Integrand size = 52, antiderivative size = 428 \[ \int \frac {\left (-q+p x^4\right ) \sqrt {q+p x^4}}{c x^4+b x^2 \left (q+p x^4\right )+a \left (q+p x^4\right )^2} \, dx=\frac {1}{4} \text {RootSum}\left [16 a p^2 q^2+16 b p q \text {$\#$1}^2-32 a p^{3/2} q^{3/2} \text {$\#$1}^2+16 c \text {$\#$1}^4-16 b \sqrt {p} \sqrt {q} \text {$\#$1}^4+24 a p q \text {$\#$1}^4+4 b \text {$\#$1}^6-8 a \sqrt {p} \sqrt {q} \text {$\#$1}^6+a \text {$\#$1}^8\&,\frac {8 p^{3/2} q^{3/2} \log (x)-8 p^{3/2} q^{3/2} \log \left (-\sqrt {q}-\sqrt {p} x^2+\sqrt {q+p x^4}-x \text {$\#$1}\right )-4 p q \log (x) \text {$\#$1}^2+4 p q \log \left (-\sqrt {q}-\sqrt {p} x^2+\sqrt {q+p x^4}-x \text {$\#$1}\right ) \text {$\#$1}^2-2 \sqrt {p} \sqrt {q} \log (x) \text {$\#$1}^4+2 \sqrt {p} \sqrt {q} \log \left (-\sqrt {q}-\sqrt {p} x^2+\sqrt {q+p x^4}-x \text {$\#$1}\right ) \text {$\#$1}^4+\log (x) \text {$\#$1}^6-\log \left (-\sqrt {q}-\sqrt {p} x^2+\sqrt {q+p x^4}-x \text {$\#$1}\right ) \text {$\#$1}^6}{-4 b p q \text {$\#$1}+8 a p^{3/2} q^{3/2} \text {$\#$1}-8 c \text {$\#$1}^3+8 b \sqrt {p} \sqrt {q} \text {$\#$1}^3-12 a p q \text {$\#$1}^3-3 b \text {$\#$1}^5+6 a \sqrt {p} \sqrt {q} \text {$\#$1}^5-a \text {$\#$1}^7}\&\right ] \]
Time = 2.90 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.35 \[ \int \frac {\left (-q+p x^4\right ) \sqrt {q+p x^4}}{c x^4+b x^2 \left (q+p x^4\right )+a \left (q+p x^4\right )^2} \, dx=\frac {\sqrt {b-\sqrt {b^2-4 a c}} \arctan \left (\frac {\sqrt {b-\sqrt {b^2-4 a c}} x}{\sqrt {2} \sqrt {a} \sqrt {q+p x^4}}\right )-\sqrt {b+\sqrt {b^2-4 a c}} \arctan \left (\frac {\sqrt {b+\sqrt {b^2-4 a c}} x}{\sqrt {2} \sqrt {a} \sqrt {q+p x^4}}\right )}{\sqrt {2} \sqrt {a} \sqrt {b^2-4 a c}} \]
(Sqrt[b - Sqrt[b^2 - 4*a*c]]*ArcTan[(Sqrt[b - Sqrt[b^2 - 4*a*c]]*x)/(Sqrt[ 2]*Sqrt[a]*Sqrt[q + p*x^4])] - Sqrt[b + Sqrt[b^2 - 4*a*c]]*ArcTan[(Sqrt[b + Sqrt[b^2 - 4*a*c]]*x)/(Sqrt[2]*Sqrt[a]*Sqrt[q + p*x^4])])/(Sqrt[2]*Sqrt[ a]*Sqrt[b^2 - 4*a*c])
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.
