3.304 \(\int \frac{x^{7/2}}{(a+b x^2)^3} \, dx\)

Optimal. Leaf size=239 \[ -\frac{5 \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{64 \sqrt{2} a^{3/4} b^{9/4}}+\frac{5 \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{64 \sqrt{2} a^{3/4} b^{9/4}}-\frac{5 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{32 \sqrt{2} a^{3/4} b^{9/4}}+\frac{5 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}+1\right )}{32 \sqrt{2} a^{3/4} b^{9/4}}-\frac{5 \sqrt{x}}{16 b^2 \left (a+b x^2\right )}-\frac{x^{5/2}}{4 b \left (a+b x^2\right )^2} \]

[Out]

-x^(5/2)/(4*b*(a + b*x^2)^2) - (5*Sqrt[x])/(16*b^2*(a + b*x^2)) - (5*ArcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1
/4)])/(32*Sqrt[2]*a^(3/4)*b^(9/4)) + (5*ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(32*Sqrt[2]*a^(3/4)*b^(
9/4)) - (5*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(64*Sqrt[2]*a^(3/4)*b^(9/4)) + (5*Log[S
qrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(64*Sqrt[2]*a^(3/4)*b^(9/4))

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Rubi [A]  time = 0.1774, antiderivative size = 239, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 8, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.533, Rules used = {288, 329, 211, 1165, 628, 1162, 617, 204} \[ -\frac{5 \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{64 \sqrt{2} a^{3/4} b^{9/4}}+\frac{5 \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{64 \sqrt{2} a^{3/4} b^{9/4}}-\frac{5 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{32 \sqrt{2} a^{3/4} b^{9/4}}+\frac{5 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}+1\right )}{32 \sqrt{2} a^{3/4} b^{9/4}}-\frac{5 \sqrt{x}}{16 b^2 \left (a+b x^2\right )}-\frac{x^{5/2}}{4 b \left (a+b x^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[x^(7/2)/(a + b*x^2)^3,x]

[Out]

-x^(5/2)/(4*b*(a + b*x^2)^2) - (5*Sqrt[x])/(16*b^2*(a + b*x^2)) - (5*ArcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1
/4)])/(32*Sqrt[2]*a^(3/4)*b^(9/4)) + (5*ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(32*Sqrt[2]*a^(3/4)*b^(
9/4)) - (5*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(64*Sqrt[2]*a^(3/4)*b^(9/4)) + (5*Log[S
qrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(64*Sqrt[2]*a^(3/4)*b^(9/4))

