\(\int \frac {1}{\sqrt {b \sqrt [3]{x}+a x}} \, dx\) [153]

Optimal result

Integrand size = 15, antiderivative size = 126 \[ \int \frac {1}{\sqrt {b \sqrt [3]{x}+a x}} \, dx=\frac {2 \sqrt {b \sqrt [3]{x}+a x}}{a}-\frac {b^{3/4} \left (\sqrt {b}+\sqrt {a} \sqrt [3]{x}\right ) \sqrt {\frac {b+a x^{2/3}}{\left (\sqrt {b}+\sqrt {a} \sqrt [3]{x}\right )^2}} \sqrt [6]{x} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{a} \sqrt [6]{x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{a^{5/4} \sqrt {b \sqrt [3]{x}+a x}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2035, 2038, 2036, 335, 226} \[ \int \frac {1}{\sqrt {b \sqrt [3]{x}+a x}} \, dx=\frac {2 \sqrt {a x+b \sqrt [3]{x}}}{a}-\frac {b^{3/4} \sqrt [6]{x} \left (\sqrt {a} \sqrt [3]{x}+\sqrt {b}\right ) \sqrt {\frac {a x^{2/3}+b}{\left (\sqrt {a} \sqrt [3]{x}+\sqrt {b}\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{a} \sqrt [6]{x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{a^{5/4} \sqrt {a x+b \sqrt [3]{x}}} \]

[In]

Int[1/Sqrt[b*x^(1/3) + a*x],x]

[Out]

(2*Sqrt[b*x^(1/3) + a*x])/a - (b^(3/4)*(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])/(a^(5/4)*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 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 2035

Int[1/Sqrt[(a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.)], x_Symbol] :> Simp[-2*(Sqrt[a*x^j + b*x^n]/(b*(n - 2)*x^(n - 1
))), x] - Dist[a*((2*n - j - 2)/(b*(n - 2))), Int[1/(x^(n - j)*Sqrt[a*x^j + b*x^n]), x], x] /; FreeQ[{a, b}, x
] && LtQ[2*(n - 1), j, n]

Rule 2036

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[(a*x^j + b*x^n)^FracPart[p]/(x^(j*FracPart[p
])*(a + b*x^(n - j))^FracPart[p]), Int[x^(j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, j, n, p}, x] &&  !I
ntegerQ[p] && NeQ[n, j] && PosQ[n - j]

Rule 2038

Int[(x_)^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[(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]] && EqQ[Simplify[m - n + 1], 0]

Rubi steps \begin{align*} \text {integral}& = \frac {2 \sqrt {b \sqrt [3]{x}+a x}}{a}-\frac {b \int \frac {1}{x^{2/3} \sqrt {b \sqrt [3]{x}+a x}} \, dx}{3 a} \\ & = \frac {2 \sqrt {b \sqrt [3]{x}+a x}}{a}-\frac {b \text {Subst}\left (\int \frac {1}{\sqrt {b x+a x^3}} \, dx,x,\sqrt [3]{x}\right )}{a} \\ & = \frac {2 \sqrt {b \sqrt [3]{x}+a x}}{a}-\frac {\left (b \sqrt {b+a x^{2/3}} \sqrt [6]{x}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {x} \sqrt {b+a x^2}} \, dx,x,\sqrt [3]{x}\right )}{a \sqrt {b \sqrt [3]{x}+a x}} \\ & = \frac {2 \sqrt {b \sqrt [3]{x}+a x}}{a}-\frac {\left (2 b \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 )}{a \sqrt {b \sqrt [3]{x}+a x}} \\ & = \frac {2 \sqrt {b \sqrt [3]{x}+a x}}{a}-\frac {b^{3/4} \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 )}{a^{5/4} \sqrt {b \sqrt [3]{x}+a x}} \\ \end{align*}

Mathematica [C] (verified)

