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

Optimal. Leaf size=107 \[ -\frac{64 a^2 \sqrt{a x+b \sqrt{x}}}{5 b^4 \sqrt{x}}-\frac{24 \sqrt{a x+b \sqrt{x}}}{5 b^2 x^{3/2}}+\frac{32 a \sqrt{a x+b \sqrt{x}}}{5 b^3 x}+\frac{4}{b x \sqrt{a x+b \sqrt{x}}} \]

[Out]

4/(b*x*Sqrt[b*Sqrt[x] + a*x]) - (24*Sqrt[b*Sqrt[x] + a*x])/(5*b^2*x^(3/2)) + (32*a*Sqrt[b*Sqrt[x] + a*x])/(5*b
^3*x) - (64*a^2*Sqrt[b*Sqrt[x] + a*x])/(5*b^4*Sqrt[x])

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Rubi [A]  time = 0.156028, antiderivative size = 107, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {2015, 2016, 2014} \[ -\frac{64 a^2 \sqrt{a x+b \sqrt{x}}}{5 b^4 \sqrt{x}}-\frac{24 \sqrt{a x+b \sqrt{x}}}{5 b^2 x^{3/2}}+\frac{32 a \sqrt{a x+b \sqrt{x}}}{5 b^3 x}+\frac{4}{b x \sqrt{a x+b \sqrt{x}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

4/(b*x*Sqrt[b*Sqrt[x] + a*x]) - (24*Sqrt[b*Sqrt[x] + a*x])/(5*b^2*x^(3/2)) + (32*a*Sqrt[b*Sqrt[x] + a*x])/(5*b
^3*x) - (64*a^2*Sqrt[b*Sqrt[x] + a*x])/(5*b^4*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^{3/2} \left (b \sqrt{x}+a x\right )^{3/2}} \, dx &=\frac{4}{b x \sqrt{b \sqrt{x}+a x}}+\frac{6 \int \frac{1}{x^2 \sqrt{b \sqrt{x}+a x}} \, dx}{b}\\ &=\frac{4}{b x \sqrt{b \sqrt{x}+a x}}-\frac{24 \sqrt{b \sqrt{x}+a x}}{5 b^2 x^{3/2}}-\frac{(24 a) \int \frac{1}{x^{3/2} \sqrt{b \sqrt{x}+a x}} \, dx}{5 b^2}\\ &=\frac{4}{b x \sqrt{b \sqrt{x}+a x}}-\frac{24 \sqrt{b \sqrt{x}+a x}}{5 b^2 x^{3/2}}+\frac{32 a \sqrt{b \sqrt{x}+a x}}{5 b^3 x}+\frac{\left (16 a^2\right ) \int \frac{1}{x \sqrt{b \sqrt{x}+a x}} \, dx}{5 b^3}\\ &=\frac{4}{b x \sqrt{b \sqrt{x}+a x}}-\frac{24 \sqrt{b \sqrt{x}+a x}}{5 b^2 x^{3/2}}+\frac{32 a \sqrt{b \sqrt{x}+a x}}{5 b^3 x}-\frac{64 a^2 \sqrt{b \sqrt{x}+a x}}{5 b^4 \sqrt{x}}\\ \end{align*}

Mathematica [A]  time = 0.0506263, size = 57, normalized size = 0.53 \[ -\frac{4 \left (8 a^2 b x+16 a^3 x^{3/2}-2 a b^2 \sqrt{x}+b^3\right )}{5 b^4 x \sqrt{a x+b \sqrt{x}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*(b^3 - 2*a*b^2*Sqrt[x] + 8*a^2*b*x + 16*a^3*x^(3/2)))/(5*b^4*x*Sqrt[b*Sqrt[x] + a*x])

