3.937 \(\int x^5 (a+b x^2+c x^4)^{3/2} \, dx\)

Optimal. Leaf size=204 \[ \frac{\left (7 b^2-4 a c\right ) \left (b+2 c x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{384 c^3}-\frac{\left (b^2-4 a c\right ) \left (7 b^2-4 a c\right ) \left (b+2 c x^2\right ) \sqrt{a+b x^2+c x^4}}{1024 c^4}+\frac{\left (b^2-4 a c\right )^2 \left (7 b^2-4 a c\right ) \tanh ^{-1}\left (\frac{b+2 c x^2}{2 \sqrt{c} \sqrt{a+b x^2+c x^4}}\right )}{2048 c^{9/2}}-\frac{7 b \left (a+b x^2+c x^4\right )^{5/2}}{120 c^2}+\frac{x^2 \left (a+b x^2+c x^4\right )^{5/2}}{12 c} \]

[Out]

-((b^2 - 4*a*c)*(7*b^2 - 4*a*c)*(b + 2*c*x^2)*Sqrt[a + b*x^2 + c*x^4])/(1024*c^4) + ((7*b^2 - 4*a*c)*(b + 2*c*
x^2)*(a + b*x^2 + c*x^4)^(3/2))/(384*c^3) - (7*b*(a + b*x^2 + c*x^4)^(5/2))/(120*c^2) + (x^2*(a + b*x^2 + c*x^
4)^(5/2))/(12*c) + ((b^2 - 4*a*c)^2*(7*b^2 - 4*a*c)*ArcTanh[(b + 2*c*x^2)/(2*Sqrt[c]*Sqrt[a + b*x^2 + c*x^4])]
)/(2048*c^(9/2))

________________________________________________________________________________________

Rubi [A]  time = 0.182342, antiderivative size = 204, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {1114, 742, 640, 612, 621, 206} \[ \frac{\left (7 b^2-4 a c\right ) \left (b+2 c x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{384 c^3}-\frac{\left (b^2-4 a c\right ) \left (7 b^2-4 a c\right ) \left (b+2 c x^2\right ) \sqrt{a+b x^2+c x^4}}{1024 c^4}+\frac{\left (b^2-4 a c\right )^2 \left (7 b^2-4 a c\right ) \tanh ^{-1}\left (\frac{b+2 c x^2}{2 \sqrt{c} \sqrt{a+b x^2+c x^4}}\right )}{2048 c^{9/2}}-\frac{7 b \left (a+b x^2+c x^4\right )^{5/2}}{120 c^2}+\frac{x^2 \left (a+b x^2+c x^4\right )^{5/2}}{12 c} \]

Antiderivative was successfully verified.

[In]

Int[x^5*(a + b*x^2 + c*x^4)^(3/2),x]

[Out]

-((b^2 - 4*a*c)*(7*b^2 - 4*a*c)*(b + 2*c*x^2)*Sqrt[a + b*x^2 + c*x^4])/(1024*c^4) + ((7*b^2 - 4*a*c)*(b + 2*c*
x^2)*(a + b*x^2 + c*x^4)^(3/2))/(384*c^3) - (7*b*(a + b*x^2 + c*x^4)^(5/2))/(120*c^2) + (x^2*(a + b*x^2 + c*x^
4)^(5/2))/(12*c) + ((b^2 - 4*a*c)^2*(7*b^2 - 4*a*c)*ArcTanh[(b + 2*c*x^2)/(2*Sqrt[c]*Sqrt[a + b*x^2 + c*x^4])]
)/(2048*c^(9/2))

Rule 1114

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[x^((m - 1)/2)*(a +
b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, p}, x] && IntegerQ[(m - 1)/2]

Rule 742

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)
*(a + b*x + c*x^2)^(p + 1))/(c*(m + 2*p + 1)), x] + Dist[1/(c*(m + 2*p + 1)), Int[(d + e*x)^(m - 2)*Simp[c*d^2
*(m + 2*p + 1) - e*(a*e*(m - 1) + b*d*(p + 1)) + e*(2*c*d - b*e)*(m + p)*x, x]*(a + b*x + c*x^2)^p, x], x] /;
FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0]
&& If[RationalQ[m], GtQ[m, 1], SumSimplerQ[m, -2]] && NeQ[m + 2*p + 1, 0] && IntQuadraticQ[a, b, c, d, e, m,
p, x]

