3.404 \(\int \frac {x^3 \tan ^{-1}(a x)^3}{(c+a^2 c x^2)^3} \, dx\)

Optimal. Leaf size=212 \[ -\frac {3 \tan ^{-1}(a x)^3}{32 a^4 c^3}-\frac {27 \tan ^{-1}(a x)}{256 a^4 c^3}+\frac {x^4 \tan ^{-1}(a x)^3}{4 c^3 \left (a^2 x^2+1\right )^2}-\frac {3 x^4 \tan ^{-1}(a x)}{32 c^3 \left (a^2 x^2+1\right )^2}-\frac {3 x^3}{128 a c^3 \left (a^2 x^2+1\right )^2}+\frac {3 x^3 \tan ^{-1}(a x)^2}{16 a c^3 \left (a^2 x^2+1\right )^2}+\frac {9 \tan ^{-1}(a x)}{32 a^4 c^3 \left (a^2 x^2+1\right )}-\frac {45 x}{256 a^3 c^3 \left (a^2 x^2+1\right )}+\frac {9 x \tan ^{-1}(a x)^2}{32 a^3 c^3 \left (a^2 x^2+1\right )} \]

[Out]

-3/128*x^3/a/c^3/(a^2*x^2+1)^2-45/256*x/a^3/c^3/(a^2*x^2+1)-27/256*arctan(a*x)/a^4/c^3-3/32*x^4*arctan(a*x)/c^
3/(a^2*x^2+1)^2+9/32*arctan(a*x)/a^4/c^3/(a^2*x^2+1)+3/16*x^3*arctan(a*x)^2/a/c^3/(a^2*x^2+1)^2+9/32*x*arctan(
a*x)^2/a^3/c^3/(a^2*x^2+1)-3/32*arctan(a*x)^3/a^4/c^3+1/4*x^4*arctan(a*x)^3/c^3/(a^2*x^2+1)^2

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Rubi [A]  time = 0.29, antiderivative size = 212, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {4944, 4940, 4936, 4930, 199, 205, 288} \[ -\frac {3 x^3}{128 a c^3 \left (a^2 x^2+1\right )^2}-\frac {45 x}{256 a^3 c^3 \left (a^2 x^2+1\right )}+\frac {x^4 \tan ^{-1}(a x)^3}{4 c^3 \left (a^2 x^2+1\right )^2}-\frac {3 x^4 \tan ^{-1}(a x)}{32 c^3 \left (a^2 x^2+1\right )^2}+\frac {3 x^3 \tan ^{-1}(a x)^2}{16 a c^3 \left (a^2 x^2+1\right )^2}+\frac {9 x \tan ^{-1}(a x)^2}{32 a^3 c^3 \left (a^2 x^2+1\right )}+\frac {9 \tan ^{-1}(a x)}{32 a^4 c^3 \left (a^2 x^2+1\right )}-\frac {3 \tan ^{-1}(a x)^3}{32 a^4 c^3}-\frac {27 \tan ^{-1}(a x)}{256 a^4 c^3} \]

Antiderivative was successfully verified.

[In]

Int[(x^3*ArcTan[a*x]^3)/(c + a^2*c*x^2)^3,x]

[Out]

(-3*x^3)/(128*a*c^3*(1 + a^2*x^2)^2) - (45*x)/(256*a^3*c^3*(1 + a^2*x^2)) - (27*ArcTan[a*x])/(256*a^4*c^3) - (
3*x^4*ArcTan[a*x])/(32*c^3*(1 + a^2*x^2)^2) + (9*ArcTan[a*x])/(32*a^4*c^3*(1 + a^2*x^2)) + (3*x^3*ArcTan[a*x]^
2)/(16*a*c^3*(1 + a^2*x^2)^2) + (9*x*ArcTan[a*x]^2)/(32*a^3*c^3*(1 + a^2*x^2)) - (3*ArcTan[a*x]^3)/(32*a^4*c^3
) + (x^4*ArcTan[a*x]^3)/(4*c^3*(1 + a^2*x^2)^2)

