\(\int \frac {-b+a x^8}{x^2 (-b+a x^4)^{3/4}} \, dx\) [1207]

Optimal result

Integrand size = 26, antiderivative size = 88 \[ \int \frac {-b+a x^8}{x^2 \left (-b+a x^4\right )^{3/4}} \, dx=\frac {\left (-4+x^4\right ) \sqrt [4]{-b+a x^4}}{4 x}-\frac {3 b \arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )}{8 a^{3/4}}+\frac {3 b \text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )}{8 a^{3/4}} \]

[Out]

1/4*(x^4-4)*(a*x^4-b)^(1/4)/x-3/8*b*arctan(a^(1/4)*x/(a*x^4-b)^(1/4))/a^(3/4)+3/8*b*arctanh(a^(1/4)*x/(a*x^4-b
)^(1/4))/a^(3/4)

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.15, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {1501, 462, 338, 304, 209, 212} \[ \int \frac {-b+a x^8}{x^2 \left (-b+a x^4\right )^{3/4}} \, dx=-\frac {3 b \arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )}{8 a^{3/4}}+\frac {3 b \text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )}{8 a^{3/4}}-\frac {\sqrt [4]{a x^4-b}}{x}+\frac {1}{4} x^3 \sqrt [4]{a x^4-b} \]

[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 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 304

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 338

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 462

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 1501

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*} \text {integral}& = \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) \text {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) \text {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) \text {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 \arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )}{8 a^{3/4}}+\frac {3 b \text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )}{8 a^{3/4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.49 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00 \[ \int \frac {-b+a x^8}{x^2 \left (-b+a x^4\right )^{3/4}} \, dx=\frac {\left (-4+x^4\right ) \sqrt [4]{-b+a x^4}}{4 x}-\frac {3 b \arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )}{8 a^{3/4}}+\frac {3 b \text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )}{8 a^{3/4}} \]

[In]

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

Maple [A] (verified)

Time = 1.27 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.12

method	result	size
pseudoelliptic	\(\frac {\frac {\left (a \,x^{4}-b \right )^{\frac {1}{4}} a^{\frac {3}{4}} \left (x^{4}-4\right )}{4}+\frac {3 b x \left (\ln \left (\frac {-a^{\frac {1}{4}} x -\left (a \,x^{4}-b \right )^{\frac {1}{4}}}{a^{\frac {1}{4}} x -\left (a \,x^{4}-b \right )^{\frac {1}{4}}}\right )+2 \arctan \left (\frac {\left (a \,x^{4}-b \right )^{\frac {1}{4}}}{a^{\frac {1}{4}} x}\right )\right )}{16}}{x \,a^{\frac {3}{4}}}\)	\(99\)

[In]

int((a*x^8-b)/x^2/(a*x^4-b)^(3/4),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [F(-1)]

Timed out. \[ \int \frac {-b+a x^8}{x^2 \left (-b+a x^4\right )^{3/4}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.32 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.43 \[ \int \frac {-b+a x^8}{x^2 \left (-b+a x^4\right )^{3/4}} \, dx=- \frac {a x^{7} e^{\frac {i \pi }{4}} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {a x^{4}}{b}} \right )}}{4 b^{\frac {3}{4}} \Gamma \left (\frac {11}{4}\right )} - b \left (\begin {cases} - \frac {\sqrt [4]{a} \sqrt [4]{-1 + \frac {b}{a x^{4}}} e^{\frac {i \pi }{4}} \Gamma \left (- \frac {1}{4}\right )}{4 b \Gamma \left (\frac {3}{4}\right )} & \text {for}\: \left |{\frac {b}{a x^{4}}}\right | > 1 \\- \frac {\sqrt [4]{a} \sqrt [4]{1 - \frac {b}{a x^{4}}} \Gamma \left (- \frac {1}{4}\right )}{4 b \Gamma \left (\frac {3}{4}\right )} & \text {otherwise} \end {cases}\right ) \]

[In]

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

[Out]

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 140 vs. \(2 (68) = 136\).

Time = 0.31 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.59 \[ \int \frac {-b+a x^8}{x^2 \left (-b+a x^4\right )^{3/4}} \, dx=\frac {1}{16} \, a {\left (\frac {3 \, {\left (\frac {2 \, b \arctan \left (\frac {{\left (a x^{4} - b\right )}^{\frac {1}{4}}}{a^{\frac {1}{4}} x}\right )}{a^{\frac {3}{4}}} - \frac {b \log \left (-\frac {a^{\frac {1}{4}} - \frac {{\left (a x^{4} - b\right )}^{\frac {1}{4}}}{x}}{a^{\frac {1}{4}} + \frac {{\left (a x^{4} - b\right )}^{\frac {1}{4}}}{x}}\right )}{a^{\frac {3}{4}}}\right )}}{a} + \frac {4 \, {\left (a x^{4} - b\right )}^{\frac {1}{4}} b}{{\left (a^{2} - \frac {{\left (a x^{4} - b\right )} a}{x^{4}}\right )} x}\right )} - \frac {{\left (a x^{4} - b\right )}^{\frac {1}{4}}}{x} \]

[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

Giac [F]

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

[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)

Mupad [F(-1)]

Timed out. \[ \int \frac {-b+a x^8}{x^2 \left (-b+a x^4\right )^{3/4}} \, dx=-\int \frac {b-a\,x^8}{x^2\,{\left (a\,x^4-b\right )}^{3/4}} \,d x \]

[In]

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

[Out]

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