3.16.32 \(\int \frac {(-b+a x^4) \sqrt [4]{-b x^2+a x^4}}{x^4 (b+a x^4)} \, dx\)

Optimal. Leaf size=106 \[ \frac {1}{2} a \text {RootSum}\left [\text {$\#$1}^8-2 \text {$\#$1}^4 a+a^2+a b\& ,\frac {\text {$\#$1} \log \left (\sqrt [4]{a x^4-b x^2}-\text {$\#$1} x\right )-\text {$\#$1} \log (x)}{\text {$\#$1}^4-a}\& \right ]-\frac {2 \left (a x^2-b\right ) \sqrt [4]{a x^4-b x^2}}{5 b x^3} \]

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Rubi [B]  time = 2.12, antiderivative size = 236, normalized size of antiderivative = 2.23, number of steps used = 17, number of rules used = 9, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.237, Rules used = {2056, 6725, 264, 1270, 1521, 271, 1529, 511, 510} \begin {gather*} \frac {2 a x \sqrt [4]{a x^4-b x^2} F_1\left (\frac {3}{4};1,-\frac {1}{4};\frac {7}{4};-\frac {\sqrt {-a} x^2}{\sqrt {b}},\frac {a x^2}{b}\right )}{3 b \sqrt [4]{1-\frac {a x^2}{b}}}+\frac {2 a x \sqrt [4]{a x^4-b x^2} F_1\left (\frac {3}{4};1,-\frac {1}{4};\frac {7}{4};\frac {\sqrt {-a} x^2}{\sqrt {b}},\frac {a x^2}{b}\right )}{3 b \sqrt [4]{1-\frac {a x^2}{b}}}-\frac {4 a \sqrt [4]{a x^4-b x^2}}{5 b x}-\frac {2 \left (b-a x^2\right ) \sqrt [4]{a x^4-b x^2}}{5 b x^3}+\frac {4 \sqrt [4]{a x^4-b x^2}}{5 x^3} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

Rule 264

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

Rule 271

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

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 1270

Int[((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_), x_Symbol] :> With[{k = Denominat
or[m]}, Dist[k/f, Subst[Int[x^(k*(m + 1) - 1)*(d + (e*x^(2*k))/f)^q*(a + (c*x^(4*k))/f)^p, x], x, (f*x)^(1/k)]
, x]] /; FreeQ[{a, c, d, e, f, p, q}, x] && FractionQ[m] && IntegerQ[p]

Rule 1521

Int[(((f_.)*(x_))^(m_)*((d_.) + (e_.)*(x_)^(n_))^(q_))/((a_) + (c_.)*(x_)^(n2_.)), x_Symbol] :> Dist[d/a, Int[
(f*x)^m*(d + e*x^n)^(q - 1), x], x] + Dist[1/(a*f^n), Int[((f*x)^(m + n)*(d + e*x^n)^(q - 1)*Simp[a*e - c*d*x^
n, x])/(a + c*x^(2*n)), x], x] /; FreeQ[{a, c, d, e, f}, x] && EqQ[n2, 2*n] && IGtQ[n, 0] &&  !IntegerQ[q] &&
GtQ[q, 0] && LtQ[m, 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 2056

Int[(u_.)*(P_)^(p_.), x_Symbol] :> With[{m = MinimumMonomialExponent[P, x]}, Dist[P^FracPart[p]/(x^(m*FracPart
[p])*Distrib[1/x^m, P]^FracPart[p]), Int[u*x^(m*p)*Distrib[1/x^m, P]^p, x], x]] /; FreeQ[p, x] &&  !IntegerQ[p
] && SumQ[P] && EveryQ[BinomialQ[#1, x] & , P] &&  !PolyQ[P, x, 2]

