Integrand size = 31, antiderivative size = 389 \[ \int \frac {A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )}{(g+h x)^5} \, dx=-\frac {B (b c-a d) n}{12 (b g-a h) (d g-c h) (g+h x)^3}-\frac {B (b c-a d) (2 b d g-b c h-a d h) n}{8 (b g-a h)^2 (d g-c h)^2 (g+h x)^2}-\frac {B (b c-a d) \left (a^2 d^2 h^2-a b d h (3 d g-c h)+b^2 \left (3 d^2 g^2-3 c d g h+c^2 h^2\right )\right ) n}{4 (b g-a h)^3 (d g-c h)^3 (g+h x)}+\frac {b^4 B n \log (a+b x)}{4 h (b g-a h)^4}-\frac {B d^4 n \log (c+d x)}{4 h (d g-c h)^4}-\frac {A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )}{4 h (g+h x)^4}-\frac {B (b c-a d) (2 b d g-b c h-a d h) \left (2 a b d^2 g h-a^2 d^2 h^2-b^2 \left (2 d^2 g^2-2 c d g h+c^2 h^2\right )\right ) n \log (g+h x)}{4 (b g-a h)^4 (d g-c h)^4} \]
-1/12*B*(-a*d+b*c)*n/(-a*h+b*g)/(-c*h+d*g)/(h*x+g)^3-1/8*B*(-a*d+b*c)*(-a* d*h-b*c*h+2*b*d*g)*n/(-a*h+b*g)^2/(-c*h+d*g)^2/(h*x+g)^2-1/4*B*(-a*d+b*c)* (a^2*d^2*h^2-a*b*d*h*(-c*h+3*d*g)+b^2*(c^2*h^2-3*c*d*g*h+3*d^2*g^2))*n/(-a *h+b*g)^3/(-c*h+d*g)^3/(h*x+g)+1/4*b^4*B*n*ln(b*x+a)/h/(-a*h+b*g)^4-1/4*B* d^4*n*ln(d*x+c)/h/(-c*h+d*g)^4+1/4*(-A-B*ln(e*(b*x+a)^n/((d*x+c)^n)))/h/(h *x+g)^4-1/4*B*(-a*d+b*c)*(-a*d*h-b*c*h+2*b*d*g)*(2*a*b*d^2*g*h-a^2*d^2*h^2 -b^2*(c^2*h^2-2*c*d*g*h+2*d^2*g^2))*n*ln(h*x+g)/(-a*h+b*g)^4/(-c*h+d*g)^4
Time = 0.65 (sec) , antiderivative size = 366, normalized size of antiderivative = 0.94 \[ \int \frac {A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )}{(g+h x)^5} \, dx=-\frac {\frac {A}{(g+h x)^4}+\frac {B \log \left (e (a+b x)^n (c+d x)^{-n}\right )}{(g+h x)^4}-B (b c-a d) n \left (-\frac {h}{3 (b g-a h) (d g-c h) (g+h x)^3}+\frac {h (-2 b d g+b c h+a d h)}{2 (b g-a h)^2 (d g-c h)^2 (g+h x)^2}-\frac {h \left (a^2 d^2 h^2+a b d h (-3 d g+c h)+b^2 \left (3 d^2 g^2-3 c d g h+c^2 h^2\right )\right )}{(b g-a h)^3 (d g-c h)^3 (g+h x)}+\frac {b^4 \log (a+b x)}{(b c-a d) (b g-a h)^4}-\frac {d^4 \log (c+d x)}{(b c-a d) (d g-c h)^4}-\frac {h (-2 b d g+b c h+a d h) \left (-2 a b d^2 g h+a^2 d^2 h^2+b^2 \left (2 d^2 g^2-2 c d g h+c^2 h^2\right )\right ) \log (g+h x)}{(b g-a h)^4 (d g-c h)^4}\right )}{4 h} \]
-1/4*(A/(g + h*x)^4 + (B*Log[(e*(a + b*x)^n)/(c + d*x)^n])/(g + h*x)^4 - B *(b*c - a*d)*n*(-1/3*h/((b*g - a*h)*(d*g - c*h)*(g + h*x)^3) + (h*(-2*b*d* g + b*c*h + a*d*h))/(2*(b*g - a*h)^2*(d*g - c*h)^2*(g + h*x)^2) - (h*(a^2* d^2*h^2 + a*b*d*h*(-3*d*g + c*h) + b^2*(3*d^2*g^2 - 3*c*d*g*h + c^2*h^2))) /((b*g - a*h)^3*(d*g - c*h)^3*(g + h*x)) + (b^4*Log[a + b*x])/((b*c - a*d) *(b*g - a*h)^4) - (d^4*Log[c + d*x])/((b*c - a*d)*(d*g - c*h)^4) - (h*(-2* b*d*g + b*c*h + a*d*h)*(-2*a*b*d^2*g*h + a^2*d^2*h^2 + b^2*(2*d^2*g^2 - 2* c*d*g*h + c^2*h^2))*Log[g + h*x])/((b*g - a*h)^4*(d*g - c*h)^4)))/h
Time = 0.75 (sec) , antiderivative size = 372, normalized size of antiderivative = 0.96, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {2948, 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.
