3.16.18 \(\int \frac {-2 b-2 a x^4+x^8}{x^4 \sqrt [4]{b+a x^4}} \, dx\)

Optimal. Leaf size=105 \[ \frac {\left (-8 a^2-b\right ) \tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4+b}}\right )}{8 a^{5/4}}+\frac {\left (-8 a^2-b\right ) \tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4+b}}\right )}{8 a^{5/4}}+\frac {\left (8 a+3 x^4\right ) \left (a x^4+b\right )^{3/4}}{12 a x^3} \]

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Rubi [A]  time = 0.07, antiderivative size = 108, normalized size of antiderivative = 1.03, number of steps used = 6, number of rules used = 6, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {1486, 451, 240, 212, 206, 203} \begin {gather*} -\frac {\left (8 a^2+b\right ) \tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4+b}}\right )}{8 a^{5/4}}-\frac {\left (8 a^2+b\right ) \tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4+b}}\right )}{8 a^{5/4}}+\frac {x \left (a x^4+b\right )^{3/4}}{4 a}+\frac {2 \left (a x^4+b\right )^{3/4}}{3 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-2*b - 2*a*x^4 + x^8)/(x^4*(b + a*x^4)^(1/4)),x]

[Out]

(2*(b + a*x^4)^(3/4))/(3*x^3) + (x*(b + a*x^4)^(3/4))/(4*a) - ((8*a^2 + b)*ArcTan[(a^(1/4)*x)/(b + a*x^4)^(1/4
)])/(8*a^(5/4)) - ((8*a^2 + b)*ArcTanh[(a^(1/4)*x)/(b + a*x^4)^(1/4)])/(8*a^(5/4))

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 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]))

Rule 1486

Int[((f_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), x_Sy
mbol] :> Simp[(c^p*(f*x)^(m + 2*n*p - n + 1)*(d + e*x^n)^(q + 1))/(e*f^(2*n*p - n + 1)*(m + 2*n*p + n*q + 1)),
x] + Dist[1/(e*(m + 2*n*p + n*q + 1)), Int[(f*x)^m*(d + e*x^n)^q*ExpandToSum[e*(m + 2*n*p + n*q + 1)*((a + b*
x^n + c*x^(2*n))^p - c^p*x^(2*n*p)) - d*c^p*(m + 2*n*p - n + 1)*x^(2*n*p - n), x], x], x] /; FreeQ[{a, b, c, d
, e, f, m, q}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && IGtQ[n, 0] && IGtQ[p, 0] && GtQ[2*n*p, n - 1] &&
!IntegerQ[q] && NeQ[m + 2*n*p + n*q + 1, 0]

Rubi steps

\begin {align*} \int \frac {-2 b-2 a x^4+x^8}{x^4 \sqrt [4]{b+a x^4}} \, dx &=\frac {x \left (b+a x^4\right )^{3/4}}{4 a}+\frac {\int \frac {-8 a b-\left (8 a^2+b\right ) x^4}{x^4 \sqrt [4]{b+a x^4}} \, dx}{4 a}\\ &=\frac {2 \left (b+a x^4\right )^{3/4}}{3 x^3}+\frac {x \left (b+a x^4\right )^{3/4}}{4 a}+\frac {\left (-8 a^2-b\right ) \int \frac {1}{\sqrt [4]{b+a x^4}} \, dx}{4 a}\\ &=\frac {2 \left (b+a x^4\right )^{3/4}}{3 x^3}+\frac {x \left (b+a x^4\right )^{3/4}}{4 a}+\frac {\left (-8 a^2-b\right ) \operatorname {Subst}\left (\int \frac {1}{1-a x^4} \, dx,x,\frac {x}{\sqrt [4]{b+a x^4}}\right )}{4 a}\\ &=\frac {2 \left (b+a x^4\right )^{3/4}}{3 x^3}+\frac {x \left (b+a x^4\right )^{3/4}}{4 a}+\frac {\left (-8 a^2-b\right ) \operatorname {Subst}\left (\int \frac {1}{1-\sqrt {a} x^2} \, dx,x,\frac {x}{\sqrt [4]{b+a x^4}}\right )}{8 a}+\frac {\left (-8 a^2-b\right ) \operatorname {Subst}\left (\int \frac {1}{1+\sqrt {a} x^2} \, dx,x,\frac {x}{\sqrt [4]{b+a x^4}}\right )}{8 a}\\ &=\frac {2 \left (b+a x^4\right )^{3/4}}{3 x^3}+\frac {x \left (b+a x^4\right )^{3/4}}{4 a}-\frac {\left (8 a^2+b\right ) \tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right )}{8 a^{5/4}}-\frac {\left (8 a^2+b\right ) \tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right )}{8 a^{5/4}}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 102, normalized size = 0.97 \begin {gather*} \frac {-3 x^3 \left (8 a^2+b\right ) \tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4+b}}\right )-3 x^3 \left (8 a^2+b\right ) \tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4+b}}\right )+2 \sqrt [4]{a} \left (8 a+3 x^4\right ) \left (a x^4+b\right )^{3/4}}{24 a^{5/4} x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2*b - 2*a*x^4 + x^8)/(x^4*(b + a*x^4)^(1/4)),x]

