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

Optimal. Leaf size=68 \[ \frac{b p q \text{PolyLog}\left (2,-\frac{h (e+f x)}{f g-e h}\right )}{h}+\frac{\log \left (\frac{f (g+h x)}{f g-e h}\right ) \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{h} \]

[Out]

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

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Rubi [A]  time = 0.112487, antiderivative size = 68, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {2394, 2393, 2391, 2445} \[ \frac{b p q \text{PolyLog}\left (2,-\frac{h (e+f x)}{f g-e h}\right )}{h}+\frac{\log \left (\frac{f (g+h x)}{f g-e h}\right ) \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{h} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.0131297, size = 67, normalized size = 0.99 \[ \frac{b p q \text{PolyLog}\left (2,\frac{h (e+f x)}{e h-f g}\right )}{h}+\frac{\log \left (\frac{f (g+h x)}{f g-e h}\right ) \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{h} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.729, size = 0, normalized size = 0. \begin{align*} \int{\frac{a+b\ln \left ( c \left ( d \left ( fx+e \right ) ^{p} \right ) ^{q} \right ) }{hx+g}}\, 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),x)

[Out]

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

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

[Out]

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

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

[Out]

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

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

[Out]

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

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

[Out]

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