3.13.8 \(\int \frac {-3 b+2 a x^8}{x^8 \sqrt [4]{-b+a x^4}} \, dx\)

Optimal. Leaf size=88 \[ a^{3/4} \tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )+a^{3/4} \tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )+\frac {\left (a x^4-b\right )^{3/4} \left (-4 a x^4-3 b\right )}{7 b x^7} \]

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Rubi [A]  time = 0.06, antiderivative size = 99, normalized size of antiderivative = 1.12, number of steps used = 8, number of rules used = 7, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {1489, 240, 212, 206, 203, 271, 264} \begin {gather*} a^{3/4} \tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )+a^{3/4} \tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )-\frac {3 \left (a x^4-b\right )^{3/4}}{7 x^7}-\frac {4 a \left (a x^4-b\right )^{3/4}}{7 b x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*(-b + a*x^4)^(3/4))/(7*x^7) - (4*a*(-b + a*x^4)^(3/4))/(7*b*x^3) + a^(3/4)*ArcTan[(a^(1/4)*x)/(-b + a*x^4)
^(1/4)] + a^(3/4)*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 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 264

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

Rule 271

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

Rule 1489

Int[((f_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), x_Symbol] :> Int[Expan
dIntegrand[(f*x)^m*(d + e*x^n)^q*(a + c*x^(2*n))^p, x], x] /; FreeQ[{a, c, d, e, f, m, q}, x] && EqQ[n2, 2*n]
&& IGtQ[n, 0] && IGtQ[p, 0]

Rubi steps

\begin {align*} \int \frac {-3 b+2 a x^8}{x^8 \sqrt [4]{-b+a x^4}} \, dx &=\int \left (\frac {2 a}{\sqrt [4]{-b+a x^4}}-\frac {3 b}{x^8 \sqrt [4]{-b+a x^4}}\right ) \, dx\\ &=(2 a) \int \frac {1}{\sqrt [4]{-b+a x^4}} \, dx-(3 b) \int \frac {1}{x^8 \sqrt [4]{-b+a x^4}} \, dx\\ &=-\frac {3 \left (-b+a x^4\right )^{3/4}}{7 x^7}-\frac {1}{7} (12 a) \int \frac {1}{x^4 \sqrt [4]{-b+a x^4}} \, dx+(2 a) \operatorname {Subst}\left (\int \frac {1}{1-a x^4} \, dx,x,\frac {x}{\sqrt [4]{-b+a x^4}}\right )\\ &=-\frac {3 \left (-b+a x^4\right )^{3/4}}{7 x^7}-\frac {4 a \left (-b+a x^4\right )^{3/4}}{7 b x^3}+a \operatorname {Subst}\left (\int \frac {1}{1-\sqrt {a} x^2} \, dx,x,\frac {x}{\sqrt [4]{-b+a x^4}}\right )+a \operatorname {Subst}\left (\int \frac {1}{1+\sqrt {a} x^2} \, dx,x,\frac {x}{\sqrt [4]{-b+a x^4}}\right )\\ &=-\frac {3 \left (-b+a x^4\right )^{3/4}}{7 x^7}-\frac {4 a \left (-b+a x^4\right )^{3/4}}{7 b x^3}+a^{3/4} \tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )+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 = 0.08, size = 88, normalized size = 1.00 \begin {gather*} a^{3/4} \tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )+a^{3/4} \tanh ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )-\frac {\left (a x^4-b\right )^{3/4} \left (4 a x^4+3 b\right )}{7 b x^7} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/7*((-b + a*x^4)^(3/4)*(3*b + 4*a*x^4))/(b*x^7) + a^(3/4)*ArcTan[(a^(1/4)*x)/(-b + a*x^4)^(1/4)] + a^(3/4)*A
rcTanh[(a^(1/4)*x)/(-b + a*x^4)^(1/4)]

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IntegrateAlgebraic [A]  time = 0.39, size = 88, normalized size = 1.00 \begin {gather*} \frac {\left (-3 b-4 a x^4\right ) \left (-b+a x^4\right )^{3/4}}{7 b x^7}+a^{3/4} \tan ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )+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[(-3*b + 2*a*x^8)/(x^8*(-b + a*x^4)^(1/4)),x]

[Out]

((-3*b - 4*a*x^4)*(-b + a*x^4)^(3/4))/(7*b*x^7) + a^(3/4)*ArcTan[(a^(1/4)*x)/(-b + a*x^4)^(1/4)] + a^(3/4)*Arc
Tanh[(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^8-3*b)/x^8/(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 {2 \, a x^{8} - 3 \, b}{{\left (a x^{4} - b\right )}^{\frac {1}{4}} x^{8}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*a*x^8-3*b)/x^8/(a*x^4-b)^(1/4),x, algorithm="giac")

