3.439 \(\int \frac {(a+b \log (c (d (e+f x)^p)^q))^3}{(g+h x)^2} \, dx\)

Optimal. Leaf size=209 \[ -\frac {6 b^2 f p^2 q^2 \text {Li}_2\left (-\frac {h (e+f x)}{f g-e h}\right ) \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{h (f g-e h)}-\frac {3 b f p q \log \left (\frac {f (g+h x)}{f g-e h}\right ) \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )^2}{h (f g-e h)}+\frac {(e+f x) \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )^3}{(g+h x) (f g-e h)}+\frac {6 b^3 f p^3 q^3 \text {Li}_3\left (-\frac {h (e+f x)}{f g-e h}\right )}{h (f g-e h)} \]

[Out]

(f*x+e)*(a+b*ln(c*(d*(f*x+e)^p)^q))^3/(-e*h+f*g)/(h*x+g)-3*b*f*p*q*(a+b*ln(c*(d*(f*x+e)^p)^q))^2*ln(f*(h*x+g)/
(-e*h+f*g))/h/(-e*h+f*g)-6*b^2*f*p^2*q^2*(a+b*ln(c*(d*(f*x+e)^p)^q))*polylog(2,-h*(f*x+e)/(-e*h+f*g))/h/(-e*h+
f*g)+6*b^3*f*p^3*q^3*polylog(3,-h*(f*x+e)/(-e*h+f*g))/h/(-e*h+f*g)

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Rubi [A]  time = 0.37, antiderivative size = 209, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {2397, 2396, 2433, 2374, 6589, 2445} \[ -\frac {6 b^2 f p^2 q^2 \text {PolyLog}\left (2,-\frac {h (e+f x)}{f g-e h}\right ) \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{h (f g-e h)}+\frac {6 b^3 f p^3 q^3 \text {PolyLog}\left (3,-\frac {h (e+f x)}{f g-e h}\right )}{h (f g-e h)}-\frac {3 b f p q \log \left (\frac {f (g+h x)}{f g-e h}\right ) \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )^2}{h (f g-e h)}+\frac {(e+f x) \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )^3}{(g+h x) (f g-e h)} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Log[c*(d*(e + f*x)^p)^q])^3/(g + h*x)^2,x]

[Out]

((e + f*x)*(a + b*Log[c*(d*(e + f*x)^p)^q])^3)/((f*g - e*h)*(g + h*x)) - (3*b*f*p*q*(a + b*Log[c*(d*(e + f*x)^
p)^q])^2*Log[(f*(g + h*x))/(f*g - e*h)])/(h*(f*g - e*h)) - (6*b^2*f*p^2*q^2*(a + b*Log[c*(d*(e + f*x)^p)^q])*P
olyLog[2, -((h*(e + f*x))/(f*g - e*h))])/(h*(f*g - e*h)) + (6*b^3*f*p^3*q^3*PolyLog[3, -((h*(e + f*x))/(f*g -
e*h))])/(h*(f*g - e*h))

Rule 2374

Int[(Log[(d_.)*((e_) + (f_.)*(x_)^(m_.))]*((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.))/(x_), x_Symbol] :> -Sim
p[(PolyLog[2, -(d*f*x^m)]*(a + b*Log[c*x^n])^p)/m, x] + Dist[(b*n*p)/m, Int[(PolyLog[2, -(d*f*x^m)]*(a + b*Log
[c*x^n])^(p - 1))/x, x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IGtQ[p, 0] && EqQ[d*e, 1]

Rule 2396

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_)/((f_.) + (g_.)*(x_)), x_Symbol] :> Simp[(Log[(e*
(f + g*x))/(e*f - d*g)]*(a + b*Log[c*(d + e*x)^n])^p)/g, x] - Dist[(b*e*n*p)/g, Int[(Log[(e*(f + g*x))/(e*f -
d*g)]*(a + b*Log[c*(d + e*x)^n])^(p - 1))/(d + e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p}, x] && NeQ[e*
f - d*g, 0] && IGtQ[p, 1]

Rule 2397

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_)/((f_.) + (g_.)*(x_))^2, x_Symbol] :> Simp[((d +
e*x)*(a + b*Log[c*(d + e*x)^n])^p)/((e*f - d*g)*(f + g*x)), x] - Dist[(b*e*n*p)/(e*f - d*g), Int[(a + b*Log[c*
(d + e*x)^n])^(p - 1)/(f + g*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n}, x] && NeQ[e*f - d*g, 0] && GtQ[p, 0
]

Rule 2433

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_.) + Log[(h_.)*((i_.) + (j_.)*(x_))^(m_.)]*
(g_.))*((k_.) + (l_.)*(x_))^(r_.), x_Symbol] :> Dist[1/e, Subst[Int[((k*x)/d)^r*(a + b*Log[c*x^n])^p*(f + g*Lo
g[h*((e*i - d*j)/e + (j*x)/e)^m]), x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, h, i, j, k, l, n, p, r},
x] && EqQ[e*k - d*l, 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]]