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 {\left (-b+a x^4\right ) \sqrt [4]{-b x^2+a x^4}}{x^4 \left (b+a x^4\right )} \, dx &=\frac {\sqrt [4]{-b x^2+a x^4} \int \frac {\sqrt [4]{-b+a x^2} \left (-b+a x^4\right )}{x^{7/2} \left (b+a x^4\right )} \, dx}{\sqrt {x} \sqrt [4]{-b+a x^2}}\\ &=\frac {\sqrt [4]{-b x^2+a x^4} \int \left (\frac {\sqrt [4]{-b+a x^2}}{x^{7/2}}-\frac {2 b \sqrt [4]{-b+a x^2}}{x^{7/2} \left (b+a x^4\right )}\right ) \, dx}{\sqrt {x} \sqrt [4]{-b+a x^2}}\\ &=\frac {\sqrt [4]{-b x^2+a x^4} \int \frac {\sqrt [4]{-b+a x^2}}{x^{7/2}} \, dx}{\sqrt {x} \sqrt [4]{-b+a x^2}}-\frac {\left (2 b \sqrt [4]{-b x^2+a x^4}\right ) \int \frac {\sqrt [4]{-b+a x^2}}{x^{7/2} \left (b+a x^4\right )} \, dx}{\sqrt {x} \sqrt [4]{-b+a x^2}}\\ &=-\frac {2 \left (b-a x^2\right ) \sqrt [4]{-b x^2+a x^4}}{5 b x^3}-\frac {\left (4 b \sqrt [4]{-b x^2+a x^4}\right ) \operatorname {Subst}\left (\int \frac {\sqrt [4]{-b+a x^4}}{x^6 \left (b+a x^8\right )} \, dx,x,\sqrt {x}\right )}{\sqrt {x} \sqrt [4]{-b+a x^2}}\\ &=-\frac {2 \left (b-a x^2\right ) \sqrt [4]{-b x^2+a x^4}}{5 b x^3}-\frac {\left (4 \sqrt [4]{-b x^2+a x^4}\right ) \operatorname {Subst}\left (\int \frac {a b+a b x^4}{x^2 \left (-b+a x^4\right )^{3/4} \left (b+a x^8\right )} \, dx,x,\sqrt {x}\right )}{\sqrt {x} \sqrt [4]{-b+a x^2}}+\frac {\left (4 b \sqrt [4]{-b x^2+a x^4}\right ) \operatorname {Subst}\left (\int \frac {1}{x^6 \left (-b+a x^4\right )^{3/4}} \, dx,x,\sqrt {x}\right )}{\sqrt {x} \sqrt [4]{-b+a x^2}}\\ &=\frac {4 \sqrt [4]{-b x^2+a x^4}}{5 x^3}-\frac {2 \left (b-a x^2\right ) \sqrt [4]{-b x^2+a x^4}}{5 b x^3}-\frac {\left (4 \sqrt [4]{-b x^2+a x^4}\right ) \operatorname {Subst}\left (\int \left (\frac {a}{x^2 \left (-b+a x^4\right )^{3/4}}-\frac {a x^2 \sqrt [4]{-b+a x^4}}{b+a x^8}\right ) \, dx,x,\sqrt {x}\right )}{\sqrt {x} \sqrt [4]{-b+a x^2}}+\frac {\left (16 a \sqrt [4]{-b x^2+a x^4}\right ) \operatorname {Subst}\left (\int \frac {1}{x^2 \left (-b+a x^4\right )^{3/4}} \, dx,x,\sqrt {x}\right )}{5 \sqrt {x} \sqrt [4]{-b+a x^2}}\\ &=\frac {4 \sqrt [4]{-b x^2+a x^4}}{5 x^3}+\frac {16 a \sqrt [4]{-b x^2+a x^4}}{5 b x}-\frac {2 \left (b-a x^2\right ) \sqrt [4]{-b x^2+a x^4}}{5 b x^3}-\frac {\left (4 a \sqrt [4]{-b x^2+a x^4}\right ) \operatorname {Subst}\left (\int \frac {1}{x^2 \left (-b+a x^4\right )^{3/4}} \, dx,x,\sqrt {x}\right )}{\sqrt {x} \sqrt [4]{-b+a x^2}}+\frac {\left (4 a \sqrt [4]{-b x^2+a x^4}\right ) \operatorname {Subst}\left (\int \frac {x^2 \sqrt [4]{-b+a x^4}}{b+a x^8} \, dx,x,\sqrt {x}\right )}{\sqrt {x} \sqrt [4]{-b+a x^2}}\\ &=\frac {4 \sqrt [4]{-b x^2+a x^4}}{5 x^3}-\frac {4 a \sqrt [4]{-b x^2+a x^4}}{5 b x}-\frac {2 \left (b-a x^2\right ) \sqrt [4]{-b x^2+a x^4}}{5 b x^3}+\frac {\left (4 a \sqrt [4]{-b x^2+a x^4}\right ) \operatorname {Subst}\left (\int \left (-\frac {a x^2 \sqrt [4]{-b+a x^4}}{2 \sqrt {-a} \sqrt {b} \left (\sqrt {-a} \sqrt {b}-a x^4\right )}-\frac {a x^2 \sqrt [4]{-b+a x^4}}{2 \sqrt {-a} \sqrt {b} \left (\sqrt {-a} \sqrt {b}+a x^4\right )}\right ) \, dx,x,\sqrt {x}\right )}{\sqrt {x} \sqrt [4]{-b+a x^2}}\\ &=\frac {4 \sqrt [4]{-b x^2+a x^4}}{5 x^3}-\frac {4 a \sqrt [4]{-b x^2+a x^4}}{5 b x}-\frac {2 \left (b-a x^2\right ) \sqrt [4]{-b x^2+a x^4}}{5 b x^3}+\frac {\left (2 \sqrt {-a} a \sqrt [4]{-b x^2+a x^4}\right ) \operatorname {Subst}\left (\int \frac {x^2 \sqrt [4]{-b+a x^4}}{\sqrt {-a} \sqrt {b}-a x^4} \, dx,x,\sqrt {x}\right )}{\sqrt {b} \sqrt {x} \sqrt [4]{-b+a x^2}}+\frac {\left (2 \sqrt {-a} a \sqrt [4]{-b x^2+a x^4}\right ) \operatorname {Subst}\left (\int \frac {x^2 \sqrt [4]{-b+a x^4}}{\sqrt {-a} \sqrt {b}+a x^4} \, dx,x,\sqrt {x}\right )}{\sqrt {b} \sqrt {x} \sqrt [4]{-b+a x^2}}\\ &=\frac {4 \sqrt [4]{-b x^2+a x^4}}{5 x^3}-\frac {4 a \sqrt [4]{-b x^2+a x^4}}{5 b x}-\frac {2 \left (b-a x^2\right ) \sqrt [4]{-b x^2+a x^4}}{5 b x^3}+\frac {\left (2 \sqrt {-a} a \sqrt [4]{-b x^2+a x^4}\right ) \operatorname {Subst}\left (\int \frac {x^2 \sqrt [4]{1-\frac {a x^4}{b}}}{\sqrt {-a} \sqrt {b}-a x^4} \, dx,x,\sqrt {x}\right )}{\sqrt {b} \sqrt {x} \sqrt [4]{1-\frac {a x^2}{b}}}+\frac {\left (2 \sqrt {-a} a \sqrt [4]{-b x^2+a x^4}\right ) \operatorname {Subst}\left (\int \frac {x^2 \sqrt [4]{1-\frac {a x^4}{b}}}{\sqrt {-a} \sqrt {b}+a x^4} \, dx,x,\sqrt {x}\right )}{\sqrt {b} \sqrt {x} \sqrt [4]{1-\frac {a x^2}{b}}}\\ &=\frac {4 \sqrt [4]{-b x^2+a x^4}}{5 x^3}-\frac {4 a \sqrt [4]{-b x^2+a x^4}}{5 b x}-\frac {2 \left (b-a x^2\right ) \sqrt [4]{-b x^2+a x^4}}{5 b x^3}+\frac {2 a x \sqrt [4]{-b x^2+a x^4} F_1\left (\frac {3}{4};1,-\frac {1}{4};\frac {7}{4};-\frac {\sqrt {-a} x^2}{\sqrt {b}},\frac {a x^2}{b}\right )}{3 b \sqrt [4]{1-\frac {a x^2}{b}}}+\frac {2 a x \sqrt [4]{-b x^2+a x^4} F_1\left (\frac {3}{4};1,-\frac {1}{4};\frac {7}{4};\frac {\sqrt {-a} x^2}{\sqrt {b}},\frac {a x^2}{b}\right )}{3 b \sqrt [4]{1-\frac {a x^2}{b}}}\\ \end {align*}

