\(\int \frac {(-q+p x^4) \sqrt {q+p x^4}}{b x^4+a (q+p x^4)^2} \, dx\) [2949]

Optimal result

Integrand size = 40, antiderivative size = 352 \[ \int \frac {\left (-q+p x^4\right ) \sqrt {q+p x^4}}{b x^4+a \left (q+p x^4\right )^2} \, dx=\frac {1}{4} \text {RootSum}\left [16 a p^2 q^2-32 a p^{3/2} q^{3/2} \text {$\#$1}^2+16 b \text {$\#$1}^4+24 a p q \text {$\#$1}^4-8 a \sqrt {p} \sqrt {q} \text {$\#$1}^6+a \text {$\#$1}^8\&,\frac {8 p^{3/2} q^{3/2} \log (x)-8 p^{3/2} q^{3/2} \log \left (\sqrt {q}+\sqrt {p} x^2+\sqrt {q+p x^4}-x \text {$\#$1}\right )-4 p q \log (x) \text {$\#$1}^2+4 p q \log \left (\sqrt {q}+\sqrt {p} x^2+\sqrt {q+p x^4}-x \text {$\#$1}\right ) \text {$\#$1}^2-2 \sqrt {p} \sqrt {q} \log (x) \text {$\#$1}^4+2 \sqrt {p} \sqrt {q} \log \left (\sqrt {q}+\sqrt {p} x^2+\sqrt {q+p x^4}-x \text {$\#$1}\right ) \text {$\#$1}^4+\log (x) \text {$\#$1}^6-\log \left (\sqrt {q}+\sqrt {p} x^2+\sqrt {q+p x^4}-x \text {$\#$1}\right ) \text {$\#$1}^6}{8 a p^{3/2} q^{3/2} \text {$\#$1}-8 b \text {$\#$1}^3-12 a p q \text {$\#$1}^3+6 a \sqrt {p} \sqrt {q} \text {$\#$1}^5-a \text {$\#$1}^7}\&\right ] \] Output:

Unintegrable
 

Mathematica [A] (verified)

Time = 1.89 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.37 \[ \int \frac {\left (-q+p x^4\right ) \sqrt {q+p x^4}}{b x^4+a \left (q+p x^4\right )^2} \, dx=-\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x \sqrt {q+p x^4}}{-\sqrt {b} x^2+\sqrt {a} \left (q+p x^4\right )}\right )+\text {arctanh}\left (\frac {\sqrt {b} x^2+\sqrt {a} \left (q+p x^4\right )}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x \sqrt {q+p x^4}}\right )}{2 \sqrt {2} a^{3/4} \sqrt [4]{b}} \] Input:

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

Output:

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

Rubi [F]

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {\left (p x^4-q\right ) \sqrt {p x^4+q}}{a \left (p x^4+q\right )^2+b x^4} \, dx\)

\(\Big \downarrow \) 2091

\(\displaystyle \int \frac {\left (p x^4-q\right ) \sqrt {p x^4+q}}{x^4 (2 a p q+b)+a p^2 x^8+a q^2}dx\)

\(\Big \downarrow \) 7279

\(\displaystyle \int \left (\frac {\sqrt {p x^4+q} \left (p-\frac {p \sqrt {4 a p q+b}}{\sqrt {b}}\right )}{-\sqrt {b} \sqrt {4 a p q+b}+2 a p^2 x^4+2 a p q+b}+\frac {\sqrt {p x^4+q} \left (\frac {p \sqrt {4 a p q+b}}{\sqrt {b}}+p\right )}{\sqrt {b} \sqrt {4 a p q+b}+2 a p^2 x^4+2 a p q+b}\right )dx\)

\(\Big \downarrow \) 7239

\(\displaystyle \int \frac {\left (p x^4-q\right ) \sqrt {p x^4+q}}{x^4 (2 a p q+b)+a p^2 x^8+a q^2}dx\)

