\(\int \frac {x^4 (-q+p x^4) \sqrt {q+p x^4}}{b x^8+a (q+p x^4)^4} \, dx\) [3047]

   Optimal result
   Rubi [F]
   Mathematica [C] (warning: unable to verify)
   Maple [C] (verified)
   Fricas [F(-1)]
   Sympy [F(-1)]
   Maxima [F]
   Giac [F(-1)]
   Mupad [F(-1)]

Optimal result

Integrand size = 43, antiderivative size = 457 \[ \int \frac {x^4 \left (-q+p x^4\right ) \sqrt {q+p x^4}}{b x^8+a \left (q+p x^4\right )^4} \, dx=-\frac {\sqrt {2+\sqrt {2}} \arctan \left (\frac {\left (\sqrt {\frac {2}{2-\sqrt {2}}} \sqrt [8]{a} \sqrt [8]{b}-\frac {2 \sqrt [8]{a} \sqrt [8]{b}}{\sqrt {2-\sqrt {2}}}\right ) x \sqrt {q+p x^4}}{-\sqrt [4]{a} q+\sqrt [4]{b} x^2-\sqrt [4]{a} p x^4}\right )}{8 a^{3/8} b^{5/8}}+\frac {\sqrt {2-\sqrt {2}} \arctan \left (\frac {\sqrt {2+\sqrt {2}} \sqrt [8]{a} \sqrt [8]{b} x \sqrt {q+p x^4}}{\sqrt [4]{a} q-\sqrt [4]{b} x^2+\sqrt [4]{a} p x^4}\right )}{8 a^{3/8} b^{5/8}}-\frac {\sqrt {2+\sqrt {2}} \text {arctanh}\left (\frac {\frac {\sqrt [8]{a} q}{\sqrt {2-\sqrt {2}} \sqrt [8]{b}}+\frac {\sqrt [8]{b} x^2}{\sqrt {2-\sqrt {2}} \sqrt [8]{a}}+\frac {\sqrt [8]{a} p x^4}{\sqrt {2-\sqrt {2}} \sqrt [8]{b}}}{x \sqrt {q+p x^4}}\right )}{8 a^{3/8} b^{5/8}}+\frac {\sqrt {2-\sqrt {2}} \text {arctanh}\left (\frac {\frac {\sqrt [8]{a} q}{\sqrt {2+\sqrt {2}} \sqrt [8]{b}}+\frac {\sqrt [8]{b} x^2}{\sqrt {2+\sqrt {2}} \sqrt [8]{a}}+\frac {\sqrt [8]{a} p x^4}{\sqrt {2+\sqrt {2}} \sqrt [8]{b}}}{x \sqrt {q+p x^4}}\right )}{8 a^{3/8} b^{5/8}} \]

[Out]

-1/8*(2+2^(1/2))^(1/2)*arctan((2^(1/2)/(2-2^(1/2))^(1/2)*a^(1/8)*b^(1/8)-2*a^(1/8)*b^(1/8)/(2-2^(1/2))^(1/2))*
x*(p*x^4+q)^(1/2)/(-a^(1/4)*q+b^(1/4)*x^2-a^(1/4)*p*x^4))/a^(3/8)/b^(5/8)+1/8*(2-2^(1/2))^(1/2)*arctan((2+2^(1
/2))^(1/2)*a^(1/8)*b^(1/8)*x*(p*x^4+q)^(1/2)/(a^(1/4)*q-b^(1/4)*x^2+a^(1/4)*p*x^4))/a^(3/8)/b^(5/8)-1/8*(2+2^(
1/2))^(1/2)*arctanh((a^(1/8)*q/(2-2^(1/2))^(1/2)/b^(1/8)+b^(1/8)*x^2/(2-2^(1/2))^(1/2)/a^(1/8)+a^(1/8)*p*x^4/(
2-2^(1/2))^(1/2)/b^(1/8))/x/(p*x^4+q)^(1/2))/a^(3/8)/b^(5/8)+1/8*(2-2^(1/2))^(1/2)*arctanh((a^(1/8)*q/(2+2^(1/
2))^(1/2)/b^(1/8)+b^(1/8)*x^2/(2+2^(1/2))^(1/2)/a^(1/8)+a^(1/8)*p*x^4/(2+2^(1/2))^(1/2)/b^(1/8))/x/(p*x^4+q)^(
1/2))/a^(3/8)/b^(5/8)

Rubi [F]

\[ \int \frac {x^4 \left (-q+p x^4\right ) \sqrt {q+p x^4}}{b x^8+a \left (q+p x^4\right )^4} \, dx=\int \frac {x^4 \left (-q+p x^4\right ) \sqrt {q+p x^4}}{b x^8+a \left (q+p x^4\right )^4} \, dx \]

[In]

Int[(x^4*(-q + p*x^4)*Sqrt[q + p*x^4])/(b*x^8 + a*(q + p*x^4)^4),x]

