\(\int \frac {a+b \log (c (d (e+f x)^p)^q)}{(g+h x)^4} \, dx\) [427]
Optimal result
Integrand size = 26, antiderivative size = 149 \[
\int \frac {a+b \log \left (c \left (d (e+f x)^p\right )^q\right )}{(g+h x)^4} \, dx=\frac {b f p q}{6 h (f g-e h) (g+h x)^2}+\frac {b f^2 p q}{3 h (f g-e h)^2 (g+h x)}+\frac {b f^3 p q \log (e+f x)}{3 h (f g-e h)^3}-\frac {a+b \log \left (c \left (d (e+f x)^p\right )^q\right )}{3 h (g+h x)^3}-\frac {b f^3 p q \log (g+h x)}{3 h (f g-e h)^3}
\]
[Out]
1/6*b*f*p*q/h/(-e*h+f*g)/(h*x+g)^2+1/3*b*f^2*p*q/h/(-e*h+f*g)^2/(h*x+g)+1/3*b*f^3*p*q*ln(f*x+e)/h/(-e*h+f*g)^3
+1/3*(-a-b*ln(c*(d*(f*x+e)^p)^q))/h/(h*x+g)^3-1/3*b*f^3*p*q*ln(h*x+g)/h/(-e*h+f*g)^3
Rubi [A] (verified)
Time = 0.12 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.00, number of
steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {2442, 46, 2495}
\[
\int \frac {a+b \log \left (c \left (d (e+f x)^p\right )^q\right )}{(g+h x)^4} \, dx=-\frac {a+b \log \left (c \left (d (e+f x)^p\right )^q\right )}{3 h (g+h x)^3}+\frac {b f^3 p q \log (e+f x)}{3 h (f g-e h)^3}-\frac {b f^3 p q \log (g+h x)}{3 h (f g-e h)^3}+\frac {b f^2 p q}{3 h (g+h x) (f g-e h)^2}+\frac {b f p q}{6 h (g+h x)^2 (f g-e h)}
\]
[In]
Int[(a + b*Log[c*(d*(e + f*x)^p)^q])/(g + h*x)^4,x]
[Out]
(b*f*p*q)/(6*h*(f*g - e*h)*(g + h*x)^2) + (b*f^2*p*q)/(3*h*(f*g - e*h)^2*(g + h*x)) + (b*f^3*p*q*Log[e + f*x])
/(3*h*(f*g - e*h)^3) - (a + b*Log[c*(d*(e + f*x)^p)^q])/(3*h*(g + h*x)^3) - (b*f^3*p*q*Log[g + h*x])/(3*h*(f*g
- e*h)^3)
Rule 46
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] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] && !(IGtQ[n, 0] && Lt
Q[m + n + 2, 0])
Rule 2442
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))*((f_.) + (g_.)*(x_))^(q_.), x_Symbol] :> Simp[(f + g*
x)^(q + 1)*((a + b*Log[c*(d + e*x)^n])/(g*(q + 1))), x] - Dist[b*e*(n/(g*(q + 1))), Int[(f + g*x)^(q + 1)/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, q}, x] && NeQ[e*f - d*g, 0] && NeQ[q, -1]
Rule 2495
Int[((a_.) + Log[(c_.)*((d_.)*((e_.) + (f_.)*(x_))^(m_.))^(n_)]*(b_.))^(p_.)*(u_.), x_Symbol] :> Subst[Int[u*(
a + b*Log[c*d^n*(e + f*x)^(m*n)])^p, x], c*d^n*(e + f*x)^(m*n), c*(d*(e + f*x)^m)^n] /; FreeQ[{a, b, c, d, e,
f, m, n, p}, x] && !IntegerQ[n] && !(EqQ[d, 1] && EqQ[m, 1]) && IntegralFreeQ[IntHide[u*(a + b*Log[c*d^n*(e
+ f*x)^(m*n)])^p, x]]
Rubi steps \begin{align*}
\text {integral}& = \text {Subst}\left (\int \frac {a+b \log \left (c d^q (e+f x)^{p q}\right )}{(g+h x)^4} \, dx,c d^q (e+f x)^{p q},c \left (d (e+f x)^p\right )^q\right ) \\ & = -\frac {a+b \log \left (c \left (d (e+f x)^p\right )^q\right )}{3 h (g+h x)^3}+\text {Subst}\left (\frac {(b f p q) \int \frac {1}{(e+f x) (g+h x)^3} \, dx}{3 h},c d^q (e+f x)^{p q},c \left (d (e+f x)^p\right )^q\right ) \\ & = -\frac {a+b \log \left (c \left (d (e+f x)^p\right )^q\right )}{3 h (g+h x)^3}+\text {Subst}\left (\frac {(b f p q) \int \left (\frac {f^3}{(f g-e h)^3 (e+f x)}-\frac {h}{(f g-e h) (g+h x)^3}-\frac {f h}{(f g-e h)^2 (g+h x)^2}-\frac {f^2 h}{(f g-e h)^3 (g+h x)}\right ) \, dx}{3 h},c d^q (e+f x)^{p q},c \left (d (e+f x)^p\right )^q\right ) \\ & = \frac {b f p q}{6 h (f g-e h) (g+h x)^2}+\frac {b f^2 p q}{3 h (f g-e h)^2 (g+h x)}+\frac {b f^3 p q \log (e+f x)}{3 h (f g-e h)^3}-\frac {a+b \log \left (c \left (d (e+f x)^p\right )^q\right )}{3 h (g+h x)^3}-\frac {b f^3 p q \log (g+h x)}{3 h (f g-e h)^3} \\
\end{align*}
Mathematica [A] (verified)
Time = 0.12 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.77
\[
\int \frac {a+b \log \left (c \left (d (e+f x)^p\right )^q\right )}{(g+h x)^4} \, dx=\frac {-2 a-2 b \log \left (c \left (d (e+f x)^p\right )^q\right )+\frac {b f p q (g+h x) \left ((f g-e h) (3 f g-e h+2 f h x)+2 f^2 (g+h x)^2 \log (e+f x)-2 f^2 (g+h x)^2 \log (g+h x)\right )}{(f g-e h)^3}}{6 h (g+h x)^3}
\]
[In]
Integrate[(a + b*Log[c*(d*(e + f*x)^p)^q])/(g + h*x)^4,x]
[Out]
(-2*a - 2*b*Log[c*(d*(e + f*x)^p)^q] + (b*f*p*q*(g + h*x)*((f*g - e*h)*(3*f*g - e*h + 2*f*h*x) + 2*f^2*(g + h*
x)^2*Log[e + f*x] - 2*f^2*(g + h*x)^2*Log[g + h*x]))/(f*g - e*h)^3)/(6*h*(g + h*x)^3)
Maple [B] (verified)
Leaf count of result is larger than twice the leaf count of optimal. \(486\) vs. \(2(142)=284\).