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 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^{7/2}}{\left (a+b x^2\right )^3} \, dx &=-\frac{x^{5/2}}{4 b \left (a+b x^2\right )^2}+\frac{5 \int \frac{x^{3/2}}{\left (a+b x^2\right )^2} \, dx}{8 b}\\ &=-\frac{x^{5/2}}{4 b \left (a+b x^2\right )^2}-\frac{5 \sqrt{x}}{16 b^2 \left (a+b x^2\right )}+\frac{5 \int \frac{1}{\sqrt{x} \left (a+b x^2\right )} \, dx}{32 b^2}\\ &=-\frac{x^{5/2}}{4 b \left (a+b x^2\right )^2}-\frac{5 \sqrt{x}}{16 b^2 \left (a+b x^2\right )}+\frac{5 \operatorname{Subst}\left (\int \frac{1}{a+b x^4} \, dx,x,\sqrt{x}\right )}{16 b^2}\\ &=-\frac{x^{5/2}}{4 b \left (a+b x^2\right )^2}-\frac{5 \sqrt{x}}{16 b^2 \left (a+b x^2\right )}+\frac{5 \operatorname{Subst}\left (\int \frac{\sqrt{a}-\sqrt{b} x^2}{a+b x^4} \, dx,x,\sqrt{x}\right )}{32 \sqrt{a} b^2}+\frac{5 \operatorname{Subst}\left (\int \frac{\sqrt{a}+\sqrt{b} x^2}{a+b x^4} \, dx,x,\sqrt{x}\right )}{32 \sqrt{a} b^2}\\ &=-\frac{x^{5/2}}{4 b \left (a+b x^2\right )^2}-\frac{5 \sqrt{x}}{16 b^2 \left (a+b x^2\right )}+\frac{5 \operatorname{Subst}\left (\int \frac{1}{\frac{\sqrt{a}}{\sqrt{b}}-\frac{\sqrt{2} \sqrt [4]{a} x}{\sqrt [4]{b}}+x^2} \, dx,x,\sqrt{x}\right )}{64 \sqrt{a} b^{5/2}}+\frac{5 \operatorname{Subst}\left (\int \frac{1}{\frac{\sqrt{a}}{\sqrt{b}}+\frac{\sqrt{2} \sqrt [4]{a} x}{\sqrt [4]{b}}+x^2} \, dx,x,\sqrt{x}\right )}{64 \sqrt{a} b^{5/2}}-\frac{5 \operatorname{Subst}\left (\int \frac{\frac{\sqrt{2} \sqrt [4]{a}}{\sqrt [4]{b}}+2 x}{-\frac{\sqrt{a}}{\sqrt{b}}-\frac{\sqrt{2} \sqrt [4]{a} x}{\sqrt [4]{b}}-x^2} \, dx,x,\sqrt{x}\right )}{64 \sqrt{2} a^{3/4} b^{9/4}}-\frac{5 \operatorname{Subst}\left (\int \frac{\frac{\sqrt{2} \sqrt [4]{a}}{\sqrt [4]{b}}-2 x}{-\frac{\sqrt{a}}{\sqrt{b}}+\frac{\sqrt{2} \sqrt [4]{a} x}{\sqrt [4]{b}}-x^2} \, dx,x,\sqrt{x}\right )}{64 \sqrt{2} a^{3/4} b^{9/4}}\\ &=-\frac{x^{5/2}}{4 b \left (a+b x^2\right )^2}-\frac{5 \sqrt{x}}{16 b^2 \left (a+b x^2\right )}-\frac{5 \log \left (\sqrt{a}-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{b} x\right )}{64 \sqrt{2} a^{3/4} b^{9/4}}+\frac{5 \log \left (\sqrt{a}+\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{b} x\right )}{64 \sqrt{2} a^{3/4} b^{9/4}}+\frac{5 \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{32 \sqrt{2} a^{3/4} b^{9/4}}-\frac{5 \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{32 \sqrt{2} a^{3/4} b^{9/4}}\\ &=-\frac{x^{5/2}}{4 b \left (a+b x^2\right )^2}-\frac{5 \sqrt{x}}{16 b^2 \left (a+b x^2\right )}-\frac{5 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{32 \sqrt{2} a^{3/4} b^{9/4}}+\frac{5 \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{32 \sqrt{2} a^{3/4} b^{9/4}}-\frac{5 \log \left (\sqrt{a}-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{b} x\right )}{64 \sqrt{2} a^{3/4} b^{9/4}}+\frac{5 \log \left (\sqrt{a}+\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{b} x\right )}{64 \sqrt{2} a^{3/4} b^{9/4}}\\ \end{align*}

Mathematica [A]  time = 0.106279, size = 242, normalized size = 1.01 \[ \frac{-\frac{15 \sqrt{2} \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{a^{3/4}}+\frac{15 \sqrt{2} \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{a^{3/4}}-\frac{30 \sqrt{2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{a^{3/4}}+\frac{30 \sqrt{2} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}+1\right )}{a^{3/4}}-\frac{256 b^{5/4} x^{5/2}}{\left (a+b x^2\right )^2}+\frac{40 \sqrt [4]{b} \sqrt{x}}{a+b x^2}-\frac{160 a \sqrt [4]{b} \sqrt{x}}{\left (a+b x^2\right )^2}}{384 b^{9/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(7/2)/(a + b*x^2)^3,x]

[Out]

