3.1.63 \(\int \frac {a+b \tanh ^{-1}(c x)}{x (d+c d x)^3} \, dx\) [63]

Optimal. Leaf size=161 \[ \frac {b}{8 d^3 (1+c x)^2}+\frac {5 b}{8 d^3 (1+c x)}-\frac {5 b \tanh ^{-1}(c x)}{8 d^3}+\frac {a+b \tanh ^{-1}(c x)}{2 d^3 (1+c x)^2}+\frac {a+b \tanh ^{-1}(c x)}{d^3 (1+c x)}+\frac {a \log (x)}{d^3}+\frac {\left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1+c x}\right )}{d^3}-\frac {b \text {PolyLog}(2,-c x)}{2 d^3}+\frac {b \text {PolyLog}(2,c x)}{2 d^3}-\frac {b \text {PolyLog}\left (2,1-\frac {2}{1+c x}\right )}{2 d^3} \]

[Out]

1/8*b/d^3/(c*x+1)^2+5/8*b/d^3/(c*x+1)-5/8*b*arctanh(c*x)/d^3+1/2*(a+b*arctanh(c*x))/d^3/(c*x+1)^2+(a+b*arctanh
(c*x))/d^3/(c*x+1)+a*ln(x)/d^3+(a+b*arctanh(c*x))*ln(2/(c*x+1))/d^3-1/2*b*polylog(2,-c*x)/d^3+1/2*b*polylog(2,
c*x)/d^3-1/2*b*polylog(2,1-2/(c*x+1))/d^3

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Rubi [A]
time = 0.17, antiderivative size = 161, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 9, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.450, Rules used = {6087, 6031, 6063, 641, 46, 213, 6055, 2449, 2352} \begin {gather*} \frac {a+b \tanh ^{-1}(c x)}{d^3 (c x+1)}+\frac {a+b \tanh ^{-1}(c x)}{2 d^3 (c x+1)^2}+\frac {\log \left (\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{d^3}+\frac {a \log (x)}{d^3}-\frac {b \text {Li}_2(-c x)}{2 d^3}+\frac {b \text {Li}_2(c x)}{2 d^3}-\frac {b \text {Li}_2\left (1-\frac {2}{c x+1}\right )}{2 d^3}+\frac {5 b}{8 d^3 (c x+1)}+\frac {b}{8 d^3 (c x+1)^2}-\frac {5 b \tanh ^{-1}(c x)}{8 d^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c*x])/(x*(d + c*d*x)^3),x]

[Out]

b/(8*d^3*(1 + c*x)^2) + (5*b)/(8*d^3*(1 + c*x)) - (5*b*ArcTanh[c*x])/(8*d^3) + (a + b*ArcTanh[c*x])/(2*d^3*(1
+ c*x)^2) + (a + b*ArcTanh[c*x])/(d^3*(1 + c*x)) + (a*Log[x])/d^3 + ((a + b*ArcTanh[c*x])*Log[2/(1 + c*x)])/d^
3 - (b*PolyLog[2, -(c*x)])/(2*d^3) + (b*PolyLog[2, c*x])/(2*d^3) - (b*PolyLog[2, 1 - 2/(1 + c*x)])/(2*d^3)

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 213

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(-1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])]
, x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 641

Int[((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a/d + (c/e)*x)^
p, x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && I
ntegerQ[m + p]))

Rule 2352

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-e^(-1))*PolyLog[2, 1 - c*x], x] /; FreeQ[{c, d, e
}, x] && EqQ[e + c*d, 0]

Rule 2449

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> Dist[-e/g, Subst[Int[Log[2*d*x]/(1 - 2*
d*x), x], x, 1/(d + e*x)], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]

Rule 6031

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))/(x_), x_Symbol] :> Simp[a*Log[x], x] + (-Simp[(b/2)*PolyLog[2, (-c)*x]
, x] + Simp[(b/2)*PolyLog[2, c*x], x]) /; FreeQ[{a, b, c}, x]