\(\displaystyle \int \frac {B \log \left (e (a+b x)^n (c+d x)^{-n}\right )+A}{(g+h x)^5} \, dx\) |
\(\Big \downarrow \) 2948 |
\(\displaystyle \frac {B n (b c-a d) \int \frac {1}{(a+b x) (c+d x) (g+h x)^4}dx}{4 h}-\frac {B \log \left (e (a+b x)^n (c+d x)^{-n}\right )+A}{4 h (g+h x)^4}\) |
\(\Big \downarrow \) 93 |
\(\displaystyle \frac {B n (b c-a d) \int \left (\frac {b^5}{(b c-a d) (b g-a h)^4 (a+b x)}-\frac {d^5}{(b c-a d) (c h-d g)^4 (c+d x)}+\frac {h^2 (2 b d g-b c h-a d h) \left (2 d^2 g^2 b^2+c^2 h^2 b^2-2 c d g h b^2-2 a d^2 g h b+a^2 d^2 h^2\right )}{(b g-a h)^4 (d g-c h)^4 (g+h x)}+\frac {h^2 \left (\left (3 d^2 g^2-3 c d h g+c^2 h^2\right ) b^2-a d h (3 d g-c h) b+a^2 d^2 h^2\right )}{(b g-a h)^3 (d g-c h)^3 (g+h x)^2}-\frac {h^2 (-2 b d g+b c h+a d h)}{(b g-a h)^2 (d g-c h)^2 (g+h x)^3}+\frac {h^2}{(b g-a h) (d g-c h) (g+h x)^4}\right )dx}{4 h}-\frac {B \log \left (e (a+b x)^n (c+d x)^{-n}\right )+A}{4 h (g+h x)^4}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {B n (b c-a d) \left (-\frac {h \left (a^2 d^2 h^2-a b d h (3 d g-c h)+b^2 \left (c^2 h^2-3 c d g h+3 d^2 g^2\right )\right )}{(g+h x) (b g-a h)^3 (d g-c h)^3}-\frac {h \log (g+h x) (-a d h-b c h+2 b d g) \left (-a^2 d^2 h^2+2 a b d^2 g h-\left (b^2 \left (c^2 h^2-2 c d g h+2 d^2 g^2\right )\right )\right )}{(b g-a h)^4 (d g-c h)^4}+\frac {b^4 \log (a+b x)}{(b c-a d) (b g-a h)^4}-\frac {d^4 \log (c+d x)}{(b c-a d) (d g-c h)^4}-\frac {h (-a d h-b c h+2 b d g)}{2 (g+h x)^2 (b g-a h)^2 (d g-c h)^2}-\frac {h}{3 (g+h x)^3 (b g-a h) (d g-c h)}\right )}{4 h}-\frac {B \log \left (e (a+b x)^n (c+d x)^{-n}\right )+A}{4 h (g+h x)^4}\) |
-1/4*(A + B*Log[(e*(a + b*x)^n)/(c + d*x)^n])/(h*(g + h*x)^4) + (B*(b*c - a*d)*n*(-1/3*h/((b*g - a*h)*(d*g - c*h)*(g + h*x)^3) - (h*(2*b*d*g - b*c*h - a*d*h))/(2*(b*g - a*h)^2*(d*g - c*h)^2*(g + h*x)^2) - (h*(a^2*d^2*h^2 - a*b*d*h*(3*d*g - c*h) + b^2*(3*d^2*g^2 - 3*c*d*g*h + c^2*h^2)))/((b*g - a *h)^3*(d*g - c*h)^3*(g + h*x)) + (b^4*Log[a + b*x])/((b*c - a*d)*(b*g - a* h)^4) - (d^4*Log[c + d*x])/((b*c - a*d)*(d*g - c*h)^4) - (h*(2*b*d*g - b*c *h - a*d*h)*(2*a*b*d^2*g*h - a^2*d^2*h^2 - b^2*(2*d^2*g^2 - 2*c*d*g*h + c^ 2*h^2))*Log[g + h*x])/((b*g - a*h)^4*(d*g - c*h)^4)))/(4*h)
3.4.2.3.1 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_))^(n_.)*((c_.) + (d_.)*(x_))^(mn_ )]*(B_.))*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> Simp[(f + g*x)^(m + 1)*( (A + B*Log[e*((a + b*x)^n/(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] / ; FreeQ[{a, b, c, d, e, f, g, A, B, m, n}, x] && EqQ[n + mn, 0] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && !(EqQ[m, -2] && IntegerQ[n])
Leaf count of result is larger than twice the leaf count of optimal. \(5230\) vs. \(2(378)=756\).
