3.329 \(\int \frac{x^{15/2}}{(b x^2+c x^4)^2} \, dx\)

Optimal. Leaf size=230 \[ \frac{5 \sqrt [4]{b} \log \left (-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{8 \sqrt{2} c^{9/4}}-\frac{5 \sqrt [4]{b} \log \left (\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{8 \sqrt{2} c^{9/4}}+\frac{5 \sqrt [4]{b} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )}{4 \sqrt{2} c^{9/4}}-\frac{5 \sqrt [4]{b} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}+1\right )}{4 \sqrt{2} c^{9/4}}-\frac{x^{5/2}}{2 c \left (b+c x^2\right )}+\frac{5 \sqrt{x}}{2 c^2} \]

[Out]

(5*Sqrt[x])/(2*c^2) - x^(5/2)/(2*c*(b + c*x^2)) + (5*b^(1/4)*ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(4
*Sqrt[2]*c^(9/4)) - (5*b^(1/4)*ArcTan[1 + (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(4*Sqrt[2]*c^(9/4)) + (5*b^(1/4)
*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(8*Sqrt[2]*c^(9/4)) - (5*b^(1/4)*Log[Sqrt[b] + Sq
rt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(8*Sqrt[2]*c^(9/4))

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Rubi [A]  time = 0.181984, antiderivative size = 230, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 10, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.526, Rules used = {1584, 288, 321, 329, 211, 1165, 628, 1162, 617, 204} \[ \frac{5 \sqrt [4]{b} \log \left (-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{8 \sqrt{2} c^{9/4}}-\frac{5 \sqrt [4]{b} \log \left (\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{8 \sqrt{2} c^{9/4}}+\frac{5 \sqrt [4]{b} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )}{4 \sqrt{2} c^{9/4}}-\frac{5 \sqrt [4]{b} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}+1\right )}{4 \sqrt{2} c^{9/4}}-\frac{x^{5/2}}{2 c \left (b+c x^2\right )}+\frac{5 \sqrt{x}}{2 c^2} \]

Antiderivative was successfully verified.

[In]

Int[x^(15/2)/(b*x^2 + c*x^4)^2,x]

[Out]

(5*Sqrt[x])/(2*c^2) - x^(5/2)/(2*c*(b + c*x^2)) + (5*b^(1/4)*ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(4
*Sqrt[2]*c^(9/4)) - (5*b^(1/4)*ArcTan[1 + (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(4*Sqrt[2]*c^(9/4)) + (5*b^(1/4)
*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(8*Sqrt[2]*c^(9/4)) - (5*b^(1/4)*Log[Sqrt[b] + Sq
rt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(8*Sqrt[2]*c^(9/4))

Rule 1584

Int[(u_.)*(x_)^(m_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(m + n*p)*(a + b*x^(q -
p))^n, x] /; FreeQ[{a, b, m, p, q}, x] && IntegerQ[n] && PosQ[q - p]

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 321

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*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 211

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Di
st[1/(2*r), Int[(r - s*x^2)/(a + b*x^4), x], x] + Dist[1/(2*r), Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[
{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ, b
]]))