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Maple [C]  time = 0.011, size = 548, normalized size = 5.1 \begin{align*} -{\frac{2}{5\,{b}^{5}}\sqrt{b\sqrt{x}+ax} \left ( 30\, \left ( b\sqrt{x}+ax \right ) ^{3/2}{a}^{9/2}{x}^{7/2}-10\,\sqrt{b\sqrt{x}+ax}{a}^{11/2}{x}^{9/2}-5\,\ln \left ( 1/2\,{\frac{2\,a\sqrt{x}+2\,\sqrt{b\sqrt{x}+ax}\sqrt{a}+b}{\sqrt{a}}} \right ){x}^{9/2}{a}^{5}b-10\,{a}^{11/2}{x}^{9/2}\sqrt{\sqrt{x} \left ( b+a\sqrt{x} \right ) }+5\,\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}^{9/2}{a}^{5}b+16\, \left ( b\sqrt{x}+ax \right ) ^{3/2}{a}^{5/2}{x}^{5/2}{b}^{2}+52\, \left ( b\sqrt{x}+ax \right ) ^{3/2}{a}^{7/2}{x}^{3}b-20\,\sqrt{b\sqrt{x}+ax}{a}^{9/2}{x}^{4}b-10\,\ln \left ( 1/2\,{\frac{2\,a\sqrt{x}+2\,\sqrt{b\sqrt{x}+ax}\sqrt{a}+b}{\sqrt{a}}} \right ){x}^{4}{a}^{4}{b}^{2}-20\,{a}^{9/2}{x}^{4}\sqrt{\sqrt{x} \left ( b+a\sqrt{x} \right ) }b-10\,{a}^{9/2}{x}^{7/2} \left ( \sqrt{x} \left ( b+a\sqrt{x} \right ) \right ) ^{3/2}+10\,\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}^{4}{a}^{4}{b}^{2}-4\, \left ( b\sqrt{x}+ax \right ) ^{3/2}{a}^{3/2}{x}^{2}{b}^{3}-10\,\sqrt{b\sqrt{x}+ax}{a}^{7/2}{x}^{7/2}{b}^{2}-5\,\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}^{3}-10\,{a}^{7/2}{x}^{7/2}\sqrt{\sqrt{x} \left ( b+a\sqrt{x} \right ) }{b}^{2}+5\,\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}^{3}+2\, \left ( b\sqrt{x}+ax \right ) ^{3/2}\sqrt{a}{x}^{3/2}{b}^{4} \right ){\frac{1}{\sqrt{\sqrt{x} \left ( b+a\sqrt{x} \right ) }}}{\frac{1}{\sqrt{a}}}{x}^{-{\frac{7}{2}}} \left ( b+a\sqrt{x} \right ) ^{-2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/5*(b*x^(1/2)+a*x)^(1/2)*(30*(b*x^(1/2)+a*x)^(3/2)*a^(9/2)*x^(7/2)-10*(b*x^(1/2)+a*x)^(1/2)*a^(11/2)*x^(9/2)
-5*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^(9/2)*a^5*b-10*a^(11/2)*x^(9/2)*(x^(1/2)*
(b+a*x^(1/2)))^(1/2)+5*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^(9/2)*a^5*b+1
6*(b*x^(1/2)+a*x)^(3/2)*a^(5/2)*x^(5/2)*b^2+52*(b*x^(1/2)+a*x)^(3/2)*a^(7/2)*x^3*b-20*(b*x^(1/2)+a*x)^(1/2)*a^
(9/2)*x^4*b-10*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^4*a^4*b^2-20*a^(9/2)*x^4*(x^(
1/2)*(b+a*x^(1/2)))^(1/2)*b-10*a^(9/2)*x^(7/2)*(x^(1/2)*(b+a*x^(1/2)))^(3/2)+10*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^4*a^4*b^2-4*(b*x^(1/2)+a*x)^(3/2)*a^(3/2)*x^2*b^3-10*(b*x^(1/2)+a*
x)^(1/2)*a^(7/2)*x^(7/2)*b^2-5*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^3
-10*a^(7/2)*x^(7/2)*(x^(1/2)*(b+a*x^(1/2)))^(1/2)*b^2+5*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^3+2*(b*x^(1/2)+a*x)^(3/2)*a^(1/2)*x^(3/2)*b^4)/(x^(1/2)*(b+a*x^(1/2)))^(1/2)/b
^5/a^(1/2)/x^(7/2)/(b+a*x^(1/2))^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^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.34268, size = 163, normalized size = 1.52 \begin{align*} \frac{4 \,{\left (8 \, a^{3} b x^{2} - 3 \, a b^{3} x -{\left (16 \, a^{4} x^{2} - 10 \, a^{2} b^{2} x - b^{4}\right )} \sqrt{x}\right )} \sqrt{a x + b \sqrt{x}}}{5 \,{\left (a^{2} b^{4} x^{3} - b^{6} x^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/5*(8*a^3*b*x^2 - 3*a*b^3*x - (16*a^4*x^2 - 10*a^2*b^2*x - b^4)*sqrt(x))*sqrt(a*x + b*sqrt(x))/(a^2*b^4*x^3 -
b^6*x^2)

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

[Out]

Integral(1/(x**(3/2)*(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^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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