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

Optimal. Leaf size=160 \[ -\frac {b g \log \left (\frac {b c-a d}{b (c+d x)}\right ) \left (B \log \left (\frac {e (a+b x)}{c+d x}\right )+A\right )}{d^2 i^2}-\frac {A g (a+b x)}{d i^2 (c+d x)}-\frac {b B g \text {Li}_2\left (\frac {d (a+b x)}{b (c+d x)}\right )}{d^2 i^2}-\frac {B g (a+b x) \log \left (\frac {e (a+b x)}{c+d x}\right )}{d i^2 (c+d x)}+\frac {B g (a+b x)}{d i^2 (c+d x)} \]

[Out]

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

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Rubi [A]  time = 0.40, antiderivative size = 222, normalized size of antiderivative = 1.39, number of steps used = 15, number of rules used = 11, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.290, Rules used = {2528, 2525, 12, 44, 2524, 2418, 2394, 2393, 2391, 2390, 2301} \[ -\frac {b B g \text {PolyLog}\left (2,\frac {b (c+d x)}{b c-a d}\right )}{d^2 i^2}+\frac {b g \log (c+d x) \left (B \log \left (\frac {e (a+b x)}{c+d x}\right )+A\right )}{d^2 i^2}+\frac {g (b c-a d) \left (B \log \left (\frac {e (a+b x)}{c+d x}\right )+A\right )}{d^2 i^2 (c+d x)}-\frac {B g (b c-a d)}{d^2 i^2 (c+d x)}-\frac {b B g \log (c+d x) \log \left (-\frac {d (a+b x)}{b c-a d}\right )}{d^2 i^2}-\frac {b B g \log (a+b x)}{d^2 i^2}+\frac {b B g \log ^2(c+d x)}{2 d^2 i^2}+\frac {b B g \log (c+d x)}{d^2 i^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

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

Rule 44

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] && L
tQ[m + n + 2, 0])

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 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 2525