Rule 1165

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

Rule 628

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

Rule 1162

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

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

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

Rubi steps

\begin{align*} \int \frac{x^{15/2}}{\left (b x^2+c x^4\right )^2} \, dx &=\int \frac{x^{7/2}}{\left (b+c x^2\right )^2} \, dx\\ &=-\frac{x^{5/2}}{2 c \left (b+c x^2\right )}+\frac{5 \int \frac{x^{3/2}}{b+c x^2} \, dx}{4 c}\\ &=\frac{5 \sqrt{x}}{2 c^2}-\frac{x^{5/2}}{2 c \left (b+c x^2\right )}-\frac{(5 b) \int \frac{1}{\sqrt{x} \left (b+c x^2\right )} \, dx}{4 c^2}\\ &=\frac{5 \sqrt{x}}{2 c^2}-\frac{x^{5/2}}{2 c \left (b+c x^2\right )}-\frac{(5 b) \operatorname{Subst}\left (\int \frac{1}{b+c x^4} \, dx,x,\sqrt{x}\right )}{2 c^2}\\ &=\frac{5 \sqrt{x}}{2 c^2}-\frac{x^{5/2}}{2 c \left (b+c x^2\right )}-\frac{\left (5 \sqrt{b}\right ) \operatorname{Subst}\left (\int \frac{\sqrt{b}-\sqrt{c} x^2}{b+c x^4} \, dx,x,\sqrt{x}\right )}{4 c^2}-\frac{\left (5 \sqrt{b}\right ) \operatorname{Subst}\left (\int \frac{\sqrt{b}+\sqrt{c} x^2}{b+c x^4} \, dx,x,\sqrt{x}\right )}{4 c^2}\\ &=\frac{5 \sqrt{x}}{2 c^2}-\frac{x^{5/2}}{2 c \left (b+c x^2\right )}-\frac{\left (5 \sqrt{b}\right ) \operatorname{Subst}\left (\int \frac{1}{\frac{\sqrt{b}}{\sqrt{c}}-\frac{\sqrt{2} \sqrt [4]{b} x}{\sqrt [4]{c}}+x^2} \, dx,x,\sqrt{x}\right )}{8 c^{5/2}}-\frac{\left (5 \sqrt{b}\right ) \operatorname{Subst}\left (\int \frac{1}{\frac{\sqrt{b}}{\sqrt{c}}+\frac{\sqrt{2} \sqrt [4]{b} x}{\sqrt [4]{c}}+x^2} \, dx,x,\sqrt{x}\right )}{8 c^{5/2}}+\frac{\left (5 \sqrt [4]{b}\right ) \operatorname{Subst}\left (\int \frac{\frac{\sqrt{2} \sqrt [4]{b}}{\sqrt [4]{c}}+2 x}{-\frac{\sqrt{b}}{\sqrt{c}}-\frac{\sqrt{2} \sqrt [4]{b} x}{\sqrt [4]{c}}-x^2} \, dx,x,\sqrt{x}\right )}{8 \sqrt{2} c^{9/4}}+\frac{\left (5 \sqrt [4]{b}\right ) \operatorname{Subst}\left (\int \frac{\frac{\sqrt{2} \sqrt [4]{b}}{\sqrt [4]{c}}-2 x}{-\frac{\sqrt{b}}{\sqrt{c}}+\frac{\sqrt{2} \sqrt [4]{b} x}{\sqrt [4]{c}}-x^2} \, dx,x,\sqrt{x}\right )}{8 \sqrt{2} c^{9/4}}\\ &=\frac{5 \sqrt{x}}{2 c^2}-\frac{x^{5/2}}{2 c \left (b+c x^2\right )}+\frac{5 \sqrt [4]{b} \log \left (\sqrt{b}-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{c} x\right )}{8 \sqrt{2} c^{9/4}}-\frac{5 \sqrt [4]{b} \log \left (\sqrt{b}+\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{c} x\right )}{8 \sqrt{2} c^{9/4}}-\frac{\left (5 \sqrt [4]{b}\right ) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )}{4 \sqrt{2} c^{9/4}}+\frac{\left (5 \sqrt [4]{b}\right ) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )}{4 \sqrt{2} c^{9/4}}\\ &=\frac{5 \sqrt{x}}{2 c^2}-\frac{x^{5/2}}{2 c \left (b+c x^2\right )}+\frac{5 \sqrt [4]{b} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )}{4 \sqrt{2} c^{9/4}}-\frac{5 \sqrt [4]{b} \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )}{4 \sqrt{2} c^{9/4}}+\frac{5 \sqrt [4]{b} \log \left (\sqrt{b}-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{c} x\right )}{8 \sqrt{2} c^{9/4}}-\frac{5 \sqrt [4]{b} \log \left (\sqrt{b}+\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{c} x\right )}{8 \sqrt{2} c^{9/4}}\\ \end{align*}