\(\Big \downarrow \) 7279

\(\displaystyle \int \left (\frac {\sqrt {p x^4+q} \left (p-\frac {p \sqrt {4 a p q+b}}{\sqrt {b}}\right )}{-\sqrt {b} \sqrt {4 a p q+b}+2 a p^2 x^4+2 a p q+b}+\frac {\sqrt {p x^4+q} \left (\frac {p \sqrt {4 a p q+b}}{\sqrt {b}}+p\right )}{\sqrt {b} \sqrt {4 a p q+b}+2 a p^2 x^4+2 a p q+b}\right )dx\)

\(\Big \downarrow \) 7239

\(\displaystyle \int \frac {\left (p x^4-q\right ) \sqrt {p x^4+q}}{x^4 (2 a p q+b)+a p^2 x^8+a q^2}dx\)

\(\Big \downarrow \) 7279

\(\displaystyle \int \left (\frac {\sqrt {p x^4+q} \left (p-\frac {p \sqrt {4 a p q+b}}{\sqrt {b}}\right )}{-\sqrt {b} \sqrt {4 a p q+b}+2 a p^2 x^4+2 a p q+b}+\frac {\sqrt {p x^4+q} \left (\frac {p \sqrt {4 a p q+b}}{\sqrt {b}}+p\right )}{\sqrt {b} \sqrt {4 a p q+b}+2 a p^2 x^4+2 a p q+b}\right )dx\)

\(\Big \downarrow \) 7239

\(\displaystyle \int \frac {\left (p x^4-q\right ) \sqrt {p x^4+q}}{x^4 (2 a p q+b)+a p^2 x^8+a q^2}dx\)

\(\Big \downarrow \) 7279

\(\displaystyle \int \left (\frac {\sqrt {p x^4+q} \left (p-\frac {p \sqrt {4 a p q+b}}{\sqrt {b}}\right )}{-\sqrt {b} \sqrt {4 a p q+b}+2 a p^2 x^4+2 a p q+b}+\frac {\sqrt {p x^4+q} \left (\frac {p \sqrt {4 a p q+b}}{\sqrt {b}}+p\right )}{\sqrt {b} \sqrt {4 a p q+b}+2 a p^2 x^4+2 a p q+b}\right )dx\)

\(\Big \downarrow \) 7239

\(\displaystyle \int \frac {\left (p x^4-q\right ) \sqrt {p x^4+q}}{x^4 (2 a p q+b)+a p^2 x^8+a q^2}dx\)

\(\Big \downarrow \) 7279

\(\displaystyle \int \left (\frac {\sqrt {p x^4+q} \left (p-\frac {p \sqrt {4 a p q+b}}{\sqrt {b}}\right )}{-\sqrt {b} \sqrt {4 a p q+b}+2 a p^2 x^4+2 a p q+b}+\frac {\sqrt {p x^4+q} \left (\frac {p \sqrt {4 a p q+b}}{\sqrt {b}}+p\right )}{\sqrt {b} \sqrt {4 a p q+b}+2 a p^2 x^4+2 a p q+b}\right )dx\)

\(\Big \downarrow \) 7239

\(\displaystyle \int \frac {\left (p x^4-q\right ) \sqrt {p x^4+q}}{x^4 (2 a p q+b)+a p^2 x^8+a q^2}dx\)

\(\Big \downarrow \) 7279

\(\displaystyle \int \left (\frac {\sqrt {p x^4+q} \left (p-\frac {p \sqrt {4 a p q+b}}{\sqrt {b}}\right )}{-\sqrt {b} \sqrt {4 a p q+b}+2 a p^2 x^4+2 a p q+b}+\frac {\sqrt {p x^4+q} \left (\frac {p \sqrt {4 a p q+b}}{\sqrt {b}}+p\right )}{\sqrt {b} \sqrt {4 a p q+b}+2 a p^2 x^4+2 a p q+b}\right )dx\)