Time = 233.55 (sec) , antiderivative size = 5231, normalized size of antiderivative = 13.45
method | result | size |
parallelrisch | \(\text {Expression too large to display}\) | \(5231\) |
risch | \(\text {Expression too large to display}\) | \(16077\) |
Timed out. \[ \int \frac {A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )}{(g+h x)^5} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )}{(g+h x)^5} \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 1912 vs. \(2 (375) = 750\).
Time = 0.36 (sec) , antiderivative size = 1912, normalized size of antiderivative = 4.92 \[ \int \frac {A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )}{(g+h x)^5} \, dx=\text {Too large to display} \]
1/24*(6*b^4*e*n*log(b*x + a)/(b^4*g^4*h - 4*a*b^3*g^3*h^2 + 6*a^2*b^2*g^2* h^3 - 4*a^3*b*g*h^4 + a^4*h^5) - 6*d^4*e*n*log(d*x + c)/(d^4*g^4*h - 4*c*d ^3*g^3*h^2 + 6*c^2*d^2*g^2*h^3 - 4*c^3*d*g*h^4 + c^4*h^5) - 6*(4*a*b^3*d^4 *e*g^3*n - 6*a^2*b^2*d^4*e*g^2*h*n + 4*a^3*b*d^4*e*g*h^2*n - a^4*d^4*e*h^3 *n - (4*c*d^3*e*g^3*n - 6*c^2*d^2*e*g^2*h*n + 4*c^3*d*e*g*h^2*n - c^4*e*h^ 3*n)*b^4)*log(h*x + g)/((d^4*g^4*h^4 - 4*c*d^3*g^3*h^5 + 6*c^2*d^2*g^2*h^6 - 4*c^3*d*g*h^7 + c^4*h^8)*a^4 - 4*(d^4*g^5*h^3 - 4*c*d^3*g^4*h^4 + 6*c^2 *d^2*g^3*h^5 - 4*c^3*d*g^2*h^6 + c^4*g*h^7)*a^3*b + 6*(d^4*g^6*h^2 - 4*c*d ^3*g^5*h^3 + 6*c^2*d^2*g^4*h^4 - 4*c^3*d*g^3*h^5 + c^4*g^2*h^6)*a^2*b^2 - 4*(d^4*g^7*h - 4*c*d^3*g^6*h^2 + 6*c^2*d^2*g^5*h^3 - 4*c^3*d*g^4*h^4 + c^4 *g^3*h^5)*a*b^3 + (d^4*g^8 - 4*c*d^3*g^7*h + 6*c^2*d^2*g^6*h^2 - 4*c^3*d*g ^5*h^3 + c^4*g^4*h^4)*b^4) - ((11*d^3*e*g^2*h^2*n - 7*c*d^2*e*g*h^3*n + 2* c^2*d*e*h^4*n)*a^3 - (31*d^3*e*g^3*h*n - 15*c*d^2*e*g^2*h^2*n + 2*c^3*e*h^ 4*n)*a^2*b + (26*d^3*e*g^4*n - 15*c^2*d*e*g^2*h^2*n + 7*c^3*e*g*h^3*n)*a*b ^2 - (26*c*d^2*e*g^4*n - 31*c^2*d*e*g^3*h*n + 11*c^3*e*g^2*h^2*n)*b^3 + 6* (3*a*b^2*d^3*e*g^2*h^2*n - 3*a^2*b*d^3*e*g*h^3*n + a^3*d^3*e*h^4*n - (3*c* d^2*e*g^2*h^2*n - 3*c^2*d*e*g*h^3*n + c^3*e*h^4*n)*b^3)*x^2 + 3*((5*d^3*e* g*h^3*n - c*d^2*e*h^4*n)*a^3 - 3*(5*d^3*e*g^2*h^2*n - c*d^2*e*g*h^3*n)*a^2 *b + (14*d^3*e*g^3*h*n - 3*c^2*d*e*g*h^3*n + c^3*e*h^4*n)*a*b^2 - (14*c*d^ 2*e*g^3*h*n - 15*c^2*d*e*g^2*h^2*n + 5*c^3*e*g*h^3*n)*b^3)*x)/((d^3*g^6...
