3.112 \(\int \frac {(c i+d i x) (A+B \log (e (\frac {a+b x}{c+d x})^n))}{a g+b g x} \, dx\)

Optimal. Leaf size=141 \[ -\frac {i (b c-a d) \log \left (-\frac {b c-a d}{d (a+b x)}\right ) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A-B n\right )}{b^2 g}+\frac {i (c+d x) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{b g}+\frac {B i n (b c-a d) \text {Li}_2\left (\frac {b c-a d}{d (a+b x)}+1\right )}{b^2 g} \]

[Out]

i*(d*x+c)*(A+B*ln(e*((b*x+a)/(d*x+c))^n))/b/g-(-a*d+b*c)*i*ln((a*d-b*c)/d/(b*x+a))*(A-B*n+B*ln(e*((b*x+a)/(d*x
+c))^n))/b^2/g+B*(-a*d+b*c)*i*n*polylog(2,1+(-a*d+b*c)/d/(b*x+a))/b^2/g

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Rubi [A]  time = 0.34, antiderivative size = 223, normalized size of antiderivative = 1.58, number of steps used = 13, number of rules used = 10, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.244, Rules used = {2528, 2486, 31, 2524, 2418, 2390, 2301, 2394, 2393, 2391} \[ \frac {B i n (b c-a d) \text {PolyLog}\left (2,-\frac {d (a+b x)}{b c-a d}\right )}{b^2 g}+\frac {i (b c-a d) \log (a+b x) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{b^2 g}+\frac {B d i (a+b x) \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{b^2 g}-\frac {B i n (b c-a d) \log ^2(a+b x)}{2 b^2 g}+\frac {B i n (b c-a d) \log (a+b x) \log \left (\frac {b (c+d x)}{b c-a d}\right )}{b^2 g}-\frac {B i n (b c-a d) \log (c+d x)}{b^2 g}+\frac {A d i x}{b g} \]

Antiderivative was successfully verified.

[In]

Int[((c*i + d*i*x)*(A + B*Log[e*((a + b*x)/(c + d*x))^n]))/(a*g + b*g*x),x]

[Out]

(A*d*i*x)/(b*g) - (B*(b*c - a*d)*i*n*Log[a + b*x]^2)/(2*b^2*g) + (B*d*i*(a + b*x)*Log[e*((a + b*x)/(c + d*x))^
n])/(b^2*g) + ((b*c - a*d)*i*Log[a + b*x]*(A + B*Log[e*((a + b*x)/(c + d*x))^n]))/(b^2*g) - (B*(b*c - a*d)*i*n
*Log[c + d*x])/(b^2*g) + (B*(b*c - a*d)*i*n*Log[a + b*x]*Log[(b*(c + d*x))/(b*c - a*d)])/(b^2*g) + (B*(b*c - a
*d)*i*n*PolyLog[2, -((d*(a + b*x))/(b*c - a*d))])/(b^2*g)

Rule 31

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

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 2418

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*(RFx_), x_Symbol] :> With[{u = ExpandIntegrand[
(a + b*Log[c*(d + e*x)^n])^p, RFx, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, d, e, n}, x] && RationalFunct
ionQ[RFx, x] && IntegerQ[p]

