3.296 \(\int \frac{(a+b \sinh ^{-1}(c x))^2}{x \sqrt{d+c^2 d x^2}} \, dx\)

Optimal. Leaf size=223 \[ -\frac{2 b \sqrt{c^2 x^2+1} \text{PolyLog}\left (2,-e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt{c^2 d x^2+d}}+\frac{2 b \sqrt{c^2 x^2+1} \text{PolyLog}\left (2,e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt{c^2 d x^2+d}}+\frac{2 b^2 \sqrt{c^2 x^2+1} \text{PolyLog}\left (3,-e^{\sinh ^{-1}(c x)}\right )}{\sqrt{c^2 d x^2+d}}-\frac{2 b^2 \sqrt{c^2 x^2+1} \text{PolyLog}\left (3,e^{\sinh ^{-1}(c x)}\right )}{\sqrt{c^2 d x^2+d}}-\frac{2 \sqrt{c^2 x^2+1} \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{\sqrt{c^2 d x^2+d}} \]

[Out]

(-2*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x])^2*ArcTanh[E^ArcSinh[c*x]])/Sqrt[d + c^2*d*x^2] - (2*b*Sqrt[1 + c^2*
x^2]*(a + b*ArcSinh[c*x])*PolyLog[2, -E^ArcSinh[c*x]])/Sqrt[d + c^2*d*x^2] + (2*b*Sqrt[1 + c^2*x^2]*(a + b*Arc
Sinh[c*x])*PolyLog[2, E^ArcSinh[c*x]])/Sqrt[d + c^2*d*x^2] + (2*b^2*Sqrt[1 + c^2*x^2]*PolyLog[3, -E^ArcSinh[c*
x]])/Sqrt[d + c^2*d*x^2] - (2*b^2*Sqrt[1 + c^2*x^2]*PolyLog[3, E^ArcSinh[c*x]])/Sqrt[d + c^2*d*x^2]

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Rubi [A]  time = 0.34232, antiderivative size = 223, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {5764, 5760, 4182, 2531, 2282, 6589} \[ -\frac{2 b \sqrt{c^2 x^2+1} \text{PolyLog}\left (2,-e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt{c^2 d x^2+d}}+\frac{2 b \sqrt{c^2 x^2+1} \text{PolyLog}\left (2,e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt{c^2 d x^2+d}}+\frac{2 b^2 \sqrt{c^2 x^2+1} \text{PolyLog}\left (3,-e^{\sinh ^{-1}(c x)}\right )}{\sqrt{c^2 d x^2+d}}-\frac{2 b^2 \sqrt{c^2 x^2+1} \text{PolyLog}\left (3,e^{\sinh ^{-1}(c x)}\right )}{\sqrt{c^2 d x^2+d}}-\frac{2 \sqrt{c^2 x^2+1} \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{\sqrt{c^2 d x^2+d}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSinh[c*x])^2/(x*Sqrt[d + c^2*d*x^2]),x]

[Out]

(-2*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x])^2*ArcTanh[E^ArcSinh[c*x]])/Sqrt[d + c^2*d*x^2] - (2*b*Sqrt[1 + c^2*
x^2]*(a + b*ArcSinh[c*x])*PolyLog[2, -E^ArcSinh[c*x]])/Sqrt[d + c^2*d*x^2] + (2*b*Sqrt[1 + c^2*x^2]*(a + b*Arc
Sinh[c*x])*PolyLog[2, E^ArcSinh[c*x]])/Sqrt[d + c^2*d*x^2] + (2*b^2*Sqrt[1 + c^2*x^2]*PolyLog[3, -E^ArcSinh[c*
x]])/Sqrt[d + c^2*d*x^2] - (2*b^2*Sqrt[1 + c^2*x^2]*PolyLog[3, E^ArcSinh[c*x]])/Sqrt[d + c^2*d*x^2]

