Integrand size = 17, antiderivative size = 248 \[ \int \frac {\sqrt {a x+b x^3}}{x^2} \, dx=\frac {4 \sqrt {b} x \left (a+b x^2\right )}{\left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {a x+b x^3}}-\frac {2 \sqrt {a x+b x^3}}{x}-\frac {4 \sqrt [4]{a} \sqrt [4]{b} \sqrt {x} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{\sqrt {a x+b x^3}}+\frac {2 \sqrt [4]{a} \sqrt [4]{b} \sqrt {x} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{\sqrt {a x+b x^3}} \]
4*x*(b*x^2+a)*b^(1/2)/(a^(1/2)+x*b^(1/2))/(b*x^3+a*x)^(1/2)-2*(b*x^3+a*x)^ (1/2)/x-4*a^(1/4)*b^(1/4)*(cos(2*arctan(b^(1/4)*x^(1/2)/a^(1/4)))^2)^(1/2) /cos(2*arctan(b^(1/4)*x^(1/2)/a^(1/4)))*EllipticE(sin(2*arctan(b^(1/4)*x^( 1/2)/a^(1/4))),1/2*2^(1/2))*(a^(1/2)+x*b^(1/2))*x^(1/2)*((b*x^2+a)/(a^(1/2 )+x*b^(1/2))^2)^(1/2)/(b*x^3+a*x)^(1/2)+2*a^(1/4)*b^(1/4)*(cos(2*arctan(b^ (1/4)*x^(1/2)/a^(1/4)))^2)^(1/2)/cos(2*arctan(b^(1/4)*x^(1/2)/a^(1/4)))*El lipticF(sin(2*arctan(b^(1/4)*x^(1/2)/a^(1/4))),1/2*2^(1/2))*(a^(1/2)+x*b^( 1/2))*x^(1/2)*((b*x^2+a)/(a^(1/2)+x*b^(1/2))^2)^(1/2)/(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 = 51, normalized size of antiderivative = 0.21 \[ \int \frac {\sqrt {a x+b x^3}}{x^2} \, dx=-\frac {2 \sqrt {x \left (a+b x^2\right )} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-\frac {1}{4},\frac {3}{4},-\frac {b x^2}{a}\right )}{x \sqrt {1+\frac {b x^2}{a}}} \]
(-2*Sqrt[x*(a + b*x^2)]*Hypergeometric2F1[-1/2, -1/4, 3/4, -((b*x^2)/a)])/ (x*Sqrt[1 + (b*x^2)/a])
Time = 0.37 (sec) , antiderivative size = 266, normalized size of antiderivative = 1.07, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.412, Rules used = {1926, 1938, 266, 834, 27, 761, 1510}
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.
\(\displaystyle \int \frac {\sqrt {a x+b x^3}}{x^2} \, dx\) |
\(\Big \downarrow \) 1926 |
\(\displaystyle 2 b \int \frac {x}{\sqrt {b x^3+a x}}dx-\frac {2 \sqrt {a x+b x^3}}{x}\) |
\(\Big \downarrow \) 1938 |
\(\displaystyle \frac {2 b \sqrt {x} \sqrt {a+b x^2} \int \frac {\sqrt {x}}{\sqrt {b x^2+a}}dx}{\sqrt {a x+b x^3}}-\frac {2 \sqrt {a x+b x^3}}{x}\) |
\(\Big \downarrow \) 266 |
\(\displaystyle \frac {4 b \sqrt {x} \sqrt {a+b x^2} \int \frac {x}{\sqrt {b x^2+a}}d\sqrt {x}}{\sqrt {a x+b x^3}}-\frac {2 \sqrt {a x+b x^3}}{x}\) |
\(\Big \downarrow \) 834 |
\(\displaystyle \frac {4 b \sqrt {x} \sqrt {a+b x^2} \left (\frac {\sqrt {a} \int \frac {1}{\sqrt {b x^2+a}}d\sqrt {x}}{\sqrt {b}}-\frac {\sqrt {a} \int \frac {\sqrt {a}-\sqrt {b} x}{\sqrt {a} \sqrt {b x^2+a}}d\sqrt {x}}{\sqrt {b}}\right )}{\sqrt {a x+b x^3}}-\frac {2 \sqrt {a x+b x^3}}{x}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {4 b \sqrt {x} \sqrt {a+b x^2} \left (\frac {\sqrt {a} \int \frac {1}{\sqrt {b x^2+a}}d\sqrt {x}}{\sqrt {b}}-\frac {\int \frac {\sqrt {a}-\sqrt {b} x}{\sqrt {b x^2+a}}d\sqrt {x}}{\sqrt {b}}\right )}{\sqrt {a x+b x^3}}-\frac {2 \sqrt {a x+b x^3}}{x}\) |
\(\Big \downarrow \) 761 |
\(\displaystyle \frac {4 b \sqrt {x} \sqrt {a+b x^2} \left (\frac {\sqrt [4]{a} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{2 b^{3/4} \sqrt {a+b x^2}}-\frac {\int \frac {\sqrt {a}-\sqrt {b} x}{\sqrt {b x^2+a}}d\sqrt {x}}{\sqrt {b}}\right )}{\sqrt {a x+b x^3}}-\frac {2 \sqrt {a x+b x^3}}{x}\) |
\(\Big \downarrow \) 1510 |
\(\displaystyle \frac {4 b \sqrt {x} \sqrt {a+b x^2} \left (\frac {\sqrt [4]{a} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{2 b^{3/4} \sqrt {a+b x^2}}-\frac {\frac {\sqrt [4]{a} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{\sqrt [4]{b} \sqrt {a+b x^2}}-\frac {\sqrt {x} \sqrt {a+b x^2}}{\sqrt {a}+\sqrt {b} x}}{\sqrt {b}}\right )}{\sqrt {a x+b x^3}}-\frac {2 \sqrt {a x+b x^3}}{x}\) |