((-160*a*b^(1/4)*Sqrt[x])/(a + b*x^2)^2 - (256*b^(5/4)*x^(5/2))/(a + b*x^2)^2 + (40*b^(1/4)*Sqrt[x])/(a + b*x^
2) - (30*Sqrt[2]*ArcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/a^(3/4) + (30*Sqrt[2]*ArcTan[1 + (Sqrt[2]*b^(1
/4)*Sqrt[x])/a^(1/4)])/a^(3/4) - (15*Sqrt[2]*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/a^(3/
4) + (15*Sqrt[2]*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/a^(3/4))/(384*b^(9/4))

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Maple [A]  time = 0.012, size = 170, normalized size = 0.7 \begin{align*} 2\,{\frac{1}{ \left ( b{x}^{2}+a \right ) ^{2}} \left ( -{\frac{9\,{x}^{5/2}}{32\,b}}-{\frac{5\,a\sqrt{x}}{32\,{b}^{2}}} \right ) }+{\frac{5\,\sqrt{2}}{128\,a{b}^{2}}\sqrt [4]{{\frac{a}{b}}}\ln \left ({ \left ( x+\sqrt [4]{{\frac{a}{b}}}\sqrt{x}\sqrt{2}+\sqrt{{\frac{a}{b}}} \right ) \left ( x-\sqrt [4]{{\frac{a}{b}}}\sqrt{x}\sqrt{2}+\sqrt{{\frac{a}{b}}} \right ) ^{-1}} \right ) }+{\frac{5\,\sqrt{2}}{64\,a{b}^{2}}\sqrt [4]{{\frac{a}{b}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}+1 \right ) }+{\frac{5\,\sqrt{2}}{64\,a{b}^{2}}\sqrt [4]{{\frac{a}{b}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}-1 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(7/2)/(b*x^2+a)^3,x)

[Out]

2*(-9/32*x^(5/2)/b-5/32*a*x^(1/2)/b^2)/(b*x^2+a)^2+5/128/b^2*(1/b*a)^(1/4)/a*2^(1/2)*ln((x+(1/b*a)^(1/4)*x^(1/
2)*2^(1/2)+(1/b*a)^(1/2))/(x-(1/b*a)^(1/4)*x^(1/2)*2^(1/2)+(1/b*a)^(1/2)))+5/64/b^2*(1/b*a)^(1/4)/a*2^(1/2)*ar
ctan(2^(1/2)/(1/b*a)^(1/4)*x^(1/2)+1)+5/64/b^2*(1/b*a)^(1/4)/a*2^(1/2)*arctan(2^(1/2)/(1/b*a)^(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^(7/2)/(b*x^2+a)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.47267, size = 599, normalized size = 2.51 \begin{align*} \frac{20 \,{\left (b^{4} x^{4} + 2 \, a b^{3} x^{2} + a^{2} b^{2}\right )} \left (-\frac{1}{a^{3} b^{9}}\right )^{\frac{1}{4}} \arctan \left (\sqrt{a^{2} b^{4} \sqrt{-\frac{1}{a^{3} b^{9}}} + x} a^{2} b^{7} \left (-\frac{1}{a^{3} b^{9}}\right )^{\frac{3}{4}} - a^{2} b^{7} \sqrt{x} \left (-\frac{1}{a^{3} b^{9}}\right )^{\frac{3}{4}}\right ) + 5 \,{\left (b^{4} x^{4} + 2 \, a b^{3} x^{2} + a^{2} b^{2}\right )} \left (-\frac{1}{a^{3} b^{9}}\right )^{\frac{1}{4}} \log \left (a b^{2} \left (-\frac{1}{a^{3} b^{9}}\right )^{\frac{1}{4}} + \sqrt{x}\right ) - 5 \,{\left (b^{4} x^{4} + 2 \, a b^{3} x^{2} + a^{2} b^{2}\right )} \left (-\frac{1}{a^{3} b^{9}}\right )^{\frac{1}{4}} \log \left (-a b^{2} \left (-\frac{1}{a^{3} b^{9}}\right )^{\frac{1}{4}} + \sqrt{x}\right ) - 4 \,{\left (9 \, b x^{2} + 5 \, a\right )} \sqrt{x}}{64 \,{\left (b^{4} x^{4} + 2 \, a b^{3} x^{2} + a^{2} b^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(7/2)/(b*x^2+a)^3,x, algorithm="fricas")

