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

Optimal. Leaf size=87 \[ \frac {\sqrt {d x-1} \sqrt {d x+1} \left (2 \left (3 a d^2+2 c\right )+3 b d^2 x\right )}{6 d^4}+\frac {b \cosh ^{-1}(d x)}{2 d^3}+\frac {c x^2 \sqrt {d x-1} \sqrt {d x+1}}{3 d^2} \]

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Rubi [A]  time = 0.15, antiderivative size = 151, normalized size of antiderivative = 1.74, number of steps used = 5, number of rules used = 5, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1610, 1809, 780, 217, 206} \begin {gather*} -\frac {\left (1-d^2 x^2\right ) \left (2 \left (3 a d^2+2 c\right )+3 b d^2 x\right )}{6 d^4 \sqrt {d x-1} \sqrt {d x+1}}+\frac {b \sqrt {d^2 x^2-1} \tanh ^{-1}\left (\frac {d x}{\sqrt {d^2 x^2-1}}\right )}{2 d^3 \sqrt {d x-1} \sqrt {d x+1}}-\frac {c x^2 \left (1-d^2 x^2\right )}{3 d^2 \sqrt {d x-1} \sqrt {d x+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x*(a + b*x + c*x^2))/(Sqrt[-1 + d*x]*Sqrt[1 + d*x]),x]

[Out]

-(c*x^2*(1 - d^2*x^2))/(3*d^2*Sqrt[-1 + d*x]*Sqrt[1 + d*x]) - ((2*(2*c + 3*a*d^2) + 3*b*d^2*x)*(1 - d^2*x^2))/
(6*d^4*Sqrt[-1 + d*x]*Sqrt[1 + d*x]) + (b*Sqrt[-1 + d^2*x^2]*ArcTanh[(d*x)/Sqrt[-1 + d^2*x^2]])/(2*d^3*Sqrt[-1
+ d*x]*Sqrt[1 + d*x])

Rule 206

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

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 780

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(((e*f + d*g)*(2*p
+ 3) + 2*e*g*(p + 1)*x)*(a + c*x^2)^(p + 1))/(2*c*(p + 1)*(2*p + 3)), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(c*
(2*p + 3)), Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, p}, x] &&  !LeQ[p, -1]

Rule 1610

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Dist[((
a + b*x)^FracPart[m]*(c + d*x)^FracPart[m])/(a*c + b*d*x^2)^FracPart[m], Int[Px*(a*c + b*d*x^2)^m*(e + f*x)^p,
x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && PolyQ[Px, x] && EqQ[b*c + a*d, 0] && EqQ[m, n] &&  !Intege
rQ[m]

Rule 1809

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], f = Coeff[Pq, x,
Expon[Pq, x]]}, Simp[(f*(c*x)^(m + q - 1)*(a + b*x^2)^(p + 1))/(b*c^(q - 1)*(m + q + 2*p + 1)), x] + Dist[1/(
b*(m + q + 2*p + 1)), Int[(c*x)^m*(a + b*x^2)^p*ExpandToSum[b*(m + q + 2*p + 1)*Pq - b*f*(m + q + 2*p + 1)*x^q
- a*f*(m + q - 1)*x^(q - 2), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, b, c, m, p}, x]
&& PolyQ[Pq, x] && ( !IGtQ[m, 0] || IGtQ[p + 1/2, -1])

