3.188 \(\int \frac {(a+b \log (c (e+f x)))^2}{(d e+d f x) (h+i x)} \, dx\)

Optimal. Leaf size=142 \[ \frac {2 b \text {Li}_2\left (-\frac {f h-e i}{i (e+f x)}\right ) (a+b \log (c (e+f x)))}{d (f h-e i)}-\frac {\log \left (\frac {f h-e i}{i (e+f x)}+1\right ) (a+b \log (c (e+f x)))^2}{d (f h-e i)}+\frac {2 b^2 \text {Li}_3\left (-\frac {f h-e i}{i (e+f x)}\right )}{d (f h-e i)} \]

[Out]

-(a+b*ln(c*(f*x+e)))^2*ln(1+(-e*i+f*h)/i/(f*x+e))/d/(-e*i+f*h)+2*b*(a+b*ln(c*(f*x+e)))*polylog(2,(e*i-f*h)/i/(
f*x+e))/d/(-e*i+f*h)+2*b^2*polylog(3,(e*i-f*h)/i/(f*x+e))/d/(-e*i+f*h)

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Rubi [A]  time = 0.38, antiderivative size = 168, normalized size of antiderivative = 1.18, number of steps used = 8, number of rules used = 8, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2411, 12, 2344, 2302, 30, 2317, 2374, 6589} \[ -\frac {2 b \text {PolyLog}\left (2,-\frac {i (e+f x)}{f h-e i}\right ) (a+b \log (c (e+f x)))}{d (f h-e i)}+\frac {2 b^2 \text {PolyLog}\left (3,-\frac {i (e+f x)}{f h-e i}\right )}{d (f h-e i)}+\frac {(a+b \log (c (e+f x)))^3}{3 b d (f h-e i)}-\frac {\log \left (\frac {f (h+i x)}{f h-e i}\right ) (a+b \log (c (e+f x)))^2}{d (f h-e i)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2302

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rule 2317

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

Rule 2344

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

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 2411

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_.) + (g_.)*(x_))^(q_.)*((h_.) + (i_.)*(x_))
^(r_.), x_Symbol] :> Dist[1/e, Subst[Int[((g*x)/e)^q*((e*h - d*i)/e + (i*x)/e)^r*(a + b*Log[c*x^n])^p, x], x,
d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, h, i, n, p, q, r}, x] && EqQ[e*f - d*g, 0] && (IGtQ[p, 0] || IGtQ[
r, 0]) && IntegerQ[2*r]

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 {(a+b \log (c (e+f x)))^2}{(h+188 x) (d e+d f x)} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {(a+b \log (c x))^2}{d x \left (\frac {-188 e+f h}{f}+\frac {188 x}{f}\right )} \, dx,x,e+f x\right )}{f}\\ &=\frac {\operatorname {Subst}\left (\int \frac {(a+b \log (c x))^2}{x \left (\frac {-188 e+f h}{f}+\frac {188 x}{f}\right )} \, dx,x,e+f x\right )}{d f}\\ &=-\frac {\operatorname {Subst}\left (\int \frac {(a+b \log (c x))^2}{x} \, dx,x,e+f x\right )}{d (188 e-f h)}+\frac {188 \operatorname {Subst}\left (\int \frac {(a+b \log (c x))^2}{\frac {-188 e+f h}{f}+\frac {188 x}{f}} \, dx,x,e+f x\right )}{d f (188 e-f h)}\\ &=\frac {\log \left (-\frac {f (h+188 x)}{188 e-f h}\right ) (a+b \log (c (e+f x)))^2}{d (188 e-f h)}-\frac {\operatorname {Subst}\left (\int x^2 \, dx,x,a+b \log (c (e+f x))\right )}{b d (188 e-f h)}-\frac {(2 b) \operatorname {Subst}\left (\int \frac {(a+b \log (c x)) \log \left (1+\frac {188 x}{-188 e+f h}\right )}{x} \, dx,x,e+f x\right )}{d (188 e-f h)}\\ &=\frac {\log \left (-\frac {f (h+188 x)}{188 e-f h}\right ) (a+b \log (c (e+f x)))^2}{d (188 e-f h)}-\frac {(a+b \log (c (e+f x)))^3}{3 b d (188 e-f h)}+\frac {2 b (a+b \log (c (e+f x))) \text {Li}_2\left (\frac {188 (e+f x)}{188 e-f h}\right )}{d (188 e-f h)}-\frac {\left (2 b^2\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (-\frac {188 x}{-188 e+f h}\right )}{x} \, dx,x,e+f x\right )}{d (188 e-f h)}\\ &=\frac {\log \left (-\frac {f (h+188 x)}{188 e-f h}\right ) (a+b \log (c (e+f x)))^2}{d (188 e-f h)}-\frac {(a+b \log (c (e+f x)))^3}{3 b d (188 e-f h)}+\frac {2 b (a+b \log (c (e+f x))) \text {Li}_2\left (\frac {188 (e+f x)}{188 e-f h}\right )}{d (188 e-f h)}-\frac {2 b^2 \text {Li}_3\left (\frac {188 (e+f x)}{188 e-f h}\right )}{d (188 e-f h)}\\ \end {align*}

