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

Optimal. Leaf size=84 \[ -\frac {32 a^2 \sqrt {a x+b \sqrt {x}}}{15 b^3 \sqrt {x}}+\frac {16 a \sqrt {a x+b \sqrt {x}}}{15 b^2 x}-\frac {4 \sqrt {a x+b \sqrt {x}}}{5 b x^{3/2}} \]

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Rubi [A]  time = 0.12, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2016, 2014} \begin {gather*} -\frac {32 a^2 \sqrt {a x+b \sqrt {x}}}{15 b^3 \sqrt {x}}+\frac {16 a \sqrt {a x+b \sqrt {x}}}{15 b^2 x}-\frac {4 \sqrt {a x+b \sqrt {x}}}{5 b x^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^2*Sqrt[b*Sqrt[x] + a*x]),x]

[Out]

(-4*Sqrt[b*Sqrt[x] + a*x])/(5*b*x^(3/2)) + (16*a*Sqrt[b*Sqrt[x] + a*x])/(15*b^2*x) - (32*a^2*Sqrt[b*Sqrt[x] +
a*x])/(15*b^3*Sqrt[x])

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 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 {1}{x^2 \sqrt {b \sqrt {x}+a x}} \, dx &=-\frac {4 \sqrt {b \sqrt {x}+a x}}{5 b x^{3/2}}-\frac {(4 a) \int \frac {1}{x^{3/2} \sqrt {b \sqrt {x}+a x}} \, dx}{5 b}\\ &=-\frac {4 \sqrt {b \sqrt {x}+a x}}{5 b x^{3/2}}+\frac {16 a \sqrt {b \sqrt {x}+a x}}{15 b^2 x}+\frac {\left (8 a^2\right ) \int \frac {1}{x \sqrt {b \sqrt {x}+a x}} \, dx}{15 b^2}\\ &=-\frac {4 \sqrt {b \sqrt {x}+a x}}{5 b x^{3/2}}+\frac {16 a \sqrt {b \sqrt {x}+a x}}{15 b^2 x}-\frac {32 a^2 \sqrt {b \sqrt {x}+a x}}{15 b^3 \sqrt {x}}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 48, normalized size = 0.57 \begin {gather*} -\frac {4 \sqrt {a x+b \sqrt {x}} \left (8 a^2 x-4 a b \sqrt {x}+3 b^2\right )}{15 b^3 x^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*Sqrt[b*Sqrt[x] + a*x]*(3*b^2 - 4*a*b*Sqrt[x] + 8*a^2*x))/(15*b^3*x^(3/2))

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IntegrateAlgebraic [A]  time = 0.17, size = 48, normalized size = 0.57 \begin {gather*} -\frac {4 \sqrt {a x+b \sqrt {x}} \left (8 a^2 x-4 a b \sqrt {x}+3 b^2\right )}{15 b^3 x^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*Sqrt[b*Sqrt[x] + a*x]*(3*b^2 - 4*a*b*Sqrt[x] + 8*a^2*x))/(15*b^3*x^(3/2))

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fricas [A]  time = 0.90, size = 42, normalized size = 0.50 \begin {gather*} \frac {4 \, {\left (4 \, a b x - {\left (8 \, a^{2} x + 3 \, b^{2}\right )} \sqrt {x}\right )} \sqrt {a x + b \sqrt {x}}}{15 \, b^{3} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/15*(4*a*b*x - (8*a^2*x + 3*b^2)*sqrt(x))*sqrt(a*x + b*sqrt(x))/(b^3*x^2)

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giac [A]  time = 0.23, size = 84, normalized size = 1.00 \begin {gather*} \frac {4 \, {\left (20 \, a {\left (\sqrt {a} \sqrt {x} - \sqrt {a x + b \sqrt {x}}\right )}^{2} + 15 \, \sqrt {a} b {\left (\sqrt {a} \sqrt {x} - \sqrt {a x + b \sqrt {x}}\right )} + 3 \, b^{2}\right )}}{15 \, {\left (\sqrt {a} \sqrt {x} - \sqrt {a x + b \sqrt {x}}\right )}^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/15*(20*a*(sqrt(a)*sqrt(x) - sqrt(a*x + b*sqrt(x)))^2 + 15*sqrt(a)*b*(sqrt(a)*sqrt(x) - sqrt(a*x + b*sqrt(x))
) + 3*b^2)/(sqrt(a)*sqrt(x) - sqrt(a*x + b*sqrt(x)))^5

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maple [C]  time = 0.07, size = 218, normalized size = 2.60 \begin {gather*} \frac {\sqrt {a x +b \sqrt {x}}\, \left (-15 a^{3} b \,x^{\frac {7}{2}} \ln \left (\frac {2 a \sqrt {x}+b +2 \sqrt {\left (a \sqrt {x}+b \right ) \sqrt {x}}\, \sqrt {a}}{2 \sqrt {a}}\right )+15 a^{3} b \,x^{\frac {7}{2}} \ln \left (\frac {2 a \sqrt {x}+b +2 \sqrt {a x +b \sqrt {x}}\, \sqrt {a}}{2 \sqrt {a}}\right )+30 \sqrt {\left (a \sqrt {x}+b \right ) \sqrt {x}}\, a^{\frac {7}{2}} x^{\frac {7}{2}}+30 \sqrt {a x +b \sqrt {x}}\, a^{\frac {7}{2}} x^{\frac {7}{2}}-60 \left (a x +b \sqrt {x}\right )^{\frac {3}{2}} a^{\frac {5}{2}} x^{\frac {5}{2}}+28 \left (a x +b \sqrt {x}\right )^{\frac {3}{2}} a^{\frac {3}{2}} b \,x^{2}-12 \left (a x +b \sqrt {x}\right )^{\frac {3}{2}} \sqrt {a}\, b^{2} x^{\frac {3}{2}}\right )}{15 \sqrt {\left (a \sqrt {x}+b \right ) \sqrt {x}}\, \sqrt {a}\, b^{4} x^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(a*x+b*x^(1/2))^(1/2)*(30*x^(7/2)*((a*x^(1/2)+b)*x^(1/2))^(1/2)*a^(7/2)-60*x^(5/2)*(a*x+b*x^(1/2))^(3/2)*
a^(5/2)+30*x^(7/2)*(a*x+b*x^(1/2))^(1/2)*a^(7/2)-15*x^(7/2)*ln(1/2*(2*a*x^(1/2)+b+2*((a*x^(1/2)+b)*x^(1/2))^(1
/2)*a^(1/2))/a^(1/2))*a^3*b+15*x^(7/2)*ln(1/2*(2*a*x^(1/2)+b+2*(a*x+b*x^(1/2))^(1/2)*a^(1/2))/a^(1/2))*a^3*b-1
2*x^(3/2)*(a*x+b*x^(1/2))^(3/2)*a^(1/2)*b^2+28*a^(3/2)*(a*x+b*x^(1/2))^(3/2)*b*x^2)/((a*x^(1/2)+b)*x^(1/2))^(1
/2)/b^4/x^(7/2)/a^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/(sqrt(a*x + b*sqrt(x))*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\,\sqrt {x}}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(1/(x^2*(a*x + b*x^(1/2))^(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 \sqrt {x}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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