\(\int \frac {(b+a x^4) \sqrt [4]{b x^2+a x^4}}{x^2} \, dx\) [1802]

Optimal result

Integrand size = 26, antiderivative size = 122 \[ \int \frac {\left (b+a x^4\right ) \sqrt [4]{b x^2+a x^4}}{x^2} \, dx=\frac {\sqrt [4]{b x^2+a x^4} \left (-32 b+b x^2+4 a x^4\right )}{16 x}+\frac {\left (-32 a b+3 b^2\right ) \arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b x^2+a x^4}}\right )}{32 a^{3/4}}+\frac {\left (32 a b-3 b^2\right ) \text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b x^2+a x^4}}\right )}{32 a^{3/4}} \]

[Out]

1/16*(a*x^4+b*x^2)^(1/4)*(4*a*x^4+b*x^2-32*b)/x+1/32*(-32*a*b+3*b^2)*arctan(a^(1/4)*x/(a*x^4+b*x^2)^(1/4))/a^(
3/4)+1/32*(32*a*b-3*b^2)*arctanh(a^(1/4)*x/(a*x^4+b*x^2)^(1/4))/a^(3/4)

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(321\) vs. \(2(122)=244\).

Time = 0.30 (sec) , antiderivative size = 321, normalized size of antiderivative = 2.63, number of steps used = 17, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {2077, 2045, 2057, 335, 338, 304, 209, 212, 2046, 2049} \[ \int \frac {\left (b+a x^4\right ) \sqrt [4]{b x^2+a x^4}}{x^2} \, dx=\frac {3 b^2 x^{3/2} \left (a x^2+b\right )^{3/4} \arctan \left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{a x^2+b}}\right )}{32 a^{3/4} \left (a x^4+b x^2\right )^{3/4}}-\frac {3 b^2 x^{3/2} \left (a x^2+b\right )^{3/4} \text {arctanh}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{a x^2+b}}\right )}{32 a^{3/4} \left (a x^4+b x^2\right )^{3/4}}-\frac {\sqrt [4]{a} b x^{3/2} \left (a x^2+b\right )^{3/4} \arctan \left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{a x^2+b}}\right )}{\left (a x^4+b x^2\right )^{3/4}}+\frac {\sqrt [4]{a} b x^{3/2} \left (a x^2+b\right )^{3/4} \text {arctanh}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{a x^2+b}}\right )}{\left (a x^4+b x^2\right )^{3/4}}+\frac {1}{16} b x \sqrt [4]{a x^4+b x^2}-\frac {2 b \sqrt [4]{a x^4+b x^2}}{x}+\frac {1}{4} a x^3 \sqrt [4]{a x^4+b x^2} \]

[In]

Int[((b + a*x^4)*(b*x^2 + a*x^4)^(1/4))/x^2,x]

[Out]

(-2*b*(b*x^2 + a*x^4)^(1/4))/x + (b*x*(b*x^2 + a*x^4)^(1/4))/16 + (a*x^3*(b*x^2 + a*x^4)^(1/4))/4 - (a^(1/4)*b
*x^(3/2)*(b + a*x^2)^(3/4)*ArcTan[(a^(1/4)*Sqrt[x])/(b + a*x^2)^(1/4)])/(b*x^2 + a*x^4)^(3/4) + (3*b^2*x^(3/2)
*(b + a*x^2)^(3/4)*ArcTan[(a^(1/4)*Sqrt[x])/(b + a*x^2)^(1/4)])/(32*a^(3/4)*(b*x^2 + a*x^4)^(3/4)) + (a^(1/4)*
b*x^(3/2)*(b + a*x^2)^(3/4)*ArcTanh[(a^(1/4)*Sqrt[x])/(b + a*x^2)^(1/4)])/(b*x^2 + a*x^4)^(3/4) - (3*b^2*x^(3/
2)*(b + a*x^2)^(3/4)*ArcTanh[(a^(1/4)*Sqrt[x])/(b + a*x^2)^(1/4)])/(32*a^(3/4)*(b*x^2 + a*x^4)^(3/4))

