3.49 \(\int \frac{(a g+b g x) (A+B \log (\frac{e (a+b x)}{c+d x}))}{(c i+d i x)^3} \, dx\)

Optimal. Leaf size=85 \[ \frac{g (a+b x)^2 \left (B \log \left (\frac{e (a+b x)}{c+d x}\right )+A\right )}{2 i^3 (c+d x)^2 (b c-a d)}-\frac{B g (a+b x)^2}{4 i^3 (c+d x)^2 (b c-a d)} \]

[Out]

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

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Rubi [B]  time = 0.292345, antiderivative size = 191, normalized size of antiderivative = 2.25, number of steps used = 10, number of rules used = 4, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {2528, 2525, 12, 44} \[ -\frac{b g \left (B \log \left (\frac{e (a+b x)}{c+d x}\right )+A\right )}{d^2 i^3 (c+d x)}+\frac{g (b c-a d) \left (B \log \left (\frac{e (a+b x)}{c+d x}\right )+A\right )}{2 d^2 i^3 (c+d x)^2}+\frac{b^2 B g \log (a+b x)}{2 d^2 i^3 (b c-a d)}-\frac{b^2 B g \log (c+d x)}{2 d^2 i^3 (b c-a d)}-\frac{B g (b c-a d)}{4 d^2 i^3 (c+d x)^2}+\frac{b B g}{2 d^2 i^3 (c+d x)} \]

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)^3,x]

[Out]

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

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{(a g+b g x) \left (A+B \log \left (\frac{e (a+b x)}{c+d x}\right )\right )}{(49 c+49 d x)^3} \, dx &=\int \left (\frac{(-b c+a d) g \left (A+B \log \left (\frac{e (a+b x)}{c+d x}\right )\right )}{117649 d (c+d x)^3}+\frac{b g \left (A+B \log \left (\frac{e (a+b x)}{c+d x}\right )\right )}{117649 d (c+d x)^2}\right ) \, dx\\ &=\frac{(b g) \int \frac{A+B \log \left (\frac{e (a+b x)}{c+d x}\right )}{(c+d x)^2} \, dx}{117649 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)^3} \, dx}{117649 d}\\ &=\frac{(b c-a d) g \left (A+B \log \left (\frac{e (a+b x)}{c+d x}\right )\right )}{235298 d^2 (c+d x)^2}-\frac{b g \left (A+B \log \left (\frac{e (a+b x)}{c+d x}\right )\right )}{117649 d^2 (c+d x)}+\frac{(b B g) \int \frac{b c-a d}{(a+b x) (c+d x)^2} \, dx}{117649 d^2}-\frac{(B (b c-a d) g) \int \frac{b c-a d}{(a+b x) (c+d x)^3} \, dx}{235298 d^2}\\ &=\frac{(b c-a d) g \left (A+B \log \left (\frac{e (a+b x)}{c+d x}\right )\right )}{235298 d^2 (c+d x)^2}-\frac{b g \left (A+B \log \left (\frac{e (a+b x)}{c+d x}\right )\right )}{117649 d^2 (c+d x)}+\frac{(b B (b c-a d) g) \int \frac{1}{(a+b x) (c+d x)^2} \, dx}{117649 d^2}-\frac{\left (B (b c-a d)^2 g\right ) \int \frac{1}{(a+b x) (c+d x)^3} \, dx}{235298 d^2}\\ &=\frac{(b c-a d) g \left (A+B \log \left (\frac{e (a+b x)}{c+d x}\right )\right )}{235298 d^2 (c+d x)^2}-\frac{b g \left (A+B \log \left (\frac{e (a+b x)}{c+d x}\right )\right )}{117649 d^2 (c+d x)}+\frac{(b B (b c-a d) g) \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}{117649 d^2}-\frac{\left (B (b c-a d)^2 g\right ) \int \left (\frac{b^3}{(b c-a d)^3 (a+b x)}-\frac{d}{(b c-a d) (c+d x)^3}-\frac{b d}{(b c-a d)^2 (c+d x)^2}-\frac{b^2 d}{(b c-a d)^3 (c+d x)}\right ) \, dx}{235298 d^2}\\ &=-\frac{B (b c-a d) g}{470596 d^2 (c+d x)^2}+\frac{b B g}{235298 d^2 (c+d x)}+\frac{b^2 B g \log (a+b x)}{235298 d^2 (b c-a d)}+\frac{(b c-a d) g \left (A+B \log \left (\frac{e (a+b x)}{c+d x}\right )\right )}{235298 d^2 (c+d x)^2}-\frac{b g \left (A+B \log \left (\frac{e (a+b x)}{c+d x}\right )\right )}{117649 d^2 (c+d x)}-\frac{b^2 B g \log (c+d x)}{235298 d^2 (b c-a d)}\\ \end{align*}

