3.10.79 \(\int \frac {-b+2 a x^4}{x^2 (-b+a x^4)^{3/4}} \, dx\)

Optimal. Leaf size=74 \[ -\frac {\sqrt [4]{a x^4-b}}{x}-\sqrt [4]{a} \tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )+\sqrt [4]{a} \tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right ) \]

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Rubi [A]  time = 0.08, antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {451, 331, 298, 203, 206} \begin {gather*} -\frac {\sqrt [4]{a x^4-b}}{x}-\sqrt [4]{a} \tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )+\sqrt [4]{a} \tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((-b + a*x^4)^(1/4)/x) - a^(1/4)*ArcTan[(a^(1/4)*x)/(-b + a*x^4)^(1/4)] + a^(1/4)*ArcTanh[(a^(1/4)*x)/(-b + 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 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 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 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 {-b+2 a x^4}{x^2 \left (-b+a x^4\right )^{3/4}} \, dx &=-\frac {\sqrt [4]{-b+a x^4}}{x}+(2 a) \int \frac {x^2}{\left (-b+a x^4\right )^{3/4}} \, dx\\ &=-\frac {\sqrt [4]{-b+a x^4}}{x}+(2 a) \operatorname {Subst}\left (\int \frac {x^2}{1-a x^4} \, dx,x,\frac {x}{\sqrt [4]{-b+a x^4}}\right )\\ &=-\frac {\sqrt [4]{-b+a x^4}}{x}+\sqrt {a} \operatorname {Subst}\left (\int \frac {1}{1-\sqrt {a} x^2} \, dx,x,\frac {x}{\sqrt [4]{-b+a x^4}}\right )-\sqrt {a} \operatorname {Subst}\left (\int \frac {1}{1+\sqrt {a} x^2} \, dx,x,\frac {x}{\sqrt [4]{-b+a x^4}}\right )\\ &=-\frac {\sqrt [4]{-b+a x^4}}{x}-\sqrt [4]{a} \tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )+\sqrt [4]{a} \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 = 74, normalized size = 1.00 \begin {gather*} -\frac {\sqrt [4]{a x^4-b}}{x}-\sqrt [4]{a} \tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )+\sqrt [4]{a} \tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.44, size = 74, normalized size = 1.00 \begin {gather*} -\frac {\sqrt [4]{-b+a x^4}}{x}-\sqrt [4]{a} \tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )+\sqrt [4]{a} \tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*a*x^4-b)/x^2/(a*x^4-b)^(3/4),x, algorithm="giac")

[Out]

integrate((2*a*x^4 - b)/((a*x^4 - b)^(3/4)*x^2), x)

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maple [F]  time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {2 a \,x^{4}-b}{x^{2} \left (a \,x^{4}-b \right )^{\frac {3}{4}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*a*x^4-b)/x^2/(a*x^4-b)^(3/4),x)

[Out]

int((2*a*x^4-b)/x^2/(a*x^4-b)^(3/4),x)

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maxima [A]  time = 0.50, size = 94, normalized size = 1.27 \begin {gather*} \frac {1}{2} \, a {\left (\frac {2 \, \arctan \left (\frac {{\left (a x^{4} - b\right )}^{\frac {1}{4}}}{a^{\frac {1}{4}} x}\right )}{a^{\frac {3}{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 {3}{4}}}\right )} - \frac {{\left (a x^{4} - b\right )}^{\frac {1}{4}}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*a*x^4-b)/x^2/(a*x^4-b)^(3/4),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} -\int \frac {b-2\,a\,x^4}{x^2\,{\left (a\,x^4-b\right )}^{3/4}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(b - 2*a*x^4)/(x^2*(a*x^4 - b)^(3/4)),x)

[Out]

-int((b - 2*a*x^4)/(x^2*(a*x^4 - b)^(3/4)), x)

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sympy [C]  time = 2.22, size = 126, normalized size = 1.70 \begin {gather*} \frac {a x^{3} e^{- \frac {3 i \pi }{4}} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {a x^{4}}{b}} \right )}}{2 b^{\frac {3}{4}} \Gamma \left (\frac {7}{4}\right )} - b \left (\begin {cases} - \frac {\sqrt [4]{a} \sqrt [4]{-1 + \frac {b}{a x^{4}}} e^{\frac {i \pi }{4}} \Gamma \left (- \frac {1}{4}\right )}{4 b \Gamma \left (\frac {3}{4}\right )} & \text {for}\: \left |{\frac {b}{a x^{4}}}\right | > 1 \\- \frac {\sqrt [4]{a} \sqrt [4]{1 - \frac {b}{a x^{4}}} \Gamma \left (- \frac {1}{4}\right )}{4 b \Gamma \left (\frac {3}{4}\right )} & \text {otherwise} \end {cases}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*a*x**4-b)/x**2/(a*x**4-b)**(3/4),x)

[Out]

a*x**3*exp(-3*I*pi/4)*gamma(3/4)*hyper((3/4, 3/4), (7/4,), a*x**4/b)/(2*b**(3/4)*gamma(7/4)) - b*Piecewise((-a
**(1/4)*(-1 + b/(a*x**4))**(1/4)*exp(I*pi/4)*gamma(-1/4)/(4*b*gamma(3/4)), Abs(b/(a*x**4)) > 1), (-a**(1/4)*(1
- b/(a*x**4))**(1/4)*gamma(-1/4)/(4*b*gamma(3/4)), True))

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