3.17 \(\int \frac{(c i+d i x)^2 (A+B \log (\frac{e (a+b x)}{c+d x}))}{(a g+b g x)^4} \, dx\)

Optimal. Leaf size=89 \[ -\frac{i^2 (c+d x)^3 \left (B \log \left (\frac{e (a+b x)}{c+d x}\right )+A\right )}{3 g^4 (a+b x)^3 (b c-a d)}-\frac{B i^2 (c+d x)^3}{9 g^4 (a+b x)^3 (b c-a d)} \]

[Out]

-(B*i^2*(c + d*x)^3)/(9*(b*c - a*d)*g^4*(a + b*x)^3) - (i^2*(c + d*x)^3*(A + B*Log[(e*(a + b*x))/(c + d*x)]))/
(3*(b*c - a*d)*g^4*(a + b*x)^3)

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Rubi [B]  time = 0.489984, antiderivative size = 287, normalized size of antiderivative = 3.22, number of steps used = 14, number of rules used = 4, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {2528, 2525, 12, 44} \[ -\frac{d^2 i^2 \left (B \log \left (\frac{e (a+b x)}{c+d x}\right )+A\right )}{b^3 g^4 (a+b x)}-\frac{d i^2 (b c-a d) \left (B \log \left (\frac{e (a+b x)}{c+d x}\right )+A\right )}{b^3 g^4 (a+b x)^2}-\frac{i^2 (b c-a d)^2 \left (B \log \left (\frac{e (a+b x)}{c+d x}\right )+A\right )}{3 b^3 g^4 (a+b x)^3}-\frac{B d^3 i^2 \log (a+b x)}{3 b^3 g^4 (b c-a d)}+\frac{B d^3 i^2 \log (c+d x)}{3 b^3 g^4 (b c-a d)}-\frac{B d i^2 (b c-a d)}{3 b^3 g^4 (a+b x)^2}-\frac{B i^2 (b c-a d)^2}{9 b^3 g^4 (a+b x)^3}-\frac{B d^2 i^2}{3 b^3 g^4 (a+b x)} \]

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)^4,x]

[Out]

-(B*(b*c - a*d)^2*i^2)/(9*b^3*g^4*(a + b*x)^3) - (B*d*(b*c - a*d)*i^2)/(3*b^3*g^4*(a + b*x)^2) - (B*d^2*i^2)/(
3*b^3*g^4*(a + b*x)) - (B*d^3*i^2*Log[a + b*x])/(3*b^3*(b*c - a*d)*g^4) - ((b*c - a*d)^2*i^2*(A + B*Log[(e*(a
+ b*x))/(c + d*x)]))/(3*b^3*g^4*(a + b*x)^3) - (d*(b*c - a*d)*i^2*(A + B*Log[(e*(a + b*x))/(c + d*x)]))/(b^3*g
^4*(a + b*x)^2) - (d^2*i^2*(A + B*Log[(e*(a + b*x))/(c + d*x)]))/(b^3*g^4*(a + b*x)) + (B*d^3*i^2*Log[c + d*x]
)/(3*b^3*(b*c - a*d)*g^4)

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]

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 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])