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Mathematica [F]  time = 2.29, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (-b+a x^4\right ) \sqrt [4]{-b x^2+a x^4}}{x^4 \left (b+a x^4\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 1.79, size = 106, normalized size = 1.00 \begin {gather*} -\frac {2 \left (-b+a x^2\right ) \sqrt [4]{-b x^2+a x^4}}{5 b x^3}+\frac {1}{2} a \text {RootSum}\left [a^2+a b-2 a \text {$\#$1}^4+\text {$\#$1}^8\&,\frac {-\log (x) \text {$\#$1}+\log \left (\sqrt [4]{-b x^2+a x^4}-x \text {$\#$1}\right ) \text {$\#$1}}{-a+\text {$\#$1}^4}\&\right ] \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(-b + a*x^2)*(-(b*x^2) + a*x^4)^(1/4))/(5*b*x^3) + (a*RootSum[a^2 + a*b - 2*a*#1^4 + #1^8 & , (-(Log[x]*#1
) + Log[(-(b*x^2) + a*x^4)^(1/4) - x*#1]*#1)/(-a + #1^4) & ])/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((a*x^4-b)*(a*x^4-b*x^2)^(1/4)/x^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 {{\left (a x^{4} - b x^{2}\right )}^{\frac {1}{4}} {\left (a x^{4} - b\right )}}{{\left (a x^{4} + b\right )} x^{4}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x^4-b)*(a*x^4-b*x^2)^(1/4)/x^4/(a*x^4+b),x, algorithm="giac")

