3.1.14 \(\int \frac {(c i+d i x)^2 (A+B \log (\frac {e (a+b x)}{c+d x}))}{a g+b g x} \, dx\) [14]

Optimal. Leaf size=276 \[ -\frac {B d (b c-a d) i^2 x}{2 b^2 g}-\frac {B (b c-a d)^2 i^2 \log \left (\frac {a+b x}{c+d x}\right )}{2 b^3 g}+\frac {d (b c-a d) i^2 (a+b x) \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{b^3 g}+\frac {i^2 (c+d x)^2 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{2 b g}-\frac {3 B (b c-a d)^2 i^2 \log (c+d x)}{2 b^3 g}-\frac {(b c-a d)^2 i^2 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right ) \log \left (1-\frac {b (c+d x)}{d (a+b x)}\right )}{b^3 g}+\frac {B (b c-a d)^2 i^2 \text {Li}_2\left (\frac {b (c+d x)}{d (a+b x)}\right )}{b^3 g} \]

[Out]

-1/2*B*d*(-a*d+b*c)*i^2*x/b^2/g-1/2*B*(-a*d+b*c)^2*i^2*ln((b*x+a)/(d*x+c))/b^3/g+d*(-a*d+b*c)*i^2*(b*x+a)*(A+B
*ln(e*(b*x+a)/(d*x+c)))/b^3/g+1/2*i^2*(d*x+c)^2*(A+B*ln(e*(b*x+a)/(d*x+c)))/b/g-3/2*B*(-a*d+b*c)^2*i^2*ln(d*x+
c)/b^3/g-(-a*d+b*c)^2*i^2*(A+B*ln(e*(b*x+a)/(d*x+c)))*ln(1-b*(d*x+c)/d/(b*x+a))/b^3/g+B*(-a*d+b*c)^2*i^2*polyl
og(2,b*(d*x+c)/d/(b*x+a))/b^3/g

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Rubi [A]
time = 0.24, antiderivative size = 276, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2562, 2389, 2379, 2438, 2351, 31, 2356, 46} \begin {gather*} \frac {B i^2 (b c-a d)^2 \text {PolyLog}\left (2,\frac {b (c+d x)}{d (a+b x)}\right )}{b^3 g}+\frac {d i^2 (a+b x) (b c-a d) \left (B \log \left (\frac {e (a+b x)}{c+d x}\right )+A\right )}{b^3 g}-\frac {i^2 (b c-a d)^2 \log \left (1-\frac {b (c+d x)}{d (a+b x)}\right ) \left (B \log \left (\frac {e (a+b x)}{c+d x}\right )+A\right )}{b^3 g}+\frac {i^2 (c+d x)^2 \left (B \log \left (\frac {e (a+b x)}{c+d x}\right )+A\right )}{2 b g}-\frac {B i^2 (b c-a d)^2 \log \left (\frac {a+b x}{c+d x}\right )}{2 b^3 g}-\frac {3 B i^2 (b c-a d)^2 \log (c+d x)}{2 b^3 g}-\frac {B d i^2 x (b c-a d)}{2 b^2 g} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(B*d*(b*c - a*d)*i^2*x)/(b^2*g) - (B*(b*c - a*d)^2*i^2*Log[(a + b*x)/(c + d*x)])/(2*b^3*g) + (d*(b*c - a*
d)*i^2*(a + b*x)*(A + B*Log[(e*(a + b*x))/(c + d*x)]))/(b^3*g) + (i^2*(c + d*x)^2*(A + B*Log[(e*(a + b*x))/(c
+ d*x)]))/(2*b*g) - (3*B*(b*c - a*d)^2*i^2*Log[c + d*x])/(2*b^3*g) - ((b*c - a*d)^2*i^2*(A + B*Log[(e*(a + b*x
))/(c + d*x)])*Log[1 - (b*(c + d*x))/(d*(a + b*x))])/(b^3*g) + (B*(b*c - a*d)^2*i^2*PolyLog[2, (b*(c + d*x))/(
d*(a + b*x))])/(b^3*g)

Rule 31

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

Rule 46

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[m + n + 2, 0])

Rule 2351

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 2356

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Simp[(d + e*x)^(q + 1)
*((a + b*Log[c*x^n])^p/(e*(q + 1))), x] - Dist[b*n*(p/(e*(q + 1))), Int[((d + e*x)^(q + 1)*(a + b*Log[c*x^n])^
(p - 1))/x, x], x] /; FreeQ[{a, b, c, d, e, n, p, q}, x] && GtQ[p, 0] && NeQ[q, -1] && (EqQ[p, 1] || (Integers
Q[2*p, 2*q] &&  !IGtQ[q, 0]) || (EqQ[p, 2] && NeQ[q, 1]))

