3.1.10 \(\int \frac {(e+f x) (A+B x+C x^2)}{\sqrt {1-d x} \sqrt {1+d x}} \, dx\)

Optimal. Leaf size=130 \[ -\frac {\sqrt {1-d^2 x^2} \left (2 \left (3 d^2 f (A f+B e)-C \left (d^2 e^2-2 f^2\right )\right )-d^2 f x (C e-3 B f)\right )}{6 d^4 f}+\frac {\sin ^{-1}(d x) \left (2 A d^2 e+B f+C e\right )}{2 d^3}-\frac {C \sqrt {1-d^2 x^2} (e+f x)^2}{3 d^2 f} \]

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Rubi [A]  time = 0.23, antiderivative size = 133, normalized size of antiderivative = 1.02, number of steps used = 4, number of rules used = 4, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.114, Rules used = {1609, 1654, 780, 216} \begin {gather*} -\frac {\sqrt {1-d^2 x^2} \left (2 \left (3 d^2 f (A f+B e)-\frac {1}{2} C \left (2 d^2 e^2-4 f^2\right )\right )-d^2 f x (C e-3 B f)\right )}{6 d^4 f}+\frac {\sin ^{-1}(d x) \left (2 A d^2 e+B f+C e\right )}{2 d^3}-\frac {C \sqrt {1-d^2 x^2} (e+f x)^2}{3 d^2 f} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((e + f*x)*(A + B*x + C*x^2))/(Sqrt[1 - d*x]*Sqrt[1 + d*x]),x]

[Out]

-(C*(e + f*x)^2*Sqrt[1 - d^2*x^2])/(3*d^2*f) - ((2*(3*d^2*f*(B*e + A*f) - (C*(2*d^2*e^2 - 4*f^2))/2) - d^2*f*(
C*e - 3*B*f)*x)*Sqrt[1 - d^2*x^2])/(6*d^4*f) + ((C*e + 2*A*d^2*e + B*f)*ArcSin[d*x])/(2*d^3)

Rule 216

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

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 1609

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

Rule 1654

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

Rubi steps

\begin {align*} \int \frac {(e+f x) \left (A+B x+C x^2\right )}{\sqrt {1-d x} \sqrt {1+d x}} \, dx &=\int \frac {(e+f x) \left (A+B x+C x^2\right )}{\sqrt {1-d^2 x^2}} \, dx\\ &=-\frac {C (e+f x)^2 \sqrt {1-d^2 x^2}}{3 d^2 f}-\frac {\int \frac {(e+f x) \left (-\left (\left (2 C+3 A d^2\right ) f^2\right )+d^2 f (C e-3 B f) x\right )}{\sqrt {1-d^2 x^2}} \, dx}{3 d^2 f^2}\\ &=-\frac {C (e+f x)^2 \sqrt {1-d^2 x^2}}{3 d^2 f}-\frac {\left (2 \left (3 d^2 f (B e+A f)-\frac {1}{2} C \left (2 d^2 e^2-4 f^2\right )\right )-d^2 f (C e-3 B f) x\right ) \sqrt {1-d^2 x^2}}{6 d^4 f}+\frac {\left (C e+2 A d^2 e+B f\right ) \int \frac {1}{\sqrt {1-d^2 x^2}} \, dx}{2 d^2}\\ &=-\frac {C (e+f x)^2 \sqrt {1-d^2 x^2}}{3 d^2 f}-\frac {\left (2 \left (3 d^2 f (B e+A f)-\frac {1}{2} C \left (2 d^2 e^2-4 f^2\right )\right )-d^2 f (C e-3 B f) x\right ) \sqrt {1-d^2 x^2}}{6 d^4 f}+\frac {\left (C e+2 A d^2 e+B f\right ) \sin ^{-1}(d x)}{2 d^3}\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 88, normalized size = 0.68 \begin {gather*} \frac {3 d \sin ^{-1}(d x) \left (2 A d^2 e+B f+C e\right )-\sqrt {1-d^2 x^2} \left (6 A d^2 f+3 B d^2 (2 e+f x)+C \left (3 d^2 e x+2 d^2 f x^2+4 f\right )\right )}{6 d^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((e + f*x)*(A + B*x + C*x^2))/(Sqrt[1 - d*x]*Sqrt[1 + d*x]),x]