Rule 6055

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-(a + b*ArcTanh[c*x])^p)
*(Log[2/(1 + e*(x/d))]/e), x] + Dist[b*c*(p/e), Int[(a + b*ArcTanh[c*x])^(p - 1)*(Log[2/(1 + e*(x/d))]/(1 - c^
2*x^2)), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[c^2*d^2 - e^2, 0]

Rule 6063

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Simp[(d + e*x)^(q + 1)*((a + b
*ArcTanh[c*x])/(e*(q + 1))), x] - Dist[b*(c/(e*(q + 1))), Int[(d + e*x)^(q + 1)/(1 - c^2*x^2), x], x] /; FreeQ
[{a, b, c, d, e, q}, x] && NeQ[q, -1]

Rule 6087

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Int[E
xpandIntegrand[(a + b*ArcTanh[c*x])^p, (f*x)^m*(d + e*x)^q, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[
p, 0] && IntegerQ[q] && (GtQ[q, 0] || NeQ[a, 0] || IntegerQ[m])

Rubi steps

\begin {align*} \int \frac {a+b \tanh ^{-1}(c x)}{x (d+c d x)^3} \, dx &=\int \left (\frac {a+b \tanh ^{-1}(c x)}{d^3 x}-\frac {c \left (a+b \tanh ^{-1}(c x)\right )}{d^3 (1+c x)^3}-\frac {c \left (a+b \tanh ^{-1}(c x)\right )}{d^3 (1+c x)^2}-\frac {c \left (a+b \tanh ^{-1}(c x)\right )}{d^3 (1+c x)}\right ) \, dx\\ &=\frac {\int \frac {a+b \tanh ^{-1}(c x)}{x} \, dx}{d^3}-\frac {c \int \frac {a+b \tanh ^{-1}(c x)}{(1+c x)^3} \, dx}{d^3}-\frac {c \int \frac {a+b \tanh ^{-1}(c x)}{(1+c x)^2} \, dx}{d^3}-\frac {c \int \frac {a+b \tanh ^{-1}(c x)}{1+c x} \, dx}{d^3}\\ &=\frac {a+b \tanh ^{-1}(c x)}{2 d^3 (1+c x)^2}+\frac {a+b \tanh ^{-1}(c x)}{d^3 (1+c x)}+\frac {a \log (x)}{d^3}+\frac {\left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1+c x}\right )}{d^3}-\frac {b \text {Li}_2(-c x)}{2 d^3}+\frac {b \text {Li}_2(c x)}{2 d^3}-\frac {(b c) \int \frac {1}{(1+c x)^2 \left (1-c^2 x^2\right )} \, dx}{2 d^3}-\frac {(b c) \int \frac {1}{(1+c x) \left (1-c^2 x^2\right )} \, dx}{d^3}-\frac {(b c) \int \frac {\log \left (\frac {2}{1+c x}\right )}{1-c^2 x^2} \, dx}{d^3}\\ &=\frac {a+b \tanh ^{-1}(c x)}{2 d^3 (1+c x)^2}+\frac {a+b \tanh ^{-1}(c x)}{d^3 (1+c x)}+\frac {a \log (x)}{d^3}+\frac {\left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1+c x}\right )}{d^3}-\frac {b \text {Li}_2(-c x)}{2 d^3}+\frac {b \text {Li}_2(c x)}{2 d^3}-\frac {b \text {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1+c x}\right )}{d^3}-\frac {(b c) \int \frac {1}{(1-c x) (1+c x)^3} \, dx}{2 d^3}-\frac {(b c) \int \frac {1}{(1-c x) (1+c x)^2} \, dx}{d^3}\\ &=\frac {a+b \tanh ^{-1}(c x)}{2 d^3 (1+c x)^2}+\frac {a+b \tanh ^{-1}(c x)}{d^3 (1+c x)}+\frac {a \log (x)}{d^3}+\frac {\left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1+c x}\right )}{d^3}-\frac {b \text {Li}_2(-c x)}{2 d^3}+\frac {b \text {Li}_2(c x)}{2 d^3}-\frac {b \text {Li}_2\left (1-\frac {2}{1+c x}\right )}{2 d^3}-\frac {(b c) \int \left (\frac {1}{2 (1+c x)^3}+\frac {1}{4 (1+c x)^2}-\frac {1}{4 \left (-1+c^2 x^2\right )}\right ) \, dx}{2 d^3}-\frac {(b c) \int \left (\frac {1}{2 (1+c x)^2}-\frac {1}{2 \left (-1+c^2 x^2\right )}\right ) \, dx}{d^3}\\ &=\frac {b}{8 d^3 (1+c x)^2}+\frac {5 b}{8 d^3 (1+c x)}+\frac {a+b \tanh ^{-1}(c x)}{2 d^3 (1+c x)^2}+\frac {a+b \tanh ^{-1}(c x)}{d^3 (1+c x)}+\frac {a \log (x)}{d^3}+\frac {\left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1+c x}\right )}{d^3}-\frac {b \text {Li}_2(-c x)}{2 d^3}+\frac {b \text {Li}_2(c x)}{2 d^3}-\frac {b \text {Li}_2\left (1-\frac {2}{1+c x}\right )}{2 d^3}+\frac {(b c) \int \frac {1}{-1+c^2 x^2} \, dx}{8 d^3}+\frac {(b c) \int \frac {1}{-1+c^2 x^2} \, dx}{2 d^3}\\ &=\frac {b}{8 d^3 (1+c x)^2}+\frac {5 b}{8 d^3 (1+c x)}-\frac {5 b \tanh ^{-1}(c x)}{8 d^3}+\frac {a+b \tanh ^{-1}(c x)}{2 d^3 (1+c x)^2}+\frac {a+b \tanh ^{-1}(c x)}{d^3 (1+c x)}+\frac {a \log (x)}{d^3}+\frac {\left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1+c x}\right )}{d^3}-\frac {b \text {Li}_2(-c x)}{2 d^3}+\frac {b \text {Li}_2(c x)}{2 d^3}-\frac {b \text {Li}_2\left (1-\frac {2}{1+c x}\right )}{2 d^3}\\ \end {align*}