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Mathematica [A]  time = 0.24, size = 189, normalized size = 1.33 \[ \frac {3 a^2 \log (e+f x)-3 a^2 \log (h+i x)-6 b \text {Li}_2\left (\frac {i (e+f x)}{e i-f h}\right ) (a+b \log (c (e+f x)))-6 a b \log (c (e+f x)) \log \left (\frac {f (h+i x)}{f h-e i}\right )+3 a b \log ^2(c (e+f x))-3 b^2 \log ^2(c (e+f x)) \log \left (\frac {f (h+i x)}{f h-e i}\right )+b^2 \log ^3(c (e+f x))+6 b^2 \text {Li}_3\left (\frac {i (e+f x)}{e i-f h}\right )}{3 d (f h-e i)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.41, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{2} \log \left (c f x + c e\right )^{2} + 2 \, a b \log \left (c f x + c e\right ) + a^{2}}{d f i x^{2} + d e h + {\left (d f h + d e i\right )} x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.08, size = 383, normalized size = 2.70 \[ \frac {b^{2} \ln \left (\frac {\left (c f x +c e \right ) i}{-c e i +c f h}+1\right ) \ln \left (c f x +c e \right )^{2}}{\left (e i -f h \right ) d}-\frac {b^{2} \ln \left (c f x +c e \right )^{3}}{3 \left (e i -f h \right ) d}+\frac {2 a b \ln \left (\frac {-c e i +c f h +\left (c f x +c e \right ) i}{-c e i +c f h}\right ) \ln \left (c f x +c e \right )}{\left (e i -f h \right ) d}-\frac {a b \ln \left (c f x +c e \right )^{2}}{\left (e i -f h \right ) d}+\frac {2 b^{2} \polylog \left (2, -\frac {\left (c f x +c e \right ) i}{-c e i +c f h}\right ) \ln \left (c f x +c e \right )}{\left (e i -f h \right ) d}+\frac {a^{2} \ln \left (-c e i +c f h +\left (c f x +c e \right ) i \right )}{\left (e i -f h \right ) d}-\frac {a^{2} \ln \left (c f x +c e \right )}{\left (e i -f h \right ) d}+\frac {2 a b \dilog \left (\frac {-c e i +c f h +\left (c f x +c e \right ) i}{-c e i +c f h}\right )}{\left (e i -f h \right ) d}-\frac {2 b^{2} \polylog \left (3, -\frac {\left (c f x +c e \right ) i}{-c e i +c f h}\right )}{\left (e i -f h \right ) d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/d*a^2/(e*i-f*h)*ln(c*f*x+c*e)+1/d*a^2/(e*i-f*h)*ln(-c*e*i+c*f*h+(c*f*x+c*e)*i)-1/3/d*b^2/(e*i-f*h)*ln(c*f*x
+c*e)^3+1/d*b^2/(e*i-f*h)*ln(c*f*x+c*e)^2*ln(1+i/(-c*e*i+c*f*h)*(c*f*x+c*e))+2/d*b^2/(e*i-f*h)*ln(c*f*x+c*e)*p
olylog(2,-i/(-c*e*i+c*f*h)*(c*f*x+c*e))-2/d*b^2/(e*i-f*h)*polylog(3,-i/(-c*e*i+c*f*h)*(c*f*x+c*e))-1/d*a*b*ln(
c*f*x+c*e)^2/(e*i-f*h)+2/d*a*b/(e*i-f*h)*dilog((-c*e*i+c*f*h+(c*f*x+c*e)*i)/(-c*e*i+c*f*h))+2/d*a*b/(e*i-f*h)*
ln(c*f*x+c*e)*ln((-c*e*i+c*f*h+(c*f*x+c*e)*i)/(-c*e*i+c*f*h))