Rubi steps

\begin{align*} \int \frac{(17 c+17 d x)^2 \left (A+B \log \left (\frac{e (a+b x)}{c+d x}\right )\right )}{(a g+b g x)^4} \, dx &=\int \left (\frac{289 (b c-a d)^2 \left (A+B \log \left (\frac{e (a+b x)}{c+d x}\right )\right )}{b^2 g^4 (a+b x)^4}+\frac{578 d (b c-a d) \left (A+B \log \left (\frac{e (a+b x)}{c+d x}\right )\right )}{b^2 g^4 (a+b x)^3}+\frac{289 d^2 \left (A+B \log \left (\frac{e (a+b x)}{c+d x}\right )\right )}{b^2 g^4 (a+b x)^2}\right ) \, dx\\ &=\frac{\left (289 d^2\right ) \int \frac{A+B \log \left (\frac{e (a+b x)}{c+d x}\right )}{(a+b x)^2} \, dx}{b^2 g^4}+\frac{(578 d (b c-a d)) \int \frac{A+B \log \left (\frac{e (a+b x)}{c+d x}\right )}{(a+b x)^3} \, dx}{b^2 g^4}+\frac{\left (289 (b c-a d)^2\right ) \int \frac{A+B \log \left (\frac{e (a+b x)}{c+d x}\right )}{(a+b x)^4} \, dx}{b^2 g^4}\\ &=-\frac{289 (b c-a d)^2 \left (A+B \log \left (\frac{e (a+b x)}{c+d x}\right )\right )}{3 b^3 g^4 (a+b x)^3}-\frac{289 d (b c-a d) \left (A+B \log \left (\frac{e (a+b x)}{c+d x}\right )\right )}{b^3 g^4 (a+b x)^2}-\frac{289 d^2 \left (A+B \log \left (\frac{e (a+b x)}{c+d x}\right )\right )}{b^3 g^4 (a+b x)}+\frac{\left (289 B d^2\right ) \int \frac{b c-a d}{(a+b x)^2 (c+d x)} \, dx}{b^3 g^4}+\frac{(289 B d (b c-a d)) \int \frac{b c-a d}{(a+b x)^3 (c+d x)} \, dx}{b^3 g^4}+\frac{\left (289 B (b c-a d)^2\right ) \int \frac{b c-a d}{(a+b x)^4 (c+d x)} \, dx}{3 b^3 g^4}\\ &=-\frac{289 (b c-a d)^2 \left (A+B \log \left (\frac{e (a+b x)}{c+d x}\right )\right )}{3 b^3 g^4 (a+b x)^3}-\frac{289 d (b c-a d) \left (A+B \log \left (\frac{e (a+b x)}{c+d x}\right )\right )}{b^3 g^4 (a+b x)^2}-\frac{289 d^2 \left (A+B \log \left (\frac{e (a+b x)}{c+d x}\right )\right )}{b^3 g^4 (a+b x)}+\frac{\left (289 B d^2 (b c-a d)\right ) \int \frac{1}{(a+b x)^2 (c+d x)} \, dx}{b^3 g^4}+\frac{\left (289 B d (b c-a d)^2\right ) \int \frac{1}{(a+b x)^3 (c+d x)} \, dx}{b^3 g^4}+\frac{\left (289 B (b c-a d)^3\right ) \int \frac{1}{(a+b x)^4 (c+d x)} \, dx}{3 b^3 g^4}\\ &=-\frac{289 (b c-a d)^2 \left (A+B \log \left (\frac{e (a+b x)}{c+d x}\right )\right )}{3 b^3 g^4 (a+b x)^3}-\frac{289 d (b c-a d) \left (A+B \log \left (\frac{e (a+b x)}{c+d x}\right )\right )}{b^3 g^4 (a+b x)^2}-\frac{289 d^2 \left (A+B \log \left (\frac{e (a+b x)}{c+d x}\right )\right )}{b^3 g^4 (a+b x)}+\frac{\left (289 B d^2 (b c-a d)\right ) \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}{b^3 g^4}+\frac{\left (289 B d (b c-a d)^2\right ) \int \left (\frac{b}{(b c-a d) (a+b x)^3}-\frac{b d}{(b c-a d)^2 (a+b x)^2}+\frac{b d^2}{(b c-a d)^3 (a+b x)}-\frac{d^3}{(b c-a d)^3 (c+d x)}\right ) \, dx}{b^3 g^4}+\frac{\left (289 B (b c-a d)^3\right ) \int \left (\frac{b}{(b c-a d) (a+b x)^4}-\frac{b d}{(b c-a d)^2 (a+b x)^3}+\frac{b d^2}{(b c-a d)^3 (a+b x)^2}-\frac{b d^3}{(b c-a d)^4 (a+b x)}+\frac{d^4}{(b c-a d)^4 (c+d x)}\right ) \, dx}{3 b^3 g^4}\\ &=-\frac{289 B (b c-a d)^2}{9 b^3 g^4 (a+b x)^3}-\frac{289 B d (b c-a d)}{3 b^3 g^4 (a+b x)^2}-\frac{289 B d^2}{3 b^3 g^4 (a+b x)}-\frac{289 B d^3 \log (a+b x)}{3 b^3 (b c-a d) g^4}-\frac{289 (b c-a d)^2 \left (A+B \log \left (\frac{e (a+b x)}{c+d x}\right )\right )}{3 b^3 g^4 (a+b x)^3}-\frac{289 d (b c-a d) \left (A+B \log \left (\frac{e (a+b x)}{c+d x}\right )\right )}{b^3 g^4 (a+b x)^2}-\frac{289 d^2 \left (A+B \log \left (\frac{e (a+b x)}{c+d x}\right )\right )}{b^3 g^4 (a+b x)}+\frac{289 B d^3 \log (c+d x)}{3 b^3 (b c-a d) g^4}\\ \end{align*}

