3.2.18 \(\int \frac {x}{(b x^{2/3}+a x)^{3/2}} \, dx\)

Optimal. Leaf size=68 \[ -\frac {16 b \sqrt {a x+b x^{2/3}}}{a^3 \sqrt [3]{x}}+\frac {8 \sqrt {a x+b x^{2/3}}}{a^2}-\frac {6 x}{a \sqrt {a x+b x^{2/3}}} \]

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Rubi [A]  time = 0.08, antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {2015, 2002, 2014} \begin {gather*} -\frac {16 b \sqrt {a x+b x^{2/3}}}{a^3 \sqrt [3]{x}}+\frac {8 \sqrt {a x+b x^{2/3}}}{a^2}-\frac {6 x}{a \sqrt {a x+b x^{2/3}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x/(b*x^(2/3) + a*x)^(3/2),x]

[Out]

(-6*x)/(a*Sqrt[b*x^(2/3) + a*x]) + (8*Sqrt[b*x^(2/3) + a*x])/a^2 - (16*b*Sqrt[b*x^(2/3) + a*x])/(a^3*x^(1/3))

Rule 2002

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(a*x^j + b*x^n)^(p + 1)/(a*(j*p + 1)*x^(j -
1)), x] - Dist[(b*(n*p + n - j + 1))/(a*(j*p + 1)), Int[x^(n - j)*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, j,
n, p}, x] &&  !IntegerQ[p] && NeQ[n, j] && ILtQ[Simplify[(n*p + n - j + 1)/(n - j)], 0] && NeQ[j*p + 1, 0]

Rule 2014

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> -Simp[(c^(j - 1)*(c*x)^(m - j
+ 1)*(a*x^j + b*x^n)^(p + 1))/(a*(n - j)*(p + 1)), x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && N
eQ[n, j] && EqQ[m + n*p + n - j + 1, 0] && (IntegerQ[j] || GtQ[c, 0])

Rule 2015

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> -Simp[(c^(j - 1)*(c*x)^(m - j
+ 1)*(a*x^j + b*x^n)^(p + 1))/(a*(n - j)*(p + 1)), x] + Dist[(c^j*(m + n*p + n - j + 1))/(a*(n - j)*(p + 1)),
Int[(c*x)^(m - j)*(a*x^j + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, j, m, n}, x] &&  !IntegerQ[p] && NeQ[n, j
] && ILtQ[Simplify[(m + n*p + n - j + 1)/(n - j)], 0] && LtQ[p, -1] && (IntegerQ[j] || GtQ[c, 0])

Rubi steps

\begin {align*} \int \frac {x}{\left (b x^{2/3}+a x\right )^{3/2}} \, dx &=-\frac {6 x}{a \sqrt {b x^{2/3}+a x}}+\frac {4 \int \frac {1}{\sqrt {b x^{2/3}+a x}} \, dx}{a}\\ &=-\frac {6 x}{a \sqrt {b x^{2/3}+a x}}+\frac {8 \sqrt {b x^{2/3}+a x}}{a^2}-\frac {(8 b) \int \frac {1}{\sqrt [3]{x} \sqrt {b x^{2/3}+a x}} \, dx}{3 a^2}\\ &=-\frac {6 x}{a \sqrt {b x^{2/3}+a x}}+\frac {8 \sqrt {b x^{2/3}+a x}}{a^2}-\frac {16 b \sqrt {b x^{2/3}+a x}}{a^3 \sqrt [3]{x}}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 60, normalized size = 0.88 \begin {gather*} \frac {2 \left (a^2 x^{2/3}-4 a b \sqrt [3]{x}-8 b^2\right ) \sqrt {a x+b x^{2/3}}}{a^3 \sqrt [3]{x} \left (a \sqrt [3]{x}+b\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(-8*b^2 - 4*a*b*x^(1/3) + a^2*x^(2/3))*Sqrt[b*x^(2/3) + a*x])/(a^3*(b + a*x^(1/3))*x^(1/3))

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IntegrateAlgebraic [A]  time = 3.97, size = 54, normalized size = 0.79 \begin {gather*} -\frac {2 \sqrt [3]{x} \left (a^2 \left (-x^{2/3}\right )+4 a b \sqrt [3]{x}+8 b^2\right )}{a^3 \sqrt {x^{2/3} \left (a \sqrt [3]{x}+b\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(8*b^2 + 4*a*b*x^(1/3) - a^2*x^(2/3))*x^(1/3))/(a^3*Sqrt[(b + a*x^(1/3))*x^(2/3)])

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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giac [A]  time = 0.22, size = 60, normalized size = 0.88 \begin {gather*} \frac {16 \, b^{\frac {3}{2}}}{a^{3}} - \frac {6 \, b^{2}}{\sqrt {a x^{\frac {1}{3}} + b} a^{3}} + \frac {2 \, {\left ({\left (a x^{\frac {1}{3}} + b\right )}^{\frac {3}{2}} a^{6} - 6 \, \sqrt {a x^{\frac {1}{3}} + b} a^{6} b\right )}}{a^{9}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

16*b^(3/2)/a^3 - 6*b^2/(sqrt(a*x^(1/3) + b)*a^3) + 2*((a*x^(1/3) + b)^(3/2)*a^6 - 6*sqrt(a*x^(1/3) + b)*a^6*b)
/a^9

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maple [A]  time = 0.05, size = 45, normalized size = 0.66 \begin {gather*} \frac {2 \left (a \,x^{\frac {1}{3}}+b \right ) \left (a^{2} x^{\frac {2}{3}}-4 a b \,x^{\frac {1}{3}}-8 b^{2}\right ) x}{\left (a x +b \,x^{\frac {2}{3}}\right )^{\frac {3}{2}} a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*x*(a*x^(1/3)+b)*(a^2*x^(2/3)-4*a*b*x^(1/3)-8*b^2)/(a*x+b*x^(2/3))^(3/2)/a^3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x/(a*x + b*x^(2/3))^(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^{2/3}\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(a*x + b*x^(2/3))^(3/2),x)

[Out]

int(x/(a*x + b*x^(2/3))^(3/2), 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 x^{\frac {2}{3}}\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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