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

Optimal. Leaf size=153 \[ -\frac {105 a^4 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt [3]{x}}{\sqrt {a x+b x^{2/3}}}\right )}{64 b^{9/2}}+\frac {105 a^3 \sqrt {a x+b x^{2/3}}}{64 b^4 x^{2/3}}-\frac {35 a^2 \sqrt {a x+b x^{2/3}}}{32 b^3 x}+\frac {7 a \sqrt {a x+b x^{2/3}}}{8 b^2 x^{4/3}}-\frac {3 \sqrt {a x+b x^{2/3}}}{4 b x^{5/3}} \]

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Rubi [A]  time = 0.24, antiderivative size = 153, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2025, 2029, 206} \begin {gather*} \frac {105 a^3 \sqrt {a x+b x^{2/3}}}{64 b^4 x^{2/3}}-\frac {35 a^2 \sqrt {a x+b x^{2/3}}}{32 b^3 x}-\frac {105 a^4 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt [3]{x}}{\sqrt {a x+b x^{2/3}}}\right )}{64 b^{9/2}}+\frac {7 a \sqrt {a x+b x^{2/3}}}{8 b^2 x^{4/3}}-\frac {3 \sqrt {a x+b x^{2/3}}}{4 b x^{5/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^2*Sqrt[b*x^(2/3) + a*x]),x]

[Out]

(-3*Sqrt[b*x^(2/3) + a*x])/(4*b*x^(5/3)) + (7*a*Sqrt[b*x^(2/3) + a*x])/(8*b^2*x^(4/3)) - (35*a^2*Sqrt[b*x^(2/3
) + a*x])/(32*b^3*x) + (105*a^3*Sqrt[b*x^(2/3) + a*x])/(64*b^4*x^(2/3)) - (105*a^4*ArcTanh[(Sqrt[b]*x^(1/3))/S
qrt[b*x^(2/3) + a*x]])/(64*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 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^2 \sqrt {b x^{2/3}+a x}} \, dx &=-\frac {3 \sqrt {b x^{2/3}+a x}}{4 b x^{5/3}}-\frac {(7 a) \int \frac {1}{x^{5/3} \sqrt {b x^{2/3}+a x}} \, dx}{8 b}\\ &=-\frac {3 \sqrt {b x^{2/3}+a x}}{4 b x^{5/3}}+\frac {7 a \sqrt {b x^{2/3}+a x}}{8 b^2 x^{4/3}}+\frac {\left (35 a^2\right ) \int \frac {1}{x^{4/3} \sqrt {b x^{2/3}+a x}} \, dx}{48 b^2}\\ &=-\frac {3 \sqrt {b x^{2/3}+a x}}{4 b x^{5/3}}+\frac {7 a \sqrt {b x^{2/3}+a x}}{8 b^2 x^{4/3}}-\frac {35 a^2 \sqrt {b x^{2/3}+a x}}{32 b^3 x}-\frac {\left (35 a^3\right ) \int \frac {1}{x \sqrt {b x^{2/3}+a x}} \, dx}{64 b^3}\\ &=-\frac {3 \sqrt {b x^{2/3}+a x}}{4 b x^{5/3}}+\frac {7 a \sqrt {b x^{2/3}+a x}}{8 b^2 x^{4/3}}-\frac {35 a^2 \sqrt {b x^{2/3}+a x}}{32 b^3 x}+\frac {105 a^3 \sqrt {b x^{2/3}+a x}}{64 b^4 x^{2/3}}+\frac {\left (35 a^4\right ) \int \frac {1}{x^{2/3} \sqrt {b x^{2/3}+a x}} \, dx}{128 b^4}\\ &=-\frac {3 \sqrt {b x^{2/3}+a x}}{4 b x^{5/3}}+\frac {7 a \sqrt {b x^{2/3}+a x}}{8 b^2 x^{4/3}}-\frac {35 a^2 \sqrt {b x^{2/3}+a x}}{32 b^3 x}+\frac {105 a^3 \sqrt {b x^{2/3}+a x}}{64 b^4 x^{2/3}}-\frac {\left (105 a^4\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 )}{64 b^4}\\ &=-\frac {3 \sqrt {b x^{2/3}+a x}}{4 b x^{5/3}}+\frac {7 a \sqrt {b x^{2/3}+a x}}{8 b^2 x^{4/3}}-\frac {35 a^2 \sqrt {b x^{2/3}+a x}}{32 b^3 x}+\frac {105 a^3 \sqrt {b x^{2/3}+a x}}{64 b^4 x^{2/3}}-\frac {105 a^4 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt [3]{x}}{\sqrt {b x^{2/3}+a x}}\right )}{64 b^{9/2}}\\ \end {align*}