Mathematica [B]  time = 0.318232, size = 315, normalized size = 3.54 \[ -\frac{i^2 \left (-9 a^2 A b d^3 x-3 a^3 A d^3+3 B (b c-a d) \left (a^2 d^2+a b d (c+3 d x)+b^2 \left (c^2+3 c d x+3 d^2 x^2\right )\right ) \log \left (\frac{e (a+b x)}{c+d x}\right )-9 a^2 b B d^3 x \log (c+d x)-3 a^2 b B d^3 x-3 a^3 B d^3 \log (c+d x)-a^3 B d^3-9 a A b^2 d^3 x^2-9 a b^2 B d^3 x^2 \log (c+d x)-3 a b^2 B d^3 x^2+3 B d^3 (a+b x)^3 \log (a+b x)+9 A b^3 c^2 d x+3 A b^3 c^3+9 A b^3 c d^2 x^2+3 b^3 B c^2 d x+b^3 B c^3+3 b^3 B c d^2 x^2-3 b^3 B d^3 x^3 \log (c+d x)\right )}{9 b^3 g^4 (a+b x)^3 (b c-a d)} \]

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)^4,x]

[Out]

-(i^2*(3*A*b^3*c^3 + b^3*B*c^3 - 3*a^3*A*d^3 - a^3*B*d^3 + 9*A*b^3*c^2*d*x + 3*b^3*B*c^2*d*x - 9*a^2*A*b*d^3*x
- 3*a^2*b*B*d^3*x + 9*A*b^3*c*d^2*x^2 + 3*b^3*B*c*d^2*x^2 - 9*a*A*b^2*d^3*x^2 - 3*a*b^2*B*d^3*x^2 + 3*B*d^3*(
a + b*x)^3*Log[a + b*x] + 3*B*(b*c - a*d)*(a^2*d^2 + a*b*d*(c + 3*d*x) + b^2*(c^2 + 3*c*d*x + 3*d^2*x^2))*Log[
(e*(a + b*x))/(c + d*x)] - 3*a^3*B*d^3*Log[c + d*x] - 9*a^2*b*B*d^3*x*Log[c + d*x] - 9*a*b^2*B*d^3*x^2*Log[c +
d*x] - 3*b^3*B*d^3*x^3*Log[c + d*x]))/(9*b^3*(b*c - a*d)*g^4*(a + b*x)^3)

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Maple [B]  time = 0.055, size = 406, normalized size = 4.6 \begin{align*}{\frac{d{e}^{3}{i}^{2}Aa}{3\, \left ( ad-bc \right ) ^{2}{g}^{4}} \left ({\frac{be}{d}}+{\frac{ae}{dx+c}}-{\frac{bec}{ \left ( dx+c \right ) d}} \right ) ^{-3}}-{\frac{{e}^{3}{i}^{2}Abc}{3\, \left ( ad-bc \right ) ^{2}{g}^{4}} \left ({\frac{be}{d}}+{\frac{ae}{dx+c}}-{\frac{bec}{ \left ( dx+c \right ) d}} \right ) ^{-3}}+{\frac{d{e}^{3}{i}^{2}Ba}{3\, \left ( ad-bc \right ) ^{2}{g}^{4}}\ln \left ({\frac{be}{d}}+{\frac{e \left ( ad-bc \right ) }{ \left ( dx+c \right ) d}} \right ) \left ({\frac{be}{d}}+{\frac{ae}{dx+c}}-{\frac{bec}{ \left ( dx+c \right ) d}} \right ) ^{-3}}-{\frac{{e}^{3}{i}^{2}Bbc}{3\, \left ( ad-bc \right ) ^{2}{g}^{4}}\ln \left ({\frac{be}{d}}+{\frac{e \left ( ad-bc \right ) }{ \left ( dx+c \right ) d}} \right ) \left ({\frac{be}{d}}+{\frac{ae}{dx+c}}-{\frac{bec}{ \left ( dx+c \right ) d}} \right ) ^{-3}}+{\frac{d{e}^{3}{i}^{2}Ba}{9\, \left ( ad-bc \right ) ^{2}{g}^{4}} \left ({\frac{be}{d}}+{\frac{ae}{dx+c}}-{\frac{bec}{ \left ( dx+c \right ) d}} \right ) ^{-3}}-{\frac{{e}^{3}{i}^{2}Bbc}{9\, \left ( ad-bc \right ) ^{2}{g}^{4}} \left ({\frac{be}{d}}+{\frac{ae}{dx+c}}-{\frac{bec}{ \left ( dx+c \right ) d}} \right ) ^{-3}} \end{align*}