Rule 2486

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

Rule 2524

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

Rule 2528

Int[((a_.) + Log[(c_.)*(RFx_)^(p_.)]*(b_.))^(n_.)*(RGx_), x_Symbol] :> With[{u = ExpandIntegrand[(a + b*Log[c*
RFx^p])^n, RGx, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, p}, x] && RationalFunctionQ[RFx, x] && RationalF
unctionQ[RGx, x] && IGtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {(112 c+112 d x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{a g+b g x} \, dx &=\int \left (\frac {112 d \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{b g}+\frac {112 (b c-a d) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{b g (a+b x)}\right ) \, dx\\ &=\frac {(112 d) \int \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \, dx}{b g}+\frac {(112 (b c-a d)) \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{a+b x} \, dx}{b g}\\ &=\frac {112 A d x}{b g}+\frac {112 (b c-a d) \log (a+b x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{b^2 g}+\frac {(112 B d) \int \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right ) \, dx}{b g}-\frac {(112 B (b c-a d) n) \int \frac {(c+d x) \left (-\frac {d (a+b x)}{(c+d x)^2}+\frac {b}{c+d x}\right ) \log (a+b x)}{a+b x} \, dx}{b^2 g}\\ &=\frac {112 A d x}{b g}+\frac {112 B d (a+b x) \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{b^2 g}+\frac {112 (b c-a d) \log (a+b x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{b^2 g}-\frac {(112 B (b c-a d) n) \int \left (\frac {b \log (a+b x)}{a+b x}-\frac {d \log (a+b x)}{c+d x}\right ) \, dx}{b^2 g}-\frac {(112 B d (b c-a d) n) \int \frac {1}{c+d x} \, dx}{b^2 g}\\ &=\frac {112 A d x}{b g}+\frac {112 B d (a+b x) \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{b^2 g}+\frac {112 (b c-a d) \log (a+b x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{b^2 g}-\frac {112 B (b c-a d) n \log (c+d x)}{b^2 g}-\frac {(112 B (b c-a d) n) \int \frac {\log (a+b x)}{a+b x} \, dx}{b g}+\frac {(112 B d (b c-a d) n) \int \frac {\log (a+b x)}{c+d x} \, dx}{b^2 g}\\ &=\frac {112 A d x}{b g}+\frac {112 B d (a+b x) \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{b^2 g}+\frac {112 (b c-a d) \log (a+b x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{b^2 g}-\frac {112 B (b c-a d) n \log (c+d x)}{b^2 g}+\frac {112 B (b c-a d) n \log (a+b x) \log \left (\frac {b (c+d x)}{b c-a d}\right )}{b^2 g}-\frac {(112 B (b c-a d) n) \operatorname {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,a+b x\right )}{b^2 g}-\frac {(112 B (b c-a d) n) \int \frac {\log \left (\frac {b (c+d x)}{b c-a d}\right )}{a+b x} \, dx}{b g}\\ &=\frac {112 A d x}{b g}-\frac {56 B (b c-a d) n \log ^2(a+b x)}{b^2 g}+\frac {112 B d (a+b x) \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{b^2 g}+\frac {112 (b c-a d) \log (a+b x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{b^2 g}-\frac {112 B (b c-a d) n \log (c+d x)}{b^2 g}+\frac {112 B (b c-a d) n \log (a+b x) \log \left (\frac {b (c+d x)}{b c-a d}\right )}{b^2 g}-\frac {(112 B (b c-a d) n) \operatorname {Subst}\left (\int \frac {\log \left (1+\frac {d x}{b c-a d}\right )}{x} \, dx,x,a+b x\right )}{b^2 g}\\ &=\frac {112 A d x}{b g}-\frac {56 B (b c-a d) n \log ^2(a+b x)}{b^2 g}+\frac {112 B d (a+b x) \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{b^2 g}+\frac {112 (b c-a d) \log (a+b x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{b^2 g}-\frac {112 B (b c-a d) n \log (c+d x)}{b^2 g}+\frac {112 B (b c-a d) n \log (a+b x) \log \left (\frac {b (c+d x)}{b c-a d}\right )}{b^2 g}+\frac {112 B (b c-a d) n \text {Li}_2\left (-\frac {d (a+b x)}{b c-a d}\right )}{b^2 g}\\ \end {align*}

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Mathematica [A]  time = 0.13, size = 172, normalized size = 1.22 \[ \frac {i \left (2 (b c-a d) \log (a+b x) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+B n \log \left (\frac {b (c+d x)}{b c-a d}\right )+A\right )+2 \left (B d (a+b x) \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+B n (a d-b c) \log (c+d x)+A b d x\right )+2 B n (b c-a d) \text {Li}_2\left (\frac {d (a+b x)}{a d-b c}\right )+B n (a d-b c) \log ^2(a+b x)\right )}{2 b^2 g} \]

Antiderivative was successfully verified.

[In]

Integrate[((c*i + d*i*x)*(A + B*Log[e*((a + b*x)/(c + d*x))^n]))/(a*g + b*g*x),x]

[Out]

(i*(B*(-(b*c) + a*d)*n*Log[a + b*x]^2 + 2*(A*b*d*x + B*d*(a + b*x)*Log[e*((a + b*x)/(c + d*x))^n] + B*(-(b*c)
+ a*d)*n*Log[c + d*x]) + 2*(b*c - a*d)*Log[a + b*x]*(A + B*Log[e*((a + b*x)/(c + d*x))^n] + B*n*Log[(b*(c + d*
x))/(b*c - a*d)]) + 2*B*(b*c - a*d)*n*PolyLog[2, (d*(a + b*x))/(-(b*c) + a*d)]))/(2*b^2*g)

