3.310 \(\int \frac{1}{x^{5/2} (a+b x^2)^3} \, dx\)

Optimal. Leaf size=251 \[ \frac{77 b^{3/4} \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{64 \sqrt{2} a^{15/4}}-\frac{77 b^{3/4} \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{64 \sqrt{2} a^{15/4}}+\frac{77 b^{3/4} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{32 \sqrt{2} a^{15/4}}-\frac{77 b^{3/4} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}+1\right )}{32 \sqrt{2} a^{15/4}}+\frac{11}{16 a^2 x^{3/2} \left (a+b x^2\right )}-\frac{77}{48 a^3 x^{3/2}}+\frac{1}{4 a x^{3/2} \left (a+b x^2\right )^2} \]

[Out]

-77/(48*a^3*x^(3/2)) + 1/(4*a*x^(3/2)*(a + b*x^2)^2) + 11/(16*a^2*x^(3/2)*(a + b*x^2)) + (77*b^(3/4)*ArcTan[1
- (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(32*Sqrt[2]*a^(15/4)) - (77*b^(3/4)*ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt[x])
/a^(1/4)])/(32*Sqrt[2]*a^(15/4)) + (77*b^(3/4)*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(64
*Sqrt[2]*a^(15/4)) - (77*b^(3/4)*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(64*Sqrt[2]*a^(15
/4))

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Rubi [A]  time = 0.184596, antiderivative size = 251, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 9, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.6, Rules used = {290, 325, 329, 211, 1165, 628, 1162, 617, 204} \[ \frac{77 b^{3/4} \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{64 \sqrt{2} a^{15/4}}-\frac{77 b^{3/4} \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{64 \sqrt{2} a^{15/4}}+\frac{77 b^{3/4} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{32 \sqrt{2} a^{15/4}}-\frac{77 b^{3/4} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}+1\right )}{32 \sqrt{2} a^{15/4}}+\frac{11}{16 a^2 x^{3/2} \left (a+b x^2\right )}-\frac{77}{48 a^3 x^{3/2}}+\frac{1}{4 a x^{3/2} \left (a+b x^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^(5/2)*(a + b*x^2)^3),x]

[Out]

-77/(48*a^3*x^(3/2)) + 1/(4*a*x^(3/2)*(a + b*x^2)^2) + 11/(16*a^2*x^(3/2)*(a + b*x^2)) + (77*b^(3/4)*ArcTan[1
- (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(32*Sqrt[2]*a^(15/4)) - (77*b^(3/4)*ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt[x])
/a^(1/4)])/(32*Sqrt[2]*a^(15/4)) + (77*b^(3/4)*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(64
*Sqrt[2]*a^(15/4)) - (77*b^(3/4)*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(64*Sqrt[2]*a^(15
/4))