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Mathematica [A]
time = 0.31, size = 147, normalized size = 0.91 \begin {gather*} \frac {\frac {16 a}{(1+c x)^2}+\frac {32 a}{1+c x}+32 a \log (x)-32 a \log (1+c x)+b \left (12 \cosh \left (2 \tanh ^{-1}(c x)\right )+\cosh \left (4 \tanh ^{-1}(c x)\right )-16 \text {PolyLog}\left (2,e^{-2 \tanh ^{-1}(c x)}\right )-12 \sinh \left (2 \tanh ^{-1}(c x)\right )+4 \tanh ^{-1}(c x) \left (6 \cosh \left (2 \tanh ^{-1}(c x)\right )+\cosh \left (4 \tanh ^{-1}(c x)\right )+8 \log \left (1-e^{-2 \tanh ^{-1}(c x)}\right )-6 \sinh \left (2 \tanh ^{-1}(c x)\right )-\sinh \left (4 \tanh ^{-1}(c x)\right )\right )-\sinh \left (4 \tanh ^{-1}(c x)\right )\right )}{32 d^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcTanh[c*x])/(x*(d + c*d*x)^3),x]

[Out]

((16*a)/(1 + c*x)^2 + (32*a)/(1 + c*x) + 32*a*Log[x] - 32*a*Log[1 + c*x] + b*(12*Cosh[2*ArcTanh[c*x]] + Cosh[4
*ArcTanh[c*x]] - 16*PolyLog[2, E^(-2*ArcTanh[c*x])] - 12*Sinh[2*ArcTanh[c*x]] + 4*ArcTanh[c*x]*(6*Cosh[2*ArcTa
nh[c*x]] + Cosh[4*ArcTanh[c*x]] + 8*Log[1 - E^(-2*ArcTanh[c*x])] - 6*Sinh[2*ArcTanh[c*x]] - Sinh[4*ArcTanh[c*x
]]) - Sinh[4*ArcTanh[c*x]]))/(32*d^3)