Mathematica [B]  time = 0.151509, size = 207, normalized size = 2.44 \[ \frac{g \left (-\frac{b \left (B \log \left (\frac{e (a+b x)}{c+d x}\right )+A\right )}{d^2 (c+d x)}+\frac{(b c-a d) \left (B \log \left (\frac{e (a+b x)}{c+d x}\right )+A\right )}{2 d^2 (c+d x)^2}-\frac{B \left (\frac{2 b^2 \log (a+b x)}{b c-a d}-\frac{2 b^2 \log (c+d x)}{b c-a d}+\frac{b c-a d}{(c+d x)^2}+\frac{2 b}{c+d x}\right )}{4 d^2}+\frac{b B \left (\frac{b \log (a+b x)}{b c-a d}-\frac{b \log (c+d x)}{b c-a d}+\frac{1}{c+d x}\right )}{d^2}\right )}{i^3} \]

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)^3,x]

[Out]

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

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Maple [B]  time = 0.05, size = 1049, normalized size = 12.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*g*x+a*g)*(A+B*ln(e*(b*x+a)/(d*x+c)))/(d*i*x+c*i)^3,x)

[Out]

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

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Maxima [B]  time = 1.25232, size = 765, normalized size = 9. \begin{align*} -\frac{1}{4} \, B b g{\left (\frac{2 \,{\left (2 \, d x + c\right )} \log \left (\frac{b e x}{d x + c} + \frac{a e}{d x + c}\right )}{d^{4} i^{3} x^{2} + 2 \, c d^{3} i^{3} x + c^{2} d^{2} i^{3}} - \frac{b c^{2} - 3 \, a c d + 2 \,{\left (b c d - 2 \, a d^{2}\right )} x}{{\left (b c d^{4} - a d^{5}\right )} i^{3} x^{2} + 2 \,{\left (b c^{2} d^{3} - a c d^{4}\right )} i^{3} x +{\left (b c^{3} d^{2} - a c^{2} d^{3}\right )} i^{3}} - \frac{2 \,{\left (b^{2} c - 2 \, a b d\right )} \log \left (b x + a\right )}{{\left (b^{2} c^{2} d^{2} - 2 \, a b c d^{3} + a^{2} d^{4}\right )} i^{3}} + \frac{2 \,{\left (b^{2} c - 2 \, a b d\right )} \log \left (d x + c\right )}{{\left (b^{2} c^{2} d^{2} - 2 \, a b c d^{3} + a^{2} d^{4}\right )} i^{3}}\right )} + \frac{1}{4} \, B a g{\left (\frac{2 \, b d x + 3 \, b c - a d}{{\left (b c d^{3} - a d^{4}\right )} i^{3} x^{2} + 2 \,{\left (b c^{2} d^{2} - a c d^{3}\right )} i^{3} x +{\left (b c^{3} d - a c^{2} d^{2}\right )} i^{3}} - \frac{2 \, \log \left (\frac{b e x}{d x + c} + \frac{a e}{d x + c}\right )}{d^{3} i^{3} x^{2} + 2 \, c d^{2} i^{3} x + c^{2} d i^{3}} + \frac{2 \, b^{2} \log \left (b x + a\right )}{{\left (b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\right )} i^{3}} - \frac{2 \, b^{2} \log \left (d x + c\right )}{{\left (b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\right )} i^{3}}\right )} - \frac{{\left (2 \, d x + c\right )} A b g}{2 \,{\left (d^{4} i^{3} x^{2} + 2 \, c d^{3} i^{3} x + c^{2} d^{2} i^{3}\right )}} - \frac{A a g}{2 \,{\left (d^{3} i^{3} x^{2} + 2 \, c d^{2} i^{3} x + c^{2} d i^{3}\right )}} \end{align*}

