Integrand size = 43, antiderivative size = 399 \[ \int \frac {x^4 \left (-2 q+p x^3\right ) \sqrt {q+p x^3}}{b x^8+a \left (q+p x^3\right )^4} \, dx=-\frac {\sqrt {2+\sqrt {2}} \arctan \left (\frac {\left (\sqrt {\frac {2}{2-\sqrt {2}}} \sqrt [8]{a} \sqrt [8]{b}-\frac {2 \sqrt [8]{a} \sqrt [8]{b}}{\sqrt {2-\sqrt {2}}}\right ) x \sqrt {q+p x^3}}{-\sqrt [4]{a} q+\sqrt [4]{b} x^2-\sqrt [4]{a} p x^3}\right )}{4 a^{3/8} b^{5/8}}+\frac {\sqrt {2-\sqrt {2}} \arctan \left (\frac {\sqrt {2+\sqrt {2}} \sqrt [8]{a} \sqrt [8]{b} x \sqrt {q+p x^3}}{\sqrt [4]{a} q-\sqrt [4]{b} x^2+\sqrt [4]{a} p x^3}\right )}{4 a^{3/8} b^{5/8}}-\frac {\sqrt {2+\sqrt {2}} \text {arctanh}\left (\frac {\sqrt {2-\sqrt {2}} \sqrt [8]{a} \sqrt [8]{b} x \sqrt {q+p x^3}}{\sqrt [4]{a} q+\sqrt [4]{b} x^2+\sqrt [4]{a} p x^3}\right )}{4 a^{3/8} b^{5/8}}+\frac {\sqrt {2-\sqrt {2}} \text {arctanh}\left (\frac {\sqrt {2+\sqrt {2}} \sqrt [8]{a} \sqrt [8]{b} x \sqrt {q+p x^3}}{\sqrt [4]{a} q+\sqrt [4]{b} x^2+\sqrt [4]{a} p x^3}\right )}{4 a^{3/8} b^{5/8}} \] Output:
-1/4*(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^3+q)^(1/2)/(-a^(1/4)*q+b^(1/4)* x^2-a^(1/4)*p*x^3))/a^(3/8)/b^(5/8)+1/4*(2-2^(1/2))^(1/2)*arctan((2+2^(1/2 ))^(1/2)*a^(1/8)*b^(1/8)*x*(p*x^3+q)^(1/2)/(a^(1/4)*q-b^(1/4)*x^2+a^(1/4)* p*x^3))/a^(3/8)/b^(5/8)-1/4*(2+2^(1/2))^(1/2)*arctanh((2-2^(1/2))^(1/2)*a^ (1/8)*b^(1/8)*x*(p*x^3+q)^(1/2)/(a^(1/4)*q+b^(1/4)*x^2+a^(1/4)*p*x^3))/a^( 3/8)/b^(5/8)+1/4*(2-2^(1/2))^(1/2)*arctanh((2+2^(1/2))^(1/2)*a^(1/8)*b^(1/ 8)*x*(p*x^3+q)^(1/2)/(a^(1/4)*q+b^(1/4)*x^2+a^(1/4)*p*x^3))/a^(3/8)/b^(5/8 )
Time = 3.05 (sec) , antiderivative size = 307, normalized size of antiderivative = 0.77 \[ \int \frac {x^4 \left (-2 q+p x^3\right ) \sqrt {q+p x^3}}{b x^8+a \left (q+p x^3\right )^4} \, dx=\frac {\sqrt {2-\sqrt {2}} \arctan \left (\frac {\sqrt {2+\sqrt {2}} \sqrt [8]{a} \sqrt [8]{b} x \sqrt {q+p x^3}}{-\sqrt [4]{b} x^2+\sqrt [4]{a} \left (q+p x^3\right )}\right )-\sqrt {2+\sqrt {2}} \arctan \left (\frac {\sqrt [8]{a} \sqrt [8]{b} x \sqrt {-\left (\left (-2+\sqrt {2}\right ) \left (q+p x^3\right )\right )}}{-\sqrt [4]{b} x^2+\sqrt [4]{a} \left (q+p x^3\right )}\right )+\sqrt {2-\sqrt {2}} \text {arctanh}\left (\frac {\sqrt {2+\sqrt {2}} \sqrt [8]{a} \sqrt [8]{b} x \sqrt {q+p x^3}}{\sqrt [4]{b} x^2+\sqrt [4]{a} \left (q+p x^3\right )}\right )-\sqrt {2+\sqrt {2}} \text {arctanh}\left (\frac {\sqrt [8]{a} \sqrt [8]{b} x \sqrt {-\left (\left (-2+\sqrt {2}\right ) \left (q+p x^3\right )\right )}}{\sqrt [4]{b} x^2+\sqrt [4]{a} \left (q+p x^3\right )}\right )}{4 a^{3/8} b^{5/8}} \] Input:
Integrate[(x^4*(-2*q + p*x^3)*Sqrt[q + p*x^3])/(b*x^8 + a*(q + p*x^3)^4),x ]
Output:
(Sqrt[2 - Sqrt[2]]*ArcTan[(Sqrt[2 + Sqrt[2]]*a^(1/8)*b^(1/8)*x*Sqrt[q + p* x^3])/(-(b^(1/4)*x^2) + a^(1/4)*(q + p*x^3))] - Sqrt[2 + Sqrt[2]]*ArcTan[( a^(1/8)*b^(1/8)*x*Sqrt[-((-2 + Sqrt[2])*(q + p*x^3))])/(-(b^(1/4)*x^2) + a ^(1/4)*(q + p*x^3))] + Sqrt[2 - Sqrt[2]]*ArcTanh[(Sqrt[2 + Sqrt[2]]*a^(1/8 )*b^(1/8)*x*Sqrt[q + p*x^3])/(b^(1/4)*x^2 + a^(1/4)*(q + p*x^3))] - Sqrt[2 + Sqrt[2]]*ArcTanh[(a^(1/8)*b^(1/8)*x*Sqrt[-((-2 + Sqrt[2])*(q + p*x^3))] )/(b^(1/4)*x^2 + a^(1/4)*(q + p*x^3))])/(4*a^(3/8)*b^(5/8))
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.
\(\displaystyle \int \frac {x^4 \left (p x^3-2 q\right ) \sqrt {p x^3+q}}{a \left (p x^3+q\right )^4+b x^8} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {p x^7 \sqrt {p x^3+q}}{a p^4 x^{12}+4 a p^3 q x^9+6 a p^2 q^2 x^6+4 a p q^3 x^3+a q^4+b x^8}-\frac {2 q x^4 \sqrt {p x^3+q}}{a p^4 x^{12}+4 a p^3 q x^9+6 a p^2 q^2 x^6+4 a p q^3 x^3+a q^4+b x^8}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle p \int \frac {x^7 \sqrt {p x^3+q}}{b x^8+a \left (p x^3+q\right )^4}dx-2 q \int \frac {x^4 \sqrt {p x^3+q}}{b x^8+a \left (p x^3+q\right )^4}dx\) |
Input:
Int[(x^4*(-2*q + p*x^3)*Sqrt[q + p*x^3])/(b*x^8 + a*(q + p*x^3)^4),x]
Output:
$Aborted
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 1.28 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.10
method | result | size |
default | \(\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (a \,\textit {\_Z}^{8}+b \right )}{\sum }\frac {\ln \left (\frac {-\textit {\_R} x +\sqrt {p \,x^{3}+q}}{x}\right )}{\textit {\_R}^{5}}}{4 a}\) | \(40\) |
pseudoelliptic | \(\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (a \,\textit {\_Z}^{8}+b \right )}{\sum }\frac {\ln \left (\frac {-\textit {\_R} x +\sqrt {p \,x^{3}+q}}{x}\right )}{\textit {\_R}^{5}}}{4 a}\) | \(40\) |
elliptic | \(\text {Expression too large to display}\) | \(1596\) |
Input:
int(x^4*(p*x^3-2*q)*(p*x^3+q)^(1/2)/(b*x^8+a*(p*x^3+q)^4),x,method=_RETURN VERBOSE)
Output:
1/4*sum(ln((-_R*x+(p*x^3+q)^(1/2))/x)/_R^5,_R=RootOf(_Z^8*a+b))/a
Result contains complex when optimal does not.