Rubi steps

\begin {align*} \int \frac {x \left (a+b x+c x^2\right )}{\sqrt {-1+d x} \sqrt {1+d x}} \, dx &=\frac {\sqrt {-1+d^2 x^2} \int \frac {x \left (a+b x+c x^2\right )}{\sqrt {-1+d^2 x^2}} \, dx}{\sqrt {-1+d x} \sqrt {1+d x}}\\ &=-\frac {c x^2 \left (1-d^2 x^2\right )}{3 d^2 \sqrt {-1+d x} \sqrt {1+d x}}+\frac {\sqrt {-1+d^2 x^2} \int \frac {x \left (2 c+3 a d^2+3 b d^2 x\right )}{\sqrt {-1+d^2 x^2}} \, dx}{3 d^2 \sqrt {-1+d x} \sqrt {1+d x}}\\ &=-\frac {c x^2 \left (1-d^2 x^2\right )}{3 d^2 \sqrt {-1+d x} \sqrt {1+d x}}-\frac {\left (2 \left (2 c+3 a d^2\right )+3 b d^2 x\right ) \left (1-d^2 x^2\right )}{6 d^4 \sqrt {-1+d x} \sqrt {1+d x}}+\frac {\left (b \sqrt {-1+d^2 x^2}\right ) \int \frac {1}{\sqrt {-1+d^2 x^2}} \, dx}{2 d^2 \sqrt {-1+d x} \sqrt {1+d x}}\\ &=-\frac {c x^2 \left (1-d^2 x^2\right )}{3 d^2 \sqrt {-1+d x} \sqrt {1+d x}}-\frac {\left (2 \left (2 c+3 a d^2\right )+3 b d^2 x\right ) \left (1-d^2 x^2\right )}{6 d^4 \sqrt {-1+d x} \sqrt {1+d x}}+\frac {\left (b \sqrt {-1+d^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{1-d^2 x^2} \, dx,x,\frac {x}{\sqrt {-1+d^2 x^2}}\right )}{2 d^2 \sqrt {-1+d x} \sqrt {1+d x}}\\ &=-\frac {c x^2 \left (1-d^2 x^2\right )}{3 d^2 \sqrt {-1+d x} \sqrt {1+d x}}-\frac {\left (2 \left (2 c+3 a d^2\right )+3 b d^2 x\right ) \left (1-d^2 x^2\right )}{6 d^4 \sqrt {-1+d x} \sqrt {1+d x}}+\frac {b \sqrt {-1+d^2 x^2} \tanh ^{-1}\left (\frac {d x}{\sqrt {-1+d^2 x^2}}\right )}{2 d^3 \sqrt {-1+d x} \sqrt {1+d x}}\\ \end {align*}

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Mathematica [A]  time = 0.35, size = 149, normalized size = 1.71 \begin {gather*} \frac {\sqrt {-(d x-1)^2} \sqrt {d x+1} \left (3 d^2 (2 a+b x)+2 c \left (d^2 x^2+2\right )\right )+6 \sqrt {d x-1} \sin ^{-1}\left (\frac {\sqrt {1-d x}}{\sqrt {2}}\right ) (d (2 a d-b)+2 c)-12 \sqrt {1-d x} \tanh ^{-1}\left (\sqrt {\frac {d x-1}{d x+1}}\right ) (d (a d-b)+c)}{6 d^4 \sqrt {1-d x}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[(x*(a + b*x + c*x^2))/(Sqrt[-1 + d*x]*Sqrt[1 + d*x]),x]

[Out]

(Sqrt[-(-1 + d*x)^2]*Sqrt[1 + d*x]*(3*d^2*(2*a + b*x) + 2*c*(2 + d^2*x^2)) + 6*(2*c + d*(-b + 2*a*d))*Sqrt[-1
+ d*x]*ArcSin[Sqrt[1 - d*x]/Sqrt[2]] - 12*(c + d*(-b + a*d))*Sqrt[1 - d*x]*ArcTanh[Sqrt[(-1 + d*x)/(1 + d*x)]]
)/(6*d^4*Sqrt[1 - d*x])

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IntegrateAlgebraic [B]  time = 0.17, size = 230, normalized size = 2.64 \begin {gather*} \frac {-\frac {6 a d^2 (d x-1)^{5/2}}{(d x+1)^{5/2}}+\frac {12 a d^2 (d x-1)^{3/2}}{(d x+1)^{3/2}}-\frac {6 a d^2 \sqrt {d x-1}}{\sqrt {d x+1}}+\frac {3 b d (d x-1)^{5/2}}{(d x+1)^{5/2}}-\frac {3 b d \sqrt {d x-1}}{\sqrt {d x+1}}-\frac {6 c (d x-1)^{5/2}}{(d x+1)^{5/2}}+\frac {4 c (d x-1)^{3/2}}{(d x+1)^{3/2}}-\frac {6 c \sqrt {d x-1}}{\sqrt {d x+1}}}{3 d^4 \left (\frac {d x-1}{d x+1}-1\right )^3}+\frac {b \tanh ^{-1}\left (\frac {\sqrt {d x-1}}{\sqrt {d x+1}}\right )}{d^3} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(x*(a + b*x + c*x^2))/(Sqrt[-1 + d*x]*Sqrt[1 + d*x]),x]