\(\Big \downarrow \) 7239

\(\displaystyle \int \frac {\left (p x^4-q\right ) \sqrt {p x^4+q}}{x^4 (2 a p q+b)+a p^2 x^8+a q^2}dx\)

\(\Big \downarrow \) 7279

\(\displaystyle \int \left (\frac {\sqrt {p x^4+q} \left (p-\frac {p \sqrt {4 a p q+b}}{\sqrt {b}}\right )}{-\sqrt {b} \sqrt {4 a p q+b}+2 a p^2 x^4+2 a p q+b}+\frac {\sqrt {p x^4+q} \left (\frac {p \sqrt {4 a p q+b}}{\sqrt {b}}+p\right )}{\sqrt {b} \sqrt {4 a p q+b}+2 a p^2 x^4+2 a p q+b}\right )dx\)

\(\Big \downarrow \) 7239

\(\displaystyle \int \frac {\left (p x^4-q\right ) \sqrt {p x^4+q}}{x^4 (2 a p q+b)+a p^2 x^8+a q^2}dx\)

\(\Big \downarrow \) 7279

\(\displaystyle \int \left (\frac {\sqrt {p x^4+q} \left (p-\frac {p \sqrt {4 a p q+b}}{\sqrt {b}}\right )}{-\sqrt {b} \sqrt {4 a p q+b}+2 a p^2 x^4+2 a p q+b}+\frac {\sqrt {p x^4+q} \left (\frac {p \sqrt {4 a p q+b}}{\sqrt {b}}+p\right )}{\sqrt {b} \sqrt {4 a p q+b}+2 a p^2 x^4+2 a p q+b}\right )dx\)

\(\Big \downarrow \) 7239

\(\displaystyle \int \frac {\left (p x^4-q\right ) \sqrt {p x^4+q}}{x^4 (2 a p q+b)+a p^2 x^8+a q^2}dx\)

\(\Big \downarrow \) 7279

\(\displaystyle \int \left (\frac {\sqrt {p x^4+q} \left (p-\frac {p \sqrt {4 a p q+b}}{\sqrt {b}}\right )}{-\sqrt {b} \sqrt {4 a p q+b}+2 a p^2 x^4+2 a p q+b}+\frac {\sqrt {p x^4+q} \left (\frac {p \sqrt {4 a p q+b}}{\sqrt {b}}+p\right )}{\sqrt {b} \sqrt {4 a p q+b}+2 a p^2 x^4+2 a p q+b}\right )dx\)

\(\Big \downarrow \) 7239

\(\displaystyle \int \frac {\left (p x^4-q\right ) \sqrt {p x^4+q}}{x^4 (2 a p q+b)+a p^2 x^8+a q^2}dx\)

\(\Big \downarrow \) 7279

\(\displaystyle \int \left (\frac {\sqrt {p x^4+q} \left (p-\frac {p \sqrt {4 a p q+b}}{\sqrt {b}}\right )}{-\sqrt {b} \sqrt {4 a p q+b}+2 a p^2 x^4+2 a p q+b}+\frac {\sqrt {p x^4+q} \left (\frac {p \sqrt {4 a p q+b}}{\sqrt {b}}+p\right )}{\sqrt {b} \sqrt {4 a p q+b}+2 a p^2 x^4+2 a p q+b}\right )dx\)

\(\Big \downarrow \) 7239

\(\displaystyle \int \frac {\left (p x^4-q\right ) \sqrt {p x^4+q}}{x^4 (2 a p q+b)+a p^2 x^8+a q^2}dx\)

\(\Big \downarrow \) 7279

\(\displaystyle \int \left (\frac {\sqrt {p x^4+q} \left (p-\frac {p \sqrt {4 a p q+b}}{\sqrt {b}}\right )}{-\sqrt {b} \sqrt {4 a p q+b}+2 a p^2 x^4+2 a p q+b}+\frac {\sqrt {p x^4+q} \left (\frac {p \sqrt {4 a p q+b}}{\sqrt {b}}+p\right )}{\sqrt {b} \sqrt {4 a p q+b}+2 a p^2 x^4+2 a p q+b}\right )dx\)

