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

Optimal. Leaf size=98 \[ \frac {\sqrt [4]{a x^4+b x^2} \left (a x^2+8 b\right )}{2 x}+\frac {7}{4} \sqrt [4]{a} b \tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4+b x^2}}\right )-\frac {7}{4} \sqrt [4]{a} b \tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4+b x^2}}\right ) \]

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Rubi [A]  time = 0.22, antiderivative size = 170, normalized size of antiderivative = 1.73, number of steps used = 8, number of rules used = 8, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {2038, 2004, 2032, 329, 331, 298, 203, 206} \begin {gather*} -\frac {7}{2} a x \sqrt [4]{a x^4+b x^2}+\frac {7 \sqrt [4]{a} b x^{3/2} \left (a x^2+b\right )^{3/4} \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{a x^2+b}}\right )}{4 \left (a x^4+b x^2\right )^{3/4}}-\frac {7 \sqrt [4]{a} b x^{3/2} \left (a x^2+b\right )^{3/4} \tanh ^{-1}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{a x^2+b}}\right )}{4 \left (a x^4+b x^2\right )^{3/4}}+\frac {4 \left (a x^4+b x^2\right )^{5/4}}{x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-7*a*x*(b*x^2 + a*x^4)^(1/4))/2 + (4*(b*x^2 + a*x^4)^(5/4))/x^3 + (7*a^(1/4)*b*x^(3/2)*(b + a*x^2)^(3/4)*ArcT
an[(a^(1/4)*Sqrt[x])/(b + a*x^2)^(1/4)])/(4*(b*x^2 + a*x^4)^(3/4)) - (7*a^(1/4)*b*x^(3/2)*(b + a*x^2)^(3/4)*Ar
cTanh[(a^(1/4)*Sqrt[x])/(b + a*x^2)^(1/4)])/(4*(b*x^2 + a*x^4)^(3/4))

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 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 329

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 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 2004

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(x*(a*x^j + b*x^n)^p)/(n*p + 1), x] + Dist[(
a*(n - j)*p)/(n*p + 1), Int[x^j*(a*x^j + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] &&  !IntegerQ[p] && LtQ[0,
j, n] && GtQ[p, 0] && NeQ[n*p + 1, 0]

Rule 2032

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[(c^IntPart[m]*(c*x)^FracP
art[m]*(a*x^j + b*x^n)^FracPart[p])/(x^(FracPart[m] + j*FracPart[p])*(a + b*x^(n - j))^FracPart[p]), Int[x^(m
+ j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[n, j] && PosQ[n
- j]

Rule 2038

Int[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + (d_.)*(x_)^(n_.)), x_Symbol] :> Sim
p[(c*e^(j - 1)*(e*x)^(m - j + 1)*(a*x^j + b*x^(j + n))^(p + 1))/(a*(m + j*p + 1)), x] + Dist[(a*d*(m + j*p + 1
) - b*c*(m + n + p*(j + n) + 1))/(a*e^n*(m + j*p + 1)), Int[(e*x)^(m + n)*(a*x^j + b*x^(j + n))^p, x], x] /; F
reeQ[{a, b, c, d, e, j, p}, x] && EqQ[jn, j + n] &&  !IntegerQ[p] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && (LtQ[m
+ j*p, -1] || (IntegersQ[m - 1/2, p - 1/2] && LtQ[p, 0] && LtQ[m, -(n*p) - 1])) && (GtQ[e, 0] || IntegersQ[j,
n]) && NeQ[m + j*p + 1, 0] && NeQ[m - n + j*p + 1, 0]

