3.213 \(\int (e+f x) \sin (a+b (c+d x)^{2/3}) \, dx\)

Optimal. Leaf size=243 \[ \frac{3 \sqrt{\frac{\pi }{2}} \cos (a) (d e-c f) \text{FresnelC}\left (\sqrt{\frac{2}{\pi }} \sqrt{b} \sqrt [3]{c+d x}\right )}{2 b^{3/2} d^2}-\frac{3 \sqrt{\frac{\pi }{2}} \sin (a) (d e-c f) S\left (\sqrt{b} \sqrt{\frac{2}{\pi }} \sqrt [3]{c+d x}\right )}{2 b^{3/2} d^2}+\frac{3 f (c+d x)^{2/3} \sin \left (a+b (c+d x)^{2/3}\right )}{b^2 d^2}+\frac{3 f \cos \left (a+b (c+d x)^{2/3}\right )}{b^3 d^2}-\frac{3 \sqrt [3]{c+d x} (d e-c f) \cos \left (a+b (c+d x)^{2/3}\right )}{2 b d^2}-\frac{3 f (c+d x)^{4/3} \cos \left (a+b (c+d x)^{2/3}\right )}{2 b d^2} \]

[Out]

(3*f*Cos[a + b*(c + d*x)^(2/3)])/(b^3*d^2) - (3*(d*e - c*f)*(c + d*x)^(1/3)*Cos[a + b*(c + d*x)^(2/3)])/(2*b*d
^2) - (3*f*(c + d*x)^(4/3)*Cos[a + b*(c + d*x)^(2/3)])/(2*b*d^2) + (3*(d*e - c*f)*Sqrt[Pi/2]*Cos[a]*FresnelC[S
qrt[b]*Sqrt[2/Pi]*(c + d*x)^(1/3)])/(2*b^(3/2)*d^2) - (3*(d*e - c*f)*Sqrt[Pi/2]*FresnelS[Sqrt[b]*Sqrt[2/Pi]*(c
+ d*x)^(1/3)]*Sin[a])/(2*b^(3/2)*d^2) + (3*f*(c + d*x)^(2/3)*Sin[a + b*(c + d*x)^(2/3)])/(b^2*d^2)

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Rubi [A]  time = 0.264173, antiderivative size = 243, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 8, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {3433, 3385, 3354, 3352, 3351, 3379, 3296, 2638} \[ \frac{3 \sqrt{\frac{\pi }{2}} \cos (a) (d e-c f) \text{FresnelC}\left (\sqrt{\frac{2}{\pi }} \sqrt{b} \sqrt [3]{c+d x}\right )}{2 b^{3/2} d^2}-\frac{3 \sqrt{\frac{\pi }{2}} \sin (a) (d e-c f) S\left (\sqrt{b} \sqrt{\frac{2}{\pi }} \sqrt [3]{c+d x}\right )}{2 b^{3/2} d^2}+\frac{3 f (c+d x)^{2/3} \sin \left (a+b (c+d x)^{2/3}\right )}{b^2 d^2}+\frac{3 f \cos \left (a+b (c+d x)^{2/3}\right )}{b^3 d^2}-\frac{3 \sqrt [3]{c+d x} (d e-c f) \cos \left (a+b (c+d x)^{2/3}\right )}{2 b d^2}-\frac{3 f (c+d x)^{4/3} \cos \left (a+b (c+d x)^{2/3}\right )}{2 b d^2} \]

Antiderivative was successfully verified.

