3.2.61 \(\int \frac {x}{(b \sqrt [3]{x}+a x)^{3/2}} \, dx\) [161]

Optimal. Leaf size=149 \[ -\frac {3 x}{a \sqrt {b \sqrt [3]{x}+a x}}+\frac {5 \sqrt {b \sqrt [3]{x}+a x}}{a^2}-\frac {5 b^{3/4} \left (\sqrt {b}+\sqrt {a} \sqrt [3]{x}\right ) \sqrt {\frac {b+a x^{2/3}}{\left (\sqrt {b}+\sqrt {a} \sqrt [3]{x}\right )^2}} \sqrt [6]{x} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt [6]{x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{2 a^{9/4} \sqrt {b \sqrt [3]{x}+a x}} \]

[Out]

-3*x/a/(b*x^(1/3)+a*x)^(1/2)+5*(b*x^(1/3)+a*x)^(1/2)/a^2-5/2*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^(9/4)/(b*x^(1/3)+a*x)^(1/
2)

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Rubi [A]
time = 0.14, antiderivative size = 149, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {2043, 2047, 2049, 2036, 335, 226} \begin {gather*} -\frac {5 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}} F\left (2 \text {ArcTan}\left (\frac {\sqrt [4]{a} \sqrt [6]{x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{2 a^{9/4} \sqrt {a x+b \sqrt [3]{x}}}+\frac {5 \sqrt {a x+b \sqrt [3]{x}}}{a^2}-\frac {3 x}{a \sqrt {a x+b \sqrt [3]{x}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x/(b*x^(1/3) + a*x)^(3/2),x]

[Out]

(-3*x)/(a*Sqrt[b*x^(1/3) + a*x]) + (5*Sqrt[b*x^(1/3) + a*x])/a^2 - (5*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])/(
2*a^(9/4)*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 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 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 2047

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*(n - j)*(p + 1))), x] - Dist[c^n*((m + j*p - n + j + 1)/(b*(n - j)*(p + 1))), I
nt[(c*x)^(m - n)*(a*x^j + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c}, x] &&  !IntegerQ[p] && LtQ[0, j, n] && (I
ntegersQ[j, n] || GtQ[c, 0]) && LtQ[p, -1] && GtQ[m + j*p + 1, n - j]

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]

Rubi steps

\begin {align*} \int \frac {x}{\left (b \sqrt [3]{x}+a x\right )^{3/2}} \, dx &=3 \text {Subst}\left (\int \frac {x^5}{\left (b x+a x^3\right )^{3/2}} \, dx,x,\sqrt [3]{x}\right )\\ &=-\frac {3 x}{a \sqrt {b \sqrt [3]{x}+a x}}+\frac {15 \text {Subst}\left (\int \frac {x^2}{\sqrt {b x+a x^3}} \, dx,x,\sqrt [3]{x}\right )}{2 a}\\ &=-\frac {3 x}{a \sqrt {b \sqrt [3]{x}+a x}}+\frac {5 \sqrt {b \sqrt [3]{x}+a x}}{a^2}-\frac {(5 b) \text {Subst}\left (\int \frac {1}{\sqrt {b x+a x^3}} \, dx,x,\sqrt [3]{x}\right )}{2 a^2}\\ &=-\frac {3 x}{a \sqrt {b \sqrt [3]{x}+a x}}+\frac {5 \sqrt {b \sqrt [3]{x}+a x}}{a^2}-\frac {\left (5 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 )}{2 a^2 \sqrt {b \sqrt [3]{x}+a x}}\\ &=-\frac {3 x}{a \sqrt {b \sqrt [3]{x}+a x}}+\frac {5 \sqrt {b \sqrt [3]{x}+a x}}{a^2}-\frac {\left (5 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^2 \sqrt {b \sqrt [3]{x}+a x}}\\ &=-\frac {3 x}{a \sqrt {b \sqrt [3]{x}+a x}}+\frac {5 \sqrt {b \sqrt [3]{x}+a x}}{a^2}-\frac {5 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 )}{2 a^{9/4} \sqrt {b \sqrt [3]{x}+a x}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 10.05, size = 82, normalized size = 0.55 \begin {gather*} \frac {\sqrt {b \sqrt [3]{x}+a x} \left (5 b+2 a x^{2/3}-5 b \sqrt {1+\frac {a x^{2/3}}{b}} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-\frac {a x^{2/3}}{b}\right )\right )}{a^2 \left (b+a x^{2/3}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x/(b*x^(1/3) + a*x)^(3/2),x]

