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

Optimal. Leaf size=212 \[ \frac {\tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x}{\sqrt {a x^4+b}+\sqrt {a} x^2+\sqrt {b}}\right )}{2 \sqrt {2} a^{3/4} b^{3/4}}+\frac {\tanh ^{-1}\left (\frac {\sqrt {6-4 \sqrt {2}} \sqrt [4]{a} \sqrt [4]{b} x}{\sqrt {a x^4+b}+\sqrt {a} x^2+\sqrt {b}}\right )}{4 \sqrt {2} a^{3/4} b^{3/4}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {6+4 \sqrt {2}} \sqrt [4]{a} \sqrt [4]{b} x}{\sqrt {a x^4+b}+\sqrt {a} x^2+\sqrt {b}}\right )}{4 \sqrt {2} a^{3/4} b^{3/4}} \]

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Rubi [A]  time = 0.30, antiderivative size = 97, normalized size of antiderivative = 0.46, number of steps used = 9, number of rules used = 6, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {490, 1211, 220, 1699, 205, 208} \begin {gather*} \frac {\tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x}{\sqrt {a x^4+b}}\right )}{4 \sqrt {2} a^{3/4} b^{3/4}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x}{\sqrt {a x^4+b}}\right )}{4 \sqrt {2} a^{3/4} b^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 205

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

Rule 208

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

Rule 220

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[((1 + q^2*x^2)*Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 490

Int[(x_)^2/(((a_) + (b_.)*(x_)^4)*Sqrt[(c_) + (d_.)*(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)*Sqrt[c + d*x^4]), x], x] - Dist[s/(2*b), In
t[1/((r - s*x^2)*Sqrt[c + d*x^4]), x], x]] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 1211

Int[1/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> Dist[1/(2*d), Int[1/Sqrt[a + c*x^4], x],
x] + Dist[1/(2*d), Int[(d - e*x^2)/((d + e*x^2)*Sqrt[a + c*x^4]), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d
^2 + a*e^2, 0] && EqQ[c*d^2 - a*e^2, 0]

Rule 1699

Int[((A_) + (B_.)*(x_)^2)/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> Dist[A, Subst[Int[1/
(d + 2*a*e*x^2), x], x, x/Sqrt[a + c*x^4]], x] /; FreeQ[{a, c, d, e, A, B}, x] && NeQ[c*d^2 + a*e^2, 0] && EqQ
[c*d^2 - a*e^2, 0] && EqQ[B*d + A*e, 0]

Rubi steps

\begin {align*} \int \frac {x^2}{\left (-b+a x^4\right ) \sqrt {b+a x^4}} \, dx &=-\frac {\int \frac {1}{\left (\sqrt {b}-\sqrt {a} x^2\right ) \sqrt {b+a x^4}} \, dx}{2 \sqrt {a}}+\frac {\int \frac {1}{\left (\sqrt {b}+\sqrt {a} x^2\right ) \sqrt {b+a x^4}} \, dx}{2 \sqrt {a}}\\ &=\frac {\int \frac {\sqrt {b}-\sqrt {a} x^2}{\left (\sqrt {b}+\sqrt {a} x^2\right ) \sqrt {b+a x^4}} \, dx}{4 \sqrt {a} \sqrt {b}}-\frac {\int \frac {\sqrt {b}+\sqrt {a} x^2}{\left (\sqrt {b}-\sqrt {a} x^2\right ) \sqrt {b+a x^4}} \, dx}{4 \sqrt {a} \sqrt {b}}\\ &=-\frac {\operatorname {Subst}\left (\int \frac {1}{\sqrt {b}-2 \sqrt {a} b x^2} \, dx,x,\frac {x}{\sqrt {b+a x^4}}\right )}{4 \sqrt {a}}+\frac {\operatorname {Subst}\left (\int \frac {1}{\sqrt {b}+2 \sqrt {a} b x^2} \, dx,x,\frac {x}{\sqrt {b+a x^4}}\right )}{4 \sqrt {a}}\\ &=\frac {\tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x}{\sqrt {b+a x^4}}\right )}{4 \sqrt {2} a^{3/4} b^{3/4}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x}{\sqrt {b+a x^4}}\right )}{4 \sqrt {2} a^{3/4} b^{3/4}}\\ \end {align*}

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Mathematica [C]  time = 0.05, size = 64, normalized size = 0.30 \begin {gather*} -\frac {x^3 \sqrt {\frac {a x^4+b}{b}} F_1\left (\frac {3}{4};\frac {1}{2},1;\frac {7}{4};-\frac {a x^4}{b},\frac {a x^4}{b}\right )}{3 b \sqrt {a x^4+b}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-1/3*(x^3*Sqrt[(b + a*x^4)/b]*AppellF1[3/4, 1/2, 1, 7/4, -((a*x^4)/b), (a*x^4)/b])/(b*Sqrt[b + a*x^4])