[In]

Int[(e + f*x)*Sin[a + b*(c + d*x)^(2/3)],x]

[Out]

(3*f*Cos[a + b*(c + d*x)^(2/3)])/(b^3*d^2) - (3*(d*e - c*f)*(c + d*x)^(1/3)*Cos[a + b*(c + d*x)^(2/3)])/(2*b*d
^2) - (3*f*(c + d*x)^(4/3)*Cos[a + b*(c + d*x)^(2/3)])/(2*b*d^2) + (3*(d*e - c*f)*Sqrt[Pi/2]*Cos[a]*FresnelC[S
qrt[b]*Sqrt[2/Pi]*(c + d*x)^(1/3)])/(2*b^(3/2)*d^2) - (3*(d*e - c*f)*Sqrt[Pi/2]*FresnelS[Sqrt[b]*Sqrt[2/Pi]*(c
+ d*x)^(1/3)]*Sin[a])/(2*b^(3/2)*d^2) + (3*f*(c + d*x)^(2/3)*Sin[a + b*(c + d*x)^(2/3)])/(b^2*d^2)

Rule 3433

Int[((g_.) + (h_.)*(x_))^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*((e_.) + (f_.)*(x_))^(n_)])^(p_.), x_Symbol] :
> Module[{k = If[FractionQ[n], Denominator[n], 1]}, Dist[k/f^(m + 1), Subst[Int[ExpandIntegrand[(a + b*Sin[c +
d*x^(k*n)])^p, x^(k - 1)*(f*g - e*h + h*x^k)^m, x], x], x, (e + f*x)^(1/k)], x]] /; FreeQ[{a, b, c, d, e, f,
g, h}, x] && IGtQ[p, 0] && IGtQ[m, 0]

Rule 3385

Int[((e_.)*(x_))^(m_.)*Sin[(c_.) + (d_.)*(x_)^(n_)], x_Symbol] :> -Simp[(e^(n - 1)*(e*x)^(m - n + 1)*Cos[c + d
*x^n])/(d*n), x] + Dist[(e^n*(m - n + 1))/(d*n), Int[(e*x)^(m - n)*Cos[c + d*x^n], x], x] /; FreeQ[{c, d, e},
x] && IGtQ[n, 0] && LtQ[n, m + 1]

Rule 3354

Int[Cos[(c_) + (d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Dist[Cos[c], Int[Cos[d*(e + f*x)^2], x], x] - Dist[
Sin[c], Int[Sin[d*(e + f*x)^2], x], x] /; FreeQ[{c, d, e, f}, x]

Rule 3352

Int[Cos[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]*FresnelC[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)])/
(f*Rt[d, 2]), x] /; FreeQ[{d, e, f}, x]

Rule 3351

Int[Sin[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]*FresnelS[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)])/
(f*Rt[d, 2]), x] /; FreeQ[{d, e, f}, x]

Rule 3379

Int[(x_)^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*Sin[c + d*x])^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IntegerQ[Simpl
ify[(m + 1)/n]] && (EqQ[p, 1] || EqQ[m, n - 1] || (IntegerQ[p] && GtQ[Simplify[(m + 1)/n], 0]))

