3.2.38 \(\int \frac {A+B \log (e (\frac {a+b x}{c+d x})^n)}{c i+d i x} \, dx\) [138]

Optimal. Leaf size=80 \[ -\frac {\left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \log \left (\frac {b c-a d}{b (c+d x)}\right )}{d i}-\frac {B n \text {Li}_2\left (\frac {d (a+b x)}{b (c+d x)}\right )}{d i} \]

[Out]

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

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Rubi [A]
time = 0.15, antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {2543, 2458, 2378, 2370, 2352} \begin {gather*} -\frac {B n \text {PolyLog}\left (2,\frac {d (a+b x)}{b (c+d x)}\right )}{d i}-\frac {\log \left (\frac {b c-a d}{b (c+d x)}\right ) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{d i} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2352

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-e^(-1))*PolyLog[2, 1 - c*x], x] /; FreeQ[{c, d, e
}, x] && EqQ[e + c*d, 0]

Rule 2370

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)/(x_))^(q_.)*(x_)^(m_.), x_Symbol] :> Int[(e + d*
x)^q*(a + b*Log[c*x^n])^p, x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && EqQ[m, q] && IntegerQ[q]

Rule 2378

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

Rule 2458

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 2543

Int[((A_.) + Log[(e_.)*(((a_.) + (b_.)*(x_))/((c_.) + (d_.)*(x_)))^(n_.)]*(B_.))/((f_.) + (g_.)*(x_)), x_Symbo
l] :> Simp[(-Log[(b*c - a*d)/(b*(c + d*x))])*((A + B*Log[e*((a + b*x)/(c + d*x))^n])/g), x] + Dist[B*n*((b*c -
a*d)/g), Int[Log[(b*c - a*d)/(b*(c + d*x))]/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f, g, A, B
, n}, x] && NeQ[b*c - a*d, 0] && EqQ[d*f - c*g, 0]

Rubi steps

\begin {align*} \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{138 c+138 d x} \, dx &=\frac {\left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \log (138 c+138 d x)}{138 d}-\frac {(B n) \int \frac {(c+d x) \left (-\frac {d (a+b x)}{(c+d x)^2}+\frac {b}{c+d x}\right ) \log (138 c+138 d x)}{a+b x} \, dx}{138 d}\\ &=\frac {\left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \log (138 c+138 d x)}{138 d}-\frac {(B n) \int \left (\frac {b \log (138 c+138 d x)}{a+b x}-\frac {d \log (138 c+138 d x)}{c+d x}\right ) \, dx}{138 d}\\ &=\frac {\left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \log (138 c+138 d x)}{138 d}+\frac {1}{138} (B n) \int \frac {\log (138 c+138 d x)}{c+d x} \, dx-\frac {(b B n) \int \frac {\log (138 c+138 d x)}{a+b x} \, dx}{138 d}\\ &=-\frac {B n \log \left (-\frac {d (a+b x)}{b c-a d}\right ) \log (138 c+138 d x)}{138 d}+\frac {\left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \log (138 c+138 d x)}{138 d}+(B n) \int \frac {\log \left (\frac {138 d (a+b x)}{-138 b c+138 a d}\right )}{138 c+138 d x} \, dx+\frac {(B n) \text {Subst}\left (\int \frac {138 \log (x)}{x} \, dx,x,138 c+138 d x\right )}{19044 d}\\ &=-\frac {B n \log \left (-\frac {d (a+b x)}{b c-a d}\right ) \log (138 c+138 d x)}{138 d}+\frac {\left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \log (138 c+138 d x)}{138 d}+\frac {(B n) \text {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,138 c+138 d x\right )}{138 d}+\frac {(B n) \text {Subst}\left (\int \frac {\log \left (1+\frac {b x}{-138 b c+138 a d}\right )}{x} \, dx,x,138 c+138 d x\right )}{138 d}\\ &=\frac {B n \log ^2(138 (c+d x))}{276 d}-\frac {B n \log \left (-\frac {d (a+b x)}{b c-a d}\right ) \log (138 c+138 d x)}{138 d}+\frac {\left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \log (138 c+138 d x)}{138 d}-\frac {B n \text {Li}_2\left (\frac {b (c+d x)}{b c-a d}\right )}{138 d}\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 101, normalized size = 1.26 \begin {gather*} \frac {\log (i (c+d x)) \left (2 A-2 B n \log \left (\frac {d (a+b x)}{-b c+a d}\right )+2 B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+B n \log (i (c+d x))\right )-2 B n \text {Li}_2\left (\frac {b (c+d x)}{b c-a d}\right )}{2 d i} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.20, size = 0, normalized size = 0.00 \[\int \frac {A +B \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right )}{d i x +c i}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*B*((2*I*n*log(b*x + a)*log(d*x + c) - I*n*log(d*x + c)^2 - 2*I*log(d*x + c)*log((b*x + a)^n) + 2*I*log(d*x
+ c)*log((d*x + c)^n))/d - 2*integrate((I*n*log(b*x + a) + I)/(d*x + c), x)) - I*A*log(I*d*x + I*c)/d