Time = 11.92 (sec) , antiderivative size = 487, normalized size of antiderivative =
3.27
| | |
| method | result | size |
| | |
| parallelrisch |
\(-\frac {-6 x b e \,f^{3} g \,h^{4} p q -6 \ln \left (c \left (d \left (f x +e \right )^{p}\right )^{q}\right ) b \,e^{2} f^{2} g \,h^{4}+6 \ln \left (c \left (d \left (f x +e \right )^{p}\right )^{q}\right ) b e \,f^{3} g^{2} h^{3}+2 \ln \left (c \left (d \left (f x +e \right )^{p}\right )^{q}\right ) b \,e^{3} f \,h^{5}-2 \ln \left (c \left (d \left (f x +e \right )^{p}\right )^{q}\right ) b \,f^{4} g^{3} h^{2}+3 b \,f^{4} g^{3} h^{2} p q +6 \ln \left (f x +e \right ) x^{2} b \,f^{4} g \,h^{4} p q -6 \ln \left (h x +g \right ) x^{2} b \,f^{4} g \,h^{4} p q +6 \ln \left (f x +e \right ) x b \,f^{4} g^{2} h^{3} p q -6 \ln \left (h x +g \right ) x b \,f^{4} g^{2} h^{3} p q +2 a \,e^{3} f \,h^{5}-2 a \,f^{4} g^{3} h^{2}-2 x^{2} b e \,f^{3} h^{5} p q +2 x^{2} b \,f^{4} g \,h^{4} p q +2 \ln \left (f x +e \right ) x^{3} b \,f^{4} h^{5} p q -2 \ln \left (h x +g \right ) x^{3} b \,f^{4} h^{5} p q +2 \ln \left (f x +e \right ) b \,f^{4} g^{3} h^{2} p q -2 \ln \left (h x +g \right ) b \,f^{4} g^{3} h^{2} p q -6 a \,e^{2} f^{2} g \,h^{4}+6 a e \,f^{3} g^{2} h^{3}+b \,e^{2} f^{2} g \,h^{4} p q -4 b e \,f^{3} g^{2} h^{3} p q +x b \,e^{2} f^{2} h^{5} p q +5 x b \,f^{4} g^{2} h^{3} p q}{6 \left (e^{3} h^{3}-3 e^{2} f g \,h^{2}+3 e \,f^{2} g^{2} h -g^{3} f^{3}\right ) \left (h x +g \right )^{3} f \,h^{3}}\) |
\(487\) |
| | |
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[In]
int((a+b*ln(c*(d*(f*x+e)^p)^q))/(h*x+g)^4,x,method=_RETURNVERBOSE)
[Out]
-1/6*(-6*x*b*e*f^3*g*h^4*p*q-6*ln(c*(d*(f*x+e)^p)^q)*b*e^2*f^2*g*h^4+6*ln(c*(d*(f*x+e)^p)^q)*b*e*f^3*g^2*h^3+2
*ln(c*(d*(f*x+e)^p)^q)*b*e^3*f*h^5-2*ln(c*(d*(f*x+e)^p)^q)*b*f^4*g^3*h^2+3*b*f^4*g^3*h^2*p*q+6*ln(f*x+e)*x^2*b
*f^4*g*h^4*p*q-6*ln(h*x+g)*x^2*b*f^4*g*h^4*p*q+6*ln(f*x+e)*x*b*f^4*g^2*h^3*p*q-6*ln(h*x+g)*x*b*f^4*g^2*h^3*p*q
+2*a*e^3*f*h^5-2*a*f^4*g^3*h^2-2*x^2*b*e*f^3*h^5*p*q+2*x^2*b*f^4*g*h^4*p*q+2*ln(f*x+e)*x^3*b*f^4*h^5*p*q-2*ln(
h*x+g)*x^3*b*f^4*h^5*p*q+2*ln(f*x+e)*b*f^4*g^3*h^2*p*q-2*ln(h*x+g)*b*f^4*g^3*h^2*p*q-6*a*e^2*f^2*g*h^4+6*a*e*f
^3*g^2*h^3+b*e^2*f^2*g*h^4*p*q-4*b*e*f^3*g^2*h^3*p*q+x*b*e^2*f^2*h^5*p*q+5*x*b*f^4*g^2*h^3*p*q)/(e^3*h^3-3*e^2
*f*g*h^2+3*e*f^2*g^2*h-f^3*g^3)/(h*x+g)^3/f/h^3
Fricas [B] (verification not implemented)
Leaf count of result is larger than twice the leaf count of optimal. 563 vs. \(2 (139) = 278\).
