3.2.34 \(\int \sqrt {b \sqrt [3]{x}+a x} \, dx\) [134]

Optimal. Leaf size=323 \[ -\frac {4 b^2 \left (b+a x^{2/3}\right ) \sqrt [3]{x}}{5 a^{3/2} \left (\sqrt {b}+\sqrt {a} \sqrt [3]{x}\right ) \sqrt {b \sqrt [3]{x}+a x}}+\frac {4 b \sqrt [3]{x} \sqrt {b \sqrt [3]{x}+a x}}{15 a}+\frac {2}{3} x \sqrt {b \sqrt [3]{x}+a x}+\frac {4 b^{9/4} \left (\sqrt {b}+\sqrt {a} \sqrt [3]{x}\right ) \sqrt {\frac {b+a x^{2/3}}{\left (\sqrt {b}+\sqrt {a} \sqrt [3]{x}\right )^2}} \sqrt [6]{x} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt [6]{x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{5 a^{7/4} \sqrt {b \sqrt [3]{x}+a x}}-\frac {2 b^{9/4} \left (\sqrt {b}+\sqrt {a} \sqrt [3]{x}\right ) \sqrt {\frac {b+a x^{2/3}}{\left (\sqrt {b}+\sqrt {a} \sqrt [3]{x}\right )^2}} \sqrt [6]{x} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt [6]{x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{5 a^{7/4} \sqrt {b \sqrt [3]{x}+a x}} \]

[Out]

-4/5*b^2*(b+a*x^(2/3))*x^(1/3)/a^(3/2)/(x^(1/3)*a^(1/2)+b^(1/2))/(b*x^(1/3)+a*x)^(1/2)+4/15*b*x^(1/3)*(b*x^(1/
3)+a*x)^(1/2)/a+2/3*x*(b*x^(1/3)+a*x)^(1/2)+4/5*b^(9/4)*x^(1/6)*(cos(2*arctan(a^(1/4)*x^(1/6)/b^(1/4)))^2)^(1/
2)/cos(2*arctan(a^(1/4)*x^(1/6)/b^(1/4)))*EllipticE(sin(2*arctan(a^(1/4)*x^(1/6)/b^(1/4))),1/2*2^(1/2))*(x^(1/
3)*a^(1/2)+b^(1/2))*((b+a*x^(2/3))/(x^(1/3)*a^(1/2)+b^(1/2))^2)^(1/2)/a^(7/4)/(b*x^(1/3)+a*x)^(1/2)-2/5*b^(9/4
)*x^(1/6)*(cos(2*arctan(a^(1/4)*x^(1/6)/b^(1/4)))^2)^(1/2)/cos(2*arctan(a^(1/4)*x^(1/6)/b^(1/4)))*EllipticF(si
n(2*arctan(a^(1/4)*x^(1/6)/b^(1/4))),1/2*2^(1/2))*(x^(1/3)*a^(1/2)+b^(1/2))*((b+a*x^(2/3))/(x^(1/3)*a^(1/2)+b^
(1/2))^2)^(1/2)/a^(7/4)/(b*x^(1/3)+a*x)^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 0.23, antiderivative size = 323, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.533, Rules used = {2029, 2043, 2049, 2057, 335, 311, 226, 1210} \begin {gather*} -\frac {2 b^{9/4} \sqrt [6]{x} \left (\sqrt {a} \sqrt [3]{x}+\sqrt {b}\right ) \sqrt {\frac {a x^{2/3}+b}{\left (\sqrt {a} \sqrt [3]{x}+\sqrt {b}\right )^2}} F\left (2 \text {ArcTan}\left (\frac {\sqrt [4]{a} \sqrt [6]{x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{5 a^{7/4} \sqrt {a x+b \sqrt [3]{x}}}+\frac {4 b^{9/4} \sqrt [6]{x} \left (\sqrt {a} \sqrt [3]{x}+\sqrt {b}\right ) \sqrt {\frac {a x^{2/3}+b}{\left (\sqrt {a} \sqrt [3]{x}+\sqrt {b}\right )^2}} E\left (2 \text {ArcTan}\left (\frac {\sqrt [4]{a} \sqrt [6]{x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{5 a^{7/4} \sqrt {a x+b \sqrt [3]{x}}}-\frac {4 b^2 \sqrt [3]{x} \left (a x^{2/3}+b\right )}{5 a^{3/2} \left (\sqrt {a} \sqrt [3]{x}+\sqrt {b}\right ) \sqrt {a x+b \sqrt [3]{x}}}+\frac {4 b \sqrt [3]{x} \sqrt {a x+b \sqrt [3]{x}}}{15 a}+\frac {2}{3} x \sqrt {a x+b \sqrt [3]{x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[b*x^(1/3) + a*x],x]

