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

Optimal. Leaf size=151 \[ \frac{b f \text{PolyLog}\left (2,-\frac{f h-e i}{i (e+f x)}\right )}{d (f h-e i)^2}-\frac{f \log \left (\frac{f h-e i}{i (e+f x)}+1\right ) (a+b \log (c (e+f x)))}{d (f h-e i)^2}-\frac{i (e+f x) (a+b \log (c (e+f x)))}{d (h+i x) (f h-e i)^2}+\frac{b f \log (h+i x)}{d (f h-e i)^2} \]

[Out]

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

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Rubi [A]  time = 0.364054, antiderivative size = 181, normalized size of antiderivative = 1.2, number of steps used = 9, number of rules used = 9, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {2411, 12, 2347, 2344, 2301, 2317, 2391, 2314, 31} \[ -\frac{b f \text{PolyLog}\left (2,-\frac{i (e+f x)}{f h-e i}\right )}{d (f h-e i)^2}+\frac{f (a+b \log (c (e+f x)))^2}{2 b d (f h-e i)^2}-\frac{f \log \left (\frac{f (h+i x)}{f h-e i}\right ) (a+b \log (c (e+f x)))}{d (f h-e i)^2}-\frac{i (e+f x) (a+b \log (c (e+f x)))}{d (h+i x) (f h-e i)^2}+\frac{b f \log (h+i x)}{d (f h-e i)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 2347

Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_))^(q_))/(x_), x_Symbol] :> Dist[1/d, Int[((
d + e*x)^(q + 1)*(a + b*Log[c*x^n])^p)/x, x], x] - Dist[e/d, Int[(d + e*x)^q*(a + b*Log[c*x^n])^p, x], x] /; F
reeQ[{a, b, c, d, e, n}, x] && IGtQ[p, 0] && LtQ[q, -1] && IntegerQ[2*q]

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]

Rule 2314

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

\begin{align*} \int \frac{a+b \log (c (e+f x))}{(h+181 x)^2 (d e+d f x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{a+b \log (c x)}{d x \left (\frac{-181 e+f h}{f}+\frac{181 x}{f}\right )^2} \, dx,x,e+f x\right )}{f}\\ &=\frac{\operatorname{Subst}\left (\int \frac{a+b \log (c x)}{x \left (\frac{-181 e+f h}{f}+\frac{181 x}{f}\right )^2} \, dx,x,e+f x\right )}{d f}\\ &=-\frac{\operatorname{Subst}\left (\int \frac{a+b \log (c x)}{x \left (\frac{-181 e+f h}{f}+\frac{181 x}{f}\right )} \, dx,x,e+f x\right )}{d (181 e-f h)}+\frac{181 \operatorname{Subst}\left (\int \frac{a+b \log (c x)}{\left (\frac{-181 e+f h}{f}+\frac{181 x}{f}\right )^2} \, dx,x,e+f x\right )}{d f (181 e-f h)}\\ &=-\frac{181 (e+f x) (a+b \log (c (e+f x)))}{d (181 e-f h)^2 (h+181 x)}-\frac{181 \operatorname{Subst}\left (\int \frac{a+b \log (c x)}{\frac{-181 e+f h}{f}+\frac{181 x}{f}} \, dx,x,e+f x\right )}{d (181 e-f h)^2}+\frac{(181 b) \operatorname{Subst}\left (\int \frac{1}{\frac{-181 e+f h}{f}+\frac{181 x}{f}} \, dx,x,e+f x\right )}{d (181 e-f h)^2}+\frac{f \operatorname{Subst}\left (\int \frac{a+b \log (c x)}{x} \, dx,x,e+f x\right )}{d (181 e-f h)^2}\\ &=\frac{b f \log (h+181 x)}{d (181 e-f h)^2}-\frac{181 (e+f x) (a+b \log (c (e+f x)))}{d (181 e-f h)^2 (h+181 x)}-\frac{f \log \left (-\frac{f (h+181 x)}{181 e-f h}\right ) (a+b \log (c (e+f x)))}{d (181 e-f h)^2}+\frac{f (a+b \log (c (e+f x)))^2}{2 b d (181 e-f h)^2}+\frac{(b f) \operatorname{Subst}\left (\int \frac{\log \left (1+\frac{181 x}{-181 e+f h}\right )}{x} \, dx,x,e+f x\right )}{d (181 e-f h)^2}\\ &=\frac{b f \log (h+181 x)}{d (181 e-f h)^2}-\frac{181 (e+f x) (a+b \log (c (e+f x)))}{d (181 e-f h)^2 (h+181 x)}-\frac{f \log \left (-\frac{f (h+181 x)}{181 e-f h}\right ) (a+b \log (c (e+f x)))}{d (181 e-f h)^2}+\frac{f (a+b \log (c (e+f x)))^2}{2 b d (181 e-f h)^2}-\frac{b f \text{Li}_2\left (\frac{181 (e+f x)}{181 e-f h}\right )}{d (181 e-f h)^2}\\ \end{align*}