Rule 5764

Int[(((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Dist
[Sqrt[1 + c^2*x^2]/Sqrt[d + e*x^2], Int[((f*x)^m*(a + b*ArcSinh[c*x])^n)/Sqrt[1 + c^2*x^2], x], x] /; FreeQ[{a
, b, c, d, e, f, m}, x] && EqQ[e, c^2*d] && GtQ[n, 0] &&  !GtQ[d, 0] && (IntegerQ[m] || EqQ[n, 1])

Rule 5760

Int[(((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Dist[1/(c^(m
+ 1)*Sqrt[d]), Subst[Int[(a + b*x)^n*Sinh[x]^m, x], x, ArcSinh[c*x]], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[
e, c^2*d] && GtQ[d, 0] && IGtQ[n, 0] && IntegerQ[m]

Rule 4182

Int[csc[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c + d*x)^m*Ar
cTanh[E^(-(I*e) + f*fz*x)])/(f*fz*I), x] + (-Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*Log[1 - E^(-(I*e) + f*
fz*x)], x], x] + Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*Log[1 + E^(-(I*e) + f*fz*x)], x], x]) /; FreeQ[{c,
d, e, f, fz}, x] && IGtQ[m, 0]

Rule 2531

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> -Simp[((
f + g*x)^m*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)])/(b*c*n*Log[F]), x] + Dist[(g*m)/(b*c*n*Log[F]), Int[(f + g*x)
^(m - 1)*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 6589

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rubi steps

\begin{align*} \int \frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{x \sqrt{d+c^2 d x^2}} \, dx &=\frac{\sqrt{1+c^2 x^2} \int \frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{x \sqrt{1+c^2 x^2}} \, dx}{\sqrt{d+c^2 d x^2}}\\ &=\frac{\sqrt{1+c^2 x^2} \operatorname{Subst}\left (\int (a+b x)^2 \text{csch}(x) \, dx,x,\sinh ^{-1}(c x)\right )}{\sqrt{d+c^2 d x^2}}\\ &=-\frac{2 \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{\sqrt{d+c^2 d x^2}}-\frac{\left (2 b \sqrt{1+c^2 x^2}\right ) \operatorname{Subst}\left (\int (a+b x) \log \left (1-e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{\sqrt{d+c^2 d x^2}}+\frac{\left (2 b \sqrt{1+c^2 x^2}\right ) \operatorname{Subst}\left (\int (a+b x) \log \left (1+e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{\sqrt{d+c^2 d x^2}}\\ &=-\frac{2 \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{\sqrt{d+c^2 d x^2}}-\frac{2 b \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right ) \text{Li}_2\left (-e^{\sinh ^{-1}(c x)}\right )}{\sqrt{d+c^2 d x^2}}+\frac{2 b \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right ) \text{Li}_2\left (e^{\sinh ^{-1}(c x)}\right )}{\sqrt{d+c^2 d x^2}}+\frac{\left (2 b^2 \sqrt{1+c^2 x^2}\right ) \operatorname{Subst}\left (\int \text{Li}_2\left (-e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{\sqrt{d+c^2 d x^2}}-\frac{\left (2 b^2 \sqrt{1+c^2 x^2}\right ) \operatorname{Subst}\left (\int \text{Li}_2\left (e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{\sqrt{d+c^2 d x^2}}\\ &=-\frac{2 \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{\sqrt{d+c^2 d x^2}}-\frac{2 b \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right ) \text{Li}_2\left (-e^{\sinh ^{-1}(c x)}\right )}{\sqrt{d+c^2 d x^2}}+\frac{2 b \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right ) \text{Li}_2\left (e^{\sinh ^{-1}(c x)}\right )}{\sqrt{d+c^2 d x^2}}+\frac{\left (2 b^2 \sqrt{1+c^2 x^2}\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2(-x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{\sqrt{d+c^2 d x^2}}-\frac{\left (2 b^2 \sqrt{1+c^2 x^2}\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2(x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{\sqrt{d+c^2 d x^2}}\\ &=-\frac{2 \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{\sqrt{d+c^2 d x^2}}-\frac{2 b \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right ) \text{Li}_2\left (-e^{\sinh ^{-1}(c x)}\right )}{\sqrt{d+c^2 d x^2}}+\frac{2 b \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right ) \text{Li}_2\left (e^{\sinh ^{-1}(c x)}\right )}{\sqrt{d+c^2 d x^2}}+\frac{2 b^2 \sqrt{1+c^2 x^2} \text{Li}_3\left (-e^{\sinh ^{-1}(c x)}\right )}{\sqrt{d+c^2 d x^2}}-\frac{2 b^2 \sqrt{1+c^2 x^2} \text{Li}_3\left (e^{\sinh ^{-1}(c x)}\right )}{\sqrt{d+c^2 d x^2}}\\ \end{align*}