[Out]

(-4*b^2*(b + a*x^(2/3))*x^(1/3))/(5*a^(3/2)*(Sqrt[b] + Sqrt[a]*x^(1/3))*Sqrt[b*x^(1/3) + a*x]) + (4*b*x^(1/3)*
Sqrt[b*x^(1/3) + a*x])/(15*a) + (2*x*Sqrt[b*x^(1/3) + a*x])/3 + (4*b^(9/4)*(Sqrt[b] + Sqrt[a]*x^(1/3))*Sqrt[(b
+ a*x^(2/3))/(Sqrt[b] + Sqrt[a]*x^(1/3))^2]*x^(1/6)*EllipticE[2*ArcTan[(a^(1/4)*x^(1/6))/b^(1/4)], 1/2])/(5*a
^(7/4)*Sqrt[b*x^(1/3) + a*x]) - (2*b^(9/4)*(Sqrt[b] + Sqrt[a]*x^(1/3))*Sqrt[(b + a*x^(2/3))/(Sqrt[b] + Sqrt[a]
*x^(1/3))^2]*x^(1/6)*EllipticF[2*ArcTan[(a^(1/4)*x^(1/6))/b^(1/4)], 1/2])/(5*a^(7/4)*Sqrt[b*x^(1/3) + a*x])

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 2043

Int[(x_)^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)
/n] - 1)*(a*x^Simplify[j/n] + b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[
n, j] && IntegerQ[Simplify[j/n]] && IntegerQ[Simplify[(m + 1)/n]] && NeQ[n^2, 1]