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)^4,x)

[Out]

1/3*e^3*d*i^2/(a*d-b*c)^2/g^4*A/(b*e/d+e/(d*x+c)*a-e/d/(d*x+c)*b*c)^3*a-1/3*e^3*i^2/(a*d-b*c)^2/g^4*A/(b*e/d+e
/(d*x+c)*a-e/d/(d*x+c)*b*c)^3*b*c+1/3*e^3*d*i^2/(a*d-b*c)^2/g^4*B/(b*e/d+e/(d*x+c)*a-e/d/(d*x+c)*b*c)^3*ln(b*e
/d+(a*d-b*c)*e/d/(d*x+c))*a-1/3*e^3*i^2/(a*d-b*c)^2/g^4*B/(b*e/d+e/(d*x+c)*a-e/d/(d*x+c)*b*c)^3*ln(b*e/d+(a*d-
b*c)*e/d/(d*x+c))*b*c+1/9*e^3*d*i^2/(a*d-b*c)^2/g^4*B/(b*e/d+e/(d*x+c)*a-e/d/(d*x+c)*b*c)^3*a-1/9*e^3*i^2/(a*d
-b*c)^2/g^4*B/(b*e/d+e/(d*x+c)*a-e/d/(d*x+c)*b*c)^3*b*c

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Maxima [B]  time = 1.58937, size = 2045, normalized size = 22.98 \begin{align*} \text{result too large to display} \end{align*}

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)^4,x, algorithm="maxima")

[Out]

