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

Optimal. Leaf size=326 \[ \frac {7 b^{9/4} \sqrt [6]{x} \left (\sqrt {a} \sqrt [3]{x}+\sqrt {b}\right ) \sqrt {\frac {a x^{2/3}+b}{\left (\sqrt {a} \sqrt [3]{x}+\sqrt {b}\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt [6]{x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{5 a^{11/4} \sqrt {a x+b \sqrt [3]{x}}}-\frac {14 b^{9/4} \sqrt [6]{x} \left (\sqrt {a} \sqrt [3]{x}+\sqrt {b}\right ) \sqrt {\frac {a x^{2/3}+b}{\left (\sqrt {a} \sqrt [3]{x}+\sqrt {b}\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt [6]{x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{5 a^{11/4} \sqrt {a x+b \sqrt [3]{x}}}+\frac {14 b^2 \sqrt [3]{x} \left (a x^{2/3}+b\right )}{5 a^{5/2} \left (\sqrt {a} \sqrt [3]{x}+\sqrt {b}\right ) \sqrt {a x+b \sqrt [3]{x}}}-\frac {14 b \sqrt [3]{x} \sqrt {a x+b \sqrt [3]{x}}}{15 a^2}+\frac {2 x \sqrt {a x+b \sqrt [3]{x}}}{3 a} \]

[Out]

14/5*b^2*(b+a*x^(2/3))*x^(1/3)/a^(5/2)/(x^(1/3)*a^(1/2)+b^(1/2))/(b*x^(1/3)+a*x)^(1/2)-14/15*b*x^(1/3)*(b*x^(1
/3)+a*x)^(1/2)/a^2+2/3*x*(b*x^(1/3)+a*x)^(1/2)/a-14/5*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^(11/4)/(b*x^(1/3)+a*x)^(1/2)+7/5
*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)))*Ellip
ticF(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^(11/4)/(b*x^(1/3)+a*x)^(1/2)

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Rubi [A]  time = 0.35, antiderivative size = 326, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.412, Rules used = {2018, 2024, 2032, 329, 305, 220, 1196} \[ \frac {14 b^2 \sqrt [3]{x} \left (a x^{2/3}+b\right )}{5 a^{5/2} \left (\sqrt {a} \sqrt [3]{x}+\sqrt {b}\right ) \sqrt {a x+b \sqrt [3]{x}}}+\frac {7 b^{9/4} \sqrt [6]{x} \left (\sqrt {a} \sqrt [3]{x}+\sqrt {b}\right ) \sqrt {\frac {a x^{2/3}+b}{\left (\sqrt {a} \sqrt [3]{x}+\sqrt {b}\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt [6]{x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{5 a^{11/4} \sqrt {a x+b \sqrt [3]{x}}}-\frac {14 b^{9/4} \sqrt [6]{x} \left (\sqrt {a} \sqrt [3]{x}+\sqrt {b}\right ) \sqrt {\frac {a x^{2/3}+b}{\left (\sqrt {a} \sqrt [3]{x}+\sqrt {b}\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt [6]{x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{5 a^{11/4} \sqrt {a x+b \sqrt [3]{x}}}-\frac {14 b \sqrt [3]{x} \sqrt {a x+b \sqrt [3]{x}}}{15 a^2}+\frac {2 x \sqrt {a x+b \sqrt [3]{x}}}{3 a} \]

Antiderivative was successfully verified.

[In]

Int[x/Sqrt[b*x^(1/3) + a*x],x]

[Out]

(14*b^2*(b + a*x^(2/3))*x^(1/3))/(5*a^(5/2)*(Sqrt[b] + Sqrt[a]*x^(1/3))*Sqrt[b*x^(1/3) + a*x]) - (14*b*x^(1/3)
*Sqrt[b*x^(1/3) + a*x])/(15*a^2) + (2*x*Sqrt[b*x^(1/3) + a*x])/(3*a) - (14*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^(11/4)*Sqrt[b*x^(1/3) + a*x]) + (7*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^(11/4)*Sqrt[b*x^(1/3)
+ a*x])

