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

Optimal. Leaf size=133 \[ -\frac {i (b c-a d) \log \left (-\frac {b c-a d}{d (a+b x)}\right ) \left (B \log \left (\frac {e (a+b x)}{c+d x}\right )+A-B\right )}{b^2 g}+\frac {i (c+d x) \left (B \log \left (\frac {e (a+b x)}{c+d x}\right )+A\right )}{b g}+\frac {B i (b c-a d) \text {Li}_2\left (\frac {b c-a d}{d (a+b x)}+1\right )}{b^2 g} \]

[Out]

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

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Rubi [A]  time = 0.35, antiderivative size = 213, normalized size of antiderivative = 1.60, number of steps used = 14, number of rules used = 11, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.290, Rules used = {2528, 2486, 31, 2524, 12, 2418, 2390, 2301, 2394, 2393, 2391} \[ \frac {B i (b c-a d) \text {PolyLog}\left (2,-\frac {d (a+b x)}{b c-a d}\right )}{b^2 g}+\frac {i (b c-a d) \log (a+b x) \left (B \log \left (\frac {e (a+b x)}{c+d x}\right )+A\right )}{b^2 g}+\frac {B d i (a+b x) \log \left (\frac {e (a+b x)}{c+d x}\right )}{b^2 g}-\frac {B i (b c-a d) \log ^2(a+b x)}{2 b^2 g}+\frac {B i (b c-a d) \log (a+b x) \log \left (\frac {b (c+d x)}{b c-a d}\right )}{b^2 g}-\frac {B i (b c-a d) \log (c+d x)}{b^2 g}+\frac {A d i x}{b g} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 {(5 c+5 d x) \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{a g+b g x} \, dx &=\int \left (\frac {5 d \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{b g}+\frac {5 (b c-a d) \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{b g (a+b x)}\right ) \, dx\\ &=\frac {(5 d) \int \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right ) \, dx}{b g}+\frac {(5 (b c-a d)) \int \frac {A+B \log \left (\frac {e (a+b x)}{c+d x}\right )}{a+b x} \, dx}{b g}\\ &=\frac {5 A d x}{b g}+\frac {5 (b c-a d) \log (a+b x) \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{b^2 g}+\frac {(5 B d) \int \log \left (\frac {e (a+b x)}{c+d x}\right ) \, dx}{b g}-\frac {(5 B (b c-a d)) \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+b x)}{e (a+b x)} \, dx}{b^2 g}\\ &=\frac {5 A d x}{b g}+\frac {5 B d (a+b x) \log \left (\frac {e (a+b x)}{c+d x}\right )}{b^2 g}+\frac {5 (b c-a d) \log (a+b x) \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{b^2 g}-\frac {(5 B d (b c-a d)) \int \frac {1}{c+d x} \, dx}{b^2 g}-\frac {(5 B (b c-a d)) \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+b x)}{a+b x} \, dx}{b^2 e g}\\ &=\frac {5 A d x}{b g}+\frac {5 B d (a+b x) \log \left (\frac {e (a+b x)}{c+d x}\right )}{b^2 g}+\frac {5 (b c-a d) \log (a+b x) \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{b^2 