Time = 0.32 (sec) , antiderivative size = 563, normalized size of antiderivative = 3.78
\[
\int \frac {a+b \log \left (c \left (d (e+f x)^p\right )^q\right )}{(g+h x)^4} \, dx=-\frac {2 \, a f^{3} g^{3} - 6 \, a e f^{2} g^{2} h + 6 \, a e^{2} f g h^{2} - 2 \, a e^{3} h^{3} - 2 \, {\left (b f^{3} g h^{2} - b e f^{2} h^{3}\right )} p q x^{2} - {\left (5 \, b f^{3} g^{2} h - 6 \, b e f^{2} g h^{2} + b e^{2} f h^{3}\right )} p q x - {\left (3 \, b f^{3} g^{3} - 4 \, b e f^{2} g^{2} h + b e^{2} f g h^{2}\right )} p q + 2 \, {\left (b f^{3} g^{3} - 3 \, b e f^{2} g^{2} h + 3 \, b e^{2} f g h^{2} - b e^{3} h^{3}\right )} q \log \left (d\right ) - 2 \, {\left (b f^{3} h^{3} p q x^{3} + 3 \, b f^{3} g h^{2} p q x^{2} + 3 \, b f^{3} g^{2} h p q x + {\left (3 \, b e f^{2} g^{2} h - 3 \, b e^{2} f g h^{2} + b e^{3} h^{3}\right )} p q\right )} \log \left (f x + e\right ) + 2 \, {\left (b f^{3} h^{3} p q x^{3} + 3 \, b f^{3} g h^{2} p q x^{2} + 3 \, b f^{3} g^{2} h p q x + b f^{3} g^{3} p q\right )} \log \left (h x + g\right ) + 2 \, {\left (b f^{3} g^{3} - 3 \, b e f^{2} g^{2} h + 3 \, b e^{2} f g h^{2} - b e^{3} h^{3}\right )} \log \left (c\right )}{6 \, {\left (f^{3} g^{6} h - 3 \, e f^{2} g^{5} h^{2} + 3 \, e^{2} f g^{4} h^{3} - e^{3} g^{3} h^{4} + {\left (f^{3} g^{3} h^{4} - 3 \, e f^{2} g^{2} h^{5} + 3 \, e^{2} f g h^{6} - e^{3} h^{7}\right )} x^{3} + 3 \, {\left (f^{3} g^{4} h^{3} - 3 \, e f^{2} g^{3} h^{4} + 3 \, e^{2} f g^{2} h^{5} - e^{3} g h^{6}\right )} x^{2} + 3 \, {\left (f^{3} g^{5} h^{2} - 3 \, e f^{2} g^{4} h^{3} + 3 \, e^{2} f g^{3} h^{4} - e^{3} g^{2} h^{5}\right )} x\right )}}
\]
[In]
integrate((a+b*log(c*(d*(f*x+e)^p)^q))/(h*x+g)^4,x, algorithm="fricas")
[Out]
-1/6*(2*a*f^3*g^3 - 6*a*e*f^2*g^2*h + 6*a*e^2*f*g*h^2 - 2*a*e^3*h^3 - 2*(b*f^3*g*h^2 - b*e*f^2*h^3)*p*q*x^2 -
(5*b*f^3*g^2*h - 6*b*e*f^2*g*h^2 + b*e^2*f*h^3)*p*q*x - (3*b*f^3*g^3 - 4*b*e*f^2*g^2*h + b*e^2*f*g*h^2)*p*q +
2*(b*f^3*g^3 - 3*b*e*f^2*g^2*h + 3*b*e^2*f*g*h^2 - b*e^3*h^3)*q*log(d) - 2*(b*f^3*h^3*p*q*x^3 + 3*b*f^3*g*h^2*
p*q*x^2 + 3*b*f^3*g^2*h*p*q*x + (3*b*e*f^2*g^2*h - 3*b*e^2*f*g*h^2 + b*e^3*h^3)*p*q)*log(f*x + e) + 2*(b*f^3*h
^3*p*q*x^3 + 3*b*f^3*g*h^2*p*q*x^2 + 3*b*f^3*g^2*h*p*q*x + b*f^3*g^3*p*q)*log(h*x + g) + 2*(b*f^3*g^3 - 3*b*e*
f^2*g^2*h + 3*b*e^2*f*g*h^2 - b*e^3*h^3)*log(c))/(f^3*g^6*h - 3*e*f^2*g^5*h^2 + 3*e^2*f*g^4*h^3 - e^3*g^3*h^4
+ (f^3*g^3*h^4 - 3*e*f^2*g^2*h^5 + 3*e^2*f*g*h^6 - e^3*h^7)*x^3 + 3*(f^3*g^4*h^3 - 3*e*f^2*g^3*h^4 + 3*e^2*f*g
^2*h^5 - e^3*g*h^6)*x^2 + 3*(f^3*g^5*h^2 - 3*e*f^2*g^4*h^3 + 3*e^2*f*g^3*h^4 - e^3*g^2*h^5)*x)
Sympy [B] (verification not implemented)
Leaf count of result is larger than twice the leaf count of optimal. 5673 vs. \(2 (133) = 266\).