[Out]

(-(Sqrt[1 - d^2*x^2]*(6*A*d^2*f + 3*B*d^2*(2*e + f*x) + C*(4*f + 3*d^2*e*x + 2*d^2*f*x^2))) + 3*d*(C*e + 2*A*d
^2*e + B*f)*ArcSin[d*x])/(6*d^4)

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IntegrateAlgebraic [B]  time = 0.26, size = 275, normalized size = 2.12 \begin {gather*} \frac {\tan ^{-1}\left (\frac {\sqrt {1-d x}}{\sqrt {d x+1}}\right ) \left (-2 A d^2 e-B f-C e\right )}{d^3}-\frac {\sqrt {1-d x} \left (\frac {12 A d^2 f (1-d x)}{d x+1}+\frac {6 A d^2 f (1-d x)^2}{(d x+1)^2}+6 A d^2 f+\frac {12 B d^2 e (1-d x)}{d x+1}+\frac {6 B d^2 e (1-d x)^2}{(d x+1)^2}+6 B d^2 e-\frac {3 B d f (1-d x)^2}{(d x+1)^2}+3 B d f-\frac {3 C d e (1-d x)^2}{(d x+1)^2}+3 C d e+\frac {4 C f (1-d x)}{d x+1}+\frac {6 C f (1-d x)^2}{(d x+1)^2}+6 C f\right )}{3 d^4 \sqrt {d x+1} \left (\frac {1-d x}{d x+1}+1\right )^3} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((e + f*x)*(A + B*x + C*x^2))/(Sqrt[1 - d*x]*Sqrt[1 + d*x]),x]

[Out]

-1/3*(Sqrt[1 - d*x]*(3*C*d*e + 6*B*d^2*e + 6*C*f + 3*B*d*f + 6*A*d^2*f - (3*C*d*e*(1 - d*x)^2)/(1 + d*x)^2 + (
6*B*d^2*e*(1 - d*x)^2)/(1 + d*x)^2 + (6*C*f*(1 - d*x)^2)/(1 + d*x)^2 - (3*B*d*f*(1 - d*x)^2)/(1 + d*x)^2 + (6*
A*d^2*f*(1 - d*x)^2)/(1 + d*x)^2 + (12*B*d^2*e*(1 - d*x))/(1 + d*x) + (4*C*f*(1 - d*x))/(1 + d*x) + (12*A*d^2*
f*(1 - d*x))/(1 + d*x)))/(d^4*Sqrt[1 + d*x]*(1 + (1 - d*x)/(1 + d*x))^3) + ((-(C*e) - 2*A*d^2*e - B*f)*ArcTan[
Sqrt[1 - d*x]/Sqrt[1 + d*x]])/d^3

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fricas [A]  time = 0.96, size = 114, normalized size = 0.88 \begin {gather*} -\frac {{\left (2 \, C d^{2} f x^{2} + 6 \, B d^{2} e + 2 \, {\left (3 \, A d^{2} + 2 \, C\right )} f + 3 \, {\left (C d^{2} e + B d^{2} f\right )} x\right )} \sqrt {d x + 1} \sqrt {-d x + 1} + 6 \, {\left (B d f + {\left (2 \, A d^{3} + C d\right )} e\right )} \arctan \left (\frac {\sqrt {d x + 1} \sqrt {-d x + 1} - 1}{d x}\right )}{6 \, d^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)*(C*x^2+B*x+A)/(-d*x+1)^(1/2)/(d*x+1)^(1/2),x, algorithm="fricas")

[Out]