-1/18*B*d^2*i^2*(6*(3*b^2*x^2 + 3*a*b*x + a^2)*log(b*e*x/(d*x + c) + a*e/(d*x + c))/(b^6*g^4*x^3 + 3*a*b^5*g^4
*x^2 + 3*a^2*b^4*g^4*x + a^3*b^3*g^4) + (11*a^2*b^2*c^2 - 7*a^3*b*c*d + 2*a^4*d^2 + 6*(3*b^4*c^2 - 3*a*b^3*c*d
+ a^2*b^2*d^2)*x^2 + 3*(9*a*b^3*c^2 - 7*a^2*b^2*c*d + 2*a^3*b*d^2)*x)/((b^8*c^2 - 2*a*b^7*c*d + a^2*b^6*d^2)*
g^4*x^3 + 3*(a*b^7*c^2 - 2*a^2*b^6*c*d + a^3*b^5*d^2)*g^4*x^2 + 3*(a^2*b^6*c^2 - 2*a^3*b^5*c*d + a^4*b^4*d^2)*
g^4*x + (a^3*b^5*c^2 - 2*a^4*b^4*c*d + a^5*b^3*d^2)*g^4) + 6*(3*b^2*c^2*d - 3*a*b*c*d^2 + a^2*d^3)*log(b*x + a
)/((b^6*c^3 - 3*a*b^5*c^2*d + 3*a^2*b^4*c*d^2 - a^3*b^3*d^3)*g^4) - 6*(3*b^2*c^2*d - 3*a*b*c*d^2 + a^2*d^3)*lo
g(d*x + c)/((b^6*c^3 - 3*a*b^5*c^2*d + 3*a^2*b^4*c*d^2 - a^3*b^3*d^3)*g^4)) - 1/18*B*c*d*i^2*(6*(3*b*x + a)*lo
g(b*e*x/(d*x + c) + a*e/(d*x + c))/(b^5*g^4*x^3 + 3*a*b^4*g^4*x^2 + 3*a^2*b^3*g^4*x + a^3*b^2*g^4) + (5*a*b^2*
c^2 - 22*a^2*b*c*d + 5*a^3*d^2 - 6*(3*b^3*c*d - a*b^2*d^2)*x^2 + 3*(3*b^3*c^2 - 16*a*b^2*c*d + 5*a^2*b*d^2)*x)
/((b^7*c^2 - 2*a*b^6*c*d + a^2*b^5*d^2)*g^4*x^3 + 3*(a*b^6*c^2 - 2*a^2*b^5*c*d + a^3*b^4*d^2)*g^4*x^2 + 3*(a^2
*b^5*c^2 - 2*a^3*b^4*c*d + a^4*b^3*d^2)*g^4*x + (a^3*b^4*c^2 - 2*a^4*b^3*c*d + a^5*b^2*d^2)*g^4) - 6*(3*b*c*d^
2 - a*d^3)*log(b*x + a)/((b^5*c^3 - 3*a*b^4*c^2*d + 3*a^2*b^3*c*d^2 - a^3*b^2*d^3)*g^4) + 6*(3*b*c*d^2 - a*d^3
)*log(d*x + c)/((b^5*c^3 - 3*a*b^4*c^2*d + 3*a^2*b^3*c*d^2 - a^3*b^2*d^3)*g^4)) - 1/18*B*c^2*i^2*((6*b^2*d^2*x
^2 + 2*b^2*c^2 - 7*a*b*c*d + 11*a^2*d^2 - 3*(b^2*c*d - 5*a*b*d^2)*x)/((b^6*c^2 - 2*a*b^5*c*d + a^2*b^4*d^2)*g^
4*x^3 + 3*(a*b^5*c^2 - 2*a^2*b^4*c*d + a^3*b^3*d^2)*g^4*x^2 + 3*(a^2*b^4*c^2 - 2*a^3*b^3*c*d + a^4*b^2*d^2)*g^
4*x + (a^3*b^3*c^2 - 2*a^4*b^2*c*d + a^5*b*d^2)*g^4) + 6*log(b*e*x/(d*x + c) + a*e/(d*x + c))/(b^4*g^4*x^3 + 3
*a*b^3*g^4*x^2 + 3*a^2*b^2*g^4*x + a^3*b*g^4) + 6*d^3*log(b*x + a)/((b^4*c^3 - 3*a*b^3*c^2*d + 3*a^2*b^2*c*d^2
- a^3*b*d^3)*g^4) - 6*d^3*log(d*x + c)/((b^4*c^3 - 3*a*b^3*c^2*d + 3*a^2*b^2*c*d^2 - a^3*b*d^3)*g^4)) - 1/3*(
3*b*x + a)*A*c*d*i^2/(b^5*g^4*x^3 + 3*a*b^4*g^4*x^2 + 3*a^2*b^3*g^4*x + a^3*b^2*g^4) - 1/3*(3*b^2*x^2 + 3*a*b*
x + a^2)*A*d^2*i^2/(b^6*g^4*x^3 + 3*a*b^5*g^4*x^2 + 3*a^2*b^4*g^4*x + a^3*b^3*g^4) - 1/3*A*c^2*i^2/(b^4*g^4*x^
3 + 3*a*b^3*g^4*x^2 + 3*a^2*b^2*g^4*x + a^3*b*g^4)

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Fricas [B]  time = 0.780945, size = 544, normalized size = 6.11 \begin{align*} -\frac{3 \,{\left ({\left (3 \, A + B\right )} b^{3} c d^{2} -{\left (3 \, A + B\right )} a b^{2} d^{3}\right )} i^{2} x^{2} + 3 \,{\left ({\left (3 \, A + B\right )} b^{3} c^{2} d -{\left (3 \, A + B\right )} a^{2} b d^{3}\right )} i^{2} x +{\left ({\left (3 \, A + B\right )} b^{3} c^{3} -{\left (3 \, A + B\right )} a^{3} d^{3}\right )} i^{2} + 3 \,{\left (B b^{3} d^{3} i^{2} x^{3} + 3 \, B b^{3} c d^{2} i^{2} x^{2} + 3 \, B b^{3} c^{2} d i^{2} x + B b^{3} c^{3} i^{2}\right )} \log \left (\frac{b e x + a e}{d x + c}\right )}{9 \,{\left ({\left (b^{7} c - a b^{6} d\right )} g^{4} x^{3} + 3 \,{\left (a b^{6} c - a^{2} b^{5} d\right )} g^{4} x^{2} + 3 \,{\left (a^{2} b^{5} c - a^{3} b^{4} d\right )} g^{4} x +{\left (a^{3} b^{4} c - a^{4} b^{3} d\right )} g^{4}\right )}} \end{align*}

