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

Optimal. Leaf size=88 \[ \frac {a^2 \text {RootSum}\left [\text {$\#$1}^8-2 \text {$\#$1}^4 a+2 a^2\& ,\frac {\log \left (\sqrt [4]{a x^4+b}-\text {$\#$1} x\right )-\log (x)}{\text {$\#$1}^3 a-\text {$\#$1}^7}\& \right ]}{4 b}+\frac {\sqrt [4]{a x^4+b}}{b x} \]

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Rubi [B]  time = 0.84, antiderivative size = 226, normalized size of antiderivative = 2.57, number of steps used = 14, number of rules used = 10, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.244, Rules used = {1586, 6725, 277, 331, 298, 203, 206, 1529, 511, 510} \begin {gather*} \frac {a x^3 \sqrt [4]{a x^4+b} F_1\left (\frac {3}{4};1,-\frac {5}{4};\frac {7}{4};-\frac {\sqrt {-a^2} x^4}{b},-\frac {a x^4}{b}\right )}{6 b^2 \sqrt [4]{\frac {a x^4}{b}+1}}+\frac {a x^3 \sqrt [4]{a x^4+b} F_1\left (\frac {3}{4};1,-\frac {5}{4};\frac {7}{4};\frac {\sqrt {-a^2} x^4}{b},-\frac {a x^4}{b}\right )}{6 b^2 \sqrt [4]{\frac {a x^4}{b}+1}}+\frac {\sqrt [4]{a x^4+b}}{b x}+\frac {\sqrt [4]{a} \tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4+b}}\right )}{2 b}-\frac {\sqrt [4]{a} \tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4+b}}\right )}{2 b} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(b + a*x^4)^(1/4)/(b*x) + (a*x^3*(b + a*x^4)^(1/4)*AppellF1[3/4, 1, -5/4, 7/4, -((Sqrt[-a^2]*x^4)/b), -((a*x^4
)/b)])/(6*b^2*(1 + (a*x^4)/b)^(1/4)) + (a*x^3*(b + a*x^4)^(1/4)*AppellF1[3/4, 1, -5/4, 7/4, (Sqrt[-a^2]*x^4)/b
, -((a*x^4)/b)])/(6*b^2*(1 + (a*x^4)/b)^(1/4)) + (a^(1/4)*ArcTan[(a^(1/4)*x)/(b + a*x^4)^(1/4)])/(2*b) - (a^(1
/4)*ArcTanh[(a^(1/4)*x)/(b + a*x^4)^(1/4)])/(2*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 277

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^p)/(c*(m +
1)), x] - Dist[(b*n*p)/(c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] &&
IGtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1] &&  !ILtQ[(m + n*p + n + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, x]

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]

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 1529

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

Rule 1586

Int[(u_.)*(Px_)^(p_.)*(Qx_)^(q_.), x_Symbol] :> Int[u*PolynomialQuotient[Px, Qx, x]^p*Qx^(p + q), x] /; FreeQ[
q, x] && PolyQ[Px, x] && PolyQ[Qx, x] && EqQ[PolynomialRemainder[Px, Qx, x], 0] && IntegerQ[p] && LtQ[p*q, 0]

