Integrand size = 40, antiderivative size = 116 \[ \int \frac {2 b+a x^3}{\left (-b+x^2+a x^3\right ) \sqrt [4]{-b x^2+a x^5}} \, dx=-\sqrt {2} \arctan \left (\frac {\sqrt {2} x \sqrt [4]{-b x^2+a x^5}}{-x^2+\sqrt {-b x^2+a x^5}}\right )-\sqrt {2} \text {arctanh}\left (\frac {\frac {x^2}{\sqrt {2}}+\frac {\sqrt {-b x^2+a x^5}}{\sqrt {2}}}{x \sqrt [4]{-b x^2+a x^5}}\right ) \] Output:
-2^(1/2)*arctan(2^(1/2)*x*(a*x^5-b*x^2)^(1/4)/(-x^2+(a*x^5-b*x^2)^(1/2)))- 2^(1/2)*arctanh((1/2*2^(1/2)*x^2+1/2*(a*x^5-b*x^2)^(1/2)*2^(1/2))/x/(a*x^5 -b*x^2)^(1/4))
Time = 3.19 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.09 \[ \int \frac {2 b+a x^3}{\left (-b+x^2+a x^3\right ) \sqrt [4]{-b x^2+a x^5}} \, dx=-\frac {\sqrt {2} \sqrt {x} \sqrt [4]{-b+a x^3} \left (\arctan \left (\frac {\sqrt {2} \sqrt {x} \sqrt [4]{-b+a x^3}}{-x+\sqrt {-b+a x^3}}\right )+\text {arctanh}\left (\frac {x+\sqrt {-b+a x^3}}{\sqrt {2} \sqrt {x} \sqrt [4]{-b+a x^3}}\right )\right )}{\sqrt [4]{-b x^2+a x^5}} \] Input:
Integrate[(2*b + a*x^3)/((-b + x^2 + a*x^3)*(-(b*x^2) + a*x^5)^(1/4)),x]
Output:
-((Sqrt[2]*Sqrt[x]*(-b + a*x^3)^(1/4)*(ArcTan[(Sqrt[2]*Sqrt[x]*(-b + a*x^3 )^(1/4))/(-x + Sqrt[-b + a*x^3])] + ArcTanh[(x + Sqrt[-b + a*x^3])/(Sqrt[2 ]*Sqrt[x]*(-b + a*x^3)^(1/4))]))/(-(b*x^2) + a*x^5)^(1/4))
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.
\(\displaystyle \int \frac {a x^3+2 b}{\left (a x^3-b+x^2\right ) \sqrt [4]{a x^5-b x^2}} \, dx\) |
\(\Big \downarrow \) 2467 |
\(\displaystyle \frac {\sqrt {x} \sqrt [4]{a x^3-b} \int -\frac {a x^3+2 b}{\sqrt {x} \left (-a x^3-x^2+b\right ) \sqrt [4]{a x^3-b}}dx}{\sqrt [4]{a x^5-b x^2}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\sqrt {x} \sqrt [4]{a x^3-b} \int \frac {a x^3+2 b}{\sqrt {x} \left (-a x^3-x^2+b\right ) \sqrt [4]{a x^3-b}}dx}{\sqrt [4]{a x^5-b x^2}}\) |
\(\Big \downarrow \) 2035 |
\(\displaystyle -\frac {2 \sqrt {x} \sqrt [4]{a x^3-b} \int \frac {a x^3+2 b}{\left (-a x^3-x^2+b\right ) \sqrt [4]{a x^3-b}}d\sqrt {x}}{\sqrt [4]{a x^5-b x^2}}\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {2 \sqrt {x} \sqrt [4]{a x^3-b} \int \left (\frac {3 b-x^2}{\left (-a x^3-x^2+b\right ) \sqrt [4]{a x^3-b}}-\frac {1}{\sqrt [4]{a x^3-b}}\right )d\sqrt {x}}{\sqrt [4]{a x^5-b x^2}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 \sqrt {x} \sqrt [4]{a x^3-b} \left (3 b \int \frac {1}{\left (-a x^3-x^2+b\right ) \sqrt [4]{a x^3-b}}d\sqrt {x}+\int \frac {x^2}{\sqrt [4]{a x^3-b} \left (a x^3+x^2-b\right )}d\sqrt {x}-\frac {\sqrt {x} \sqrt [4]{1-\frac {a x^3}{b}} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{4},\frac {7}{6},\frac {a x^3}{b}\right )}{\sqrt [4]{a x^3-b}}\right )}{\sqrt [4]{a x^5-b x^2}}\) |