Rule 209

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

Rule 212

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

Rule 304

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

Rule 335

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 338

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + (m + 1)/n), Subst[Int[x^m/(1 - b*x^n)^(
p + (m + 1)/n + 1), x], x, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[
p, -2^(-1)] && IntegersQ[m, p + (m + 1)/n]

Rule 2045

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

Rule 2046

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a*x^j + b
*x^n)^p/(c*(m + n*p + 1))), x] + Dist[a*(n - j)*(p/(c^j*(m + n*p + 1))), Int[(c*x)^(m + j)*(a*x^j + b*x^n)^(p
- 1), x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[p] && LtQ[0, j, n] && (IntegersQ[j, n] || GtQ[c, 0]) && G
tQ[p, 0] && NeQ[m + n*p + 1, 0]

Rule 2049

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

Rule 2057

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[c^IntPart[m]*(c*x)^FracPa
rt[m]*((a*x^j + b*x^n)^FracPart[p]/(x^(FracPart[m] + j*FracPart[p])*(a + b*x^(n - j))^FracPart[p])), Int[x^(m
+ j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[n, j] && PosQ[n
- j]

Rule 2077

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(c*x)
^m*Pq*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] && (PolyQ[Pq, x] || PolyQ[Pq, x^n]) &&  !In
tegerQ[p] && NeQ[n, j]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {b \sqrt [4]{b x^2+a x^4}}{x^2}+a x^2 \sqrt [4]{b x^2+a x^4}\right ) \, dx \\ & = a \int x^2 \sqrt [4]{b x^2+a x^4} \, dx+b \int \frac {\sqrt [4]{b x^2+a x^4}}{x^2} \, dx \\ & = -\frac {2 b \sqrt [4]{b x^2+a x^4}}{x}+\frac {1}{4} a x^3 \sqrt [4]{b x^2+a x^4}+\frac {1}{8} (a b) \int \frac {x^4}{\left (b x^2+a x^4\right )^{3/4}} \, dx+(a b) \int \frac {x^2}{\left (b x^2+a x^4\right )^{3/4}} \, dx \\ & = -\frac {2 b \sqrt [4]{b x^2+a x^4}}{x}+\frac {1}{16} b x \sqrt [4]{b x^2+a x^4}+\frac {1}{4} a x^3 \sqrt [4]{b x^2+a x^4}-\frac {1}{32} \left (3 b^2\right ) \int \frac {x^2}{\left (b x^2+a x^4\right )^{3/4}} \, dx+\frac {\left (a b x^{3/2} \left (b+a x^2\right )^{3/4}\right ) \int \frac {\sqrt {x}}{\left (b+a x^2\right )^{3/4}} \, dx}{\left (b x^2+a x^4\right )^{3/4}} \\ & = -\frac {2 b \sqrt [4]{b x^2+a x^4}}{x}+\frac {1}{16} b x \sqrt [4]{b x^2+a x^4}+\frac {1}{4} a x^3 \sqrt [4]{b x^2+a x^4}+\frac {\left (2 a b x^{3/2} \left (b+a x^2\right )^{3/4}\right ) \text {Subst}\left (\int \frac {x^2}{\left (b+a x^4\right )^{3/4}} \, dx,x,\sqrt {x}\right )}{\left (b x^2+a x^4\right )^{3/4}}-\frac {\left (3 b^2 x^{3/2} \left (b+a x^2\right )^{3/4}\right ) \int \frac {\sqrt {x}}{\left (b+a x^2\right )^{3/4}} \, dx}{32 \left (b x^2+a x^4\right )^{3/4}} \\ & = -\frac {2 b \sqrt [4]{b x^2+a x^4}}{x}+\frac {1}{16} b x \sqrt [4]{b x^2+a x^4}+\frac {1}{4} a x^3 \sqrt [4]{b x^2+a x^4}+\frac {\left (2 a b x^{3/2} \left (b+a x^2\right )^{3/4}\right ) \text {Subst}\left (\int \frac {x^2}{1-a x^4} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{b+a x^2}}\right )}{\left (b x^2+a x^4\right )^{3/4}}-\frac {\left (3 b^2 x^{3/2} \left (b+a x^2\right )^{3/4}\right ) \text {Subst}\left (\int \frac {x^2}{\left (b+a x^4\right )^{3/4}} \, dx,x,\sqrt {x}\right )}{16 \left (b x^2+a x^4\right )^{3/4}} \\ & = -\frac {2 b \sqrt [4]{b x^2+a x^4}}{x}+\frac {1}{16} b x \sqrt [4]{b x^2+a x^4}+\frac {1}{4} a x^3 \sqrt [4]{b x^2+a x^4}+\frac {\left (\sqrt {a} b x^{3/2} \left (b+a x^2\right )^{3/4}\right ) \text {Subst}\left (\int \frac {1}{1-\sqrt {a} x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{b+a x^2}}\right )}{\left (b x^2+a x^4\right )^{3/4}}-\frac {\left (\sqrt {a} b x^{3/2} \left (b+a x^2\right )^{3/4}\right ) \text {Subst}\left (\int \frac {1}{1+\sqrt {a} x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{b+a x^2}}\right )}{\left (b x^2+a x^4\right )^{3/4}}-\frac {\left (3 b^2 x^{3/2} \left (b+a x^2\right )^{3/4}\right ) \text {Subst}\left (\int \frac {x^2}{1-a x^4} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{b+a x^2}}\right )}{16 \left (b x^2+a x^4\right )^{3/4}} \\ & = -\frac {2 b \sqrt [4]{b x^2+a x^4}}{x}+\frac {1}{16} b x \sqrt [4]{b x^2+a x^4}+\frac {1}{4} a x^3 \sqrt [4]{b x^2+a x^4}-\frac {\sqrt [4]{a} b x^{3/2} \left (b+a x^2\right )^{3/4} \arctan \left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{b+a x^2}}\right )}{\left (b x^2+a x^4\right )^{3/4}}+\frac {\sqrt [4]{a} b x^{3/2} \left (b+a x^2\right )^{3/4} \text {arctanh}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{b+a x^2}}\right )}{\left (b x^2+a x^4\right )^{3/4}}-\frac {\left (3 b^2 x^{3/2} \left (b+a x^2\right )^{3/4}\right ) \text {Subst}\left (\int \frac {1}{1-\sqrt {a} x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{b+a x^2}}\right )}{32 \sqrt {a} \left (b x^2+a x^4\right )^{3/4}}+\frac {\left (3 b^2 x^{3/2} \left (b+a x^2\right )^{3/4}\right ) \text {Subst}\left (\int \frac {1}{1+\sqrt {a} x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{b+a x^2}}\right )}{32 \sqrt {a} \left (b x^2+a x^4\right )^{3/4}} \\ & = -\frac {2 b \sqrt [4]{b x^2+a x^4}}{x}+\frac {1}{16} b x \sqrt [4]{b x^2+a x^4}+\frac {1}{4} a x^3 \sqrt [4]{b x^2+a x^4}-\frac {\sqrt [4]{a} b x^{3/2} \left (b+a x^2\right )^{3/4} \arctan \left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{b+a x^2}}\right )}{\left (b x^2+a x^4\right )^{3/4}}+\frac {3 b^2 x^{3/2} \left (b+a x^2\right )^{3/4} \arctan \left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{b+a x^2}}\right )}{32 a^{3/4} \left (b x^2+a x^4\right )^{3/4}}+\frac {\sqrt [4]{a} b x^{3/2} \left (b+a x^2\right )^{3/4} \text {arctanh}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{b+a x^2}}\right )}{\left (b x^2+a x^4\right )^{3/4}}-\frac {3 b^2 x^{3/2} \left (b+a x^2\right )^{3/4} \text {arctanh}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{b+a x^2}}\right )}{32 a^{3/4} \left (b x^2+a x^4\right )^{3/4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.44 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.18 \[ \int \frac {\left (b+a x^4\right ) \sqrt [4]{b x^2+a x^4}}{x^2} \, dx=\frac {x \left (b+a x^2\right )^{3/4} \left (2 a^{3/4} \sqrt [4]{b+a x^2} \left (4 a x^4+b \left (-32+x^2\right )\right )-(32 a-3 b) b \sqrt {x} \arctan \left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{b+a x^2}}\right )+(32 a-3 b) b \sqrt {x} \text {arctanh}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{b+a x^2}}\right )\right )}{32 a^{3/4} \left (x^2 \left (b+a x^2\right )\right )^{3/4}} \]