-1/6*((2*C*d^2*f*x^2 + 6*B*d^2*e + 2*(3*A*d^2 + 2*C)*f + 3*(C*d^2*e + B*d^2*f)*x)*sqrt(d*x + 1)*sqrt(-d*x + 1)
+ 6*(B*d*f + (2*A*d^3 + C*d)*e)*arctan((sqrt(d*x + 1)*sqrt(-d*x + 1) - 1)/(d*x)))/d^4

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giac [A]  time = 1.31, size = 146, normalized size = 1.12 \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 f}{d^{3}} + \frac {3 \, B d^{10} f + 3 \, C d^{10} e - 4 \, C d^{9} f}{d^{12}}\right )} + \frac {3 \, {\left (2 \, A d^{11} f + 2 \, B d^{11} e - B d^{10} f - C d^{10} e + 2 \, C d^{9} f\right )}}{d^{12}}\right )} - \frac {6 \, {\left (2 \, A d^{2} e + B f + C e\right )} \arcsin \left (\frac {1}{2} \, \sqrt {2} \sqrt {d x + 1}\right )}{d^{2}}}{6 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)*(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*f/d^3 + (3*B*d^10*f + 3*C*d^10*e - 4*C*d^9*f)/d^1
2) + 3*(2*A*d^11*f + 2*B*d^11*e - B*d^10*f - C*d^10*e + 2*C*d^9*f)/d^12) - 6*(2*A*d^2*e + B*f + C*e)*arcsin(1/
2*sqrt(2)*sqrt(d*x + 1))/d^2)/d

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maple [C]  time = 0.02, size = 235, normalized size = 1.81 \begin {gather*} -\frac {\sqrt {-d x +1}\, \sqrt {d x +1}\, \left (2 \sqrt {-d^{2} x^{2}+1}\, C \,d^{2} f \,x^{2} \mathrm {csgn}\relax (d )-6 A \,d^{3} e \arctan \left (\frac {d x \,\mathrm {csgn}\relax (d )}{\sqrt {-d^{2} x^{2}+1}}\right )+3 \sqrt {-d^{2} x^{2}+1}\, B \,d^{2} f x \,\mathrm {csgn}\relax (d )+3 \sqrt {-d^{2} x^{2}+1}\, C \,d^{2} e x \,\mathrm {csgn}\relax (d )+6 \sqrt {-d^{2} x^{2}+1}\, A \,d^{2} f \,\mathrm {csgn}\relax (d )+6 \sqrt {-d^{2} x^{2}+1}\, B \,d^{2} e \,\mathrm {csgn}\relax (d )-3 B d f \arctan \left (\frac {d x \,\mathrm {csgn}\relax (d )}{\sqrt {-d^{2} x^{2}+1}}\right )-3 C d e \arctan \left (\frac {d x \,\mathrm {csgn}\relax (d )}{\sqrt {-d^{2} x^{2}+1}}\right )+4 \sqrt {-d^{2} x^{2}+1}\, C f \,\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((f*x+e)*(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*(-d^2*x^2+1)^(1/2)*C*d^2*f*x^2*csgn(d)+3*(-d^2*x^2+1)^(1/2)*B*d^2*f*x*csg
n(d)+3*(-d^2*x^2+1)^(1/2)*C*d^2*e*x*csgn(d)+6*(-d^2*x^2+1)^(1/2)*A*d^2*f*csgn(d)-6*A*d^3*e*arctan(1/(-d^2*x^2+
1)^(1/2)*d*x*csgn(d))+6*(-d^2*x^2+1)^(1/2)*B*d^2*e*csgn(d)-3*B*d*f*arctan(1/(-d^2*x^2+1)^(1/2)*d*x*csgn(d))+4*
(-d^2*x^2+1)^(1/2)*C*f*csgn(d)-3*C*d*e*arctan(1/(-d^2*x^2+1)^(1/2)*d*x*csgn(d)))*csgn(d)/d^4/(-d^2*x^2+1)^(1/2
)