Rule 6589

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rubi steps

\begin {align*} \int \frac {\left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )^3}{(g+h x)^2} \, dx &=\operatorname {Subst}\left (\int \frac {\left (a+b \log \left (c d^q (e+f x)^{p q}\right )\right )^3}{(g+h x)^2} \, dx,c d^q (e+f x)^{p q},c \left (d (e+f x)^p\right )^q\right )\\ &=\frac {(e+f x) \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )^3}{(f g-e h) (g+h x)}-\operatorname {Subst}\left (\frac {(3 b f p q) \int \frac {\left (a+b \log \left (c d^q (e+f x)^{p q}\right )\right )^2}{g+h x} \, dx}{f g-e h},c d^q (e+f x)^{p q},c \left (d (e+f x)^p\right )^q\right )\\ &=\frac {(e+f x) \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )^3}{(f g-e h) (g+h x)}-\frac {3 b f p q \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )^2 \log \left (\frac {f (g+h x)}{f g-e h}\right )}{h (f g-e h)}+\operatorname {Subst}\left (\frac {\left (6 b^2 f^2 p^2 q^2\right ) \int \frac {\left (a+b \log \left (c d^q (e+f x)^{p q}\right )\right ) \log \left (\frac {f (g+h x)}{f g-e h}\right )}{e+f x} \, dx}{h (f g-e h)},c d^q (e+f x)^{p q},c \left (d (e+f x)^p\right )^q\right )\\ &=\frac {(e+f x) \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )^3}{(f g-e h) (g+h x)}-\frac {3 b f p q \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )^2 \log \left (\frac {f (g+h x)}{f g-e h}\right )}{h (f g-e h)}+\operatorname {Subst}\left (\frac {\left (6 b^2 f p^2 q^2\right ) \operatorname {Subst}\left (\int \frac {\left (a+b \log \left (c d^q x^{p q}\right )\right ) \log \left (\frac {f \left (\frac {f g-e h}{f}+\frac {h x}{f}\right )}{f g-e h}\right )}{x} \, dx,x,e+f x\right )}{h (f g-e h)},c d^q (e+f x)^{p q},c \left (d (e+f x)^p\right )^q\right )\\ &=\frac {(e+f x) \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )^3}{(f g-e h) (g+h x)}-\frac {3 b f p q \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )^2 \log \left (\frac {f (g+h x)}{f g-e h}\right )}{h (f g-e h)}-\frac {6 b^2 f p^2 q^2 \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right ) \text {Li}_2\left (-\frac {h (e+f x)}{f g-e h}\right )}{h (f g-e h)}+\operatorname {Subst}\left (\frac {\left (6 b^3 f p^3 q^3\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (-\frac {h x}{f g-e h}\right )}{x} \, dx,x,e+f x\right )}{h (f g-e h)},c d^q (e+f x)^{p q},c \left (d (e+f x)^p\right )^q\right )\\ &=\frac {(e+f x) \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )^3}{(f g-e h) (g+h x)}-\frac {3 b f p q \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )^2 \log \left (\frac {f (g+h x)}{f g-e h}\right )}{h (f g-e h)}-\frac {6 b^2 f p^2 q^2 \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right ) \text {Li}_2\left (-\frac {h (e+f x)}{f g-e h}\right )}{h (f g-e h)}+\frac {6 b^3 f p^3 q^3 \text {Li}_3\left (-\frac {h (e+f x)}{f g-e h}\right )}{h (f g-e h)}\\ \end {align*}

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Mathematica [B]  time = 0.55, size = 444, normalized size = 2.12 \[ \frac {3 b^2 p^2 q^2 \left (\log (e+f x) \left (h (e+f x) \log (e+f x)-2 f (g+h x) \log \left (\frac {f (g+h x)}{f g-e h}\right )\right )-2 f (g+h x) \text {Li}_2\left (\frac {h (e+f x)}{e h-f g}\right )\right ) \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )-b p q \log (e+f x)\right )-3 b p q (f g-e h) \log (e+f x) \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )-b p q \log (e+f x)\right )^2+3 b f p q (g+h x) \log (e+f x) \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )-b p q \log (e+f x)\right )^2-3 b f p q (g+h x) \log (g+h x) \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )-b p q \log (e+f x)\right )^2-(f g-e h) \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )-b p q \log (e+f x)\right )^3+b^3 p^3 q^3 \left (6 f (g+h x) \text {Li}_3\left (\frac {h (e+f x)}{e h-f g}\right )-6 f (g+h x) \log (e+f x) \text {Li}_2\left (\frac {h (e+f x)}{e h-f g}\right )+\log ^2(e+f x) \left (h (e+f x) \log (e+f x)-3 f (g+h x) \log \left (\frac {f (g+h x)}{f g-e h}\right )\right )\right )}{h (g+h x) (f g-e h)} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Log[c*(d*(e + f*x)^p)^q])^3/(g + h*x)^2,x]

