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

Optimal. Leaf size=122 \[ \frac {\left (32 a b+3 b^2\right ) \tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4+b x^2}}\right )}{32 a^{3/4}}+\frac {\left (-32 a b-3 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4+b x^2}}\right )}{32 a^{3/4}}+\frac {\sqrt [4]{a x^4+b x^2} \left (4 a x^4+b x^2+32 b\right )}{16 x} \]

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Rubi [B]  time = 0.49, antiderivative size = 321, normalized size of antiderivative = 2.63, number of steps used = 17, number of rules used = 10, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {2052, 2020, 2032, 329, 331, 298, 203, 206, 2021, 2024} \begin {gather*} \frac {3 b^2 x^{3/2} \left (a x^2+b\right )^{3/4} \tan ^{-1}\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} \tanh ^{-1}\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 {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 {\sqrt [4]{a} b x^{3/2} \left (a x^2+b\right )^{3/4} \tan ^{-1}\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} \tanh ^{-1}\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}{4} a x^3 \sqrt [4]{a x^4+b x^2} \end {gather*}

Antiderivative was successfully verified.

[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 203

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

Rule 206

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

Rule 298

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 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 331

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 2020

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 2021

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 2024

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 2032

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[(c^IntPart[m]*(c*x)^FracP
art[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 2052

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*} \int \frac {\left (-b+a x^4\right ) \sqrt [4]{b x^2+a x^4}}{x^2} \, dx &=\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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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} \tan ^{-1}\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} \tanh ^{-1}\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 ) \operatorname {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 ) \operatorname {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} \tan ^{-1}\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} \tan ^{-1}\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} \tanh ^{-1}\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} \tanh ^{-1}\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*}

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Mathematica [C]  time = 0.15, size = 108, normalized size = 0.89 \begin {gather*} \frac {\sqrt [4]{x^2 \left (a x^2+b\right )} \left (x^2 \left (\left (a x^2+b\right ) \sqrt [4]{\frac {a x^2}{b}+1}-b \, _2F_1\left (-\frac {1}{4},\frac {3}{4};\frac {7}{4};-\frac {a x^2}{b}\right )\right )+8 b \, _2F_1\left (-\frac {1}{4},-\frac {1}{4};\frac {3}{4};-\frac {a x^2}{b}\right )\right )}{4 x \sqrt [4]{\frac {a x^2}{b}+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((x^2*(b + a*x^2))^(1/4)*(8*b*Hypergeometric2F1[-1/4, -1/4, 3/4, -((a*x^2)/b)] + x^2*((b + a*x^2)*(1 + (a*x^2)
/b)^(1/4) - b*Hypergeometric2F1[-1/4, 3/4, 7/4, -((a*x^2)/b)])))/(4*x*(1 + (a*x^2)/b)^(1/4))

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IntegrateAlgebraic [A]  time = 0.54, size = 122, normalized size = 1.00 \begin {gather*} \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 ) \tan ^{-1}\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 ) \tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b x^2+a x^4}}\right )}{32 a^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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giac [B]  time = 0.38, size = 277, normalized size = 2.27 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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maple [F]  time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {\left (a \,x^{4}-b \right ) \left (a \,x^{4}+b \,x^{2}\right )^{\frac {1}{4}}}{x^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (a x^{4} + b x^{2}\right )}^{\frac {1}{4}} {\left (a x^{4} - b\right )}}{x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} -\int \frac {\left (b-a\,x^4\right )\,{\left (a\,x^4+b\,x^2\right )}^{1/4}}{x^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt [4]{x^{2} \left (a x^{2} + b\right )} \left (a x^{4} - b\right )}{x^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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