Rule 3296

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> -Simp[((c + d*x)^m*Cos[e + f*x])/f, x] +
Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin{align*} \int (e+f x) \sin \left (a+b (c+d x)^{2/3}\right ) \, dx &=\frac{3 \operatorname{Subst}\left (\int \left ((d e-c f) x^2 \sin \left (a+b x^2\right )+f x^5 \sin \left (a+b x^2\right )\right ) \, dx,x,\sqrt [3]{c+d x}\right )}{d^2}\\ &=\frac{(3 f) \operatorname{Subst}\left (\int x^5 \sin \left (a+b x^2\right ) \, dx,x,\sqrt [3]{c+d x}\right )}{d^2}+\frac{(3 (d e-c f)) \operatorname{Subst}\left (\int x^2 \sin \left (a+b x^2\right ) \, dx,x,\sqrt [3]{c+d x}\right )}{d^2}\\ &=-\frac{3 (d e-c f) \sqrt [3]{c+d x} \cos \left (a+b (c+d x)^{2/3}\right )}{2 b d^2}+\frac{(3 f) \operatorname{Subst}\left (\int x^2 \sin (a+b x) \, dx,x,(c+d x)^{2/3}\right )}{2 d^2}+\frac{(3 (d e-c f)) \operatorname{Subst}\left (\int \cos \left (a+b x^2\right ) \, dx,x,\sqrt [3]{c+d x}\right )}{2 b d^2}\\ &=-\frac{3 (d e-c f) \sqrt [3]{c+d x} \cos \left (a+b (c+d x)^{2/3}\right )}{2 b d^2}-\frac{3 f (c+d x)^{4/3} \cos \left (a+b (c+d x)^{2/3}\right )}{2 b d^2}+\frac{(3 f) \operatorname{Subst}\left (\int x \cos (a+b x) \, dx,x,(c+d x)^{2/3}\right )}{b d^2}+\frac{(3 (d e-c f) \cos (a)) \operatorname{Subst}\left (\int \cos \left (b x^2\right ) \, dx,x,\sqrt [3]{c+d x}\right )}{2 b d^2}-\frac{(3 (d e-c f) \sin (a)) \operatorname{Subst}\left (\int \sin \left (b x^2\right ) \, dx,x,\sqrt [3]{c+d x}\right )}{2 b d^2}\\ &=-\frac{3 (d e-c f) \sqrt [3]{c+d x} \cos \left (a+b (c+d x)^{2/3}\right )}{2 b d^2}-\frac{3 f (c+d x)^{4/3} \cos \left (a+b (c+d x)^{2/3}\right )}{2 b d^2}+\frac{3 (d e-c f) \sqrt{\frac{\pi }{2}} \cos (a) C\left (\sqrt{b} \sqrt{\frac{2}{\pi }} \sqrt [3]{c+d x}\right )}{2 b^{3/2} d^2}-\frac{3 (d e-c f) \sqrt{\frac{\pi }{2}} S\left (\sqrt{b} \sqrt{\frac{2}{\pi }} \sqrt [3]{c+d x}\right ) \sin (a)}{2 b^{3/2} d^2}+\frac{3 f (c+d x)^{2/3} \sin \left (a+b (c+d x)^{2/3}\right )}{b^2 d^2}-\frac{(3 f) \operatorname{Subst}\left (\int \sin (a+b x) \, dx,x,(c+d x)^{2/3}\right )}{b^2 d^2}\\ &=\frac{3 f \cos \left (a+b (c+d x)^{2/3}\right )}{b^3 d^2}-\frac{3 (d e-c f) \sqrt [3]{c+d x} \cos \left (a+b (c+d x)^{2/3}\right )}{2 b d^2}-\frac{3 f (c+d x)^{4/3} \cos \left (a+b (c+d x)^{2/3}\right )}{2 b d^2}+\frac{3 (d e-c f) \sqrt{\frac{\pi }{2}} \cos (a) C\left (\sqrt{b} \sqrt{\frac{2}{\pi }} \sqrt [3]{c+d x}\right )}{2 b^{3/2} d^2}-\frac{3 (d e-c f) \sqrt{\frac{\pi }{2}} S\left (\sqrt{b} \sqrt{\frac{2}{\pi }} \sqrt [3]{c+d x}\right ) \sin (a)}{2 b^{3/2} d^2}+\frac{3 f (c+d x)^{2/3} \sin \left (a+b (c+d x)^{2/3}\right )}{b^2 d^2}\\ \end{align*}

Mathematica [A]  time = 0.824316, size = 213, normalized size = 0.88 \[ \frac{3 \left (\sqrt{2 \pi } b^{3/2} \cos (a) (d e-c f) \text{FresnelC}\left (\sqrt{\frac{2}{\pi }} \sqrt{b} \sqrt [3]{c+d x}\right )-\sqrt{2 \pi } b^{3/2} \sin (a) (d e-c f) S\left (\sqrt{b} \sqrt{\frac{2}{\pi }} \sqrt [3]{c+d x}\right )-2 b^2 d e \sqrt [3]{c+d x} \cos \left (a+b (c+d x)^{2/3}\right )-2 b^2 d f x \sqrt [3]{c+d x} \cos \left (a+b (c+d x)^{2/3}\right )+4 b f (c+d x)^{2/3} \sin \left (a+b (c+d x)^{2/3}\right )+4 f \cos \left (a+b (c+d x)^{2/3}\right )\right )}{4 b^3 d^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(e + f*x)*Sin[a + b*(c + d*x)^(2/3)],x]