\(\displaystyle \int \frac {\left (p x^4-q\right ) \sqrt {p x^4+q}}{a \left (p x^4+q\right )^2+b x^2 \left (p x^4+q\right )+c x^4} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {p x^4 \sqrt {p x^4+q}}{c x^4 \left (\frac {2 a p q}{c}+1\right )+a p^2 x^8+a q^2+b p x^6+b q x^2}+\frac {q \sqrt {p x^4+q}}{-c x^4 \left (\frac {2 a p q}{c}+1\right )-a p^2 x^8-a q^2-b p x^6-b q x^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle q \int \frac {\sqrt {p x^4+q}}{-a p^2 x^8-b p x^6-c \left (\frac {2 a p q}{c}+1\right ) x^4-b q x^2-a q^2}dx+p \int \frac {x^4 \sqrt {p x^4+q}}{a p^2 x^8+b p x^6+c \left (\frac {2 a p q}{c}+1\right ) x^4+b q x^2+a q^2}dx\) |
3.31.22.3.1 Defintions of rubi rules used
Time = 0.46 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.35
method | result | size |
pseudoelliptic | \(\frac {\sqrt {2}\, \left (\frac {\left (b +\sqrt {-4 a c +b^{2}}\right ) \arctan \left (\frac {\sqrt {p \,x^{4}+q}\, \sqrt {2}\, a}{x \sqrt {\left (b +\sqrt {-4 a c +b^{2}}\right ) a}}\right )}{\sqrt {\left (b +\sqrt {-4 a c +b^{2}}\right ) a}}-\frac {\left (-b +\sqrt {-4 a c +b^{2}}\right ) \operatorname {arctanh}\left (\frac {\sqrt {p \,x^{4}+q}\, \sqrt {2}\, a}{x \sqrt {\left (-b +\sqrt {-4 a c +b^{2}}\right ) a}}\right )}{\sqrt {\left (-b +\sqrt {-4 a c +b^{2}}\right ) a}}\right )}{2 \sqrt {-4 a c +b^{2}}}\) | \(150\) |
default | \(8 a \left (-\frac {\left (-b +\sqrt {-4 a c +b^{2}}\right ) \operatorname {arctanh}\left (\frac {\sqrt {p \,x^{4}+q}\, \sqrt {2}\, a}{x \sqrt {\left (-b +\sqrt {-4 a c +b^{2}}\right ) a}}\right )}{16 \sqrt {-4 a c +b^{2}}\, a \sqrt {\left (-b +\sqrt {-4 a c +b^{2}}\right ) a}}+\frac {\left (b +\sqrt {-4 a c +b^{2}}\right ) \arctan \left (\frac {\sqrt {p \,x^{4}+q}\, \sqrt {2}\, a}{x \sqrt {\left (b +\sqrt {-4 a c +b^{2}}\right ) a}}\right )}{16 a \sqrt {-4 a c +b^{2}}\, \sqrt {\left (b +\sqrt {-4 a c +b^{2}}\right ) a}}\right ) \sqrt {2}\) | \(168\) |
elliptic | \(8 a \left (-\frac {\left (-b +\sqrt {-4 a c +b^{2}}\right ) \operatorname {arctanh}\left (\frac {\sqrt {p \,x^{4}+q}\, \sqrt {2}\, a}{x \sqrt {\left (-b +\sqrt {-4 a c +b^{2}}\right ) a}}\right )}{16 \sqrt {-4 a c +b^{2}}\, a \sqrt {\left (-b +\sqrt {-4 a c +b^{2}}\right ) a}}+\frac {\left (b +\sqrt {-4 a c +b^{2}}\right ) \arctan \left (\frac {\sqrt {p \,x^{4}+q}\, \sqrt {2}\, a}{x \sqrt {\left (b +\sqrt {-4 a c +b^{2}}\right ) a}}\right )}{16 a \sqrt {-4 a c +b^{2}}\, \sqrt {\left (b +\sqrt {-4 a c +b^{2}}\right ) a}}\right ) \sqrt {2}\) | \(168\) |
1/2*2^(1/2)/(-4*a*c+b^2)^(1/2)*((b+(-4*a*c+b^2)^(1/2))/((b+(-4*a*c+b^2)^(1 /2))*a)^(1/2)*arctan((p*x^4+q)^(1/2)*2^(1/2)/x*a/((b+(-4*a*c+b^2)^(1/2))*a )^(1/2))-(-b+(-4*a*c+b^2)^(1/2))/((-b+(-4*a*c+b^2)^(1/2))*a)^(1/2)*arctanh ((p*x^4+q)^(1/2)*2^(1/2)/x*a/((-b+(-4*a*c+b^2)^(1/2))*a)^(1/2)))
Result contains higher order function than in optimal. Order 3 vs. order 1.