Time = 6.33 (sec) , antiderivative size = 3667, normalized size of antiderivative = 9.19 \[ \int \frac {x^4 \left (-2 q+p x^3\right ) \sqrt {q+p x^3}}{b x^8+a \left (q+p x^3\right )^4} \, dx=\text {Too large to display} \] Input:
integrate(x^4*(p*x^3-2*q)*(p*x^3+q)^(1/2)/(b*x^8+a*(p*x^3+q)^4),x, algorit hm="fricas")
Output:
Too large to include
\[ \int \frac {x^4 \left (-2 q+p x^3\right ) \sqrt {q+p x^3}}{b x^8+a \left (q+p x^3\right )^4} \, dx=\int \frac {x^{4} \left (p x^{3} - 2 q\right ) \sqrt {p x^{3} + q}}{a p^{4} x^{12} + 4 a p^{3} q x^{9} + 6 a p^{2} q^{2} x^{6} + 4 a p q^{3} x^{3} + a q^{4} + b x^{8}}\, dx \] Input:
integrate(x**4*(p*x**3-2*q)*(p*x**3+q)**(1/2)/(b*x**8+a*(p*x**3+q)**4),x)
Output:
Integral(x**4*(p*x**3 - 2*q)*sqrt(p*x**3 + q)/(a*p**4*x**12 + 4*a*p**3*q*x **9 + 6*a*p**2*q**2*x**6 + 4*a*p*q**3*x**3 + a*q**4 + b*x**8), x)
\[ \int \frac {x^4 \left (-2 q+p x^3\right ) \sqrt {q+p x^3}}{b x^8+a \left (q+p x^3\right )^4} \, dx=\int { \frac {\sqrt {p x^{3} + q} {\left (p x^{3} - 2 \, q\right )} x^{4}}{b x^{8} + {\left (p x^{3} + q\right )}^{4} a} \,d x } \] Input:
integrate(x^4*(p*x^3-2*q)*(p*x^3+q)^(1/2)/(b*x^8+a*(p*x^3+q)^4),x, algorit hm="maxima")
Output:
integrate(sqrt(p*x^3 + q)*(p*x^3 - 2*q)*x^4/(b*x^8 + (p*x^3 + q)^4*a), x)
Timed out. \[ \int \frac {x^4 \left (-2 q+p x^3\right ) \sqrt {q+p x^3}}{b x^8+a \left (q+p x^3\right )^4} \, dx=\text {Timed out} \] Input:
integrate(x^4*(p*x^3-2*q)*(p*x^3+q)^(1/2)/(b*x^8+a*(p*x^3+q)^4),x, algorit hm="giac")
Output:
Timed out
Timed out. \[ \int \frac {x^4 \left (-2 q+p x^3\right ) \sqrt {q+p x^3}}{b x^8+a \left (q+p x^3\right )^4} \, dx=\text {Hanged} \] Input:
int(-(x^4*(q + p*x^3)^(1/2)*(2*q - p*x^3))/(a*(q + p*x^3)^4 + b*x^8),x)
Output:
\text{Hanged}
\[ \int \frac {x^4 \left (-2 q+p x^3\right ) \sqrt {q+p x^3}}{b x^8+a \left (q+p x^3\right )^4} \, dx=\left (\int \frac {\sqrt {p \,x^{3}+q}\, x^{7}}{a \,p^{4} x^{12}+4 a \,p^{3} q \,x^{9}+6 a \,p^{2} q^{2} x^{6}+b \,x^{8}+4 a p \,q^{3} x^{3}+a \,q^{4}}d x \right ) p -2 \left (\int \frac {\sqrt {p \,x^{3}+q}\, x^{4}}{a \,p^{4} x^{12}+4 a \,p^{3} q \,x^{9}+6 a \,p^{2} q^{2} x^{6}+b \,x^{8}+4 a p \,q^{3} x^{3}+a \,q^{4}}d x \right ) q \] Input:
int(x^4*(p*x^3-2*q)*(p*x^3+q)^(1/2)/(b*x^8+a*(p*x^3+q)^4),x)
Output:
int((sqrt(p*x**3 + q)*x**7)/(a*p**4*x**12 + 4*a*p**3*q*x**9 + 6*a*p**2*q** 2*x**6 + 4*a*p*q**3*x**3 + a*q**4 + b*x**8),x)*p - 2*int((sqrt(p*x**3 + q) *x**4)/(a*p**4*x**12 + 4*a*p**3*q*x**9 + 6*a*p**2*q**2*x**6 + 4*a*p*q**3*x **3 + a*q**4 + b*x**8),x)*q