[Out]

(3*(4*f*Cos[a + b*(c + d*x)^(2/3)] - 2*b^2*d*e*(c + d*x)^(1/3)*Cos[a + b*(c + d*x)^(2/3)] - 2*b^2*d*f*x*(c + d
*x)^(1/3)*Cos[a + b*(c + d*x)^(2/3)] + b^(3/2)*(d*e - c*f)*Sqrt[2*Pi]*Cos[a]*FresnelC[Sqrt[b]*Sqrt[2/Pi]*(c +
d*x)^(1/3)] - b^(3/2)*(d*e - c*f)*Sqrt[2*Pi]*FresnelS[Sqrt[b]*Sqrt[2/Pi]*(c + d*x)^(1/3)]*Sin[a] + 4*b*f*(c +
d*x)^(2/3)*Sin[a + b*(c + d*x)^(2/3)]))/(4*b^3*d^2)

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Maple [A]  time = 0.009, size = 175, normalized size = 0.7 \begin{align*} 3\,{\frac{1}{{d}^{2}} \left ( -1/2\,{\frac{f \left ( dx+c \right ) ^{4/3}\cos \left ( a+b \left ( dx+c \right ) ^{2/3} \right ) }{b}}+2\,{\frac{f}{b} \left ( 1/2\,{\frac{ \left ( dx+c \right ) ^{2/3}\sin \left ( a+b \left ( dx+c \right ) ^{2/3} \right ) }{b}}+1/2\,{\frac{\cos \left ( a+b \left ( dx+c \right ) ^{2/3} \right ) }{{b}^{2}}} \right ) }-1/2\,{\frac{ \left ( -cf+de \right ) \sqrt [3]{dx+c}\cos \left ( a+b \left ( dx+c \right ) ^{2/3} \right ) }{b}}+1/4\,{\frac{ \left ( -cf+de \right ) \sqrt{2}\sqrt{\pi }}{{b}^{3/2}} \left ( \cos \left ( a \right ){\it FresnelC} \left ({\frac{\sqrt [3]{dx+c}\sqrt{b}\sqrt{2}}{\sqrt{\pi }}} \right ) -\sin \left ( a \right ){\it FresnelS} \left ({\frac{\sqrt [3]{dx+c}\sqrt{b}\sqrt{2}}{\sqrt{\pi }}} \right ) \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x+e)*sin(a+b*(d*x+c)^(2/3)),x)

[Out]

3/d^2*(-1/2*f/b*(d*x+c)^(4/3)*cos(a+b*(d*x+c)^(2/3))+2*f/b*(1/2/b*(d*x+c)^(2/3)*sin(a+b*(d*x+c)^(2/3))+1/2/b^2
*cos(a+b*(d*x+c)^(2/3)))-1/2*(-c*f+d*e)/b*(d*x+c)^(1/3)*cos(a+b*(d*x+c)^(2/3))+1/4*(-c*f+d*e)/b^(3/2)*2^(1/2)*
Pi^(1/2)*(cos(a)*FresnelC((d*x+c)^(1/3)*b^(1/2)*2^(1/2)/Pi^(1/2))-sin(a)*FresnelS((d*x+c)^(1/3)*b^(1/2)*2^(1/2
)/Pi^(1/2))))

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Maxima [C]  time = 1.88412, size = 832, normalized size = 3.42 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)*sin(a+b*(d*x+c)^(2/3)),x, algorithm="maxima")

[Out]