[Out]

((-6*c*(-1 + d*x)^(5/2))/(1 + d*x)^(5/2) + (3*b*d*(-1 + d*x)^(5/2))/(1 + d*x)^(5/2) - (6*a*d^2*(-1 + d*x)^(5/2
))/(1 + d*x)^(5/2) + (4*c*(-1 + d*x)^(3/2))/(1 + d*x)^(3/2) + (12*a*d^2*(-1 + d*x)^(3/2))/(1 + d*x)^(3/2) - (6
*c*Sqrt[-1 + d*x])/Sqrt[1 + d*x] - (3*b*d*Sqrt[-1 + d*x])/Sqrt[1 + d*x] - (6*a*d^2*Sqrt[-1 + d*x])/Sqrt[1 + d*
x])/(3*d^4*(-1 + (-1 + d*x)/(1 + d*x))^3) + (b*ArcTanh[Sqrt[-1 + d*x]/Sqrt[1 + d*x]])/d^3

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fricas [A]  time = 0.91, size = 73, normalized size = 0.84 \begin {gather*} -\frac {3 \, b d \log \left (-d x + \sqrt {d x + 1} \sqrt {d x - 1}\right ) - {\left (2 \, c d^{2} x^{2} + 3 \, b d^{2} x + 6 \, a d^{2} + 4 \, c\right )} \sqrt {d x + 1} \sqrt {d x - 1}}{6 \, d^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(3*b*d*log(-d*x + sqrt(d*x + 1)*sqrt(d*x - 1)) - (2*c*d^2*x^2 + 3*b*d^2*x + 6*a*d^2 + 4*c)*sqrt(d*x + 1)*
sqrt(d*x - 1))/d^4

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giac [A]  time = 1.30, size = 105, normalized size = 1.21 \begin {gather*} \frac {\sqrt {d x + 1} \sqrt {d x - 1} {\left ({\left (d x + 1\right )} {\left (\frac {2 \, {\left (d x + 1\right )} c}{d^{3}} + \frac {3 \, b d^{10} - 4 \, c d^{9}}{d^{12}}\right )} + \frac {3 \, {\left (2 \, a d^{11} - b d^{10} + 2 \, c d^{9}\right )}}{d^{12}}\right )} - \frac {6 \, b \log \left (\sqrt {d x + 1} - \sqrt {d x - 1}\right )}{d^{2}}}{6 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(sqrt(d*x + 1)*sqrt(d*x - 1)*((d*x + 1)*(2*(d*x + 1)*c/d^3 + (3*b*d^10 - 4*c*d^9)/d^12) + 3*(2*a*d^11 - b*
d^10 + 2*c*d^9)/d^12) - 6*b*log(sqrt(d*x + 1) - sqrt(d*x - 1))/d^2)/d

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maple [C]  time = 0.03, size = 137, normalized size = 1.57 \begin {gather*} \frac {\sqrt {d x -1}\, \sqrt {d x +1}\, \left (2 \sqrt {d^{2} x^{2}-1}\, c \,d^{2} x^{2} \mathrm {csgn}\relax (d )+3 \sqrt {d^{2} x^{2}-1}\, b \,d^{2} x \,\mathrm {csgn}\relax (d )+6 \sqrt {d^{2} x^{2}-1}\, a \,d^{2} \mathrm {csgn}\relax (d )+3 b d \ln \left (\left (d x +\sqrt {d^{2} x^{2}-1}\, \mathrm {csgn}\relax (d )\right ) \mathrm {csgn}\relax (d )\right )+4 \sqrt {d^{2} x^{2}-1}\, c \,\mathrm {csgn}\relax (d )\right ) \mathrm {csgn}\relax (d )}{6 \sqrt {d^{2} x^{2}-1}\, d^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(d*x-1)^(1/2)*(d*x+1)^(1/2)*(2*csgn(d)*x^2*c*d^2*(d^2*x^2-1)^(1/2)+3*csgn(d)*(d^2*x^2-1)^(1/2)*x*b*d^2+6*c
sgn(d)*(d^2*x^2-1)^(1/2)*a*d^2+4*csgn(d)*(d^2*x^2-1)^(1/2)*c+3*ln((csgn(d)*(d^2*x^2-1)^(1/2)+d*x)*csgn(d))*b*d
)*csgn(d)/d^4/(d^2*x^2-1)^(1/2)

