Integrand size = 17, antiderivative size = 115 \[ \int \frac {x^3}{\left (a x+b x^3\right )^{3/2}} \, dx=-\frac {x}{b \sqrt {a x+b x^3}}+\frac {\sqrt {x} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{2 \sqrt [4]{a} b^{5/4} \sqrt {a x+b x^3}} \] Output:
-x/b/(b*x^3+a*x)^(1/2)+1/2*x^(1/2)*(a^(1/2)+b^(1/2)*x)*((b*x^2+a)/(a^(1/2) +b^(1/2)*x)^2)^(1/2)*InverseJacobiAM(2*arctan(b^(1/4)*x^(1/2)/a^(1/4)),1/2 *2^(1/2))/a^(1/4)/b^(5/4)/(b*x^3+a*x)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.02 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.47 \[ \int \frac {x^3}{\left (a x+b x^3\right )^{3/2}} \, dx=\frac {x \left (-1+\sqrt {1+\frac {b x^2}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\frac {b x^2}{a}\right )\right )}{b \sqrt {x \left (a+b x^2\right )}} \] Input:
Integrate[x^3/(a*x + b*x^3)^(3/2),x]
Output:
(x*(-1 + Sqrt[1 + (b*x^2)/a]*Hypergeometric2F1[1/4, 1/2, 5/4, -((b*x^2)/a) ]))/(b*Sqrt[x*(a + b*x^2)])
Time = 0.40 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {1928, 1917, 266, 761}
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.
| \(\displaystyle \int \frac {x^3}{\left (a x+b x^3\right )^{3/2}} \, dx\) |
| \(\Big \downarrow \) 1928 |
| \(\displaystyle \frac {\int \frac {1}{\sqrt {b x^3+a x}}dx}{2 b}-\frac {x}{b \sqrt {a x+b x^3}}\) |
| \(\Big \downarrow \) 1917 |
| \(\displaystyle \frac {\sqrt {x} \sqrt {a+b x^2} \int \frac {1}{\sqrt {x} \sqrt {b x^2+a}}dx}{2 b \sqrt {a x+b x^3}}-\frac {x}{b \sqrt {a x+b x^3}}\) |
| \(\Big \downarrow \) 266 |
| \(\displaystyle \frac {\sqrt {x} \sqrt {a+b x^2} \int \frac {1}{\sqrt {b x^2+a}}d\sqrt {x}}{b \sqrt {a x+b x^3}}-\frac {x}{b \sqrt {a x+b x^3}}\) |
| \(\Big \downarrow \) 761 |
| \(\displaystyle \frac {\sqrt {x} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{2 \sqrt [4]{a} b^{5/4} \sqrt {a x+b x^3}}-\frac {x}{b \sqrt {a x+b x^3}}\) |
Input:
Int[x^3/(a*x + b*x^3)^(3/2),x]
Output:
-(x/(b*Sqrt[a*x + b*x^3])) + (Sqrt[x]*(Sqrt[a] + Sqrt[b]*x)*Sqrt[(a + b*x^ 2)/(Sqrt[a] + Sqrt[b]*x)^2]*EllipticF[2*ArcTan[(b^(1/4)*Sqrt[x])/a^(1/4)], 1/2])/(2*a^(1/4)*b^(5/4)*Sqrt[a*x + b*x^3])
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De nominator[m]}, Simp[k/c Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) ^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I ntBinomialQ[a, b, c, 2, m, p, x]
Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[( 1 + q^2*x^2)*(Sqrt[(a + b*x^4)/(a*(1 + q^2*x^2)^2)]/(2*q*Sqrt[a + b*x^4]))* EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]
Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(a*x^j + b*x^n)^FracPart[p]/(x^(j*FracPart[p])*(a + b*x^(n - j))^FracPart[p]) Int[ x^(j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, j, n, p}, x] && !Integ erQ[p] && NeQ[n, j] && PosQ[n - j]
Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol ] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a*x^j + b*x^n)^(p + 1)/(b*(n - j)*( p + 1))), x] - Simp[c^n*((m + j*p - n + j + 1)/(b*(n - j)*(p + 1))) Int[( c*x)^(m - n)*(a*x^j + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c}, x] && !In tegerQ[p] && LtQ[0, j, n] && (IntegersQ[j, n] || GtQ[c, 0]) && LtQ[p, -1] & & GtQ[m + j*p + 1, n - j]