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maxima [B]  time = 0.73, size = 331, normalized size = 2.33 \[ a^{2} {\left (\frac {\log \left (f x + e\right )}{d f h - d e i} - \frac {\log \left (i x + h\right )}{d f h - d e i}\right )} - \frac {{\left (\log \left (f x + e\right )^{2} \log \left (\frac {f i x + e i}{f h - e i} + 1\right ) + 2 \, {\rm Li}_2\left (-\frac {f i x + e i}{f h - e i}\right ) \log \left (f x + e\right ) - 2 \, {\rm Li}_{3}(-\frac {f i x + e i}{f h - e i})\right )} b^{2}}{{\left (f h - e i\right )} d} - \frac {2 \, {\left (b^{2} \log \relax (c) + a b\right )} {\left (\log \left (f x + e\right ) \log \left (\frac {f i x + e i}{f h - e i} + 1\right ) + {\rm Li}_2\left (-\frac {f i x + e i}{f h - e i}\right )\right )}}{{\left (f h - e i\right )} d} - \frac {{\left (b^{2} \log \relax (c)^{2} + 2 \, a b \log \relax (c)\right )} \log \left (i x + h\right )}{{\left (f h - e i\right )} d} + \frac {b^{2} \log \left (f x + e\right )^{3} + 3 \, {\left (b^{2} \log \relax (c) + a b\right )} \log \left (f x + e\right )^{2} + 3 \, {\left (b^{2} \log \relax (c)^{2} + 2 \, a b \log \relax (c)\right )} \log \left (f x + e\right )}{3 \, {\left (f h - e i\right )} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^2*(log(f*x + e)/(d*f*h - d*e*i) - log(i*x + h)/(d*f*h - d*e*i)) - (log(f*x + e)^2*log((f*i*x + e*i)/(f*h - e
*i) + 1) + 2*dilog(-(f*i*x + e*i)/(f*h - e*i))*log(f*x + e) - 2*polylog(3, -(f*i*x + e*i)/(f*h - e*i)))*b^2/((
f*h - e*i)*d) - 2*(b^2*log(c) + a*b)*(log(f*x + e)*log((f*i*x + e*i)/(f*h - e*i) + 1) + dilog(-(f*i*x + e*i)/(
f*h - e*i)))/((f*h - e*i)*d) - (b^2*log(c)^2 + 2*a*b*log(c))*log(i*x + h)/((f*h - e*i)*d) + 1/3*(b^2*log(f*x +
e)^3 + 3*(b^2*log(c) + a*b)*log(f*x + e)^2 + 3*(b^2*log(c)^2 + 2*a*b*log(c))*log(f*x + e))/((f*h - e*i)*d)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+b\,\ln \left (c\,\left (e+f\,x\right )\right )\right )}^2}{\left (h+i\,x\right )\,\left (d\,e+d\,f\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*log(c*(e + f*x)))^2/((h + i*x)*(d*e + d*f*x)),x)

[Out]

int((a + b*log(c*(e + f*x)))^2/((h + i*x)*(d*e + d*f*x)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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