3.2.20 \(\int \frac {1}{x (b x^{2/3}+a x)^{3/2}} \, dx\)

Optimal. Leaf size=146 \[ \frac {105 a^3 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt [3]{x}}{\sqrt {a x+b x^{2/3}}}\right )}{8 b^{9/2}}-\frac {105 a^2 \sqrt {a x+b x^{2/3}}}{8 b^4 x^{2/3}}+\frac {35 a \sqrt {a x+b x^{2/3}}}{4 b^3 x}-\frac {7 \sqrt {a x+b x^{2/3}}}{b^2 x^{4/3}}+\frac {6}{b x^{2/3} \sqrt {a x+b x^{2/3}}} \]

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Rubi [A]  time = 0.24, antiderivative size = 146, 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 = {2023, 2025, 2029, 206} \begin {gather*} -\frac {105 a^2 \sqrt {a x+b x^{2/3}}}{8 b^4 x^{2/3}}+\frac {105 a^3 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt [3]{x}}{\sqrt {a x+b x^{2/3}}}\right )}{8 b^{9/2}}+\frac {35 a \sqrt {a x+b x^{2/3}}}{4 b^3 x}-\frac {7 \sqrt {a x+b x^{2/3}}}{b^2 x^{4/3}}+\frac {6}{b x^{2/3} \sqrt {a x+b x^{2/3}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x*(b*x^(2/3) + a*x)^(3/2)),x]

[Out]

6/(b*x^(2/3)*Sqrt[b*x^(2/3) + a*x]) - (7*Sqrt[b*x^(2/3) + a*x])/(b^2*x^(4/3)) + (35*a*Sqrt[b*x^(2/3) + a*x])/(
4*b^3*x) - (105*a^2*Sqrt[b*x^(2/3) + a*x])/(8*b^4*x^(2/3)) + (105*a^3*ArcTanh[(Sqrt[b]*x^(1/3))/Sqrt[b*x^(2/3)
+ a*x]])/(8*b^(9/2))

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2023

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, m}, x] &&  !IntegerQ[p] && LtQ[0, j, n] &
& (IntegersQ[j, n] || GtQ[c, 0]) && LtQ[p, -1]

Rule 2025

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, m, p}, x] &&  !IntegerQ[p] && LtQ[0, j,
n] && (IntegersQ[j, n] || GtQ[c, 0]) && LtQ[m + j*p + 1, 0]

Rule 2029

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

Rubi steps

\begin {align*} \int \frac {1}{x \left (b x^{2/3}+a x\right )^{3/2}} \, dx &=\frac {6}{b x^{2/3} \sqrt {b x^{2/3}+a x}}+\frac {7 \int \frac {1}{x^{5/3} \sqrt {b x^{2/3}+a x}} \, dx}{b}\\ &=\frac {6}{b x^{2/3} \sqrt {b x^{2/3}+a x}}-\frac {7 \sqrt {b x^{2/3}+a x}}{b^2 x^{4/3}}-\frac {(35 a) \int \frac {1}{x^{4/3} \sqrt {b x^{2/3}+a x}} \, dx}{6 b^2}\\ &=\frac {6}{b x^{2/3} \sqrt {b x^{2/3}+a x}}-\frac {7 \sqrt {b x^{2/3}+a x}}{b^2 x^{4/3}}+\frac {35 a \sqrt {b x^{2/3}+a x}}{4 b^3 x}+\frac {\left (35 a^2\right ) \int \frac {1}{x \sqrt {b x^{2/3}+a x}} \, dx}{8 b^3}\\ &=\frac {6}{b x^{2/3} \sqrt {b x^{2/3}+a x}}-\frac {7 \sqrt {b x^{2/3}+a x}}{b^2 x^{4/3}}+\frac {35 a \sqrt {b x^{2/3}+a x}}{4 b^3 x}-\frac {105 a^2 \sqrt {b x^{2/3}+a x}}{8 b^4 x^{2/3}}-\frac {\left (35 a^3\right ) \int \frac {1}{x^{2/3} \sqrt {b x^{2/3}+a x}} \, dx}{16 b^4}\\ &=\frac {6}{b x^{2/3} \sqrt {b x^{2/3}+a x}}-\frac {7 \sqrt {b x^{2/3}+a x}}{b^2 x^{4/3}}+\frac {35 a \sqrt {b x^{2/3}+a x}}{4 b^3 x}-\frac {105 a^2 \sqrt {b x^{2/3}+a x}}{8 b^4 x^{2/3}}+\frac {\left (105 a^3\right ) \operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {\sqrt [3]{x}}{\sqrt {b x^{2/3}+a x}}\right )}{8 b^4}\\ &=\frac {6}{b x^{2/3} \sqrt {b x^{2/3}+a x}}-\frac {7 \sqrt {b x^{2/3}+a x}}{b^2 x^{4/3}}+\frac {35 a \sqrt {b x^{2/3}+a x}}{4 b^3 x}-\frac {105 a^2 \sqrt {b x^{2/3}+a x}}{8 b^4 x^{2/3}}+\frac {105 a^3 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt [3]{x}}{\sqrt {b x^{2/3}+a x}}\right )}{8 b^{9/2}}\\ \end {align*}