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maxima [A]  time = 1.31, size = 131, normalized size = 1.01 \begin {gather*} -\frac {\sqrt {-d^{2} x^{2} + 1} C f x^{2}}{3 \, d^{2}} + \frac {A e \arcsin \left (d x\right )}{d} - \frac {\sqrt {-d^{2} x^{2} + 1} B e}{d^{2}} - \frac {\sqrt {-d^{2} x^{2} + 1} A f}{d^{2}} - \frac {\sqrt {-d^{2} x^{2} + 1} {\left (C e + B f\right )} x}{2 \, d^{2}} + \frac {{\left (C e + B f\right )} \arcsin \left (d x\right )}{2 \, d^{3}} - \frac {2 \, \sqrt {-d^{2} x^{2} + 1} C f}{3 \, d^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)*(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*f*x^2/d^2 + A*e*arcsin(d*x)/d - sqrt(-d^2*x^2 + 1)*B*e/d^2 - sqrt(-d^2*x^2 + 1)*A*f/
d^2 - 1/2*sqrt(-d^2*x^2 + 1)*(C*e + B*f)*x/d^2 + 1/2*(C*e + B*f)*arcsin(d*x)/d^3 - 2/3*sqrt(-d^2*x^2 + 1)*C*f/
d^4

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mupad [B]  time = 12.86, size = 492, normalized size = 3.78 \begin {gather*} \frac {\frac {2\,B\,f\,\left (\sqrt {1-d\,x}-1\right )}{\sqrt {d\,x+1}-1}-\frac {14\,B\,f\,{\left (\sqrt {1-d\,x}-1\right )}^3}{{\left (\sqrt {d\,x+1}-1\right )}^3}+\frac {14\,B\,f\,{\left (\sqrt {1-d\,x}-1\right )}^5}{{\left (\sqrt {d\,x+1}-1\right )}^5}-\frac {2\,B\,f\,{\left (\sqrt {1-d\,x}-1\right )}^7}{{\left (\sqrt {d\,x+1}-1\right )}^7}}{d^3\,{\left (\frac {{\left (\sqrt {1-d\,x}-1\right )}^2}{{\left (\sqrt {d\,x+1}-1\right )}^2}+1\right )}^4}-\frac {\sqrt {1-d\,x}\,\left (\frac {2\,C\,f}{3\,d^4}+\frac {2\,C\,f\,x}{3\,d^3}+\frac {C\,f\,x^3}{3\,d}+\frac {C\,f\,x^2}{3\,d^2}\right )}{\sqrt {d\,x+1}}+\frac {\frac {2\,C\,e\,\left (\sqrt {1-d\,x}-1\right )}{\sqrt {d\,x+1}-1}-\frac {14\,C\,e\,{\left (\sqrt {1-d\,x}-1\right )}^3}{{\left (\sqrt {d\,x+1}-1\right )}^3}+\frac {14\,C\,e\,{\left (\sqrt {1-d\,x}-1\right )}^5}{{\left (\sqrt {d\,x+1}-1\right )}^5}-\frac {2\,C\,e\,{\left (\sqrt {1-d\,x}-1\right )}^7}{{\left (\sqrt {d\,x+1}-1\right )}^7}}{d^3\,{\left (\frac {{\left (\sqrt {1-d\,x}-1\right )}^2}{{\left (\sqrt {d\,x+1}-1\right )}^2}+1\right )}^4}-\frac {\left (\frac {A\,f}{d^2}+\frac {A\,f\,x}{d}\right )\,\sqrt {1-d\,x}}{\sqrt {d\,x+1}}-\frac {\left (\frac {B\,e}{d^2}+\frac {B\,e\,x}{d}\right )\,\sqrt {1-d\,x}}{\sqrt {d\,x+1}}-\frac {4\,A\,e\,\mathrm {atan}\left (\frac {d\,\left (\sqrt {1-d\,x}-1\right )}{\left (\sqrt {d\,x+1}-1\right )\,\sqrt {d^2}}\right )}{\sqrt {d^2}}-\frac {2\,B\,f\,\mathrm {atan}\left (\frac {\sqrt {1-d\,x}-1}{\sqrt {d\,x+1}-1}\right )}{d^3}-\frac {2\,C\,e\,\mathrm {atan}\left (\frac {\sqrt {1-d\,x}-1}{\sqrt {d\,x+1}-1}\right )}{d^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((e + f*x)*(A + B*x + C*x^2))/((1 - d*x)^(1/2)*(d*x + 1)^(1/2)),x)