Mathematica [A]  time = 0.880241, size = 266, normalized size = 1.19 \[ \frac{2 a b \sqrt{c^2 x^2+1} \left (\text{PolyLog}\left (2,-e^{-\sinh ^{-1}(c x)}\right )-\text{PolyLog}\left (2,e^{-\sinh ^{-1}(c x)}\right )+\sinh ^{-1}(c x) \left (\log \left (1-e^{-\sinh ^{-1}(c x)}\right )-\log \left (e^{-\sinh ^{-1}(c x)}+1\right )\right )\right )}{\sqrt{c^2 d x^2+d}}+\frac{b^2 \sqrt{c^2 x^2+1} \left (2 \sinh ^{-1}(c x) \text{PolyLog}\left (2,-e^{-\sinh ^{-1}(c x)}\right )-2 \sinh ^{-1}(c x) \text{PolyLog}\left (2,e^{-\sinh ^{-1}(c x)}\right )+2 \text{PolyLog}\left (3,-e^{-\sinh ^{-1}(c x)}\right )-2 \text{PolyLog}\left (3,e^{-\sinh ^{-1}(c x)}\right )+\sinh ^{-1}(c x)^2 \log \left (1-e^{-\sinh ^{-1}(c x)}\right )-\sinh ^{-1}(c x)^2 \log \left (e^{-\sinh ^{-1}(c x)}+1\right )\right )}{\sqrt{c^2 d x^2+d}}-\frac{a^2 \log \left (\sqrt{d} \sqrt{c^2 d x^2+d}+d\right )}{\sqrt{d}}+\frac{a^2 \log (c x)}{\sqrt{d}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*ArcSinh[c*x])^2/(x*Sqrt[d + c^2*d*x^2]),x]

[Out]

(a^2*Log[c*x])/Sqrt[d] - (a^2*Log[d + Sqrt[d]*Sqrt[d + c^2*d*x^2]])/Sqrt[d] + (2*a*b*Sqrt[1 + c^2*x^2]*(ArcSin
h[c*x]*(Log[1 - E^(-ArcSinh[c*x])] - Log[1 + E^(-ArcSinh[c*x])]) + PolyLog[2, -E^(-ArcSinh[c*x])] - PolyLog[2,
E^(-ArcSinh[c*x])]))/Sqrt[d + c^2*d*x^2] + (b^2*Sqrt[1 + c^2*x^2]*(ArcSinh[c*x]^2*Log[1 - E^(-ArcSinh[c*x])]
- ArcSinh[c*x]^2*Log[1 + E^(-ArcSinh[c*x])] + 2*ArcSinh[c*x]*PolyLog[2, -E^(-ArcSinh[c*x])] - 2*ArcSinh[c*x]*P
olyLog[2, E^(-ArcSinh[c*x])] + 2*PolyLog[3, -E^(-ArcSinh[c*x])] - 2*PolyLog[3, E^(-ArcSinh[c*x])]))/Sqrt[d + c
^2*d*x^2]