[Out]

(-3*b*(f*g - e*h)*p*q*Log[e + f*x]*(a - b*p*q*Log[e + f*x] + b*Log[c*(d*(e + f*x)^p)^q])^2 + 3*b*f*p*q*(g + h*
x)*Log[e + f*x]*(a - b*p*q*Log[e + f*x] + b*Log[c*(d*(e + f*x)^p)^q])^2 - (f*g - e*h)*(a - b*p*q*Log[e + f*x]
+ b*Log[c*(d*(e + f*x)^p)^q])^3 - 3*b*f*p*q*(g + h*x)*(a - b*p*q*Log[e + f*x] + b*Log[c*(d*(e + f*x)^p)^q])^2*
Log[g + h*x] + 3*b^2*p^2*q^2*(a - b*p*q*Log[e + f*x] + b*Log[c*(d*(e + f*x)^p)^q])*(Log[e + f*x]*(h*(e + f*x)*
Log[e + f*x] - 2*f*(g + h*x)*Log[(f*(g + h*x))/(f*g - e*h)]) - 2*f*(g + h*x)*PolyLog[2, (h*(e + f*x))/(-(f*g)
+ e*h)]) + b^3*p^3*q^3*(Log[e + f*x]^2*(h*(e + f*x)*Log[e + f*x] - 3*f*(g + h*x)*Log[(f*(g + h*x))/(f*g - e*h)
]) - 6*f*(g + h*x)*Log[e + f*x]*PolyLog[2, (h*(e + f*x))/(-(f*g) + e*h)] + 6*f*(g + h*x)*PolyLog[3, (h*(e + f*
x))/(-(f*g) + e*h)]))/(h*(f*g - e*h)*(g + h*x))

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fricas [F]  time = 0.49, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{3} \log \left (\left ({\left (f x + e\right )}^{p} d\right )^{q} c\right )^{3} + 3 \, a b^{2} \log \left (\left ({\left (f x + e\right )}^{p} d\right )^{q} c\right )^{2} + 3 \, a^{2} b \log \left (\left ({\left (f x + e\right )}^{p} d\right )^{q} c\right ) + a^{3}}{h^{2} x^{2} + 2 \, g h x + g^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*(d*(f*x+e)^p)^q))^3/(h*x+g)^2,x, algorithm="fricas")

[Out]

integral((b^3*log(((f*x + e)^p*d)^q*c)^3 + 3*a*b^2*log(((f*x + e)^p*d)^q*c)^2 + 3*a^2*b*log(((f*x + e)^p*d)^q*
c) + a^3)/(h^2*x^2 + 2*g*h*x + g^2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \log \left (\left ({\left (f x + e\right )}^{p} d\right )^{q} c\right ) + a\right )}^{3}}{{\left (h x + g\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*(d*(f*x+e)^p)^q))^3/(h*x+g)^2,x, algorithm="giac")

[Out]

integrate((b*log(((f*x + e)^p*d)^q*c) + a)^3/(h*x + g)^2, x)

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maple [F]  time = 0.35, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \ln \left (c \left (d \left (f x +e \right )^{p}\right )^{q}\right )+a \right )^{3}}{\left (h x +g \right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*ln(c*(d*(f*x+e)^p)^q)+a)^3/(h*x+g)^2,x)

[Out]

