3.2.40 \(\int \frac {A+B \log (e (\frac {a+b x}{c+d x})^n)}{(a g+b g x)^2 (c i+d i x)} \, dx\) [140]

Optimal. Leaf size=181 \[ -\frac {b B n (c+d x)}{(b c-a d)^2 g^2 i (a+b x)}-\frac {b (c+d x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{(b c-a d)^2 g^2 i (a+b x)}-\frac {d \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \log \left (\frac {a+b x}{c+d x}\right )}{(b c-a d)^2 g^2 i}+\frac {B d n \log ^2\left (\frac {a+b x}{c+d x}\right )}{2 (b c-a d)^2 g^2 i} \]

[Out]

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

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Rubi [A]
time = 0.12, antiderivative size = 181, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.116, Rules used = {2561, 45, 2372, 14, 2338} \begin {gather*} -\frac {d \log \left (\frac {a+b x}{c+d x}\right ) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{g^2 i (b c-a d)^2}-\frac {b (c+d x) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{g^2 i (a+b x) (b c-a d)^2}-\frac {b B n (c+d x)}{g^2 i (a+b x) (b c-a d)^2}+\frac {B d n \log ^2\left (\frac {a+b x}{c+d x}\right )}{2 g^2 i (b c-a d)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2338

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 2372

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(x_)^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> With[{u = I
ntHide[x^m*(d + e*x^r)^q, x]}, Dist[a + b*Log[c*x^n], u, x] - Dist[b*n, Int[SimplifyIntegrand[u/x, x], x], x]]
/; FreeQ[{a, b, c, d, e, n, r}, x] && IGtQ[q, 0] && IntegerQ[m] &&  !(EqQ[q, 1] && EqQ[m, -1])