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Maple [B]  time = 0.234, size = 564, normalized size = 2.5 \begin{align*} -{{a}^{2}\ln \left ({\frac{1}{x} \left ( 2\,d+2\,\sqrt{d}\sqrt{{c}^{2}d{x}^{2}+d} \right ) } \right ){\frac{1}{\sqrt{d}}}}-{\frac{{b}^{2} \left ({\it Arcsinh} \left ( cx \right ) \right ) ^{2}}{d}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }\ln \left ( 1+cx+\sqrt{{c}^{2}{x}^{2}+1} \right ){\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}}-2\,{\frac{{b}^{2}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }{\it Arcsinh} \left ( cx \right ){\it polylog} \left ( 2,-cx-\sqrt{{c}^{2}{x}^{2}+1} \right ) }{\sqrt{{c}^{2}{x}^{2}+1}d}}+2\,{\frac{{b}^{2}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }{\it polylog} \left ( 3,-cx-\sqrt{{c}^{2}{x}^{2}+1} \right ) }{\sqrt{{c}^{2}{x}^{2}+1}d}}+{\frac{{b}^{2} \left ({\it Arcsinh} \left ( cx \right ) \right ) ^{2}}{d}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }\ln \left ( 1-cx-\sqrt{{c}^{2}{x}^{2}+1} \right ){\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}}+2\,{\frac{{b}^{2}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }{\it Arcsinh} \left ( cx \right ){\it polylog} \left ( 2,cx+\sqrt{{c}^{2}{x}^{2}+1} \right ) }{\sqrt{{c}^{2}{x}^{2}+1}d}}-2\,{\frac{{b}^{2}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }{\it polylog} \left ( 3,cx+\sqrt{{c}^{2}{x}^{2}+1} \right ) }{\sqrt{{c}^{2}{x}^{2}+1}d}}-2\,{\frac{ab\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }{\it Arcsinh} \left ( cx \right ) \ln \left ( 1+cx+\sqrt{{c}^{2}{x}^{2}+1} \right ) }{\sqrt{{c}^{2}{x}^{2}+1}d}}-2\,{\frac{ab\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }{\it polylog} \left ( 2,-cx-\sqrt{{c}^{2}{x}^{2}+1} \right ) }{\sqrt{{c}^{2}{x}^{2}+1}d}}+2\,{\frac{ab\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }{\it Arcsinh} \left ( cx \right ) \ln \left ( 1-cx-\sqrt{{c}^{2}{x}^{2}+1} \right ) }{\sqrt{{c}^{2}{x}^{2}+1}d}}+2\,{\frac{ab\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }{\it polylog} \left ( 2,cx+\sqrt{{c}^{2}{x}^{2}+1} \right ) }{\sqrt{{c}^{2}{x}^{2}+1}d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsinh(c*x))^2/x/(c^2*d*x^2+d)^(1/2),x)

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(c*x))^2/x/(c^2*d*x^2+d)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{c^{2} d x^{2} + d}{\left (b^{2} \operatorname{arsinh}\left (c x\right )^{2} + 2 \, a b \operatorname{arsinh}\left (c x\right ) + a^{2}\right )}}{c^{2} d x^{3} + d x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(c*x))^2/x/(c^2*d*x^2+d)^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(c^2*d*x^2 + d)*(b^2*arcsinh(c*x)^2 + 2*a*b*arcsinh(c*x) + a^2)/(c^2*d*x^3 + d*x), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b \operatorname{asinh}{\left (c x \right )}\right )^{2}}{x \sqrt{d \left (c^{2} x^{2} + 1\right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asinh(c*x))**2/x/(c**2*d*x**2+d)**(1/2),x)

[Out]

Integral((a + b*asinh(c*x))**2/(x*sqrt(d*(c**2*x**2 + 1))), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \operatorname{arsinh}\left (c x\right ) + a\right )}^{2}}{\sqrt{c^{2} d x^{2} + d} x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(c*x))^2/x/(c^2*d*x^2+d)^(1/2),x, algorithm="giac")

[Out]

integrate((b*arcsinh(c*x) + a)^2/(sqrt(c^2*d*x^2 + d)*x), x)