Leaf count of result is larger than twice the leaf count of optimal. 3325 vs. \(2 (375) = 750\).
Time = 2.55 (sec) , antiderivative size = 3325, normalized size of antiderivative = 8.55 \[ \int \frac {A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )}{(g+h x)^5} \, dx=\text {Too large to display} \]
1/4*B*b^5*n*log(abs(b*x + a))/(b^5*g^4*h - 4*a*b^4*g^3*h^2 + 6*a^2*b^3*g^2 *h^3 - 4*a^3*b^2*g*h^4 + a^4*b*h^5) - 1/4*B*d^5*n*log(abs(d*x + c))/(d^5*g ^4*h - 4*c*d^4*g^3*h^2 + 6*c^2*d^3*g^2*h^3 - 4*c^3*d^2*g*h^4 + c^4*d*h^5) - 1/4*B*n*log(b*x + a)/(h^5*x^4 + 4*g*h^4*x^3 + 6*g^2*h^3*x^2 + 4*g^3*h^2* x + g^4*h) + 1/4*B*n*log(d*x + c)/(h^5*x^4 + 4*g*h^4*x^3 + 6*g^2*h^3*x^2 + 4*g^3*h^2*x + g^4*h) + 1/4*(4*B*b^4*c*d^3*g^3*n - 4*B*a*b^3*d^4*g^3*n - 6 *B*b^4*c^2*d^2*g^2*h*n + 6*B*a^2*b^2*d^4*g^2*h*n + 4*B*b^4*c^3*d*g*h^2*n - 4*B*a^3*b*d^4*g*h^2*n - B*b^4*c^4*h^3*n + B*a^4*d^4*h^3*n)*log(h*x + g)/( b^4*d^4*g^8 - 4*b^4*c*d^3*g^7*h - 4*a*b^3*d^4*g^7*h + 6*b^4*c^2*d^2*g^6*h^ 2 + 16*a*b^3*c*d^3*g^6*h^2 + 6*a^2*b^2*d^4*g^6*h^2 - 4*b^4*c^3*d*g^5*h^3 - 24*a*b^3*c^2*d^2*g^5*h^3 - 24*a^2*b^2*c*d^3*g^5*h^3 - 4*a^3*b*d^4*g^5*h^3 + b^4*c^4*g^4*h^4 + 16*a*b^3*c^3*d*g^4*h^4 + 36*a^2*b^2*c^2*d^2*g^4*h^4 + 16*a^3*b*c*d^3*g^4*h^4 + a^4*d^4*g^4*h^4 - 4*a*b^3*c^4*g^3*h^5 - 24*a^2*b ^2*c^3*d*g^3*h^5 - 24*a^3*b*c^2*d^2*g^3*h^5 - 4*a^4*c*d^3*g^3*h^5 + 6*a^2* b^2*c^4*g^2*h^6 + 16*a^3*b*c^3*d*g^2*h^6 + 6*a^4*c^2*d^2*g^2*h^6 - 4*a^3*b *c^4*g*h^7 - 4*a^4*c^3*d*g*h^7 + a^4*c^4*h^8) - 1/24*(18*B*b^3*c*d^2*g^2*h ^4*n*x^3 - 18*B*a*b^2*d^3*g^2*h^4*n*x^3 - 18*B*b^3*c^2*d*g*h^5*n*x^3 + 18* B*a^2*b*d^3*g*h^5*n*x^3 + 6*B*b^3*c^3*h^6*n*x^3 - 6*B*a^3*d^3*h^6*n*x^3 + 60*B*b^3*c*d^2*g^3*h^3*n*x^2 - 60*B*a*b^2*d^3*g^3*h^3*n*x^2 - 63*B*b^3*c^2 *d*g^2*h^4*n*x^2 + 63*B*a^2*b*d^3*g^2*h^4*n*x^2 + 21*B*b^3*c^3*g*h^5*n*...