(-2*Sqrt[a*x + b*x^3])/x + (4*b*Sqrt[x]*Sqrt[a + b*x^2]*(-((-((Sqrt[x]*Sqr t[a + b*x^2])/(Sqrt[a] + Sqrt[b]*x)) + (a^(1/4)*(Sqrt[a] + Sqrt[b]*x)*Sqrt [(a + b*x^2)/(Sqrt[a] + Sqrt[b]*x)^2]*EllipticE[2*ArcTan[(b^(1/4)*Sqrt[x]) /a^(1/4)], 1/2])/(b^(1/4)*Sqrt[a + b*x^2]))/Sqrt[b]) + (a^(1/4)*(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*b^(3/4)*Sqrt[a + b*x^2])))/Sqrt[a*x + b*x^3]
3.1.43.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 2]}, S imp[1/q Int[1/Sqrt[a + b*x^4], x], x] - Simp[1/q Int[(1 - q*x^2)/Sqrt[a + b*x^4], x], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]
Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, Simp[(-d)*x*(Sqrt[a + c*x^4]/(a*(1 + q^2*x^2))), x] + Simp[d* (1 + q^2*x^2)*(Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]/(q*Sqrt[a + c*x^4]))*E llipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e }, x] && PosQ[c/a]
Int[((c_.)*(x_))^(m_)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a*x^j + b*x^n)^p/(c*(m + j*p + 1))), x] - Simp[b*p *((n - j)/(c^n*(m + j*p + 1))) Int[(c*x)^(m + n)*(a*x^j + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && !IntegerQ[p] && LtQ[0, j, n] && (Integer sQ[j, n] || GtQ[c, 0]) && GtQ[p, 0] && LtQ[m + j*p + 1, 0]
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 = 2.13 (sec) , antiderivative size = 177, normalized size of antiderivative = 0.71
method | result | size |
default | \(-\frac {2 \left (b \,x^{2}+a \right )}{\sqrt {x \left (b \,x^{2}+a \right )}}+\frac {2 \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, \left (-\frac {2 \sqrt {-a b}\, E\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}+\frac {\sqrt {-a b}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}\right )}{\sqrt {b \,x^{3}+a x}}\) | \(177\) |
risch | \(-\frac {2 \left (b \,x^{2}+a \right )}{\sqrt {x \left (b \,x^{2}+a \right )}}+\frac {2 \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, \left (-\frac {2 \sqrt {-a b}\, E\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}+\frac {\sqrt {-a b}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}\right )}{\sqrt {b \,x^{3}+a x}}\) | \(177\) |
elliptic | \(-\frac {2 \left (b \,x^{2}+a \right )}{\sqrt {x \left (b \,x^{2}+a \right )}}+\frac {2 \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, \left (-\frac {2 \sqrt {-a b}\, E\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}+\frac {\sqrt {-a b}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}\right )}{\sqrt {b \,x^{3}+a x}}\) | \(177\) |
-2*(b*x^2+a)/(x*(b*x^2+a))^(1/2)+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)*(-x/(-a*b)^(1 /2)*b)^(1/2)/(b*x^3+a*x)^(1/2)*(-2*(-a*b)^(1/2)/b*EllipticE(((x+(-a*b)^(1/ 2)/b)/(-a*b)^(1/2)*b)^(1/2),1/2*2^(1/2))+(-a*b)^(1/2)/b*EllipticF(((x+(-a* b)^(1/2)/b)/(-a*b)^(1/2)*b)^(1/2),1/2*2^(1/2)))
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.08 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.16 \[ \int \frac {\sqrt {a x+b x^3}}{x^2} \, dx=-\frac {2 \, {\left (2 \, \sqrt {b} x {\rm weierstrassZeta}\left (-\frac {4 \, a}{b}, 0, {\rm weierstrassPInverse}\left (-\frac {4 \, a}{b}, 0, x\right )\right ) + \sqrt {b x^{3} + a x}\right )}}{x} \]
-2*(2*sqrt(b)*x*weierstrassZeta(-4*a/b, 0, weierstrassPInverse(-4*a/b, 0, x)) + sqrt(b*x^3 + a*x))/x
\[ \int \frac {\sqrt {a x+b x^3}}{x^2} \, dx=\int \frac {\sqrt {x \left (a + b x^{2}\right )}}{x^{2}}\, dx \]
\[ \int \frac {\sqrt {a x+b x^3}}{x^2} \, dx=\int { \frac {\sqrt {b x^{3} + a x}}{x^{2}} \,d x } \]
\[ \int \frac {\sqrt {a x+b x^3}}{x^2} \, dx=\int { \frac {\sqrt {b x^{3} + a x}}{x^{2}} \,d x } \]
Timed out. \[ \int \frac {\sqrt {a x+b x^3}}{x^2} \, dx=\int \frac {\sqrt {b\,x^3+a\,x}}{x^2} \,d x \]