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Maple [A]
time = 0.23, size = 264, normalized size = 1.64

method	result	size
derivativedivides	\(\frac {a}{2 d^{3} \left (c x +1\right )^{2}}+\frac {a}{d^{3} \left (c x +1\right )}-\frac {a \ln \left (c x +1\right )}{d^{3}}+\frac {a \ln \left (c x \right )}{d^{3}}+\frac {b \arctanh \left (c x \right )}{2 d^{3} \left (c x +1\right )^{2}}+\frac {b \arctanh \left (c x \right )}{d^{3} \left (c x +1\right )}-\frac {b \arctanh \left (c x \right ) \ln \left (c x +1\right )}{d^{3}}+\frac {b \arctanh \left (c x \right ) \ln \left (c x \right )}{d^{3}}-\frac {b \dilog \left (c x \right )}{2 d^{3}}-\frac {b \dilog \left (c x +1\right )}{2 d^{3}}-\frac {b \ln \left (c x \right ) \ln \left (c x +1\right )}{2 d^{3}}-\frac {b \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (c x +1\right )}{2 d^{3}}+\frac {b \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (\frac {c x}{2}+\frac {1}{2}\right )}{2 d^{3}}+\frac {b \dilog \left (\frac {c x}{2}+\frac {1}{2}\right )}{2 d^{3}}+\frac {b \ln \left (c x +1\right )^{2}}{4 d^{3}}+\frac {b}{8 d^{3} \left (c x +1\right )^{2}}+\frac {5 b}{8 d^{3} \left (c x +1\right )}-\frac {5 b \ln \left (c x +1\right )}{16 d^{3}}+\frac {5 b \ln \left (c x -1\right )}{16 d^{3}}\)	\(264\)
default	\(\frac {a}{2 d^{3} \left (c x +1\right )^{2}}+\frac {a}{d^{3} \left (c x +1\right )}-\frac {a \ln \left (c x +1\right )}{d^{3}}+\frac {a \ln \left (c x \right )}{d^{3}}+\frac {b \arctanh \left (c x \right )}{2 d^{3} \left (c x +1\right )^{2}}+\frac {b \arctanh \left (c x \right )}{d^{3} \left (c x +1\right )}-\frac {b \arctanh \left (c x \right ) \ln \left (c x +1\right )}{d^{3}}+\frac {b \arctanh \left (c x \right ) \ln \left (c x \right )}{d^{3}}-\frac {b \dilog \left (c x \right )}{2 d^{3}}-\frac {b \dilog \left (c x +1\right )}{2 d^{3}}-\frac {b \ln \left (c x \right ) \ln \left (c x +1\right )}{2 d^{3}}-\frac {b \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (c x +1\right )}{2 d^{3}}+\frac {b \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (\frac {c x}{2}+\frac {1}{2}\right )}{2 d^{3}}+\frac {b \dilog \left (\frac {c x}{2}+\frac {1}{2}\right )}{2 d^{3}}+\frac {b \ln \left (c x +1\right )^{2}}{4 d^{3}}+\frac {b}{8 d^{3} \left (c x +1\right )^{2}}+\frac {5 b}{8 d^{3} \left (c x +1\right )}-\frac {5 b \ln \left (c x +1\right )}{16 d^{3}}+\frac {5 b \ln \left (c x -1\right )}{16 d^{3}}\)	\(264\)
risch	\(-\frac {5 b \ln \left (-c x -1\right )}{16 d^{3}}-\frac {b \ln \left (-c x +1\right ) c x}{4 d^{3} \left (-c x -1\right )}+\frac {b \ln \left (-c x +1\right )}{4 d^{3} \left (-c x -1\right )}-\frac {b}{8 d^{3} \left (-c x -1\right )}+\frac {b \ln \left (-c x +1\right ) x^{2} c^{2}}{16 d^{3} \left (-c x -1\right )^{2}}+\frac {b \ln \left (-c x +1\right ) c x}{8 d^{3} \left (-c x -1\right )^{2}}-\frac {3 b \ln \left (-c x +1\right )}{16 d^{3} \left (-c x -1\right )^{2}}+\frac {\dilog \left (-c x +1\right ) b}{2 d^{3}}-\frac {b \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (\frac {c x}{2}+\frac {1}{2}\right )}{2 d^{3}}+\frac {b \ln \left (\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (-c x +1\right )}{2 d^{3}}-\frac {b \dilog \left (-\frac {c x}{2}+\frac {1}{2}\right )}{2 d^{3}}+\frac {a \ln \left (-c x \right )}{d^{3}}+\frac {a}{2 d^{3} \left (-c x -1\right )^{2}}-\frac {a \ln \left (-c x -1\right )}{d^{3}}-\frac {a}{d^{3} \left (-c x -1\right )}-\frac {b \ln \left (c x +1\right )^{2}}{4 d^{3}}+\frac {b \ln \left (c x +1\right )}{4 d^{3} \left (c x +1\right )^{2}}+\frac {b}{8 d^{3} \left (c x +1\right )^{2}}-\frac {b \dilog \left (c x +1\right )}{2 d^{3}}+\frac {b \ln \left (c x +1\right )}{2 d^{3} \left (c x +1\right )}+\frac {b}{2 d^{3} \left (c x +1\right )}\)	\(351\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctanh(c*x))/x/(c*d*x+d)^3,x,method=_RETURNVERBOSE)

