3.15.99 \(\int \frac {x^2}{(b+a x^4)^{3/4} (-b^2+a^2 x^8)} \, dx\)

Optimal. Leaf size=104 \[ \frac {\tan ^{-1}\left (\frac {\sqrt [4]{2} \sqrt [4]{a} x}{\sqrt [4]{a x^4+b}}\right )}{4\ 2^{3/4} a^{3/4} b^2}-\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{2} \sqrt [4]{a} x}{\sqrt [4]{a x^4+b}}\right )}{4\ 2^{3/4} a^{3/4} b^2}-\frac {x^3}{6 b^2 \left (a x^4+b\right )^{3/4}} \]

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Rubi [C]  time = 4.06, antiderivative size = 122, normalized size of antiderivative = 1.17, number of steps used = 3, number of rules used = 3, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {1479, 511, 510} \begin {gather*} -\frac {4 x^3 \Gamma \left (\frac {7}{4}\right ) \left (11 \left (-4 a^2 x^8-3 a b x^4+7 b^2\right ) \, _2F_1\left (1,1;\frac {11}{4};-\frac {2 a x^4}{b-a x^4}\right )-32 a x^4 \left (a x^4+b\right ) \, _2F_1\left (2,2;\frac {15}{4};-\frac {2 a x^4}{b-a x^4}\right )\right )}{693 b^2 \Gamma \left (\frac {3}{4}\right ) \left (b-a x^4\right )^2 \left (a x^4+b\right )^{3/4}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-4*x^3*Gamma[7/4]*(11*(7*b^2 - 3*a*b*x^4 - 4*a^2*x^8)*Hypergeometric2F1[1, 1, 11/4, (-2*a*x^4)/(b - a*x^4)] -
32*a*x^4*(b + a*x^4)*Hypergeometric2F1[2, 2, 15/4, (-2*a*x^4)/(b - a*x^4)]))/(693*b^2*(b - a*x^4)^2*(b + a*x^
4)^(3/4)*Gamma[3/4])

Rule 510

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(a^p*c^q
*(e*x)^(m + 1)*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, -((b*x^n)/a), -((d*x^n)/c)])/(e*(m + 1)), x] /; Free
Q[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a
, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rule 511

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[(a^IntPa
rt[p]*(a + b*x^n)^FracPart[p])/(1 + (b*x^n)/a)^FracPart[p], Int[(e*x)^m*(1 + (b*x^n)/a)^p*(c + d*x^n)^q, x], x
] /; FreeQ[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] &&  !(IntegerQ[
p] || GtQ[a, 0])

Rule 1479

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(n_))^(q_.)*((a_) + (c_.)*(x_)^(n2_))^(p_.), x_Symbol] :> Int[(f*x)^
m*(d + e*x^n)^(q + p)*(a/d + (c*x^n)/e)^p, x] /; FreeQ[{a, c, d, e, f, q, m, n, q}, x] && EqQ[n2, 2*n] && EqQ[
c*d^2 + a*e^2, 0] && IntegerQ[p]

Rubi steps

\begin {align*} \int \frac {x^2}{\left (b+a x^4\right )^{3/4} \left (-b^2+a^2 x^8\right )} \, dx &=\int \frac {x^2}{\left (-b+a x^4\right ) \left (b+a x^4\right )^{7/4}} \, dx\\ &=\frac {\left (1+\frac {a x^4}{b}\right )^{3/4} \int \frac {x^2}{\left (-b+a x^4\right ) \left (1+\frac {a x^4}{b}\right )^{7/4}} \, dx}{b \left (b+a x^4\right )^{3/4}}\\ &=-\frac {4 x^3 \Gamma \left (\frac {7}{4}\right ) \left (11 \left (7 b^2-3 a b x^4-4 a^2 x^8\right ) \, _2F_1\left (1,1;\frac {11}{4};-\frac {2 a x^4}{b-a x^4}\right )-32 a x^4 \left (b+a x^4\right ) \, _2F_1\left (2,2;\frac {15}{4};-\frac {2 a x^4}{b-a x^4}\right )\right )}{693 b^2 \left (b-a x^4\right )^2 \left (b+a x^4\right )^{3/4} \Gamma \left (\frac {3}{4}\right )}\\ \end {align*}

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Mathematica [C]  time = 0.12, size = 93, normalized size = 0.89 \begin {gather*} -\frac {x^3 \left (\left (\frac {a x^4}{b}+1\right )^{3/4} \, _2F_1\left (\frac {3}{4},\frac {3}{4};\frac {7}{4};-\frac {2 a x^4}{b-a x^4}\right )+\left (1-\frac {a x^4}{b}\right )^{3/4}\right )}{6 b^2 \left (a x^4+b\right )^{3/4} \left (1-\frac {a x^4}{b}\right )^{3/4}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-1/6*(x^3*((1 - (a*x^4)/b)^(3/4) + (1 + (a*x^4)/b)^(3/4)*Hypergeometric2F1[3/4, 3/4, 7/4, (-2*a*x^4)/(b - a*x^
4)]))/(b^2*(b + a*x^4)^(3/4)*(1 - (a*x^4)/b)^(3/4))

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IntegrateAlgebraic [A]  time = 0.52, size = 104, normalized size = 1.00 \begin {gather*} -\frac {x^3}{6 b^2 \left (b+a x^4\right )^{3/4}}+\frac {\tan ^{-1}\left (\frac {\sqrt [4]{2} \sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right )}{4\ 2^{3/4} a^{3/4} b^2}-\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{2} \sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right )}{4\ 2^{3/4} a^{3/4} b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/6*x^3/(b^2*(b + a*x^4)^(3/4)) + ArcTan[(2^(1/4)*a^(1/4)*x)/(b + a*x^4)^(1/4)]/(4*2^(3/4)*a^(3/4)*b^2) - Arc
Tanh[(2^(1/4)*a^(1/4)*x)/(b + a*x^4)^(1/4)]/(4*2^(3/4)*a^(3/4)*b^2)

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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^2*x^8-b^2),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^{2} x^{8} - b^{2}\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^2*x^8-b^2),x, algorithm="giac")

[Out]

integrate(x^2/((a^2*x^8 - b^2)*(a*x^4 + b)^(3/4)), x)

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maple [F]  time = 0.04, size = 0, normalized size = 0.00 \[\int \frac {x^{2}}{\left (a \,x^{4}+b \right )^{\frac {3}{4}} \left (a^{2} x^{8}-b^{2}\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(a*x^4+b)^(3/4)/(a^2*x^8-b^2),x)

[Out]

int(x^2/(a*x^4+b)^(3/4)/(a^2*x^8-b^2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{2}}{{\left (a^{2} x^{8} - b^{2}\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^2*x^8-b^2),x, algorithm="maxima")

[Out]

integrate(x^2/((a^2*x^8 - b^2)*(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 (b^2-a^2\,x^8\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^2 - a^2*x^8)*(b + a*x^4)^(3/4)),x)

[Out]

-int(x^2/((b^2 - a^2*x^8)*(b + a*x^4)^(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 ) \left (a x^{4} + b\right )^{\frac {7}{4}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2/(a*x**4+b)**(3/4)/(a**2*x**8-b**2),x)

[Out]

Integral(x**2/((a*x**4 - b)*(a*x**4 + b)**(7/4)), x)

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