Input:
Int[(2*b + a*x^3)/((-b + x^2 + a*x^3)*(-(b*x^2) + a*x^5)^(1/4)),x]
Output:
$Aborted
Time = 1.11 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.28
method | result | size |
pseudoelliptic | \(\frac {\sqrt {2}\, \left (\ln \left (\frac {-\left (x^{2} \left (a \,x^{3}-b \right )\right )^{\frac {1}{4}} \sqrt {2}\, x +x^{2}+\sqrt {x^{2} \left (a \,x^{3}-b \right )}}{\left (x^{2} \left (a \,x^{3}-b \right )\right )^{\frac {1}{4}} \sqrt {2}\, x +x^{2}+\sqrt {x^{2} \left (a \,x^{3}-b \right )}}\right )+2 \arctan \left (\frac {\left (x^{2} \left (a \,x^{3}-b \right )\right )^{\frac {1}{4}} \sqrt {2}+x}{x}\right )+2 \arctan \left (\frac {\left (x^{2} \left (a \,x^{3}-b \right )\right )^{\frac {1}{4}} \sqrt {2}-x}{x}\right )\right )}{2}\) | \(148\) |
Input:
int((a*x^3+2*b)/(a*x^3+x^2-b)/(a*x^5-b*x^2)^(1/4),x,method=_RETURNVERBOSE)
Output:
1/2*2^(1/2)*(ln((-(x^2*(a*x^3-b))^(1/4)*2^(1/2)*x+x^2+(x^2*(a*x^3-b))^(1/2 ))/((x^2*(a*x^3-b))^(1/4)*2^(1/2)*x+x^2+(x^2*(a*x^3-b))^(1/2)))+2*arctan(( (x^2*(a*x^3-b))^(1/4)*2^(1/2)+x)/x)+2*arctan(((x^2*(a*x^3-b))^(1/4)*2^(1/2 )-x)/x))
Timed out. \[ \int \frac {2 b+a x^3}{\left (-b+x^2+a x^3\right ) \sqrt [4]{-b x^2+a x^5}} \, dx=\text {Timed out} \] Input:
integrate((a*x^3+2*b)/(a*x^3+x^2-b)/(a*x^5-b*x^2)^(1/4),x, algorithm="fric as")
Output:
Timed out
\[ \int \frac {2 b+a x^3}{\left (-b+x^2+a x^3\right ) \sqrt [4]{-b x^2+a x^5}} \, dx=\int \frac {a x^{3} + 2 b}{\sqrt [4]{x^{2} \left (a x^{3} - b\right )} \left (a x^{3} - b + x^{2}\right )}\, dx \] Input:
integrate((a*x**3+2*b)/(a*x**3+x**2-b)/(a*x**5-b*x**2)**(1/4),x)
Output:
Integral((a*x**3 + 2*b)/((x**2*(a*x**3 - b))**(1/4)*(a*x**3 - b + x**2)), x)
\[ \int \frac {2 b+a x^3}{\left (-b+x^2+a x^3\right ) \sqrt [4]{-b x^2+a x^5}} \, dx=\int { \frac {a x^{3} + 2 \, b}{{\left (a x^{5} - b x^{2}\right )}^{\frac {1}{4}} {\left (a x^{3} + x^{2} - b\right )}} \,d x } \] Input:
integrate((a*x^3+2*b)/(a*x^3+x^2-b)/(a*x^5-b*x^2)^(1/4),x, algorithm="maxi ma")
Output:
integrate((a*x^3 + 2*b)/((a*x^5 - b*x^2)^(1/4)*(a*x^3 + x^2 - b)), x)
