3.472 \(\int \frac {A+B x}{(e x)^{5/2} (a+c x^2)^{3/2}} \, dx\)

Optimal. Leaf size=357 \[ \frac {\sqrt [4]{c} \sqrt {x} \left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} \left (9 \sqrt {a} B-5 A \sqrt {c}\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{6 a^{9/4} e^2 \sqrt {e x} \sqrt {a+c x^2}}-\frac {3 B \sqrt [4]{c} \sqrt {x} \left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{a^{7/4} e^2 \sqrt {e x} \sqrt {a+c x^2}}-\frac {5 A \sqrt {a+c x^2}}{3 a^2 e (e x)^{3/2}}-\frac {3 B \sqrt {a+c x^2}}{a^2 e^2 \sqrt {e x}}+\frac {3 B \sqrt {c} x \sqrt {a+c x^2}}{a^2 e^2 \sqrt {e x} \left (\sqrt {a}+\sqrt {c} x\right )}+\frac {A+B x}{a e (e x)^{3/2} \sqrt {a+c x^2}} \]

[Out]

(B*x+A)/a/e/(e*x)^(3/2)/(c*x^2+a)^(1/2)-5/3*A*(c*x^2+a)^(1/2)/a^2/e/(e*x)^(3/2)-3*B*(c*x^2+a)^(1/2)/a^2/e^2/(e
*x)^(1/2)+3*B*x*c^(1/2)*(c*x^2+a)^(1/2)/a^2/e^2/(a^(1/2)+x*c^(1/2))/(e*x)^(1/2)-3*B*c^(1/4)*(cos(2*arctan(c^(1
/4)*x^(1/2)/a^(1/4)))^2)^(1/2)/cos(2*arctan(c^(1/4)*x^(1/2)/a^(1/4)))*EllipticE(sin(2*arctan(c^(1/4)*x^(1/2)/a
^(1/4))),1/2*2^(1/2))*(a^(1/2)+x*c^(1/2))*x^(1/2)*((c*x^2+a)/(a^(1/2)+x*c^(1/2))^2)^(1/2)/a^(7/4)/e^2/(e*x)^(1
/2)/(c*x^2+a)^(1/2)+1/6*c^(1/4)*(cos(2*arctan(c^(1/4)*x^(1/2)/a^(1/4)))^2)^(1/2)/cos(2*arctan(c^(1/4)*x^(1/2)/
a^(1/4)))*EllipticF(sin(2*arctan(c^(1/4)*x^(1/2)/a^(1/4))),1/2*2^(1/2))*(9*B*a^(1/2)-5*A*c^(1/2))*(a^(1/2)+x*c
^(1/2))*x^(1/2)*((c*x^2+a)/(a^(1/2)+x*c^(1/2))^2)^(1/2)/a^(9/4)/e^2/(e*x)^(1/2)/(c*x^2+a)^(1/2)

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Rubi [A]  time = 0.42, antiderivative size = 357, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {823, 835, 842, 840, 1198, 220, 1196} \[ \frac {\sqrt [4]{c} \sqrt {x} \left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} \left (9 \sqrt {a} B-5 A \sqrt {c}\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{6 a^{9/4} e^2 \sqrt {e x} \sqrt {a+c x^2}}-\frac {5 A \sqrt {a+c x^2}}{3 a^2 e (e x)^{3/2}}-\frac {3 B \sqrt {a+c x^2}}{a^2 e^2 \sqrt {e x}}+\frac {3 B \sqrt {c} x \sqrt {a+c x^2}}{a^2 e^2 \sqrt {e x} \left (\sqrt {a}+\sqrt {c} x\right )}-\frac {3 B \sqrt [4]{c} \sqrt {x} \left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{a^{7/4} e^2 \sqrt {e x} \sqrt {a+c x^2}}+\frac {A+B x}{a e (e x)^{3/2} \sqrt {a+c x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/((e*x)^(5/2)*(a + c*x^2)^(3/2)),x]

[Out]