Mathematica [A]  time = 0.109156, size = 221, normalized size = 0.96 \[ \frac{\frac{32 c^{5/4} x^{5/2}}{b+c x^2}+\frac{40 b \sqrt [4]{c} \sqrt{x}}{b+c x^2}+5 \sqrt{2} \sqrt [4]{b} \log \left (-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )-5 \sqrt{2} \sqrt [4]{b} \log \left (\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )+10 \sqrt{2} \sqrt [4]{b} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )-10 \sqrt{2} \sqrt [4]{b} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}+1\right )}{16 c^{9/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(15/2)/(b*x^2 + c*x^4)^2,x]

[Out]

((40*b*c^(1/4)*Sqrt[x])/(b + c*x^2) + (32*c^(5/4)*x^(5/2))/(b + c*x^2) + 10*Sqrt[2]*b^(1/4)*ArcTan[1 - (Sqrt[2
]*c^(1/4)*Sqrt[x])/b^(1/4)] - 10*Sqrt[2]*b^(1/4)*ArcTan[1 + (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)] + 5*Sqrt[2]*b^(
1/4)*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x] - 5*Sqrt[2]*b^(1/4)*Log[Sqrt[b] + Sqrt[2]*b^(1
/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(16*c^(9/4))

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Maple [A]  time = 0.061, size = 158, normalized size = 0.7 \begin{align*} 2\,{\frac{\sqrt{x}}{{c}^{2}}}+{\frac{b}{2\,{c}^{2} \left ( c{x}^{2}+b \right ) }\sqrt{x}}-{\frac{5\,\sqrt{2}}{16\,{c}^{2}}\sqrt [4]{{\frac{b}{c}}}\ln \left ({ \left ( x+\sqrt [4]{{\frac{b}{c}}}\sqrt{x}\sqrt{2}+\sqrt{{\frac{b}{c}}} \right ) \left ( x-\sqrt [4]{{\frac{b}{c}}}\sqrt{x}\sqrt{2}+\sqrt{{\frac{b}{c}}} \right ) ^{-1}} \right ) }-{\frac{5\,\sqrt{2}}{8\,{c}^{2}}\sqrt [4]{{\frac{b}{c}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}}+1 \right ) }-{\frac{5\,\sqrt{2}}{8\,{c}^{2}}\sqrt [4]{{\frac{b}{c}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}}-1 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(15/2)/(c*x^4+b*x^2)^2,x)

[Out]

2*x^(1/2)/c^2+1/2*b/c^2*x^(1/2)/(c*x^2+b)-5/16/c^2*(b/c)^(1/4)*2^(1/2)*ln((x+(b/c)^(1/4)*x^(1/2)*2^(1/2)+(b/c)
^(1/2))/(x-(b/c)^(1/4)*x^(1/2)*2^(1/2)+(b/c)^(1/2)))-5/8/c^2*(b/c)^(1/4)*2^(1/2)*arctan(2^(1/2)/(b/c)^(1/4)*x^
(1/2)+1)-5/8/c^2*(b/c)^(1/4)*2^(1/2)*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)-1)

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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^(15/2)/(c*x^4+b*x^2)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.36376, size = 443, normalized size = 1.93 \begin{align*} -\frac{20 \,{\left (c^{3} x^{2} + b c^{2}\right )} \left (-\frac{b}{c^{9}}\right )^{\frac{1}{4}} \arctan \left (\frac{\sqrt{c^{4} \sqrt{-\frac{b}{c^{9}}} + x} c^{7} \left (-\frac{b}{c^{9}}\right )^{\frac{3}{4}} - c^{7} \sqrt{x} \left (-\frac{b}{c^{9}}\right )^{\frac{3}{4}}}{b}\right ) + 5 \,{\left (c^{3} x^{2} + b c^{2}\right )} \left (-\frac{b}{c^{9}}\right )^{\frac{1}{4}} \log \left (5 \, c^{2} \left (-\frac{b}{c^{9}}\right )^{\frac{1}{4}} + 5 \, \sqrt{x}\right ) - 5 \,{\left (c^{3} x^{2} + b c^{2}\right )} \left (-\frac{b}{c^{9}}\right )^{\frac{1}{4}} \log \left (-5 \, c^{2} \left (-\frac{b}{c^{9}}\right )^{\frac{1}{4}} + 5 \, \sqrt{x}\right ) - 4 \,{\left (4 \, c x^{2} + 5 \, b\right )} \sqrt{x}}{8 \,{\left (c^{3} x^{2} + b c^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(15/2)/(c*x^4+b*x^2)^2,x, algorithm="fricas")