\[ \int \frac {2 b+a x^3}{\left (-b+x^2+a x^3\right ) \sqrt [4]{-b x^2+a x^5}} \, dx=\int { \frac {a x^{3} + 2 \, b}{{\left (a x^{5} - b x^{2}\right )}^{\frac {1}{4}} {\left (a x^{3} + x^{2} - b\right )}} \,d x } \] Input:
integrate((a*x^3+2*b)/(a*x^3+x^2-b)/(a*x^5-b*x^2)^(1/4),x, algorithm="giac ")
Output:
integrate((a*x^3 + 2*b)/((a*x^5 - b*x^2)^(1/4)*(a*x^3 + x^2 - b)), x)
Timed out. \[ \int \frac {2 b+a x^3}{\left (-b+x^2+a x^3\right ) \sqrt [4]{-b x^2+a x^5}} \, dx=\int \frac {a\,x^3+2\,b}{{\left (a\,x^5-b\,x^2\right )}^{1/4}\,\left (a\,x^3+x^2-b\right )} \,d x \] Input:
int((2*b + a*x^3)/((a*x^5 - b*x^2)^(1/4)*(a*x^3 - b + x^2)),x)
Output:
int((2*b + a*x^3)/((a*x^5 - b*x^2)^(1/4)*(a*x^3 - b + x^2)), x)
\[ \int \frac {2 b+a x^3}{\left (-b+x^2+a x^3\right ) \sqrt [4]{-b x^2+a x^5}} \, dx =\text {Too large to display} \] Input:
int((a*x^3+2*b)/(a*x^3+x^2-b)/(a*x^5-b*x^2)^(1/4),x)
Output:
(20*sqrt(x)*(a*x**3 - b)**(5/4)*a*x + 60*sqrt(x)*(a*x**3 - b)**(5/4) + 12* sqrt(x)*(a*x**3 - b)**(1/4)*a*x**3 - 12*sqrt(x)*(a*x**3 - b)**(1/4)*b - 66 *sqrt(a*x**3 - b)*int((a*x**3 - b)**(3/4)/(sqrt(x)*a**3*x**9 - 3*sqrt(x)*a **2*b*x**6 + sqrt(x)*a**2*x**8 + 3*sqrt(x)*a*b**2*x**3 - 2*sqrt(x)*a*b*x** 5 - sqrt(x)*b**3 + sqrt(x)*b**2*x**2),x)*a*b**2*x**3 + 66*sqrt(a*x**3 - b) *int((a*x**3 - b)**(3/4)/(sqrt(x)*a**3*x**9 - 3*sqrt(x)*a**2*b*x**6 + sqrt (x)*a**2*x**8 + 3*sqrt(x)*a*b**2*x**3 - 2*sqrt(x)*a*b*x**5 - sqrt(x)*b**3 + sqrt(x)*b**2*x**2),x)*b**3 - 15*sqrt(a*x**3 - b)*int((sqrt(x)*(a*x**3 - b)**(3/4)*x**6)/(a**3*x**9 - 3*a**2*b*x**6 + a**2*x**8 + 3*a*b**2*x**3 - 2 *a*b*x**5 - b**3 + b**2*x**2),x)*a**4*x**3 + 15*sqrt(a*x**3 - b)*int((sqrt (x)*(a*x**3 - b)**(3/4)*x**6)/(a**3*x**9 - 3*a**2*b*x**6 + a**2*x**8 + 3*a *b**2*x**3 - 2*a*b*x**5 - b**3 + b**2*x**2),x)*a**3*b + 45*sqrt(a*x**3 - b )*int((sqrt(x)*(a*x**3 - b)**(3/4)*x**3)/(a**3*x**9 - 3*a**2*b*x**6 + a**2 *x**8 + 3*a*b**2*x**3 - 2*a*b*x**5 - b**3 + b**2*x**2),x)*a**3*b*x**3 - 45 *sqrt(a*x**3 - b)*int((sqrt(x)*(a*x**3 - b)**(3/4)*x**3)/(a**3*x**9 - 3*a* *2*b*x**6 + a**2*x**8 + 3*a*b**2*x**3 - 2*a*b*x**5 - b**3 + b**2*x**2),x)* a**2*b**2 + 81*sqrt(a*x**3 - b)*int((sqrt(x)*(a*x**3 - b)**(3/4)*x**2)/(a* *3*x**9 - 3*a**2*b*x**6 + a**2*x**8 + 3*a*b**2*x**3 - 2*a*b*x**5 - b**3 + b**2*x**2),x)*a**2*b*x**3 - 81*sqrt(a*x**3 - b)*int((sqrt(x)*(a*x**3 - b)* *(3/4)*x**2)/(a**3*x**9 - 3*a**2*b*x**6 + a**2*x**8 + 3*a*b**2*x**3 - 2...