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)^3,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 0.476337, size = 366, normalized size = 4.31 \begin{align*} -\frac{2 \,{\left ({\left (2 \, A - B\right )} b^{2} c d -{\left (2 \, A - B\right )} a b d^{2}\right )} g x +{\left ({\left (2 \, A - B\right )} b^{2} c^{2} -{\left (2 \, A - B\right )} a^{2} d^{2}\right )} g - 2 \,{\left (B b^{2} d^{2} g x^{2} + 2 \, B a b d^{2} g x + B a^{2} d^{2} g\right )} \log \left (\frac{b e x + a e}{d x + c}\right )}{4 \,{\left ({\left (b c d^{4} - a d^{5}\right )} i^{3} x^{2} + 2 \,{\left (b c^{2} d^{3} - a c d^{4}\right )} i^{3} x +{\left (b c^{3} d^{2} - a c^{2} d^{3}\right )} i^{3}\right )}} \end{align*}

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)^3,x, algorithm="fricas")

[Out]

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

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Sympy [B]  time = 6.02962, size = 382, normalized size = 4.49 \begin{align*} \frac{B b^{2} g \log{\left (x + \frac{- \frac{B a^{2} b^{2} d^{2} g}{a d - b c} + \frac{2 B a b^{3} c d g}{a d - b c} + B a b^{2} d g - \frac{B b^{4} c^{2} g}{a d - b c} + B b^{3} c g}{2 B b^{3} d g} \right )}}{2 d^{2} i^{3} \left (a d - b c\right )} - \frac{B b^{2} g \log{\left (x + \frac{\frac{B a^{2} b^{2} d^{2} g}{a d - b c} - \frac{2 B a b^{3} c d g}{a d - b c} + B a b^{2} d g + \frac{B b^{4} c^{2} g}{a d - b c} + B b^{3} c g}{2 B b^{3} d g} \right )}}{2 d^{2} i^{3} \left (a d - b c\right )} - \frac{2 A a d g + 2 A b c g - B a d g - B b c g + x \left (4 A b d g - 2 B b d g\right )}{4 c^{2} d^{2} i^{3} + 8 c d^{3} i^{3} x + 4 d^{4} i^{3} x^{2}} + \frac{\left (- B a d g - B b c g - 2 B b d g x\right ) \log{\left (\frac{e \left (a + b x\right )}{c + d x} \right )}}{2 c^{2} d^{2} i^{3} + 4 c d^{3} i^{3} x + 2 d^{4} i^{3} x^{2}} \end{align*}

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)**3,x)

[Out]

B*b**2*g*log(x + (-B*a**2*b**2*d**2*g/(a*d - b*c) + 2*B*a*b**3*c*d*g/(a*d - b*c) + B*a*b**2*d*g - B*b**4*c**2*
g/(a*d - b*c) + B*b**3*c*g)/(2*B*b**3*d*g))/(2*d**2*i**3*(a*d - b*c)) - B*b**2*g*log(x + (B*a**2*b**2*d**2*g/(
a*d - b*c) - 2*B*a*b**3*c*d*g/(a*d - b*c) + B*a*b**2*d*g + B*b**4*c**2*g/(a*d - b*c) + B*b**3*c*g)/(2*B*b**3*d
*g))/(2*d**2*i**3*(a*d - b*c)) - (2*A*a*d*g + 2*A*b*c*g - B*a*d*g - B*b*c*g + x*(4*A*b*d*g - 2*B*b*d*g))/(4*c*
*2*d**2*i**3 + 8*c*d**3*i**3*x + 4*d**4*i**3*x**2) + (-B*a*d*g - B*b*c*g - 2*B*b*d*g*x)*log(e*(a + b*x)/(c + d
*x))/(2*c**2*d**2*i**3 + 4*c*d**3*i**3*x + 2*d**4*i**3*x**2)

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Giac [B]  time = 1.38173, size = 258, normalized size = 3.04 \begin{align*} -\frac{B b^{2} g \log \left (b x + a\right )}{2 \,{\left (b c d^{2} i - a d^{3} i\right )}} + \frac{B b^{2} g \log \left (d x + c\right )}{2 \,{\left (b c d^{2} i - a d^{3} i\right )}} - \frac{{\left (2 \, B b d g i x + B b c g i + B a d g i\right )} \log \left (\frac{b x + a}{d x + c}\right )}{2 \,{\left (d^{4} x^{2} + 2 \, c d^{3} x + c^{2} d^{2}\right )}} - \frac{4 \, A b d g i x + 2 \, B b d g i x + 2 \, A b c g i + B b c g i + 2 \, A a d g i + B a d g i}{4 \,{\left (d^{4} x^{2} + 2 \, c d^{3} x + c^{2} d^{2}\right )}} \end{align*}

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)^3,x, algorithm="giac")

[Out]

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