[Out]

integrate((a*x^4 - b*x^2)^(1/4)*(a*x^4 - b)/((a*x^4 + b)*x^4), x)

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maple [F]  time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {\left (a \,x^{4}-b \right ) \left (a \,x^{4}-b \,x^{2}\right )^{\frac {1}{4}}}{x^{4} \left (a \,x^{4}+b \right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x^4-b)*(a*x^4-b*x^2)^(1/4)/x^4/(a*x^4+b),x)

[Out]

int((a*x^4-b)*(a*x^4-b*x^2)^(1/4)/x^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 {{\left (a x^{4} - b x^{2}\right )}^{\frac {1}{4}} {\left (a x^{4} - b\right )}}{{\left (a x^{4} + b\right )} x^{4}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x^4-b)*(a*x^4-b*x^2)^(1/4)/x^4/(a*x^4+b),x, algorithm="maxima")

[Out]

integrate((a*x^4 - b*x^2)^(1/4)*(a*x^4 - b)/((a*x^4 + b)*x^4), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((b - a*x^4)*(a*x^4 - b*x^2)^(1/4))/(x^4*(b + a*x^4)),x)

[Out]

int(-((b - a*x^4)*(a*x^4 - b*x^2)^(1/4))/(x^4*(b + a*x^4)), x)

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sympy [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*x**4-b)*(a*x**4-b*x**2)**(1/4)/x**4/(a*x**4+b),x)

[Out]

Timed out

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