Rubi steps

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

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Mathematica [C]  time = 0.07, size = 71, normalized size = 0.72 \begin {gather*} \frac {2 \sqrt [4]{x^2 \left (a x^2+b\right )} \left (6 \left (a x^2+b\right )-\frac {7 a x^2 \, _2F_1\left (-\frac {1}{4},\frac {3}{4};\frac {7}{4};-\frac {a x^2}{b}\right )}{\sqrt [4]{\frac {a x^2}{b}+1}}\right )}{3 x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(x^2*(b + a*x^2))^(1/4)*(6*(b + a*x^2) - (7*a*x^2*Hypergeometric2F1[-1/4, 3/4, 7/4, -((a*x^2)/b)])/(1 + (a*
x^2)/b)^(1/4)))/(3*x)

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IntegrateAlgebraic [A]  time = 0.36, size = 98, normalized size = 1.00 \begin {gather*} \frac {\left (8 b+a x^2\right ) \sqrt [4]{b x^2+a x^4}}{2 x}+\frac {7}{4} \sqrt [4]{a} b \tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b x^2+a x^4}}\right )-\frac {7}{4} \sqrt [4]{a} b \tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b x^2+a x^4}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((8*b + a*x^2)*(b*x^2 + a*x^4)^(1/4))/(2*x) + (7*a^(1/4)*b*ArcTan[(a^(1/4)*x)/(b*x^2 + a*x^4)^(1/4)])/4 - (7*a
^(1/4)*b*ArcTanh[(a^(1/4)*x)/(b*x^2 + a*x^4)^(1/4)])/4

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

[Out]

Timed out

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giac [B]  time = 0.20, size = 221, normalized size = 2.26 \begin {gather*} \frac {8 \, {\left (a + \frac {b}{x^{2}}\right )}^{\frac {1}{4}} a b x^{2} - 14 \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} b^{2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} + 2 \, {\left (a + \frac {b}{x^{2}}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right ) - 14 \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} b^{2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} - 2 \, {\left (a + \frac {b}{x^{2}}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right ) - 7 \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} b^{2} \log \left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} {\left (a + \frac {b}{x^{2}}\right )}^{\frac {1}{4}} + \sqrt {-a} + \sqrt {a + \frac {b}{x^{2}}}\right ) + 7 \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} b^{2} \log \left (-\sqrt {2} \left (-a\right )^{\frac {1}{4}} {\left (a + \frac {b}{x^{2}}\right )}^{\frac {1}{4}} + \sqrt {-a} + \sqrt {a + \frac {b}{x^{2}}}\right ) + 64 \, {\left (a + \frac {b}{x^{2}}\right )}^{\frac {1}{4}} b^{2}}{16 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(8*(a + b/x^2)^(1/4)*a*b*x^2 - 14*sqrt(2)*(-a)^(1/4)*b^2*arctan(1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) + 2*(a +
b/x^2)^(1/4))/(-a)^(1/4)) - 14*sqrt(2)*(-a)^(1/4)*b^2*arctan(-1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) - 2*(a + b/x^2)^
(1/4))/(-a)^(1/4)) - 7*sqrt(2)*(-a)^(1/4)*b^2*log(sqrt(2)*(-a)^(1/4)*(a + b/x^2)^(1/4) + sqrt(-a) + sqrt(a + b
/x^2)) + 7*sqrt(2)*(-a)^(1/4)*b^2*log(-sqrt(2)*(-a)^(1/4)*(a + b/x^2)^(1/4) + sqrt(-a) + sqrt(a + b/x^2)) + 64
*(a + b/x^2)^(1/4)*b^2)/b

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.28, size = 89, normalized size = 0.91 \begin {gather*} \frac {2\,a\,x\,{\left (a\,x^4+b\,x^2\right )}^{1/4}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {3}{4};\ \frac {7}{4};\ -\frac {a\,x^2}{b}\right )}{3\,{\left (\frac {a\,x^2}{b}+1\right )}^{1/4}}+\frac {4\,b\,{\left (a\,x^4+b\,x^2\right )}^{1/4}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},-\frac {1}{4};\ \frac {3}{4};\ -\frac {a\,x^2}{b}\right )}{x\,{\left (\frac {a\,x^2}{b}+1\right )}^{1/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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