Rule 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

Rule 205

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

Rule 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 4930

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

Rule 4936

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_)^2)/((d_) + (e_.)*(x_)^2)^2, x_Symbol] :> Simp[(a + b*ArcTan
[c*x])^(p + 1)/(2*b*c^3*d^2*(p + 1)), x] + (Dist[(b*p)/(2*c), Int[(x*(a + b*ArcTan[c*x])^(p - 1))/(d + e*x^2)^
2, x], x] - Simp[(x*(a + b*ArcTan[c*x])^p)/(2*c^2*d*(d + e*x^2)), x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c
^2*d] && GtQ[p, 0]

Rule 4940

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> Simp[(b
*p*(f*x)^m*(d + e*x^2)^(q + 1)*(a + b*ArcTan[c*x])^(p - 1))/(c*d*m^2), x] + (Dist[(f^2*(m - 1))/(c^2*d*m), Int
[(f*x)^(m - 2)*(d + e*x^2)^(q + 1)*(a + b*ArcTan[c*x])^p, x], x] - Dist[(b^2*p*(p - 1))/m^2, Int[(f*x)^m*(d +
e*x^2)^q*(a + b*ArcTan[c*x])^(p - 2), x], x] - Simp[(f*(f*x)^(m - 1)*(d + e*x^2)^(q + 1)*(a + b*ArcTan[c*x])^p
)/(c^2*d*m), x]) /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[e, c^2*d] && EqQ[m + 2*q + 2, 0] && LtQ[q, -1] && G
tQ[p, 1]

Rule 4944

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> Simp
[((f*x)^(m + 1)*(d + e*x^2)^(q + 1)*(a + b*ArcTan[c*x])^p)/(d*f*(m + 1)), x] - Dist[(b*c*p)/(f*(m + 1)), Int[(
f*x)^(m + 1)*(d + e*x^2)^q*(a + b*ArcTan[c*x])^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && EqQ[e,
c^2*d] && EqQ[m + 2*q + 3, 0] && GtQ[p, 0] && NeQ[m, -1]