Rule 2379

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

Rule 2389

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 2438

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 2562

Int[((A_.) + Log[(e_.)*((a_.) + (b_.)*(x_))^(n_.)*((c_.) + (d_.)*(x_))^(mn_)]*(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] && EqQ[n + mn, 0] && IGtQ[n, 0] && 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 {(14 c+14 d x)^2 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{a g+b g x} \, dx &=\int \left (\frac {196 d (b c-a d) \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{b^2 g}+\frac {14 d (14 c+14 d x) \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{b g}+\frac {196 (b c-a d)^2 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{b^2 (a g+b g x)}\right ) \, dx\\ &=\frac {\left (196 (b c-a d)^2\right ) \int \frac {A+B \log \left (\frac {e (a+b x)}{c+d x}\right )}{a g+b g x} \, dx}{b^2}+\frac {(14 d) \int (14 c+14 d x) \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right ) \, dx}{b g}+\frac {(196 d (b c-a d)) \int \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right ) \, dx}{b^2 g}\\ &=\frac {196 A d (b c-a d) x}{b^2 g}+\frac {98 (c+d x)^2 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{b g}+\frac {196 (b c-a d)^2 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right ) \log (a g+b g x)}{b^3 g}-\frac {B \int \frac {196 (b c-a d) (c+d x)}{a+b x} \, dx}{2 b g}+\frac {(196 B d (b c-a d)) \int \log \left (\frac {e (a+b x)}{c+d x}\right ) \, dx}{b^2 g}-\frac {\left (196 B (b c-a d)^2\right ) \int \frac {(c+d x) \left (-\frac {d e (a+b x)}{(c+d x)^2}+\frac {b e}{c+d x}\right ) \log (a g+b g x)}{e (a+b x)} \, dx}{b^3 g}\\ &=\frac {196 A d (b c-a d) x}{b^2 g}+\frac {196 B d (b c-a d) (a+b x) \log \left (\frac {e (a+b x)}{c+d x}\right )}{b^3 g}+\frac {98 (c+d x)^2 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{b g}+\frac {196 (b c-a d)^2 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right ) \log (a g+b g x)}{b^3 g}-\frac {(98 B (b c-a d)) \int \frac {c+d x}{a+b x} \, dx}{b g}-\frac {\left (196 B d (b c-a d)^2\right ) \int \frac {1}{c+d x} \, dx}{b^3 g}-\frac {\left (196 B (b c-a d)^2\right ) \int \frac {(c+d x) \left (-\frac {d e (a+b x)}{(c+d x)^2}+\frac {b e}{c+d x}\right ) \log (a g+b g x)}{a+b x} \, dx}{b^3 e g}\\ &=\frac {196 A d (b c-a d) x}{b^2 g}+\frac {196 B d (b c-a d) (a+b x) \log \left (\frac {e (a+b x)}{c+d x}\right )}{b^3 g}+\frac {98 (c+d x)^2 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{b g}-\frac {196 B (b c-a d)^2 \log (c+d x)}{b^3 g}+\frac {196 (b c-a d)^2 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right ) \log (a g+b g x)}{b^3 g}-\frac {(98 B (b c-a d)) \int \left (\frac {d}{b}+\frac {b c-a d}{b (a+b x)}\right ) \, dx}{b g}-\frac {\left (196 B (b c-a d)^2\right ) \int \left (\frac {b e \log (a g+b g x)}{a+b x}-\frac {d e \log (a g+b g x)}{c+d x}\right ) \, dx}{b^3 e g}\\ &=\frac {196 A d (b c-a d) x}{b^2 g}-\frac {98 B d (b c-a d) x}{b^2 g}-\frac {98 B (b c-a d)^2 \log (a+b x)}{b^3 g}+\frac {196 B d (b c-a d) (a+b x) \log \left (\frac {e (a+b x)}{c+d x}\right )}{b^3 g}+\frac {98 (c+d x)^2 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{b g}-\frac {196 B (b c-a d)^2 \log (c+d x)}{b^3 g}+\frac {196 (b c-a d)^2 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right ) \log (a g+b g x)}{b^3 g}-\frac {\left (196 B (b c-a d)^2\right ) \int \frac {\log (a g+b g x)}{a+b x} \, dx}{b^2 g}+\frac {\left (196 B d (b c-a d)^2\right ) \int \frac {\log (a g+b g x)}{c+d x} \, dx}{b^3 g}\\ &=\frac {196 A d (b c-a d) x}{b^2 g}-\frac {98 B d (b c-a d) x}{b^2 g}-\frac {98 B (b c-a d)^2 \log (a+b x)}{b^3 g}+\frac {196 B d (b c-a d) (a+b x) \log \left (\frac {e (a+b x)}{c+d x}\right )}{b^3 g}+\frac {98 (c+d x)^2 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{b g}-\frac {196 B (b c-a d)^2 \log (c+d x)}{b^3 g}+\frac {196 (b c-a d)^2 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right ) \log (a g+b g x)}{b^3 g}+\frac {196 B (b c-a d)^2 \log \left (\frac {b (c+d x)}{b c-a d}\right ) \log (a g+b g x)}{b^3 g}-\frac {\left (196 B (b c-a d)^2\right ) \int \frac {\log \left (\frac {b g (c+d x)}{b c g-a d g}\right )}{a g+b g x} \, dx}{b^2}-\frac {\left (196 B (b c-a d)^2\right ) \text {Subst}\left (\int \frac {g \log (x)}{x} \, dx,x,a g+b g x\right )}{b^3 g^2}\\ &=\frac {196 A d (b c-a d) x}{b^2 g}-\frac {98 B d (b c-a d) x}{b^2 g}-\frac {98 B (b c-a d)^2 \log (a+b x)}{b^3 g}+\frac {196 B d (b c-a d) (a+b x) \log \left (\frac {e (a+b x)}{c+d x}\right )}{b^3 g}+\frac {98 (c+d x)^2 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{b g}-\frac {196 B (b c-a d)^2 \log (c+d x)}{b^3 g}+\frac {196 (b c-a d)^2 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right ) \log (a g+b g x)}{b^3 g}+\frac {196 B (b c-a d)^2 \log \left (\frac {b (c+d x)}{b c-a d}\right ) \log (a g+b g x)}{b^3 g}-\frac {\left (196 B (b c-a d)^2\right ) \text {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,a g+b g x\right )}{b^3 g}-\frac {\left (196 B (b c-a d)^2\right ) \text {Subst}\left (\int \frac {\log \left (1+\frac {d x}{b c g-a d g}\right )}{x} \, dx,x,a g+b g x\right )}{b^3 g}\\ &=\frac {196 A d (b c-a d) x}{b^2 g}-\frac {98 B d (b c-a d) x}{b^2 g}-\frac {98 B (b c-a d)^2 \log (a+b x)}{b^3 g}-\frac {98 B (b c-a d)^2 \log ^2(g (a+b x))}{b^3 g}+\frac {196 B d (b c-a d) (a+b x) \log \left (\frac {e (a+b x)}{c+d x}\right )}{b^3 g}+\frac {98 (c+d x)^2 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{b g}-\frac {196 B (b c-a d)^2 \log (c+d x)}{b^3 g}+\frac {196 (b c-a d)^2 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right ) \log (a g+b g x)}{b^3 g}+\frac {196 B (b c-a d)^2 \log \left (\frac {b (c+d x)}{b c-a d}\right ) \log (a g+b g x)}{b^3 g}+\frac {196 B (b c-a d)^2 \text {Li}_2\left (-\frac {d (a+b x)}{b c-a d}\right )}{b^3 g}\\ \end {align*}

