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

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

[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])

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Rubi [A]  time = 0.11658, antiderivative size = 84, normalized size of antiderivative = 1., 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} \[ -\frac{32 a^2 \sqrt{a x+b \sqrt{x}}}{15 b^3 \sqrt{x}}+\frac{16 a \sqrt{a x+b \sqrt{x}}}{15 b^2 x}-\frac{4 \sqrt{a x+b \sqrt{x}}}{5 b x^{3/2}} \]

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 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])

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])

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*}

Mathematica [A]  time = 0.0481187, size = 48, normalized size = 0.57 \[ -\frac{4 \sqrt{a x+b \sqrt{x}} \left (8 a^2 x-4 a b \sqrt{x}+3 b^2\right )}{15 b^3 x^{3/2}} \]

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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Maple [C]  time = 0.01, size = 218, normalized size = 2.6 \begin{align*} -{\frac{1}{15\,{b}^{4}}\sqrt{b\sqrt{x}+ax} \left ( 60\, \left ( b\sqrt{x}+ax \right ) ^{3/2}{a}^{5/2}{x}^{5/2}-30\,\sqrt{b\sqrt{x}+ax}{a}^{7/2}{x}^{7/2}-15\,\ln \left ( 1/2\,{\frac{2\,a\sqrt{x}+2\,\sqrt{b\sqrt{x}+ax}\sqrt{a}+b}{\sqrt{a}}} \right ){x}^{7/2}{a}^{3}b-30\,{a}^{7/2}{x}^{7/2}\sqrt{\sqrt{x} \left ( b+a\sqrt{x} \right ) }+15\,\ln \left ( 1/2\,{\frac{2\,\sqrt{\sqrt{x} \left ( b+a\sqrt{x} \right ) }\sqrt{a}+2\,a\sqrt{x}+b}{\sqrt{a}}} \right ){x}^{7/2}{a}^{3}b+12\, \left ( b\sqrt{x}+ax \right ) ^{3/2}\sqrt{a}{x}^{3/2}{b}^{2}-28\,{a}^{3/2} \left ( b\sqrt{x}+ax \right ) ^{3/2}b{x}^{2} \right ){\frac{1}{\sqrt{\sqrt{x} \left ( b+a\sqrt{x} \right ) }}}{\frac{1}{\sqrt{a}}}{x}^{-{\frac{7}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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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Fricas [A]  time = 2.31516, size = 103, normalized size = 1.23 \begin{align*} \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{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x^{2} \sqrt{a x + b \sqrt{x}}}\, dx \end{align*}

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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Giac [A]  time = 1.25171, size = 113, normalized size = 1.35 \begin{align*} \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{align*}

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