g}-\frac {5 B (b c-a d) \log (c+d x)}{b^2 g}-\frac {(5 B (b c-a d)) \int \left (\frac {b e \log (a+b x)}{a+b x}-\frac {d e \log (a+b x)}{c+d x}\right ) \, dx}{b^2 e g}\\ &=\frac {5 A d x}{b g}+\frac {5 B d (a+b x) \log \left (\frac {e (a+b x)}{c+d x}\right )}{b^2 g}+\frac {5 (b c-a d) \log (a+b x) \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{b^2 g}-\frac {5 B (b c-a d) \log (c+d x)}{b^2 g}-\frac {(5 B (b c-a d)) \int \frac {\log (a+b x)}{a+b x} \, dx}{b g}+\frac {(5 B d (b c-a d)) \int \frac {\log (a+b x)}{c+d x} \, dx}{b^2 g}\\ &=\frac {5 A d x}{b g}+\frac {5 B d (a+b x) \log \left (\frac {e (a+b x)}{c+d x}\right )}{b^2 g}+\frac {5 (b c-a d) \log (a+b x) \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{b^2 g}-\frac {5 B (b c-a d) \log (c+d x)}{b^2 g}+\frac {5 B (b c-a d) \log (a+b x) \log \left (\frac {b (c+d x)}{b c-a d}\right )}{b^2 g}-\frac {(5 B (b c-a d)) \operatorname {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,a+b x\right )}{b^2 g}-\frac {(5 B (b c-a d)) \int \frac {\log \left (\frac {b (c+d x)}{b c-a d}\right )}{a+b x} \, dx}{b g}\\ &=\frac {5 A d x}{b g}-\frac {5 B (b c-a d) \log ^2(a+b x)}{2 b^2 g}+\frac {5 B d (a+b x) \log \left (\frac {e (a+b x)}{c+d x}\right )}{b^2 g}+\frac {5 (b c-a d) \log (a+b x) \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{b^2 g}-\frac {5 B (b c-a d) \log (c+d x)}{b^2 g}+\frac {5 B (b c-a d) \log (a+b x) \log \left (\frac {b (c+d x)}{b c-a d}\right )}{b^2 g}-\frac {(5 B (b c-a d)) \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 {5 A d x}{b g}-\frac {5 B (b c-a d) \log ^2(a+b x)}{2 b^2 g}+\frac {5 B d (a+b x) \log \left (\frac {e (a+b x)}{c+d x}\right )}{b^2 g}+\frac {5 (b c-a d) \log (a+b x) \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{b^2 g}-\frac {5 B (b c-a d) \log (c+d x)}{b^2 g}+\frac {5 B (b c-a d) \log (a+b x) \log \left (\frac {b (c+d x)}{b c-a d}\right )}{b^2 g}+\frac {5 B (b c-a d) \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 = 0.12, size = 164, normalized size = 1.23 \[ \frac {i \left (2 (b c-a d) \log (a+b x) \left (B \log \left (\frac {e (a+b x)}{c+d x}\right )+B \log \left (\frac {b (c+d x)}{b c-a d}\right )+A\right )+2 \left (B d (a+b x) \log \left (\frac {e (a+b x)}{c+d x}\right )+\log (c+d x) (a B d-b B c)+A b d x\right )+2 B (b c-a d) \text {Li}_2\left (\frac {d (a+b x)}{a d-b c}\right )+\log ^2(a+b x) (a B d-b B c)\right )}{2 b^2 g} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.92, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {A d i x + A c i + {\left (B d i x + B c i\right )} \log \left (\frac {b e x + a e}{d x + c}\right )}{b g x + a g}, x\right ) \]