[Out]

1/64*(20*(b^4*x^4 + 2*a*b^3*x^2 + a^2*b^2)*(-1/(a^3*b^9))^(1/4)*arctan(sqrt(a^2*b^4*sqrt(-1/(a^3*b^9)) + x)*a^
2*b^7*(-1/(a^3*b^9))^(3/4) - a^2*b^7*sqrt(x)*(-1/(a^3*b^9))^(3/4)) + 5*(b^4*x^4 + 2*a*b^3*x^2 + a^2*b^2)*(-1/(
a^3*b^9))^(1/4)*log(a*b^2*(-1/(a^3*b^9))^(1/4) + sqrt(x)) - 5*(b^4*x^4 + 2*a*b^3*x^2 + a^2*b^2)*(-1/(a^3*b^9))
^(1/4)*log(-a*b^2*(-1/(a^3*b^9))^(1/4) + sqrt(x)) - 4*(9*b*x^2 + 5*a)*sqrt(x))/(b^4*x^4 + 2*a*b^3*x^2 + a^2*b^
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**(7/2)/(b*x**2+a)**3,x)

[Out]

Timed out

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Giac [A]  time = 2.435, size = 282, normalized size = 1.18 \begin{align*} \frac{5 \, \sqrt{2} \left (a b^{3}\right )^{\frac{1}{4}} \arctan \left (\frac{\sqrt{2}{\left (\sqrt{2} \left (\frac{a}{b}\right )^{\frac{1}{4}} + 2 \, \sqrt{x}\right )}}{2 \, \left (\frac{a}{b}\right )^{\frac{1}{4}}}\right )}{64 \, a b^{3}} + \frac{5 \, \sqrt{2} \left (a b^{3}\right )^{\frac{1}{4}} \arctan \left (-\frac{\sqrt{2}{\left (\sqrt{2} \left (\frac{a}{b}\right )^{\frac{1}{4}} - 2 \, \sqrt{x}\right )}}{2 \, \left (\frac{a}{b}\right )^{\frac{1}{4}}}\right )}{64 \, a b^{3}} + \frac{5 \, \sqrt{2} \left (a b^{3}\right )^{\frac{1}{4}} \log \left (\sqrt{2} \sqrt{x} \left (\frac{a}{b}\right )^{\frac{1}{4}} + x + \sqrt{\frac{a}{b}}\right )}{128 \, a b^{3}} - \frac{5 \, \sqrt{2} \left (a b^{3}\right )^{\frac{1}{4}} \log \left (-\sqrt{2} \sqrt{x} \left (\frac{a}{b}\right )^{\frac{1}{4}} + x + \sqrt{\frac{a}{b}}\right )}{128 \, a b^{3}} - \frac{9 \, b x^{\frac{5}{2}} + 5 \, a \sqrt{x}}{16 \,{\left (b x^{2} + a\right )}^{2} b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(7/2)/(b*x^2+a)^3,x, algorithm="giac")

[Out]

5/64*sqrt(2)*(a*b^3)^(1/4)*arctan(1/2*sqrt(2)*(sqrt(2)*(a/b)^(1/4) + 2*sqrt(x))/(a/b)^(1/4))/(a*b^3) + 5/64*sq
rt(2)*(a*b^3)^(1/4)*arctan(-1/2*sqrt(2)*(sqrt(2)*(a/b)^(1/4) - 2*sqrt(x))/(a/b)^(1/4))/(a*b^3) + 5/128*sqrt(2)
*(a*b^3)^(1/4)*log(sqrt(2)*sqrt(x)*(a/b)^(1/4) + x + sqrt(a/b))/(a*b^3) - 5/128*sqrt(2)*(a*b^3)^(1/4)*log(-sqr
t(2)*sqrt(x)*(a/b)^(1/4) + x + sqrt(a/b))/(a*b^3) - 1/16*(9*b*x^(5/2) + 5*a*sqrt(x))/((b*x^2 + a)^2*b^2)