Time = 63.37 (sec) , antiderivative size = 5673, normalized size of antiderivative = 38.07
\[
\int \frac {a+b \log \left (c \left (d (e+f x)^p\right )^q\right )}{(g+h x)^4} \, dx=\text {Too large to display}
\]
[In]
integrate((a+b*ln(c*(d*(f*x+e)**p)**q))/(h*x+g)**4,x)
[Out]
Piecewise(((a*x + b*e*log(c*(d*(e + f*x)**p)**q)/f - b*p*q*x + b*x*log(c*(d*(e + f*x)**p)**q))/g**4, Eq(h, 0))
, (-3*a/(9*g**3*h + 27*g**2*h**2*x + 27*g*h**3*x**2 + 9*h**4*x**3) - b*p*q/(9*g**3*h + 27*g**2*h**2*x + 27*g*h
**3*x**2 + 9*h**4*x**3) - 3*b*log(c*(d*(f*g/h + f*x)**p)**q)/(9*g**3*h + 27*g**2*h**2*x + 27*g*h**3*x**2 + 9*h
**4*x**3), Eq(e, f*g/h)), (-2*a*e**3*h**3/(6*e**3*g**3*h**4 + 18*e**3*g**2*h**5*x + 18*e**3*g*h**6*x**2 + 6*e*
*3*h**7*x**3 - 18*e**2*f*g**4*h**3 - 54*e**2*f*g**3*h**4*x - 54*e**2*f*g**2*h**5*x**2 - 18*e**2*f*g*h**6*x**3
+ 18*e*f**2*g**5*h**2 + 54*e*f**2*g**4*h**3*x + 54*e*f**2*g**3*h**4*x**2 + 18*e*f**2*g**2*h**5*x**3 - 6*f**3*g
**6*h - 18*f**3*g**5*h**2*x - 18*f**3*g**4*h**3*x**2 - 6*f**3*g**3*h**4*x**3) + 6*a*e**2*f*g*h**2/(6*e**3*g**3
*h**4 + 18*e**3*g**2*h**5*x + 18*e**3*g*h**6*x**2 + 6*e**3*h**7*x**3 - 18*e**2*f*g**4*h**3 - 54*e**2*f*g**3*h*
*4*x - 54*e**2*f*g**2*h**5*x**2 - 18*e**2*f*g*h**6*x**3 + 18*e*f**2*g**5*h**2 + 54*e*f**2*g**4*h**3*x + 54*e*f
**2*g**3*h**4*x**2 + 18*e*f**2*g**2*h**5*x**3 - 6*f**3*g**6*h - 18*f**3*g**5*h**2*x - 18*f**3*g**4*h**3*x**2 -
6*f**3*g**3*h**4*x**3) - 6*a*e*f**2*g**2*h/(6*e**3*g**3*h**4 + 18*e**3*g**2*h**5*x + 18*e**3*g*h**6*x**2 + 6*
e**3*h**7*x**3 - 18*e**2*f*g**4*h**3 - 54*e**2*f*g**3*h**4*x - 54*e**2*f*g**2*h**5*x**2 - 18*e**2*f*g*h**6*x**
3 + 18*e*f**2*g**5*h**2 + 54*e*f**2*g**4*h**3*x + 54*e*f**2*g**3*h**4*x**2 + 18*e*f**2*g**2*h**5*x**3 - 6*f**3
*g**6*h - 18*f**3*g**5*h**2*x - 18*f**3*g**4*h**3*x**2 - 6*f**3*g**3*h**4*x**3) + 2*a*f**3*g**3/(6*e**3*g**3*h
**4 + 18*e**3*g**2*h**5*x + 18*e**3*g*h**6*x**2 + 6*e**3*h**7*x**3 - 18*e**2*f*g**4*h**3 - 54*e**2*f*g**3*h**4
*x - 54*e**2*f*g**2*h**5*x**2 - 18*e**2*f*g*h**6*x**3 + 18*e*f**2*g**5*h**2 + 54*e*f**2*g**4*h**3*x + 54*e*f**
2*g**3*h**4*x**2 + 18*e*f**2*g**2*h**5*x**3 - 6*f**3*g**6*h - 18*f**3*g**5*h**2*x - 18*f**3*g**4*h**3*x**2 - 6
*f**3*g**3*h**4*x**3) - 2*b*e**3*h**3*log(c*(d*(e + f*x)**p)**q)/(6*e**3*g**3*h**4 + 18*e**3*g**2*h**5*x + 18*
e**3*g*h**6*x**2 + 6*e**3*h**7*x**3 - 18*e**2*f*g**4*h**3 - 54*e**2*f*g**3*h**4*x - 54*e**2*f*g**2*h**5*x**2 -
18*e**2*f*g*h**6*x**3 + 18*e*f**2*g**5*h**2 + 54*e*f**2*g**4*h**3*x + 54*e*f**2*g**3*h**4*x**2 + 18*e*f**2*g*
*2*h**5*x**3 - 6*f**3*g**6*h - 18*f**3*g**5*h**2*x - 18*f**3*g**4*h**3*x**2 - 6*f**3*g**3*h**4*x**3) - b*e**2*
f*g*h**2*p*q/(6*e**3*g**3*h**4 + 18*e**3*g**2*h**5*x + 18*e**3*g*h**6*x**2 + 6*e**3*h**7*x**3 - 18*e**2*f*g**4
*h**3 - 54*e**2*f*g**3*h**4*x - 54*e**2*f*g**2*h**5*x**2 - 18*e**2*f*g*h**6*x**3 + 18*e*f**2*g**5*h**2 + 54*e*
f**2*g**4*h**3*x + 54*e*f**2*g**3*h**4*x**2 + 18*e*f**2*g**2*h**5*x**3 - 6*f**3*g**6*h - 18*f**3*g**5*h**2*x -
18*f**3*g**4*h**3*x**2 - 6*f**3*g**3*h**4*x**3) + 6*b*e**2*f*g*h**2*log(c*(d*(e + f*x)**p)**q)/(6*e**3*g**3*h
**4 + 18*e**3*g**2*h**5*x + 18*e**3*g*h**6*x**2 + 6*e**3*h**7*x**3 - 18*e**2*f*g**4*h**3 - 54*e**2*f*g**3*h**4
*x - 54*e**2*f*g**2*h**5*x**2 - 18*e**2*f*g*h**6*x**3 + 18*e*f**2*g**5*h**2 + 54*e*f**2*g**4*h**3*x + 54*e*f**