Time = 0.46 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.13
| method | result | size |
| default | \(-\frac {x}{b \sqrt {\left (x^{2}+\frac {a}{b}\right ) b x}}+\frac {\sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{2 b^{2} \sqrt {b \,x^{3}+a x}}\) | \(130\) |
| elliptic | \(-\frac {x}{b \sqrt {\left (x^{2}+\frac {a}{b}\right ) b x}}+\frac {\sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{2 b^{2} \sqrt {b \,x^{3}+a x}}\) | \(130\) |
Input:
int(x^3/(b*x^3+a*x)^(3/2),x,method=_RETURNVERBOSE)
Output:
-1/b*x/((x^2+a/b)*b*x)^(1/2)+1/2/b^2*(-a*b)^(1/2)*((x+(-a*b)^(1/2)/b)/(-a* b)^(1/2)*b)^(1/2)*(-2*(x-(-a*b)^(1/2)/b)/(-a*b)^(1/2)*b)^(1/2)*(-1/(-a*b)^ (1/2)*b*x)^(1/2)/(b*x^3+a*x)^(1/2)*EllipticF(((x+(-a*b)^(1/2)/b)/(-a*b)^(1 /2)*b)^(1/2),1/2*2^(1/2))
Time = 0.09 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.44 \[ \int \frac {x^3}{\left (a x+b x^3\right )^{3/2}} \, dx=\frac {{\left (b x^{2} + a\right )} \sqrt {b} {\rm weierstrassPInverse}\left (-\frac {4 \, a}{b}, 0, x\right ) - \sqrt {b x^{3} + a x} b}{b^{3} x^{2} + a b^{2}} \] Input:
integrate(x^3/(b*x^3+a*x)^(3/2),x, algorithm="fricas")
Output:
((b*x^2 + a)*sqrt(b)*weierstrassPInverse(-4*a/b, 0, x) - sqrt(b*x^3 + a*x) *b)/(b^3*x^2 + a*b^2)
\[ \int \frac {x^3}{\left (a x+b x^3\right )^{3/2}} \, dx=\int \frac {x^{3}}{\left (x \left (a + b x^{2}\right )\right )^{\frac {3}{2}}}\, dx \] Input:
integrate(x**3/(b*x**3+a*x)**(3/2),x)
Output:
Integral(x**3/(x*(a + b*x**2))**(3/2), x)
\[ \int \frac {x^3}{\left (a x+b x^3\right )^{3/2}} \, dx=\int { \frac {x^{3}}{{\left (b x^{3} + a x\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate(x^3/(b*x^3+a*x)^(3/2),x, algorithm="maxima")
Output:
integrate(x^3/(b*x^3 + a*x)^(3/2), x)
\[ \int \frac {x^3}{\left (a x+b x^3\right )^{3/2}} \, dx=\int { \frac {x^{3}}{{\left (b x^{3} + a x\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate(x^3/(b*x^3+a*x)^(3/2),x, algorithm="giac")
Output:
integrate(x^3/(b*x^3 + a*x)^(3/2), x)
Timed out. \[ \int \frac {x^3}{\left (a x+b x^3\right )^{3/2}} \, dx=\int \frac {x^3}{{\left (b\,x^3+a\,x\right )}^{3/2}} \,d x \] Input:
int(x^3/(a*x + b*x^3)^(3/2),x)
Output:
int(x^3/(a*x + b*x^3)^(3/2), x)
\[ \int \frac {x^3}{\left (a x+b x^3\right )^{3/2}} \, dx=\frac {-2 \sqrt {x}\, \sqrt {b \,x^{2}+a}+\left (\int \frac {\sqrt {x}\, \sqrt {b \,x^{2}+a}}{b^{2} x^{5}+2 a b \,x^{3}+a^{2} x}d x \right ) a^{2}+\left (\int \frac {\sqrt {x}\, \sqrt {b \,x^{2}+a}}{b^{2} x^{5}+2 a b \,x^{3}+a^{2} x}d x \right ) a b \,x^{2}}{b \left (b \,x^{2}+a \right )} \] Input:
int(x^3/(b*x^3+a*x)^(3/2),x)
Output:
( - 2*sqrt(x)*sqrt(a + b*x**2) + int((sqrt(x)*sqrt(a + b*x**2))/(a**2*x + 2*a*b*x**3 + b**2*x**5),x)*a**2 + int((sqrt(x)*sqrt(a + b*x**2))/(a**2*x + 2*a*b*x**3 + b**2*x**5),x)*a*b*x**2)/(b*(a + b*x**2))