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fricas [F]  time = 0.81, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {A d i x + A c i + {\left (B d i x + B c i\right )} \log \left (e \left (\frac {b x + a}{d x + c}\right )^{n}\right )}{b g x + a g}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*i*x+c*i)*(A+B*log(e*((b*x+a)/(d*x+c))^n))/(b*g*x+a*g),x, algorithm="fricas")

[Out]

integral((A*d*i*x + A*c*i + (B*d*i*x + B*c*i)*log(e*((b*x + a)/(d*x + c))^n))/(b*g*x + a*g), x)

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*i*x+c*i)*(A+B*log(e*((b*x+a)/(d*x+c))^n))/(b*g*x+a*g),x, algorithm="giac")

[Out]

Timed out

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maple [F]  time = 0.41, size = 0, normalized size = 0.00 \[ \int \frac {\left (d i x +c i \right ) \left (B \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right )+A \right )}{b g x +a g}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*i*x+c*i)*(B*ln(e*((b*x+a)/(d*x+c))^n)+A)/(b*g*x+a*g),x)

[Out]

int((d*i*x+c*i)*(B*ln(e*((b*x+a)/(d*x+c))^n)+A)/(b*g*x+a*g),x)

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maxima [A]  time = 4.58, size = 276, normalized size = 1.96 \[ A d i {\left (\frac {x}{b g} - \frac {a \log \left (b x + a\right )}{b^{2} g}\right )} - \frac {B c i n \log \left (d x + c\right )}{b g} + \frac {A c i \log \left (b g x + a g\right )}{b g} + \frac {{\left (b c i n - a d i n\right )} {\left (\log \left (b x + a\right ) \log \left (\frac {b d x + a d}{b c - a d} + 1\right ) + {\rm Li}_2\left (-\frac {b d x + a d}{b c - a d}\right )\right )} B}{b^{2} g} + \frac {2 \, B b d i x \log \relax (e) - {\left (b c i n - a d i n\right )} B \log \left (b x + a\right )^{2} + 2 \, {\left (b c i \log \relax (e) + {\left (i n - i \log \relax (e)\right )} a d\right )} B \log \left (b x + a\right ) + 2 \, {\left (B b d i x + {\left (b c i - a d i\right )} B \log \left (b x + a\right )\right )} \log \left ({\left (b x + a\right )}^{n}\right ) - 2 \, {\left (B b d i x + {\left (b c i - a d i\right )} B \log \left (b x + a\right )\right )} \log \left ({\left (d x + c\right )}^{n}\right )}{2 \, b^{2} g} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*i*x+c*i)*(A+B*log(e*((b*x+a)/(d*x+c))^n))/(b*g*x+a*g),x, algorithm="maxima")

[Out]

A*d*i*(x/(b*g) - a*log(b*x + a)/(b^2*g)) - B*c*i*n*log(d*x + c)/(b*g) + A*c*i*log(b*g*x + a*g)/(b*g) + (b*c*i*
n - a*d*i*n)*(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)))*B/(b^2*g) +
1/2*(2*B*b*d*i*x*log(e) - (b*c*i*n - a*d*i*n)*B*log(b*x + a)^2 + 2*(b*c*i*log(e) + (i*n - i*log(e))*a*d)*B*lo
g(b*x + a) + 2*(B*b*d*i*x + (b*c*i - a*d*i)*B*log(b*x + a))*log((b*x + a)^n) - 2*(B*b*d*i*x + (b*c*i - a*d*i)*
B*log(b*x + a))*log((d*x + c)^n))/(b^2*g)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((c*i + d*i*x)*(A + B*log(e*((a + b*x)/(c + d*x))^n)))/(a*g + b*g*x),x)

[Out]

int(((c*i + d*i*x)*(A + B*log(e*((a + b*x)/(c + d*x))^n)))/(a*g + b*g*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*i*x+c*i)*(A+B*ln(e*((b*x+a)/(d*x+c))**n))/(b*g*x+a*g),x)

[Out]

i*(Integral(A*c/(a + b*x), x) + Integral(A*d*x/(a + b*x), x) + Integral(B*c*log(e*(a/(c + d*x) + b*x/(c + d*x)
)**n)/(a + b*x), x) + Integral(B*d*x*log(e*(a/(c + d*x) + b*x/(c + d*x))**n)/(a + b*x), x))/g

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