Rule 6725

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {-b^2+a^2 x^8}{x^2 \left (b+a x^4\right )^{3/4} \left (b^2+a^2 x^8\right )} \, dx &=\int \frac {\left (-b+a x^4\right ) \sqrt [4]{b+a x^4}}{x^2 \left (b^2+a^2 x^8\right )} \, dx\\ &=\int \left (-\frac {\sqrt [4]{b+a x^4}}{b x^2}+\frac {a x^2 \left (b+a x^4\right )^{5/4}}{b \left (b^2+a^2 x^8\right )}\right ) \, dx\\ &=-\frac {\int \frac {\sqrt [4]{b+a x^4}}{x^2} \, dx}{b}+\frac {a \int \frac {x^2 \left (b+a x^4\right )^{5/4}}{b^2+a^2 x^8} \, dx}{b}\\ &=\frac {\sqrt [4]{b+a x^4}}{b x}-\frac {a \int \frac {x^2}{\left (b+a x^4\right )^{3/4}} \, dx}{b}+\frac {a \int \left (-\frac {a^2 x^2 \left (b+a x^4\right )^{5/4}}{2 \sqrt {-a^2} b \left (\sqrt {-a^2} b-a^2 x^4\right )}-\frac {a^2 x^2 \left (b+a x^4\right )^{5/4}}{2 \sqrt {-a^2} b \left (\sqrt {-a^2} b+a^2 x^4\right )}\right ) \, dx}{b}\\ &=\frac {\sqrt [4]{b+a x^4}}{b x}+\frac {\left (a \sqrt {-a^2}\right ) \int \frac {x^2 \left (b+a x^4\right )^{5/4}}{\sqrt {-a^2} b-a^2 x^4} \, dx}{2 b^2}+\frac {\left (a \sqrt {-a^2}\right ) \int \frac {x^2 \left (b+a x^4\right )^{5/4}}{\sqrt {-a^2} b+a^2 x^4} \, dx}{2 b^2}-\frac {a \operatorname {Subst}\left (\int \frac {x^2}{1-a x^4} \, dx,x,\frac {x}{\sqrt [4]{b+a x^4}}\right )}{b}\\ &=\frac {\sqrt [4]{b+a x^4}}{b x}-\frac {\sqrt {a} \operatorname {Subst}\left (\int \frac {1}{1-\sqrt {a} x^2} \, dx,x,\frac {x}{\sqrt [4]{b+a x^4}}\right )}{2 b}+\frac {\sqrt {a} \operatorname {Subst}\left (\int \frac {1}{1+\sqrt {a} x^2} \, dx,x,\frac {x}{\sqrt [4]{b+a x^4}}\right )}{2 b}+\frac {\left (a \sqrt {-a^2} \sqrt [4]{b+a x^4}\right ) \int \frac {x^2 \left (1+\frac {a x^4}{b}\right )^{5/4}}{\sqrt {-a^2} b-a^2 x^4} \, dx}{2 b \sqrt [4]{1+\frac {a x^4}{b}}}+\frac {\left (a \sqrt {-a^2} \sqrt [4]{b+a x^4}\right ) \int \frac {x^2 \left (1+\frac {a x^4}{b}\right )^{5/4}}{\sqrt {-a^2} b+a^2 x^4} \, dx}{2 b \sqrt [4]{1+\frac {a x^4}{b}}}\\ &=\frac {\sqrt [4]{b+a x^4}}{b x}+\frac {a x^3 \sqrt [4]{b+a x^4} F_1\left (\frac {3}{4};1,-\frac {5}{4};\frac {7}{4};-\frac {\sqrt {-a^2} x^4}{b},-\frac {a x^4}{b}\right )}{6 b^2 \sqrt [4]{1+\frac {a x^4}{b}}}+\frac {a x^3 \sqrt [4]{b+a x^4} F_1\left (\frac {3}{4};1,-\frac {5}{4};\frac {7}{4};\frac {\sqrt {-a^2} x^4}{b},-\frac {a x^4}{b}\right )}{6 b^2 \sqrt [4]{1+\frac {a x^4}{b}}}+\frac {\sqrt [4]{a} \tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right )}{2 b}-\frac {\sqrt [4]{a} \tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right )}{2 b}\\ \end {align*}

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Mathematica [F]  time = 0.22, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {-b^2+a^2 x^8}{x^2 \left (b+a x^4\right )^{3/4} \left (b^2+a^2 x^8\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.63, size = 87, normalized size = 0.99 \begin {gather*} \frac {\sqrt [4]{b+a x^4}}{b x}+\frac {a^2 \text {RootSum}\left [2 a^2-2 a \text {$\#$1}^4+\text {$\#$1}^8\&,\frac {\log (x)-\log \left (\sqrt [4]{b+a x^4}-x \text {$\#$1}\right )}{-a \text {$\#$1}^3+\text {$\#$1}^7}\&\right ]}{4 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(b + a*x^4)^(1/4)/(b*x) + (a^2*RootSum[2*a^2 - 2*a*#1^4 + #1^8 & , (Log[x] - Log[(b + a*x^4)^(1/4) - x*#1])/(-
(a*#1^3) + #1^7) & ])/(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((a^2*x^8-b^2)/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 {a^{2} x^{8} - b^{2}}{{\left (a^{2} x^{8} + b^{2}\right )} {\left (a x^{4} + b\right )}^{\frac {3}{4}} x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {a^{2} x^{8}-b^{2}}{x^{2} \left (a \,x^{4}+b \right )^{\frac {3}{4}} \left (a^{2} x^{8}+b^{2}\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((a^2*x^8-b^2)/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 {a^{2} x^{8} - b^{2}}{{\left (a^{2} x^{8} + b^{2}\right )} {\left (a x^{4} + b\right )}^{\frac {3}{4}} x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int -\frac {b^2-a^2\,x^8}{x^2\,\left (a^2\,x^8+b^2\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(-(b^2 - a^2*x^8)/(x^2*(b^2 + a^2*x^8)*(b + a*x^4)^(3/4)),x)

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a*x**4 - b)*(a*x**4 + b)**(1/4)/(x**2*(a**2*x**8 + b**2)), x)

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