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Mathematica [A]
time = 0.13, size = 252, normalized size = 0.91 \begin {gather*} \frac {i^2 \left (2 A b d (b c-a d) x-B (b c-a d) (b d x+(b c-a d) \log (a+b x))+2 B d (b c-a d) (a+b x) \log \left (\frac {e (a+b x)}{c+d x}\right )+b^2 (c+d x)^2 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )+2 (b c-a d)^2 \log (g (a+b x)) \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )-2 B (b c-a d)^2 \log (c+d x)+B (b c-a d)^2 \left (-\log (g (a+b x)) \left (\log (g (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 )\right )}{2 b^3 g} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(i^2*(2*A*b*d*(b*c - a*d)*x - B*(b*c - a*d)*(b*d*x + (b*c - a*d)*Log[a + b*x]) + 2*B*d*(b*c - a*d)*(a + b*x)*L
og[(e*(a + b*x))/(c + d*x)] + b^2*(c + d*x)^2*(A + B*Log[(e*(a + b*x))/(c + d*x)]) + 2*(b*c - a*d)^2*Log[g*(a
+ b*x)]*(A + B*Log[(e*(a + b*x))/(c + d*x)]) - 2*B*(b*c - a*d)^2*Log[c + d*x] + B*(b*c - a*d)^2*(-(Log[g*(a +
b*x)]*(Log[g*(a + b*x)] - 2*Log[(b*(c + d*x))/(b*c - a*d)])) + 2*PolyLog[2, (d*(a + b*x))/(-(b*c) + a*d)])))/(
2*b^3*g)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1689\) vs. \(2(268)=536\).
time = 1.38, size = 1690, normalized size = 6.12