[Out]

(2*a^(1/4)*(8*a + 3*x^4)*(b + a*x^4)^(3/4) - 3*(8*a^2 + b)*x^3*ArcTan[(a^(1/4)*x)/(b + a*x^4)^(1/4)] - 3*(8*a^
2 + b)*x^3*ArcTanh[(a^(1/4)*x)/(b + a*x^4)^(1/4)])/(24*a^(5/4)*x^3)

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IntegrateAlgebraic [A]  time = 0.50, size = 105, normalized size = 1.00 \begin {gather*} \frac {\left (8 a+3 x^4\right ) \left (b+a x^4\right )^{3/4}}{12 a x^3}+\frac {\left (-8 a^2-b\right ) \tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right )}{8 a^{5/4}}+\frac {\left (-8 a^2-b\right ) \tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right )}{8 a^{5/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(-2*b - 2*a*x^4 + x^8)/(x^4*(b + a*x^4)^(1/4)),x]

[Out]

((8*a + 3*x^4)*(b + a*x^4)^(3/4))/(12*a*x^3) + ((-8*a^2 - b)*ArcTan[(a^(1/4)*x)/(b + a*x^4)^(1/4)])/(8*a^(5/4)
) + ((-8*a^2 - b)*ArcTanh[(a^(1/4)*x)/(b + a*x^4)^(1/4)])/(8*a^(5/4))

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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((x^8-2*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 {x^{8} - 2 \, 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((x^8-2*a*x^4-2*b)/x^4/(a*x^4+b)^(1/4),x, algorithm="giac")

[Out]

integrate((x^8 - 2*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 \[\int \frac {x^{8}-2 a \,x^{4}-2 b}{x^{4} \left (a \,x^{4}+b \right )^{\frac {1}{4}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^8-2*a*x^4-2*b)/x^4/(a*x^4+b)^(1/4),x)

[Out]

int((x^8-2*a*x^4-2*b)/x^4/(a*x^4+b)^(1/4),x)

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maxima [B]  time = 0.42, size = 192, normalized size = 1.83 \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 {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 {b {\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 )}}{16 \, a} + \frac {2 \, {\left (a x^{4} + b\right )}^{\frac {3}{4}}}{3 \, x^{3}} - \frac {{\left (a x^{4} + b\right )}^{\frac {3}{4}} b}{4 \, {\left (a^{2} - \frac {{\left (a x^{4} + b\right )} a}{x^{4}}\right )} x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^8-2*a*x^4-2*b)/x^4/(a*x^4+b)^(1/4),x, algorithm="maxima")

[Out]

1/2*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)) + 1/16*b*(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))/a + 2/3*(a*x^4 + b)^(3/4)/x^3 - 1/4*(a*x^4 + b)^(3/4)*b
/((a^2 - (a*x^4 + b)*a/x^4)*x^3)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int -\frac {-x^8+2\,a\,x^4+2\,b}{x^4\,{\left (a\,x^4+b\right )}^{1/4}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*b + 2*a*x^4 - x^8)/(x^4*(b + a*x^4)^(1/4)),x)

[Out]

int(-(2*b + 2*a*x^4 - x^8)/(x^4*(b + a*x^4)^(1/4)), x)

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sympy [C]  time = 2.95, size = 107, normalized size = 1.02 \begin {gather*} - \frac {a^{\frac {3}{4}} \left (1 + \frac {b}{a x^{4}}\right )^{\frac {3}{4}} \Gamma \left (- \frac {3}{4}\right )}{2 \Gamma \left (\frac {1}{4}\right )} - \frac {a x \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} e^{i \pi }}{b}} \right )}}{2 \sqrt [4]{b} \Gamma \left (\frac {5}{4}\right )} + \frac {x^{5} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {a x^{4} e^{i \pi }}{b}} \right )}}{4 \sqrt [4]{b} \Gamma \left (\frac {9}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**8-2*a*x**4-2*b)/x**4/(a*x**4+b)**(1/4),x)

[Out]

-a**(3/4)*(1 + b/(a*x**4))**(3/4)*gamma(-3/4)/(2*gamma(1/4)) - a*x*gamma(1/4)*hyper((1/4, 1/4), (5/4,), a*x**4
*exp_polar(I*pi)/b)/(2*b**(1/4)*gamma(5/4)) + x**5*gamma(5/4)*hyper((1/4, 5/4), (9/4,), a*x**4*exp_polar(I*pi)
/b)/(4*b**(1/4)*gamma(9/4))

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