[Out]

q*Defer[Int][(x^4*Sqrt[q + p*x^4])/(-(b*x^8) - a*(q + p*x^4)^4), x] + p*Defer[Int][(x^8*Sqrt[q + p*x^4])/(b*x^
8 + a*(q + p*x^4)^4), x]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {q x^4 \sqrt {q+p x^4}}{-a q^4-4 a p q^3 x^4-b \left (1+\frac {6 a p^2 q^2}{b}\right ) x^8-4 a p^3 q x^{12}-a p^4 x^{16}}+\frac {p x^8 \sqrt {q+p x^4}}{a q^4+4 a p q^3 x^4+b \left (1+\frac {6 a p^2 q^2}{b}\right ) x^8+4 a p^3 q x^{12}+a p^4 x^{16}}\right ) \, dx \\ & = p \int \frac {x^8 \sqrt {q+p x^4}}{a q^4+4 a p q^3 x^4+b \left (1+\frac {6 a p^2 q^2}{b}\right ) x^8+4 a p^3 q x^{12}+a p^4 x^{16}} \, dx+q \int \frac {x^4 \sqrt {q+p x^4}}{-a q^4-4 a p q^3 x^4-b \left (1+\frac {6 a p^2 q^2}{b}\right ) x^8-4 a p^3 q x^{12}-a p^4 x^{16}} \, dx \\ & = p \int \frac {x^8 \sqrt {q+p x^4}}{b x^8+a \left (q+p x^4\right )^4} \, dx+q \int \frac {x^4 \sqrt {q+p x^4}}{-b x^8-a \left (q+p x^4\right )^4} \, dx \\ \end{align*}

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.

Time = 17.59 (sec) , antiderivative size = 15065, normalized size of antiderivative = 32.96 \[ \int \frac {x^4 \left (-q+p x^4\right ) \sqrt {q+p x^4}}{b x^8+a \left (q+p x^4\right )^4} \, dx=\text {Result too large to show} \]

[In]

Integrate[(x^4*(-q + p*x^4)*Sqrt[q + p*x^4])/(b*x^8 + a*(q + p*x^4)^4),x]

[Out]

Result too large to show

Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.70 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.09

method result size
default \(\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (a \,\textit {\_Z}^{8}+b \right )}{\sum }\frac {\ln \left (\frac {-\textit {\_R} x +\sqrt {p \,x^{4}+q}}{x}\right )}{\textit {\_R}^{5}}}{8 a}\) \(40\)
pseudoelliptic \(\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (a \,\textit {\_Z}^{8}+b \right )}{\sum }\frac {\ln \left (\frac {-\textit {\_R} x +\sqrt {p \,x^{4}+q}}{x}\right )}{\textit {\_R}^{5}}}{8 a}\) \(40\)
elliptic \(\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (16 a \,\textit {\_Z}^{8}+b \right )}{\sum }\frac {\ln \left (\frac {\sqrt {p \,x^{4}+q}\, \sqrt {2}}{2 x}-\textit {\_R} \right )}{\textit {\_R}^{5}}\right ) \sqrt {2}}{64 a}\) \(47\)

[In]

int(x^4*(p*x^4-q)*(p*x^4+q)^(1/2)/(b*x^8+a*(p*x^4+q)^4),x,method=_RETURNVERBOSE)

[Out]

1/8*sum(ln((-_R*x+(p*x^4+q)^(1/2))/x)/_R^5,_R=RootOf(_Z^8*a+b))/a

Fricas [F(-1)]

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

[In]

integrate(x^4*(p*x^4-q)*(p*x^4+q)^(1/2)/(b*x^8+a*(p*x^4+q)^4),x, algorithm="fricas")

[Out]

Timed out

Sympy [F(-1)]

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

[In]

integrate(x**4*(p*x**4-q)*(p*x**4+q)**(1/2)/(b*x**8+a*(p*x**4+q)**4),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {x^4 \left (-q+p x^4\right ) \sqrt {q+p x^4}}{b x^8+a \left (q+p x^4\right )^4} \, dx=\int { \frac {\sqrt {p x^{4} + q} {\left (p x^{4} - q\right )} x^{4}}{b x^{8} + {\left (p x^{4} + q\right )}^{4} a} \,d x } \]

[In]

integrate(x^4*(p*x^4-q)*(p*x^4+q)^(1/2)/(b*x^8+a*(p*x^4+q)^4),x, algorithm="maxima")

[Out]

integrate(sqrt(p*x^4 + q)*(p*x^4 - q)*x^4/(b*x^8 + (p*x^4 + q)^4*a), x)

Giac [F(-1)]

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

[In]

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

[Out]

Timed out

Mupad [F(-1)]

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

[In]

int(-(x^4*(q + p*x^4)^(1/2)*(q - p*x^4))/(a*(q + p*x^4)^4 + b*x^8),x)

[Out]

-int((x^4*(q + p*x^4)^(1/2)*(q - p*x^4))/(a*(q + p*x^4)^4 + b*x^8), x)