Integrand size = 17, antiderivative size = 212 \[ \int \frac {x}{\sqrt {a x^2+b x^5}} \, dx=\frac {2 \sqrt {2+\sqrt {3}} x \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt {\frac {a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x}{\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x}\right ),-7-4 \sqrt {3}\right )}{\sqrt [4]{3} \sqrt [3]{b} \sqrt {\frac {\sqrt [3]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \sqrt {a x^2+b x^5}} \] Output:
2/3*(1/2*6^(1/2)+1/2*2^(1/2))*x*(a^(1/3)+b^(1/3)*x)*((a^(2/3)-a^(1/3)*b^(1 /3)*x+b^(2/3)*x^2)/((1+3^(1/2))*a^(1/3)+b^(1/3)*x)^2)^(1/2)*EllipticF(((1- 3^(1/2))*a^(1/3)+b^(1/3)*x)/((1+3^(1/2))*a^(1/3)+b^(1/3)*x),I*3^(1/2)+2*I) *3^(3/4)/b^(1/3)/(a^(1/3)*(a^(1/3)+b^(1/3)*x)/((1+3^(1/2))*a^(1/3)+b^(1/3) *x)^2)^(1/2)/(b*x^5+a*x^2)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.02 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.25 \[ \int \frac {x}{\sqrt {a x^2+b x^5}} \, dx=\frac {x^2 \sqrt {1+\frac {b x^3}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},-\frac {b x^3}{a}\right )}{\sqrt {x^2 \left (a+b x^3\right )}} \] Input:
Integrate[x/Sqrt[a*x^2 + b*x^5],x]
Output:
(x^2*Sqrt[1 + (b*x^3)/a]*Hypergeometric2F1[1/3, 1/2, 4/3, -((b*x^3)/a)])/S qrt[x^2*(a + b*x^3)]
Time = 0.40 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {1938, 759}
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.
\(\displaystyle \int \frac {x}{\sqrt {a x^2+b x^5}} \, dx\) |
\(\Big \downarrow \) 1938 |
\(\displaystyle \frac {x \sqrt {a+b x^3} \int \frac {1}{\sqrt {b x^3+a}}dx}{\sqrt {a x^2+b x^5}}\) |
\(\Big \downarrow \) 759 |
\(\displaystyle \frac {2 \sqrt {2+\sqrt {3}} x \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt {\frac {a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [3]{b} x+\left (1-\sqrt {3}\right ) \sqrt [3]{a}}{\sqrt [3]{b} x+\left (1+\sqrt {3}\right ) \sqrt [3]{a}}\right ),-7-4 \sqrt {3}\right )}{\sqrt [4]{3} \sqrt [3]{b} \sqrt {\frac {\sqrt [3]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \sqrt {a x^2+b x^5}}\) |
Input:
Int[x/Sqrt[a*x^2 + b*x^5],x]
Output:
(2*Sqrt[2 + Sqrt[3]]*x*(a^(1/3) + b^(1/3)*x)*Sqrt[(a^(2/3) - a^(1/3)*b^(1/ 3)*x + b^(2/3)*x^2)/((1 + Sqrt[3])*a^(1/3) + b^(1/3)*x)^2]*EllipticF[ArcSi n[((1 - Sqrt[3])*a^(1/3) + b^(1/3)*x)/((1 + Sqrt[3])*a^(1/3) + b^(1/3)*x)] , -7 - 4*Sqrt[3]])/(3^(1/4)*b^(1/3)*Sqrt[(a^(1/3)*(a^(1/3) + b^(1/3)*x))/( (1 + Sqrt[3])*a^(1/3) + b^(1/3)*x)^2]*Sqrt[a*x^2 + b*x^5])
Int[1/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[2*Sqrt[2 + Sqrt[3]]*(s + r*x)*(Sqrt[(s^2 - r*s *x + r^2*x^2)/((1 + Sqrt[3])*s + r*x)^2]/(3^(1/4)*r*Sqrt[a + b*x^3]*Sqrt[s* ((s + r*x)/((1 + Sqrt[3])*s + r*x)^2)]))*EllipticF[ArcSin[((1 - Sqrt[3])*s + r*x)/((1 + Sqrt[3])*s + r*x)], -7 - 4*Sqrt[3]], x]] /; FreeQ[{a, b}, x] & & PosQ[a]
Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol ] :> Simp[c^IntPart[m]*(c*x)^FracPart[m]*((a*x^j + b*x^n)^FracPart[p]/(x^(F racPart[m] + j*FracPart[p])*(a + b*x^(n - j))^FracPart[p])) Int[x^(m + j* p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] && !Inte gerQ[p] && NeQ[n, j] && PosQ[n - j]