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

Time = 10.04 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.63 \[ \int \frac {1}{\sqrt {b \sqrt [3]{x}+a x}} \, dx=\frac {2 \sqrt {b \sqrt [3]{x}+a x} \left (b+a x^{2/3}-b \sqrt {1+\frac {a x^{2/3}}{b}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\frac {a x^{2/3}}{b}\right )\right )}{a \left (b+a x^{2/3}\right )} \]

[In]

Integrate[1/Sqrt[b*x^(1/3) + a*x],x]

[Out]

(2*Sqrt[b*x^(1/3) + a*x]*(b + a*x^(2/3) - b*Sqrt[1 + (a*x^(2/3))/b]*Hypergeometric2F1[1/4, 1/2, 5/4, -((a*x^(2
/3))/b)]))/(a*(b + a*x^(2/3)))

Maple [A] (verified)

Time = 2.01 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.01

method	result	size
default	\(\frac {-b \sqrt {-a b}\, \sqrt {\frac {a \,x^{\frac {1}{3}}+\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (a \,x^{\frac {1}{3}}-\sqrt {-a b}\right )}{\sqrt {-a b}}}\, \sqrt {-\frac {x^{\frac {1}{3}} a}{\sqrt {-a b}}}\, F\left (\sqrt {\frac {a \,x^{\frac {1}{3}}+\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )+2 a b \,x^{\frac {1}{3}}+2 a^{2} x}{\sqrt {x^{\frac {1}{3}} \left (b +a \,x^{\frac {2}{3}}\right )}\, a^{2}}\)	\(127\)
derivativedivides	\(\frac {2 \sqrt {b \,x^{\frac {1}{3}}+a x}}{a}-\frac {b \sqrt {-a b}\, \sqrt {\frac {\left (x^{\frac {1}{3}}+\frac {\sqrt {-a b}}{a}\right ) a}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x^{\frac {1}{3}}-\frac {\sqrt {-a b}}{a}\right ) a}{\sqrt {-a b}}}\, \sqrt {-\frac {x^{\frac {1}{3}} a}{\sqrt {-a b}}}\, F\left (\sqrt {\frac {\left (x^{\frac {1}{3}}+\frac {\sqrt {-a b}}{a}\right ) a}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{a^{2} \sqrt {b \,x^{\frac {1}{3}}+a x}}\)	\(135\)

[In]

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

[Out]

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

Fricas [F]

\[ \int \frac {1}{\sqrt {b \sqrt [3]{x}+a x}} \, dx=\int { \frac {1}{\sqrt {a x + b x^{\frac {1}{3}}}} \,d x } \]

[In]

integrate(1/(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^3 + b^3*x), x)

Sympy [F]

\[ \int \frac {1}{\sqrt {b \sqrt [3]{x}+a x}} \, dx=\int \frac {1}{\sqrt {a x + b \sqrt [3]{x}}}\, dx \]

[In]

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

[Out]

Integral(1/sqrt(a*x + b*x**(1/3)), x)

Maxima [F]

\[ \int \frac {1}{\sqrt {b \sqrt [3]{x}+a x}} \, dx=\int { \frac {1}{\sqrt {a x + b x^{\frac {1}{3}}}} \,d x } \]

[In]

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

[Out]

integrate(1/sqrt(a*x + b*x^(1/3)), x)

Giac [F]

\[ \int \frac {1}{\sqrt {b \sqrt [3]{x}+a x}} \, dx=\int { \frac {1}{\sqrt {a x + b x^{\frac {1}{3}}}} \,d x } \]

[In]

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

[Out]

integrate(1/sqrt(a*x + b*x^(1/3)), x)

Mupad [B] (verification not implemented)

Time = 9.26 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.32 \[ \int \frac {1}{\sqrt {b \sqrt [3]{x}+a x}} \, dx=\frac {2\,x\,\sqrt {\frac {b}{a\,x^{2/3}}+1}\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ -\frac {b}{a\,x^{2/3}}\right )}{\sqrt {a\,x+b\,x^{1/3}}} \]

[In]

int(1/(a*x + b*x^(1/3))^(1/2),x)

[Out]

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