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

Optimal. Leaf size=107 \[ \frac {\log (a+b x) \log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{b}-\frac {q r \text {Li}_2\left (-\frac {d (a+b x)}{b c-a d}\right )}{b}-\frac {q r \log (a+b x) \log \left (\frac {b (c+d x)}{b c-a d}\right )}{b}-\frac {p r \log ^2(a+b x)}{2 b} \]

[Out]

-1/2*p*r*ln(b*x+a)^2/b-q*r*ln(b*x+a)*ln(b*(d*x+c)/(-a*d+b*c))/b+ln(b*x+a)*ln(e*(f*(b*x+a)^p*(d*x+c)^q)^r)/b-q*
r*polylog(2,-d*(b*x+a)/(-a*d+b*c))/b

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Rubi [A]  time = 0.08, antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {2494, 2390, 2301, 2394, 2393, 2391} \[ -\frac {q r \text {PolyLog}\left (2,-\frac {d (a+b x)}{b c-a d}\right )}{b}+\frac {\log (a+b x) \log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{b}-\frac {q r \log (a+b x) \log \left (\frac {b (c+d x)}{b c-a d}\right )}{b}-\frac {p r \log ^2(a+b x)}{2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2301

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

Rule 2390

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_) + (g_.)*(x_))^(q_.), x_Symbol] :> Dist[1/
e, Subst[Int[((f*x)/d)^q*(a + b*Log[c*x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p, q}, x]
&& EqQ[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 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 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 2494

Int[Log[(e_.)*((f_.)*((a_.) + (b_.)*(x_))^(p_.)*((c_.) + (d_.)*(x_))^(q_.))^(r_.)]/((g_.) + (h_.)*(x_)), x_Sym
bol] :> Simp[(Log[g + h*x]*Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r])/h, x] + (-Dist[(b*p*r)/h, Int[Log[g + h*x]/(a
+ b*x), x], x] - Dist[(d*q*r)/h, Int[Log[g + h*x]/(c + d*x), x], x]) /; FreeQ[{a, b, c, d, e, f, g, h, p, q,
r}, x] && NeQ[b*c - a*d, 0]

Rubi steps

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

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Mathematica [A]  time = 0.11, size = 93, normalized size = 0.87 \[ -\frac {\log (a+b x) \left (-2 \log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )+2 q r \log \left (\frac {b (c+d x)}{b c-a d}\right )+p r \log (a+b x)\right )+2 q r \text {Li}_2\left (\frac {d (a+b x)}{a d-b c}\right )}{2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(Log[a + b*x]*(p*r*Log[a + b*x] + 2*q*r*Log[(b*(c + d*x))/(b*c - a*d)] - 2*Log[e*(f*(a + b*x)^p*(c + d*x)
^q)^r]) + 2*q*r*PolyLog[2, (d*(a + b*x))/(-(b*c) + a*d)])/b

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fricas [F]  time = 0.43, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\log \left (\left ({\left (b x + a\right )}^{p} {\left (d x + c\right )}^{q} f\right )^{r} e\right )}{b x + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.33, size = 0, normalized size = 0.00 \[ \int \frac {\ln \left (e \left (f \left (b x +a \right )^{p} \left (d x +c \right )^{q}\right )^{r}\right )}{b x +a}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.80, size = 164, normalized size = 1.53 \[ -\frac {{\left (\frac {2 \, {\left (\log \left (b x + a\right ) \log \left (\frac {b d x + a d}{b c - a d} + 1\right ) + {\rm Li}_2\left (-\frac {b d x + a d}{b c - a d}\right )\right )} f q}{b} - \frac {f p \log \left (b x + a\right )^{2} + 2 \, f q \log \left (b x + a\right ) \log \left (d x + c\right )}{b}\right )} r}{2 \, f} - \frac {{\left (f p \log \left (b x + a\right ) + f q \log \left (d x + c\right )\right )} r \log \left (b x + a\right )}{b f} + \frac {\log \left (\left ({\left (b x + a\right )}^{p} {\left (d x + c\right )}^{q} f\right )^{r} e\right ) \log \left (b x + a\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*(log(b*x + a)*log((b*d*x + a*d)/(b*c - a*d) + 1) + dilog(-(b*d*x + a*d)/(b*c - a*d)))*f*q/b - (f*p*log
(b*x + a)^2 + 2*f*q*log(b*x + a)*log(d*x + c))/b)*r/f - (f*p*log(b*x + a) + f*q*log(d*x + c))*r*log(b*x + a)/(
b*f) + log(((b*x + a)^p*(d*x + c)^q*f)^r*e)*log(b*x + a)/b

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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