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)))/(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((b*e*x + a*e)/(d*x + c)))/(b*g*x + a*g), x)

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

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)))/(b*g*x+a*g),x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.12, size = 1044, normalized size = 7.85 \[ \frac {B \,a^{2} d^{2} e i \ln \left (\frac {b e}{d}+\frac {\left (a d -b c \right ) e}{\left (d x +c \right ) d}\right )}{\left (\frac {a d e}{d x +c}-\frac {b c e}{d x +c}\right ) \left (d x +c \right ) b^{2} g}-\frac {2 B a c d e i \ln \left (\frac {b e}{d}+\frac {\left (a d -b c \right ) e}{\left (d x +c \right ) d}\right )}{\left (\frac {a d e}{d x +c}-\frac {b c e}{d x +c}\right ) \left (d x +c \right ) b g}+\frac {B \,c^{2} e i \ln \left (\frac {b e}{d}+\frac {\left (a d -b c \right ) e}{\left (d x +c \right ) d}\right )}{\left (\frac {a d e}{d x +c}-\frac {b c e}{d x +c}\right ) \left (d x +c \right ) g}+\frac {B a d e i \ln \left (\frac {b e}{d}+\frac {\left (a d -b c \right ) e}{\left (d x +c \right ) d}\right )}{\left (\frac {a d e}{d x +c}-\frac {b c e}{d x +c}\right ) b g}-\frac {B c e i \ln \left (\frac {b e}{d}+\frac {\left (a d -b c \right ) e}{\left (d x +c \right ) d}\right )}{\left (\frac {a d e}{d x +c}-\frac {b c e}{d x +c}\right ) g}+\frac {A a d e i}{\left (\frac {a d e}{d x +c}-\frac {b c e}{d x +c}\right ) b g}-\frac {A c e i}{\left (\frac {a d e}{d x +c}-\frac {b c e}{d x +c}\right ) g}+\frac {B a d i \ln \left (-\frac {-b e +\left (\frac {b e}{d}+\frac {\left (a d -b c \right ) e}{\left (d x +c \right ) d}\right ) d}{b e}\right ) \ln \left (\frac {b e}{d}+\frac {\left (a d -b c \right ) e}{\left (d x +c \right ) d}\right )}{b^{2} g}-\frac {B a d i \ln \left (\frac {b e}{d}+\frac {\left (a d -b c \right ) e}{\left (d x +c \right ) d}\right )^{2}}{2 b^{2} g}-\frac {B c i \ln \left (-\frac {-b e +\left (\frac {b e}{d}+\frac {\left (a d -b c \right ) e}{\left (d x +c \right ) d}\right ) d}{b e}\right ) \ln \left (\frac {b e}{d}+\frac {\left (a d -b c \right ) e}{\left (d x +c \right ) d}\right )}{b g}+\frac {B c i \ln \left (\frac {b e}{d}+\frac {\left (a d -b c \right ) e}{\left (d x +c \right ) d}\right )^{2}}{2 b g}+\frac {A a d i \ln \left (-b e +\left (\frac {b e}{d}+\frac {\left (a d -b c \right ) e}{\left (d x +c \right ) d}\right ) d \right )}{b^{2} g}-\frac {A a d i \ln \left (\frac {b e}{d}+\frac {\left (a d -b c \right ) e}{\left (d x +c \right ) d}\right )}{b^{2} g}-\frac {A c i \ln \left (-b e +\left (\frac {b e}{d}+\frac {\left (a d -b c \right ) e}{\left (d x +c \right ) d}\right ) d \right )}{b g}+\frac {A c i \ln \left (\frac {b e}{d}+\frac {\left (a d -b c \right ) e}{\left (d x +c \right ) d}\right )}{b g}+\frac {B a d i \dilog \left (-\frac {-b e +\left (\frac {b e}{d}+\frac {\left (a d -b c \right ) e}{\left (d x +c \right ) d}\right ) d}{b e}\right )}{b^{2} g}-\frac {B a d i \ln \left (-b e +\left (\frac {b e}{d}+\frac {\left (a d -b c \right ) e}{\left (d x +c \right ) d}\right ) d \right )}{b^{2} g}-\frac {B c i \dilog \left (-\frac {-b e +\left (\frac {b e}{d}+\frac {\left (a d -b c \right ) e}{\left (d x +c \right ) d}\right ) d}{b e}\right )}{b g}+\frac {B c i \ln \left (-b e +\left (\frac {b e}{d}+\frac {\left (a d -b c \right ) e}{\left (d x +c \right ) d}\right ) d \right )}{b g} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.84, size = 241, normalized size = 1.81 \[ A d i {\left (\frac {x}{b g} - \frac {a \log \left (b x + a\right )}{b^{2} g}\right )} + \frac {A c i \log \left (b g x + a g\right )}{b g} - \frac {B c i \log \left (d x + c\right )}{b g} + \frac {{\left (b c i - a d i\right )} {\left (\log \left (b x + a\right ) \log \left (\frac {b d x + a d}{b c - a d} + 1\right ) + {\rm Li}_2\left (-\frac {b d x + a d}{b c - a d}\right )\right )} B}{b^{2} g} + \frac {2 \, B b d i x \log \relax (e) + {\left (b c i - a d i\right )} B \log \left (b x + a\right )^{2} + 2 \, {\left (B b d i x + {\left (b c i \log \relax (e) - {\left (i \log \relax (e) - i\right )} a d\right )} B\right )} \log \left (b x + a\right ) - 2 \, {\left (B b d i x + {\left (b c i - a d i\right )} B \log \left (b x + a\right )\right )} \log \left (d x + c\right )}{2 \, b^{2} g} \]

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)))/(b*g*x+a*g),x, algorithm="maxima")

[Out]

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

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

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))))/(a*g + b*g*x),x)

[Out]

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

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

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

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