Rubi steps

\begin {align*} \int \frac {x^3 \tan ^{-1}(a x)^3}{\left (c+a^2 c x^2\right )^3} \, dx &=\frac {x^4 \tan ^{-1}(a x)^3}{4 c^3 \left (1+a^2 x^2\right )^2}-\frac {1}{4} (3 a) \int \frac {x^4 \tan ^{-1}(a x)^2}{\left (c+a^2 c x^2\right )^3} \, dx\\ &=-\frac {3 x^4 \tan ^{-1}(a x)}{32 c^3 \left (1+a^2 x^2\right )^2}+\frac {3 x^3 \tan ^{-1}(a x)^2}{16 a c^3 \left (1+a^2 x^2\right )^2}+\frac {x^4 \tan ^{-1}(a x)^3}{4 c^3 \left (1+a^2 x^2\right )^2}+\frac {1}{32} (3 a) \int \frac {x^4}{\left (c+a^2 c x^2\right )^3} \, dx-\frac {9 \int \frac {x^2 \tan ^{-1}(a x)^2}{\left (c+a^2 c x^2\right )^2} \, dx}{16 a c}\\ &=-\frac {3 x^3}{128 a c^3 \left (1+a^2 x^2\right )^2}-\frac {3 x^4 \tan ^{-1}(a x)}{32 c^3 \left (1+a^2 x^2\right )^2}+\frac {3 x^3 \tan ^{-1}(a x)^2}{16 a c^3 \left (1+a^2 x^2\right )^2}+\frac {9 x \tan ^{-1}(a x)^2}{32 a^3 c^3 \left (1+a^2 x^2\right )}-\frac {3 \tan ^{-1}(a x)^3}{32 a^4 c^3}+\frac {x^4 \tan ^{-1}(a x)^3}{4 c^3 \left (1+a^2 x^2\right )^2}-\frac {9 \int \frac {x \tan ^{-1}(a x)}{\left (c+a^2 c x^2\right )^2} \, dx}{16 a^2 c}+\frac {9 \int \frac {x^2}{\left (c+a^2 c x^2\right )^2} \, dx}{128 a c}\\ &=-\frac {3 x^3}{128 a c^3 \left (1+a^2 x^2\right )^2}-\frac {9 x}{256 a^3 c^3 \left (1+a^2 x^2\right )}-\frac {3 x^4 \tan ^{-1}(a x)}{32 c^3 \left (1+a^2 x^2\right )^2}+\frac {9 \tan ^{-1}(a x)}{32 a^4 c^3 \left (1+a^2 x^2\right )}+\frac {3 x^3 \tan ^{-1}(a x)^2}{16 a c^3 \left (1+a^2 x^2\right )^2}+\frac {9 x \tan ^{-1}(a x)^2}{32 a^3 c^3 \left (1+a^2 x^2\right )}-\frac {3 \tan ^{-1}(a x)^3}{32 a^4 c^3}+\frac {x^4 \tan ^{-1}(a x)^3}{4 c^3 \left (1+a^2 x^2\right )^2}+\frac {9 \int \frac {1}{c+a^2 c x^2} \, dx}{256 a^3 c^2}-\frac {9 \int \frac {1}{\left (c+a^2 c x^2\right )^2} \, dx}{32 a^3 c}\\ &=-\frac {3 x^3}{128 a c^3 \left (1+a^2 x^2\right )^2}-\frac {45 x}{256 a^3 c^3 \left (1+a^2 x^2\right )}+\frac {9 \tan ^{-1}(a x)}{256 a^4 c^3}-\frac {3 x^4 \tan ^{-1}(a x)}{32 c^3 \left (1+a^2 x^2\right )^2}+\frac {9 \tan ^{-1}(a x)}{32 a^4 c^3 \left (1+a^2 x^2\right )}+\frac {3 x^3 \tan ^{-1}(a x)^2}{16 a c^3 \left (1+a^2 x^2\right )^2}+\frac {9 x \tan ^{-1}(a x)^2}{32 a^3 c^3 \left (1+a^2 x^2\right )}-\frac {3 \tan ^{-1}(a x)^3}{32 a^4 c^3}+\frac {x^4 \tan ^{-1}(a x)^3}{4 c^3 \left (1+a^2 x^2\right )^2}-\frac {9 \int \frac {1}{c+a^2 c x^2} \, dx}{64 a^3 c^2}\\ &=-\frac {3 x^3}{128 a c^3 \left (1+a^2 x^2\right )^2}-\frac {45 x}{256 a^3 c^3 \left (1+a^2 x^2\right )}-\frac {27 \tan ^{-1}(a x)}{256 a^4 c^3}-\frac {3 x^4 \tan ^{-1}(a x)}{32 c^3 \left (1+a^2 x^2\right )^2}+\frac {9 \tan ^{-1}(a x)}{32 a^4 c^3 \left (1+a^2 x^2\right )}+\frac {3 x^3 \tan ^{-1}(a x)^2}{16 a c^3 \left (1+a^2 x^2\right )^2}+\frac {9 x \tan ^{-1}(a x)^2}{32 a^3 c^3 \left (1+a^2 x^2\right )}-\frac {3 \tan ^{-1}(a x)^3}{32 a^4 c^3}+\frac {x^4 \tan ^{-1}(a x)^3}{4 c^3 \left (1+a^2 x^2\right )^2}\\ \end {align*}

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Mathematica [A]  time = 0.27, size = 105, normalized size = 0.50 \[ \frac {-3 a x \left (17 a^2 x^2+15\right )+24 a x \left (5 a^2 x^2+3\right ) \tan ^{-1}(a x)^2+8 \left (5 a^4 x^4-6 a^2 x^2-3\right ) \tan ^{-1}(a x)^3+\left (-51 a^4 x^4+18 a^2 x^2+45\right ) \tan ^{-1}(a x)}{256 a^4 c^3 \left (a^2 x^2+1\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^3*ArcTan[a*x]^3)/(c + a^2*c*x^2)^3,x]

[Out]

(-3*a*x*(15 + 17*a^2*x^2) + (45 + 18*a^2*x^2 - 51*a^4*x^4)*ArcTan[a*x] + 24*a*x*(3 + 5*a^2*x^2)*ArcTan[a*x]^2
+ 8*(-3 - 6*a^2*x^2 + 5*a^4*x^4)*ArcTan[a*x]^3)/(256*a^4*c^3*(1 + a^2*x^2)^2)