Rule 640

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(a + b*x + c*x^2)^(p +
1))/(2*c*(p + 1)), x] + Dist[(2*c*d - b*e)/(2*c), Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}
, x] && NeQ[2*c*d - b*e, 0] && NeQ[p, -1]

Rule 612

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b + 2*c*x)*(a + b*x + c*x^2)^p)/(2*c*(2*p +
1)), x] - Dist[(p*(b^2 - 4*a*c))/(2*c*(2*p + 1)), Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x]
&& NeQ[b^2 - 4*a*c, 0] && GtQ[p, 0] && IntegerQ[4*p]

Rule 621

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(4*c - x^2), x], x, (b + 2*c*x)
/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

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])

Rubi steps

\begin{align*} \int x^5 \left (a+b x^2+c x^4\right )^{3/2} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int x^2 \left (a+b x+c x^2\right )^{3/2} \, dx,x,x^2\right )\\ &=\frac{x^2 \left (a+b x^2+c x^4\right )^{5/2}}{12 c}+\frac{\operatorname{Subst}\left (\int \left (-a-\frac{7 b x}{2}\right ) \left (a+b x+c x^2\right )^{3/2} \, dx,x,x^2\right )}{12 c}\\ &=-\frac{7 b \left (a+b x^2+c x^4\right )^{5/2}}{120 c^2}+\frac{x^2 \left (a+b x^2+c x^4\right )^{5/2}}{12 c}+\frac{\left (7 b^2-4 a c\right ) \operatorname{Subst}\left (\int \left (a+b x+c x^2\right )^{3/2} \, dx,x,x^2\right )}{48 c^2}\\ &=\frac{\left (7 b^2-4 a c\right ) \left (b+2 c x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{384 c^3}-\frac{7 b \left (a+b x^2+c x^4\right )^{5/2}}{120 c^2}+\frac{x^2 \left (a+b x^2+c x^4\right )^{5/2}}{12 c}-\frac{\left (\left (b^2-4 a c\right ) \left (7 b^2-4 a c\right )\right ) \operatorname{Subst}\left (\int \sqrt{a+b x+c x^2} \, dx,x,x^2\right )}{256 c^3}\\ &=-\frac{\left (b^2-4 a c\right ) \left (7 b^2-4 a c\right ) \left (b+2 c x^2\right ) \sqrt{a+b x^2+c x^4}}{1024 c^4}+\frac{\left (7 b^2-4 a c\right ) \left (b+2 c x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{384 c^3}-\frac{7 b \left (a+b x^2+c x^4\right )^{5/2}}{120 c^2}+\frac{x^2 \left (a+b x^2+c x^4\right )^{5/2}}{12 c}+\frac{\left (\left (b^2-4 a c\right )^2 \left (7 b^2-4 a c\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x+c x^2}} \, dx,x,x^2\right )}{2048 c^4}\\ &=-\frac{\left (b^2-4 a c\right ) \left (7 b^2-4 a c\right ) \left (b+2 c x^2\right ) \sqrt{a+b x^2+c x^4}}{1024 c^4}+\frac{\left (7 b^2-4 a c\right ) \left (b+2 c x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{384 c^3}-\frac{7 b \left (a+b x^2+c x^4\right )^{5/2}}{120 c^2}+\frac{x^2 \left (a+b x^2+c x^4\right )^{5/2}}{12 c}+\frac{\left (\left (b^2-4 a c\right )^2 \left (7 b^2-4 a c\right )\right ) \operatorname{Subst}\left (\int \frac{1}{4 c-x^2} \, dx,x,\frac{b+2 c x^2}{\sqrt{a+b x^2+c x^4}}\right )}{1024 c^4}\\ &=-\frac{\left (b^2-4 a c\right ) \left (7 b^2-4 a c\right ) \left (b+2 c x^2\right ) \sqrt{a+b x^2+c x^4}}{1024 c^4}+\frac{\left (7 b^2-4 a c\right ) \left (b+2 c x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{384 c^3}-\frac{7 b \left (a+b x^2+c x^4\right )^{5/2}}{120 c^2}+\frac{x^2 \left (a+b x^2+c x^4\right )^{5/2}}{12 c}+\frac{\left (b^2-4 a c\right )^2 \left (7 b^2-4 a c\right ) \tanh ^{-1}\left (\frac{b+2 c x^2}{2 \sqrt{c} \sqrt{a+b x^2+c x^4}}\right )}{2048 c^{9/2}}\\ \end{align*}

