\(\int \frac {x^2}{(b+a x^4)^{3/4} (-b^2+a^2 x^8)} \, dx\) [1499]

Optimal result

Integrand size = 30, antiderivative size = 104 \[ \int \frac {x^2}{\left (b+a x^4\right )^{3/4} \left (-b^2+a^2 x^8\right )} \, dx=-\frac {x^3}{6 b^2 \left (b+a x^4\right )^{3/4}}+\frac {\arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right )}{4\ 2^{3/4} a^{3/4} b^2}-\frac {\text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right )}{4\ 2^{3/4} a^{3/4} b^2} \]

[Out]

-1/6*x^3/b^2/(a*x^4+b)^(3/4)+1/8*arctan(2^(1/4)*a^(1/4)*x/(a*x^4+b)^(1/4))*2^(1/4)/a^(3/4)/b^2-1/8*arctanh(2^(
1/4)*a^(1/4)*x/(a*x^4+b)^(1/4))*2^(1/4)/a^(3/4)/b^2

Rubi [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 2.41 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.17, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {1493, 525, 524} \[ \int \frac {x^2}{\left (b+a x^4\right )^{3/4} \left (-b^2+a^2 x^8\right )} \, dx=-\frac {4 x^3 \operatorname {Gamma}\left (\frac {7}{4}\right ) \left (11 \left (-4 a^2 x^8-3 a b x^4+7 b^2\right ) \operatorname {Hypergeometric2F1}\left (1,1,\frac {11}{4},-\frac {2 a x^4}{b-a x^4}\right )-32 a x^4 \left (a x^4+b\right ) \operatorname {Hypergeometric2F1}\left (2,2,\frac {15}{4},-\frac {2 a x^4}{b-a x^4}\right )\right )}{693 b^2 \operatorname {Gamma}\left (\frac {3}{4}\right ) \left (b-a x^4\right )^2 \left (a x^4+b\right )^{3/4}} \]

[In]

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

[Out]

(-4*x^3*Gamma[7/4]*(11*(7*b^2 - 3*a*b*x^4 - 4*a^2*x^8)*Hypergeometric2F1[1, 1, 11/4, (-2*a*x^4)/(b - a*x^4)] -
32*a*x^4*(b + a*x^4)*Hypergeometric2F1[2, 2, 15/4, (-2*a*x^4)/(b - a*x^4)]))/(693*b^2*(b - a*x^4)^2*(b + a*x^
4)^(3/4)*Gamma[3/4])

Rule 524

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*
((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; Free
Q[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a
, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rule 525

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

Rule 1493

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {x^2}{\left (-b+a x^4\right ) \left (b+a x^4\right )^{7/4}} \, dx \\ & = \frac {\left (1+\frac {a x^4}{b}\right )^{3/4} \int \frac {x^2}{\left (-b+a x^4\right ) \left (1+\frac {a x^4}{b}\right )^{7/4}} \, dx}{b \left (b+a x^4\right )^{3/4}} \\ & = -\frac {4 x^3 \operatorname {Gamma}\left (\frac {7}{4}\right ) \left (11 \left (7 b^2-3 a b x^4-4 a^2 x^8\right ) \operatorname {Hypergeometric2F1}\left (1,1,\frac {11}{4},-\frac {2 a x^4}{b-a x^4}\right )-32 a x^4 \left (b+a x^4\right ) \operatorname {Hypergeometric2F1}\left (2,2,\frac {15}{4},-\frac {2 a x^4}{b-a x^4}\right )\right )}{693 b^2 \left (b-a x^4\right )^2 \left (b+a x^4\right )^{3/4} \operatorname {Gamma}\left (\frac {3}{4}\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.51 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.92 \[ \int \frac {x^2}{\left (b+a x^4\right )^{3/4} \left (-b^2+a^2 x^8\right )} \, dx=-\frac {\frac {4 x^3}{\left (b+a x^4\right )^{3/4}}-\frac {3 \sqrt [4]{2} \arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right )}{a^{3/4}}+\frac {3 \sqrt [4]{2} \text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right )}{a^{3/4}}}{24 b^2} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.47 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.19

method	result	size
pseudoelliptic	\(-\frac {a \,x^{3}+\frac {3 \arctan \left (\frac {2^{\frac {3}{4}} \left (a \,x^{4}+b \right )^{\frac {1}{4}}}{2 a^{\frac {1}{4}} x}\right ) 2^{\frac {1}{4}} a^{\frac {1}{4}} \left (a \,x^{4}+b \right )^{\frac {3}{4}}}{4}+\frac {3 \ln \left (\frac {-x 2^{\frac {1}{4}} a^{\frac {1}{4}}-\left (a \,x^{4}+b \right )^{\frac {1}{4}}}{x 2^{\frac {1}{4}} a^{\frac {1}{4}}-\left (a \,x^{4}+b \right )^{\frac {1}{4}}}\right ) 2^{\frac {1}{4}} a^{\frac {1}{4}} \left (a \,x^{4}+b \right )^{\frac {3}{4}}}{8}}{6 \left (a \,x^{4}+b \right )^{\frac {3}{4}} a \,b^{2}}\)	\(124\)

[In]

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

[Out]

-1/6/(a*x^4+b)^(3/4)*(a*x^3+3/4*arctan(1/2*2^(3/4)/a^(1/4)/x*(a*x^4+b)^(1/4))*2^(1/4)*a^(1/4)*(a*x^4+b)^(3/4)+
3/8*ln((-x*2^(1/4)*a^(1/4)-(a*x^4+b)^(1/4))/(x*2^(1/4)*a^(1/4)-(a*x^4+b)^(1/4)))*2^(1/4)*a^(1/4)*(a*x^4+b)^(3/
4))/a/b^2

Fricas [F(-1)]

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

[In]

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

[Out]

Timed out

Sympy [F]

\[ \int \frac {x^2}{\left (b+a x^4\right )^{3/4} \left (-b^2+a^2 x^8\right )} \, dx=\int \frac {x^{2}}{\left (a x^{4} - b\right ) \left (a x^{4} + b\right )^{\frac {7}{4}}}\, dx \]

[In]

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

[Out]

Integral(x**2/((a*x**4 - b)*(a*x**4 + b)**(7/4)), x)

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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