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Mathematica [C]  time = 0.09, size = 48, normalized size = 0.33 \begin {gather*} -\frac {6 a^3 \sqrt [3]{x} \, _2F_1\left (-\frac {1}{2},4;\frac {1}{2};\frac {\sqrt [3]{x} a}{b}+1\right )}{b^4 \sqrt {a x+b x^{2/3}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-6*a^3*x^(1/3)*Hypergeometric2F1[-1/2, 4, 1/2, 1 + (a*x^(1/3))/b])/(b^4*Sqrt[b*x^(2/3) + a*x])

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IntegrateAlgebraic [A]  time = 4.34, size = 128, normalized size = 0.88 \begin {gather*} \frac {\sqrt [3]{x} \sqrt {a \sqrt [3]{x}+b} \left (\frac {105 a^3 \tanh ^{-1}\left (\frac {\sqrt {a \sqrt [3]{x}+b}}{\sqrt {b}}\right )}{8 b^{9/2}}+\frac {-105 a^3 x-35 a^2 b x^{2/3}+14 a b^2 \sqrt [3]{x}-8 b^3}{8 b^4 x \sqrt {a \sqrt [3]{x}+b}}\right )}{\sqrt {x^{2/3} \left (a \sqrt [3]{x}+b\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[b + a*x^(1/3)]*x^(1/3)*((-8*b^3 + 14*a*b^2*x^(1/3) - 35*a^2*b*x^(2/3) - 105*a^3*x)/(8*b^4*Sqrt[b + a*x^(
1/3)]*x) + (105*a^3*ArcTanh[Sqrt[b + a*x^(1/3)]/Sqrt[b]])/(8*b^(9/2))))/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(1/x/(b*x^(2/3)+a*x)^(3/2),x, algorithm="fricas")

[Out]

Timed out

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giac [A]  time = 0.26, size = 105, normalized size = 0.72 \begin {gather*} -\frac {105 \, a^{3} \arctan \left (\frac {\sqrt {a x^{\frac {1}{3}} + b}}{\sqrt {-b}}\right )}{8 \, \sqrt {-b} b^{4}} - \frac {6 \, a^{3}}{\sqrt {a x^{\frac {1}{3}} + b} b^{4}} - \frac {57 \, {\left (a x^{\frac {1}{3}} + b\right )}^{\frac {5}{2}} a^{3} - 136 \, {\left (a x^{\frac {1}{3}} + b\right )}^{\frac {3}{2}} a^{3} b + 87 \, \sqrt {a x^{\frac {1}{3}} + b} a^{3} b^{2}}{8 \, a^{3} b^{4} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-105/8*a^3*arctan(sqrt(a*x^(1/3) + b)/sqrt(-b))/(sqrt(-b)*b^4) - 6*a^3/(sqrt(a*x^(1/3) + b)*b^4) - 1/8*(57*(a*
x^(1/3) + b)^(5/2)*a^3 - 136*(a*x^(1/3) + b)^(3/2)*a^3*b + 87*sqrt(a*x^(1/3) + b)*a^3*b^2)/(a^3*b^4*x)

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maple [A]  time = 0.06, size = 88, normalized size = 0.60 \begin {gather*} -\frac {\left (a \,x^{\frac {1}{3}}+b \right ) \left (-105 \sqrt {a \,x^{\frac {1}{3}}+b}\, a^{3} x \arctanh \left (\frac {\sqrt {a \,x^{\frac {1}{3}}+b}}{\sqrt {b}}\right )+105 a^{3} \sqrt {b}\, x +35 a^{2} b^{\frac {3}{2}} x^{\frac {2}{3}}-14 a \,b^{\frac {5}{2}} x^{\frac {1}{3}}+8 b^{\frac {7}{2}}\right )}{8 \left (a x +b \,x^{\frac {2}{3}}\right )^{\frac {3}{2}} b^{\frac {9}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(a*x^(1/3)+b)*(-14*b^(5/2)*x^(1/3)*a+35*b^(3/2)*x^(2/3)*a^2+105*x*a^3*b^(1/2)+8*b^(7/2)-105*arctanh((a*x^
(1/3)+b)^(1/2)/b^(1/2))*(a*x^(1/3)+b)^(1/2)*x*a^3)/(a*x+b*x^(2/3))^(3/2)/b^(9/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{{\left (a x + b x^{\frac {2}{3}}\right )}^{\frac {3}{2}} x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

[Out]

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

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