2*g**3*h**4*x**2 + 18*e*f**2*g**2*h**5*x**3 - 6*f**3*g**6*h - 18*f**3*g**5*h**2*x - 18*f**3*g**4*h**3*x**2 - 6
*f**3*g**3*h**4*x**3) - b*e**2*f*h**3*p*q*x/(6*e**3*g**3*h**4 + 18*e**3*g**2*h**5*x + 18*e**3*g*h**6*x**2 + 6*
e**3*h**7*x**3 - 18*e**2*f*g**4*h**3 - 54*e**2*f*g**3*h**4*x - 54*e**2*f*g**2*h**5*x**2 - 18*e**2*f*g*h**6*x**
3 + 18*e*f**2*g**5*h**2 + 54*e*f**2*g**4*h**3*x + 54*e*f**2*g**3*h**4*x**2 + 18*e*f**2*g**2*h**5*x**3 - 6*f**3
*g**6*h - 18*f**3*g**5*h**2*x - 18*f**3*g**4*h**3*x**2 - 6*f**3*g**3*h**4*x**3) + 4*b*e*f**2*g**2*h*p*q/(6*e**
3*g**3*h**4 + 18*e**3*g**2*h**5*x + 18*e**3*g*h**6*x**2 + 6*e**3*h**7*x**3 - 18*e**2*f*g**4*h**3 - 54*e**2*f*g
**3*h**4*x - 54*e**2*f*g**2*h**5*x**2 - 18*e**2*f*g*h**6*x**3 + 18*e*f**2*g**5*h**2 + 54*e*f**2*g**4*h**3*x +
54*e*f**2*g**3*h**4*x**2 + 18*e*f**2*g**2*h**5*x**3 - 6*f**3*g**6*h - 18*f**3*g**5*h**2*x - 18*f**3*g**4*h**3*
x**2 - 6*f**3*g**3*h**4*x**3) - 6*b*e*f**2*g**2*h*log(c*(d*(e + f*x)**p)**q)/(6*e**3*g**3*h**4 + 18*e**3*g**2*
h**5*x + 18*e**3*g*h**6*x**2 + 6*e**3*h**7*x**3 - 18*e**2*f*g**4*h**3 - 54*e**2*f*g**3*h**4*x - 54*e**2*f*g**2
*h**5*x**2 - 18*e**2*f*g*h**6*x**3 + 18*e*f**2*g**5*h**2 + 54*e*f**2*g**4*h**3*x + 54*e*f**2*g**3*h**4*x**2 +
18*e*f**2*g**2*h**5*x**3 - 6*f**3*g**6*h - 18*f**3*g**5*h**2*x - 18*f**3*g**4*h**3*x**2 - 6*f**3*g**3*h**4*x**
3) + 6*b*e*f**2*g*h**2*p*q*x/(6*e**3*g**3*h**4 + 18*e**3*g**2*h**5*x + 18*e**3*g*h**6*x**2 + 6*e**3*h**7*x**3
- 18*e**2*f*g**4*h**3 - 54*e**2*f*g**3*h**4*x - 54*e**2*f*g**2*h**5*x**2 - 18*e**2*f*g*h**6*x**3 + 18*e*f**2*g
**5*h**2 + 54*e*f**2*g**4*h**3*x + 54*e*f**2*g**3*h**4*x**2 + 18*e*f**2*g**2*h**5*x**3 - 6*f**3*g**6*h - 18*f*
*3*g**5*h**2*x - 18*f**3*g**4*h**3*x**2 - 6*f**3*g**3*h**4*x**3) + 2*b*e*f**2*h**3*p*q*x**2/(6*e**3*g**3*h**4
+ 18*e**3*g**2*h**5*x + 18*e**3*g*h**6*x**2 + 6*e**3*h**7*x**3 - 18*e**2*f*g**4*h**3 - 54*e**2*f*g**3*h**4*x -
54*e**2*f*g**2*h**5*x**2 - 18*e**2*f*g*h**6*x**3 + 18*e*f**2*g**5*h**2 + 54*e*f**2*g**4*h**3*x + 54*e*f**2*g*
*3*h**4*x**2 + 18*e*f**2*g**2*h**5*x**3 - 6*f**3*g**6*h - 18*f**3*g**5*h**2*x - 18*f**3*g**4*h**3*x**2 - 6*f**
3*g**3*h**4*x**3) + 2*b*f**3*g**3*p*q*log(g/h + x)/(6*e**3*g**3*h**4 + 18*e**3*g**2*h**5*x + 18*e**3*g*h**6*x*
*2 + 6*e**3*h**7*x**3 - 18*e**2*f*g**4*h**3 - 54*e**2*f*g**3*h**4*x - 54*e**2*f*g**2*h**5*x**2 - 18*e**2*f*g*h
**6*x**3 + 18*e*f**2*g**5*h**2 + 54*e*f**2*g**4*h**3*x + 54*e*f**2*g**3*h**4*x**2 + 18*e*f**2*g**2*h**5*x**3 -
6*f**3*g**6*h - 18*f**3*g**5*h**2*x - 18*f**3*g**4*h**3*x**2 - 6*f**3*g**3*h**4*x**3) - 3*b*f**3*g**3*p*q/(6*
e**3*g**3*h**4 + 18*e**3*g**2*h**5*x + 18*e**3*g*h**6*x**2 + 6*e**3*h**7*x**3 - 18*e**2*f*g**4*h**3 - 54*e**2*
f*g**3*h**4*x - 54*e**2*f*g**2*h**5*x**2 - 18*e**2*f*g*h**6*x**3 + 18*e*f**2*g**5*h**2 + 54*e*f**2*g**4*h**3*x
+ 54*e*f**2*g**3*h**4*x**2 + 18*e*f**2*g**2*h**5*x**3 - 6*f**3*g**6*h - 18*f**3*g**5*h**2*x - 18*f**3*g**4*h*
*3*x**2 - 6*f**3*g**3*h**4*x**3) + 6*b*f**3*g**2*h*p*q*x*log(g/h + x)/(6*e**3*g**3*h**4 + 18*e**3*g**2*h**5*x
+ 18*e**3*g*h**6*x**2 + 6*e**3*h**7*x**3 - 18*e**2*f*g**4*h**3 - 54*e**2*f*g**3*h**4*x - 54*e**2*f*g**2*h**5*x