Rule 290

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

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && 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{1}{x^{5/2} \left (a+b x^2\right )^3} \, dx &=\frac{1}{4 a x^{3/2} \left (a+b x^2\right )^2}+\frac{11 \int \frac{1}{x^{5/2} \left (a+b x^2\right )^2} \, dx}{8 a}\\ &=\frac{1}{4 a x^{3/2} \left (a+b x^2\right )^2}+\frac{11}{16 a^2 x^{3/2} \left (a+b x^2\right )}+\frac{77 \int \frac{1}{x^{5/2} \left (a+b x^2\right )} \, dx}{32 a^2}\\ &=-\frac{77}{48 a^3 x^{3/2}}+\frac{1}{4 a x^{3/2} \left (a+b x^2\right )^2}+\frac{11}{16 a^2 x^{3/2} \left (a+b x^2\right )}-\frac{(77 b) \int \frac{1}{\sqrt{x} \left (a+b x^2\right )} \, dx}{32 a^3}\\ &=-\frac{77}{48 a^3 x^{3/2}}+\frac{1}{4 a x^{3/2} \left (a+b x^2\right )^2}+\frac{11}{16 a^2 x^{3/2} \left (a+b x^2\right )}-\frac{(77 b) \operatorname{Subst}\left (\int \frac{1}{a+b x^4} \, dx,x,\sqrt{x}\right )}{16 a^3}\\ &=-\frac{77}{48 a^3 x^{3/2}}+\frac{1}{4 a x^{3/2} \left (a+b x^2\right )^2}+\frac{11}{16 a^2 x^{3/2} \left (a+b x^2\right )}-\frac{(77 b) \operatorname{Subst}\left (\int \frac{\sqrt{a}-\sqrt{b} x^2}{a+b x^4} \, dx,x,\sqrt{x}\right )}{32 a^{7/2}}-\frac{(77 b) \operatorname{Subst}\left (\int \frac{\sqrt{a}+\sqrt{b} x^2}{a+b x^4} \, dx,x,\sqrt{x}\right )}{32 a^{7/2}}\\ &=-\frac{77}{48 a^3 x^{3/2}}+\frac{1}{4 a x^{3/2} \left (a+b x^2\right )^2}+\frac{11}{16 a^2 x^{3/2} \left (a+b x^2\right )}-\frac{\left (77 \sqrt{b}\right ) \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 a^{7/2}}-\frac{\left (77 \sqrt{b}\right ) \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 a^{7/2}}+\frac{\left (77 b^{3/4}\right ) \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^{15/4}}+\frac{\left (77 b^{3/4}\right ) \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^{15/4}}\\ &=-\frac{77}{48 a^3 x^{3/2}}+\frac{1}{4 a x^{3/2} \left (a+b x^2\right )^2}+\frac{11}{16 a^2 x^{3/2} \left (a+b x^2\right )}+\frac{77 b^{3/4} \log \left (\sqrt{a}-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{b} x\right )}{64 \sqrt{2} a^{15/4}}-\frac{77 b^{3/4} \log \left (\sqrt{a}+\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{b} x\right )}{64 \sqrt{2} a^{15/4}}-\frac{\left (77 b^{3/4}\right ) \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^{15/4}}+\frac{\left (77 b^{3/4}\right ) \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^{15/4}}\\ &=-\frac{77}{48 a^3 x^{3/2}}+\frac{1}{4 a x^{3/2} \left (a+b x^2\right )^2}+\frac{11}{16 a^2 x^{3/2} \left (a+b x^2\right )}+\frac{77 b^{3/4} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{32 \sqrt{2} a^{15/4}}-\frac{77 b^{3/4} \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{32 \sqrt{2} a^{15/4}}+\frac{77 b^{3/4} \log \left (\sqrt{a}-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{b} x\right )}{64 \sqrt{2} a^{15/4}}-\frac{77 b^{3/4} \log \left (\sqrt{a}+\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{b} x\right )}{64 \sqrt{2} a^{15/4}}\\ \end{align*}

Mathematica [C]  time = 0.0065073, size = 29, normalized size = 0.12 \[ -\frac{2 \, _2F_1\left (-\frac{3}{4},3;\frac{1}{4};-\frac{b x^2}{a}\right )}{3 a^3 x^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^(5/2)*(a + b*x^2)^3),x]

[Out]

(-2*Hypergeometric2F1[-3/4, 3, 1/4, -((b*x^2)/a)])/(3*a^3*x^(3/2))

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Maple [A]  time = 0.016, size = 181, normalized size = 0.7 \begin{align*} -{\frac{15\,{b}^{2}}{16\,{a}^{3} \left ( b{x}^{2}+a \right ) ^{2}}{x}^{{\frac{5}{2}}}}-{\frac{19\,b}{16\,{a}^{2} \left ( b{x}^{2}+a \right ) ^{2}}\sqrt{x}}-{\frac{77\,b\sqrt{2}}{128\,{a}^{4}}\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{77\,b\sqrt{2}}{64\,{a}^{4}}\sqrt [4]{{\frac{a}{b}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}+1 \right ) }-{\frac{77\,b\sqrt{2}}{64\,{a}^{4}}\sqrt [4]{{\frac{a}{b}}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}-1 \right ) }-{\frac{2}{3\,{a}^{3}}{x}^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^(5/2)/(b*x^2+a)^3,x)

[Out]

-15/16/a^3*b^2/(b*x^2+a)^2*x^(5/2)-19/16/a^2*b/(b*x^2+a)^2*x^(1/2)-77/128/a^4*b*(1/b*a)^(1/4)*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)))-77/64/a^4*b*(1/b*a
)^(1/4)*2^(1/2)*arctan(2^(1/2)/(1/b*a)^(1/4)*x^(1/2)+1)-77/64/a^4*b*(1/b*a)^(1/4)*2^(1/2)*arctan(2^(1/2)/(1/b*
a)^(1/4)*x^(1/2)-1)-2/3/a^3/x^(3/2)