(A + B*x)/(a*e*(e*x)^(3/2)*Sqrt[a + c*x^2]) - (5*A*Sqrt[a + c*x^2])/(3*a^2*e*(e*x)^(3/2)) - (3*B*Sqrt[a + c*x^
2])/(a^2*e^2*Sqrt[e*x]) + (3*B*Sqrt[c]*x*Sqrt[a + c*x^2])/(a^2*e^2*Sqrt[e*x]*(Sqrt[a] + Sqrt[c]*x)) - (3*B*c^(
1/4)*Sqrt[x]*(Sqrt[a] + Sqrt[c]*x)*Sqrt[(a + c*x^2)/(Sqrt[a] + Sqrt[c]*x)^2]*EllipticE[2*ArcTan[(c^(1/4)*Sqrt[
x])/a^(1/4)], 1/2])/(a^(7/4)*e^2*Sqrt[e*x]*Sqrt[a + c*x^2]) + ((9*Sqrt[a]*B - 5*A*Sqrt[c])*c^(1/4)*Sqrt[x]*(Sq
rt[a] + Sqrt[c]*x)*Sqrt[(a + c*x^2)/(Sqrt[a] + Sqrt[c]*x)^2]*EllipticF[2*ArcTan[(c^(1/4)*Sqrt[x])/a^(1/4)], 1/
2])/(6*a^(9/4)*e^2*Sqrt[e*x]*Sqrt[a + c*x^2])

Rule 220

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[((1 + q^2*x^2)*Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 823

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(
m + 1)*(f*a*c*e - a*g*c*d + c*(c*d*f + a*e*g)*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), x] + Di
st[1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Simp[f*(c^2*d^2*(2*p + 3) + a*c*e^2*
(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f + a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x]
&& NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 835

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((e*f - d*g)
*(d + e*x)^(m + 1)*(a + c*x^2)^(p + 1))/((m + 1)*(c*d^2 + a*e^2)), x] + Dist[1/((m + 1)*(c*d^2 + a*e^2)), Int[
(d + e*x)^(m + 1)*(a + c*x^2)^p*Simp[(c*d*f + a*e*g)*(m + 1) - c*(e*f - d*g)*(m + 2*p + 3)*x, x], x], x] /; Fr
eeQ[{a, c, d, e, f, g, p}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[m, -1] && (IntegerQ[m] || IntegerQ[p] || Integer
sQ[2*m, 2*p])

Rule 840

Int[((f_) + (g_.)*(x_))/(Sqrt[x_]*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[2, Subst[Int[(f + g*x^2)/Sqrt[
a + c*x^4], x], x, Sqrt[x]], x] /; FreeQ[{a, c, f, g}, x]

Rule 842

Int[((f_) + (g_.)*(x_))/(Sqrt[(e_)*(x_)]*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[Sqrt[x]/Sqrt[e*x], Int[
(f + g*x)/(Sqrt[x]*Sqrt[a + c*x^2]), x], x] /; FreeQ[{a, c, e, f, g}, x]

Rule 1196

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, -Simp[(d*x*Sqrt[a + c
*x^4])/(a*(1 + q^2*x^2)), x] + Simp[(d*(1 + q^2*x^2)*Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]*EllipticE[2*ArcTan[
q*x], 1/2])/(q*Sqrt[a + c*x^4]), x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rule 1198

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 2]}, Dist[(e + d*q)/q, Int
[1/Sqrt[a + c*x^4], x], x] - Dist[e/q, Int[(1 - q*x^2)/Sqrt[a + c*x^4], x], x] /; NeQ[e + d*q, 0]] /; FreeQ[{a
, c, d, e}, x] && PosQ[c/a]