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)^4,x, algorithm="fricas")

[Out]

-1/9*(3*((3*A + B)*b^3*c*d^2 - (3*A + B)*a*b^2*d^3)*i^2*x^2 + 3*((3*A + B)*b^3*c^2*d - (3*A + B)*a^2*b*d^3)*i^
2*x + ((3*A + B)*b^3*c^3 - (3*A + B)*a^3*d^3)*i^2 + 3*(B*b^3*d^3*i^2*x^3 + 3*B*b^3*c*d^2*i^2*x^2 + 3*B*b^3*c^2
*d*i^2*x + B*b^3*c^3*i^2)*log((b*e*x + a*e)/(d*x + c)))/((b^7*c - a*b^6*d)*g^4*x^3 + 3*(a*b^6*c - a^2*b^5*d)*g
^4*x^2 + 3*(a^2*b^5*c - a^3*b^4*d)*g^4*x + (a^3*b^4*c - a^4*b^3*d)*g^4)

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Sympy [B]  time = 26.5454, size = 610, normalized size = 6.85 \begin{align*} - \frac{B d^{3} i^{2} \log{\left (x + \frac{- \frac{B a^{2} d^{5} i^{2}}{a d - b c} + \frac{2 B a b c d^{4} i^{2}}{a d - b c} + B a d^{4} i^{2} - \frac{B b^{2} c^{2} d^{3} i^{2}}{a d - b c} + B b c d^{3} i^{2}}{2 B b d^{4} i^{2}} \right )}}{3 b^{3} g^{4} \left (a d - b c\right )} + \frac{B d^{3} i^{2} \log{\left (x + \frac{\frac{B a^{2} d^{5} i^{2}}{a d - b c} - \frac{2 B a b c d^{4} i^{2}}{a d - b c} + B a d^{4} i^{2} + \frac{B b^{2} c^{2} d^{3} i^{2}}{a d - b c} + B b c d^{3} i^{2}}{2 B b d^{4} i^{2}} \right )}}{3 b^{3} g^{4} \left (a d - b c\right )} - \frac{3 A a^{2} d^{2} i^{2} + 3 A a b c d i^{2} + 3 A b^{2} c^{2} i^{2} + B a^{2} d^{2} i^{2} + B a b c d i^{2} + B b^{2} c^{2} i^{2} + x^{2} \left (9 A b^{2} d^{2} i^{2} + 3 B b^{2} d^{2} i^{2}\right ) + x \left (9 A a b d^{2} i^{2} + 9 A b^{2} c d i^{2} + 3 B a b d^{2} i^{2} + 3 B b^{2} c d i^{2}\right )}{9 a^{3} b^{3} g^{4} + 27 a^{2} b^{4} g^{4} x + 27 a b^{5} g^{4} x^{2} + 9 b^{6} g^{4} x^{3}} + \frac{\left (- B a^{2} d^{2} i^{2} - B a b c d i^{2} - 3 B a b d^{2} i^{2} x - B b^{2} c^{2} i^{2} - 3 B b^{2} c d i^{2} x - 3 B b^{2} d^{2} i^{2} x^{2}\right ) \log{\left (\frac{e \left (a + b x\right )}{c + d x} \right )}}{3 a^{3} b^{3} g^{4} + 9 a^{2} b^{4} g^{4} x + 9 a b^{5} g^{4} x^{2} + 3 b^{6} g^{4} x^{3}} \end{align*}

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)**4,x)

[Out]

