\(\int \frac {x^6}{(b+a x^4)^{3/4} (-b^2+a^2 x^8)} \, dx\) [1569]

Optimal result

Integrand size = 30, antiderivative size = 107 \[ \int \frac {x^6}{\left (b+a x^4\right )^{3/4} \left (-b^2+a^2 x^8\right )} \, dx=\frac {x^3}{6 a b \left (b+a x^4\right )^{3/4}}+\frac {\arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right )}{4\ 2^{3/4} a^{7/4} b}-\frac {\text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right )}{4\ 2^{3/4} a^{7/4} b} \]

[Out]

1/6*x^3/a/b/(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^(7/4)/b-1/8*arctanh(2^(1/4
)*a^(1/4)*x/(a*x^4+b)^(1/4))*2^(1/4)/a^(7/4)/b

Rubi [C] (verified)

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

Time = 0.05 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.41, 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} \[ \int \frac {x^6}{\left (b+a x^4\right )^{3/4} \left (-b^2+a^2 x^8\right )} \, dx=-\frac {x^7 \operatorname {Hypergeometric2F1}\left (1,\frac {7}{4},\frac {11}{4},\frac {2 a x^4}{a x^4+b}\right )}{7 b \left (a x^4+b\right )^{7/4}} \]

[In]

Int[x^6/((b + a*x^4)^(3/4)*(-b^2 + a^2*x^8)),x]

[Out]

-1/7*(x^7*Hypergeometric2F1[1, 7/4, 11/4, (2*a*x^4)/(b + a*x^4)])/(b*(b + a*x^4)^(7/4))

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^6}{\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^6}{\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 {x^7 \operatorname {Hypergeometric2F1}\left (1,\frac {7}{4},\frac {11}{4},\frac {2 a x^4}{b+a x^4}\right )}{7 b \left (b+a x^4\right )^{7/4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.71 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.90 \[ \int \frac {x^6}{\left (b+a x^4\right )^{3/4} \left (-b^2+a^2 x^8\right )} \, dx=\frac {\frac {4 a^{3/4} x^3}{\left (b+a x^4\right )^{3/4}}+3 \sqrt [4]{2} \arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right )-3 \sqrt [4]{2} \text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right )}{24 a^{7/4} b} \]

[In]

Integrate[x^6/((b + a*x^4)^(3/4)*(-b^2 + a^2*x^8)),x]

[Out]

((4*a^(3/4)*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)] - 3*2^(1/4)*ArcTa
nh[(2^(1/4)*a^(1/4)*x)/(b + a*x^4)^(1/4)])/(24*a^(7/4)*b)

Maple [A] (verified)

Time = 0.40 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.17

method	result	size
pseudoelliptic	\(\frac {-6 \arctan \left (\frac {2^{\frac {3}{4}} \left (a \,x^{4}+b \right )^{\frac {1}{4}}}{2 a^{\frac {1}{4}} x}\right ) 2^{\frac {1}{4}} a^{\frac {1}{4}} \left (a \,x^{4}+b \right )^{\frac {3}{4}}-3 \ln \left (\frac {-x 2^{\frac {1}{4}} a^{\frac {1}{4}}-\left (a \,x^{4}+b \right )^{\frac {1}{4}}}{x 2^{\frac {1}{4}} a^{\frac {1}{4}}-\left (a \,x^{4}+b \right )^{\frac {1}{4}}}\right ) 2^{\frac {1}{4}} a^{\frac {1}{4}} \left (a \,x^{4}+b \right )^{\frac {3}{4}}+8 a \,x^{3}}{48 a^{2} \left (a \,x^{4}+b \right )^{\frac {3}{4}} b}\)	\(125\)

[In]

int(x^6/(a*x^4+b)^(3/4)/(a^2*x^8-b^2),x,method=_RETURNVERBOSE)

[Out]

1/48*(-6*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*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)+8*a*x^3)/a^2/(a*x^4+b)
^(3/4)/b

Fricas [F(-1)]

Timed out. \[ \int \frac {x^6}{\left (b+a x^4\right )^{3/4} \left (-b^2+a^2 x^8\right )} \, dx=\text {Timed out} \]

[In]

integrate(x^6/(a*x^4+b)^(3/4)/(a^2*x^8-b^2),x, algorithm="fricas")

[Out]

Timed out

Sympy [F(-1)]

Timed out. \[ \int \frac {x^6}{\left (b+a x^4\right )^{3/4} \left (-b^2+a^2 x^8\right )} \, dx=\text {Timed out} \]

[In]

integrate(x**6/(a*x**4+b)**(3/4)/(a**2*x**8-b**2),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {x^6}{\left (b+a x^4\right )^{3/4} \left (-b^2+a^2 x^8\right )} \, dx=\int { \frac {x^{6}}{{\left (a^{2} x^{8} - b^{2}\right )} {\left (a x^{4} + b\right )}^{\frac {3}{4}}} \,d x } \]

[In]

integrate(x^6/(a*x^4+b)^(3/4)/(a^2*x^8-b^2),x, algorithm="maxima")

[Out]

integrate(x^6/((a^2*x^8 - b^2)*(a*x^4 + b)^(3/4)), x)

Giac [F]

\[ \int \frac {x^6}{\left (b+a x^4\right )^{3/4} \left (-b^2+a^2 x^8\right )} \, dx=\int { \frac {x^{6}}{{\left (a^{2} x^{8} - b^{2}\right )} {\left (a x^{4} + b\right )}^{\frac {3}{4}}} \,d x } \]

[In]

integrate(x^6/(a*x^4+b)^(3/4)/(a^2*x^8-b^2),x, algorithm="giac")

[Out]

integrate(x^6/((a^2*x^8 - b^2)*(a*x^4 + b)^(3/4)), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {x^6}{\left (b+a x^4\right )^{3/4} \left (-b^2+a^2 x^8\right )} \, dx=-\int \frac {x^6}{\left (b^2-a^2\,x^8\right )\,{\left (a\,x^4+b\right )}^{3/4}} \,d x \]

[In]

int(-x^6/((b^2 - a^2*x^8)*(b + a*x^4)^(3/4)),x)

[Out]

-int(x^6/((b^2 - a^2*x^8)*(b + a*x^4)^(3/4)), x)