\(\int \frac {x^2 (-b+a x^4)}{(b+a x^4)^{3/4}} \, dx\) [1016]

Optimal result

Integrand size = 24, antiderivative size = 77 \[ \int \frac {x^2 \left (-b+a x^4\right )}{\left (b+a x^4\right )^{3/4}} \, dx=\frac {1}{4} x^3 \sqrt [4]{b+a x^4}+\frac {7 b \arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right )}{8 a^{3/4}}-\frac {7 b \text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right )}{8 a^{3/4}} \]

[Out]

1/4*x^3*(a*x^4+b)^(1/4)+7/8*b*arctan(a^(1/4)*x/(a*x^4+b)^(1/4))/a^(3/4)-7/8*b*arctanh(a^(1/4)*x/(a*x^4+b)^(1/4
))/a^(3/4)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {470, 338, 304, 209, 212} \[ \int \frac {x^2 \left (-b+a x^4\right )}{\left (b+a x^4\right )^{3/4}} \, dx=\frac {7 b \arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4+b}}\right )}{8 a^{3/4}}-\frac {7 b \text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4+b}}\right )}{8 a^{3/4}}+\frac {1}{4} x^3 \sqrt [4]{a x^4+b} \]

[In]

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

[Out]

(x^3*(b + a*x^4)^(1/4))/4 + (7*b*ArcTan[(a^(1/4)*x)/(b + a*x^4)^(1/4)])/(8*a^(3/4)) - (7*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 470

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

Rubi steps \begin{align*} \text {integral}& = \frac {1}{4} x^3 \sqrt [4]{b+a x^4}-\frac {1}{4} (7 b) \int \frac {x^2}{\left (b+a x^4\right )^{3/4}} \, dx \\ & = \frac {1}{4} x^3 \sqrt [4]{b+a x^4}-\frac {1}{4} (7 b) \text {Subst}\left (\int \frac {x^2}{1-a x^4} \, dx,x,\frac {x}{\sqrt [4]{b+a x^4}}\right ) \\ & = \frac {1}{4} x^3 \sqrt [4]{b+a x^4}-\frac {(7 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 {(7 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 {1}{4} x^3 \sqrt [4]{b+a x^4}+\frac {7 b \arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right )}{8 a^{3/4}}-\frac {7 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 = 0.37 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00 \[ \int \frac {x^2 \left (-b+a x^4\right )}{\left (b+a x^4\right )^{3/4}} \, dx=\frac {1}{4} x^3 \sqrt [4]{b+a x^4}+\frac {7 b \arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right )}{8 a^{3/4}}-\frac {7 b \text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right )}{8 a^{3/4}} \]

[In]

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

[Out]

(x^3*(b + a*x^4)^(1/4))/4 + (7*b*ArcTan[(a^(1/4)*x)/(b + a*x^4)^(1/4)])/(8*a^(3/4)) - (7*b*ArcTanh[(a^(1/4)*x)
/(b + a*x^4)^(1/4)])/(8*a^(3/4))

Maple [A] (verified)

Time = 1.32 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.10

method	result	size
pseudoelliptic	\(-\frac {7 \left (-\frac {4 \left (a \,x^{4}+b \right )^{\frac {1}{4}} x^{3} a^{\frac {3}{4}}}{7}+\ln \left (\frac {-a^{\frac {1}{4}} x -\left (a \,x^{4}+b \right )^{\frac {1}{4}}}{a^{\frac {1}{4}} x -\left (a \,x^{4}+b \right )^{\frac {1}{4}}}\right ) b +2 \arctan \left (\frac {\left (a \,x^{4}+b \right )^{\frac {1}{4}}}{a^{\frac {1}{4}} x}\right ) b \right )}{16 a^{\frac {3}{4}}}\)	\(85\)

[In]

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

[Out]

-7/16/a^(3/4)*(-4/7*(a*x^4+b)^(1/4)*x^3*a^(3/4)+ln((-a^(1/4)*x-(a*x^4+b)^(1/4))/(a^(1/4)*x-(a*x^4+b)^(1/4)))*b
+2*arctan(1/a^(1/4)/x*(a*x^4+b)^(1/4))*b)