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(1690\)
default	\(\text {Expression too large to display}\)	\(1690\)
risch	\(\text {Expression too large to display}\)	\(3116\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/d^2*e*(a*d-b*c)*(-A*d^3*i^2/g/b^2/(b*e-(b*e/d+(a*d-b*c)*e/d/(d*x+c))*d)*a+A*d^2*i^2/g/b/(b*e-(b*e/d+(a*d-b*
c)*e/d/(d*x+c))*d)*c+A*d^3/e*i^2/g/b^3*ln(b*e-(b*e/d+(a*d-b*c)*e/d/(d*x+c))*d)*a-A*d^2/e*i^2/g/b^2*ln(b*e-(b*e
/d+(a*d-b*c)*e/d/(d*x+c))*d)*c-1/2*A*d^3*e*i^2/g/b/(b*e-(b*e/d+(a*d-b*c)*e/d/(d*x+c))*d)^2*a+1/2*A*d^2*e*i^2/g
/(b*e-(b*e/d+(a*d-b*c)*e/d/(d*x+c))*d)^2*c-A*d^3/e*i^2/g/b^3*ln(b*e/d+(a*d-b*c)*e/d/(d*x+c))*a+A*d^2/e*i^2/g/b
^2*ln(b*e/d+(a*d-b*c)*e/d/(d*x+c))*c-3/2*B*d^3/e*i^2/g/b^3*ln(b*e-(b*e/d+(a*d-b*c)*e/d/(d*x+c))*d)*a+3/2*B*d^2
/e*i^2/g/b^2*ln(b*e-(b*e/d+(a*d-b*c)*e/d/(d*x+c))*d)*c-B*d^4/e*i^2/g/b^3*ln(b*e/d+(a*d-b*c)*e/d/(d*x+c))*(b*e/
d+(a*d-b*c)*e/d/(d*x+c))/(b*e-(b*e/d+(a*d-b*c)*e/d/(d*x+c))*d)*a+B*d^3/e*i^2/g/b^2*ln(b*e/d+(a*d-b*c)*e/d/(d*x
+c))*(b*e/d+(a*d-b*c)*e/d/(d*x+c))/(b*e-(b*e/d+(a*d-b*c)*e/d/(d*x+c))*d)*c+1/2*B*d^3*i^2/g/b^2/(b*e-(b*e/d+(a*
d-b*c)*e/d/(d*x+c))*d)*a-1/2*B*d^2*i^2/g/b/(b*e-(b*e/d+(a*d-b*c)*e/d/(d*x+c))*d)*c-B*d^4*i^2/g/b^2*ln(b*e/d+(a
*d-b*c)*e/d/(d*x+c))*(b*e/d+(a*d-b*c)*e/d/(d*x+c))/(b*e-(b*e/d+(a*d-b*c)*e/d/(d*x+c))*d)^2*a+B*d^3*i^2/g/b*ln(
b*e/d+(a*d-b*c)*e/d/(d*x+c))*(b*e/d+(a*d-b*c)*e/d/(d*x+c))/(b*e-(b*e/d+(a*d-b*c)*e/d/(d*x+c))*d)^2*c+1/2*B*d^5
/e*i^2/g/b^3*ln(b*e/d+(a*d-b*c)*e/d/(d*x+c))*(b*e/d+(a*d-b*c)*e/d/(d*x+c))^2/(b*e-(b*e/d+(a*d-b*c)*e/d/(d*x+c)
)*d)^2*a-1/2*B*d^4/e*i^2/g/b^2*ln(b*e/d+(a*d-b*c)*e/d/(d*x+c))*(b*e/d+(a*d-b*c)*e/d/(d*x+c))^2/(b*e-(b*e/d+(a*
d-b*c)*e/d/(d*x+c))*d)^2*c-1/2*B*d^3/e*i^2/g*ln(b*e/d+(a*d-b*c)*e/d/(d*x+c))^2/b^3*a+1/2*B*d^2/e*i^2/g*ln(b*e/
d+(a*d-b*c)*e/d/(d*x+c))^2/b^2*c+B*d^3/e*i^2/g/b^3*dilog(-(-b*e+(b*e/d+(a*d-b*c)*e/d/(d*x+c))*d)/b/e)*a-B*d^2/
e*i^2/g/b^2*dilog(-(-b*e+(b*e/d+(a*d-b*c)*e/d/(d*x+c))*d)/b/e)*c+B*d^3/e*i^2/g/b^3*ln(b*e/d+(a*d-b*c)*e/d/(d*x
+c))*ln(-(-b*e+(b*e/d+(a*d-b*c)*e/d/(d*x+c))*d)/b/e)*a-B*d^2/e*i^2/g/b^2*ln(b*e/d+(a*d-b*c)*e/d/(d*x+c))*ln(-(
-b*e+(b*e/d+(a*d-b*c)*e/d/(d*x+c))*d)/b/e)*c)

