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

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

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Rubi [A]  time = 1.24, antiderivative size = 197, normalized size of antiderivative = 1.86, number of steps used = 14, number of rules used = 9, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {2056, 1270, 1521, 271, 264, 6725, 1529, 511, 510} \begin {gather*} -\frac {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^2 \sqrt [4]{\frac {a x^2}{b}+1}}-\frac {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^2 \sqrt [4]{\frac {a x^2}{b}+1}}+\frac {2 a \sqrt [4]{a x^4+b x^2}}{5 b^2 x}+\frac {2 \sqrt [4]{a x^4+b x^2}}{5 b x^3} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2*(b*x^2 + a*x^4)^(1/4))/(5*b*x^3) + (2*a*(b*x^2 + a*x^4)^(1/4))/(5*b^2*x) - (a*x*(b*x^2 + a*x^4)^(1/4)*Appel
lF1[3/4, 1, -1/4, 7/4, -((Sqrt[a]*x^2)/Sqrt[b]), -((a*x^2)/b)])/(3*b^2*(1 + (a*x^2)/b)^(1/4)) - (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^2*(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 {\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}}{x^{7/2} \left (-b+a x^4\right )} \, dx}{\sqrt {x} \sqrt [4]{b+a x^2}}\\ &=\frac {\left (2 \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 {\left (2 \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 {\left (2 \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 )}{b \sqrt {x} \sqrt [4]{b+a x^2}}\\ &=\frac {2 \sqrt [4]{b x^2+a x^4}}{5 b x^3}-\frac {\left (2 \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 )}{b \sqrt {x} \sqrt [4]{b+a x^2}}+\frac {\left (8 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 b \sqrt {x} \sqrt [4]{b+a x^2}}\\ &=\frac {2 \sqrt [4]{b x^2+a x^4}}{5 b x^3}-\frac {8 a \sqrt [4]{b x^2+a x^4}}{5 b^2 x}-\frac {\left (2 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 )}{b \sqrt {x} \sqrt [4]{b+a x^2}}+\frac {\left (2 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 )}{b \sqrt {x} \sqrt [4]{b+a x^2}}\\ &=\frac {2 \sqrt [4]{b x^2+a x^4}}{5 b x^3}+\frac {2 a \sqrt [4]{b x^2+a x^4}}{5 b^2 x}+\frac {\left (2 a \sqrt [4]{b x^2+a x^4}\right ) \operatorname {Subst}\left (\int \left (-\frac {\sqrt {a} x^2 \sqrt [4]{b+a x^4}}{2 \sqrt {b} \left (\sqrt {a} \sqrt {b}-a x^4\right )}-\frac {\sqrt {a} x^2 \sqrt [4]{b+a x^4}}{2 \sqrt {b} \left (\sqrt {a} \sqrt {b}+a x^4\right )}\right ) \, dx,x,\sqrt {x}\right )}{b \sqrt {x} \sqrt [4]{b+a x^2}}\\ &=\frac {2 \sqrt [4]{b x^2+a x^4}}{5 b x^3}+\frac {2 a \sqrt [4]{b x^2+a x^4}}{5 b^2 x}-\frac {\left (a^{3/2} \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 )}{b^{3/2} \sqrt {x} \sqrt [4]{b+a x^2}}-\frac {\left (a^{3/2} \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 )}{b^{3/2} \sqrt {x} \sqrt [4]{b+a x^2}}\\ &=\frac {2 \sqrt [4]{b x^2+a x^4}}{5 b x^3}+\frac {2 a \sqrt [4]{b x^2+a x^4}}{5 b^2 x}-\frac {\left (a^{3/2} \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 )}{b^{3/2} \sqrt {x} \sqrt [4]{1+\frac {a x^2}{b}}}-\frac {\left (a^{3/2} \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 )}{b^{3/2} \sqrt {x} \sqrt [4]{1+\frac {a x^2}{b}}}\\ &=\frac {2 \sqrt [4]{b x^2+a x^4}}{5 b x^3}+\frac {2 a \sqrt [4]{b x^2+a x^4}}{5 b^2 x}-\frac {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^2 \sqrt [4]{1+\frac {a x^2}{b}}}-\frac {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^2 \sqrt [4]{1+\frac {a x^2}{b}}}\\ \end {align*}

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Mathematica [F]  time = 1.49, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\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*x^2 + a*x^4)^(1/4)/(x^4*(-b + a*x^4)),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.00, 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^2 x^3}+\frac {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 ]}{4 b} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(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^2*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) & ])/(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*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 )} x^{4}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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)*x^4), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((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 )} x^{4}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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)*x^4), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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