Mathematica [A]  time = 0.154471, size = 141, normalized size = 0.93 \[ \frac{-2 b f \text{PolyLog}\left (2,\frac{i (e+f x)}{e i-f h}\right )-2 f \log \left (\frac{f (h+i x)}{f h-e i}\right ) (a+b \log (c (e+f x)))+\frac{2 (f h-e i) (a+b \log (c (e+f x)))}{h+i x}+\frac{f (a+b \log (c (e+f x)))^2}{b}-2 b f (\log (e+f x)-\log (h+i x))}{2 d (f h-e i)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.563, size = 355, normalized size = 2.4 \begin{align*}{\frac{af\ln \left ( cfx+ce \right ) }{d \left ( ei-fh \right ) ^{2}}}-{\frac{acf}{d \left ( ei-fh \right ) \left ( cfix+hcf \right ) }}-{\frac{af\ln \left ( -cei+hcf+ \left ( cfx+ce \right ) i \right ) }{d \left ( ei-fh \right ) ^{2}}}+{\frac{bf \left ( \ln \left ( cfx+ce \right ) \right ) ^{2}}{2\,d \left ( ei-fh \right ) ^{2}}}-{\frac{bf}{d \left ( ei-fh \right ) ^{2}}{\it dilog} \left ({\frac{-cei+hcf+ \left ( cfx+ce \right ) i}{-cei+hcf}} \right ) }-{\frac{bf\ln \left ( cfx+ce \right ) }{d \left ( ei-fh \right ) ^{2}}\ln \left ({\frac{-cei+hcf+ \left ( cfx+ce \right ) i}{-cei+hcf}} \right ) }+{\frac{bf\ln \left ( -cei+hcf+ \left ( cfx+ce \right ) i \right ) }{d \left ( ei-fh \right ) ^{2}}}-{\frac{c{f}^{2}bi\ln \left ( cfx+ce \right ) x}{d \left ( ei-fh \right ) ^{2} \left ( cfix+hcf \right ) }}-{\frac{cfbi\ln \left ( cfx+ce \right ) e}{d \left ( ei-fh \right ) ^{2} \left ( cfix+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)^2,x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a{\left (\frac{f \log \left (f x + e\right )}{d f^{2} h^{2} - 2 \, d e f h i + d e^{2} i^{2}} - \frac{f \log \left (i x + h\right )}{d f^{2} h^{2} - 2 \, d e f h i + d e^{2} i^{2}} + \frac{1}{d f h^{2} - d e h i +{\left (d f h i - d e i^{2}\right )} x}\right )} + b \int \frac{\log \left (f x + e\right ) + \log \left (c\right )}{d f i^{2} x^{3} + d e h^{2} +{\left (2 \, f h i + e i^{2}\right )} d x^{2} +{\left (f h^{2} + 2 \, e h 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)^2,x, algorithm="maxima")

[Out]

a*(f*log(f*x + e)/(d*f^2*h^2 - 2*d*e*f*h*i + d*e^2*i^2) - f*log(i*x + h)/(d*f^2*h^2 - 2*d*e*f*h*i + d*e^2*i^2)
+ 1/(d*f*h^2 - d*e*h*i + (d*f*h*i - d*e*i^2)*x)) + b*integrate((log(f*x + e) + log(c))/(d*f*i^2*x^3 + d*e*h^2
+ (2*f*h*i + e*i^2)*d*x^2 + (f*h^2 + 2*e*h*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^{2} x^{3} + d e h^{2} +{\left (2 \, d f h i + d e i^{2}\right )} x^{2} +{\left (d f h^{2} + 2 \, d e h 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)^2,x, algorithm="fricas")

[Out]

integral((b*log(c*f*x + c*e) + a)/(d*f*i^2*x^3 + d*e*h^2 + (2*d*f*h*i + d*e*i^2)*x^2 + (d*f*h^2 + 2*d*e*h*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)**2,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 )}^{2}}\,{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)^2,x, algorithm="giac")

[Out]

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