3.12.84 \(\int \frac {x^2}{(-b+a x^4)^{3/4} (b+a x^4)} \, dx\) [1184]

Optimal. Leaf size=87 \[ -\frac {\text {ArcTan}\left (\frac {\sqrt [4]{2} \sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )}{2\ 2^{3/4} a^{3/4} b}+\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{2} \sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )}{2\ 2^{3/4} a^{3/4} b} \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {508, 304, 209, 212} \begin {gather*} \frac {\tanh ^{-1}\left (\frac {\sqrt [4]{2} \sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )}{2\ 2^{3/4} a^{3/4} b}-\frac {\text {ArcTan}\left (\frac {\sqrt [4]{2} \sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )}{2\ 2^{3/4} a^{3/4} b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 508

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> With[{k = Denominato
r[p]}, Dist[k*(a^(p + (m + 1)/n)/n), Subst[Int[x^(k*((m + 1)/n) - 1)*((c - (b*c - a*d)*x^k)^q/(1 - b*x^k)^(p +
q + (m + 1)/n + 1)), x], x, x^(n/k)/(a + b*x^n)^(1/k)], x]] /; FreeQ[{a, b, c, d}, x] && IGtQ[n, 0] && Ration
alQ[m, p] && IntegersQ[p + (m + 1)/n, q] && LtQ[-1, p, 0]

Rubi steps

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

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Mathematica [A]
time = 0.35, size = 72, normalized size = 0.83 \begin {gather*} \frac {-\text {ArcTan}\left (\frac {\sqrt [4]{2} \sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )+\tanh ^{-1}\left (\frac {\sqrt [4]{2} \sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )}{2\ 2^{3/4} a^{3/4} b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F(-1)] Timed out
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(x^2/(a*x^4-b)^(3/4)/(a*x^4+b),x, algorithm="fricas")

[Out]

Timed out

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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 )^{\frac {3}{4}} \left (a x^{4} + b\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^2}{\left (a\,x^4+b\right )\,{\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)*(a*x^4 - b)^(3/4)),x)

[Out]

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

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