**2 - 18*e**2*f*g*h**6*x**3 + 18*e*f**2*g**5*h**2 + 54*e*f**2*g**4*h**3*x + 54*e*f**2*g**3*h**4*x**2 + 18*e*f*
*2*g**2*h**5*x**3 - 6*f**3*g**6*h - 18*f**3*g**5*h**2*x - 18*f**3*g**4*h**3*x**2 - 6*f**3*g**3*h**4*x**3) - 5*
b*f**3*g**2*h*p*q*x/(6*e**3*g**3*h**4 + 18*e**3*g**2*h**5*x + 18*e**3*g*h**6*x**2 + 6*e**3*h**7*x**3 - 18*e**2
*f*g**4*h**3 - 54*e**2*f*g**3*h**4*x - 54*e**2*f*g**2*h**5*x**2 - 18*e**2*f*g*h**6*x**3 + 18*e*f**2*g**5*h**2
+ 54*e*f**2*g**4*h**3*x + 54*e*f**2*g**3*h**4*x**2 + 18*e*f**2*g**2*h**5*x**3 - 6*f**3*g**6*h - 18*f**3*g**5*h
**2*x - 18*f**3*g**4*h**3*x**2 - 6*f**3*g**3*h**4*x**3) - 6*b*f**3*g**2*h*x*log(c*(d*(e + f*x)**p)**q)/(6*e**3
*g**3*h**4 + 18*e**3*g**2*h**5*x + 18*e**3*g*h**6*x**2 + 6*e**3*h**7*x**3 - 18*e**2*f*g**4*h**3 - 54*e**2*f*g*
*3*h**4*x - 54*e**2*f*g**2*h**5*x**2 - 18*e**2*f*g*h**6*x**3 + 18*e*f**2*g**5*h**2 + 54*e*f**2*g**4*h**3*x + 5
4*e*f**2*g**3*h**4*x**2 + 18*e*f**2*g**2*h**5*x**3 - 6*f**3*g**6*h - 18*f**3*g**5*h**2*x - 18*f**3*g**4*h**3*x
**2 - 6*f**3*g**3*h**4*x**3) + 6*b*f**3*g*h**2*p*q*x**2*log(g/h + x)/(6*e**3*g**3*h**4 + 18*e**3*g**2*h**5*x +
18*e**3*g*h**6*x**2 + 6*e**3*h**7*x**3 - 18*e**2*f*g**4*h**3 - 54*e**2*f*g**3*h**4*x - 54*e**2*f*g**2*h**5*x*
*2 - 18*e**2*f*g*h**6*x**3 + 18*e*f**2*g**5*h**2 + 54*e*f**2*g**4*h**3*x + 54*e*f**2*g**3*h**4*x**2 + 18*e*f**
2*g**2*h**5*x**3 - 6*f**3*g**6*h - 18*f**3*g**5*h**2*x - 18*f**3*g**4*h**3*x**2 - 6*f**3*g**3*h**4*x**3) - 2*b
*f**3*g*h**2*p*q*x**2/(6*e**3*g**3*h**4 + 18*e**3*g**2*h**5*x + 18*e**3*g*h**6*x**2 + 6*e**3*h**7*x**3 - 18*e*
*2*f*g**4*h**3 - 54*e**2*f*g**3*h**4*x - 54*e**2*f*g**2*h**5*x**2 - 18*e**2*f*g*h**6*x**3 + 18*e*f**2*g**5*h**
2 + 54*e*f**2*g**4*h**3*x + 54*e*f**2*g**3*h**4*x**2 + 18*e*f**2*g**2*h**5*x**3 - 6*f**3*g**6*h - 18*f**3*g**5
*h**2*x - 18*f**3*g**4*h**3*x**2 - 6*f**3*g**3*h**4*x**3) - 6*b*f**3*g*h**2*x**2*log(c*(d*(e + f*x)**p)**q)/(6
*e**3*g**3*h**4 + 18*e**3*g**2*h**5*x + 18*e**3*g*h**6*x**2 + 6*e**3*h**7*x**3 - 18*e**2*f*g**4*h**3 - 54*e**2
*f*g**3*h**4*x - 54*e**2*f*g**2*h**5*x**2 - 18*e**2*f*g*h**6*x**3 + 18*e*f**2*g**5*h**2 + 54*e*f**2*g**4*h**3*
x + 54*e*f**2*g**3*h**4*x**2 + 18*e*f**2*g**2*h**5*x**3 - 6*f**3*g**6*h - 18*f**3*g**5*h**2*x - 18*f**3*g**4*h
**3*x**2 - 6*f**3*g**3*h**4*x**3) + 2*b*f**3*h**3*p*q*x**3*log(g/h + x)/(6*e**3*g**3*h**4 + 18*e**3*g**2*h**5*
x + 18*e**3*g*h**6*x**2 + 6*e**3*h**7*x**3 - 18*e**2*f*g**4*h**3 - 54*e**2*f*g**3*h**4*x - 54*e**2*f*g**2*h**5
*x**2 - 18*e**2*f*g*h**6*x**3 + 18*e*f**2*g**5*h**2 + 54*e*f**2*g**4*h**3*x + 54*e*f**2*g**3*h**4*x**2 + 18*e*
f**2*g**2*h**5*x**3 - 6*f**3*g**6*h - 18*f**3*g**5*h**2*x - 18*f**3*g**4*h**3*x**2 - 6*f**3*g**3*h**4*x**3) -
2*b*f**3*h**3*x**3*log(c*(d*(e + f*x)**p)**q)/(6*e**3*g**3*h**4 + 18*e**3*g**2*h**5*x + 18*e**3*g*h**6*x**2 +
6*e**3*h**7*x**3 - 18*e**2*f*g**4*h**3 - 54*e**2*f*g**3*h**4*x - 54*e**2*f*g**2*h**5*x**2 - 18*e**2*f*g*h**6*x
**3 + 18*e*f**2*g**5*h**2 + 54*e*f**2*g**4*h**3*x + 54*e*f**2*g**3*h**4*x**2 + 18*e*f**2*g**2*h**5*x**3 - 6*f*
*3*g**6*h - 18*f**3*g**5*h**2*x - 18*f**3*g**4*h**3*x**2 - 6*f**3*g**3*h**4*x**3), True))
Maxima [B] (verification not implemented)
Leaf count of result is larger than twice the leaf count of optimal. 306 vs. \(2 (139) = 278\).