Mathematica [A]  time = 0.159113, size = 175, normalized size = 0.86 \[ \frac{\frac{\left (7 b^2-4 a c\right ) \left (2 \sqrt{c} \left (b+2 c x^2\right ) \sqrt{a+b x^2+c x^4} \left (4 c \left (5 a+2 c x^4\right )-3 b^2+8 b c x^2\right )+3 \left (b^2-4 a c\right )^2 \tanh ^{-1}\left (\frac{b+2 c x^2}{2 \sqrt{c} \sqrt{a+b x^2+c x^4}}\right )\right )}{512 c^{7/2}}+x^2 \left (a+b x^2+c x^4\right )^{5/2}-\frac{7 b \left (a+b x^2+c x^4\right )^{5/2}}{10 c}}{12 c} \]

Antiderivative was successfully verified.

[In]

Integrate[x^5*(a + b*x^2 + c*x^4)^(3/2),x]

[Out]

((-7*b*(a + b*x^2 + c*x^4)^(5/2))/(10*c) + x^2*(a + b*x^2 + c*x^4)^(5/2) + ((7*b^2 - 4*a*c)*(2*Sqrt[c]*(b + 2*
c*x^2)*Sqrt[a + b*x^2 + c*x^4]*(-3*b^2 + 8*b*c*x^2 + 4*c*(5*a + 2*c*x^4)) + 3*(b^2 - 4*a*c)^2*ArcTanh[(b + 2*c
*x^2)/(2*Sqrt[c]*Sqrt[a + b*x^2 + c*x^4])]))/(512*c^(7/2)))/(12*c)

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Maple [B]  time = 0.174, size = 432, normalized size = 2.1 \begin{align*}{\frac{7\,a{x}^{6}}{48}\sqrt{c{x}^{4}+b{x}^{2}+a}}-{\frac{7\,{b}^{5}}{1024\,{c}^{4}}\sqrt{c{x}^{4}+b{x}^{2}+a}}+{\frac{7\,{b}^{6}}{2048}\ln \left ({ \left ({\frac{b}{2}}+c{x}^{2} \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{4}+b{x}^{2}+a} \right ){c}^{-{\frac{9}{2}}}}+{\frac{13\,b{x}^{8}}{120}\sqrt{c{x}^{4}+b{x}^{2}+a}}+{\frac{9\,{b}^{2}{a}^{2}}{128}\ln \left ({ \left ({\frac{b}{2}}+c{x}^{2} \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{4}+b{x}^{2}+a} \right ){c}^{-{\frac{5}{2}}}}-{\frac{27\,b{a}^{2}}{320\,{c}^{2}}\sqrt{c{x}^{4}+b{x}^{2}+a}}-{\frac{{a}^{3}}{32}\ln \left ({ \left ({\frac{b}{2}}+c{x}^{2} \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{4}+b{x}^{2}+a} \right ){c}^{-{\frac{3}{2}}}}+{\frac{c{x}^{10}}{12}\sqrt{c{x}^{4}+b{x}^{2}+a}}+{\frac{3\,ab{x}^{4}}{160\,c}\sqrt{c{x}^{4}+b{x}^{2}+a}}-{\frac{9\,a{b}^{2}{x}^{2}}{320\,{c}^{2}}\sqrt{c{x}^{4}+b{x}^{2}+a}}+{\frac{19\,a{b}^{3}}{384\,{c}^{3}}\sqrt{c{x}^{4}+b{x}^{2}+a}}-{\frac{15\,a{b}^{4}}{512}\ln \left ({ \left ({\frac{b}{2}}+c{x}^{2} \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{4}+b{x}^{2}+a} \right ){c}^{-{\frac{7}{2}}}}+{\frac{{a}^{2}{x}^{2}}{32\,c}\sqrt{c{x}^{4}+b{x}^{2}+a}}+{\frac{7\,{b}^{4}{x}^{2}}{1536\,{c}^{3}}\sqrt{c{x}^{4}+b{x}^{2}+a}}+{\frac{{b}^{2}{x}^{6}}{320\,c}\sqrt{c{x}^{4}+b{x}^{2}+a}}-{\frac{7\,{b}^{3}{x}^{4}}{1920\,{c}^{2}}\sqrt{c{x}^{4}+b{x}^{2}+a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(c*x^4+b*x^2+a)^(3/2),x)