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IntegrateAlgebraic [A]  time = 0.49, size = 97, normalized size = 0.46 \begin {gather*} \frac {\tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x}{\sqrt {b+a x^4}}\right )}{4 \sqrt {2} a^{3/4} b^{3/4}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x}{\sqrt {b+a x^4}}\right )}{4 \sqrt {2} a^{3/4} b^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 1.53, size = 288, normalized size = 1.36 \begin {gather*} -\frac {1}{2} \, \left (\frac {1}{4}\right )^{\frac {1}{4}} \left (\frac {1}{a^{3} b^{3}}\right )^{\frac {1}{4}} \arctan \left (\frac {2 \, \left (\frac {1}{4}\right )^{\frac {3}{4}} \sqrt {a x^{4} + b} a^{2} b^{2} \left (\frac {1}{a^{3} b^{3}}\right )^{\frac {3}{4}} - \frac {2 \, \left (\frac {1}{4}\right )^{\frac {3}{4}} a^{3} b^{2} x^{2} \left (\frac {1}{a^{3} b^{3}}\right )^{\frac {3}{4}} + \left (\frac {1}{4}\right )^{\frac {1}{4}} a b \left (\frac {1}{a^{3} b^{3}}\right )^{\frac {1}{4}}}{\sqrt {a}}}{x}\right ) - \frac {1}{8} \, \left (\frac {1}{4}\right )^{\frac {1}{4}} \left (\frac {1}{a^{3} b^{3}}\right )^{\frac {1}{4}} \log \left (\frac {4 \, \left (\frac {1}{4}\right )^{\frac {3}{4}} a^{2} b^{3} x \left (\frac {1}{a^{3} b^{3}}\right )^{\frac {3}{4}} + 2 \, \left (\frac {1}{4}\right )^{\frac {1}{4}} a b x^{3} \left (\frac {1}{a^{3} b^{3}}\right )^{\frac {1}{4}} + \sqrt {a x^{4} + b} {\left (a b^{2} \sqrt {\frac {1}{a^{3} b^{3}}} + x^{2}\right )}}{a x^{4} - b}\right ) + \frac {1}{8} \, \left (\frac {1}{4}\right )^{\frac {1}{4}} \left (\frac {1}{a^{3} b^{3}}\right )^{\frac {1}{4}} \log \left (-\frac {4 \, \left (\frac {1}{4}\right )^{\frac {3}{4}} a^{2} b^{3} x \left (\frac {1}{a^{3} b^{3}}\right )^{\frac {3}{4}} + 2 \, \left (\frac {1}{4}\right )^{\frac {1}{4}} a b x^{3} \left (\frac {1}{a^{3} b^{3}}\right )^{\frac {1}{4}} - \sqrt {a x^{4} + b} {\left (a b^{2} \sqrt {\frac {1}{a^{3} b^{3}}} + x^{2}\right )}}{a x^{4} - b}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.19, size = 108, normalized size = 0.51

method	result	size
default	\(\frac {\left (-\frac {\left (a b \right )^{\frac {1}{4}} \ln \left (\frac {\frac {\sqrt {a \,x^{4}+b}\, \sqrt {2}}{2 x}+\left (a b \right )^{\frac {1}{4}}}{\frac {\sqrt {a \,x^{4}+b}\, \sqrt {2}}{2 x}-\left (a b \right )^{\frac {1}{4}}}\right )}{8 a b}-\frac {\left (a b \right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {a \,x^{4}+b}\, \sqrt {2}}{2 x \left (a b \right )^{\frac {1}{4}}}\right )}{4 a b}\right ) \sqrt {2}}{2}\)	\(108\)
elliptic	\(\frac {\left (-\frac {\left (a b \right )^{\frac {1}{4}} \ln \left (\frac {\frac {\sqrt {a \,x^{4}+b}\, \sqrt {2}}{2 x}+\left (a b \right )^{\frac {1}{4}}}{\frac {\sqrt {a \,x^{4}+b}\, \sqrt {2}}{2 x}-\left (a b \right )^{\frac {1}{4}}}\right )}{8 a b}-\frac {\left (a b \right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {a \,x^{4}+b}\, \sqrt {2}}{2 x \left (a b \right )^{\frac {1}{4}}}\right )}{4 a b}\right ) \sqrt {2}}{2}\)	\(108\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(-1/8*(a*b)^(1/4)/a/b*ln((1/2*(a*x^4+b)^(1/2)*2^(1/2)/x+(a*b)^(1/4))/(1/2*(a*x^4+b)^(1/2)*2^(1/2)/x-(a*b)^
(1/4)))-1/4*(a*b)^(1/4)/a/b*arctan(1/2*(a*x^4+b)^(1/2)*2^(1/2)/x/(a*b)^(1/4)))*2^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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