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fricas [A]  time = 0.82, size = 117, normalized size = 0.55 \[ -\frac {51 \, a^{3} x^{3} - 8 \, {\left (5 \, a^{4} x^{4} - 6 \, a^{2} x^{2} - 3\right )} \arctan \left (a x\right )^{3} - 24 \, {\left (5 \, a^{3} x^{3} + 3 \, a x\right )} \arctan \left (a x\right )^{2} + 45 \, a x + 3 \, {\left (17 \, a^{4} x^{4} - 6 \, a^{2} x^{2} - 15\right )} \arctan \left (a x\right )}{256 \, {\left (a^{8} c^{3} x^{4} + 2 \, a^{6} c^{3} x^{2} + a^{4} c^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/256*(51*a^3*x^3 - 8*(5*a^4*x^4 - 6*a^2*x^2 - 3)*arctan(a*x)^3 - 24*(5*a^3*x^3 + 3*a*x)*arctan(a*x)^2 + 45*a
*x + 3*(17*a^4*x^4 - 6*a^2*x^2 - 15)*arctan(a*x))/(a^8*c^3*x^4 + 2*a^6*c^3*x^2 + a^4*c^3)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

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maple [A]  time = 0.06, size = 220, normalized size = 1.04 \[ -\frac {\arctan \left (a x \right )^{3}}{2 a^{4} c^{3} \left (a^{2} x^{2}+1\right )}+\frac {\arctan \left (a x \right )^{3}}{4 a^{4} c^{3} \left (a^{2} x^{2}+1\right )^{2}}+\frac {15 x^{3} \arctan \left (a x \right )^{2}}{32 a \,c^{3} \left (a^{2} x^{2}+1\right )^{2}}+\frac {9 \arctan \left (a x \right )^{2} x}{32 a^{3} c^{3} \left (a^{2} x^{2}+1\right )^{2}}+\frac {5 \arctan \left (a x \right )^{3}}{32 a^{4} c^{3}}+\frac {15 \arctan \left (a x \right )}{32 a^{4} c^{3} \left (a^{2} x^{2}+1\right )}-\frac {3 \arctan \left (a x \right )}{32 a^{4} c^{3} \left (a^{2} x^{2}+1\right )^{2}}-\frac {51 x^{3}}{256 a \,c^{3} \left (a^{2} x^{2}+1\right )^{2}}-\frac {45 x}{256 a^{3} c^{3} \left (a^{2} x^{2}+1\right )^{2}}-\frac {51 \arctan \left (a x \right )}{256 a^{4} c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*arctan(a*x)^3/(a^2*c*x^2+c)^3,x)

[Out]

-1/2/a^4/c^3*arctan(a*x)^3/(a^2*x^2+1)+1/4/a^4/c^3*arctan(a*x)^3/(a^2*x^2+1)^2+15/32*x^3*arctan(a*x)^2/a/c^3/(
a^2*x^2+1)^2+9/32/a^3/c^3*arctan(a*x)^2*x/(a^2*x^2+1)^2+5/32*arctan(a*x)^3/a^4/c^3+15/32*arctan(a*x)/a^4/c^3/(
a^2*x^2+1)-3/32/a^4/c^3*arctan(a*x)/(a^2*x^2+1)^2-51/256*x^3/a/c^3/(a^2*x^2+1)^2-45/256/a^3/c^3/(a^2*x^2+1)^2*
x-51/256*arctan(a*x)/a^4/c^3

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maxima [A]  time = 0.48, size = 289, normalized size = 1.36 \[ \frac {3}{32} \, a {\left (\frac {5 \, a^{2} x^{3} + 3 \, x}{a^{8} c^{3} x^{4} + 2 \, a^{6} c^{3} x^{2} + a^{4} c^{3}} + \frac {5 \, \arctan \left (a x\right )}{a^{5} c^{3}}\right )} \arctan \left (a x\right )^{2} - \frac {{\left (2 \, a^{2} x^{2} + 1\right )} \arctan \left (a x\right )^{3}}{4 \, {\left (a^{8} c^{3} x^{4} + 2 \, a^{6} c^{3} x^{2} + a^{4} c^{3}\right )}} - \frac {1}{256} \, {\left (\frac {{\left (51 \, a^{3} x^{3} - 40 \, {\left (a^{4} x^{4} + 2 \, a^{2} x^{2} + 1\right )} \arctan \left (a x\right )^{3} + 45 \, a x + 51 \, {\left (a^{4} x^{4} + 2 \, a^{2} x^{2} + 1\right )} \arctan \left (a x\right )\right )} a^{2}}{a^{11} c^{3} x^{4} + 2 \, a^{9} c^{3} x^{2} + a^{7} c^{3}} - \frac {24 \, {\left (5 \, a^{2} x^{2} - 5 \, {\left (a^{4} x^{4} + 2 \, a^{2} x^{2} + 1\right )} \arctan \left (a x\right )^{2} + 4\right )} a \arctan \left (a x\right )}{a^{10} c^{3} x^{4} + 2 \, a^{8} c^{3} x^{2} + a^{6} c^{3}}\right )} a \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arctan(a*x)^3/(a^2*c*x^2+c)^3,x, algorithm="maxima")

