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

Optimal. Leaf size=255 \[ \frac {2 a^{5/4} \sqrt {x} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 b^{3/4} \sqrt {a x+b x^3}}-\frac {4 a^{5/4} \sqrt {x} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 b^{3/4} \sqrt {a x+b x^3}}+\frac {2}{5} x \sqrt {a x+b x^3}+\frac {4 a x \left (a+b x^2\right )}{5 \sqrt {b} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {a x+b x^3}} \]

[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]  time = 0.17, antiderivative size = 255, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {2004, 2032, 329, 305, 220, 1196} \[ \frac {2 a^{5/4} \sqrt {x} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 b^{3/4} \sqrt {a x+b x^3}}-\frac {4 a^{5/4} \sqrt {x} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 b^{3/4} \sqrt {a x+b x^3}}+\frac {4 a x \left (a+b x^2\right )}{5 \sqrt {b} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {a x+b x^3}}+\frac {2}{5} x \sqrt {a x+b x^3} \]

Antiderivative was successfully verified.

[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 220

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)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 305

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 329

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 1196

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)]*EllipticE[2*ArcTan[
q*x], 1/2])/(q*Sqrt[a + c*x^4]), x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rule 2004

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 2032

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[(c^IntPart[m]*(c*x)^FracP
art[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*} \int \sqrt {a x+b x^3} \, dx &=\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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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]  time = 0.01, size = 51, normalized size = 0.20 \[ \frac {2 x \sqrt {x \left (a+b x^2\right )} \, _2F_1\left (-\frac {1}{2},\frac {3}{4};\frac {7}{4};-\frac {b x^2}{a}\right )}{3 \sqrt {\frac {b x^2}{a}+1}} \]

Antiderivative was successfully verified.

[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])

________________________________________________________________________________________

fricas [F]  time = 0.61, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {b x^{3} + a x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(b*x^3 + a*x), x)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {b x^{3} + a x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(b*x^3 + a*x), x)

________________________________________________________________________________________

maple [A]  time = 0.07, size = 175, normalized size = 0.69 \[ \frac {2 \sqrt {b \,x^{3}+a x}\, x}{5}+\frac {2 \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \left (-\frac {2 \sqrt {-a b}\, \EllipticE \left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}+\frac {\sqrt {-a b}\, \EllipticF \left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}\right ) a}{5 \sqrt {b \,x^{3}+a x}\, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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)*(-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)))

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {b x^{3} + a x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(b*x^3 + a*x), x)

________________________________________________________________________________________

mupad [B]  time = 5.05, size = 40, normalized size = 0.16 \[ \frac {2\,x\,\sqrt {b\,x^3+a\,x}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{2},\frac {3}{4};\ \frac {7}{4};\ -\frac {b\,x^2}{a}\right )}{3\,\sqrt {\frac {b\,x^2}{a}+1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {a x + b x^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(a*x + b*x**3), x)

________________________________________________________________________________________