3.5.27 \(\int \frac {a+b \log (c (d (e+f x)^p)^q)}{(g+h x)^4} \, dx\) [427]
Optimal. Leaf size=149 \[ \frac {b f p q}{6 h (f g-e h) (g+h x)^2}+\frac {b f^2 p q}{3 h (f g-e h)^2 (g+h x)}+\frac {b f^3 p q \log (e+f x)}{3 h (f g-e h)^3}-\frac {a+b \log \left (c \left (d (e+f x)^p\right )^q\right )}{3 h (g+h x)^3}-\frac {b f^3 p q \log (g+h x)}{3 h (f g-e h)^3} \]
[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
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Rubi [A]
time = 0.12, antiderivative size = 149, normalized size of antiderivative = 1.00, number of steps
used = 4, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {2442, 46, 2495}
\begin {gather*} -\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)} \end {gather*}
Antiderivative was successfully verified.
[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*} \int \frac {a+b \log \left (c \left (d (e+f x)^p\right )^q\right )}{(g+h x)^4} \, dx &=\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*}
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Mathematica [A]
time = 0.10, size = 297, normalized size = 1.99 \begin {gather*} \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^3-3 b f^3 g^3 p q+4 b e f^2 g^2 h p q-b e^2 f g h^2 p q-5 b f^3 g^2 h p q x+6 b e f^2 g h^2 p q x-b e^2 f h^3 p q x-2 b f^3 g h^2 p q x^2+2 b e f^2 h^3 p q x^2-2 b f^3 p q (g+h x)^3 \log (e+f x)+2 b (f g-e h)^3 \log \left (c \left (d (e+f x)^p\right )^q\right )+2 b f^3 g^3 p q \log (g+h x)+6 b f^3 g^2 h p q x \log (g+h x)+6 b f^3 g h^2 p q x^2 \log (g+h x)+2 b f^3 h^3 p q x^3 \log (g+h x)}{6 h (-f g+e h)^3 (g+h x)^3} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[(a + b*Log[c*(d*(e + f*x)^p)^q])/(g + h*x)^4,x]
[Out]
(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 - 3*b*f^3*g^3*p*q + 4*b*e*f^2*g^2*h*p*q - b*e^2
*f*g*h^2*p*q - 5*b*f^3*g^2*h*p*q*x + 6*b*e*f^2*g*h^2*p*q*x - b*e^2*f*h^3*p*q*x - 2*b*f^3*g*h^2*p*q*x^2 + 2*b*e
*f^2*h^3*p*q*x^2 - 2*b*f^3*p*q*(g + h*x)^3*Log[e + f*x] + 2*b*(f*g - e*h)^3*Log[c*(d*(e + f*x)^p)^q] + 2*b*f^3
*g^3*p*q*Log[g + h*x] + 6*b*f^3*g^2*h*p*q*x*Log[g + h*x] + 6*b*f^3*g*h^2*p*q*x^2*Log[g + h*x] + 2*b*f^3*h^3*p*
q*x^3*Log[g + h*x])/(6*h*(-(f*g) + e*h)^3*(g + h*x)^3)
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Maple [F]
time = 0.20, size = 0, normalized size = 0.00 \[\int \frac {a +b \ln \left (c \left (d \left (f x +e \right )^{p}\right )^{q}\right )}{\left (h x +g \right )^{4}}\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+b*ln(c*(d*(f*x+e)^p)^q))/(h*x+g)^4,x)
[Out]
int((a+b*ln(c*(d*(f*x+e)^p)^q))/(h*x+g)^4,x)
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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 307 vs.
\(2 (145) = 290\).
time = 0.30, size = 307, normalized size = 2.06 \begin {gather*} \frac {1}{6} \, {\left (\frac {2 \, f^{2} \log \left (f x + e\right )}{f^{3} g^{3} h - 3 \, f^{2} g^{2} h^{2} e + 3 \, f g h^{3} e^{2} - h^{4} e^{3}} - \frac {2 \, f^{2} \log \left (h x + g\right )}{f^{3} g^{3} h - 3 \, f^{2} g^{2} h^{2} e + 3 \, f g h^{3} e^{2} - h^{4} e^{3}} + \frac {2 \, f h x + 3 \, f g - h e}{f^{2} g^{4} h - 2 \, f g^{3} h^{2} e + g^{2} h^{3} e^{2} + {\left (f^{2} g^{2} h^{3} - 2 \, f g h^{4} e + h^{5} e^{2}\right )} x^{2} + 2 \, {\left (f^{2} g^{3} h^{2} - 2 \, f g^{2} h^{3} e + g h^{4} e^{2}\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 )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[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*f^2*g^2*h^2*e + 3*f*g*h^3*e^2 - h^4*e^3) - 2*f^2*log(h*x + g)/(f^3*g^3*
h - 3*f^2*g^2*h^2*e + 3*f*g*h^3*e^2 - h^4*e^3) + (2*f*h*x + 3*f*g - h*e)/(f^2*g^4*h - 2*f*g^3*h^2*e + g^2*h^3*
e^2 + (f^2*g^2*h^3 - 2*f*g*h^4*e + h^5*e^2)*x^2 + 2*(f^2*g^3*h^2 - 2*f*g^2*h^3*e + g*h^4*e^2)*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)
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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 563 vs.
