3.8.40 \(\int \frac {x^2}{(b+a x^4)^{3/4}} \, dx\)

Optimal. Leaf size=57 \[ \frac {\tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4+b}}\right )}{2 a^{3/4}}-\frac {\tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4+b}}\right )}{2 a^{3/4}} \]

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Rubi [A]  time = 0.02, antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {331, 298, 203, 206} \begin {gather*} \frac {\tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4+b}}\right )}{2 a^{3/4}}-\frac {\tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4+b}}\right )}{2 a^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2/(b + a*x^4)^(3/4),x]

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 50, normalized size = 0.88 \begin {gather*} \frac {\tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4+b}}\right )-\tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4+b}}\right )}{2 a^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2/(b + a*x^4)^(3/4),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.29, size = 57, normalized size = 1.00 \begin {gather*} -\frac {\tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right )}{2 a^{3/4}}+\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right )}{2 a^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^2/(b + a*x^4)^(3/4),x]

[Out]

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

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fricas [B]  time = 0.48, size = 134, normalized size = 2.35 \begin {gather*} -\frac {1}{a^{3}}^{\frac {1}{4}} \arctan \left (\frac {a^{2} x \sqrt {\frac {a^{2} \sqrt {\frac {1}{a^{3}}} x^{2} + \sqrt {a x^{4} + b}}{x^{2}}} \frac {1}{a^{3}}^{\frac {3}{4}} - {\left (a x^{4} + b\right )}^{\frac {1}{4}} a^{2} \frac {1}{a^{3}}^{\frac {3}{4}}}{x}\right ) + \frac {1}{4} \, \frac {1}{a^{3}}^{\frac {1}{4}} \log \left (\frac {a \frac {1}{a^{3}}^{\frac {1}{4}} x + {\left (a x^{4} + b\right )}^{\frac {1}{4}}}{x}\right ) - \frac {1}{4} \, \frac {1}{a^{3}}^{\frac {1}{4}} \log \left (-\frac {a \frac {1}{a^{3}}^{\frac {1}{4}} x - {\left (a x^{4} + b\right )}^{\frac {1}{4}}}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(a*x^4+b)^(3/4),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{2}}{{\left (a x^{4} + b\right )}^{\frac {3}{4}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(a*x^4+b)^(3/4),x, algorithm="giac")

[Out]

integrate(x^2/(a*x^4 + b)^(3/4), x)

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maple [F]  time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {x^{2}}{\left (a \,x^{4}+b \right )^{\frac {3}{4}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(a*x^4+b)^(3/4),x)

[Out]

int(x^2/(a*x^4+b)^(3/4),x)

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maxima [A]  time = 0.42, size = 68, normalized size = 1.19 \begin {gather*} \frac {\arctan \left (\frac {{\left (a x^{4} + b\right )}^{\frac {1}{4}}}{a^{\frac {1}{4}} x}\right )}{2 \, a^{\frac {3}{4}}} - \frac {\log \left (-\frac {a^{\frac {1}{4}} - \frac {{\left (a x^{4} + b\right )}^{\frac {1}{4}}}{x}}{a^{\frac {1}{4}} + \frac {{\left (a x^{4} + b\right )}^{\frac {1}{4}}}{x}}\right )}{4 \, a^{\frac {3}{4}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(a*x^4+b)^(3/4),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {x^2}{{\left (a\,x^4+b\right )}^{3/4}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(b + a*x^4)^(3/4),x)

[Out]

int(x^2/(b + a*x^4)^(3/4), x)

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sympy [C]  time = 0.88, size = 37, normalized size = 0.65 \begin {gather*} \frac {x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {a x^{4} e^{i \pi }}{b}} \right )}}{4 b^{\frac {3}{4}} \Gamma \left (\frac {7}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2/(a*x**4+b)**(3/4),x)

[Out]

x**3*gamma(3/4)*hyper((3/4, 3/4), (7/4,), a*x**4*exp_polar(I*pi)/b)/(4*b**(3/4)*gamma(7/4))

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