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

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

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Rubi [A]  time = 0.06, 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 = {494, 298, 203, 206} \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 {\tan ^{-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} \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 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 494

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 &=\operatorname {Subst}\left (\int \frac {x^2}{b-2 a b x^4} \, dx,x,\frac {x}{\sqrt [4]{-b+a x^4}}\right )\\ &=\frac {\operatorname {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 {\operatorname {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 [C]  time = 0.04, size = 77, normalized size = 0.89 \begin {gather*} \frac {x^3 \left (1-\frac {a x^4}{b}\right )^{3/4} \, _2F_1\left (\frac {3}{4},\frac {3}{4};\frac {7}{4};\frac {2 a x^4}{a x^4+b}\right )}{3 b \left (a x^4-b\right )^{3/4} \left (\frac {a x^4}{b}+1\right )^{3/4}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.54, size = 87, normalized size = 1.00 \begin {gather*} -\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 {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[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)

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

[Out]

Timed out

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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 )} {\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)/(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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maple [F]  time = 0.04, 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*} \int \frac {x^{2}}{{\left (a x^{4} + b\right )} {\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)/(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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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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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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