Time = 10.63 (sec) , antiderivative size = 2570, normalized size of antiderivative = 6.61 \[ \int \frac {A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )}{(g+h x)^5} \, dx=\text {Too large to display} \]
((x*(13*B*a^3*d^3*g^2*h^4*n - 13*B*b^3*c^3*g^2*h^4*n - B*a^2*b*c^3*h^6*n + B*a^3*c^2*d*h^6*n + 5*B*a*b^2*c^3*g*h^5*n - 5*B*a^3*c*d^2*g*h^5*n + 34*B* a*b^2*d^3*g^4*h^2*n - 38*B*a^2*b*d^3*g^3*h^3*n - 34*B*b^3*c*d^2*g^4*h^2*n + 38*B*b^3*c^2*d*g^3*h^3*n - 12*B*a*b^2*c^2*d*g^2*h^4*n + 12*B*a^2*b*c*d^2 *g^2*h^4*n))/(3*(a^3*c^3*h^6 + b^3*d^3*g^6 - a^3*d^3*g^3*h^3 - b^3*c^3*g^3 *h^3 - 3*a^2*b*c^3*g*h^5 - 3*a*b^2*d^3*g^5*h - 3*a^3*c^2*d*g*h^5 - 3*b^3*c *d^2*g^5*h + 3*a*b^2*c^3*g^2*h^4 + 3*a^2*b*d^3*g^4*h^2 + 3*a^3*c*d^2*g^2*h ^4 + 3*b^3*c^2*d*g^4*h^2 + 9*a*b^2*c*d^2*g^4*h^2 - 9*a*b^2*c^2*d*g^3*h^3 - 9*a^2*b*c*d^2*g^3*h^3 + 9*a^2*b*c^2*d*g^2*h^4)) - (6*A*a^3*c^3*h^6 + 6*A* b^3*d^3*g^6 - 6*A*a^3*d^3*g^3*h^3 - 6*A*b^3*c^3*g^3*h^3 + 18*A*a*b^2*c^3*g ^2*h^4 + 18*A*a^2*b*d^3*g^4*h^2 + 18*A*a^3*c*d^2*g^2*h^4 + 18*A*b^3*c^2*d* g^4*h^2 - 11*B*a^3*d^3*g^3*h^3*n + 11*B*b^3*c^3*g^3*h^3*n - 18*A*a^2*b*c^3 *g*h^5 - 18*A*a*b^2*d^3*g^5*h - 18*A*a^3*c^2*d*g*h^5 - 18*A*b^3*c*d^2*g^5* h + 2*B*a^2*b*c^3*g*h^5*n - 26*B*a*b^2*d^3*g^5*h*n - 2*B*a^3*c^2*d*g*h^5*n + 26*B*b^3*c*d^2*g^5*h*n + 54*A*a*b^2*c*d^2*g^4*h^2 - 54*A*a*b^2*c^2*d*g^ 3*h^3 - 54*A*a^2*b*c*d^2*g^3*h^3 + 54*A*a^2*b*c^2*d*g^2*h^4 - 7*B*a*b^2*c^ 3*g^2*h^4*n + 31*B*a^2*b*d^3*g^4*h^2*n + 7*B*a^3*c*d^2*g^2*h^4*n - 31*B*b^ 3*c^2*d*g^4*h^2*n + 15*B*a*b^2*c^2*d*g^3*h^3*n - 15*B*a^2*b*c*d^2*g^3*h^3* n)/(6*(a^3*c^3*h^6 + b^3*d^3*g^6 - a^3*d^3*g^3*h^3 - b^3*c^3*g^3*h^3 - 3*a ^2*b*c^3*g*h^5 - 3*a*b^2*d^3*g^5*h - 3*a^3*c^2*d*g*h^5 - 3*b^3*c*d^2*g^...