Rule 2561

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

Rubi steps

\begin {align*} \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{(140 c+140 d x) (a g+b g x)^2} \, dx &=\int \left (\frac {b \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{140 (b c-a d) g^2 (a+b x)^2}-\frac {b d \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{140 (b c-a d)^2 g^2 (a+b x)}+\frac {d^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{140 (b c-a d)^2 g^2 (c+d x)}\right ) \, dx\\ &=-\frac {(b d) \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{a+b x} \, dx}{140 (b c-a d)^2 g^2}+\frac {d^2 \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{c+d x} \, dx}{140 (b c-a d)^2 g^2}+\frac {b \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{(a+b x)^2} \, dx}{140 (b c-a d) g^2}\\ &=-\frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{140 (b c-a d) g^2 (a+b x)}-\frac {d \log (a+b x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{140 (b c-a d)^2 g^2}+\frac {d \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \log (c+d x)}{140 (b c-a d)^2 g^2}+\frac {(B 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}{140 (b c-a d)^2 g^2}-\frac {(B d n) \int \frac {(c+d x) \left (-\frac {d (a+b x)}{(c+d x)^2}+\frac {b}{c+d x}\right ) \log (c+d x)}{a+b x} \, dx}{140 (b c-a d)^2 g^2}+\frac {(B n) \int \frac {b c-a d}{(a+b x)^2 (c+d x)} \, dx}{140 (b c-a d) g^2}\\ &=-\frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{140 (b c-a d) g^2 (a+b x)}-\frac {d \log (a+b x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{140 (b c-a d)^2 g^2}+\frac {d \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \log (c+d x)}{140 (b c-a d)^2 g^2}+\frac {(B n) \int \frac {1}{(a+b x)^2 (c+d x)} \, dx}{140 g^2}+\frac {(B d n) \int \left (\frac {b \log (a+b x)}{a+b x}-\frac {d \log (a+b x)}{c+d x}\right ) \, dx}{140 (b c-a d)^2 g^2}-\frac {(B d n) \int \left (\frac {b \log (c+d x)}{a+b x}-\frac {d \log (c+d x)}{c+d x}\right ) \, dx}{140 (b c-a d)^2 g^2}\\ &=-\frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{140 (b c-a d) g^2 (a+b x)}-\frac {d \log (a+b x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{140 (b c-a d)^2 g^2}+\frac {d \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \log (c+d x)}{140 (b c-a d)^2 g^2}+\frac {(B n) \int \left (\frac {b}{(b c-a d) (a+b x)^2}-\frac {b d}{(b c-a d)^2 (a+b x)}+\frac {d^2}{(b c-a d)^2 (c+d x)}\right ) \, dx}{140 g^2}+\frac {(b B d n) \int \frac {\log (a+b x)}{a+b x} \, dx}{140 (b c-a d)^2 g^2}-\frac {(b B d n) \int \frac {\log (c+d x)}{a+b x} \, dx}{140 (b c-a d)^2 g^2}-\frac {\left (B d^2 n\right ) \int \frac {\log (a+b x)}{c+d x} \, dx}{140 (b c-a d)^2 g^2}+\frac {\left (B d^2 n\right ) \int \frac {\log (c+d x)}{c+d x} \, dx}{140 (b c-a d)^2 g^2}\\ &=-\frac {B n}{140 (b c-a d) g^2 (a+b x)}-\frac {B d n \log (a+b x)}{140 (b c-a d)^2 g^2}-\frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{140 (b c-a d) g^2 (a+b x)}-\frac {d \log (a+b x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{140 (b c-a d)^2 g^2}+\frac {B d n \log (c+d x)}{140 (b c-a d)^2 g^2}-\frac {B d n \log \left (-\frac {d (a+b x)}{b c-a d}\right ) \log (c+d x)}{140 (b c-a d)^2 g^2}+\frac {d \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \log (c+d x)}{140 (b c-a d)^2 g^2}-\frac {B d n \log (a+b x) \log \left (\frac {b (c+d x)}{b c-a d}\right )}{140 (b c-a d)^2 g^2}+\frac {(B d n) \text {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,a+b x\right )}{140 (b c-a d)^2 g^2}+\frac {(B d n) \text {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,c+d x\right )}{140 (b c-a d)^2 g^2}+\frac {(b B d n) \int \frac {\log \left (\frac {b (c+d x)}{b c-a d}\right )}{a+b x} \, dx}{140 (b c-a d)^2 g^2}+\frac {\left (B d^2 n\right ) \int \frac {\log \left (\frac {d (a+b x)}{-b c+a d}\right )}{c+d x} \, dx}{140 (b c-a d)^2 g^2}\\ &=-\frac {B n}{140 (b c-a d) g^2 (a+b x)}-\frac {B d n \log (a+b x)}{140 (b c-a d)^2 g^2}+\frac {B d n \log ^2(a+b x)}{280 (b c-a d)^2 g^2}-\frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{140 (b c-a d) g^2 (a+b x)}-\frac {d \log (a+b x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{140 (b c-a d)^2 g^2}+\frac {B d n \log (c+d x)}{140 (b c-a d)^2 g^2}-\frac {B d n \log \left (-\frac {d (a+b x)}{b c-a d}\right ) \log (c+d x)}{140 (b c-a d)^2 g^2}+\frac {d \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \log (c+d x)}{140 (b c-a d)^2 g^2}+\frac {B d n \log ^2(c+d x)}{280 (b c-a d)^2 g^2}-\frac {B d n \log (a+b x) \log \left (\frac {b (c+d x)}{b c-a d}\right )}{140 (b c-a d)^2 g^2}+\frac {(B d n) \text {Subst}\left (\int \frac {\log \left (1+\frac {d x}{b c-a d}\right )}{x} \, dx,x,a+b x\right )}{140 (b c-a d)^2 g^2}+\frac {(B d n) \text {Subst}\left (\int \frac {\log \left (1+\frac {b x}{-b c+a d}\right )}{x} \, dx,x,c+d x\right )}{140 (b c-a d)^2 g^2}\\ &=-\frac {B n}{140 (b c-a d) g^2 (a+b x)}-\frac {B d n \log (a+b x)}{140 (b c-a d)^2 g^2}+\frac {B d n \log ^2(a+b x)}{280 (b c-a d)^2 g^2}-\frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{140 (b c-a d) g^2 (a+b x)}-\frac {d \log (a+b x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{140 (b c-a d)^2 g^2}+\frac {B d n \log (c+d x)}{140 (b c-a d)^2 g^2}-\frac {B d n \log \left (-\frac {d (a+b x)}{b c-a d}\right ) \log (c+d x)}{140 (b c-a d)^2 g^2}+\frac {d \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \log (c+d x)}{140 (b c-a d)^2 g^2}+\frac {B d n \log ^2(c+d x)}{280 (b c-a d)^2 g^2}-\frac {B d n \log (a+b x) \log \left (\frac {b (c+d x)}{b c-a d}\right )}{140 (b c-a d)^2 g^2}-\frac {B d n \text {Li}_2\left (-\frac {d (a+b x)}{b c-a d}\right )}{140 (b c-a d)^2 g^2}-\frac {B d n \text {Li}_2\left (\frac {b (c+d x)}{b c-a d}\right )}{140 (b c-a d)^2 g^2}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
time = 0.20, size = 304, normalized size = 1.68 \begin {gather*} -\frac {2 (b c-a d) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )+2 d (a+b x) \log (a+b x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )-2 d (a+b x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \log (c+d x)+2 B n (b c-a d+d (a+b x) \log (a+b x)-d (a+b x) \log (c+d x))-B d n (a+b x) \left (\log (a+b x) \left (\log (a+b x)-2 \log \left (\frac {b (c+d x)}{b c-a d}\right )\right )-2 \text {Li}_2\left (\frac {d (a+b x)}{-b c+a d}\right )\right )+B d n (a+b x) \left (\left (2 \log \left (\frac {d (a+b x)}{-b c+a d}\right )-\log (c+d x)\right ) \log (c+d x)+2 \text {Li}_2\left (\frac {b (c+d x)}{b c-a d}\right )\right )}{2 (b c-a d)^2 g^2 i (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(2*(b*c - a*d)*(A + B*Log[e*((a + b*x)/(c + d*x))^n]) + 2*d*(a + b*x)*Log[a + b*x]*(A + B*Log[e*((a + b*x
)/(c + d*x))^n]) - 2*d*(a + b*x)*(A + B*Log[e*((a + b*x)/(c + d*x))^n])*Log[c + d*x] + 2*B*n*(b*c - a*d + d*(a
+ b*x)*Log[a + b*x] - d*(a + b*x)*Log[c + d*x]) - B*d*n*(a + b*x)*(Log[a + b*x]*(Log[a + b*x] - 2*Log[(b*(c +
d*x))/(b*c - a*d)]) - 2*PolyLog[2, (d*(a + b*x))/(-(b*c) + a*d)]) + B*d*n*(a + b*x)*((2*Log[(d*(a + b*x))/(-(
b*c) + a*d)] - Log[c + d*x])*Log[c + d*x] + 2*PolyLog[2, (b*(c + d*x))/(b*c - a*d)]))/((b*c - a*d)^2*g^2*i*(a
+ b*x))

