\(\int \frac {1}{x^3 \sqrt {a x+b x^3}} \, dx\) [64]

Optimal result

Integrand size = 17, antiderivative size = 286 \[ \int \frac {1}{x^3 \sqrt {a x+b x^3}} \, dx=-\frac {6 b^{3/2} x \left (a+b x^2\right )}{5 a^2 \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {a x+b x^3}}-\frac {2 \sqrt {a x+b x^3}}{5 a x^3}+\frac {6 b \sqrt {a x+b x^3}}{5 a^2 x}+\frac {6 b^{5/4} \sqrt {x} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 a^{7/4} \sqrt {a x+b x^3}}-\frac {3 b^{5/4} \sqrt {x} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{5 a^{7/4} \sqrt {a x+b x^3}} \]

[Out]

-6/5*b^(3/2)*x*(b*x^2+a)/a^2/(a^(1/2)+x*b^(1/2))/(b*x^3+a*x)^(1/2)-2/5*(b*x^3+a*x)^(1/2)/a/x^3+6/5*b*(b*x^3+a*
x)^(1/2)/a^2/x+6/5*b^(5/4)*(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)/a^(7/4)/(b*x^3+a*x)^(1/2)-3/5*b^(5/4)*(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)/a^(7/4)/(b*x^3+a*x)^(1/2)

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 286, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {2050, 2057, 335, 311, 226, 1210} \[ \int \frac {1}{x^3 \sqrt {a x+b x^3}} \, dx=-\frac {3 b^{5/4} \sqrt {x} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{5 a^{7/4} \sqrt {a x+b x^3}}+\frac {6 b^{5/4} \sqrt {x} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 a^{7/4} \sqrt {a x+b x^3}}-\frac {6 b^{3/2} x \left (a+b x^2\right )}{5 a^2 \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {a x+b x^3}}+\frac {6 b \sqrt {a x+b x^3}}{5 a^2 x}-\frac {2 \sqrt {a x+b x^3}}{5 a x^3} \]

[In]

Int[1/(x^3*Sqrt[a*x + b*x^3]),x]

[Out]

(-6*b^(3/2)*x*(a + b*x^2))/(5*a^2*(Sqrt[a] + Sqrt[b]*x)*Sqrt[a*x + b*x^3]) - (2*Sqrt[a*x + b*x^3])/(5*a*x^3) +
(6*b*Sqrt[a*x + b*x^3])/(5*a^2*x) + (6*b^(5/4)*Sqrt[x]*(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])/(5*a^(7/4)*Sqrt[a*x + b*x^3]) - (3*b^(5/4)*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])/(5*a^(7/4)*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 311

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 2]}, Dist[1/q, Int[1/Sqrt[a + b*x^4], x],
x] - Dist[1/q, Int[(1 - q*x^2)/Sqrt[a + b*x^4], x], 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 1210

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
]))*EllipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rule 2050

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[c^(j - 1)*(c*x)^(m - j +
1)*((a*x^j + b*x^n)^(p + 1)/(a*(m + j*p + 1))), x] - Dist[b*((m + n*p + n - j + 1)/(a*c^(n - j)*(m + j*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]) && LtQ[m + j*p + 1, 0]

Rule 2057

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[c^IntPart[m]*(c*x)^FracPa
rt[m]*((a*x^j + b*x^n)^FracPart[p]/(x^(FracPart[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] &&  !IntegerQ[p] && NeQ[n, j] && PosQ[n
- j]

Rubi steps \begin{align*} \text {integral}& = -\frac {2 \sqrt {a x+b x^3}}{5 a x^3}-\frac {(3 b) \int \frac {1}{x \sqrt {a x+b x^3}} \, dx}{5 a} \\ & = -\frac {2 \sqrt {a x+b x^3}}{5 a x^3}+\frac {6 b \sqrt {a x+b x^3}}{5 a^2 x}-\frac {\left (3 b^2\right ) \int \frac {x}{\sqrt {a x+b x^3}} \, dx}{5 a^2} \\ & = -\frac {2 \sqrt {a x+b x^3}}{5 a x^3}+\frac {6 b \sqrt {a x+b x^3}}{5 a^2 x}-\frac {\left (3 b^2 \sqrt {x} \sqrt {a+b x^2}\right ) \int \frac {\sqrt {x}}{\sqrt {a+b x^2}} \, dx}{5 a^2 \sqrt {a x+b x^3}} \\ & = -\frac {2 \sqrt {a x+b x^3}}{5 a x^3}+\frac {6 b \sqrt {a x+b x^3}}{5 a^2 x}-\frac {\left (6 b^2 \sqrt {x} \sqrt {a+b x^2}\right ) \text {Subst}\left (\int \frac {x^2}{\sqrt {a+b x^4}} \, dx,x,\sqrt {x}\right )}{5 a^2 \sqrt {a x+b x^3}} \\ & = -\frac {2 \sqrt {a x+b x^3}}{5 a x^3}+\frac {6 b \sqrt {a x+b x^3}}{5 a^2 x}-\frac {\left (6 b^{3/2} \sqrt {x} \sqrt {a+b x^2}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x^4}} \, dx,x,\sqrt {x}\right )}{5 a^{3/2} \sqrt {a x+b x^3}}+\frac {\left (6 b^{3/2} \sqrt {x} \sqrt {a+b x^2}\right ) \text {Subst}\left (\int \frac {1-\frac {\sqrt {b} x^2}{\sqrt {a}}}{\sqrt {a+b x^4}} \, dx,x,\sqrt {x}\right )}{5 a^{3/2} \sqrt {a x+b x^3}} \\ & = -\frac {6 b^{3/2} x \left (a+b x^2\right )}{5 a^2 \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {a x+b x^3}}-\frac {2 \sqrt {a x+b x^3}}{5 a x^3}+\frac {6 b \sqrt {a x+b x^3}}{5 a^2 x}+\frac {6 b^{5/4} \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 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 a^{7/4} \sqrt {a x+b x^3}}-\frac {3 b^{5/4} \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 )}{5 a^{7/4} \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 = 10.02 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.19 \[ \int \frac {1}{x^3 \sqrt {a x+b x^3}} \, dx=-\frac {2 \sqrt {1+\frac {b x^2}{a}} \operatorname {Hypergeometric2F1}\left (-\frac {5}{4},\frac {1}{2},-\frac {1}{4},-\frac {b x^2}{a}\right )}{5 x^2 \sqrt {x \left (a+b x^2\right )}} \]

