3.427 \(\int \frac{a+b \log (c (d (e+f x)^p)^q)}{(g+h x)^4} \, dx\)

Optimal. Leaf size=149 \[ -\frac{a+b \log \left (c \left (d (e+f x)^p\right )^q\right )}{3 h (g+h x)^3}+\frac{b f^2 p q}{3 h (g+h x) (f g-e h)^2}+\frac{b f^3 p q \log (e+f x)}{3 h (f g-e h)^3}-\frac{b f^3 p q \log (g+h x)}{3 h (f g-e h)^3}+\frac{b f p q}{6 h (g+h x)^2 (f g-e h)} \]

[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)

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Rubi [A]  time = 0.166919, antiderivative size = 149, normalized size of antiderivative = 1., 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 = {2395, 44, 2445} \[ -\frac{a+b \log \left (c \left (d (e+f x)^p\right )^q\right )}{3 h (g+h x)^3}+\frac{b f^2 p q}{3 h (g+h x) (f g-e h)^2}+\frac{b f^3 p q \log (e+f x)}{3 h (f g-e h)^3}-\frac{b f^3 p q \log (g+h x)}{3 h (f g-e h)^3}+\frac{b f p q}{6 h (g+h x)^2 (f g-e h)} \]

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 2395

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 44

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] && L
tQ[m + n + 2, 0])

Rule 2445

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 &=\operatorname{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}+\operatorname{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}+\operatorname{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]  time = 0.195175, size = 115, normalized size = 0.77 \[ \frac{-2 a-2 b \log \left (c \left (d (e+f x)^p\right )^q\right )+\frac{b f p q (g+h x) \left (2 f^2 (g+h x)^2 \log (e+f x)+(f g-e h) (-e h+3 f g+2 f h x)-2 f^2 (g+h x)^2 \log (g+h x)\right )}{(f g-e h)^3}}{6 h (g+h x)^3} \]

Antiderivative was successfully verified.

[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)

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Maple [F]  time = 0.661, size = 0, normalized size = 0. \begin{align*} \int{\frac{a+b\ln \left ( c \left ( d \left ( fx+e \right ) ^{p} \right ) ^{q} \right ) }{ \left ( hx+g \right ) ^{4}}}\, dx \end{align*}

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]  time = 1.2077, size = 413, normalized size = 2.77 \begin{align*} \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 )}} \end{align*}

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*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)

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Fricas [B]  time = 2.1103, size = 1169, normalized size = 7.85 \begin{align*} -\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 )}} \end{align*}

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*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)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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]

Timed out

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Giac [B]  time = 1.18124, size = 868, normalized size = 5.83 \begin{align*} \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{align*}

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)