\(\int \sqrt {a x+b x^3} \, dx\) [41]

Optimal result

Integrand size = 13, antiderivative size = 255 \[ \int \sqrt {a x+b x^3} \, dx=\frac {4 a x \left (a+b x^2\right )}{5 \sqrt {b} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {a x+b x^3}}+\frac {2}{5} x \sqrt {a x+b x^3}-\frac {4 a^{5/4} \sqrt {x} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 b^{3/4} \sqrt {a x+b x^3}}+\frac {2 a^{5/4} \sqrt {x} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{5 b^{3/4} \sqrt {a x+b x^3}} \]

[Out]

4/5*a*x*(b*x^2+a)/b^(1/2)/(a^(1/2)+x*b^(1/2))/(b*x^3+a*x)^(1/2)+2/5*x*(b*x^3+a*x)^(1/2)-4/5*a^(5/4)*(cos(2*arc
tan(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^(3/4)/(b*x
^3+a*x)^(1/2)+2/5*a^(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)/b^(3/4)/(b*x^3+a*x)^(1/2)

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 255, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {2029, 2057, 335, 311, 226, 1210} \[ \int \sqrt {a x+b x^3} \, dx=\frac {2 a^{5/4} \sqrt {x} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{5 b^{3/4} \sqrt {a x+b x^3}}-\frac {4 a^{5/4} \sqrt {x} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 b^{3/4} \sqrt {a x+b x^3}}+\frac {2}{5} x \sqrt {a x+b x^3}+\frac {4 a x \left (a+b x^2\right )}{5 \sqrt {b} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {a x+b x^3}} \]

[In]

Int[Sqrt[a*x + b*x^3],x]

[Out]

(4*a*x*(a + b*x^2))/(5*Sqrt[b]*(Sqrt[a] + Sqrt[b]*x)*Sqrt[a*x + b*x^3]) + (2*x*Sqrt[a*x + b*x^3])/5 - (4*a^(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*b^(3/4)*Sqrt[a*x + b*x^3]) + (2*a^(5/4)*Sqrt[x]*(Sqrt[a] + Sqrt[b]*x)*Sqrt[(a + b*x^2)/(S
qrt[a] + Sqrt[b]*x)^2]*EllipticF[2*ArcTan[(b^(1/4)*Sqrt[x])/a^(1/4)], 1/2])/(5*b^(3/4)*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 2029

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[x*((a*x^j + b*x^n)^p/(n*p + 1)), x] + Dist[a
*(n - j)*(p/(n*p + 1)), Int[x^j*(a*x^j + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] &&  !IntegerQ[p] && LtQ[0,
j, n] && GtQ[p, 0] && NeQ[n*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}{5} x \sqrt {a x+b x^3}+\frac {1}{5} (2 a) \int \frac {x}{\sqrt {a x+b x^3}} \, dx \\ & = \frac {2}{5} x \sqrt {a x+b x^3}+\frac {\left (2 a \sqrt {x} \sqrt {a+b x^2}\right ) \int \frac {\sqrt {x}}{\sqrt {a+b x^2}} \, dx}{5 \sqrt {a x+b x^3}} \\ & = \frac {2}{5} x \sqrt {a x+b x^3}+\frac {\left (4 a \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 \sqrt {a x+b x^3}} \\ & = \frac {2}{5} x \sqrt {a x+b x^3}+\frac {\left (4 a^{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 \sqrt {b} \sqrt {a x+b x^3}}-\frac {\left (4 a^{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 \sqrt {b} \sqrt {a x+b x^3}} \\ & = \frac {4 a x \left (a+b x^2\right )}{5 \sqrt {b} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {a x+b x^3}}+\frac {2}{5} x \sqrt {a x+b x^3}-\frac {4 a^{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 b^{3/4} \sqrt {a x+b x^3}}+\frac {2 a^{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 b^{3/4} \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 = 0.02 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.20 \[ \int \sqrt {a x+b x^3} \, dx=\frac {2 x \sqrt {x \left (a+b x^2\right )} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {3}{4},\frac {7}{4},-\frac {b x^2}{a}\right )}{3 \sqrt {1+\frac {b x^2}{a}}} \]

[In]

Integrate[Sqrt[a*x + b*x^3],x]

[Out]

(2*x*Sqrt[x*(a + b*x^2)]*Hypergeometric2F1[-1/2, 3/4, 7/4, -((b*x^2)/a)])/(3*Sqrt[1 + (b*x^2)/a])

Maple [A] (verified)

Time = 2.22 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.69

method	result	size
default	\(\frac {2 x \sqrt {b \,x^{3}+a x}}{5}+\frac {2 a \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, \left (-\frac {2 \sqrt {-a b}\, E\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}+\frac {\sqrt {-a b}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}\right )}{5 b \sqrt {b \,x^{3}+a x}}\)	\(175\)
elliptic	\(\frac {2 x \sqrt {b \,x^{3}+a x}}{5}+\frac {2 a \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, \left (-\frac {2 \sqrt {-a b}\, E\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}+\frac {\sqrt {-a b}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}\right )}{5 b \sqrt {b \,x^{3}+a x}}\)	\(175\)
risch	\(\frac {2 x^{2} \left (b \,x^{2}+a \right )}{5 \sqrt {x \left (b \,x^{2}+a \right )}}+\frac {2 a \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, \left (-\frac {2 \sqrt {-a b}\, E\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}+\frac {\sqrt {-a b}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}\right )}{5 b \sqrt {b \,x^{3}+a x}}\)	\(184\)

[In]

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

[Out]

2/5*x*(b*x^3+a*x)^(1/2)+2/5*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)*(-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 = 0.14 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.17 \[ \int \sqrt {a x+b x^3} \, dx=\frac {2 \, {\left (\sqrt {b x^{3} + a x} b x - 2 \, a \sqrt {b} {\rm weierstrassZeta}\left (-\frac {4 \, a}{b}, 0, {\rm weierstrassPInverse}\left (-\frac {4 \, a}{b}, 0, x\right )\right )\right )}}{5 \, b} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

Integral(sqrt(a*x + b*x**3), x)

Maxima [F]

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

[In]

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

[Out]

integrate(sqrt(b*x^3 + a*x), x)

Giac [F]

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

[In]

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

[Out]

integrate(sqrt(b*x^3 + a*x), x)

Mupad [B] (verification not implemented)

Time = 11.14 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.16 \[ \int \sqrt {a x+b x^3} \, dx=\frac {2\,x\,\sqrt {b\,x^3+a\,x}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{2},\frac {3}{4};\ \frac {7}{4};\ -\frac {b\,x^2}{a}\right )}{3\,\sqrt {\frac {b\,x^2}{a}+1}} \]

[In]

int((a*x + b*x^3)^(1/2),x)

[Out]

(2*x*(a*x + b*x^3)^(1/2)*hypergeom([-1/2, 3/4], 7/4, -(b*x^2)/a))/(3*((b*x^2)/a + 1)^(1/2))