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

Optimal. Leaf size=90 \[ -\frac {6 \left (a x^4+b x^2\right )^{3/4}}{x \left (a x^2+b\right )}+\frac {2 \tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4+b x^2}}\right )}{\sqrt [4]{a}}+\frac {2 \tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4+b x^2}}\right )}{\sqrt [4]{a}} \]

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Rubi [A]  time = 0.21, antiderivative size = 141, normalized size of antiderivative = 1.57, number of steps used = 7, number of rules used = 7, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2056, 452, 329, 240, 212, 206, 203} \begin {gather*} -\frac {6 x}{\sqrt [4]{a x^4+b x^2}}+\frac {2 \sqrt {x} \sqrt [4]{a x^2+b} \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{a x^2+b}}\right )}{\sqrt [4]{a} \sqrt [4]{a x^4+b x^2}}+\frac {2 \sqrt {x} \sqrt [4]{a x^2+b} \tanh ^{-1}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{a x^2+b}}\right )}{\sqrt [4]{a} \sqrt [4]{a x^4+b x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-6*x)/(b*x^2 + a*x^4)^(1/4) + (2*Sqrt[x]*(b + a*x^2)^(1/4)*ArcTan[(a^(1/4)*Sqrt[x])/(b + a*x^2)^(1/4)])/(a^(1
/4)*(b*x^2 + a*x^4)^(1/4)) + (2*Sqrt[x]*(b + a*x^2)^(1/4)*ArcTanh[(a^(1/4)*Sqrt[x])/(b + a*x^2)^(1/4)])/(a^(1/
4)*(b*x^2 + a*x^4)^(1/4))

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 212

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b), 2]
]}, Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&
!GtQ[a/b, 0]

Rule 240

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + 1/n), Subst[Int[1/(1 - b*x^n)^(p + 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)] && IntegerQ[p
+ 1/n]

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 452

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

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

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Mathematica [C]  time = 0.08, size = 66, normalized size = 0.73 \begin {gather*} \frac {2 x \left (2 a x^2 \sqrt [4]{\frac {a x^2}{b}+1} \, _2F_1\left (\frac {5}{4},\frac {5}{4};\frac {9}{4};-\frac {a x^2}{b}\right )-5 b\right )}{5 b \sqrt [4]{x^2 \left (a x^2+b\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x*(-5*b + 2*a*x^2*(1 + (a*x^2)/b)^(1/4)*Hypergeometric2F1[5/4, 5/4, 9/4, -((a*x^2)/b)]))/(5*b*(x^2*(b + a*x
^2))^(1/4))

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IntegrateAlgebraic [A]  time = 0.28, size = 90, normalized size = 1.00 \begin {gather*} -\frac {6 \left (b x^2+a x^4\right )^{3/4}}{x \left (b+a x^2\right )}+\frac {2 \tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b x^2+a x^4}}\right )}{\sqrt [4]{a}}+\frac {2 \tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b x^2+a x^4}}\right )}{\sqrt [4]{a}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-6*(b*x^2 + a*x^4)^(3/4))/(x*(b + a*x^2)) + (2*ArcTan[(a^(1/4)*x)/(b*x^2 + a*x^4)^(1/4)])/a^(1/4) + (2*ArcTan
h[(a^(1/4)*x)/(b*x^2 + a*x^4)^(1/4)])/a^(1/4)

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

[Out]

Timed out

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giac [B]  time = 0.22, size = 195, normalized size = 2.17 \begin {gather*} \frac {\sqrt {2} \left (-a\right )^{\frac {3}{4}} \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 )}{a} + \frac {\sqrt {2} \left (-a\right )^{\frac {3}{4}} \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 )}{a} - \frac {\sqrt {2} \left (-a\right )^{\frac {3}{4}} \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 )}{2 \, a} + \frac {\sqrt {2} \left (-a\right )^{\frac {3}{4}} \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 )}{2 \, a} - \frac {6}{{\left (a + \frac {b}{x^{2}}\right )}^{\frac {1}{4}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(2)*(-a)^(3/4)*arctan(1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) + 2*(a + b/x^2)^(1/4))/(-a)^(1/4))/a + sqrt(2)*(-a)^
(3/4)*arctan(-1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) - 2*(a + b/x^2)^(1/4))/(-a)^(1/4))/a - 1/2*sqrt(2)*(-a)^(3/4)*lo
g(sqrt(2)*(-a)^(1/4)*(a + b/x^2)^(1/4) + sqrt(-a) + sqrt(a + b/x^2))/a + 1/2*sqrt(2)*(-a)^(3/4)*log(-sqrt(2)*(
-a)^(1/4)*(a + b/x^2)^(1/4) + sqrt(-a) + sqrt(a + b/x^2))/a - 6/(a + b/x^2)^(1/4)

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maple [F]  time = 0.04, size = 0, normalized size = 0.00 \[\int \frac {2 a \,x^{2}-b}{\left (a \,x^{2}+b \right ) \left (a \,x^{4}+b \,x^{2}\right )^{\frac {1}{4}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {2 \, a x^{2} - b}{{\left (a x^{4} + b x^{2}\right )}^{\frac {1}{4}} {\left (a x^{2} + b\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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