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Mathematica [C]  time = 0.06, size = 48, normalized size = 0.31 \begin {gather*} -\frac {6 a^4 \sqrt {a x+b x^{2/3}} \, _2F_1\left (\frac {1}{2},5;\frac {3}{2};\frac {\sqrt [3]{x} a}{b}+1\right )}{b^5 \sqrt [3]{x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-6*a^4*Sqrt[b*x^(2/3) + a*x]*Hypergeometric2F1[1/2, 5, 3/2, 1 + (a*x^(1/3))/b])/(b^5*x^(1/3))

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IntegrateAlgebraic [A]  time = 0.21, size = 101, normalized size = 0.66 \begin {gather*} \frac {\sqrt {a x+b x^{2/3}} \left (105 a^3 x-70 a^2 b x^{2/3}+56 a b^2 \sqrt [3]{x}-48 b^3\right )}{64 b^4 x^{5/3}}-\frac {105 a^4 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt [3]{x}}{\sqrt {a x+b x^{2/3}}}\right )}{64 b^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[b*x^(2/3) + a*x]*(-48*b^3 + 56*a*b^2*x^(1/3) - 70*a^2*b*x^(2/3) + 105*a^3*x))/(64*b^4*x^(5/3)) - (105*a^
4*ArcTanh[(Sqrt[b]*x^(1/3))/Sqrt[b*x^(2/3) + a*x]])/(64*b^(9/2))

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

[Out]

Timed out

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giac [A]  time = 0.26, size = 109, normalized size = 0.71 \begin {gather*} \frac {\frac {105 \, a^{5} \arctan \left (\frac {\sqrt {a x^{\frac {1}{3}} + b}}{\sqrt {-b}}\right )}{\sqrt {-b} b^{4}} + \frac {105 \, {\left (a x^{\frac {1}{3}} + b\right )}^{\frac {7}{2}} a^{5} - 385 \, {\left (a x^{\frac {1}{3}} + b\right )}^{\frac {5}{2}} a^{5} b + 511 \, {\left (a x^{\frac {1}{3}} + b\right )}^{\frac {3}{2}} a^{5} b^{2} - 279 \, \sqrt {a x^{\frac {1}{3}} + b} a^{5} b^{3}}{a^{4} b^{4} x^{\frac {4}{3}}}}{64 \, a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64*(105*a^5*arctan(sqrt(a*x^(1/3) + b)/sqrt(-b))/(sqrt(-b)*b^4) + (105*(a*x^(1/3) + b)^(7/2)*a^5 - 385*(a*x^
(1/3) + b)^(5/2)*a^5*b + 511*(a*x^(1/3) + b)^(3/2)*a^5*b^2 - 279*sqrt(a*x^(1/3) + b)*a^5*b^3)/(a^4*b^4*x^(4/3)
))/a

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maple [A]  time = 0.05, size = 126, normalized size = 0.82 \begin {gather*} -\frac {\sqrt {a \,x^{\frac {1}{3}}+b}\, \left (105 a^{4} b \,x^{\frac {7}{3}} \arctanh \left (\frac {\sqrt {a \,x^{\frac {1}{3}}+b}}{\sqrt {b}}\right )-105 \sqrt {a \,x^{\frac {1}{3}}+b}\, a^{3} b^{\frac {3}{2}} x^{2}+70 \sqrt {a \,x^{\frac {1}{3}}+b}\, a^{2} b^{\frac {5}{2}} x^{\frac {5}{3}}-56 \sqrt {a \,x^{\frac {1}{3}}+b}\, a \,b^{\frac {7}{2}} x^{\frac {4}{3}}+48 \sqrt {a \,x^{\frac {1}{3}}+b}\, b^{\frac {9}{2}} x \right )}{64 \sqrt {a x +b \,x^{\frac {2}{3}}}\, b^{\frac {11}{2}} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/64*(a*x^(1/3)+b)^(1/2)*(105*x^(7/3)*arctanh((a*x^(1/3)+b)^(1/2)/b^(1/2))*a^4*b+70*x^(5/3)*(a*x^(1/3)+b)^(1/
2)*b^(5/2)*a^2-56*x^(4/3)*(a*x^(1/3)+b)^(1/2)*b^(7/2)*a+48*(a*x^(1/3)+b)^(1/2)*b^(9/2)*x-105*x^2*(a*x^(1/3)+b)
^(1/2)*b^(3/2)*a^3)/x^2/(a*x+b*x^(2/3))^(1/2)/b^(11/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x^{2} \sqrt {a x + b x^{\frac {2}{3}}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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