Time = 0.21 (sec) , antiderivative size = 306, normalized size of antiderivative = 2.05
\[
\int \frac {a+b \log \left (c \left (d (e+f x)^p\right )^q\right )}{(g+h x)^4} \, dx=\frac {1}{6} \, {\left (\frac {2 \, f^{2} \log \left (f x + e\right )}{f^{3} g^{3} h - 3 \, e f^{2} g^{2} h^{2} + 3 \, e^{2} f g h^{3} - e^{3} h^{4}} - \frac {2 \, f^{2} \log \left (h x + g\right )}{f^{3} g^{3} h - 3 \, e f^{2} g^{2} h^{2} + 3 \, e^{2} f g h^{3} - e^{3} h^{4}} + \frac {2 \, f h x + 3 \, f g - e h}{f^{2} g^{4} h - 2 \, e f g^{3} h^{2} + e^{2} g^{2} h^{3} + {\left (f^{2} g^{2} h^{3} - 2 \, e f g h^{4} + e^{2} h^{5}\right )} x^{2} + 2 \, {\left (f^{2} g^{3} h^{2} - 2 \, e f g^{2} h^{3} + e^{2} g h^{4}\right )} x}\right )} b f p q - \frac {b \log \left (\left ({\left (f x + e\right )}^{p} d\right )^{q} c\right )}{3 \, {\left (h^{4} x^{3} + 3 \, g h^{3} x^{2} + 3 \, g^{2} h^{2} x + g^{3} h\right )}} - \frac {a}{3 \, {\left (h^{4} x^{3} + 3 \, g h^{3} x^{2} + 3 \, g^{2} h^{2} x + g^{3} h\right )}}
\]
[In]
integrate((a+b*log(c*(d*(f*x+e)^p)^q))/(h*x+g)^4,x, algorithm="maxima")
[Out]
1/6*(2*f^2*log(f*x + e)/(f^3*g^3*h - 3*e*f^2*g^2*h^2 + 3*e^2*f*g*h^3 - e^3*h^4) - 2*f^2*log(h*x + g)/(f^3*g^3*
h - 3*e*f^2*g^2*h^2 + 3*e^2*f*g*h^3 - e^3*h^4) + (2*f*h*x + 3*f*g - e*h)/(f^2*g^4*h - 2*e*f*g^3*h^2 + e^2*g^2*
h^3 + (f^2*g^2*h^3 - 2*e*f*g*h^4 + e^2*h^5)*x^2 + 2*(f^2*g^3*h^2 - 2*e*f*g^2*h^3 + e^2*g*h^4)*x))*b*f*p*q - 1/
3*b*log(((f*x + e)^p*d)^q*c)/(h^4*x^3 + 3*g*h^3*x^2 + 3*g^2*h^2*x + g^3*h) - 1/3*a/(h^4*x^3 + 3*g*h^3*x^2 + 3*
g^2*h^2*x + g^3*h)
Giac [B] (verification not implemented)
Leaf count of result is larger than twice the leaf count of optimal. 443 vs. \(2 (139) = 278\).