-3/16*((8*(d*x + c)^(1/3)*abs(b)*cos((d*x + c)^(2/3)*b + a) - sqrt(pi)*(((cos(1/4*pi + 1/2*arctan2(0, b)) + co
s(-1/4*pi + 1/2*arctan2(0, b)) - I*sin(1/4*pi + 1/2*arctan2(0, b)) + I*sin(-1/4*pi + 1/2*arctan2(0, b)))*cos(a
) - (I*cos(1/4*pi + 1/2*arctan2(0, b)) + I*cos(-1/4*pi + 1/2*arctan2(0, b)) + sin(1/4*pi + 1/2*arctan2(0, b))
- sin(-1/4*pi + 1/2*arctan2(0, b)))*sin(a))*erf((d*x + c)^(1/3)*sqrt(I*b)) + ((cos(1/4*pi + 1/2*arctan2(0, b))
+ cos(-1/4*pi + 1/2*arctan2(0, b)) + I*sin(1/4*pi + 1/2*arctan2(0, b)) - I*sin(-1/4*pi + 1/2*arctan2(0, b)))*
cos(a) - (-I*cos(1/4*pi + 1/2*arctan2(0, b)) - I*cos(-1/4*pi + 1/2*arctan2(0, b)) + sin(1/4*pi + 1/2*arctan2(0
, b)) - sin(-1/4*pi + 1/2*arctan2(0, b)))*sin(a))*erf((d*x + c)^(1/3)*sqrt(-I*b)))*sqrt(abs(b)))*e/(b*abs(b))
- (8*(d*x + c)^(1/3)*abs(b)*cos((d*x + c)^(2/3)*b + a) - sqrt(pi)*(((cos(1/4*pi + 1/2*arctan2(0, b)) + cos(-1/
4*pi + 1/2*arctan2(0, b)) - I*sin(1/4*pi + 1/2*arctan2(0, b)) + I*sin(-1/4*pi + 1/2*arctan2(0, b)))*cos(a) - (
I*cos(1/4*pi + 1/2*arctan2(0, b)) + I*cos(-1/4*pi + 1/2*arctan2(0, b)) + sin(1/4*pi + 1/2*arctan2(0, b)) - sin
(-1/4*pi + 1/2*arctan2(0, b)))*sin(a))*erf((d*x + c)^(1/3)*sqrt(I*b)) + ((cos(1/4*pi + 1/2*arctan2(0, b)) + co
s(-1/4*pi + 1/2*arctan2(0, b)) + I*sin(1/4*pi + 1/2*arctan2(0, b)) - I*sin(-1/4*pi + 1/2*arctan2(0, b)))*cos(a
) - (-I*cos(1/4*pi + 1/2*arctan2(0, b)) - I*cos(-1/4*pi + 1/2*arctan2(0, b)) + sin(1/4*pi + 1/2*arctan2(0, b))
- sin(-1/4*pi + 1/2*arctan2(0, b)))*sin(a))*erf((d*x + c)^(1/3)*sqrt(-I*b)))*sqrt(abs(b)))*c*f/(b*d*abs(b)) -
8*(2*(d*x + c)^(2/3)*b*sin((d*x + c)^(2/3)*b + a) - ((d*x + c)^(4/3)*b^2 - 2)*cos((d*x + c)^(2/3)*b + a))*f/(
b^3*d))/d