[Out]

7/48*a*x^6*(c*x^4+b*x^2+a)^(1/2)-7/1024*b^5/c^4*(c*x^4+b*x^2+a)^(1/2)+7/2048*b^6/c^(9/2)*ln((1/2*b+c*x^2)/c^(1
/2)+(c*x^4+b*x^2+a)^(1/2))+13/120*b*x^8*(c*x^4+b*x^2+a)^(1/2)+9/128*a^2*b^2/c^(5/2)*ln((1/2*b+c*x^2)/c^(1/2)+(
c*x^4+b*x^2+a)^(1/2))-27/320*a^2*b/c^2*(c*x^4+b*x^2+a)^(1/2)-1/32*a^3/c^(3/2)*ln((1/2*b+c*x^2)/c^(1/2)+(c*x^4+
b*x^2+a)^(1/2))+1/12*c*x^10*(c*x^4+b*x^2+a)^(1/2)+3/160*a*b*x^4/c*(c*x^4+b*x^2+a)^(1/2)-9/320*a*b^2/c^2*x^2*(c
*x^4+b*x^2+a)^(1/2)+19/384*a*b^3/c^3*(c*x^4+b*x^2+a)^(1/2)-15/512*a*b^4/c^(7/2)*ln((1/2*b+c*x^2)/c^(1/2)+(c*x^
4+b*x^2+a)^(1/2))+1/32*a^2*x^2/c*(c*x^4+b*x^2+a)^(1/2)+7/1536*b^4/c^3*x^2*(c*x^4+b*x^2+a)^(1/2)+1/320*b^2*x^6/
c*(c*x^4+b*x^2+a)^(1/2)-7/1920*b^3/c^2*x^4*(c*x^4+b*x^2+a)^(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(x^5*(c*x^4+b*x^2+a)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.73758, size = 1061, normalized size = 5.2 \begin{align*} \left [-\frac{15 \,{\left (7 \, b^{6} - 60 \, a b^{4} c + 144 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} \sqrt{c} \log \left (-8 \, c^{2} x^{4} - 8 \, b c x^{2} - b^{2} + 4 \, \sqrt{c x^{4} + b x^{2} + a}{\left (2 \, c x^{2} + b\right )} \sqrt{c} - 4 \, a c\right ) - 4 \,{\left (1280 \, c^{6} x^{10} + 1664 \, b c^{5} x^{8} + 16 \,{\left (3 \, b^{2} c^{4} + 140 \, a c^{5}\right )} x^{6} - 105 \, b^{5} c + 760 \, a b^{3} c^{2} - 1296 \, a^{2} b c^{3} - 8 \,{\left (7 \, b^{3} c^{3} - 36 \, a b c^{4}\right )} x^{4} + 2 \,{\left (35 \, b^{4} c^{2} - 216 \, a b^{2} c^{3} + 240 \, a^{2} c^{4}\right )} x^{2}\right )} \sqrt{c x^{4} + b x^{2} + a}}{61440 \, c^{5}}, -\frac{15 \,{\left (7 \, b^{6} - 60 \, a b^{4} c + 144 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{c x^{4} + b x^{2} + a}{\left (2 \, c x^{2} + b\right )} \sqrt{-c}}{2 \,{\left (c^{2} x^{4} + b c x^{2} + a c\right )}}\right ) - 2 \,{\left (1280 \, c^{6} x^{10} + 1664 \, b c^{5} x^{8} + 16 \,{\left (3 \, b^{2} c^{4} + 140 \, a c^{5}\right )} x^{6} - 105 \, b^{5} c + 760 \, a b^{3} c^{2} - 1296 \, a^{2} b c^{3} - 8 \,{\left (7 \, b^{3} c^{3} - 36 \, a b c^{4}\right )} x^{4} + 2 \,{\left (35 \, b^{4} c^{2} - 216 \, a b^{2} c^{3} + 240 \, a^{2} c^{4}\right )} x^{2}\right )} \sqrt{c x^{4} + b x^{2} + a}}{30720 \, c^{5}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(c*x^4+b*x^2+a)^(3/2),x, algorithm="fricas")

[Out]