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

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 {(a g+b g x) \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{(41 c+41 d x)^2} \, dx &=\int \left (\frac {(-b c+a d) g \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{1681 d (c+d x)^2}+\frac {b g \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{1681 d (c+d x)}\right ) \, dx\\ &=\frac {(b g) \int \frac {A+B \log \left (\frac {e (a+b x)}{c+d x}\right )}{c+d x} \, dx}{1681 d}-\frac {((b c-a d) g) \int \frac {A+B \log \left (\frac {e (a+b x)}{c+d x}\right )}{(c+d x)^2} \, dx}{1681 d}\\ &=\frac {(b c-a d) g \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{1681 d^2 (c+d x)}+\frac {b g \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right ) \log (c+d x)}{1681 d^2}-\frac {(b B g) \int \frac {(c+d x) \left (-\frac {d e (a+b x)}{(c+d x)^2}+\frac {b e}{c+d x}\right ) \log (c+d x)}{e (a+b x)} \, dx}{1681 d^2}-\frac {(B (b c-a d) g) \int \frac {b c-a d}{(a+b x) (c+d x)^2} \, dx}{1681 d^2}\\ &=\frac {(b c-a d) g \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{1681 d^2 (c+d x)}+\frac {b g \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right ) \log (c+d x)}{1681 d^2}-\frac {\left (B (b c-a d)^2 g\right ) \int \frac {1}{(a+b x) (c+d x)^2} \, dx}{1681 d^2}-\frac {(b B g) \int \frac {(c+d x) \left (-\frac {d e (a+b x)}{(c+d x)^2}+\frac {b e}{c+d x}\right ) \log (c+d x)}{a+b x} \, dx}{1681 d^2 e}\\ &=\frac {(b c-a d) g \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{1681 d^2 (c+d x)}+\frac {b g \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right ) \log (c+d x)}{1681 d^2}-\frac {\left (B (b c-a d)^2 g\right ) \int \left (\frac {b^2}{(b c-a d)^2 (a+b x)}-\frac {d}{(b c-a d) (c+d x)^2}-\frac {b d}{(b c-a d)^2 (c+d x)}\right ) \, dx}{1681 d^2}-\frac {(b B g) \int \left (\frac {b e \log (c+d x)}{a+b x}-\frac {d e \log (c+d x)}{c+d x}\right ) \, dx}{1681 d^2 e}\\ &=-\frac {B (b c-a d) g}{1681 d^2 (c+d x)}-\frac {b B g \log (a+b x)}{1681 d^2}+\frac {(b c-a d) g \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{1681 d^2 (c+d x)}+\frac {b B g \log (c+d x)}{1681 d^2}+\frac {b g \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right ) \log (c+d x)}{1681 d^2}-\frac {\left (b^2 B g\right ) \int \frac {\log (c+d x)}{a+b x} \, dx}{1681 d^2}+\frac {(b B g) \int \frac {\log (c+d x)}{c+d x} \, dx}{1681 d}\\ &=-\frac {B (b c-a d) g}{1681 d^2 (c+d x)}-\frac {b B g \log (a+b x)}{1681 d^2}+\frac {(b c-a d) g \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{1681 d^2 (c+d x)}+\frac {b B g \log (c+d x)}{1681 d^2}-\frac {b B g \log \left (-\frac {d (a+b x)}{b c-a d}\right ) \log (c+d x)}{1681 d^2}+\frac {b g \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right ) \log (c+d x)}{1681 d^2}+\frac {(b B g) \operatorname {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,c+d x\right )}{1681 d^2}+\frac {(b B g) \int \frac {\log \left (\frac {d (a+b x)}{-b c+a d}\right )}{c+d x} \, dx}{1681 d}\\ &=-\frac {B (b c-a d) g}{1681 d^2 (c+d x)}-\frac {b B g \log (a+b x)}{1681 d^2}+\frac {(b c-a d) g \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{1681 d^2 (c+d x)}+\frac {b B g \log (c+d x)}{1681 d^2}-\frac {b B g \log \left (-\frac {d (a+b x)}{b c-a d}\right ) \log (c+d x)}{1681 d^2}+\frac {b g \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right ) \log (c+d x)}{1681 d^2}+\frac {b B g \log ^2(c+d x)}{3362 d^2}+\frac {(b B g) \operatorname {Subst}\left (\int \frac {\log \left (1+\frac {b x}{-b c+a d}\right )}{x} \, dx,x,c+d x\right )}{1681 d^2}\\ &=-\frac {B (b c-a d) g}{1681 d^2 (c+d x)}-\frac {b B g \log (a+b x)}{1681 d^2}+\frac {(b c-a d) g \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{1681 d^2 (c+d x)}+\frac {b B g \log (c+d x)}{1681 d^2}-\frac {b B g \log \left (-\frac {d (a+b x)}{b c-a d}\right ) \log (c+d x)}{1681 d^2}+\frac {b g \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right ) \log (c+d x)}{1681 d^2}+\frac {b B g \log ^2(c+d x)}{3362 d^2}-\frac {b B g \text {Li}_2\left (\frac {b (c+d x)}{b c-a d}\right )}{1681 d^2}\\ \end {align*}