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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(1/x^(5/2)/(b*x^2+a)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.48447, size = 645, normalized size = 2.57 \begin{align*} -\frac{924 \,{\left (a^{3} b^{2} x^{6} + 2 \, a^{4} b x^{4} + a^{5} x^{2}\right )} \left (-\frac{b^{3}}{a^{15}}\right )^{\frac{1}{4}} \arctan \left (-\frac{a^{11} b \sqrt{x} \left (-\frac{b^{3}}{a^{15}}\right )^{\frac{3}{4}} - \sqrt{a^{8} \sqrt{-\frac{b^{3}}{a^{15}}} + b^{2} x} a^{11} \left (-\frac{b^{3}}{a^{15}}\right )^{\frac{3}{4}}}{b^{3}}\right ) + 231 \,{\left (a^{3} b^{2} x^{6} + 2 \, a^{4} b x^{4} + a^{5} x^{2}\right )} \left (-\frac{b^{3}}{a^{15}}\right )^{\frac{1}{4}} \log \left (77 \, a^{4} \left (-\frac{b^{3}}{a^{15}}\right )^{\frac{1}{4}} + 77 \, b \sqrt{x}\right ) - 231 \,{\left (a^{3} b^{2} x^{6} + 2 \, a^{4} b x^{4} + a^{5} x^{2}\right )} \left (-\frac{b^{3}}{a^{15}}\right )^{\frac{1}{4}} \log \left (-77 \, a^{4} \left (-\frac{b^{3}}{a^{15}}\right )^{\frac{1}{4}} + 77 \, b \sqrt{x}\right ) + 4 \,{\left (77 \, b^{2} x^{4} + 121 \, a b x^{2} + 32 \, a^{2}\right )} \sqrt{x}}{192 \,{\left (a^{3} b^{2} x^{6} + 2 \, a^{4} b x^{4} + a^{5} x^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(5/2)/(b*x^2+a)^3,x, algorithm="fricas")

[Out]

-1/192*(924*(a^3*b^2*x^6 + 2*a^4*b*x^4 + a^5*x^2)*(-b^3/a^15)^(1/4)*arctan(-(a^11*b*sqrt(x)*(-b^3/a^15)^(3/4)
- sqrt(a^8*sqrt(-b^3/a^15) + b^2*x)*a^11*(-b^3/a^15)^(3/4))/b^3) + 231*(a^3*b^2*x^6 + 2*a^4*b*x^4 + a^5*x^2)*(
-b^3/a^15)^(1/4)*log(77*a^4*(-b^3/a^15)^(1/4) + 77*b*sqrt(x)) - 231*(a^3*b^2*x^6 + 2*a^4*b*x^4 + a^5*x^2)*(-b^
3/a^15)^(1/4)*log(-77*a^4*(-b^3/a^15)^(1/4) + 77*b*sqrt(x)) + 4*(77*b^2*x^4 + 121*a*b*x^2 + 32*a^2)*sqrt(x))/(
a^3*b^2*x^6 + 2*a^4*b*x^4 + a^5*x^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(1/x**(5/2)/(b*x**2+a)**3,x)

[Out]

Timed out

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Giac [A]  time = 2.11764, size = 281, normalized size = 1.12 \begin{align*} -\frac{77 \, \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^{4}} - \frac{77 \, \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^{4}} - \frac{77 \, \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^{4}} + \frac{77 \, \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^{4}} - \frac{15 \, b^{2} x^{\frac{5}{2}} + 19 \, a b \sqrt{x}}{16 \,{\left (b x^{2} + a\right )}^{2} a^{3}} - \frac{2}{3 \, a^{3} x^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(5/2)/(b*x^2+a)^3,x, algorithm="giac")

[Out]

-77/64*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^4 - 77/64*sqr
t(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^4 - 77/128*sqrt(2)*(a*
b^3)^(1/4)*log(sqrt(2)*sqrt(x)*(a/b)^(1/4) + x + sqrt(a/b))/a^4 + 77/128*sqrt(2)*(a*b^3)^(1/4)*log(-sqrt(2)*sq
rt(x)*(a/b)^(1/4) + x + sqrt(a/b))/a^4 - 1/16*(15*b^2*x^(5/2) + 19*a*b*sqrt(x))/((b*x^2 + a)^2*a^3) - 2/3/(a^3
*x^(3/2))