\(\Big \downarrow \) 7239

\(\displaystyle \int \frac {\left (p x^4-q\right ) \sqrt {p x^4+q}}{x^4 (2 a p q+b)+a p^2 x^8+a q^2}dx\)

\(\Big \downarrow \) 7279

\(\displaystyle \int \left (\frac {\sqrt {p x^4+q} \left (p-\frac {p \sqrt {4 a p q+b}}{\sqrt {b}}\right )}{-\sqrt {b} \sqrt {4 a p q+b}+2 a p^2 x^4+2 a p q+b}+\frac {\sqrt {p x^4+q} \left (\frac {p \sqrt {4 a p q+b}}{\sqrt {b}}+p\right )}{\sqrt {b} \sqrt {4 a p q+b}+2 a p^2 x^4+2 a p q+b}\right )dx\)

\(\Big \downarrow \) 7239

\(\displaystyle \int \frac {\left (p x^4-q\right ) \sqrt {p x^4+q}}{x^4 (2 a p q+b)+a p^2 x^8+a q^2}dx\)

\(\Big \downarrow \) 7279

\(\displaystyle \int \left (\frac {\sqrt {p x^4+q} \left (p-\frac {p \sqrt {4 a p q+b}}{\sqrt {b}}\right )}{-\sqrt {b} \sqrt {4 a p q+b}+2 a p^2 x^4+2 a p q+b}+\frac {\sqrt {p x^4+q} \left (\frac {p \sqrt {4 a p q+b}}{\sqrt {b}}+p\right )}{\sqrt {b} \sqrt {4 a p q+b}+2 a p^2 x^4+2 a p q+b}\right )dx\)

\(\Big \downarrow \) 7239

\(\displaystyle \int \frac {\left (p x^4-q\right ) \sqrt {p x^4+q}}{x^4 (2 a p q+b)+a p^2 x^8+a q^2}dx\)

\(\Big \downarrow \) 7279

\(\displaystyle \int \left (\frac {\sqrt {p x^4+q} \left (p-\frac {p \sqrt {4 a p q+b}}{\sqrt {b}}\right )}{-\sqrt {b} \sqrt {4 a p q+b}+2 a p^2 x^4+2 a p q+b}+\frac {\sqrt {p x^4+q} \left (\frac {p \sqrt {4 a p q+b}}{\sqrt {b}}+p\right )}{\sqrt {b} \sqrt {4 a p q+b}+2 a p^2 x^4+2 a p q+b}\right )dx\)

\(\Big \downarrow \) 7239

\(\displaystyle \int \frac {\left (p x^4-q\right ) \sqrt {p x^4+q}}{x^4 (2 a p q+b)+a p^2 x^8+a q^2}dx\)

\(\Big \downarrow \) 7279

\(\displaystyle \int \left (\frac {\sqrt {p x^4+q} \left (p-\frac {p \sqrt {4 a p q+b}}{\sqrt {b}}\right )}{-\sqrt {b} \sqrt {4 a p q+b}+2 a p^2 x^4+2 a p q+b}+\frac {\sqrt {p x^4+q} \left (\frac {p \sqrt {4 a p q+b}}{\sqrt {b}}+p\right )}{\sqrt {b} \sqrt {4 a p q+b}+2 a p^2 x^4+2 a p q+b}\right )dx\)

Input:

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

Output:

$Aborted
 

Maple [N/A] (verified)