\(2 (145) = 290\).
time = 0.40, size = 563, normalized size = 3.78 \begin {gather*} \frac {2 \, b f^{3} g h^{2} p q x^{2} + 5 \, b f^{3} g^{2} h p q x + 3 \, b f^{3} g^{3} p q - 2 \, a f^{3} g^{3} + 2 \, a h^{3} e^{3} + {\left (b f h^{3} p q x + b f g h^{2} p q - 6 \, a f g h^{2}\right )} e^{2} - 2 \, {\left (b f^{2} h^{3} p q x^{2} + 3 \, b f^{2} g h^{2} p q x + 2 \, b f^{2} g^{2} h p q - 3 \, a f^{2} g^{2} h\right )} e + 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 + 3 \, b f^{2} g^{2} h p q e - 3 \, b f g h^{2} p q e^{2} + b h^{3} p q e^{3}\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 f^{2} g^{2} h e + 3 \, b f g h^{2} e^{2} - b h^{3} e^{3}\right )} \log \left (c\right ) - 2 \, {\left (b f^{3} g^{3} q - 3 \, b f^{2} g^{2} h q e + 3 \, b f g h^{2} q e^{2} - b h^{3} q e^{3}\right )} \log \left (d\right )}{6 \, {\left (f^{3} g^{3} h^{4} x^{3} + 3 \, f^{3} g^{4} h^{3} x^{2} + 3 \, f^{3} g^{5} h^{2} x + f^{3} g^{6} h - {\left (h^{7} x^{3} + 3 \, g h^{6} x^{2} + 3 \, g^{2} h^{5} x + g^{3} h^{4}\right )} e^{3} + 3 \, {\left (f g h^{6} x^{3} + 3 \, f g^{2} h^{5} x^{2} + 3 \, f g^{3} h^{4} x + f g^{4} h^{3}\right )} e^{2} - 3 \, {\left (f^{2} g^{2} h^{5} x^{3} + 3 \, f^{2} g^{3} h^{4} x^{2} + 3 \, f^{2} g^{4} h^{3} x + f^{2} g^{5} h^{2}\right )} e\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*log(c*(d*(f*x+e)^p)^q))/(h*x+g)^4,x, algorithm="fricas")
[Out]
1/6*(2*b*f^3*g*h^2*p*q*x^2 + 5*b*f^3*g^2*h*p*q*x + 3*b*f^3*g^3*p*q - 2*a*f^3*g^3 + 2*a*h^3*e^3 + (b*f*h^3*p*q*
x + b*f*g*h^2*p*q - 6*a*f*g*h^2)*e^2 - 2*(b*f^2*h^3*p*q*x^2 + 3*b*f^2*g*h^2*p*q*x + 2*b*f^2*g^2*h*p*q - 3*a*f^
2*g^2*h)*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 + 3*b*f^2*g^2*h*p*q*e - 3*b*f*
g*h^2*p*q*e^2 + b*h^3*p*q*e^3)*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*f^2*g^2*h*e + 3*b*f*g*h^2*e^2 - b*h^3*e^3)*log(c) - 2*(b
*f^3*g^3*q - 3*b*f^2*g^2*h*q*e + 3*b*f*g*h^2*q*e^2 - b*h^3*q*e^3)*log(d))/(f^3*g^3*h^4*x^3 + 3*f^3*g^4*h^3*x^2
+ 3*f^3*g^5*h^2*x + f^3*g^6*h - (h^7*x^3 + 3*g*h^6*x^2 + 3*g^2*h^5*x + g^3*h^4)*e^3 + 3*(f*g*h^6*x^3 + 3*f*g^
2*h^5*x^2 + 3*f*g^3*h^4*x + f*g^4*h^3)*e^2 - 3*(f^2*g^2*h^5*x^3 + 3*f^2*g^3*h^4*x^2 + 3*f^2*g^4*h^3*x + f^2*g^
5*h^2)*e)
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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 5673 vs.
\(2 (133) = 266\).
time = 72.23, size = 5673, normalized size = 38.07 \begin {gather*} \text {Too large to display} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[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**...
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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 643 vs.
