\(\int \frac {x^2}{\sqrt {a x+b x^3}} \, dx\) [59]

Optimal result

Integrand size = 17, antiderivative size = 116 \[ \int \frac {x^2}{\sqrt {a x+b x^3}} \, dx=\frac {2 \sqrt {a x+b x^3}}{3 b}-\frac {a^{3/4} \sqrt {x} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{3 b^{5/4} \sqrt {a x+b x^3}} \]

[Out]

2/3*(b*x^3+a*x)^(1/2)/b-1/3*a^(3/4)*(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)))*EllipticF(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^(5/4)/(b*x^3+a*x)^(1/2)

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {2049, 2036, 335, 226} \[ \int \frac {x^2}{\sqrt {a x+b x^3}} \, dx=\frac {2 \sqrt {a x+b x^3}}{3 b}-\frac {a^{3/4} \sqrt {x} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{3 b^{5/4} \sqrt {a x+b x^3}} \]

[In]

Int[x^2/Sqrt[a*x + b*x^3],x]

[Out]

(2*Sqrt[a*x + b*x^3])/(3*b) - (a^(3/4)*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])/(3*b^(5/4)*Sqrt[a*x + b*x^3])

Rule 226

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]

Rule 335

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 2036

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[(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] &&  !I
ntegerQ[p] && NeQ[n, j] && PosQ[n - j]

Rule 2049

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*(m + n*p + 1))), x] - Dist[a*c^(n - j)*((m + j*p - n + j + 1)/(b*(m + n*p + 1))
), Int[(c*x)^(m - (n - j))*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IntegerQ[p] && LtQ[0, j
, n] && (IntegersQ[j, n] || GtQ[c, 0]) && GtQ[m + j*p + 1 - n + j, 0] && NeQ[m + n*p + 1, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {2 \sqrt {a x+b x^3}}{3 b}-\frac {a \int \frac {1}{\sqrt {a x+b x^3}} \, dx}{3 b} \\ & = \frac {2 \sqrt {a x+b x^3}}{3 b}-\frac {\left (a \sqrt {x} \sqrt {a+b x^2}\right ) \int \frac {1}{\sqrt {x} \sqrt {a+b x^2}} \, dx}{3 b \sqrt {a x+b x^3}} \\ & = \frac {2 \sqrt {a x+b x^3}}{3 b}-\frac {\left (2 a \sqrt {x} \sqrt {a+b x^2}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x^4}} \, dx,x,\sqrt {x}\right )}{3 b \sqrt {a x+b x^3}} \\ & = \frac {2 \sqrt {a x+b x^3}}{3 b}-\frac {a^{3/4} \sqrt {x} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{3 b^{5/4} \sqrt {a x+b x^3}} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 10.03 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.55 \[ \int \frac {x^2}{\sqrt {a x+b x^3}} \, dx=\frac {2 x \left (a+b x^2-a \sqrt {1+\frac {b x^2}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\frac {b x^2}{a}\right )\right )}{3 b \sqrt {x \left (a+b x^2\right )}} \]

[In]

Integrate[x^2/Sqrt[a*x + b*x^3],x]

[Out]

(2*x*(a + b*x^2 - a*Sqrt[1 + (b*x^2)/a]*Hypergeometric2F1[1/4, 1/2, 5/4, -((b*x^2)/a)]))/(3*b*Sqrt[x*(a + b*x^
2)])

Maple [A] (verified)

Time = 2.12 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.09

method	result	size
default	\(\frac {2 \sqrt {b \,x^{3}+a x}}{3 b}-\frac {a \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{3 b^{2} \sqrt {b \,x^{3}+a x}}\)	\(127\)
elliptic	\(\frac {2 \sqrt {b \,x^{3}+a x}}{3 b}-\frac {a \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{3 b^{2} \sqrt {b \,x^{3}+a x}}\)	\(127\)
risch	\(\frac {2 x \left (b \,x^{2}+a \right )}{3 b \sqrt {x \left (b \,x^{2}+a \right )}}-\frac {a \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{3 b^{2} \sqrt {b \,x^{3}+a x}}\)	\(135\)

[In]

int(x^2/(b*x^3+a*x)^(1/2),x,method=_RETURNVERBOSE)

[Out]

2/3*(b*x^3+a*x)^(1/2)/b-1/3*a/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)*(-x/(-a*b)^(1/2)*b)^(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))

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.09 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.29 \[ \int \frac {x^2}{\sqrt {a x+b x^3}} \, dx=-\frac {2 \, {\left (a \sqrt {b} {\rm weierstrassPInverse}\left (-\frac {4 \, a}{b}, 0, x\right ) - \sqrt {b x^{3} + a x} b\right )}}{3 \, b^{2}} \]

[In]

integrate(x^2/(b*x^3+a*x)^(1/2),x, algorithm="fricas")

[Out]

-2/3*(a*sqrt(b)*weierstrassPInverse(-4*a/b, 0, x) - sqrt(b*x^3 + a*x)*b)/b^2

Sympy [F]

\[ \int \frac {x^2}{\sqrt {a x+b x^3}} \, dx=\int \frac {x^{2}}{\sqrt {x \left (a + b x^{2}\right )}}\, dx \]

[In]

integrate(x**2/(b*x**3+a*x)**(1/2),x)

[Out]

Integral(x**2/sqrt(x*(a + b*x**2)), x)

Maxima [F]

\[ \int \frac {x^2}{\sqrt {a x+b x^3}} \, dx=\int { \frac {x^{2}}{\sqrt {b x^{3} + a x}} \,d x } \]

[In]

integrate(x^2/(b*x^3+a*x)^(1/2),x, algorithm="maxima")

[Out]

integrate(x^2/sqrt(b*x^3 + a*x), x)

Giac [F]

\[ \int \frac {x^2}{\sqrt {a x+b x^3}} \, dx=\int { \frac {x^{2}}{\sqrt {b x^{3} + a x}} \,d x } \]

[In]

integrate(x^2/(b*x^3+a*x)^(1/2),x, algorithm="giac")

[Out]

integrate(x^2/sqrt(b*x^3 + a*x), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {x^2}{\sqrt {a x+b x^3}} \, dx=\int \frac {x^2}{\sqrt {b\,x^3+a\,x}} \,d x \]

[In]

int(x^2/(a*x + b*x^3)^(1/2),x)

[Out]

int(x^2/(a*x + b*x^3)^(1/2), x)