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Fricas [A]  time = 1.87867, size = 450, normalized size = 1.85 \begin{align*} \frac{3 \,{\left (\sqrt{2} \pi{\left (b d e - b c f\right )} \sqrt{\frac{b}{\pi }} \cos \left (a\right ) \operatorname{C}\left (\sqrt{2}{\left (d x + c\right )}^{\frac{1}{3}} \sqrt{\frac{b}{\pi }}\right ) - \sqrt{2} \pi{\left (b d e - b c f\right )} \sqrt{\frac{b}{\pi }} \operatorname{S}\left (\sqrt{2}{\left (d x + c\right )}^{\frac{1}{3}} \sqrt{\frac{b}{\pi }}\right ) \sin \left (a\right ) + 4 \,{\left (d x + c\right )}^{\frac{2}{3}} b f \sin \left ({\left (d x + c\right )}^{\frac{2}{3}} b + a\right ) - 2 \,{\left ({\left (b^{2} d f x + b^{2} d e\right )}{\left (d x + c\right )}^{\frac{1}{3}} - 2 \, f\right )} \cos \left ({\left (d x + c\right )}^{\frac{2}{3}} b + a\right )\right )}}{4 \, b^{3} d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)*sin(a+b*(d*x+c)^(2/3)),x, algorithm="fricas")

[Out]

3/4*(sqrt(2)*pi*(b*d*e - b*c*f)*sqrt(b/pi)*cos(a)*fresnel_cos(sqrt(2)*(d*x + c)^(1/3)*sqrt(b/pi)) - sqrt(2)*pi
*(b*d*e - b*c*f)*sqrt(b/pi)*fresnel_sin(sqrt(2)*(d*x + c)^(1/3)*sqrt(b/pi))*sin(a) + 4*(d*x + c)^(2/3)*b*f*sin
((d*x + c)^(2/3)*b + a) - 2*((b^2*d*f*x + b^2*d*e)*(d*x + c)^(1/3) - 2*f)*cos((d*x + c)^(2/3)*b + a))/(b^3*d^2
)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (e + f x\right ) \sin{\left (a + b \left (c + d x\right )^{\frac{2}{3}} \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)*sin(a+b*(d*x+c)**(2/3)),x)

[Out]

Integral((e + f*x)*sin(a + b*(c + d*x)**(2/3)), x)

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Giac [C]  time = 1.2604, size = 549, normalized size = 2.26 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)*sin(a+b*(d*x+c)^(2/3)),x, algorithm="giac")

[Out]

-3/8*((sqrt(2)*sqrt(pi)*erf(-1/2*sqrt(2)*(d*x + c)^(1/3)*(-I*b/abs(b) + 1)*sqrt(abs(b)))*e^(I*a)/(b*(-I*b/abs(
b) + 1)*sqrt(abs(b))) + sqrt(2)*sqrt(pi)*erf(-1/2*sqrt(2)*(d*x + c)^(1/3)*(I*b/abs(b) + 1)*sqrt(abs(b)))*e^(-I
*a)/(b*(I*b/abs(b) + 1)*sqrt(abs(b))) + 2*(d*x + c)^(1/3)*e^(I*(d*x + c)^(2/3)*b + I*a)/b + 2*(d*x + c)^(1/3)*
e^(-I*(d*x + c)^(2/3)*b - I*a)/b)*e - (sqrt(2)*sqrt(pi)*c*erf(-1/2*sqrt(2)*(d*x + c)^(1/3)*(-I*b/abs(b) + 1)*s
qrt(abs(b)))*e^(I*a)/(b*(-I*b/abs(b) + 1)*sqrt(abs(b))) + sqrt(2)*sqrt(pi)*c*erf(-1/2*sqrt(2)*(d*x + c)^(1/3)*
(I*b/abs(b) + 1)*sqrt(abs(b)))*e^(-I*a)/(b*(I*b/abs(b) + 1)*sqrt(abs(b))) + 2*I*(I*(d*x + c)^(4/3)*b^2 - I*(d*
x + c)^(1/3)*b^2*c - 2*(d*x + c)^(2/3)*b - 2*I)*e^(I*(d*x + c)^(2/3)*b + I*a)/b^3 + 2*I*(I*(d*x + c)^(4/3)*b^2
- I*(d*x + c)^(1/3)*b^2*c + 2*(d*x + c)^(2/3)*b - 2*I)*e^(-I*(d*x + c)^(2/3)*b - I*a)/b^3)*f/d)/d