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 2018

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 2024

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 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 \frac {x}{\sqrt {b \sqrt [3]{x}+a x}} \, dx &=3 \operatorname {Subst}\left (\int \frac {x^5}{\sqrt {b x+a x^3}} \, dx,x,\sqrt [3]{x}\right )\\ &=\frac {2 x \sqrt {b \sqrt [3]{x}+a x}}{3 a}-\frac {(7 b) \operatorname {Subst}\left (\int \frac {x^3}{\sqrt {b x+a x^3}} \, dx,x,\sqrt [3]{x}\right )}{3 a}\\ &=-\frac {14 b \sqrt [3]{x} \sqrt {b \sqrt [3]{x}+a x}}{15 a^2}+\frac {2 x \sqrt {b \sqrt [3]{x}+a x}}{3 a}+\frac {\left (7 b^2\right ) \operatorname {Subst}\left (\int \frac {x}{\sqrt {b x+a x^3}} \, dx,x,\sqrt [3]{x}\right )}{5 a^2}\\ &=-\frac {14 b \sqrt [3]{x} \sqrt {b \sqrt [3]{x}+a x}}{15 a^2}+\frac {2 x \sqrt {b \sqrt [3]{x}+a x}}{3 a}+\frac {\left (7 b^2 \sqrt {b+a x^{2/3}} \sqrt [6]{x}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {x}}{\sqrt {b+a x^2}} \, dx,x,\sqrt [3]{x}\right )}{5 a^2 \sqrt {b \sqrt [3]{x}+a x}}\\ &=-\frac {14 b \sqrt [3]{x} \sqrt {b \sqrt [3]{x}+a x}}{15 a^2}+\frac {2 x \sqrt {b \sqrt [3]{x}+a x}}{3 a}+\frac {\left (14 b^2 \sqrt {b+a x^{2/3}} \sqrt [6]{x}\right ) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {b+a x^4}} \, dx,x,\sqrt [6]{x}\right )}{5 a^2 \sqrt {b \sqrt [3]{x}+a x}}\\ &=-\frac {14 b \sqrt [3]{x} \sqrt {b \sqrt [3]{x}+a x}}{15 a^2}+\frac {2 x \sqrt {b \sqrt [3]{x}+a x}}{3 a}+\frac {\left (14 b^{5/2} \sqrt {b+a x^{2/3}} \sqrt [6]{x}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {b+a x^4}} \, dx,x,\sqrt [6]{x}\right )}{5 a^{5/2} \sqrt {b \sqrt [3]{x}+a x}}-\frac {\left (14 b^{5/2} \sqrt {b+a x^{2/3}} \sqrt [6]{x}\right ) \operatorname {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^{5/2} \sqrt {b \sqrt [3]{x}+a x}}\\ &=\frac {14 b^2 \left (b+a x^{2/3}\right ) \sqrt [3]{x}}{5 a^{5/2} \left (\sqrt {b}+\sqrt {a} \sqrt [3]{x}\right ) \sqrt {b \sqrt [3]{x}+a x}}-\frac {14 b \sqrt [3]{x} \sqrt {b \sqrt [3]{x}+a x}}{15 a^2}+\frac {2 x \sqrt {b \sqrt [3]{x}+a x}}{3 a}-\frac {14 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^{11/4} \sqrt {b \sqrt [3]{x}+a x}}+\frac {7 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^{11/4} \sqrt {b \sqrt [3]{x}+a x}}\\ \end {align*}

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Mathematica [C]  time = 0.09, size = 106, normalized size = 0.33 \[ \frac {2 \sqrt {a x+b \sqrt [3]{x}} \left (5 a^2 x^{5/3}+7 b^2 \sqrt [3]{x} \sqrt {\frac {a x^{2/3}}{b}+1} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-\frac {a x^{2/3}}{b}\right )-2 a b x-7 b^2 \sqrt [3]{x}\right )}{15 a^2 \left (a x^{2/3}+b\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[x/Sqrt[b*x^(1/3) + a*x],x]

[Out]

(2*Sqrt[b*x^(1/3) + a*x]*(-7*b^2*x^(1/3) - 2*a*b*x + 5*a^2*x^(5/3) + 7*b^2*Sqrt[1 + (a*x^(2/3))/b]*x^(1/3)*Hyp
ergeometric2F1[1/2, 3/4, 7/4, -((a*x^(2/3))/b)]))/(15*a^2*(b + a*x^(2/3)))

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fricas [F]  time = 7.89, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (a^{2} x^{2} - a b x^{\frac {4}{3}} + b^{2} x^{\frac {2}{3}}\right )} \sqrt {a x + b x^{\frac {1}{3}}}}{a^{3} x^{2} + b^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((a^2*x^2 - a*b*x^(4/3) + b^2*x^(2/3))*sqrt(a*x + b*x^(1/3))/(a^3*x^2 + b^3), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x}{\sqrt {a x + b x^{\frac {1}{3}}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x/sqrt(a*x + b*x^(1/3)), x)

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maple [A]  time = 0.06, size = 228, normalized size = 0.70 \[ -\frac {-10 a^{3} x^{2}+4 a^{2} b \,x^{\frac {4}{3}}-42 \sqrt {\frac {a \,x^{\frac {1}{3}}+\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (a \,x^{\frac {1}{3}}-\sqrt {-a b}\right )}{\sqrt {-a b}}}\, \sqrt {-\frac {a \,x^{\frac {1}{3}}}{\sqrt {-a b}}}\, b^{3} \EllipticE \left (\sqrt {\frac {a \,x^{\frac {1}{3}}+\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )+21 \sqrt {\frac {a \,x^{\frac {1}{3}}+\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (a \,x^{\frac {1}{3}}-\sqrt {-a b}\right )}{\sqrt {-a b}}}\, \sqrt {-\frac {a \,x^{\frac {1}{3}}}{\sqrt {-a b}}}\, b^{3} \EllipticF \left (\sqrt {\frac {a \,x^{\frac {1}{3}}+\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )+14 a \,b^{2} x^{\frac {2}{3}}}{15 \sqrt {\left (a \,x^{\frac {2}{3}}+b \right ) x^{\frac {1}{3}}}\, a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15/a^3*(-42*b^3*((a*x^(1/3)+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2)*(-2*(a*x^(1/3)-(-a*b)^(1/2))/(-a*b)^(1/2))^(1
/2)*(-1/(-a*b)^(1/2)*a*x^(1/3))^(1/2)*EllipticE(((a*x^(1/3)+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2),1/2*2^(1/2))+21*
b^3*((a*x^(1/3)+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2)*(-2*(a*x^(1/3)-(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2)*(-1/(-a*b)^
(1/2)*a*x^(1/3))^(1/2)*EllipticF(((a*x^(1/3)+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2),1/2*2^(1/2))+14*a*b^2*x^(2/3)+4
*a^2*b*x^(4/3)-10*a^3*x^2)/((a*x^(2/3)+b)*x^(1/3))^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x}{\sqrt {a x + b x^{\frac {1}{3}}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x/sqrt(a*x + b*x^(1/3)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x}{\sqrt {a\,x+b\,x^{1/3}}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(a*x + b*x^(1/3))^(1/2),x)

[Out]

int(x/(a*x + b*x^(1/3))^(1/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x/sqrt(a*x + b*x**(1/3)), x)

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