[Out]

3/32*a*((5*a^2*x^3 + 3*x)/(a^8*c^3*x^4 + 2*a^6*c^3*x^2 + a^4*c^3) + 5*arctan(a*x)/(a^5*c^3))*arctan(a*x)^2 - 1
/4*(2*a^2*x^2 + 1)*arctan(a*x)^3/(a^8*c^3*x^4 + 2*a^6*c^3*x^2 + a^4*c^3) - 1/256*((51*a^3*x^3 - 40*(a^4*x^4 +
2*a^2*x^2 + 1)*arctan(a*x)^3 + 45*a*x + 51*(a^4*x^4 + 2*a^2*x^2 + 1)*arctan(a*x))*a^2/(a^11*c^3*x^4 + 2*a^9*c^
3*x^2 + a^7*c^3) - 24*(5*a^2*x^2 - 5*(a^4*x^4 + 2*a^2*x^2 + 1)*arctan(a*x)^2 + 4)*a*arctan(a*x)/(a^10*c^3*x^4
+ 2*a^8*c^3*x^2 + a^6*c^3))*a

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mupad [B]  time = 0.59, size = 205, normalized size = 0.97 \[ \frac {{\mathrm {atan}\left (a\,x\right )}^2\,\left (\frac {9\,x}{32\,a^5\,c^3}+\frac {15\,x^3}{32\,a^3\,c^3}\right )}{\frac {1}{a^2}+2\,x^2+a^2\,x^4}-{\mathrm {atan}\left (a\,x\right )}^3\,\left (\frac {\frac {1}{4\,a^6\,c^3}+\frac {x^2}{2\,a^4\,c^3}}{\frac {1}{a^2}+2\,x^2+a^2\,x^4}-\frac {5}{32\,a^4\,c^3}\right )-\frac {\frac {51\,a^2\,x^3}{8}+\frac {45\,x}{8}}{32\,a^7\,c^3\,x^4+64\,a^5\,c^3\,x^2+32\,a^3\,c^3}-\frac {51\,\mathrm {atan}\left (a\,x\right )}{256\,a^4\,c^3}+\frac {\mathrm {atan}\left (a\,x\right )\,\left (\frac {3}{8\,a^6\,c^3}+\frac {15\,x^2}{32\,a^4\,c^3}\right )}{\frac {1}{a^2}+2\,x^2+a^2\,x^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3*atan(a*x)^3)/(c + a^2*c*x^2)^3,x)

[Out]

(atan(a*x)^2*((9*x)/(32*a^5*c^3) + (15*x^3)/(32*a^3*c^3)))/(1/a^2 + 2*x^2 + a^2*x^4) - atan(a*x)^3*((1/(4*a^6*
c^3) + x^2/(2*a^4*c^3))/(1/a^2 + 2*x^2 + a^2*x^4) - 5/(32*a^4*c^3)) - ((45*x)/8 + (51*a^2*x^3)/8)/(32*a^3*c^3
+ 64*a^5*c^3*x^2 + 32*a^7*c^3*x^4) - (51*atan(a*x))/(256*a^4*c^3) + (atan(a*x)*(3/(8*a^6*c^3) + (15*x^2)/(32*a
^4*c^3)))/(1/a^2 + 2*x^2 + a^2*x^4)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {x^{3} \operatorname {atan}^{3}{\left (a x \right )}}{a^{6} x^{6} + 3 a^{4} x^{4} + 3 a^{2} x^{2} + 1}\, dx}{c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*atan(a*x)**3/(a**2*c*x**2+c)**3,x)

[Out]

Integral(x**3*atan(a*x)**3/(a**6*x**6 + 3*a**4*x**4 + 3*a**2*x**2 + 1), x)/c**3

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