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

Optimal. Leaf size=160 \[ \frac {512 b^4 \sqrt {a x+b x^{2/3}}}{21 a^6 \sqrt [3]{x}}-\frac {256 b^3 \sqrt {a x+b x^{2/3}}}{21 a^5}+\frac {64 b^2 \sqrt [3]{x} \sqrt {a x+b x^{2/3}}}{7 a^4}-\frac {160 b x^{2/3} \sqrt {a x+b x^{2/3}}}{21 a^3}+\frac {20 x \sqrt {a x+b x^{2/3}}}{3 a^2}-\frac {6 x^2}{a \sqrt {a x+b x^{2/3}}} \]

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Rubi [A]  time = 0.24, antiderivative size = 160, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {2015, 2016, 2002, 2014} \begin {gather*} \frac {512 b^4 \sqrt {a x+b x^{2/3}}}{21 a^6 \sqrt [3]{x}}-\frac {256 b^3 \sqrt {a x+b x^{2/3}}}{21 a^5}+\frac {64 b^2 \sqrt [3]{x} \sqrt {a x+b x^{2/3}}}{7 a^4}-\frac {160 b x^{2/3} \sqrt {a x+b x^{2/3}}}{21 a^3}+\frac {20 x \sqrt {a x+b x^{2/3}}}{3 a^2}-\frac {6 x^2}{a \sqrt {a x+b x^{2/3}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2/(b*x^(2/3) + a*x)^(3/2),x]

[Out]

(-6*x^2)/(a*Sqrt[b*x^(2/3) + a*x]) - (256*b^3*Sqrt[b*x^(2/3) + a*x])/(21*a^5) + (512*b^4*Sqrt[b*x^(2/3) + a*x]
)/(21*a^6*x^(1/3)) + (64*b^2*x^(1/3)*Sqrt[b*x^(2/3) + a*x])/(7*a^4) - (160*b*x^(2/3)*Sqrt[b*x^(2/3) + a*x])/(2
1*a^3) + (20*x*Sqrt[b*x^(2/3) + a*x])/(3*a^2)

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])

Rule 2016

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*(m + j*p + 1)), x] - Dist[(b*(m + n*p + n - j + 1))/(a*c^(n - j)*(m + j*p + 1)
), Int[(c*x)^(m + n - j)*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[
n, j] && ILtQ[Simplify[(m + n*p + n - j + 1)/(n - j)], 0] && NeQ[m + j*p + 1, 0] && (IntegersQ[j, n] || GtQ[c,
0])

Rubi steps

\begin {align*} \int \frac {x^2}{\left (b x^{2/3}+a x\right )^{3/2}} \, dx &=-\frac {6 x^2}{a \sqrt {b x^{2/3}+a x}}+\frac {10 \int \frac {x}{\sqrt {b x^{2/3}+a x}} \, dx}{a}\\ &=-\frac {6 x^2}{a \sqrt {b x^{2/3}+a x}}+\frac {20 x \sqrt {b x^{2/3}+a x}}{3 a^2}-\frac {(80 b) \int \frac {x^{2/3}}{\sqrt {b x^{2/3}+a x}} \, dx}{9 a^2}\\ &=-\frac {6 x^2}{a \sqrt {b x^{2/3}+a x}}-\frac {160 b x^{2/3} \sqrt {b x^{2/3}+a x}}{21 a^3}+\frac {20 x \sqrt {b x^{2/3}+a x}}{3 a^2}+\frac {\left (160 b^2\right ) \int \frac {\sqrt [3]{x}}{\sqrt {b x^{2/3}+a x}} \, dx}{21 a^3}\\ &=-\frac {6 x^2}{a \sqrt {b x^{2/3}+a x}}+\frac {64 b^2 \sqrt [3]{x} \sqrt {b x^{2/3}+a x}}{7 a^4}-\frac {160 b x^{2/3} \sqrt {b x^{2/3}+a x}}{21 a^3}+\frac {20 x \sqrt {b x^{2/3}+a x}}{3 a^2}-\frac {\left (128 b^3\right ) \int \frac {1}{\sqrt {b x^{2/3}+a x}} \, dx}{21 a^4}\\ &=-\frac {6 x^2}{a \sqrt {b x^{2/3}+a x}}-\frac {256 b^3 \sqrt {b x^{2/3}+a x}}{21 a^5}+\frac {64 b^2 \sqrt [3]{x} \sqrt {b x^{2/3}+a x}}{7 a^4}-\frac {160 b x^{2/3} \sqrt {b x^{2/3}+a x}}{21 a^3}+\frac {20 x \sqrt {b x^{2/3}+a x}}{3 a^2}+\frac {\left (256 b^4\right ) \int \frac {1}{\sqrt [3]{x} \sqrt {b x^{2/3}+a x}} \, dx}{63 a^5}\\ &=-\frac {6 x^2}{a \sqrt {b x^{2/3}+a x}}-\frac {256 b^3 \sqrt {b x^{2/3}+a x}}{21 a^5}+\frac {512 b^4 \sqrt {b x^{2/3}+a x}}{21 a^6 \sqrt [3]{x}}+\frac {64 b^2 \sqrt [3]{x} \sqrt {b x^{2/3}+a x}}{7 a^4}-\frac {160 b x^{2/3} \sqrt {b x^{2/3}+a x}}{21 a^3}+\frac {20 x \sqrt {b x^{2/3}+a x}}{3 a^2}\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 85, normalized size = 0.53 \begin {gather*} \frac {14 a^5 x^2-20 a^4 b x^{5/3}+32 a^3 b^2 x^{4/3}-64 a^2 b^3 x+256 a b^4 x^{2/3}+512 b^5 \sqrt [3]{x}}{21 a^6 \sqrt {a x+b x^{2/3}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(512*b^5*x^(1/3) + 256*a*b^4*x^(2/3) - 64*a^2*b^3*x + 32*a^3*b^2*x^(4/3) - 20*a^4*b*x^(5/3) + 14*a^5*x^2)/(21*
a^6*Sqrt[b*x^(2/3) + a*x])