[In]

Integrate[1/(x^3*Sqrt[a*x + b*x^3]),x]

[Out]

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

Maple [A] (verified)

Time = 2.25 (sec) , antiderivative size = 195, normalized size of antiderivative = 0.68

method	result	size
risch	\(-\frac {2 \left (b \,x^{2}+a \right ) \left (-3 b \,x^{2}+a \right )}{5 a^{2} x^{2} \sqrt {x \left (b \,x^{2}+a \right )}}-\frac {3 b \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, \left (-\frac {2 \sqrt {-a b}\, E\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}+\frac {\sqrt {-a b}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}\right )}{5 a^{2} \sqrt {b \,x^{3}+a x}}\)	\(195\)
default	\(-\frac {2 \sqrt {b \,x^{3}+a x}}{5 a \,x^{3}}+\frac {6 \left (b \,x^{2}+a \right ) b}{5 a^{2} \sqrt {x \left (b \,x^{2}+a \right )}}-\frac {3 b \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, \left (-\frac {2 \sqrt {-a b}\, E\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}+\frac {\sqrt {-a b}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}\right )}{5 a^{2} \sqrt {b \,x^{3}+a x}}\)	\(204\)
elliptic	\(-\frac {2 \sqrt {b \,x^{3}+a x}}{5 a \,x^{3}}+\frac {6 \left (b \,x^{2}+a \right ) b}{5 a^{2} \sqrt {x \left (b \,x^{2}+a \right )}}-\frac {3 b \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, \left (-\frac {2 \sqrt {-a b}\, E\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}+\frac {\sqrt {-a b}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}\right )}{5 a^{2} \sqrt {b \,x^{3}+a x}}\)	\(204\)

[In]

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

[Out]

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

Fricas [C] (verification not implemented)

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

Time = 0.08 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.20 \[ \int \frac {1}{x^3 \sqrt {a x+b x^3}} \, dx=\frac {2 \, {\left (3 \, b^{\frac {3}{2}} x^{3} {\rm weierstrassZeta}\left (-\frac {4 \, a}{b}, 0, {\rm weierstrassPInverse}\left (-\frac {4 \, a}{b}, 0, x\right )\right ) + \sqrt {b x^{3} + a x} {\left (3 \, b x^{2} - a\right )}\right )}}{5 \, a^{2} x^{3}} \]

[In]

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

[Out]

2/5*(3*b^(3/2)*x^3*weierstrassZeta(-4*a/b, 0, weierstrassPInverse(-4*a/b, 0, x)) + sqrt(b*x^3 + a*x)*(3*b*x^2
- a))/(a^2*x^3)

Sympy [F]

\[ \int \frac {1}{x^3 \sqrt {a x+b x^3}} \, dx=\int \frac {1}{x^{3} \sqrt {x \left (a + b x^{2}\right )}}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {1}{x^3 \sqrt {a x+b x^3}} \, dx=\int { \frac {1}{\sqrt {b x^{3} + a x} x^{3}} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {1}{x^3 \sqrt {a x+b x^3}} \, dx=\int { \frac {1}{\sqrt {b x^{3} + a x} x^{3}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x^3 \sqrt {a x+b x^3}} \, dx=\int \frac {1}{x^3\,\sqrt {b\,x^3+a\,x}} \,d x \]

[In]

int(1/(x^3*(a*x + b*x^3)^(1/2)),x)

[Out]

int(1/(x^3*(a*x + b*x^3)^(1/2)), x)