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((-I*B*n*log((b*x + a)/(d*x + c)) - I*A - I*B)/(d*x + c), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 563 vs. \(2 (74) = 148\).
time = 52.91, size = 563, normalized size = 7.04 \begin {gather*} -\frac {1}{2} \, {\left (\frac {{\left (i \, B b^{3} c^{3} n - 3 i \, B a b^{2} c^{2} d n + 3 i \, B a^{2} b c d^{2} n - i \, B a^{3} d^{3} n\right )} \log \left (\frac {b x + a}{d x + c}\right )}{b^{2} d - \frac {2 \, {\left (b x + a\right )} b d^{2}}{d x + c} + \frac {{\left (b x + a\right )}^{2} d^{3}}{{\left (d x + c\right )}^{2}}} + \frac {-i \, B b^{4} c^{3} n + 3 i \, B a b^{3} c^{2} d n + \frac {{\left (i \, b x + i \, a\right )} B b^{3} c^{3} d n}{d x + c} - 3 i \, B a^{2} b^{2} c d^{2} n - \frac {3 \, {\left (i \, b x + i \, a\right )} B a b^{2} c^{2} d^{2} n}{d x + c} + i \, B a^{3} b d^{3} n - \frac {3 \, {\left (-i \, b x - i \, a\right )} B a^{2} b c d^{3} n}{d x + c} + \frac {{\left (-i \, b x - i \, a\right )} B a^{3} d^{4} n}{d x + c} + i \, A b^{4} c^{3} + i \, B b^{4} c^{3} - 3 i \, A a b^{3} c^{2} d - 3 i \, B a b^{3} c^{2} d + 3 i \, A a^{2} b^{2} c d^{2} + 3 i \, B a^{2} b^{2} c d^{2} - i \, A a^{3} b d^{3} - i \, B a^{3} b d^{3}}{b^{3} d - \frac {2 \, {\left (b x + a\right )} b^{2} d^{2}}{d x + c} + \frac {{\left (b x + a\right )}^{2} b d^{3}}{{\left (d x + c\right )}^{2}}} - \frac {{\left (-i \, B b^{3} c^{3} n + 3 i \, B a b^{2} c^{2} d n - 3 i \, B a^{2} b c d^{2} n + i \, B a^{3} d^{3} n\right )} \log \left (-b + \frac {{\left (b x + a\right )} d}{d x + c}\right )}{b^{2} d} - \frac {{\left (i \, B b^{3} c^{3} n - 3 i \, B a b^{2} c^{2} d n + 3 i \, B a^{2} b c d^{2} n - i \, B a^{3} d^{3} n\right )} \log \left (\frac {b x + a}{d x + c}\right )}{b^{2} d}\right )} {\left (\frac {b c}{{\left (b c - a d\right )}^{2}} - \frac {a d}{{\left (b c - a d\right )}^{2}}\right )}^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*((I*B*b^3*c^3*n - 3*I*B*a*b^2*c^2*d*n + 3*I*B*a^2*b*c*d^2*n - I*B*a^3*d^3*n)*log((b*x + a)/(d*x + c))/(b^
2*d - 2*(b*x + a)*b*d^2/(d*x + c) + (b*x + a)^2*d^3/(d*x + c)^2) + (-I*B*b^4*c^3*n + 3*I*B*a*b^3*c^2*d*n + (I*
b*x + I*a)*B*b^3*c^3*d*n/(d*x + c) - 3*I*B*a^2*b^2*c*d^2*n - 3*(I*b*x + I*a)*B*a*b^2*c^2*d^2*n/(d*x + c) + I*B
*a^3*b*d^3*n - 3*(-I*b*x - I*a)*B*a^2*b*c*d^3*n/(d*x + c) + (-I*b*x - I*a)*B*a^3*d^4*n/(d*x + c) + I*A*b^4*c^3
+ I*B*b^4*c^3 - 3*I*A*a*b^3*c^2*d - 3*I*B*a*b^3*c^2*d + 3*I*A*a^2*b^2*c*d^2 + 3*I*B*a^2*b^2*c*d^2 - I*A*a^3*b
*d^3 - I*B*a^3*b*d^3)/(b^3*d - 2*(b*x + a)*b^2*d^2/(d*x + c) + (b*x + a)^2*b*d^3/(d*x + c)^2) - (-I*B*b^3*c^3*
n + 3*I*B*a*b^2*c^2*d*n - 3*I*B*a^2*b*c*d^2*n + I*B*a^3*d^3*n)*log(-b + (b*x + a)*d/(d*x + c))/(b^2*d) - (I*B*
b^3*c^3*n - 3*I*B*a*b^2*c^2*d*n + 3*I*B*a^2*b*c*d^2*n - I*B*a^3*d^3*n)*log((b*x + a)/(d*x + c))/(b^2*d))*(b*c/
(b*c - a*d)^2 - a*d/(b*c - a*d)^2)^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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