Rule 2049

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n +
1)*((a*x^j + b*x^n)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^(n - j)*((m + j*p - n + j + 1)/(b*(m + n*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]) && GtQ[m + j*p + 1 - n + j, 0] && NeQ[m + 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*} \int \sqrt {b \sqrt [3]{x}+a x} \, dx &=\frac {2}{3} x \sqrt {b \sqrt [3]{x}+a x}+\frac {1}{9} (2 b) \int \frac {\sqrt [3]{x}}{\sqrt {b \sqrt [3]{x}+a x}} \, dx\\ &=\frac {2}{3} x \sqrt {b \sqrt [3]{x}+a x}+\frac {1}{3} (2 b) \text {Subst}\left (\int \frac {x^3}{\sqrt {b x+a x^3}} \, dx,x,\sqrt [3]{x}\right )\\ &=\frac {4 b \sqrt [3]{x} \sqrt {b \sqrt [3]{x}+a x}}{15 a}+\frac {2}{3} x \sqrt {b \sqrt [3]{x}+a x}-\frac {\left (2 b^2\right ) \text {Subst}\left (\int \frac {x}{\sqrt {b x+a x^3}} \, dx,x,\sqrt [3]{x}\right )}{5 a}\\ &=\frac {4 b \sqrt [3]{x} \sqrt {b \sqrt [3]{x}+a x}}{15 a}+\frac {2}{3} x \sqrt {b \sqrt [3]{x}+a x}-\frac {\left (2 b^2 \sqrt {b+a x^{2/3}} \sqrt [6]{x}\right ) \text {Subst}\left (\int \frac {\sqrt {x}}{\sqrt {b+a x^2}} \, dx,x,\sqrt [3]{x}\right )}{5 a \sqrt {b \sqrt [3]{x}+a x}}\\ &=\frac {4 b \sqrt [3]{x} \sqrt {b \sqrt [3]{x}+a x}}{15 a}+\frac {2}{3} x \sqrt {b \sqrt [3]{x}+a x}-\frac {\left (4 b^2 \sqrt {b+a x^{2/3}} \sqrt [6]{x}\right ) \text {Subst}\left (\int \frac {x^2}{\sqrt {b+a x^4}} \, dx,x,\sqrt [6]{x}\right )}{5 a \sqrt {b \sqrt [3]{x}+a x}}\\ &=\frac {4 b \sqrt [3]{x} \sqrt {b \sqrt [3]{x}+a x}}{15 a}+\frac {2}{3} x \sqrt {b \sqrt [3]{x}+a x}-\frac {\left (4 b^{5/2} \sqrt {b+a x^{2/3}} \sqrt [6]{x}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {b+a x^4}} \, dx,x,\sqrt [6]{x}\right )}{5 a^{3/2} \sqrt {b \sqrt [3]{x}+a x}}+\frac {\left (4 b^{5/2} \sqrt {b+a x^{2/3}} \sqrt [6]{x}\right ) \text {Subst}\left (\int \frac {1-\frac {\sqrt {a} x^2}{\sqrt {b}}}{\sqrt {b+a x^4}} \, dx,x,\sqrt [6]{x}\right )}{5 a^{3/2} \sqrt {b \sqrt [3]{x}+a x}}\\ &=-\frac {4 b^2 \left (b+a x^{2/3}\right ) \sqrt [3]{x}}{5 a^{3/2} \left (\sqrt {b}+\sqrt {a} \sqrt [3]{x}\right ) \sqrt {b \sqrt [3]{x}+a x}}+\frac {4 b \sqrt [3]{x} \sqrt {b \sqrt [3]{x}+a x}}{15 a}+\frac {2}{3} x \sqrt {b \sqrt [3]{x}+a x}+\frac {4 b^{9/4} \left (\sqrt {b}+\sqrt {a} \sqrt [3]{x}\right ) \sqrt {\frac {b+a x^{2/3}}{\left (\sqrt {b}+\sqrt {a} \sqrt [3]{x}\right )^2}} \sqrt [6]{x} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt [6]{x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{5 a^{7/4} \sqrt {b \sqrt [3]{x}+a x}}-\frac {2 b^{9/4} \left (\sqrt {b}+\sqrt {a} \sqrt [3]{x}\right ) \sqrt {\frac {b+a x^{2/3}}{\left (\sqrt {b}+\sqrt {a} \sqrt [3]{x}\right )^2}} \sqrt [6]{x} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt [6]{x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{5 a^{7/4} \sqrt {b \sqrt [3]{x}+a x}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 10.04, size = 94, normalized size = 0.29 \begin {gather*} \frac {2 \sqrt [3]{x} \sqrt {b \sqrt [3]{x}+a x} \left (\left (b+a x^{2/3}\right ) \sqrt {1+\frac {a x^{2/3}}{b}}-b \, _2F_1\left (-\frac {1}{2},\frac {3}{4};\frac {7}{4};-\frac {a x^{2/3}}{b}\right )\right )}{3 a \sqrt {1+\frac {a x^{2/3}}{b}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[b*x^(1/3) + a*x],x]

[Out]

