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

Optimal. Leaf size=87 \[ \frac{b \text{PolyLog}\left (2,-\frac{f h-e i}{i (e+f x)}\right )}{d (f h-e i)}-\frac{\log \left (\frac{f h-e i}{i (e+f x)}+1\right ) (a+b \log (c (e+f x)))}{d (f h-e i)} \]

[Out]

-(((a + b*Log[c*(e + f*x)])*Log[1 + (f*h - e*i)/(i*(e + f*x))])/(d*(f*h - e*i))) + (b*PolyLog[2, -((f*h - e*i)
/(i*(e + f*x)))])/(d*(f*h - e*i))

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Rubi [A]  time = 0.234306, antiderivative size = 116, normalized size of antiderivative = 1.33, number of steps used = 6, number of rules used = 6, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2411, 12, 2344, 2301, 2317, 2391} \[ -\frac{b \text{PolyLog}\left (2,-\frac{i (e+f x)}{f h-e i}\right )}{d (f h-e i)}+\frac{(a+b \log (c (e+f x)))^2}{2 b d (f h-e i)}-\frac{\log \left (\frac{f (h+i x)}{f h-e i}\right ) (a+b \log (c (e+f x)))}{d (f h-e i)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 12

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

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 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 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 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]

Rubi steps

\begin{align*} \int \frac{a+b \log (c (e+f x))}{(h+180 x) (d e+d f x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{a+b \log (c x)}{d x \left (\frac{-180 e+f h}{f}+\frac{180 x}{f}\right )} \, dx,x,e+f x\right )}{f}\\ &=\frac{\operatorname{Subst}\left (\int \frac{a+b \log (c x)}{x \left (\frac{-180 e+f h}{f}+\frac{180 x}{f}\right )} \, dx,x,e+f x\right )}{d f}\\ &=-\frac{\operatorname{Subst}\left (\int \frac{a+b \log (c x)}{x} \, dx,x,e+f x\right )}{d (180 e-f h)}+\frac{180 \operatorname{Subst}\left (\int \frac{a+b \log (c x)}{\frac{-180 e+f h}{f}+\frac{180 x}{f}} \, dx,x,e+f x\right )}{d f (180 e-f h)}\\ &=\frac{\log \left (-\frac{f (h+180 x)}{180 e-f h}\right ) (a+b \log (c (e+f x)))}{d (180 e-f h)}-\frac{(a+b \log (c (e+f x)))^2}{2 b d (180 e-f h)}-\frac{b \operatorname{Subst}\left (\int \frac{\log \left (1+\frac{180 x}{-180 e+f h}\right )}{x} \, dx,x,e+f x\right )}{d (180 e-f h)}\\ &=\frac{\log \left (-\frac{f (h+180 x)}{180 e-f h}\right ) (a+b \log (c (e+f x)))}{d (180 e-f h)}-\frac{(a+b \log (c (e+f x)))^2}{2 b d (180 e-f h)}+\frac{b \text{Li}_2\left (\frac{180 (e+f x)}{180 e-f h}\right )}{d (180 e-f h)}\\ \end{align*}

Mathematica [A]  time = 0.0646556, size = 91, normalized size = 1.05 \[ \frac{(a+b \log (c (e+f x))) \left (a+b \log (c (e+f x))-2 b \log \left (\frac{f (h+i x)}{f h-e i}\right )\right )-2 b^2 \text{PolyLog}\left (2,\frac{i (e+f x)}{e i-f h}\right )}{2 b d (f h-e i)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.489, size = 197, normalized size = 2.3 \begin{align*} -{\frac{a\ln \left ( cfx+ce \right ) }{d \left ( ei-fh \right ) }}+{\frac{a\ln \left ( -cei+hcf+ \left ( cfx+ce \right ) i \right ) }{d \left ( ei-fh \right ) }}-{\frac{b \left ( \ln \left ( cfx+ce \right ) \right ) ^{2}}{2\,d \left ( ei-fh \right ) }}+{\frac{b}{d \left ( ei-fh \right ) }{\it dilog} \left ({\frac{-cei+hcf+ \left ( cfx+ce \right ) i}{-cei+hcf}} \right ) }+{\frac{b\ln \left ( cfx+ce \right ) }{d \left ( ei-fh \right ) }\ln \left ({\frac{-cei+hcf+ \left ( cfx+ce \right ) i}{-cei+hcf}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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