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Maple [F]
time = 0.19, size = 0, normalized size = 0.00 \[\int \frac {A +B \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right )}{\left (b g x +a g \right )^{2} \left (d i x +c i \right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 425 vs. \(2 (169) = 338\).
time = 0.31, size = 425, normalized size = 2.35 \begin {gather*} B {\left (\frac {1}{{\left (-i \, b^{2} c + i \, a b d\right )} g^{2} x + {\left (-i \, a b c + i \, a^{2} d\right )} g^{2}} - \frac {d \log \left (b x + a\right )}{{\left (i \, b^{2} c^{2} - 2 i \, a b c d + i \, a^{2} d^{2}\right )} g^{2}} + \frac {d \log \left (d x + c\right )}{{\left (i \, b^{2} c^{2} - 2 i \, a b c d + i \, a^{2} d^{2}\right )} g^{2}}\right )} \log \left ({\left (\frac {b x}{d x + c} + \frac {a}{d x + c}\right )}^{n} e\right ) - \frac {{\left ({\left (i \, b d x + i \, a d\right )} \log \left (b x + a\right )^{2} + {\left (i \, b d x + i \, a d\right )} \log \left (d x + c\right )^{2} - 2 i \, b c + 2 i \, a d - 2 \, {\left (i \, b d x + i \, a d\right )} \log \left (b x + a\right ) - 2 \, {\left (-i \, b d x - i \, a d + {\left (i \, b d x + i \, a d\right )} \log \left (b x + a\right )\right )} \log \left (d x + c\right )\right )} B n}{2 \, {\left (a b^{2} c^{2} g^{2} - 2 \, a^{2} b c d g^{2} + a^{3} d^{2} g^{2} + {\left (b^{3} c^{2} g^{2} - 2 \, a b^{2} c d g^{2} + a^{2} b d^{2} g^{2}\right )} x\right )}} + A {\left (\frac {1}{{\left (-i \, b^{2} c + i \, a b d\right )} g^{2} x + {\left (-i \, a b c + i \, a^{2} d\right )} g^{2}} - \frac {d \log \left (b x + a\right )}{{\left (i \, b^{2} c^{2} - 2 i \, a b c d + i \, a^{2} d^{2}\right )} g^{2}} + \frac {d \log \left (d x + c\right )}{{\left (i \, b^{2} c^{2} - 2 i \, a b c d + i \, a^{2} d^{2}\right )} g^{2}}\right )} \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))/(b*g*x+a*g)^2/(d*i*x+c*i),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.42, size = 183, normalized size = 1.01 \begin {gather*} -\frac {2 \, {\left (-i \, A - i \, B\right )} b c + 2 \, {\left (i \, A + i \, B\right )} a d - {\left (i \, B b d n x + i \, B a d n\right )} \log \left (\frac {b x + a}{d x + c}\right )^{2} + 2 \, {\left (-i \, B b c + i \, B a d\right )} n + 2 \, {\left (-i \, B b c n + {\left (-i \, A - i \, B\right )} a d + {\left (-i \, B b d n + {\left (-i \, A - i \, B\right )} b d\right )} x\right )} \log \left (\frac {b x + a}{d x + c}\right )}{2 \, {\left ({\left (b^{3} c^{2} - 2 \, a b^{2} c d + a^{2} b d^{2}\right )} g^{2} x + {\left (a b^{2} c^{2} - 2 \, a^{2} b c d + a^{3} d^{2}\right )} g^{2}\right )}} \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))/(b*g*x+a*g)^2/(d*i*x+c*i),x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \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))/(b*g*x+a*g)**2/(d*i*x+c*i),x)

