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

Optimal. Leaf size=144 \[ -\frac{2 b^2 f p^2 q^2 \text{PolyLog}\left (2,-\frac{h (e+f x)}{f g-e h}\right )}{h (f g-e h)}-\frac{2 b f p q \log \left (\frac{f (g+h x)}{f g-e h}\right ) \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{h (f g-e h)}+\frac{(e+f x) \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )^2}{(g+h x) (f g-e h)} \]

[Out]

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

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Rubi [A]  time = 0.200409, antiderivative size = 144, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.179, Rules used = {2397, 2394, 2393, 2391, 2445} \[ -\frac{2 b^2 f p^2 q^2 \text{PolyLog}\left (2,-\frac{h (e+f x)}{f g-e h}\right )}{h (f g-e h)}-\frac{2 b f p q \log \left (\frac{f (g+h x)}{f g-e h}\right ) \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{h (f g-e h)}+\frac{(e+f x) \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )^2}{(g+h x) (f g-e h)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 2394

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

Rule 2393

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

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

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{\left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )^2}{(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 )^2}{(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 )^2}{(f g-e h) (g+h x)}-\operatorname{Subst}\left (\frac{(2 b f p q) \int \frac{a+b \log \left (c d^q (e+f x)^{p q}\right )}{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 )^2}{(f g-e h) (g+h x)}-\frac{2 b f p q \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right ) \log \left (\frac{f (g+h x)}{f g-e h}\right )}{h (f g-e h)}+\operatorname{Subst}\left (\frac{\left (2 b^2 f^2 p^2 q^2\right ) \int \frac{\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 )^2}{(f g-e h) (g+h x)}-\frac{2 b f p q \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right ) \log \left (\frac{f (g+h x)}{f g-e h}\right )}{h (f g-e h)}+\operatorname{Subst}\left (\frac{\left (2 b^2 f p^2 q^2\right ) \operatorname{Subst}\left (\int \frac{\log \left (1+\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 )^2}{(f g-e h) (g+h x)}-\frac{2 b f p q \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right ) \log \left (\frac{f (g+h x)}{f g-e h}\right )}{h (f g-e h)}-\frac{2 b^2 f p^2 q^2 \text{Li}_2\left (-\frac{h (e+f x)}{f g-e h}\right )}{h (f g-e h)}\\ \end{align*}

Mathematica [A]  time = 0.229298, size = 200, normalized size = 1.39 \[ \frac{2 b^2 f p^2 q^2 (g+h x) \text{PolyLog}\left (2,\frac{h (e+f x)}{e h-f g}\right )-2 b f p q (g+h x) \log (e+f x) \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )+\left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right ) \left (a (f g-e h)+b (f g-e h) \log \left (c \left (d (e+f x)^p\right )^q\right )+2 b f p q (g+h x) \log \left (\frac{f (g+h x)}{f g-e h}\right )\right )+b^2 f p^2 q^2 (g+h x) \log ^2(e+f x)}{h (g+h x) (e h-f g)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.661, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( a+b\ln \left ( c \left ( d \left ( fx+e \right ) ^{p} \right ) ^{q} \right ) \right ) ^{2}}{ \left ( hx+g \right ) ^{2}}}\, 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))^2/(h*x+g)^2,x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} 2 \, a 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 )} - b^{2}{\left (\frac{\log \left ({\left ({\left (f x + e\right )}^{p}\right )}^{q}\right )^{2}}{h^{2} x + g h} - \int \frac{e h \log \left (c\right )^{2} + 2 \, e h \log \left (c\right ) \log \left (d^{q}\right ) + e h \log \left (d^{q}\right )^{2} +{\left (f h \log \left (c\right )^{2} + 2 \, f h \log \left (c\right ) \log \left (d^{q}\right ) + f h \log \left (d^{q}\right )^{2}\right )} x + 2 \,{\left (f g p q + e h \log \left (c\right ) + e h \log \left (d^{q}\right ) +{\left (f h p q + f h \log \left (c\right ) + f h \log \left (d^{q}\right )\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}\right )} - \frac{2 \, a b \log \left (\left ({\left (f x + e\right )}^{p} d\right )^{q} c\right )}{h^{2} x + g h} - \frac{a^{2}}{h^{2} x + g h} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b^{2} \log \left (\left ({\left (f x + e\right )}^{p} d\right )^{q} c\right )^{2} + 2 \, a b \log \left (\left ({\left (f x + e\right )}^{p} d\right )^{q} c\right ) + a^{2}}{h^{2} x^{2} + 2 \, g h x + g^{2}}, 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))^2/(h*x+g)^2,x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b \log{\left (c \left (d \left (e + f x\right )^{p}\right )^{q} \right )}\right )^{2}}{\left (g + h x\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \log \left (\left ({\left (f x + e\right )}^{p} d\right )^{q} c\right ) + a\right )}^{2}}{{\left (h x + g\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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