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

Optimal. Leaf size=79 \[ \frac{32 a \sqrt{a x+b \sqrt{x}}}{3 b^3 \sqrt{x}}-\frac{16 \sqrt{a x+b \sqrt{x}}}{3 b^2 x}+\frac{4}{b \sqrt{x} \sqrt{a x+b \sqrt{x}}} \]

[Out]

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

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Rubi [A]  time = 0.122479, antiderivative size = 79, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {2015, 2016, 2014} \[ \frac{32 a \sqrt{a x+b \sqrt{x}}}{3 b^3 \sqrt{x}}-\frac{16 \sqrt{a x+b \sqrt{x}}}{3 b^2 x}+\frac{4}{b \sqrt{x} \sqrt{a x+b \sqrt{x}}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x*(b*Sqrt[x] + a*x)^(3/2)),x]

[Out]

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

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])

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

Mathematica [A]  time = 0.0524482, size = 48, normalized size = 0.61 \[ \frac{4 \left (8 a^2 x+4 a b \sqrt{x}-b^2\right )}{3 b^3 \sqrt{x} \sqrt{a x+b \sqrt{x}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.011, size = 524, normalized size = 6.6 \begin{align*}{\frac{1}{3\,{b}^{4}}\sqrt{b\sqrt{x}+ax} \left ( 24\, \left ( b\sqrt{x}+ax \right ) ^{3/2}{a}^{7/2}{x}^{5/2}-6\,\sqrt{b\sqrt{x}+ax}{a}^{9/2}{x}^{7/2}-3\,\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}^{4}b-6\,{a}^{9/2}{x}^{7/2}\sqrt{\sqrt{x} \left ( b+a\sqrt{x} \right ) }+3\,\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}^{4}b+44\, \left ( b\sqrt{x}+ax \right ) ^{3/2}{a}^{5/2}{x}^{2}b-12\,\sqrt{b\sqrt{x}+ax}{a}^{7/2}{x}^{3}b-6\,\ln \left ( 1/2\,{\frac{2\,a\sqrt{x}+2\,\sqrt{b\sqrt{x}+ax}\sqrt{a}+b}{\sqrt{a}}} \right ){x}^{3}{a}^{3}{b}^{2}-12\,{a}^{7/2}{x}^{3}\sqrt{\sqrt{x} \left ( b+a\sqrt{x} \right ) }b-12\,{a}^{7/2}{x}^{5/2} \left ( \sqrt{x} \left ( b+a\sqrt{x} \right ) \right ) ^{3/2}+6\,\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}^{3}{a}^{3}{b}^{2}+16\, \left ( b\sqrt{x}+ax \right ) ^{3/2}{a}^{3/2}{x}^{3/2}{b}^{2}-6\,\sqrt{b\sqrt{x}+ax}{a}^{5/2}{x}^{5/2}{b}^{2}-3\,\ln \left ( 1/2\,{\frac{2\,a\sqrt{x}+2\,\sqrt{b\sqrt{x}+ax}\sqrt{a}+b}{\sqrt{a}}} \right ){x}^{5/2}{a}^{2}{b}^{3}-6\,{a}^{5/2}{x}^{5/2}\sqrt{\sqrt{x} \left ( b+a\sqrt{x} \right ) }{b}^{2}+3\,\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}^{5/2}{a}^{2}{b}^{3}-4\, \left ( b\sqrt{x}+ax \right ) ^{3/2}\sqrt{a}x{b}^{3} \right ){\frac{1}{\sqrt{\sqrt{x} \left ( b+a\sqrt{x} \right ) }}}{\frac{1}{\sqrt{a}}} \left ( b+a\sqrt{x} \right ) ^{-2}{x}^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(b*x^(1/2)+a*x)^(1/2)*(24*(b*x^(1/2)+a*x)^(3/2)*a^(7/2)*x^(5/2)-6*(b*x^(1/2)+a*x)^(1/2)*a^(9/2)*x^(7/2)-3*
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^4*b-6*a^(9/2)*x^(7/2)*(x^(1/2)*(b+a*
x^(1/2)))^(1/2)+3*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^4*b+44*(b*
x^(1/2)+a*x)^(3/2)*a^(5/2)*x^2*b-12*(b*x^(1/2)+a*x)^(1/2)*a^(7/2)*x^3*b-6*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^3*a^3*b^2-12*a^(7/2)*x^3*(x^(1/2)*(b+a*x^(1/2)))^(1/2)*b-12*a^(7/2)*x^(5/2)*(x^(
1/2)*(b+a*x^(1/2)))^(3/2)+6*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^3*a^3*b^
2+16*(b*x^(1/2)+a*x)^(3/2)*a^(3/2)*x^(3/2)*b^2-6*(b*x^(1/2)+a*x)^(1/2)*a^(5/2)*x^(5/2)*b^2-3*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^(5/2)*a^2*b^3-6*a^(5/2)*x^(5/2)*(x^(1/2)*(b+a*x^(1/2)))^(1/2)
*b^2+3*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^(5/2)*a^2*b^3-4*(b*x^(1/2)+a*
x)^(3/2)*a^(1/2)*x*b^3)/(x^(1/2)*(b+a*x^(1/2)))^(1/2)/b^4/a^(1/2)/(b+a*x^(1/2))^2/x^(5/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.87056, size = 132, normalized size = 1.67 \begin{align*} -\frac{4 \,{\left (4 \, a^{2} b x - b^{3} -{\left (8 \, a^{3} x - 5 \, a b^{2}\right )} \sqrt{x}\right )} \sqrt{a x + b \sqrt{x}}}{3 \,{\left (a^{2} b^{3} x^{2} - b^{5} x\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/3*(4*a^2*b*x - b^3 - (8*a^3*x - 5*a*b^2)*sqrt(x))*sqrt(a*x + b*sqrt(x))/(a^2*b^3*x^2 - b^5*x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x \left (a x + b \sqrt{x}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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