[Out]

-1/8*(20*(c^3*x^2 + b*c^2)*(-b/c^9)^(1/4)*arctan((sqrt(c^4*sqrt(-b/c^9) + x)*c^7*(-b/c^9)^(3/4) - c^7*sqrt(x)*
(-b/c^9)^(3/4))/b) + 5*(c^3*x^2 + b*c^2)*(-b/c^9)^(1/4)*log(5*c^2*(-b/c^9)^(1/4) + 5*sqrt(x)) - 5*(c^3*x^2 + b
*c^2)*(-b/c^9)^(1/4)*log(-5*c^2*(-b/c^9)^(1/4) + 5*sqrt(x)) - 4*(4*c*x^2 + 5*b)*sqrt(x))/(c^3*x^2 + b*c^2)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(15/2)/(c*x**4+b*x**2)**2,x)

[Out]

Timed out

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Giac [A]  time = 1.23465, size = 265, normalized size = 1.15 \begin{align*} -\frac{5 \, \sqrt{2} \left (b c^{3}\right )^{\frac{1}{4}} \arctan \left (\frac{\sqrt{2}{\left (\sqrt{2} \left (\frac{b}{c}\right )^{\frac{1}{4}} + 2 \, \sqrt{x}\right )}}{2 \, \left (\frac{b}{c}\right )^{\frac{1}{4}}}\right )}{8 \, c^{3}} - \frac{5 \, \sqrt{2} \left (b c^{3}\right )^{\frac{1}{4}} \arctan \left (-\frac{\sqrt{2}{\left (\sqrt{2} \left (\frac{b}{c}\right )^{\frac{1}{4}} - 2 \, \sqrt{x}\right )}}{2 \, \left (\frac{b}{c}\right )^{\frac{1}{4}}}\right )}{8 \, c^{3}} - \frac{5 \, \sqrt{2} \left (b c^{3}\right )^{\frac{1}{4}} \log \left (\sqrt{2} \sqrt{x} \left (\frac{b}{c}\right )^{\frac{1}{4}} + x + \sqrt{\frac{b}{c}}\right )}{16 \, c^{3}} + \frac{5 \, \sqrt{2} \left (b c^{3}\right )^{\frac{1}{4}} \log \left (-\sqrt{2} \sqrt{x} \left (\frac{b}{c}\right )^{\frac{1}{4}} + x + \sqrt{\frac{b}{c}}\right )}{16 \, c^{3}} + \frac{b \sqrt{x}}{2 \,{\left (c x^{2} + b\right )} c^{2}} + \frac{2 \, \sqrt{x}}{c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(15/2)/(c*x^4+b*x^2)^2,x, algorithm="giac")

[Out]

-5/8*sqrt(2)*(b*c^3)^(1/4)*arctan(1/2*sqrt(2)*(sqrt(2)*(b/c)^(1/4) + 2*sqrt(x))/(b/c)^(1/4))/c^3 - 5/8*sqrt(2)
*(b*c^3)^(1/4)*arctan(-1/2*sqrt(2)*(sqrt(2)*(b/c)^(1/4) - 2*sqrt(x))/(b/c)^(1/4))/c^3 - 5/16*sqrt(2)*(b*c^3)^(
1/4)*log(sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/c^3 + 5/16*sqrt(2)*(b*c^3)^(1/4)*log(-sqrt(2)*sqrt(x)*(b
/c)^(1/4) + x + sqrt(b/c))/c^3 + 1/2*b*sqrt(x)/((c*x^2 + b)*c^2) + 2*sqrt(x)/c^2