[Out]

Timed out

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Giac [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))/(b*g*x+a*g)^2/(d*i*x+c*i),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 6.08, size = 239, normalized size = 1.32 \begin {gather*} \frac {A}{g^2\,i\,\left (a\,d-b\,c\right )\,\left (a+b\,x\right )}+\frac {B\,\ln \left (e\,{\left (\frac {a+b\,x}{c+d\,x}\right )}^n\right )}{g^2\,i\,\left (a\,d-b\,c\right )\,\left (a+b\,x\right )}+\frac {B\,n}{g^2\,i\,\left (a\,d-b\,c\right )\,\left (a+b\,x\right )}-\frac {B\,d\,{\ln \left (e\,{\left (\frac {a+b\,x}{c+d\,x}\right )}^n\right )}^2}{2\,g^2\,i\,n\,{\left (a\,d-b\,c\right )}^2}+\frac {A\,d\,\mathrm {atan}\left (\frac {a\,d\,1{}\mathrm {i}+b\,c\,1{}\mathrm {i}+b\,d\,x\,2{}\mathrm {i}}{a\,d-b\,c}\right )\,2{}\mathrm {i}}{g^2\,i\,{\left (a\,d-b\,c\right )}^2}+\frac {B\,d\,n\,\mathrm {atan}\left (\frac {a\,d\,1{}\mathrm {i}+b\,c\,1{}\mathrm {i}+b\,d\,x\,2{}\mathrm {i}}{a\,d-b\,c}\right )\,2{}\mathrm {i}}{g^2\,i\,{\left (a\,d-b\,c\right )}^2} \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))/((a*g + b*g*x)^2*(c*i + d*i*x)),x)

[Out]

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

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