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

________________________________________________________________________________________

Maple [A]
time = 0.35, size = 207, normalized size = 0.64

method	result	size
derivativedivides	\(\frac {2 x \sqrt {b \,x^{\frac {1}{3}}+a x}}{3}+\frac {4 b \,x^{\frac {1}{3}} \sqrt {b \,x^{\frac {1}{3}}+a x}}{15 a}-\frac {2 b^{2} \sqrt {-a b}\, \sqrt {\frac {\left (x^{\frac {1}{3}}+\frac {\sqrt {-a b}}{a}\right ) a}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x^{\frac {1}{3}}-\frac {\sqrt {-a b}}{a}\right ) a}{\sqrt {-a b}}}\, \sqrt {-\frac {x^{\frac {1}{3}} a}{\sqrt {-a b}}}\, \left (-\frac {2 \sqrt {-a b}\, \EllipticE \left (\sqrt {\frac {\left (x^{\frac {1}{3}}+\frac {\sqrt {-a b}}{a}\right ) a}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{a}+\frac {\sqrt {-a b}\, \EllipticF \left (\sqrt {\frac {\left (x^{\frac {1}{3}}+\frac {\sqrt {-a b}}{a}\right ) a}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{a}\right )}{5 a^{2} \sqrt {b \,x^{\frac {1}{3}}+a x}}\)	\(207\)
default	\(\frac {2 x \sqrt {b \,x^{\frac {1}{3}}+a x}}{3}+\frac {4 b \,x^{\frac {1}{3}} \sqrt {b \,x^{\frac {1}{3}}+a x}}{15 a}-\frac {2 b^{2} \sqrt {-a b}\, \sqrt {\frac {\left (x^{\frac {1}{3}}+\frac {\sqrt {-a b}}{a}\right ) a}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x^{\frac {1}{3}}-\frac {\sqrt {-a b}}{a}\right ) a}{\sqrt {-a b}}}\, \sqrt {-\frac {x^{\frac {1}{3}} a}{\sqrt {-a b}}}\, \left (-\frac {2 \sqrt {-a b}\, \EllipticE \left (\sqrt {\frac {\left (x^{\frac {1}{3}}+\frac {\sqrt {-a b}}{a}\right ) a}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{a}+\frac {\sqrt {-a b}\, \EllipticF \left (\sqrt {\frac {\left (x^{\frac {1}{3}}+\frac {\sqrt {-a b}}{a}\right ) a}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{a}\right )}{5 a^{2} \sqrt {b \,x^{\frac {1}{3}}+a x}}\)	\(207\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*x*(b*x^(1/3)+a*x)^(1/2)+4/15*b*x^(1/3)*(b*x^(1/3)+a*x)^(1/2)/a-2/5*b^2/a^2*(-a*b)^(1/2)*((x^(1/3)+1/a*(-a*
b)^(1/2))*a/(-a*b)^(1/2))^(1/2)*(-2*(x^(1/3)-1/a*(-a*b)^(1/2))*a/(-a*b)^(1/2))^(1/2)*(-x^(1/3)*a/(-a*b)^(1/2))
^(1/2)/(b*x^(1/3)+a*x)^(1/2)*(-2/a*(-a*b)^(1/2)*EllipticE(((x^(1/3)+1/a*(-a*b)^(1/2))*a/(-a*b)^(1/2))^(1/2),1/
2*2^(1/2))+1/a*(-a*b)^(1/2)*EllipticF(((x^(1/3)+1/a*(-a*b)^(1/2))*a/(-a*b)^(1/2))^(1/2),1/2*2^(1/2)))

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(a*x + b*x^(1/3)), x)

________________________________________________________________________________________

Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(a*x + b*x^(1/3)), x)

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt {a x + b \sqrt [3]{x}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(a*x + b*x**(1/3)), x)

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(a*x + b*x^(1/3)), x)

________________________________________________________________________________________

Mupad [B]
time = 5.18, size = 40, normalized size = 0.12 \begin {gather*} \frac {6\,x\,\sqrt {a\,x+b\,x^{1/3}}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{2},\frac {7}{4};\ \frac {11}{4};\ -\frac {a\,x^{2/3}}{b}\right )}{7\,\sqrt {\frac {a\,x^{2/3}}{b}+1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x + b*x^(1/3))^(1/2),x)

[Out]

(6*x*(a*x + b*x^(1/3))^(1/2)*hypergeom([-1/2, 7/4], 11/4, -(a*x^(2/3))/b))/(7*((a*x^(2/3))/b + 1)^(1/2))

________________________________________________________________________________________