3.12.7 \(\int \frac {-b+a x^8}{x^2 (b+a x^4)^{3/4}} \, dx\)

Optimal. Leaf size=82 \[ \frac {3 b \tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4+b}}\right )}{8 a^{3/4}}-\frac {3 b \tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4+b}}\right )}{8 a^{3/4}}+\frac {\left (x^4+4\right ) \sqrt [4]{a x^4+b}}{4 x} \]

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Rubi [A]  time = 0.06, antiderivative size = 92, normalized size of antiderivative = 1.12, number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1487, 451, 331, 298, 203, 206} \begin {gather*} \frac {3 b \tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4+b}}\right )}{8 a^{3/4}}-\frac {3 b \tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4+b}}\right )}{8 a^{3/4}}+\frac {\sqrt [4]{a x^4+b}}{x}+\frac {1}{4} x^3 \sqrt [4]{a x^4+b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-b + a*x^8)/(x^2*(b + a*x^4)^(3/4)),x]

[Out]

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

Rule 1487

Int[((f_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), x_Symbol] :> 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 + 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, c, d, e, f, m, q}, x] && EqQ[n2,
2*n] && 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 {-b+a x^8}{x^2 \left (b+a x^4\right )^{3/4}} \, dx &=\frac {1}{4} x^3 \sqrt [4]{b+a x^4}+\frac {\int \frac {-4 a b-3 a b x^4}{x^2 \left (b+a x^4\right )^{3/4}} \, dx}{4 a}\\ &=\frac {\sqrt [4]{b+a x^4}}{x}+\frac {1}{4} x^3 \sqrt [4]{b+a x^4}-\frac {1}{4} (3 b) \int \frac {x^2}{\left (b+a x^4\right )^{3/4}} \, dx\\ &=\frac {\sqrt [4]{b+a x^4}}{x}+\frac {1}{4} x^3 \sqrt [4]{b+a x^4}-\frac {1}{4} (3 b) \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}+\frac {1}{4} x^3 \sqrt [4]{b+a x^4}-\frac {(3 b) \operatorname {Subst}\left (\int \frac {1}{1-\sqrt {a} x^2} \, dx,x,\frac {x}{\sqrt [4]{b+a x^4}}\right )}{8 \sqrt {a}}+\frac {(3 b) \operatorname {Subst}\left (\int \frac {1}{1+\sqrt {a} x^2} \, dx,x,\frac {x}{\sqrt [4]{b+a x^4}}\right )}{8 \sqrt {a}}\\ &=\frac {\sqrt [4]{b+a x^4}}{x}+\frac {1}{4} x^3 \sqrt [4]{b+a x^4}+\frac {3 b \tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right )}{8 a^{3/4}}-\frac {3 b \tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right )}{8 a^{3/4}}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 82, normalized size = 1.00 \begin {gather*} \frac {2 a^{3/4} \left (x^4+4\right ) \sqrt [4]{a x^4+b}+3 b x \tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4+b}}\right )-3 b x \tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4+b}}\right )}{8 a^{3/4} x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-b + a*x^8)/(x^2*(b + a*x^4)^(3/4)),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.56, size = 82, normalized size = 1.00 \begin {gather*} \frac {\left (4+x^4\right ) \sqrt [4]{b+a x^4}}{4 x}+\frac {3 b \tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right )}{8 a^{3/4}}-\frac {3 b \tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right )}{8 a^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(-b + a*x^8)/(x^2*(b + a*x^4)^(3/4)),x]

[Out]

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

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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^8-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 {a x^{8} - 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((a*x^8-b)/x^2/(a*x^4+b)^(3/4),x, algorithm="giac")

[Out]

integrate((a*x^8 - 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 \[\int \frac {a \,x^{8}-b}{x^{2} \left (a \,x^{4}+b \right )^{\frac {3}{4}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x^8-b)/x^2/(a*x^4+b)^(3/4),x)

[Out]

int((a*x^8-b)/x^2/(a*x^4+b)^(3/4),x)

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maxima [B]  time = 0.42, size = 127, normalized size = 1.55 \begin {gather*} -\frac {1}{16} \, a {\left (\frac {3 \, {\left (\frac {2 \, b \arctan \left (\frac {{\left (a x^{4} + b\right )}^{\frac {1}{4}}}{a^{\frac {1}{4}} x}\right )}{a^{\frac {3}{4}}} - \frac {b \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 )}}{a} + \frac {4 \, {\left (a x^{4} + b\right )}^{\frac {1}{4}} b}{{\left (a^{2} - \frac {{\left (a x^{4} + b\right )} a}{x^{4}}\right )} x}\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((a*x^8-b)/x^2/(a*x^4+b)^(3/4),x, algorithm="maxima")

[Out]

-1/16*a*(3*(2*b*arctan((a*x^4 + b)^(1/4)/(a^(1/4)*x))/a^(3/4) - b*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 + 4*(a*x^4 + b)^(1/4)*b/((a^2 - (a*x^4 + b)*a/x^4)*x)) + (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-a\,x^8}{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 - a*x^8)/(x^2*(b + a*x^4)^(3/4)),x)

[Out]

-int((b - a*x^8)/(x^2*(b + a*x^4)^(3/4)), x)

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sympy [C]  time = 2.37, size = 70, normalized size = 0.85 \begin {gather*} - \frac {\sqrt [4]{a} \sqrt [4]{1 + \frac {b}{a x^{4}}} \Gamma \left (- \frac {1}{4}\right )}{4 \Gamma \left (\frac {3}{4}\right )} + \frac {a x^{7} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {a x^{4} e^{i \pi }}{b}} \right )}}{4 b^{\frac {3}{4}} \Gamma \left (\frac {11}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x**8-b)/x**2/(a*x**4+b)**(3/4),x)

[Out]

-a**(1/4)*(1 + b/(a*x**4))**(1/4)*gamma(-1/4)/(4*gamma(3/4)) + a*x**7*gamma(7/4)*hyper((3/4, 7/4), (11/4,), a*
x**4*exp_polar(I*pi)/b)/(4*b**(3/4)*gamma(11/4))

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