[Out]

integrate((2*a*x^8 - 3*b)/((a*x^4 - b)^(1/4)*x^8), x)

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maple [F]  time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {2 a \,x^{8}-3 b}{x^{8} \left (a \,x^{4}-b \right )^{\frac {1}{4}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*a*x^8-3*b)/x^8/(a*x^4-b)^(1/4),x)

[Out]

int((2*a*x^8-3*b)/x^8/(a*x^4-b)^(1/4),x)

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maxima [A]  time = 0.43, size = 116, normalized size = 1.32 \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 {\frac {7 \, {\left (a x^{4} - b\right )}^{\frac {3}{4}} a}{x^{3}} - \frac {3 \, {\left (a x^{4} - b\right )}^{\frac {7}{4}}}{x^{7}}}{7 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*a*x^8-3*b)/x^8/(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)) - 1/7*(7*(a*x^4 - b)^(3/4)*a/x^3 - 3*(a*x^4 - b)^(7/4)/x^7)/b

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mupad [B]  time = 1.30, size = 71, normalized size = 0.81 \begin {gather*} \frac {2\,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 {{\left (a\,x^4-b\right )}^{3/4}\,\left (4\,a\,x^4+3\,b\right )}{7\,b\,x^7} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(3*b - 2*a*x^8)/(x^8*(a*x^4 - b)^(1/4)),x)

[Out]

(2*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) - ((a*x^4 - b)^(3/4)*(3*
b + 4*a*x^4))/(7*b*x^7)

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sympy [C]  time = 2.52, size = 320, normalized size = 3.64 \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 )}}{2 \sqrt [4]{b} \Gamma \left (\frac {5}{4}\right )} - 3 b \left (\begin {cases} - \frac {a^{\frac {7}{4}} \left (-1 + \frac {b}{a x^{4}}\right )^{\frac {3}{4}} e^{- \frac {i \pi }{4}} \Gamma \left (- \frac {7}{4}\right )}{4 b^{2} \Gamma \left (\frac {1}{4}\right )} - \frac {3 a^{\frac {3}{4}} \left (-1 + \frac {b}{a x^{4}}\right )^{\frac {3}{4}} e^{- \frac {i \pi }{4}} \Gamma \left (- \frac {7}{4}\right )}{16 b x^{4} \Gamma \left (\frac {1}{4}\right )} & \text {for}\: \left |{\frac {b}{a x^{4}}}\right | > 1 \\\frac {4 a^{\frac {15}{4}} x^{8} \left (1 - \frac {b}{a x^{4}}\right )^{\frac {3}{4}} \Gamma \left (- \frac {7}{4}\right )}{16 a^{2} b^{2} x^{8} \Gamma \left (\frac {1}{4}\right ) - 16 a b^{3} x^{4} \Gamma \left (\frac {1}{4}\right )} - \frac {a^{\frac {11}{4}} b x^{4} \left (1 - \frac {b}{a x^{4}}\right )^{\frac {3}{4}} \Gamma \left (- \frac {7}{4}\right )}{16 a^{2} b^{2} x^{8} \Gamma \left (\frac {1}{4}\right ) - 16 a b^{3} x^{4} \Gamma \left (\frac {1}{4}\right )} - \frac {3 a^{\frac {7}{4}} b^{2} \left (1 - \frac {b}{a x^{4}}\right )^{\frac {3}{4}} \Gamma \left (- \frac {7}{4}\right )}{16 a^{2} b^{2} x^{8} \Gamma \left (\frac {1}{4}\right ) - 16 a b^{3} x^{4} \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((2*a*x**8-3*b)/x**8/(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)/(2*b**(1/4)*gamma(5/4)) - 3*b*Piecewise((-a**(
7/4)*(-1 + b/(a*x**4))**(3/4)*exp(-I*pi/4)*gamma(-7/4)/(4*b**2*gamma(1/4)) - 3*a**(3/4)*(-1 + b/(a*x**4))**(3/
4)*exp(-I*pi/4)*gamma(-7/4)/(16*b*x**4*gamma(1/4)), Abs(b/(a*x**4)) > 1), (4*a**(15/4)*x**8*(1 - b/(a*x**4))**
(3/4)*gamma(-7/4)/(16*a**2*b**2*x**8*gamma(1/4) - 16*a*b**3*x**4*gamma(1/4)) - a**(11/4)*b*x**4*(1 - b/(a*x**4
))**(3/4)*gamma(-7/4)/(16*a**2*b**2*x**8*gamma(1/4) - 16*a*b**3*x**4*gamma(1/4)) - 3*a**(7/4)*b**2*(1 - b/(a*x
**4))**(3/4)*gamma(-7/4)/(16*a**2*b**2*x**8*gamma(1/4) - 16*a*b**3*x**4*gamma(1/4)), True))

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