[In]

Integrate[((b + a*x^4)*(b*x^2 + a*x^4)^(1/4))/x^2,x]

[Out]

(x*(b + a*x^2)^(3/4)*(2*a^(3/4)*(b + a*x^2)^(1/4)*(4*a*x^4 + b*(-32 + x^2)) - (32*a - 3*b)*b*Sqrt[x]*ArcTan[(a
^(1/4)*Sqrt[x])/(b + a*x^2)^(1/4)] + (32*a - 3*b)*b*Sqrt[x]*ArcTanh[(a^(1/4)*Sqrt[x])/(b + a*x^2)^(1/4)]))/(32
*a^(3/4)*(x^2*(b + a*x^2))^(3/4))

Maple [A] (verified)

Time = 0.34 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.05

method	result	size
pseudoelliptic	\(\frac {\frac {b x \left (a -\frac {3 b}{32}\right ) \ln \left (\frac {-a^{\frac {1}{4}} x -\left (x^{2} \left (a \,x^{2}+b \right )\right )^{\frac {1}{4}}}{a^{\frac {1}{4}} x -\left (x^{2} \left (a \,x^{2}+b \right )\right )^{\frac {1}{4}}}\right )}{2}+b x \left (a -\frac {3 b}{32}\right ) \arctan \left (\frac {\left (x^{2} \left (a \,x^{2}+b \right )\right )^{\frac {1}{4}}}{a^{\frac {1}{4}} x}\right )+\frac {\left (x^{2} \left (a \,x^{2}+b \right )\right )^{\frac {1}{4}} \left (b \left (x^{2}-32\right ) a^{\frac {3}{4}}+4 a^{\frac {7}{4}} x^{4}\right )}{16}}{a^{\frac {3}{4}} x}\)	\(128\)

[In]

int((a*x^4+b)*(a*x^4+b*x^2)^(1/4)/x^2,x,method=_RETURNVERBOSE)

[Out]

1/a^(3/4)*(1/2*b*x*(a-3/32*b)*ln((-a^(1/4)*x-(x^2*(a*x^2+b))^(1/4))/(a^(1/4)*x-(x^2*(a*x^2+b))^(1/4)))+b*x*(a-
3/32*b)*arctan(1/a^(1/4)*(x^2*(a*x^2+b))^(1/4)/x)+1/16*(x^2*(a*x^2+b))^(1/4)*(b*(x^2-32)*a^(3/4)+4*a^(7/4)*x^4
))/x

Fricas [F(-1)]

Timed out. \[ \int \frac {\left (b+a x^4\right ) \sqrt [4]{b x^2+a x^4}}{x^2} \, dx=\text {Timed out} \]

[In]

integrate((a*x^4+b)*(a*x^4+b*x^2)^(1/4)/x^2,x, algorithm="fricas")

[Out]

Timed out

Sympy [F]

\[ \int \frac {\left (b+a x^4\right ) \sqrt [4]{b x^2+a x^4}}{x^2} \, dx=\int \frac {\sqrt [4]{x^{2} \left (a x^{2} + b\right )} \left (a x^{4} + b\right )}{x^{2}}\, dx \]

[In]

integrate((a*x**4+b)*(a*x**4+b*x**2)**(1/4)/x**2,x)

[Out]

Integral((x**2*(a*x**2 + b))**(1/4)*(a*x**4 + b)/x**2, x)

Maxima [F]

\[ \int \frac {\left (b+a x^4\right ) \sqrt [4]{b x^2+a x^4}}{x^2} \, dx=\int { \frac {{\left (a x^{4} + b x^{2}\right )}^{\frac {1}{4}} {\left (a x^{4} + b\right )}}{x^{2}} \,d x } \]

[In]

integrate((a*x^4+b)*(a*x^4+b*x^2)^(1/4)/x^2,x, algorithm="maxima")

[Out]

integrate((a*x^4 + b*x^2)^(1/4)*(a*x^4 + b)/x^2, x)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 277 vs. \(2 (102) = 204\).

