3.22.61 \(\int \frac {\sqrt [4]{b x^2+a x^4} (-b+2 a x^4)}{-2 b+a x^4} \, dx\) [2161]

Optimal. Leaf size=158 \[ x \sqrt [4]{b x^2+a x^4}-\frac {b \text {ArcTan}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b x^2+a x^4}}\right )}{2 a^{3/4}}+\frac {b \tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b x^2+a x^4}}\right )}{2 a^{3/4}}+\frac {3}{8} b \text {RootSum}\left [2 a^2-a b-4 a \text {$\#$1}^4+2 \text {$\#$1}^8\& ,\frac {-\log (x) \text {$\#$1}+\log \left (\sqrt [4]{b x^2+a x^4}-x \text {$\#$1}\right ) \text {$\#$1}}{-a+\text {$\#$1}^4}\& \right ] \]

[Out]

Unintegrable

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Rubi [A]
time = 0.70, antiderivative size = 293, normalized size of antiderivative = 1.85, number of steps used = 16, number of rules used = 12, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.324, Rules used = {2081, 6857, 285, 335, 338, 304, 209, 212, 1284, 1543, 525, 524} \begin {gather*} -\frac {b \sqrt [4]{a x^4+b x^2} \text {ArcTan}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{a x^2+b}}\right )}{2 a^{3/4} \sqrt {x} \sqrt [4]{a x^2+b}}+\frac {b \sqrt [4]{a x^4+b x^2} \tanh ^{-1}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{a x^2+b}}\right )}{2 a^{3/4} \sqrt {x} \sqrt [4]{a x^2+b}}-\frac {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 {2} \sqrt {b}},-\frac {a x^2}{b}\right )}{2 \sqrt [4]{\frac {a x^2}{b}+1}}-\frac {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 {2} \sqrt {b}},-\frac {a x^2}{b}\right )}{2 \sqrt [4]{\frac {a x^2}{b}+1}}+x \sqrt [4]{a x^4+b x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

x*(b*x^2 + a*x^4)^(1/4) - (x*(b*x^2 + a*x^4)^(1/4)*AppellF1[3/4, 1, -1/4, 7/4, -((Sqrt[a]*x^2)/(Sqrt[2]*Sqrt[b
])), -((a*x^2)/b)])/(2*(1 + (a*x^2)/b)^(1/4)) - (x*(b*x^2 + a*x^4)^(1/4)*AppellF1[3/4, 1, -1/4, 7/4, (Sqrt[a]*
x^2)/(Sqrt[2]*Sqrt[b]), -((a*x^2)/b)])/(2*(1 + (a*x^2)/b)^(1/4)) - (b*(b*x^2 + a*x^4)^(1/4)*ArcTan[(a^(1/4)*Sq
rt[x])/(b + a*x^2)^(1/4)])/(2*a^(3/4)*Sqrt[x]*(b + a*x^2)^(1/4)) + (b*(b*x^2 + a*x^4)^(1/4)*ArcTanh[(a^(1/4)*S
qrt[x])/(b + a*x^2)^(1/4)])/(2*a^(3/4)*Sqrt[x]*(b + a*x^2)^(1/4))

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 285

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

Rule 304

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 335

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 338

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 524

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)/(e*(m + 1)))*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, (-b)*(x^n/a), (-d)*(x^n/c)], 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 525

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[a^IntPar
t[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 1284

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 1543

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 2081

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 6857

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

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Mathematica [A]
time = 0.00, size = 197, normalized size = 1.25 \begin {gather*} \frac {x^{3/2} \left (b+a x^2\right )^{3/4} \left (8 a^{3/4} x^{3/2} \sqrt [4]{b+a x^2}-4 b \text {ArcTan}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{b+a x^2}}\right )+4 b \tanh ^{-1}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{b+a x^2}}\right )+3 a^{3/4} b \text {RootSum}\left [2 a^2-a b-4 a \text {$\#$1}^4+2 \text {$\#$1}^8\&,\frac {-\log \left (\sqrt {x}\right ) \text {$\#$1}+\log \left (\sqrt [4]{b+a x^2}-\sqrt {x} \text {$\#$1}\right ) \text {$\#$1}}{-a+\text {$\#$1}^4}\&\right ]\right )}{8 a^{3/4} \left (x^2 \left (b+a x^2\right )\right )^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x^(3/2)*(b + a*x^2)^(3/4)*(8*a^(3/4)*x^(3/2)*(b + a*x^2)^(1/4) - 4*b*ArcTan[(a^(1/4)*Sqrt[x])/(b + a*x^2)^(1/
4)] + 4*b*ArcTanh[(a^(1/4)*Sqrt[x])/(b + a*x^2)^(1/4)] + 3*a^(3/4)*b*RootSum[2*a^2 - a*b - 4*a*#1^4 + 2*#1^8 &
, (-(Log[Sqrt[x]]*#1) + Log[(b + a*x^2)^(1/4) - Sqrt[x]*#1]*#1)/(-a + #1^4) & ]))/(8*a^(3/4)*(x^2*(b + a*x^2)
)^(3/4))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F(-1)] Timed out
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)*(2*a*x^4-b)/(a*x^4-2*b),x, algorithm="fricas")

[Out]

Timed out

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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 (2 a x^{4} - b\right )}{a x^{4} - 2 b}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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