[Out]

1/2*a/d^3/(c*x+1)^2+a/d^3/(c*x+1)-a/d^3*ln(c*x+1)+a/d^3*ln(c*x)+1/2*b/d^3*arctanh(c*x)/(c*x+1)^2+b/d^3*arctanh
(c*x)/(c*x+1)-b/d^3*arctanh(c*x)*ln(c*x+1)+b/d^3*arctanh(c*x)*ln(c*x)-1/2*b/d^3*dilog(c*x)-1/2*b/d^3*dilog(c*x
+1)-1/2*b/d^3*ln(c*x)*ln(c*x+1)-1/2*b/d^3*ln(-1/2*c*x+1/2)*ln(c*x+1)+1/2*b/d^3*ln(-1/2*c*x+1/2)*ln(1/2*c*x+1/2
)+1/2*b/d^3*dilog(1/2*c*x+1/2)+1/4*b/d^3*ln(c*x+1)^2+1/8*b/d^3/(c*x+1)^2+5/8*b/d^3/(c*x+1)-5/16*b/d^3*ln(c*x+1
)+5/16*b/d^3*ln(c*x-1)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x))/x/(c*d*x+d)^3,x, algorithm="maxima")

[Out]

1/2*a*((2*c*x + 3)/(c^2*d^3*x^2 + 2*c*d^3*x + d^3) - 2*log(c*x + 1)/d^3 + 2*log(x)/d^3) + 1/2*b*integrate((log
(c*x + 1) - log(-c*x + 1))/(c^3*d^3*x^4 + 3*c^2*d^3*x^3 + 3*c*d^3*x^2 + d^3*x), x)

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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((a+b*arctanh(c*x))/x/(c*d*x+d)^3,x, algorithm="fricas")

[Out]

integral((b*arctanh(c*x) + a)/(c^3*d^3*x^4 + 3*c^2*d^3*x^3 + 3*c*d^3*x^2 + d^3*x), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {a}{c^{3} x^{4} + 3 c^{2} x^{3} + 3 c x^{2} + x}\, dx + \int \frac {b \operatorname {atanh}{\left (c x \right )}}{c^{3} x^{4} + 3 c^{2} x^{3} + 3 c x^{2} + x}\, dx}{d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atanh(c*x))/x/(c*d*x+d)**3,x)

[Out]

(Integral(a/(c**3*x**4 + 3*c**2*x**3 + 3*c*x**2 + x), x) + Integral(b*atanh(c*x)/(c**3*x**4 + 3*c**2*x**3 + 3*
c*x**2 + x), x))/d**3

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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((a+b*arctanh(c*x))/x/(c*d*x+d)^3,x, algorithm="giac")

[Out]

integrate((b*arctanh(c*x) + a)/((c*d*x + d)^3*x), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {a+b\,\mathrm {atanh}\left (c\,x\right )}{x\,{\left (d+c\,d\,x\right )}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atanh(c*x))/(x*(d + c*d*x)^3),x)

[Out]

int((a + b*atanh(c*x))/(x*(d + c*d*x)^3), x)

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