Time = 0.96 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.45

method	result	size
pseudoelliptic	\(\frac {\sqrt {2}\, \left (\ln \left (\frac {p \,x^{4}-\left (\frac {b}{a}\right )^{\frac {1}{4}} \sqrt {p \,x^{4}+q}\, \sqrt {2}\, x +\sqrt {\frac {b}{a}}\, x^{2}+q}{p \,x^{4}+\left (\frac {b}{a}\right )^{\frac {1}{4}} \sqrt {p \,x^{4}+q}\, \sqrt {2}\, x +\sqrt {\frac {b}{a}}\, x^{2}+q}\right )+2 \arctan \left (\frac {\sqrt {p \,x^{4}+q}\, \sqrt {2}}{\left (\frac {b}{a}\right )^{\frac {1}{4}} x}+1\right )-2 \arctan \left (-\frac {\sqrt {p \,x^{4}+q}\, \sqrt {2}}{\left (\frac {b}{a}\right )^{\frac {1}{4}} x}+1\right )\right )}{8 \left (\frac {b}{a}\right )^{\frac {1}{4}} a}\)	\(157\)
default	\(\frac {\left (\ln \left (\frac {\frac {p \,x^{4}+q}{2 x^{2}}-\frac {\left (\frac {b}{a}\right )^{\frac {1}{4}} \sqrt {p \,x^{4}+q}\, \sqrt {2}}{2 x}+\frac {\sqrt {\frac {b}{a}}}{2}}{\frac {p \,x^{4}+q}{2 x^{2}}+\frac {\left (\frac {b}{a}\right )^{\frac {1}{4}} \sqrt {p \,x^{4}+q}\, \sqrt {2}}{2 x}+\frac {\sqrt {\frac {b}{a}}}{2}}\right )+2 \arctan \left (\frac {\sqrt {p \,x^{4}+q}\, \sqrt {2}}{\left (\frac {b}{a}\right )^{\frac {1}{4}} x}+1\right )+2 \arctan \left (\frac {\sqrt {p \,x^{4}+q}\, \sqrt {2}}{\left (\frac {b}{a}\right )^{\frac {1}{4}} x}-1\right )\right ) \sqrt {2}}{8 a \left (\frac {b}{a}\right )^{\frac {1}{4}}}\)	\(169\)
elliptic	\(\frac {\left (\ln \left (\frac {\frac {p \,x^{4}+q}{2 x^{2}}-\frac {\left (\frac {b}{a}\right )^{\frac {1}{4}} \sqrt {p \,x^{4}+q}\, \sqrt {2}}{2 x}+\frac {\sqrt {\frac {b}{a}}}{2}}{\frac {p \,x^{4}+q}{2 x^{2}}+\frac {\left (\frac {b}{a}\right )^{\frac {1}{4}} \sqrt {p \,x^{4}+q}\, \sqrt {2}}{2 x}+\frac {\sqrt {\frac {b}{a}}}{2}}\right )+2 \arctan \left (\frac {\sqrt {p \,x^{4}+q}\, \sqrt {2}}{\left (\frac {b}{a}\right )^{\frac {1}{4}} x}+1\right )+2 \arctan \left (\frac {\sqrt {p \,x^{4}+q}\, \sqrt {2}}{\left (\frac {b}{a}\right )^{\frac {1}{4}} x}-1\right )\right ) \sqrt {2}}{8 a \left (\frac {b}{a}\right )^{\frac {1}{4}}}\)	\(169\)

Input:

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

Output:

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

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 3 vs. order 1.

Time = 3.13 (sec) , antiderivative size = 607, normalized size of antiderivative = 1.72 \[ \int \frac {\left (-q+p x^4\right ) \sqrt {q+p x^4}}{b x^4+a \left (q+p x^4\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

