\(\int \frac {\sqrt {a x+b x^3}}{x} \, dx\) [42]

Optimal result

Integrand size = 17, antiderivative size = 113 \[ \int \frac {\sqrt {a x+b x^3}}{x} \, dx=\frac {2}{3} \sqrt {a x+b x^3}+\frac {2 a^{3/4} \sqrt {x} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{3 \sqrt [4]{b} \sqrt {a x+b x^3}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 113, 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 = {2046, 2036, 335, 226} \[ \int \frac {\sqrt {a x+b x^3}}{x} \, dx=\frac {2 a^{3/4} \sqrt {x} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{3 \sqrt [4]{b} \sqrt {a x+b x^3}}+\frac {2}{3} \sqrt {a x+b x^3} \]

[In]

Int[Sqrt[a*x + b*x^3]/x,x]

[Out]

(2*Sqrt[a*x + b*x^3])/3 + (2*a^(3/4)*Sqrt[x]*(Sqrt[a] + Sqrt[b]*x)*Sqrt[(a + b*x^2)/(Sqrt[a] + Sqrt[b]*x)^2]*E
llipticF[2*ArcTan[(b^(1/4)*Sqrt[x])/a^(1/4)], 1/2])/(3*b^(1/4)*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 2046

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

Rubi steps \begin{align*} \text {integral}& = \frac {2}{3} \sqrt {a x+b x^3}+\frac {1}{3} (2 a) \int \frac {1}{\sqrt {a x+b x^3}} \, dx \\ & = \frac {2}{3} \sqrt {a x+b x^3}+\frac {\left (2 a \sqrt {x} \sqrt {a+b x^2}\right ) \int \frac {1}{\sqrt {x} \sqrt {a+b x^2}} \, dx}{3 \sqrt {a x+b x^3}} \\ & = \frac {2}{3} \sqrt {a x+b x^3}+\frac {\left (4 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 \sqrt {a x+b x^3}} \\ & = \frac {2}{3} \sqrt {a x+b x^3}+\frac {2 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 \sqrt [4]{b} \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 = 0.01 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.42 \[ \int \frac {\sqrt {a x+b x^3}}{x} \, dx=\frac {2 \sqrt {x \left (a+b x^2\right )} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{4},\frac {5}{4},-\frac {b x^2}{a}\right )}{\sqrt {1+\frac {b x^2}{a}}} \]

[In]

Integrate[Sqrt[a*x + b*x^3]/x,x]

[Out]

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

Maple [A] (verified)

Time = 2.18 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.10

method	result	size
default	\(\frac {2 \sqrt {b \,x^{3}+a x}}{3}+\frac {2 a \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{3 b \sqrt {b \,x^{3}+a x}}\)	\(124\)
elliptic	\(\frac {2 \sqrt {b \,x^{3}+a x}}{3}+\frac {2 a \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{3 b \sqrt {b \,x^{3}+a x}}\)	\(124\)
risch	\(\frac {2 x \left (b \,x^{2}+a \right )}{3 \sqrt {x \left (b \,x^{2}+a \right )}}+\frac {2 a \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{3 b \sqrt {b \,x^{3}+a x}}\)	\(132\)

[In]

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

[Out]

2/3*(b*x^3+a*x)^(1/2)+2/3*a*(-a*b)^(1/2)/b*((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 = 0.30 \[ \int \frac {\sqrt {a x+b x^3}}{x} \, dx=\frac {2 \, {\left (2 \, a \sqrt {b} {\rm weierstrassPInverse}\left (-\frac {4 \, a}{b}, 0, x\right ) + \sqrt {b x^{3} + a x} b\right )}}{3 \, b} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

Integral(sqrt(x*(a + b*x**2))/x, x)

Maxima [F]

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

[In]

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

[Out]

integrate(sqrt(b*x^3 + a*x)/x, x)

Giac [F]

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

[In]

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

[Out]

integrate(sqrt(b*x^3 + a*x)/x, x)

Mupad [F(-1)]

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

[In]

int((a*x + b*x^3)^(1/2)/x,x)

[Out]

int((a*x + b*x^3)^(1/2)/x, x)