[Out]

((2*B*f*((1 - d*x)^(1/2) - 1))/((d*x + 1)^(1/2) - 1) - (14*B*f*((1 - d*x)^(1/2) - 1)^3)/((d*x + 1)^(1/2) - 1)^
3 + (14*B*f*((1 - d*x)^(1/2) - 1)^5)/((d*x + 1)^(1/2) - 1)^5 - (2*B*f*((1 - d*x)^(1/2) - 1)^7)/((d*x + 1)^(1/2
) - 1)^7)/(d^3*(((1 - d*x)^(1/2) - 1)^2/((d*x + 1)^(1/2) - 1)^2 + 1)^4) - ((1 - d*x)^(1/2)*((2*C*f)/(3*d^4) +
(2*C*f*x)/(3*d^3) + (C*f*x^3)/(3*d) + (C*f*x^2)/(3*d^2)))/(d*x + 1)^(1/2) + ((2*C*e*((1 - d*x)^(1/2) - 1))/((d
*x + 1)^(1/2) - 1) - (14*C*e*((1 - d*x)^(1/2) - 1)^3)/((d*x + 1)^(1/2) - 1)^3 + (14*C*e*((1 - d*x)^(1/2) - 1)^
5)/((d*x + 1)^(1/2) - 1)^5 - (2*C*e*((1 - d*x)^(1/2) - 1)^7)/((d*x + 1)^(1/2) - 1)^7)/(d^3*(((1 - d*x)^(1/2) -
1)^2/((d*x + 1)^(1/2) - 1)^2 + 1)^4) - (((A*f)/d^2 + (A*f*x)/d)*(1 - d*x)^(1/2))/(d*x + 1)^(1/2) - (((B*e)/d^
2 + (B*e*x)/d)*(1 - d*x)^(1/2))/(d*x + 1)^(1/2) - (4*A*e*atan((d*((1 - d*x)^(1/2) - 1))/(((d*x + 1)^(1/2) - 1)
*(d^2)^(1/2))))/(d^2)^(1/2) - (2*B*f*atan(((1 - d*x)^(1/2) - 1)/((d*x + 1)^(1/2) - 1)))/d^3 - (2*C*e*atan(((1
- d*x)^(1/2) - 1)/((d*x + 1)^(1/2) - 1)))/d^3

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sympy [C]  time = 158.08, size = 617, normalized size = 4.75

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)*(C*x**2+B*x+A)/(-d*x+1)**(1/2)/(d*x+1)**(1/2),x)

[Out]

-I*A*e*meijerg(((1/4, 3/4), (1/2, 1/2, 1, 1)), ((0, 1/4, 1/2, 3/4, 1, 0), ()), 1/(d**2*x**2))/(4*pi**(3/2)*d)
+ A*e*meijerg(((-1/2, -1/4, 0, 1/4, 1/2, 1), ()), ((-1/4, 1/4), (-1/2, 0, 0, 0)), exp_polar(-2*I*pi)/(d**2*x**
2))/(4*pi**(3/2)*d) - I*A*f*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) - A*f*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) - I*B*e*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) - B*e*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) - I*B*f*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) + B*f*me
ijerg(((-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) - I*C*e*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) + C*e*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) - I*C*f*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) - C*f*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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