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

Optimal. Leaf size=81 \[ \frac {1}{2} a^{3/4} \tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )+\frac {1}{2} a^{3/4} \tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )-\frac {2 \left (a x^4-b\right )^{3/4}}{3 x^3} \]

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Rubi [A]  time = 0.04, antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {451, 240, 212, 206, 203} \begin {gather*} \frac {1}{2} a^{3/4} \tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )+\frac {1}{2} a^{3/4} \tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )-\frac {2 \left (a x^4-b\right )^{3/4}}{3 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 451

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(c*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] + Dist[d/e^n, Int[(e*x)^(m + n)*(a + b*x^n)^p, 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] && (IntegerQ[n] || GtQ[e, 0]) && (
(GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1]))

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 81, normalized size = 1.00 \begin {gather*} \frac {1}{2} a^{3/4} \tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )+\frac {1}{2} a^{3/4} \tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )-\frac {2 \left (a x^4-b\right )^{3/4}}{3 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.29, size = 81, normalized size = 1.00 \begin {gather*} -\frac {2 \left (-b+a x^4\right )^{3/4}}{3 x^3}+\frac {1}{2} a^{3/4} \tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )+\frac {1}{2} a^{3/4} \tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(-b + a*x^4)^(3/4))/(3*x^3) + (a^(3/4)*ArcTan[(a^(1/4)*x)/(-b + a*x^4)^(1/4)])/2 + (a^(3/4)*ArcTanh[(a^(1/
4)*x)/(-b + a*x^4)^(1/4)])/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^4-2*b)/x^4/(a*x^4-b)^(1/4),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 {a x^{4} - 2 \, b}{{\left (a x^{4} - b\right )}^{\frac {1}{4}} x^{4}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.43, size = 93, normalized size = 1.15 \begin {gather*} -\frac {1}{4} \, a {\left (\frac {2 \, \arctan \left (\frac {{\left (a x^{4} - b\right )}^{\frac {1}{4}}}{a^{\frac {1}{4}} x}\right )}{a^{\frac {1}{4}}} + \frac {\log \left (-\frac {a^{\frac {1}{4}} - \frac {{\left (a x^{4} - b\right )}^{\frac {1}{4}}}{x}}{a^{\frac {1}{4}} + \frac {{\left (a x^{4} - b\right )}^{\frac {1}{4}}}{x}}\right )}{a^{\frac {1}{4}}}\right )} - \frac {2 \, {\left (a x^{4} - b\right )}^{\frac {3}{4}}}{3 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.15, size = 57, normalized size = 0.70 \begin {gather*} \frac {a\,x\,{\left (1-\frac {a\,x^4}{b}\right )}^{1/4}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{4},\frac {1}{4};\ \frac {5}{4};\ \frac {a\,x^4}{b}\right )}{{\left (a\,x^4-b\right )}^{1/4}}-\frac {2\,{\left (a\,x^4-b\right )}^{3/4}}{3\,x^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a*x*(1 - (a*x^4)/b)^(1/4)*hypergeom([1/4, 1/4], 5/4, (a*x^4)/b))/(a*x^4 - b)^(1/4) - (2*(a*x^4 - b)^(3/4))/(3
*x^3)

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sympy [C]  time = 2.60, size = 126, normalized size = 1.56 \begin {gather*} \frac {a x e^{- \frac {i \pi }{4}} \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {1}{4} \\ \frac {5}{4} \end {matrix}\middle | {\frac {a x^{4}}{b}} \right )}}{4 \sqrt [4]{b} \Gamma \left (\frac {5}{4}\right )} - 2 b \left (\begin {cases} - \frac {a^{\frac {3}{4}} \left (-1 + \frac {b}{a x^{4}}\right )^{\frac {3}{4}} e^{\frac {3 i \pi }{4}} \Gamma \left (- \frac {3}{4}\right )}{4 b \Gamma \left (\frac {1}{4}\right )} & \text {for}\: \left |{\frac {b}{a x^{4}}}\right | > 1 \\- \frac {a^{\frac {3}{4}} \left (1 - \frac {b}{a x^{4}}\right )^{\frac {3}{4}} \Gamma \left (- \frac {3}{4}\right )}{4 b \Gamma \left (\frac {1}{4}\right )} & \text {otherwise} \end {cases}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*x*exp(-I*pi/4)*gamma(1/4)*hyper((1/4, 1/4), (5/4,), a*x**4/b)/(4*b**(1/4)*gamma(5/4)) - 2*b*Piecewise((-a**(
3/4)*(-1 + b/(a*x**4))**(3/4)*exp(3*I*pi/4)*gamma(-3/4)/(4*b*gamma(1/4)), Abs(b/(a*x**4)) > 1), (-a**(3/4)*(1
- b/(a*x**4))**(3/4)*gamma(-3/4)/(4*b*gamma(1/4)), True))

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