-B*d**3*i**2*log(x + (-B*a**2*d**5*i**2/(a*d - b*c) + 2*B*a*b*c*d**4*i**2/(a*d - b*c) + B*a*d**4*i**2 - B*b**2
*c**2*d**3*i**2/(a*d - b*c) + B*b*c*d**3*i**2)/(2*B*b*d**4*i**2))/(3*b**3*g**4*(a*d - b*c)) + B*d**3*i**2*log(
x + (B*a**2*d**5*i**2/(a*d - b*c) - 2*B*a*b*c*d**4*i**2/(a*d - b*c) + B*a*d**4*i**2 + B*b**2*c**2*d**3*i**2/(a
*d - b*c) + B*b*c*d**3*i**2)/(2*B*b*d**4*i**2))/(3*b**3*g**4*(a*d - b*c)) - (3*A*a**2*d**2*i**2 + 3*A*a*b*c*d*
i**2 + 3*A*b**2*c**2*i**2 + B*a**2*d**2*i**2 + B*a*b*c*d*i**2 + B*b**2*c**2*i**2 + x**2*(9*A*b**2*d**2*i**2 +
3*B*b**2*d**2*i**2) + x*(9*A*a*b*d**2*i**2 + 9*A*b**2*c*d*i**2 + 3*B*a*b*d**2*i**2 + 3*B*b**2*c*d*i**2))/(9*a*
*3*b**3*g**4 + 27*a**2*b**4*g**4*x + 27*a*b**5*g**4*x**2 + 9*b**6*g**4*x**3) + (-B*a**2*d**2*i**2 - B*a*b*c*d*
i**2 - 3*B*a*b*d**2*i**2*x - B*b**2*c**2*i**2 - 3*B*b**2*c*d*i**2*x - 3*B*b**2*d**2*i**2*x**2)*log(e*(a + b*x)
/(c + d*x))/(3*a**3*b**3*g**4 + 9*a**2*b**4*g**4*x + 9*a*b**5*g**4*x**2 + 3*b**6*g**4*x**3)

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Giac [B]  time = 1.31054, size = 463, normalized size = 5.2 \begin{align*} \frac{B d^{3} \log \left (b x + a\right )}{3 \,{\left (b^{4} c g^{4} - a b^{3} d g^{4}\right )}} - \frac{B d^{3} \log \left (d x + c\right )}{3 \,{\left (b^{4} c g^{4} - a b^{3} d g^{4}\right )}} + \frac{{\left (3 \, B b^{2} d^{2} x^{2} + 3 \, B b^{2} c d x + 3 \, B a b d^{2} x + B b^{2} c^{2} + B a b c d + B a^{2} d^{2}\right )} \log \left (\frac{b x + a}{d x + c}\right )}{3 \,{\left (b^{6} g^{4} x^{3} + 3 \, a b^{5} g^{4} x^{2} + 3 \, a^{2} b^{4} g^{4} x + a^{3} b^{3} g^{4}\right )}} + \frac{9 \, A b^{2} d^{2} x^{2} + 12 \, B b^{2} d^{2} x^{2} + 9 \, A b^{2} c d x + 12 \, B b^{2} c d x + 9 \, A a b d^{2} x + 12 \, B a b d^{2} x + 3 \, A b^{2} c^{2} + 4 \, B b^{2} c^{2} + 3 \, A a b c d + 4 \, B a b c d + 3 \, A a^{2} d^{2} + 4 \, B a^{2} d^{2}}{9 \,{\left (b^{6} g^{4} x^{3} + 3 \, a b^{5} g^{4} x^{2} + 3 \, a^{2} b^{4} g^{4} x + a^{3} b^{3} g^{4}\right )}} \end{align*}

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)^4,x, algorithm="giac")

[Out]

1/3*B*d^3*log(b*x + a)/(b^4*c*g^4 - a*b^3*d*g^4) - 1/3*B*d^3*log(d*x + c)/(b^4*c*g^4 - a*b^3*d*g^4) + 1/3*(3*B
*b^2*d^2*x^2 + 3*B*b^2*c*d*x + 3*B*a*b*d^2*x + B*b^2*c^2 + B*a*b*c*d + B*a^2*d^2)*log((b*x + a)/(d*x + c))/(b^
6*g^4*x^3 + 3*a*b^5*g^4*x^2 + 3*a^2*b^4*g^4*x + a^3*b^3*g^4) + 1/9*(9*A*b^2*d^2*x^2 + 12*B*b^2*d^2*x^2 + 9*A*b
^2*c*d*x + 12*B*b^2*c*d*x + 9*A*a*b*d^2*x + 12*B*a*b*d^2*x + 3*A*b^2*c^2 + 4*B*b^2*c^2 + 3*A*a*b*c*d + 4*B*a*b
*c*d + 3*A*a^2*d^2 + 4*B*a^2*d^2)/(b^6*g^4*x^3 + 3*a*b^5*g^4*x^2 + 3*a^2*b^4*g^4*x + a^3*b^3*g^4)