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

Optimal. Leaf size=211 \[ -\frac {1}{2} \text {RootSum}\left [\text {$\#$1}^8-\text {$\#$1}^4 a+b\& ,\frac {\text {$\#$1}^4 a^2 \log \left (\sqrt [4]{a x^4-b x^2}-\text {$\#$1} x\right )+\text {$\#$1}^4 \left (-a^2\right ) \log (x)-\text {$\#$1}^4 b \log \left (\sqrt [4]{a x^4-b x^2}-\text {$\#$1} x\right )+\text {$\#$1}^4 b \log (x)-a b \log \left (\sqrt [4]{a x^4-b x^2}-\text {$\#$1} x\right )+a b \log (x)}{\text {$\#$1}^3 a-2 \text {$\#$1}^7}\& \right ]+a^{5/4} \left (-\tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4-b x^2}}\right )\right )+a^{5/4} \tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4-b x^2}}\right ) \]

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Rubi [A]  time = 0.54, antiderivative size = 200, normalized size of antiderivative = 0.95, number of steps used = 8, number of rules used = 5, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.128, Rules used = {2056, 1269, 1528, 511, 510} \begin {gather*} \frac {4 b x \sqrt [4]{a x^4-b x^2} F_1\left (\frac {3}{4};1,-\frac {5}{4};\frac {7}{4};\frac {2 x^2}{a-\sqrt {a^2-4 b}},\frac {a x^2}{b}\right )}{3 \left (-a \sqrt {a^2-4 b}+a^2-4 b\right ) \sqrt [4]{1-\frac {a x^2}{b}}}+\frac {4 b x \sqrt [4]{a x^4-b x^2} F_1\left (\frac {3}{4};1,-\frac {5}{4};\frac {7}{4};\frac {2 x^2}{a+\sqrt {a^2-4 b}},\frac {a x^2}{b}\right )}{3 \left (a \sqrt {a^2-4 b}+a^2-4 b\right ) \sqrt [4]{1-\frac {a x^2}{b}}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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 1269

Int[((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> With
[{k = Denominator[m]}, Dist[k/f, Subst[Int[x^(k*(m + 1) - 1)*(d + (e*x^(2*k))/f^2)^q*(a + (b*x^(2*k))/f^k + (c
*x^(4*k))/f^4)^p, x], x, (f*x)^(1/k)], x]] /; FreeQ[{a, b, c, d, e, f, p, q}, x] && NeQ[b^2 - 4*a*c, 0] && Fra
ctionQ[m] && IntegerQ[p]

Rule 1528

Int[(((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(n_))^(q_))/((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_)), x_Symbol]
:> Int[ExpandIntegrand[(d + e*x^n)^q, (f*x)^m/(a + b*x^n + c*x^(2*n)), x], x] /; FreeQ[{a, b, c, d, e, f, q,
n}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && IGtQ[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]

Rubi steps

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

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Mathematica [F]  time = 0.09, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (-b+a x^2\right ) \sqrt [4]{-b x^2+a x^4}}{b-a x^2+x^4} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.00, size = 212, normalized size = 1.00 \begin {gather*} -a^{5/4} \tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b x^2+a x^4}}\right )+a^{5/4} \tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b x^2+a x^4}}\right )-\frac {1}{2} \text {RootSum}\left [b-a \text {$\#$1}^4+\text {$\#$1}^8\&,\frac {-a b \log (x)+a b \log \left (\sqrt [4]{-b x^2+a x^4}-x \text {$\#$1}\right )+a^2 \log (x) \text {$\#$1}^4-b \log (x) \text {$\#$1}^4-a^2 \log \left (\sqrt [4]{-b x^2+a x^4}-x \text {$\#$1}\right ) \text {$\#$1}^4+b \log \left (\sqrt [4]{-b x^2+a x^4}-x \text {$\#$1}\right ) \text {$\#$1}^4}{-a \text {$\#$1}^3+2 \text {$\#$1}^7}\&\right ] \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^(5/4)*ArcTan[(a^(1/4)*x)/(-(b*x^2) + a*x^4)^(1/4)]) + a^(5/4)*ArcTanh[(a^(1/4)*x)/(-(b*x^2) + a*x^4)^(1/4)
] - RootSum[b - a*#1^4 + #1^8 & , (-(a*b*Log[x]) + a*b*Log[(-(b*x^2) + a*x^4)^(1/4) - x*#1] + a^2*Log[x]*#1^4
- b*Log[x]*#1^4 - a^2*Log[(-(b*x^2) + a*x^4)^(1/4) - x*#1]*#1^4 + b*Log[(-(b*x^2) + a*x^4)^(1/4) - x*#1]*#1^4)
/(-(a*#1^3) + 2*#1^7) & ]/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^2-b)*(a*x^4-b*x^2)^(1/4)/(x^4-a*x^2+b),x, algorithm="fricas")

[Out]

Timed out

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, need to choose a branch for the root of a polynomial with parameters. This might be wrong.The choice was
done assuming [a,b]=[8,-70]Warning, need to choose a branch for the root of a polynomial with parameters. This
might be wrong.The choice was done assuming [a,b]=[24,84]Evaluation time: 2.42Unable to convert to real 1/4 E
rror: Bad Argument Value

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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