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maxima [A]  time = 0.43, size = 100, normalized size = 1.15 \begin {gather*} \frac {\sqrt {d^{2} x^{2} - 1} c x^{2}}{3 \, d^{2}} + \frac {\sqrt {d^{2} x^{2} - 1} b x}{2 \, d^{2}} + \frac {\sqrt {d^{2} x^{2} - 1} a}{d^{2}} + \frac {b \log \left (2 \, d^{2} x + 2 \, \sqrt {d^{2} x^{2} - 1} d\right )}{2 \, d^{3}} + \frac {2 \, \sqrt {d^{2} x^{2} - 1} c}{3 \, d^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 12.35, size = 318, normalized size = 3.66 \begin {gather*} \frac {\sqrt {d\,x-1}\,\left (\frac {2\,c}{3\,d^4}+\frac {c\,x^3}{3\,d}+\frac {c\,x^2}{3\,d^2}+\frac {2\,c\,x}{3\,d^3}\right )}{\sqrt {d\,x+1}}+\frac {2\,b\,\mathrm {atanh}\left (\frac {\sqrt {d\,x-1}-\mathrm {i}}{\sqrt {d\,x+1}-1}\right )}{d^3}-\frac {\frac {14\,b\,{\left (\sqrt {d\,x-1}-\mathrm {i}\right )}^3}{{\left (\sqrt {d\,x+1}-1\right )}^3}+\frac {14\,b\,{\left (\sqrt {d\,x-1}-\mathrm {i}\right )}^5}{{\left (\sqrt {d\,x+1}-1\right )}^5}+\frac {2\,b\,{\left (\sqrt {d\,x-1}-\mathrm {i}\right )}^7}{{\left (\sqrt {d\,x+1}-1\right )}^7}+\frac {2\,b\,\left (\sqrt {d\,x-1}-\mathrm {i}\right )}{\sqrt {d\,x+1}-1}}{d^3-\frac {4\,d^3\,{\left (\sqrt {d\,x-1}-\mathrm {i}\right )}^2}{{\left (\sqrt {d\,x+1}-1\right )}^2}+\frac {6\,d^3\,{\left (\sqrt {d\,x-1}-\mathrm {i}\right )}^4}{{\left (\sqrt {d\,x+1}-1\right )}^4}-\frac {4\,d^3\,{\left (\sqrt {d\,x-1}-\mathrm {i}\right )}^6}{{\left (\sqrt {d\,x+1}-1\right )}^6}+\frac {d^3\,{\left (\sqrt {d\,x-1}-\mathrm {i}\right )}^8}{{\left (\sqrt {d\,x+1}-1\right )}^8}}+\frac {a\,\sqrt {d\,x-1}\,\sqrt {d\,x+1}}{d^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*b*atanh(((d*x - 1)^(1/2) - 1i)/((d*x + 1)^(1/2) - 1)))/d^3 - ((14*b*((d*x - 1)^(1/2) - 1i)^3)/((d*x + 1)^(1
/2) - 1)^3 + (14*b*((d*x - 1)^(1/2) - 1i)^5)/((d*x + 1)^(1/2) - 1)^5 + (2*b*((d*x - 1)^(1/2) - 1i)^7)/((d*x +
1)^(1/2) - 1)^7 + (2*b*((d*x - 1)^(1/2) - 1i))/((d*x + 1)^(1/2) - 1))/(d^3 - (4*d^3*((d*x - 1)^(1/2) - 1i)^2)/
((d*x + 1)^(1/2) - 1)^2 + (6*d^3*((d*x - 1)^(1/2) - 1i)^4)/((d*x + 1)^(1/2) - 1)^4 - (4*d^3*((d*x - 1)^(1/2) -
1i)^6)/((d*x + 1)^(1/2) - 1)^6 + (d^3*((d*x - 1)^(1/2) - 1i)^8)/((d*x + 1)^(1/2) - 1)^8) + ((d*x - 1)^(1/2)*(
(2*c)/(3*d^4) + (c*x^3)/(3*d) + (c*x^2)/(3*d^2) + (2*c*x)/(3*d^3)))/(d*x + 1)^(1/2) + (a*(d*x - 1)^(1/2)*(d*x
+ 1)^(1/2))/d^2