Time = 0.31 (sec) , antiderivative size = 277, normalized size of antiderivative = 2.27 \[ \int \frac {\left (b+a x^4\right ) \sqrt [4]{b x^2+a x^4}}{x^2} \, dx=\frac {\frac {8 \, {\left ({\left (a + \frac {b}{x^{2}}\right )}^{\frac {5}{4}} b^{3} + 3 \, {\left (a + \frac {b}{x^{2}}\right )}^{\frac {1}{4}} a b^{3}\right )} x^{4}}{b^{2}} - 256 \, {\left (a + \frac {b}{x^{2}}\right )}^{\frac {1}{4}} b^{2} - \frac {2 \, \sqrt {2} {\left (32 \, a b^{2} - 3 \, b^{3}\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} + 2 \, {\left (a + \frac {b}{x^{2}}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{\left (-a\right )^{\frac {3}{4}}} - \frac {2 \, \sqrt {2} {\left (32 \, a b^{2} - 3 \, b^{3}\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} - 2 \, {\left (a + \frac {b}{x^{2}}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{\left (-a\right )^{\frac {3}{4}}} - \frac {\sqrt {2} {\left (32 \, a b^{2} - 3 \, b^{3}\right )} \log \left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} {\left (a + \frac {b}{x^{2}}\right )}^{\frac {1}{4}} + \sqrt {-a} + \sqrt {a + \frac {b}{x^{2}}}\right )}{\left (-a\right )^{\frac {3}{4}}} + \frac {\sqrt {2} {\left (32 \, a b^{2} - 3 \, b^{3}\right )} \log \left (-\sqrt {2} \left (-a\right )^{\frac {1}{4}} {\left (a + \frac {b}{x^{2}}\right )}^{\frac {1}{4}} + \sqrt {-a} + \sqrt {a + \frac {b}{x^{2}}}\right )}{\left (-a\right )^{\frac {3}{4}}}}{128 \, b} \]

[In]

integrate((a*x^4+b)*(a*x^4+b*x^2)^(1/4)/x^2,x, algorithm="giac")

[Out]

1/128*(8*((a + b/x^2)^(5/4)*b^3 + 3*(a + b/x^2)^(1/4)*a*b^3)*x^4/b^2 - 256*(a + b/x^2)^(1/4)*b^2 - 2*sqrt(2)*(
32*a*b^2 - 3*b^3)*arctan(1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) + 2*(a + b/x^2)^(1/4))/(-a)^(1/4))/(-a)^(3/4) - 2*sqr
t(2)*(32*a*b^2 - 3*b^3)*arctan(-1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) - 2*(a + b/x^2)^(1/4))/(-a)^(1/4))/(-a)^(3/4)
- sqrt(2)*(32*a*b^2 - 3*b^3)*log(sqrt(2)*(-a)^(1/4)*(a + b/x^2)^(1/4) + sqrt(-a) + sqrt(a + b/x^2))/(-a)^(3/4)
+ sqrt(2)*(32*a*b^2 - 3*b^3)*log(-sqrt(2)*(-a)^(1/4)*(a + b/x^2)^(1/4) + sqrt(-a) + sqrt(a + b/x^2))/(-a)^(3/
4))/b

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (b+a x^4\right ) \sqrt [4]{b x^2+a x^4}}{x^2} \, dx=\int \frac {\left (a\,x^4+b\right )\,{\left (a\,x^4+b\,x^2\right )}^{1/4}}{x^2} \,d x \]

[In]

int(((b + a*x^4)*(a*x^4 + b*x^2)^(1/4))/x^2,x)

[Out]

int(((b + a*x^4)*(a*x^4 + b*x^2)^(1/4))/x^2, x)