\(\int \sqrt {a x+b x^3} \, dx\) [48]

Optimal result

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

4/5*a*x*(b*x^2+a)/b^(1/2)/(a^(1/2)+b^(1/2)*x)/(b*x^3+a*x)^(1/2)+2/5*x*(b*x 
^3+a*x)^(1/2)-4/5*a^(5/4)*x^(1/2)*(a^(1/2)+b^(1/2)*x)*((b*x^2+a)/(a^(1/2)+ 
b^(1/2)*x)^2)^(1/2)*EllipticE(sin(2*arctan(b^(1/4)*x^(1/2)/a^(1/4))),1/2*2 
^(1/2))/b^(3/4)/(b*x^3+a*x)^(1/2)+2/5*a^(5/4)*x^(1/2)*(a^(1/2)+b^(1/2)*x)* 
((b*x^2+a)/(a^(1/2)+b^(1/2)*x)^2)^(1/2)*InverseJacobiAM(2*arctan(b^(1/4)*x 
^(1/2)/a^(1/4)),1/2*2^(1/2))/b^(3/4)/(b*x^3+a*x)^(1/2)
 

Mathematica [C] (verified)

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

Time = 8.64 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.20 \[ \int \sqrt {a x+b x^3} \, dx=\frac {2 x \sqrt {x \left (a+b x^2\right )} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {3}{4},\frac {7}{4},-\frac {b x^2}{a}\right )}{3 \sqrt {1+\frac {b x^2}{a}}} \] Input:

Integrate[Sqrt[a*x + b*x^3],x]
 

Output:

(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])
 

Rubi [A] (verified)

Time = 0.60 (sec) , antiderivative size = 268, normalized size of antiderivative = 1.05, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.538, Rules used = {1910, 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.

\(\displaystyle \int \sqrt {a x+b x^3} \, dx\)

\(\Big \downarrow \) 1910

\(\displaystyle \frac {2}{5} a \int \frac {x}{\sqrt {b x^3+a x}}dx+\frac {2}{5} x \sqrt {a x+b x^3}\)

\(\Big \downarrow \) 1938

\(\displaystyle \frac {2 a \sqrt {x} \sqrt {a+b x^2} \int \frac {\sqrt {x}}{\sqrt {b x^2+a}}dx}{5 \sqrt {a x+b x^3}}+\frac {2}{5} x \sqrt {a x+b x^3}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {4 a \sqrt {x} \sqrt {a+b x^2} \int \frac {x}{\sqrt {b x^2+a}}d\sqrt {x}}{5 \sqrt {a x+b x^3}}+\frac {2}{5} x \sqrt {a x+b x^3}\)

\(\Big \downarrow \) 834

\(\displaystyle \frac {4 a \sqrt {x} \sqrt {a+b x^2} \left (\frac {\sqrt {a} \int \frac {1}{\sqrt {b x^2+a}}d\sqrt {x}}{\sqrt {b}}-\frac {\sqrt {a} \int \frac {\sqrt {a}-\sqrt {b} x}{\sqrt {a} \sqrt {b x^2+a}}d\sqrt {x}}{\sqrt {b}}\right )}{5 \sqrt {a x+b x^3}}+\frac {2}{5} x \sqrt {a x+b x^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {4 a \sqrt {x} \sqrt {a+b x^2} \left (\frac {\sqrt {a} \int \frac {1}{\sqrt {b x^2+a}}d\sqrt {x}}{\sqrt {b}}-\frac {\int \frac {\sqrt {a}-\sqrt {b} x}{\sqrt {b x^2+a}}d\sqrt {x}}{\sqrt {b}}\right )}{5 \sqrt {a x+b x^3}}+\frac {2}{5} x \sqrt {a x+b x^3}\)

\(\Big \downarrow \) 761

\(\displaystyle \frac {4 a \sqrt {x} \sqrt {a+b x^2} \left (\frac {\sqrt [4]{a} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{2 b^{3/4} \sqrt {a+b x^2}}-\frac {\int \frac {\sqrt {a}-\sqrt {b} x}{\sqrt {b x^2+a}}d\sqrt {x}}{\sqrt {b}}\right )}{5 \sqrt {a x+b x^3}}+\frac {2}{5} x \sqrt {a x+b x^3}\)

\(\Big \downarrow \) 1510

\(\displaystyle \frac {4 a \sqrt {x} \sqrt {a+b x^2} \left (\frac {\sqrt [4]{a} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{2 b^{3/4} \sqrt {a+b x^2}}-\frac {\frac {\sqrt [4]{a} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{\sqrt [4]{b} \sqrt {a+b x^2}}-\frac {\sqrt {x} \sqrt {a+b x^2}}{\sqrt {a}+\sqrt {b} x}}{\sqrt {b}}\right )}{5 \sqrt {a x+b x^3}}+\frac {2}{5} x \sqrt {a x+b x^3}\)

Input:

Int[Sqrt[a*x + b*x^3],x]
 

Output:

(2*x*Sqrt[a*x + b*x^3])/5 + (4*a*Sqrt[x]*Sqrt[a + b*x^2]*(-((-((Sqrt[x]*Sq 
rt[a + b*x^2])/(Sqrt[a] + Sqrt[b]*x)) + (a^(1/4)*(Sqrt[a] + Sqrt[b]*x)*Sqr 
t[(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])))/(5*Sqrt[a* 
x + b*x^3])
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 266

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]
 

rule 761

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 834

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]
 

rule 1510

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]
 

rule 1910

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[x*((a*x^j 
 + b*x^n)^p/(n*p + 1)), x] + Simp[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 1938

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]
 

Maple [A] (verified)

Time = 0.50 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.69

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

Input:

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

Output:

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)*(-1/(-a*b)^(1/2) 
*b*x)^(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 [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.17 \[ \int \sqrt {a x+b x^3} \, dx=\frac {2 \, {\left (\sqrt {b x^{3} + a x} b x - 2 \, a \sqrt {b} {\rm weierstrassZeta}\left (-\frac {4 \, a}{b}, 0, {\rm weierstrassPInverse}\left (-\frac {4 \, a}{b}, 0, x\right )\right )\right )}}{5 \, b} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

Integral(sqrt(a*x + b*x**3), x)
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [B] (verification not implemented)

Time = 8.96 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.16 \[ \int \sqrt {a x+b x^3} \, dx=\frac {2\,x\,\sqrt {b\,x^3+a\,x}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{2},\frac {3}{4};\ \frac {7}{4};\ -\frac {b\,x^2}{a}\right )}{3\,\sqrt {\frac {b\,x^2}{a}+1}} \] Input:

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

Output:

(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))
 

Reduce [F]

\[ \int \sqrt {a x+b x^3} \, dx=\frac {2 \sqrt {x}\, \sqrt {b \,x^{2}+a}\, x}{5}+\frac {2 \left (\int \frac {\sqrt {x}\, \sqrt {b \,x^{2}+a}}{b \,x^{2}+a}d x \right ) a}{5} \] Input:

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

Output:

(2*(sqrt(x)*sqrt(a + b*x**2)*x + int((sqrt(x)*sqrt(a + b*x**2))/(a + b*x** 
2),x)*a))/5