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sympy [C]  time = 78.83, size = 308, normalized size = 3.54 \begin {gather*} \frac {a {G_{6, 6}^{6, 2}\left (\begin {matrix} - \frac {1}{4}, \frac {1}{4} & 0, 0, \frac {1}{2}, 1 \\- \frac {1}{2}, - \frac {1}{4}, 0, \frac {1}{4}, \frac {1}{2}, 0 & \end {matrix} \middle | {\frac {1}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} d^{2}} + \frac {i a {G_{6, 6}^{2, 6}\left (\begin {matrix} -1, - \frac {3}{4}, - \frac {1}{2}, - \frac {1}{4}, 0, 1 & \\- \frac {3}{4}, - \frac {1}{4} & -1, - \frac {1}{2}, - \frac {1}{2}, 0 \end {matrix} \middle | {\frac {e^{2 i \pi }}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} d^{2}} + \frac {b {G_{6, 6}^{6, 2}\left (\begin {matrix} - \frac {3}{4}, - \frac {1}{4} & - \frac {1}{2}, - \frac {1}{2}, 0, 1 \\-1, - \frac {3}{4}, - \frac {1}{2}, - \frac {1}{4}, 0, 0 & \end {matrix} \middle | {\frac {1}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} d^{3}} - \frac {i b {G_{6, 6}^{2, 6}\left (\begin {matrix} - \frac {3}{2}, - \frac {5}{4}, -1, - \frac {3}{4}, - \frac {1}{2}, 1 & \\- \frac {5}{4}, - \frac {3}{4} & - \frac {3}{2}, -1, -1, 0 \end {matrix} \middle | {\frac {e^{2 i \pi }}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} d^{3}} + \frac {c {G_{6, 6}^{6, 2}\left (\begin {matrix} - \frac {5}{4}, - \frac {3}{4} & -1, -1, - \frac {1}{2}, 1 \\- \frac {3}{2}, - \frac {5}{4}, -1, - \frac {3}{4}, - \frac {1}{2}, 0 & \end {matrix} \middle | {\frac {1}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} d^{4}} + \frac {i c {G_{6, 6}^{2, 6}\left (\begin {matrix} -2, - \frac {7}{4}, - \frac {3}{2}, - \frac {5}{4}, -1, 1 & \\- \frac {7}{4}, - \frac {5}{4} & -2, - \frac {3}{2}, - \frac {3}{2}, 0 \end {matrix} \middle | {\frac {e^{2 i \pi }}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} d^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*meijerg(((-1/4, 1/4), (0, 0, 1/2, 1)), ((-1/2, -1/4, 0, 1/4, 1/2, 0), ()), 1/(d**2*x**2))/(4*pi**(3/2)*d**2)
+ I*a*meijerg(((-1, -3/4, -1/2, -1/4, 0, 1), ()), ((-3/4, -1/4), (-1, -1/2, -1/2, 0)), exp_polar(2*I*pi)/(d**
2*x**2))/(4*pi**(3/2)*d**2) + b*meijerg(((-3/4, -1/4), (-1/2, -1/2, 0, 1)), ((-1, -3/4, -1/2, -1/4, 0, 0), ())
, 1/(d**2*x**2))/(4*pi**(3/2)*d**3) - I*b*meijerg(((-3/2, -5/4, -1, -3/4, -1/2, 1), ()), ((-5/4, -3/4), (-3/2,
-1, -1, 0)), exp_polar(2*I*pi)/(d**2*x**2))/(4*pi**(3/2)*d**3) + c*meijerg(((-5/4, -3/4), (-1, -1, -1/2, 1)),
((-3/2, -5/4, -1, -3/4, -1/2, 0), ()), 1/(d**2*x**2))/(4*pi**(3/2)*d**4) + I*c*meijerg(((-2, -7/4, -3/2, -5/4
, -1, 1), ()), ((-7/4, -5/4), (-2, -3/2, -3/2, 0)), exp_polar(2*I*pi)/(d**2*x**2))/(4*pi**(3/2)*d**4)

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