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IntegrateAlgebraic [A]  time = 4.18, size = 91, normalized size = 0.57 \begin {gather*} \frac {2 \sqrt [3]{x} \left (7 a^5 x^{5/3}-10 a^4 b x^{4/3}+16 a^3 b^2 x-32 a^2 b^3 x^{2/3}+128 a b^4 \sqrt [3]{x}+256 b^5\right )}{21 a^6 \sqrt {x^{2/3} \left (a \sqrt [3]{x}+b\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x^(1/3)*(256*b^5 + 128*a*b^4*x^(1/3) - 32*a^2*b^3*x^(2/3) + 16*a^3*b^2*x - 10*a^4*b*x^(4/3) + 7*a^5*x^(5/3)
))/(21*a^6*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^2/(b*x^(2/3)+a*x)^(3/2),x, algorithm="fricas")

[Out]

Timed out

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giac [A]  time = 0.23, size = 112, normalized size = 0.70 \begin {gather*} -\frac {512 \, b^{\frac {9}{2}}}{21 \, a^{6}} + \frac {6 \, b^{5}}{\sqrt {a x^{\frac {1}{3}} + b} a^{6}} + \frac {2 \, {\left (7 \, {\left (a x^{\frac {1}{3}} + b\right )}^{\frac {9}{2}} a^{48} - 45 \, {\left (a x^{\frac {1}{3}} + b\right )}^{\frac {7}{2}} a^{48} b + 126 \, {\left (a x^{\frac {1}{3}} + b\right )}^{\frac {5}{2}} a^{48} b^{2} - 210 \, {\left (a x^{\frac {1}{3}} + b\right )}^{\frac {3}{2}} a^{48} b^{3} + 315 \, \sqrt {a x^{\frac {1}{3}} + b} a^{48} b^{4}\right )}}{21 \, a^{54}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-512/21*b^(9/2)/a^6 + 6*b^5/(sqrt(a*x^(1/3) + b)*a^6) + 2/21*(7*(a*x^(1/3) + b)^(9/2)*a^48 - 45*(a*x^(1/3) + b
)^(7/2)*a^48*b + 126*(a*x^(1/3) + b)^(5/2)*a^48*b^2 - 210*(a*x^(1/3) + b)^(3/2)*a^48*b^3 + 315*sqrt(a*x^(1/3)
+ b)*a^48*b^4)/a^54

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maple [A]  time = 0.05, size = 77, normalized size = 0.48 \begin {gather*} \frac {2 \left (a \,x^{\frac {1}{3}}+b \right ) \left (7 a^{5} x^{\frac {5}{3}}-10 a^{4} b \,x^{\frac {4}{3}}+16 a^{3} b^{2} x -32 a^{2} b^{3} x^{\frac {2}{3}}+128 a \,b^{4} x^{\frac {1}{3}}+256 b^{5}\right ) x}{21 \left (a x +b \,x^{\frac {2}{3}}\right )^{\frac {3}{2}} a^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/21*x*(a*x^(1/3)+b)*(7*a^5*x^(5/3)-10*a^4*b*x^(4/3)+16*a^3*b^2*x-32*a^2*b^3*x^(2/3)+128*a*b^4*x^(1/3)+256*b^5
)/(a*x+b*x^(2/3))^(3/2)/a^6

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

[Out]

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

[Out]

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

[Out]

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

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