1/8*(-1/(a^3*b))^(1/4)*log(1/2*(2*(a^2*b*p*x^6 + a^2*b*q*x^2)*(-1/(a^3*b)) 
^(3/4) + 2*(p*x^5 - a*b*x^3*sqrt(-1/(a^3*b)) + q*x)*sqrt(p*x^4 + q) - (a*p 
^2*x^8 + (2*a*p*q - b)*x^4 + a*q^2)*(-1/(a^3*b))^(1/4))/(a*p^2*x^8 + (2*a* 
p*q + b)*x^4 + a*q^2)) - 1/8*(-1/(a^3*b))^(1/4)*log(-1/2*(2*(a^2*b*p*x^6 + 
 a^2*b*q*x^2)*(-1/(a^3*b))^(3/4) - 2*(p*x^5 - a*b*x^3*sqrt(-1/(a^3*b)) + q 
*x)*sqrt(p*x^4 + q) - (a*p^2*x^8 + (2*a*p*q - b)*x^4 + a*q^2)*(-1/(a^3*b)) 
^(1/4))/(a*p^2*x^8 + (2*a*p*q + b)*x^4 + a*q^2)) + 1/8*I*(-1/(a^3*b))^(1/4 
)*log(-1/2*(2*(I*a^2*b*p*x^6 + I*a^2*b*q*x^2)*(-1/(a^3*b))^(3/4) - 2*(p*x^ 
5 + a*b*x^3*sqrt(-1/(a^3*b)) + q*x)*sqrt(p*x^4 + q) + (I*a*p^2*x^8 + I*(2* 
a*p*q - b)*x^4 + I*a*q^2)*(-1/(a^3*b))^(1/4))/(a*p^2*x^8 + (2*a*p*q + b)*x 
^4 + a*q^2)) - 1/8*I*(-1/(a^3*b))^(1/4)*log(-1/2*(2*(-I*a^2*b*p*x^6 - I*a^ 
2*b*q*x^2)*(-1/(a^3*b))^(3/4) - 2*(p*x^5 + a*b*x^3*sqrt(-1/(a^3*b)) + q*x) 
*sqrt(p*x^4 + q) + (-I*a*p^2*x^8 - I*(2*a*p*q - b)*x^4 - I*a*q^2)*(-1/(a^3 
*b))^(1/4))/(a*p^2*x^8 + (2*a*p*q + b)*x^4 + a*q^2))
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (-q+p x^4\right ) \sqrt {q+p x^4}}{b x^4+a \left (q+p x^4\right )^2} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [N/A]

Not integrable

Time = 0.09 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.11 \[ \int \frac {\left (-q+p x^4\right ) \sqrt {q+p x^4}}{b x^4+a \left (q+p x^4\right )^2} \, dx=\int { \frac {\sqrt {p x^{4} + q} {\left (p x^{4} - q\right )}}{b x^{4} + {\left (p x^{4} + q\right )}^{2} a} \,d x } \] Input:

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

Output:

integrate(sqrt(p*x^4 + q)*(p*x^4 - q)/(b*x^4 + (p*x^4 + q)^2*a), x)
 

Giac [F(-1)]

Timed out. \[ \int \frac {\left (-q+p x^4\right ) \sqrt {q+p x^4}}{b x^4+a \left (q+p x^4\right )^2} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Mupad [N/A]

Not integrable

Time = 9.37 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.11 \[ \int \frac {\left (-q+p x^4\right ) \sqrt {q+p x^4}}{b x^4+a \left (q+p x^4\right )^2} \, dx=\int -\frac {\sqrt {p\,x^4+q}\,\left (q-p\,x^4\right )}{a\,{\left (p\,x^4+q\right )}^2+b\,x^4} \,d x \] Input:

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

Output:

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

Reduce [N/A]

Not integrable

Time = 0.24 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.25 \[ \int \frac {\left (-q+p x^4\right ) \sqrt {q+p x^4}}{b x^4+a \left (q+p x^4\right )^2} \, dx=-\left (\int \frac {\sqrt {p \,x^{4}+q}}{a \,p^{2} x^{8}+2 a p q \,x^{4}+b \,x^{4}+a \,q^{2}}d x \right ) q +\left (\int \frac {\sqrt {p \,x^{4}+q}\, x^{4}}{a \,p^{2} x^{8}+2 a p q \,x^{4}+b \,x^{4}+a \,q^{2}}d x \right ) p \] Input:

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

Output:

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