Time = 0.37 (sec) , antiderivative size = 443, normalized size of antiderivative = 2.97
\[
\int \frac {a+b \log \left (c \left (d (e+f x)^p\right )^q\right )}{(g+h x)^4} \, dx=\frac {b f^{3} p q \log \left (f x + e\right )}{3 \, {\left (f^{3} g^{3} h - 3 \, e f^{2} g^{2} h^{2} + 3 \, e^{2} f g h^{3} - e^{3} h^{4}\right )}} - \frac {b f^{3} p q \log \left (h x + g\right )}{3 \, {\left (f^{3} g^{3} h - 3 \, e f^{2} g^{2} h^{2} + 3 \, e^{2} f g h^{3} - e^{3} h^{4}\right )}} - \frac {b p q \log \left (f x + e\right )}{3 \, {\left (h^{4} x^{3} + 3 \, g h^{3} x^{2} + 3 \, g^{2} h^{2} x + g^{3} h\right )}} + \frac {2 \, b f^{2} h^{2} p q x^{2} + 5 \, b f^{2} g h p q x - b e f h^{2} p q x + 3 \, b f^{2} g^{2} p q - b e f g h p q - 2 \, b f^{2} g^{2} q \log \left (d\right ) + 4 \, b e f g h q \log \left (d\right ) - 2 \, b e^{2} h^{2} q \log \left (d\right ) - 2 \, b f^{2} g^{2} \log \left (c\right ) + 4 \, b e f g h \log \left (c\right ) - 2 \, b e^{2} h^{2} \log \left (c\right ) - 2 \, a f^{2} g^{2} + 4 \, a e f g h - 2 \, a e^{2} h^{2}}{6 \, {\left (f^{2} g^{2} h^{4} x^{3} - 2 \, e f g h^{5} x^{3} + e^{2} h^{6} x^{3} + 3 \, f^{2} g^{3} h^{3} x^{2} - 6 \, e f g^{2} h^{4} x^{2} + 3 \, e^{2} g h^{5} x^{2} + 3 \, f^{2} g^{4} h^{2} x - 6 \, e f g^{3} h^{3} x + 3 \, e^{2} g^{2} h^{4} x + f^{2} g^{5} h - 2 \, e f g^{4} h^{2} + e^{2} g^{3} h^{3}\right )}}
\]
[In]
integrate((a+b*log(c*(d*(f*x+e)^p)^q))/(h*x+g)^4,x, algorithm="giac")
[Out]
1/3*b*f^3*p*q*log(f*x + e)/(f^3*g^3*h - 3*e*f^2*g^2*h^2 + 3*e^2*f*g*h^3 - e^3*h^4) - 1/3*b*f^3*p*q*log(h*x + g
)/(f^3*g^3*h - 3*e*f^2*g^2*h^2 + 3*e^2*f*g*h^3 - e^3*h^4) - 1/3*b*p*q*log(f*x + e)/(h^4*x^3 + 3*g*h^3*x^2 + 3*
g^2*h^2*x + g^3*h) + 1/6*(2*b*f^2*h^2*p*q*x^2 + 5*b*f^2*g*h*p*q*x - b*e*f*h^2*p*q*x + 3*b*f^2*g^2*p*q - b*e*f*
g*h*p*q - 2*b*f^2*g^2*q*log(d) + 4*b*e*f*g*h*q*log(d) - 2*b*e^2*h^2*q*log(d) - 2*b*f^2*g^2*log(c) + 4*b*e*f*g*
h*log(c) - 2*b*e^2*h^2*log(c) - 2*a*f^2*g^2 + 4*a*e*f*g*h - 2*a*e^2*h^2)/(f^2*g^2*h^4*x^3 - 2*e*f*g*h^5*x^3 +
e^2*h^6*x^3 + 3*f^2*g^3*h^3*x^2 - 6*e*f*g^2*h^4*x^2 + 3*e^2*g*h^5*x^2 + 3*f^2*g^4*h^2*x - 6*e*f*g^3*h^3*x + 3*
e^2*g^2*h^4*x + f^2*g^5*h - 2*e*f*g^4*h^2 + e^2*g^3*h^3)
Mupad [B] (verification not implemented)
Time = 3.74 (sec) , antiderivative size = 293, normalized size of antiderivative = 1.97
\[
\int \frac {a+b \log \left (c \left (d (e+f x)^p\right )^q\right )}{(g+h x)^4} \, dx=\frac {2\,a\,e\,f\,g}{3\,{\left (g+h\,x\right )}^3\,{\left (e\,h-f\,g\right )}^2}-\frac {a\,e^2\,h}{3\,{\left (g+h\,x\right )}^3\,{\left (e\,h-f\,g\right )}^2}-\frac {b\,\ln \left (c\,{\left (d\,{\left (e+f\,x\right )}^p\right )}^q\right )}{3\,h\,{\left (g+h\,x\right )}^3}-\frac {a\,f^2\,g^2}{3\,h\,{\left (g+h\,x\right )}^3\,{\left (e\,h-f\,g\right )}^2}+\frac {b\,f^2\,h\,p\,q\,x^2}{3\,{\left (g+h\,x\right )}^3\,{\left (e\,h-f\,g\right )}^2}-\frac {b\,e\,f\,g\,p\,q}{6\,{\left (g+h\,x\right )}^3\,{\left (e\,h-f\,g\right )}^2}+\frac {b\,f^2\,g^2\,p\,q}{2\,h\,{\left (g+h\,x\right )}^3\,{\left (e\,h-f\,g\right )}^2}+\frac {5\,b\,f^2\,g\,p\,q\,x}{6\,{\left (g+h\,x\right )}^3\,{\left (e\,h-f\,g\right )}^2}-\frac {b\,e\,f\,h\,p\,q\,x}{6\,{\left (g+h\,x\right )}^3\,{\left (e\,h-f\,g\right )}^2}+\frac {b\,f^3\,p\,q\,\mathrm {atan}\left (\frac {e\,h\,1{}\mathrm {i}+f\,g\,1{}\mathrm {i}+f\,h\,x\,2{}\mathrm {i}}{e\,h-f\,g}\right )\,2{}\mathrm {i}}{3\,h\,{\left (e\,h-f\,g\right )}^3}
\]
[In]
int((a + b*log(c*(d*(e + f*x)^p)^q))/(g + h*x)^4,x)
[Out]
(2*a*e*f*g)/(3*(g + h*x)^3*(e*h - f*g)^2) - (a*e^2*h)/(3*(g + h*x)^3*(e*h - f*g)^2) - (b*log(c*(d*(e + f*x)^p)
^q))/(3*h*(g + h*x)^3) - (a*f^2*g^2)/(3*h*(g + h*x)^3*(e*h - f*g)^2) + (b*f^3*p*q*atan((e*h*1i + f*g*1i + f*h*
x*2i)/(e*h - f*g))*2i)/(3*h*(e*h - f*g)^3) + (b*f^2*h*p*q*x^2)/(3*(g + h*x)^3*(e*h - f*g)^2) - (b*e*f*g*p*q)/(
6*(g + h*x)^3*(e*h - f*g)^2) + (b*f^2*g^2*p*q)/(2*h*(g + h*x)^3*(e*h - f*g)^2) + (5*b*f^2*g*p*q*x)/(6*(g + h*x
)^3*(e*h - f*g)^2) - (b*e*f*h*p*q*x)/(6*(g + h*x)^3*(e*h - f*g)^2)