g*n - 2*B*a^3*c^2*d*f*g^5*n + 26*B*b^3*c*d^2*f^5* g*n + 54*A*a*b^2*c*d^2*f^4*g^2 - 54*A*a*b^2*c^2*d*f^3*g^3 - 54*A*a^2*b*c*d ^2*f^3*g^3 + 54*A*a^2*b*c^2*d*f^2*g^4 - 7*B*a*b^2*c^3*f^2*g^4*n + 31*B*a^2 *b*d^3*f^4*g^2*n + 7*B*a^3*c*d^2*f^2*g^4*n - 31*B*b^3*c^2*d*f^4*g^2*n + 15 *B*a*b^2*c^2*d*f^3*g^3*n - 15*B*a^2*b*c*d^2*f^3*g^3*n)/(6*(a^3*c^3*g^6 + b ^3*d^3*f^6 - a^3*d^3*f^3*g^3 - b^3*c^3*f^3*g^3 - 3*a^2*b*c^3*f*g^5 - 3*a*b ^2*d^3*f^5*g - 3*a^3*c^2*d*f*g^5 - 3*b^3*c*d^2*f^5*g + 3*a*b^2*c^3*f^2*g^4 + 3*a^2*b*d^3*f^4*g^2 + 3*a^3*c*d^2*f^2*g^4 + 3*b^3*c^2*d*f^4*g^2 + 9*a*b ^2*c*d^2*f^4*g^2 - 9*a*b^2*c^2*d*f^3*g^3 - 9*a^2*b*c*d^2*f^3*g^3 + 9*a^...