[Out]

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

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Maple [A]
time = 0.36, size = 184, normalized size = 1.23

method	result	size
derivativedivides	\(\frac {3 x^{\frac {1}{3}} b}{a^{2} \sqrt {\left (x^{\frac {2}{3}}+\frac {b}{a}\right ) x^{\frac {1}{3}} a}}+\frac {2 \sqrt {b \,x^{\frac {1}{3}}+a x}}{a^{2}}-\frac {5 b \sqrt {-a b}\, \sqrt {\frac {\left (x^{\frac {1}{3}}+\frac {\sqrt {-a b}}{a}\right ) a}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x^{\frac {1}{3}}-\frac {\sqrt {-a b}}{a}\right ) a}{\sqrt {-a b}}}\, \sqrt {-\frac {x^{\frac {1}{3}} a}{\sqrt {-a b}}}\, \EllipticF \left (\sqrt {\frac {\left (x^{\frac {1}{3}}+\frac {\sqrt {-a b}}{a}\right ) a}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{2 a^{3} \sqrt {b \,x^{\frac {1}{3}}+a x}}\)	\(160\)
default	\(\frac {-5 \sqrt {x^{\frac {1}{3}} \left (b +a \,x^{\frac {2}{3}}\right )}\, \sqrt {-a b}\, \sqrt {\frac {a \,x^{\frac {1}{3}}+\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (a \,x^{\frac {1}{3}}-\sqrt {-a b}\right )}{\sqrt {-a b}}}\, \sqrt {-\frac {x^{\frac {1}{3}} a}{\sqrt {-a b}}}\, \EllipticF \left (\sqrt {\frac {a \,x^{\frac {1}{3}}+\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) b +6 \sqrt {b \,x^{\frac {1}{3}}+a x}\, x^{\frac {1}{3}} a b +4 x^{\frac {1}{3}} \sqrt {x^{\frac {1}{3}} \left (b +a \,x^{\frac {2}{3}}\right )}\, a b +4 \sqrt {x^{\frac {1}{3}} \left (b +a \,x^{\frac {2}{3}}\right )}\, a^{2} x}{2 x^{\frac {1}{3}} \left (b +a \,x^{\frac {2}{3}}\right ) a^{3}}\)	\(184\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(-5*(x^(1/3)*(b+a*x^(2/3)))^(1/2)*(-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))*b+6*(b*x^(1/3)+a*x)^(1/2)*x^(1/3)*a*b+4*x^(1/3)*(x^(1/3)*(b+a*x^(2/3)))^(1/2)*a*b+4
*(x^(1/3)*(b+a*x^(2/3)))^(1/2)*a^2*x)/x^(1/3)/(b+a*x^(2/3))/a^3

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

[Out]

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

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

[Out]

integral((a^4*x^3 + 3*a^2*b^2*x^(5/3) - 2*a*b^3*x - (2*a^3*b*x^2 - b^4)*x^(1/3))*sqrt(a*x + b*x^(1/3))/(a^6*x^
4 + 2*a^3*b^3*x^2 + b^6), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x}{\left (a x + b \sqrt [3]{x}\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x}{{\left (a\,x+b\,x^{1/3}\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(a*x + b*x^(1/3))^(3/2),x)

[Out]

int(x/(a*x + b*x^(1/3))^(3/2), x)

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