Time = 89.38 (sec) , antiderivative size = 2083, normalized size of antiderivative = 4.87 \[ \int \frac {\left (-q+p x^4\right ) \sqrt {q+p x^4}}{c x^4+b x^2 \left (q+p x^4\right )+a \left (q+p x^4\right )^2} \, dx=\text {Too large to display} \]
integrate((p*x^4-q)*(p*x^4+q)^(1/2)/(c*x^4+b*x^2*(p*x^4+q)+a*(p*x^4+q)^2), x, algorithm="fricas")
-1/4*sqrt(1/2)*sqrt(-(b + (a*b^2 - 4*a^2*c)/sqrt(a^2*b^2 - 4*a^3*c))/(a*b^ 2 - 4*a^2*c))*log(-(sqrt(1/2)*((a*b^2 - 4*a^2*c)*p^2*x^8 - (b^3 - 4*a*b*c) *p*x^6 - (b^2*c - 4*a*c^2 - 2*(a*b^2 - 4*a^2*c)*p*q)*x^4 - (b^3 - 4*a*b*c) *q*x^2 + (a*b^2 - 4*a^2*c)*q^2 + ((a^2*b^3 - 4*a^3*b*c)*p^2*x^8 - (a*b^4 - 8*a^2*b^2*c + 16*a^3*c^2)*p*x^6 - (a*b^3*c - 4*a^2*b*c^2 - 2*(a^2*b^3 - 4 *a^3*b*c)*p*q)*x^4 - (a*b^4 - 8*a^2*b^2*c + 16*a^3*c^2)*q*x^2 + (a^2*b^3 - 4*a^3*b*c)*q^2)/sqrt(a^2*b^2 - 4*a^3*c))*sqrt(-(b + (a*b^2 - 4*a^2*c)/sqr t(a^2*b^2 - 4*a^3*c))/(a*b^2 - 4*a^2*c)) + 2*((b^2 - 2*a*c)*p*x^5 + b*c*x^ 3 + (b^2 - 2*a*c)*q*x + ((a*b^3 - 4*a^2*b*c)*p*x^5 + (a*b^2*c - 4*a^2*c^2) *x^3 + (a*b^3 - 4*a^2*b*c)*q*x)/sqrt(a^2*b^2 - 4*a^3*c))*sqrt(p*x^4 + q))/ (a*p^2*x^8 + b*p*x^6 + (2*a*p*q + c)*x^4 + b*q*x^2 + a*q^2)) + 1/4*sqrt(1/ 2)*sqrt(-(b + (a*b^2 - 4*a^2*c)/sqrt(a^2*b^2 - 4*a^3*c))/(a*b^2 - 4*a^2*c) )*log((sqrt(1/2)*((a*b^2 - 4*a^2*c)*p^2*x^8 - (b^3 - 4*a*b*c)*p*x^6 - (b^2 *c - 4*a*c^2 - 2*(a*b^2 - 4*a^2*c)*p*q)*x^4 - (b^3 - 4*a*b*c)*q*x^2 + (a*b ^2 - 4*a^2*c)*q^2 + ((a^2*b^3 - 4*a^3*b*c)*p^2*x^8 - (a*b^4 - 8*a^2*b^2*c + 16*a^3*c^2)*p*x^6 - (a*b^3*c - 4*a^2*b*c^2 - 2*(a^2*b^3 - 4*a^3*b*c)*p*q )*x^4 - (a*b^4 - 8*a^2*b^2*c + 16*a^3*c^2)*q*x^2 + (a^2*b^3 - 4*a^3*b*c)*q ^2)/sqrt(a^2*b^2 - 4*a^3*c))*sqrt(-(b + (a*b^2 - 4*a^2*c)/sqrt(a^2*b^2 - 4 *a^3*c))/(a*b^2 - 4*a^2*c)) - 2*((b^2 - 2*a*c)*p*x^5 + b*c*x^3 + (b^2 - 2* a*c)*q*x + ((a*b^3 - 4*a^2*b*c)*p*x^5 + (a*b^2*c - 4*a^2*c^2)*x^3 + (a*...
Timed out. \[ \int \frac {\left (-q+p x^4\right ) \sqrt {q+p x^4}}{c x^4+b x^2 \left (q+p x^4\right )+a \left (q+p x^4\right )^2} \, dx=\text {Timed out} \]
Not integrable
Time = 0.25 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.12 \[ \int \frac {\left (-q+p x^4\right ) \sqrt {q+p x^4}}{c x^4+b x^2 \left (q+p x^4\right )+a \left (q+p x^4\right )^2} \, dx=\int { \frac {\sqrt {p x^{4} + q} {\left (p x^{4} - q\right )}}{c x^{4} + {\left (p x^{4} + q\right )} b x^{2} + {\left (p x^{4} + q\right )}^{2} a} \,d x } \]
integrate((p*x^4-q)*(p*x^4+q)^(1/2)/(c*x^4+b*x^2*(p*x^4+q)+a*(p*x^4+q)^2), x, algorithm="maxima")
Timed out. \[ \int \frac {\left (-q+p x^4\right ) \sqrt {q+p x^4}}{c x^4+b x^2 \left (q+p x^4\right )+a \left (q+p x^4\right )^2} \, dx=\text {Timed out} \]
integrate((p*x^4-q)*(p*x^4+q)^(1/2)/(c*x^4+b*x^2*(p*x^4+q)+a*(p*x^4+q)^2), x, algorithm="giac")
Not integrable
Time = 7.33 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.12 \[ \int \frac {\left (-q+p x^4\right ) \sqrt {q+p x^4}}{c x^4+b x^2 \left (q+p x^4\right )+a \left (q+p x^4\right )^2} \, dx=\int -\frac {\sqrt {p\,x^4+q}\,\left (q-p\,x^4\right )}{a\,{\left (p\,x^4+q\right )}^2+c\,x^4+b\,x^2\,\left (p\,x^4+q\right )} \,d x \]