Time = 0.54 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.09
method | result | size |
default | \(-\frac {i x \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}} \sqrt {-\frac {i \left (i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}-2 b x -\left (-a \,b^{2}\right )^{\frac {1}{3}}\right ) \sqrt {3}}{\left (-a \,b^{2}\right )^{\frac {1}{3}}}}\, \sqrt {-\frac {2 \left (-b x +\left (-a \,b^{2}\right )^{\frac {1}{3}}\right )}{\left (-a \,b^{2}\right )^{\frac {1}{3}} \left (i \sqrt {3}-3\right )}}\, \sqrt {-\frac {i \left (i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}+2 b x +\left (-a \,b^{2}\right )^{\frac {1}{3}}\right ) \sqrt {3}}{\left (-a \,b^{2}\right )^{\frac {1}{3}}}}\, \operatorname {EllipticF}\left (\frac {\sqrt {3}\, \sqrt {2}\, \sqrt {-\frac {i \left (i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}-2 b x -\left (-a \,b^{2}\right )^{\frac {1}{3}}\right ) \sqrt {3}}{\left (-a \,b^{2}\right )^{\frac {1}{3}}}}}{6}, \sqrt {2}\, \sqrt {\frac {i \sqrt {3}}{i \sqrt {3}-3}}\right )}{3 \sqrt {b \,x^{5}+a \,x^{2}}\, b}\) | \(231\) |
Input:
int(x/(b*x^5+a*x^2)^(1/2),x,method=_RETURNVERBOSE)
Output:
-1/3*I/(b*x^5+a*x^2)^(1/2)*x*3^(1/2)/b*(-a*b^2)^(1/3)*(-I*(I*3^(1/2)*(-a*b ^2)^(1/3)-2*b*x-(-a*b^2)^(1/3))*3^(1/2)/(-a*b^2)^(1/3))^(1/2)*(-2*(-b*x+(- a*b^2)^(1/3))/(-a*b^2)^(1/3)/(I*3^(1/2)-3))^(1/2)*(-I*(I*3^(1/2)*(-a*b^2)^ (1/3)+2*b*x+(-a*b^2)^(1/3))*3^(1/2)/(-a*b^2)^(1/3))^(1/2)*EllipticF(1/6*3^ (1/2)*2^(1/2)*(-I*(I*3^(1/2)*(-a*b^2)^(1/3)-2*b*x-(-a*b^2)^(1/3))*3^(1/2)/ (-a*b^2)^(1/3))^(1/2),2^(1/2)*(I*3^(1/2)/(I*3^(1/2)-3))^(1/2))
Time = 0.07 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.07 \[ \int \frac {x}{\sqrt {a x^2+b x^5}} \, dx=\frac {2 \, {\rm weierstrassPInverse}\left (0, -\frac {4 \, a}{b}, x\right )}{\sqrt {b}} \] Input:
integrate(x/(b*x^5+a*x^2)^(1/2),x, algorithm="fricas")
Output:
2*weierstrassPInverse(0, -4*a/b, x)/sqrt(b)
\[ \int \frac {x}{\sqrt {a x^2+b x^5}} \, dx=\int \frac {x}{\sqrt {x^{2} \left (a + b x^{3}\right )}}\, dx \] Input:
integrate(x/(b*x**5+a*x**2)**(1/2),x)
Output:
Integral(x/sqrt(x**2*(a + b*x**3)), x)
\[ \int \frac {x}{\sqrt {a x^2+b x^5}} \, dx=\int { \frac {x}{\sqrt {b x^{5} + a x^{2}}} \,d x } \] Input:
integrate(x/(b*x^5+a*x^2)^(1/2),x, algorithm="maxima")
Output:
integrate(x/sqrt(b*x^5 + a*x^2), x)
\[ \int \frac {x}{\sqrt {a x^2+b x^5}} \, dx=\int { \frac {x}{\sqrt {b x^{5} + a x^{2}}} \,d x } \] Input:
integrate(x/(b*x^5+a*x^2)^(1/2),x, algorithm="giac")
Output:
integrate(x/sqrt(b*x^5 + a*x^2), x)
Timed out. \[ \int \frac {x}{\sqrt {a x^2+b x^5}} \, dx=\int \frac {x}{\sqrt {b\,x^5+a\,x^2}} \,d x \] Input:
int(x/(a*x^2 + b*x^5)^(1/2),x)
Output:
int(x/(a*x^2 + b*x^5)^(1/2), x)
\[ \int \frac {x}{\sqrt {a x^2+b x^5}} \, dx=\int \frac {\sqrt {b \,x^{3}+a}}{b \,x^{3}+a}d x \] Input:
int(x/(b*x^5+a*x^2)^(1/2),x)
Output:
int(sqrt(a + b*x**3)/(a + b*x**3),x)