\(2 (145) = 290\).
time = 6.19, size = 643, normalized size = 4.32 \begin {gather*} \frac {2 \, b f^{3} h^{3} p q x^{3} \log \left (f x + e\right ) - 2 \, b f^{3} h^{3} p q x^{3} \log \left (h x + g\right ) + 6 \, b f^{3} g h^{2} p q x^{2} \log \left (f x + e\right ) - 6 \, b f^{3} g h^{2} p q x^{2} \log \left (h x + g\right ) + 2 \, b f^{3} g h^{2} p q x^{2} - 2 \, b f^{2} h^{3} p q x^{2} e + 6 \, b f^{3} g^{2} h p q x \log \left (f x + e\right ) - 6 \, b f^{3} g^{2} h p q x \log \left (h x + g\right ) + 5 \, b f^{3} g^{2} h p q x - 6 \, b f^{2} g h^{2} p q x e + 6 \, b f^{2} g^{2} h p q e \log \left (f x + e\right ) - 2 \, b f^{3} g^{3} p q \log \left (h x + g\right ) + 3 \, b f^{3} g^{3} p q + b f h^{3} p q x e^{2} - 4 \, b f^{2} g^{2} h p q e - 6 \, b f g h^{2} p q e^{2} \log \left (f x + e\right ) - 2 \, b f^{3} g^{3} q \log \left (d\right ) + 6 \, b f^{2} g^{2} h q e \log \left (d\right ) + b f g h^{2} p q e^{2} + 2 \, b h^{3} p q e^{3} \log \left (f x + e\right ) - 2 \, b f^{3} g^{3} \log \left (c\right ) + 6 \, b f^{2} g^{2} h e \log \left (c\right ) - 6 \, b f g h^{2} q e^{2} \log \left (d\right ) - 2 \, a f^{3} g^{3} + 6 \, a f^{2} g^{2} h e - 6 \, b f g h^{2} e^{2} \log \left (c\right ) + 2 \, b h^{3} q e^{3} \log \left (d\right ) - 6 \, a f g h^{2} e^{2} + 2 \, b h^{3} e^{3} \log \left (c\right ) + 2 \, a h^{3} e^{3}}{6 \, {\left (f^{3} g^{3} h^{4} x^{3} - 3 \, f^{2} g^{2} h^{5} x^{3} e + 3 \, f^{3} g^{4} h^{3} x^{2} + 3 \, f g h^{6} x^{3} e^{2} - 9 \, f^{2} g^{3} h^{4} x^{2} e + 3 \, f^{3} g^{5} h^{2} x - h^{7} x^{3} e^{3} + 9 \, f g^{2} h^{5} x^{2} e^{2} - 9 \, f^{2} g^{4} h^{3} x e + f^{3} g^{6} h - 3 \, g h^{6} x^{2} e^{3} + 9 \, f g^{3} h^{4} x e^{2} - 3 \, f^{2} g^{5} h^{2} e - 3 \, g^{2} h^{5} x e^{3} + 3 \, f g^{4} h^{3} e^{2} - g^{3} h^{4} e^{3}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*log(c*(d*(f*x+e)^p)^q))/(h*x+g)^4,x, algorithm="giac")
[Out]
1/6*(2*b*f^3*h^3*p*q*x^3*log(f*x + e) - 2*b*f^3*h^3*p*q*x^3*log(h*x + g) + 6*b*f^3*g*h^2*p*q*x^2*log(f*x + e)
- 6*b*f^3*g*h^2*p*q*x^2*log(h*x + g) + 2*b*f^3*g*h^2*p*q*x^2 - 2*b*f^2*h^3*p*q*x^2*e + 6*b*f^3*g^2*h*p*q*x*log
(f*x + e) - 6*b*f^3*g^2*h*p*q*x*log(h*x + g) + 5*b*f^3*g^2*h*p*q*x - 6*b*f^2*g*h^2*p*q*x*e + 6*b*f^2*g^2*h*p*q
*e*log(f*x + e) - 2*b*f^3*g^3*p*q*log(h*x + g) + 3*b*f^3*g^3*p*q + b*f*h^3*p*q*x*e^2 - 4*b*f^2*g^2*h*p*q*e - 6
*b*f*g*h^2*p*q*e^2*log(f*x + e) - 2*b*f^3*g^3*q*log(d) + 6*b*f^2*g^2*h*q*e*log(d) + b*f*g*h^2*p*q*e^2 + 2*b*h^
3*p*q*e^3*log(f*x + e) - 2*b*f^3*g^3*log(c) + 6*b*f^2*g^2*h*e*log(c) - 6*b*f*g*h^2*q*e^2*log(d) - 2*a*f^3*g^3
+ 6*a*f^2*g^2*h*e - 6*b*f*g*h^2*e^2*log(c) + 2*b*h^3*q*e^3*log(d) - 6*a*f*g*h^2*e^2 + 2*b*h^3*e^3*log(c) + 2*a
*h^3*e^3)/(f^3*g^3*h^4*x^3 - 3*f^2*g^2*h^5*x^3*e + 3*f^3*g^4*h^3*x^2 + 3*f*g*h^6*x^3*e^2 - 9*f^2*g^3*h^4*x^2*e
+ 3*f^3*g^5*h^2*x - h^7*x^3*e^3 + 9*f*g^2*h^5*x^2*e^2 - 9*f^2*g^4*h^3*x*e + f^3*g^6*h - 3*g*h^6*x^2*e^3 + 9*f
*g^3*h^4*x*e^2 - 3*f^2*g^5*h^2*e - 3*g^2*h^5*x*e^3 + 3*f*g^4*h^3*e^2 - g^3*h^4*e^3)
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Mupad [B]
time = 2.48, size = 293, normalized size = 1.97 \begin {gather*} \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} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[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)
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