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Maxima [A]
time = 0.33, size = 395, normalized size = 1.43 \begin {gather*} -2 \, A c d {\left (\frac {x}{b g} - \frac {a \log \left (b x + a\right )}{b^{2} g}\right )} - \frac {1}{2} \, A d^{2} {\left (\frac {2 \, a^{2} \log \left (b x + a\right )}{b^{3} g} + \frac {b x^{2} - 2 \, a x}{b^{2} g}\right )} - \frac {A c^{2} \log \left (b g x + a g\right )}{b g} + \frac {{\left (3 \, b c^{2} - 2 \, a c d\right )} B \log \left (d x + c\right )}{2 \, b^{2} g} - \frac {{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\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^{3} g} - \frac {B b^{2} d^{2} x^{2} + {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} B \log \left (b x + a\right )^{2} + {\left (3 \, b^{2} c d - a b d^{2}\right )} B x + {\left (B b^{2} d^{2} x^{2} + 2 \, {\left (2 \, b^{2} c d - a b d^{2}\right )} B x + {\left (2 \, b^{2} c^{2} - a^{2} d^{2}\right )} B\right )} \log \left (b x + a\right ) - {\left (B b^{2} d^{2} x^{2} + 2 \, {\left (2 \, b^{2} c d - a b d^{2}\right )} B x + 2 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} B \log \left (b x + a\right )\right )} \log \left (d x + c\right )}{2 \, b^{3} g} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*i*x+c*i)^2*(A+B*log(e*(b*x+a)/(d*x+c)))/(b*g*x+a*g),x, algorithm="maxima")

[Out]

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

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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((d*i*x+c*i)^2*(A+B*log(e*(b*x+a)/(d*x+c)))/(b*g*x+a*g),x, algorithm="fricas")

[Out]

integral(-(A*d^2*x^2 + 2*A*c*d*x + A*c^2 + (B*d^2*x^2 + 2*B*c*d*x + B*c^2)*log((b*x + a)*e/(d*x + c)))/(b*g*x
+ a*g), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {i^{2} \left (\int \frac {A c^{2}}{a + b x}\, dx + \int \frac {A d^{2} x^{2}}{a + b x}\, dx + \int \frac {B c^{2} \log {\left (\frac {a e}{c + d x} + \frac {b e x}{c + d x} \right )}}{a + b x}\, dx + \int \frac {2 A c d x}{a + b x}\, dx + \int \frac {B d^{2} x^{2} \log {\left (\frac {a e}{c + d x} + \frac {b e x}{c + d x} \right )}}{a + b x}\, dx + \int \frac {2 B c d x \log {\left (\frac {a e}{c + d x} + \frac {b e x}{c + d x} \right )}}{a + b x}\, dx\right )}{g} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

i**2*(Integral(A*c**2/(a + b*x), x) + Integral(A*d**2*x**2/(a + b*x), x) + Integral(B*c**2*log(a*e/(c + d*x) +
b*e*x/(c + d*x))/(a + b*x), x) + Integral(2*A*c*d*x/(a + b*x), x) + Integral(B*d**2*x**2*log(a*e/(c + d*x) +
b*e*x/(c + d*x))/(a + b*x), x) + Integral(2*B*c*d*x*log(a*e/(c + d*x) + b*e*x/(c + d*x))/(a + b*x), x))/g

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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((d*i*x+c*i)^2*(A+B*log(e*(b*x+a)/(d*x+c)))/(b*g*x+a*g),x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (c\,i+d\,i\,x\right )}^2\,\left (A+B\,\ln \left (\frac {e\,\left (a+b\,x\right )}{c+d\,x}\right )\right )}{a\,g+b\,g\,x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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