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Mathematica [A]  time = 0.16, size = 175, normalized size = 1.09 \[ \frac {g \left (2 b \log (c+d x) \left (B \log \left (\frac {e (a+b x)}{c+d x}\right )+A\right )+\frac {2 (b c-a d) \left (B \log \left (\frac {e (a+b x)}{c+d x}\right )+A\right )}{c+d x}-b B \left (2 \text {Li}_2\left (\frac {b (c+d x)}{b c-a d}\right )+\log (c+d x) \left (2 \log \left (\frac {d (a+b x)}{a d-b c}\right )-\log (c+d x)\right )\right )-2 B \left (\frac {b c-a d}{c+d x}+b \log (a+b x)-b \log (c+d x)\right )\right )}{2 d^2 i^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.86, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {A b g x + A a g + {\left (B b g x + B a g\right )} \log \left (\frac {b e x + a e}{d x + c}\right )}{d^{2} i^{2} x^{2} + 2 \, c d i^{2} x + c^{2} i^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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maple [B]  time = 0.06, size = 978, normalized size = 6.11 \[ -\frac {B \,a^{2} g \ln \left (\frac {b e}{d}+\frac {\left (a d -b c \right ) e}{\left (d x +c \right ) d}\right )}{\left (a d -b c \right ) \left (d x +c \right ) i^{2}}+\frac {2 B a b c g \ln \left (\frac {b e}{d}+\frac {\left (a d -b c \right ) e}{\left (d x +c \right ) d}\right )}{\left (a d -b c \right ) \left (d x +c \right ) d \,i^{2}}-\frac {B a b g \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 )}{\left (a d -b c \right ) d \,i^{2}}-\frac {B \,b^{2} c^{2} g \ln \left (\frac {b e}{d}+\frac {\left (a d -b c \right ) e}{\left (d x +c \right ) d}\right )}{\left (a d -b c \right ) \left (d x +c \right ) d^{2} i^{2}}+\frac {B \,b^{2} c g \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 )}{\left (a d -b c \right ) d^{2} i^{2}}-\frac {A \,a^{2} g}{\left (a d -b c \right ) \left (d x +c \right ) i^{2}}+\frac {2 A a b c g}{\left (a d -b c \right ) \left (d x +c \right ) d \,i^{2}}-\frac {A a b g \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 )}{\left (a d -b c \right ) d \,i^{2}}-\frac {A \,b^{2} c^{2} g}{\left (a d -b c \right ) \left (d x +c \right ) d^{2} i^{2}}+\frac {A \,b^{2} c g \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 )}{\left (a d -b c \right ) d^{2} i^{2}}+\frac {B \,a^{2} g}{\left (a d -b c \right ) \left (d x +c \right ) i^{2}}-\frac {2 B a b c g}{\left (a d -b c \right ) \left (d x +c \right ) d \,i^{2}}-\frac {B a b g \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 )}{\left (a d -b c \right ) d \,i^{2}}-\frac {B a b g \ln \left (\frac {b e}{d}+\frac {\left (a d -b c \right ) e}{\left (d x +c \right ) d}\right )}{\left (a d -b c \right ) d \,i^{2}}+\frac {B \,b^{2} c^{2} g}{\left (a d -b c \right ) \left (d x +c \right ) d^{2} i^{2}}+\frac {B \,b^{2} c g \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 )}{\left (a d -b c \right ) d^{2} i^{2}}+\frac {B \,b^{2} c g \ln \left (\frac {b e}{d}+\frac {\left (a d -b c \right ) e}{\left (d x +c \right ) d}\right )}{\left (a d -b c \right ) d^{2} i^{2}}-\frac {A a b g}{\left (a d -b c \right ) d \,i^{2}}+\frac {A \,b^{2} c g}{\left (a d -b c \right ) d^{2} i^{2}}+\frac {B a b g}{\left (a d -b c \right ) d \,i^{2}}-\frac {B \,b^{2} c g}{\left (a d -b c \right ) d^{2} i^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {1}{2} \, B b g {\left (\frac {{\left (d x + c\right )} \log \left (d x + c\right )^{2} + 2 \, c \log \left (d x + c\right )}{d^{3} i^{2} x + c d^{2} i^{2}} - 2 \, \int \frac {d x \log \left (b x + a\right ) + d x \log \relax (e) + c}{d^{3} i^{2} x^{2} + 2 \, c d^{2} i^{2} x + c^{2} d i^{2}}\,{d x}\right )} + A b g {\left (\frac {c}{d^{3} i^{2} x + c d^{2} i^{2}} + \frac {\log \left (d x + c\right )}{d^{2} i^{2}}\right )} - B a g {\left (\frac {\log \left (\frac {b e x}{d x + c} + \frac {a e}{d x + c}\right )}{d^{2} i^{2} x + c d i^{2}} - \frac {1}{d^{2} i^{2} x + c d i^{2}} - \frac {b \log \left (b x + a\right )}{{\left (b c d - a d^{2}\right )} i^{2}} + \frac {b \log \left (d x + c\right )}{{\left (b c d - a d^{2}\right )} i^{2}}\right )} - \frac {A a g}{d^{2} i^{2} x + c d i^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\left (a\,g+b\,g\,x\right )\,\left (A+B\,\ln \left (\frac {e\,\left (a+b\,x\right )}{c+d\,x}\right )\right )}{{\left (c\,i+d\,i\,x\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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