3.6.26 \(\int \frac {a+b \log (c (d (e+f x)^p)^q)}{g+h x} \, dx\) [526]

Optimal. Leaf size=68 \[ \frac {\left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right ) \log \left (\frac {f (g+h x)}{f g-e h}\right )}{h}+\frac {b p q \text {Li}_2\left (-\frac {h (e+f x)}{f g-e h}\right )}{h} \]

[Out]

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

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Rubi [A]
time = 0.08, antiderivative size = 68, normalized size of antiderivative = 1.00, 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 = {2441, 2440, 2438, 2495} \begin {gather*} \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} \end {gather*}

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 2438

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 2440

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 2441

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 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} \, dx &=\text {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}-\text {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}-\text {Subst}\left (\frac {(b p q) \text {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*}

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Mathematica [A]
time = 0.01, size = 67, normalized size = 0.99 \begin {gather*} \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 {gather*}

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 = 0.03, size = 0, normalized size = 0.00 \[\int \frac {a +b \ln \left (c \left (d \left (f x +e \right )^{p}\right )^{q}\right )}{h x +g}\, dx\]

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),x)

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \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),x, algorithm="maxima")

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \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),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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a + b \log {\left (c \left (d \left (e + f x\right )^{p}\right )^{q} \right )}}{g + h x}\, dx \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),x)

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \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),x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {a+b\,\ln \left (c\,{\left (d\,{\left (e+f\,x\right )}^p\right )}^q\right )}{g+h\,x} \,d x \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),x)

[Out]

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

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