int((b*ln(c*(d*(f*x+e)^p)^q)+a)^3/(h*x+g)^2,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ 3 \, a^{2} b f p q {\left (\frac {\log \left (f x + e\right )}{f g h - e h^{2}} - \frac {\log \left (h x + g\right )}{f g h - e h^{2}}\right )} - \frac {b^{3} \log \left ({\left ({\left (f x + e\right )}^{p}\right )}^{q}\right )^{3}}{h^{2} x + g h} - \frac {3 \, a^{2} b \log \left (\left ({\left (f x + e\right )}^{p} d\right )^{q} c\right )}{h^{2} x + g h} - \frac {a^{3}}{h^{2} x + g h} + \int \frac {3 \, {\left (e h q^{2} \log \relax (d)^{2} + 2 \, e h q \log \relax (c) \log \relax (d) + e h \log \relax (c)^{2}\right )} a b^{2} + {\left (e h q^{3} \log \relax (d)^{3} + 3 \, e h q^{2} \log \relax (c) \log \relax (d)^{2} + 3 \, e h q \log \relax (c)^{2} \log \relax (d) + e h \log \relax (c)^{3}\right )} b^{3} + 3 \, {\left (a b^{2} e h + {\left (f g p q + e h q \log \relax (d) + e h \log \relax (c)\right )} b^{3} + {\left (a b^{2} f h + {\left (f h p q + f h q \log \relax (d) + f h \log \relax (c)\right )} b^{3}\right )} x\right )} \log \left ({\left ({\left (f x + e\right )}^{p}\right )}^{q}\right )^{2} + {\left (3 \, {\left (f h q^{2} \log \relax (d)^{2} + 2 \, f h q \log \relax (c) \log \relax (d) + f h \log \relax (c)^{2}\right )} a b^{2} + {\left (f h q^{3} \log \relax (d)^{3} + 3 \, f h q^{2} \log \relax (c) \log \relax (d)^{2} + 3 \, f h q \log \relax (c)^{2} \log \relax (d) + f h \log \relax (c)^{3}\right )} b^{3}\right )} x + 3 \, {\left (2 \, {\left (e h q \log \relax (d) + e h \log \relax (c)\right )} a b^{2} + {\left (e h q^{2} \log \relax (d)^{2} + 2 \, e h q \log \relax (c) \log \relax (d) + e h \log \relax (c)^{2}\right )} b^{3} + {\left (2 \, {\left (f h q \log \relax (d) + f h \log \relax (c)\right )} a b^{2} + {\left (f h q^{2} \log \relax (d)^{2} + 2 \, f h q \log \relax (c) \log \relax (d) + f h \log \relax (c)^{2}\right )} b^{3}\right )} x\right )} \log \left ({\left ({\left (f x + e\right )}^{p}\right )}^{q}\right )}{f h^{3} x^{3} + e g^{2} h + {\left (2 \, f g h^{2} + e h^{3}\right )} x^{2} + {\left (f g^{2} h + 2 \, e g h^{2}\right )} x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*(d*(f*x+e)^p)^q))^3/(h*x+g)^2,x, algorithm="maxima")

[Out]

3*a^2*b*f*p*q*(log(f*x + e)/(f*g*h - e*h^2) - log(h*x + g)/(f*g*h - e*h^2)) - b^3*log(((f*x + e)^p)^q)^3/(h^2*
x + g*h) - 3*a^2*b*log(((f*x + e)^p*d)^q*c)/(h^2*x + g*h) - a^3/(h^2*x + g*h) + integrate((3*(e*h*q^2*log(d)^2
+ 2*e*h*q*log(c)*log(d) + e*h*log(c)^2)*a*b^2 + (e*h*q^3*log(d)^3 + 3*e*h*q^2*log(c)*log(d)^2 + 3*e*h*q*log(c
)^2*log(d) + e*h*log(c)^3)*b^3 + 3*(a*b^2*e*h + (f*g*p*q + e*h*q*log(d) + e*h*log(c))*b^3 + (a*b^2*f*h + (f*h*
p*q + f*h*q*log(d) + f*h*log(c))*b^3)*x)*log(((f*x + e)^p)^q)^2 + (3*(f*h*q^2*log(d)^2 + 2*f*h*q*log(c)*log(d)
+ f*h*log(c)^2)*a*b^2 + (f*h*q^3*log(d)^3 + 3*f*h*q^2*log(c)*log(d)^2 + 3*f*h*q*log(c)^2*log(d) + f*h*log(c)^
3)*b^3)*x + 3*(2*(e*h*q*log(d) + e*h*log(c))*a*b^2 + (e*h*q^2*log(d)^2 + 2*e*h*q*log(c)*log(d) + e*h*log(c)^2)
*b^3 + (2*(f*h*q*log(d) + f*h*log(c))*a*b^2 + (f*h*q^2*log(d)^2 + 2*f*h*q*log(c)*log(d) + f*h*log(c)^2)*b^3)*x
)*log(((f*x + e)^p)^q))/(f*h^3*x^3 + e*g^2*h + (2*f*g*h^2 + e*h^3)*x^2 + (f*g^2*h + 2*e*g*h^2)*x), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+b\,\ln \left (c\,{\left (d\,{\left (e+f\,x\right )}^p\right )}^q\right )\right )}^3}{{\left (g+h\,x\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*log(c*(d*(e + f*x)^p)^q))^3/(g + h*x)^2,x)

[Out]

int((a + b*log(c*(d*(e + f*x)^p)^q))^3/(g + h*x)^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b \log {\left (c \left (d \left (e + f x\right )^{p}\right )^{q} \right )}\right )^{3}}{\left (g + h x\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*ln(c*(d*(f*x+e)**p)**q))**3/(h*x+g)**2,x)

[Out]

Integral((a + b*log(c*(d*(e + f*x)**p)**q))**3/(g + h*x)**2, x)

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