Rubi steps

\begin {align*} \int \frac {A+B x}{(e x)^{5/2} \left (a+c x^2\right )^{3/2}} \, dx &=\frac {A+B x}{a e (e x)^{3/2} \sqrt {a+c x^2}}-\frac {\int \frac {-\frac {5}{2} a A c e^2-\frac {3}{2} a B c e^2 x}{(e x)^{5/2} \sqrt {a+c x^2}} \, dx}{a^2 c e^2}\\ &=\frac {A+B x}{a e (e x)^{3/2} \sqrt {a+c x^2}}-\frac {5 A \sqrt {a+c x^2}}{3 a^2 e (e x)^{3/2}}+\frac {2 \int \frac {\frac {9}{4} a^2 B c e^3-\frac {5}{4} a A c^2 e^3 x}{(e x)^{3/2} \sqrt {a+c x^2}} \, dx}{3 a^3 c e^4}\\ &=\frac {A+B x}{a e (e x)^{3/2} \sqrt {a+c x^2}}-\frac {5 A \sqrt {a+c x^2}}{3 a^2 e (e x)^{3/2}}-\frac {3 B \sqrt {a+c x^2}}{a^2 e^2 \sqrt {e x}}-\frac {4 \int \frac {\frac {5}{8} a^2 A c^2 e^4-\frac {9}{8} a^2 B c^2 e^4 x}{\sqrt {e x} \sqrt {a+c x^2}} \, dx}{3 a^4 c e^6}\\ &=\frac {A+B x}{a e (e x)^{3/2} \sqrt {a+c x^2}}-\frac {5 A \sqrt {a+c x^2}}{3 a^2 e (e x)^{3/2}}-\frac {3 B \sqrt {a+c x^2}}{a^2 e^2 \sqrt {e x}}-\frac {\left (4 \sqrt {x}\right ) \int \frac {\frac {5}{8} a^2 A c^2 e^4-\frac {9}{8} a^2 B c^2 e^4 x}{\sqrt {x} \sqrt {a+c x^2}} \, dx}{3 a^4 c e^6 \sqrt {e x}}\\ &=\frac {A+B x}{a e (e x)^{3/2} \sqrt {a+c x^2}}-\frac {5 A \sqrt {a+c x^2}}{3 a^2 e (e x)^{3/2}}-\frac {3 B \sqrt {a+c x^2}}{a^2 e^2 \sqrt {e x}}-\frac {\left (8 \sqrt {x}\right ) \operatorname {Subst}\left (\int \frac {\frac {5}{8} a^2 A c^2 e^4-\frac {9}{8} a^2 B c^2 e^4 x^2}{\sqrt {a+c x^4}} \, dx,x,\sqrt {x}\right )}{3 a^4 c e^6 \sqrt {e x}}\\ &=\frac {A+B x}{a e (e x)^{3/2} \sqrt {a+c x^2}}-\frac {5 A \sqrt {a+c x^2}}{3 a^2 e (e x)^{3/2}}-\frac {3 B \sqrt {a+c x^2}}{a^2 e^2 \sqrt {e x}}-\frac {\left (3 B \sqrt {c} \sqrt {x}\right ) \operatorname {Subst}\left (\int \frac {1-\frac {\sqrt {c} x^2}{\sqrt {a}}}{\sqrt {a+c x^4}} \, dx,x,\sqrt {x}\right )}{a^{3/2} e^2 \sqrt {e x}}+\frac {\left (\left (9 \sqrt {a} B-5 A \sqrt {c}\right ) \sqrt {c} \sqrt {x}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+c x^4}} \, dx,x,\sqrt {x}\right )}{3 a^2 e^2 \sqrt {e x}}\\ &=\frac {A+B x}{a e (e x)^{3/2} \sqrt {a+c x^2}}-\frac {5 A \sqrt {a+c x^2}}{3 a^2 e (e x)^{3/2}}-\frac {3 B \sqrt {a+c x^2}}{a^2 e^2 \sqrt {e x}}+\frac {3 B \sqrt {c} x \sqrt {a+c x^2}}{a^2 e^2 \sqrt {e x} \left (\sqrt {a}+\sqrt {c} x\right )}-\frac {3 B \sqrt [4]{c} \sqrt {x} \left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{a^{7/4} e^2 \sqrt {e x} \sqrt {a+c x^2}}+\frac {\left (9 \sqrt {a} B-5 A \sqrt {c}\right ) \sqrt [4]{c} \sqrt {x} \left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{6 a^{9/4} e^2 \sqrt {e x} \sqrt {a+c x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.06, size = 107, normalized size = 0.30 \[ \frac {x \left (3 \left (-3 B x \sqrt {\frac {c x^2}{a}+1} \, _2F_1\left (-\frac {1}{4},\frac {1}{2};\frac {3}{4};-\frac {c x^2}{a}\right )+A+B x\right )-5 A \sqrt {\frac {c x^2}{a}+1} \, _2F_1\left (-\frac {3}{4},\frac {1}{2};\frac {1}{4};-\frac {c x^2}{a}\right )\right )}{3 a (e x)^{5/2} \sqrt {a+c x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x)/((e*x)^(5/2)*(a + c*x^2)^(3/2)),x]

[Out]

(x*(-5*A*Sqrt[1 + (c*x^2)/a]*Hypergeometric2F1[-3/4, 1/2, 1/4, -((c*x^2)/a)] + 3*(A + B*x - 3*B*x*Sqrt[1 + (c*
x^2)/a]*Hypergeometric2F1[-1/4, 1/2, 3/4, -((c*x^2)/a)])))/(3*a*(e*x)^(5/2)*Sqrt[a + c*x^2])