[-1/61440*(15*(7*b^6 - 60*a*b^4*c + 144*a^2*b^2*c^2 - 64*a^3*c^3)*sqrt(c)*log(-8*c^2*x^4 - 8*b*c*x^2 - b^2 + 4
*sqrt(c*x^4 + b*x^2 + a)*(2*c*x^2 + b)*sqrt(c) - 4*a*c) - 4*(1280*c^6*x^10 + 1664*b*c^5*x^8 + 16*(3*b^2*c^4 +
140*a*c^5)*x^6 - 105*b^5*c + 760*a*b^3*c^2 - 1296*a^2*b*c^3 - 8*(7*b^3*c^3 - 36*a*b*c^4)*x^4 + 2*(35*b^4*c^2 -
216*a*b^2*c^3 + 240*a^2*c^4)*x^2)*sqrt(c*x^4 + b*x^2 + a))/c^5, -1/30720*(15*(7*b^6 - 60*a*b^4*c + 144*a^2*b^
2*c^2 - 64*a^3*c^3)*sqrt(-c)*arctan(1/2*sqrt(c*x^4 + b*x^2 + a)*(2*c*x^2 + b)*sqrt(-c)/(c^2*x^4 + b*c*x^2 + a*
c)) - 2*(1280*c^6*x^10 + 1664*b*c^5*x^8 + 16*(3*b^2*c^4 + 140*a*c^5)*x^6 - 105*b^5*c + 760*a*b^3*c^2 - 1296*a^
2*b*c^3 - 8*(7*b^3*c^3 - 36*a*b*c^4)*x^4 + 2*(35*b^4*c^2 - 216*a*b^2*c^3 + 240*a^2*c^4)*x^2)*sqrt(c*x^4 + b*x^
2 + a))/c^5]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{5} \left (a + b x^{2} + c x^{4}\right )^{\frac{3}{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5*(c*x**4+b*x**2+a)**(3/2),x)

[Out]

Integral(x**5*(a + b*x**2 + c*x**4)**(3/2), x)

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Giac [A]  time = 1.27394, size = 311, normalized size = 1.52 \begin{align*} \frac{1}{15360} \, \sqrt{c x^{4} + b x^{2} + a}{\left (2 \,{\left (4 \,{\left (2 \,{\left (8 \,{\left (10 \, c x^{2} + 13 \, b\right )} x^{2} + \frac{3 \, b^{2} c^{8} + 140 \, a c^{9}}{c^{9}}\right )} x^{2} - \frac{7 \, b^{3} c^{7} - 36 \, a b c^{8}}{c^{9}}\right )} x^{2} + \frac{35 \, b^{4} c^{6} - 216 \, a b^{2} c^{7} + 240 \, a^{2} c^{8}}{c^{9}}\right )} x^{2} - \frac{105 \, b^{5} c^{5} - 760 \, a b^{3} c^{6} + 1296 \, a^{2} b c^{7}}{c^{9}}\right )} - \frac{{\left (7 \, b^{6} c^{5} - 60 \, a b^{4} c^{6} + 144 \, a^{2} b^{2} c^{7} - 64 \, a^{3} c^{8}\right )} \log \left ({\left | -2 \,{\left (\sqrt{c} x^{2} - \sqrt{c x^{4} + b x^{2} + a}\right )} \sqrt{c} - b \right |}\right )}{2048 \, c^{\frac{19}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(c*x^4+b*x^2+a)^(3/2),x, algorithm="giac")

[Out]

1/15360*sqrt(c*x^4 + b*x^2 + a)*(2*(4*(2*(8*(10*c*x^2 + 13*b)*x^2 + (3*b^2*c^8 + 140*a*c^9)/c^9)*x^2 - (7*b^3*
c^7 - 36*a*b*c^8)/c^9)*x^2 + (35*b^4*c^6 - 216*a*b^2*c^7 + 240*a^2*c^8)/c^9)*x^2 - (105*b^5*c^5 - 760*a*b^3*c^
6 + 1296*a^2*b*c^7)/c^9) - 1/2048*(7*b^6*c^5 - 60*a*b^4*c^6 + 144*a^2*b^2*c^7 - 64*a^3*c^8)*log(abs(-2*(sqrt(c
)*x^2 - sqrt(c*x^4 + b*x^2 + a))*sqrt(c) - b))/c^(19/2)