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.27 (sec) , antiderivative size = 184, normalized size of antiderivative = 2.39 \[ \int \frac {x^2 \left (-b+a x^4\right )}{\left (b+a x^4\right )^{3/4}} \, dx=\frac {1}{4} \, {\left (a x^{4} + b\right )}^{\frac {1}{4}} x^{3} - \frac {7}{16} \, \left (\frac {b^{4}}{a^{3}}\right )^{\frac {1}{4}} \log \left (\frac {7 \, {\left (a \left (\frac {b^{4}}{a^{3}}\right )^{\frac {1}{4}} x + {\left (a x^{4} + b\right )}^{\frac {1}{4}} b\right )}}{x}\right ) + \frac {7}{16} \, \left (\frac {b^{4}}{a^{3}}\right )^{\frac {1}{4}} \log \left (-\frac {7 \, {\left (a \left (\frac {b^{4}}{a^{3}}\right )^{\frac {1}{4}} x - {\left (a x^{4} + b\right )}^{\frac {1}{4}} b\right )}}{x}\right ) + \frac {7}{16} i \, \left (\frac {b^{4}}{a^{3}}\right )^{\frac {1}{4}} \log \left (-\frac {7 \, {\left (i \, a \left (\frac {b^{4}}{a^{3}}\right )^{\frac {1}{4}} x - {\left (a x^{4} + b\right )}^{\frac {1}{4}} b\right )}}{x}\right ) - \frac {7}{16} i \, \left (\frac {b^{4}}{a^{3}}\right )^{\frac {1}{4}} \log \left (-\frac {7 \, {\left (-i \, a \left (\frac {b^{4}}{a^{3}}\right )^{\frac {1}{4}} x - {\left (a x^{4} + b\right )}^{\frac {1}{4}} b\right )}}{x}\right ) \]

[In]

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

[Out]

1/4*(a*x^4 + b)^(1/4)*x^3 - 7/16*(b^4/a^3)^(1/4)*log(7*(a*(b^4/a^3)^(1/4)*x + (a*x^4 + b)^(1/4)*b)/x) + 7/16*(
b^4/a^3)^(1/4)*log(-7*(a*(b^4/a^3)^(1/4)*x - (a*x^4 + b)^(1/4)*b)/x) + 7/16*I*(b^4/a^3)^(1/4)*log(-7*(I*a*(b^4
/a^3)^(1/4)*x - (a*x^4 + b)^(1/4)*b)/x) - 7/16*I*(b^4/a^3)^(1/4)*log(-7*(-I*a*(b^4/a^3)^(1/4)*x - (a*x^4 + b)^
(1/4)*b)/x)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.66 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.01 \[ \int \frac {x^2 \left (-b+a x^4\right )}{\left (b+a x^4\right )^{3/4}} \, dx=\frac {a x^{7} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {a x^{4} e^{i \pi }}{b}} \right )}}{4 b^{\frac {3}{4}} \Gamma \left (\frac {11}{4}\right )} - \frac {\sqrt [4]{b} x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {a x^{4} e^{i \pi }}{b}} \right )}}{4 \Gamma \left (\frac {7}{4}\right )} \]

[In]

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

[Out]

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 185 vs. \(2 (57) = 114\).

Time = 0.27 (sec) , antiderivative size = 185, normalized size of antiderivative = 2.40 \[ \int \frac {x^2 \left (-b+a x^4\right )}{\left (b+a x^4\right )^{3/4}} \, dx=-\frac {1}{4} \, b {\left (\frac {2 \, \arctan \left (\frac {{\left (a x^{4} + b\right )}^{\frac {1}{4}}}{a^{\frac {1}{4}} x}\right )}{a^{\frac {3}{4}}} - \frac {\log \left (-\frac {a^{\frac {1}{4}} - \frac {{\left (a x^{4} + b\right )}^{\frac {1}{4}}}{x}}{a^{\frac {1}{4}} + \frac {{\left (a x^{4} + b\right )}^{\frac {1}{4}}}{x}}\right )}{a^{\frac {3}{4}}}\right )} - \frac {1}{16} \, a {\left (\frac {3 \, {\left (\frac {2 \, b \arctan \left (\frac {{\left (a x^{4} + b\right )}^{\frac {1}{4}}}{a^{\frac {1}{4}} x}\right )}{a^{\frac {3}{4}}} - \frac {b \log \left (-\frac {a^{\frac {1}{4}} - \frac {{\left (a x^{4} + b\right )}^{\frac {1}{4}}}{x}}{a^{\frac {1}{4}} + \frac {{\left (a x^{4} + b\right )}^{\frac {1}{4}}}{x}}\right )}{a^{\frac {3}{4}}}\right )}}{a} + \frac {4 \, {\left (a x^{4} + b\right )}^{\frac {1}{4}} b}{{\left (a^{2} - \frac {{\left (a x^{4} + b\right )} a}{x^{4}}\right )} x}\right )} \]

[In]

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

[Out]

-1/4*b*(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)) - 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))

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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