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fricas [F]  time = 0.95, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {c x^{2} + a} {\left (B x + A\right )} \sqrt {e x}}{c^{2} e^{3} x^{7} + 2 \, a c e^{3} x^{5} + a^{2} e^{3} x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(e*x)^(5/2)/(c*x^2+a)^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(c*x^2 + a)*(B*x + A)*sqrt(e*x)/(c^2*e^3*x^7 + 2*a*c*e^3*x^5 + a^2*e^3*x^3), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {B x + A}{{\left (c x^{2} + a\right )}^{\frac {3}{2}} \left (e x\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(e*x)^(5/2)/(c*x^2+a)^(3/2),x, algorithm="giac")

[Out]

integrate((B*x + A)/((c*x^2 + a)^(3/2)*(e*x)^(5/2)), x)

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maple [A]  time = 0.11, size = 307, normalized size = 0.86 \[ -\frac {18 B c \,x^{3}+10 A c \,x^{2}-18 \sqrt {2}\, \sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {\frac {-c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {-\frac {c x}{\sqrt {-a c}}}\, B a x \EllipticE \left (\sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}, \frac {\sqrt {2}}{2}\right )+9 \sqrt {2}\, \sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {\frac {-c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {-\frac {c x}{\sqrt {-a c}}}\, B a x \EllipticF \left (\sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}, \frac {\sqrt {2}}{2}\right )+5 \sqrt {2}\, \sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {\frac {-c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {-\frac {c x}{\sqrt {-a c}}}\, \sqrt {-a c}\, A x \EllipticF \left (\sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}, \frac {\sqrt {2}}{2}\right )+12 B a x +4 A a}{6 \sqrt {c \,x^{2}+a}\, \sqrt {e x}\, a^{2} e^{2} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)/(e*x)^(5/2)/(c*x^2+a)^(3/2),x)

[Out]

-1/6/x*(5*A*2^(1/2)*((c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2)*((-c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2)*(-1/(-a*c
)^(1/2)*c*x)^(1/2)*EllipticF(((c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2),1/2*2^(1/2))*(-a*c)^(1/2)*x+9*B*2^(1/2)*(
(c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2)*((-c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2)*(-1/(-a*c)^(1/2)*c*x)^(1/2)*El
lipticF(((c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2),1/2*2^(1/2))*x*a-18*B*2^(1/2)*((c*x+(-a*c)^(1/2))/(-a*c)^(1/2)
)^(1/2)*((-c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2)*(-1/(-a*c)^(1/2)*c*x)^(1/2)*EllipticE(((c*x+(-a*c)^(1/2))/(-a
*c)^(1/2))^(1/2),1/2*2^(1/2))*x*a+18*B*c*x^3+10*A*c*x^2+12*B*a*x+4*A*a)/(c*x^2+a)^(1/2)/a^2/e^2/(e*x)^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {B x + A}{{\left (c x^{2} + a\right )}^{\frac {3}{2}} \left (e x\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(e*x)^(5/2)/(c*x^2+a)^(3/2),x, algorithm="maxima")

[Out]

integrate((B*x + A)/((c*x^2 + a)^(3/2)*(e*x)^(5/2)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {A+B\,x}{{\left (e\,x\right )}^{5/2}\,{\left (c\,x^2+a\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x)/((e*x)^(5/2)*(a + c*x^2)^(3/2)),x)

[Out]

int((A + B*x)/((e*x)^(5/2)*(a + c*x^2)^(3/2)), x)

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sympy [C]  time = 65.58, size = 100, normalized size = 0.28 \[ \frac {A \Gamma \left (- \frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, \frac {3}{2} \\ \frac {1}{4} \end {matrix}\middle | {\frac {c x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {3}{2}} e^{\frac {5}{2}} x^{\frac {3}{2}} \Gamma \left (\frac {1}{4}\right )} + \frac {B \Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {3}{2} \\ \frac {3}{4} \end {matrix}\middle | {\frac {c x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {3}{2}} e^{\frac {5}{2}} \sqrt {x} \Gamma \left (\frac {3}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(e*x)**(5/2)/(c*x**2+a)**(3/2),x)

[Out]

A*gamma(-3/4)*hyper((-3/4, 3/2), (1/4,), c*x**2*exp_polar(I*pi)/a)/(2*a**(3/2)*e**(5/2)*x**(3/2)*gamma(1/4)) +
B*gamma(-1/4)*hyper((-1/4, 3/2), (3/4,), c*x**2*exp_polar(I*pi)/a)/(2*a**(3/2)*e**(5/2)*sqrt(x)*gamma(3/4))

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