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

Optimal. Leaf size=137 \[ \frac{512 a^3 \sqrt{a x+b \sqrt{x}}}{35 b^5 \sqrt{x}}-\frac{256 a^2 \sqrt{a x+b \sqrt{x}}}{35 b^4 x}+\frac{192 a \sqrt{a x+b \sqrt{x}}}{35 b^3 x^{3/2}}-\frac{32 \sqrt{a x+b \sqrt{x}}}{7 b^2 x^2}+\frac{4}{b x^{3/2} \sqrt{a x+b \sqrt{x}}} \]

[Out]

4/(b*x^(3/2)*Sqrt[b*Sqrt[x] + a*x]) - (32*Sqrt[b*Sqrt[x] + a*x])/(7*b^2*x^2) + (192*a*Sqrt[b*Sqrt[x] + a*x])/(
35*b^3*x^(3/2)) - (256*a^2*Sqrt[b*Sqrt[x] + a*x])/(35*b^4*x) + (512*a^3*Sqrt[b*Sqrt[x] + a*x])/(35*b^5*Sqrt[x]
)

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Rubi [A]  time = 0.201879, antiderivative size = 137, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {2015, 2016, 2014} \[ \frac{512 a^3 \sqrt{a x+b \sqrt{x}}}{35 b^5 \sqrt{x}}-\frac{256 a^2 \sqrt{a x+b \sqrt{x}}}{35 b^4 x}+\frac{192 a \sqrt{a x+b \sqrt{x}}}{35 b^3 x^{3/2}}-\frac{32 \sqrt{a x+b \sqrt{x}}}{7 b^2 x^2}+\frac{4}{b x^{3/2} \sqrt{a x+b \sqrt{x}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

4/(b*x^(3/2)*Sqrt[b*Sqrt[x] + a*x]) - (32*Sqrt[b*Sqrt[x] + a*x])/(7*b^2*x^2) + (192*a*Sqrt[b*Sqrt[x] + a*x])/(
35*b^3*x^(3/2)) - (256*a^2*Sqrt[b*Sqrt[x] + a*x])/(35*b^4*x) + (512*a^3*Sqrt[b*Sqrt[x] + a*x])/(35*b^5*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^2 \left (b \sqrt{x}+a x\right )^{3/2}} \, dx &=\frac{4}{b x^{3/2} \sqrt{b \sqrt{x}+a x}}+\frac{8 \int \frac{1}{x^{5/2} \sqrt{b \sqrt{x}+a x}} \, dx}{b}\\ &=\frac{4}{b x^{3/2} \sqrt{b \sqrt{x}+a x}}-\frac{32 \sqrt{b \sqrt{x}+a x}}{7 b^2 x^2}-\frac{(48 a) \int \frac{1}{x^2 \sqrt{b \sqrt{x}+a x}} \, dx}{7 b^2}\\ &=\frac{4}{b x^{3/2} \sqrt{b \sqrt{x}+a x}}-\frac{32 \sqrt{b \sqrt{x}+a x}}{7 b^2 x^2}+\frac{192 a \sqrt{b \sqrt{x}+a x}}{35 b^3 x^{3/2}}+\frac{\left (192 a^2\right ) \int \frac{1}{x^{3/2} \sqrt{b \sqrt{x}+a x}} \, dx}{35 b^3}\\ &=\frac{4}{b x^{3/2} \sqrt{b \sqrt{x}+a x}}-\frac{32 \sqrt{b \sqrt{x}+a x}}{7 b^2 x^2}+\frac{192 a \sqrt{b \sqrt{x}+a x}}{35 b^3 x^{3/2}}-\frac{256 a^2 \sqrt{b \sqrt{x}+a x}}{35 b^4 x}-\frac{\left (128 a^3\right ) \int \frac{1}{x \sqrt{b \sqrt{x}+a x}} \, dx}{35 b^4}\\ &=\frac{4}{b x^{3/2} \sqrt{b \sqrt{x}+a x}}-\frac{32 \sqrt{b \sqrt{x}+a x}}{7 b^2 x^2}+\frac{192 a \sqrt{b \sqrt{x}+a x}}{35 b^3 x^{3/2}}-\frac{256 a^2 \sqrt{b \sqrt{x}+a x}}{35 b^4 x}+\frac{512 a^3 \sqrt{b \sqrt{x}+a x}}{35 b^5 \sqrt{x}}\\ \end{align*}

Mathematica [A]  time = 0.0594126, size = 72, normalized size = 0.53 \[ \frac{4 \left (-16 a^2 b^2 x+64 a^3 b x^{3/2}+128 a^4 x^2+8 a b^3 \sqrt{x}-5 b^4\right )}{35 b^5 x^{3/2} \sqrt{a x+b \sqrt{x}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*(-5*b^4 + 8*a*b^3*Sqrt[x] - 16*a^2*b^2*x + 64*a^3*b*x^(3/2) + 128*a^4*x^2))/(35*b^5*x^(3/2)*Sqrt[b*Sqrt[x]
+ a*x])

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Maple [C]  time = 0.01, size = 570, normalized size = 4.2 \begin{align*}{\frac{1}{35\,{b}^{6}}\sqrt{b\sqrt{x}+ax} \left ( 560\, \left ( b\sqrt{x}+ax \right ) ^{3/2}{a}^{11/2}{x}^{9/2}-210\,\sqrt{b\sqrt{x}+ax}{a}^{13/2}{x}^{11/2}-105\,\ln \left ( 1/2\,{\frac{2\,a\sqrt{x}+2\,\sqrt{b\sqrt{x}+ax}\sqrt{a}+b}{\sqrt{a}}} \right ){x}^{11/2}{a}^{6}b-210\,{a}^{13/2}\sqrt{\sqrt{x} \left ( b+a\sqrt{x} \right ) }{x}^{11/2}+105\,\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}^{11/2}{a}^{6}b+256\, \left ( b\sqrt{x}+ax \right ) ^{3/2}{a}^{7/2}{x}^{7/2}{b}^{2}+932\, \left ( b\sqrt{x}+ax \right ) ^{3/2}{a}^{9/2}{x}^{4}b-420\,\sqrt{b\sqrt{x}+ax}{a}^{11/2}{x}^{5}b-210\,\ln \left ( 1/2\,{\frac{2\,a\sqrt{x}+2\,\sqrt{b\sqrt{x}+ax}\sqrt{a}+b}{\sqrt{a}}} \right ){x}^{5}{a}^{5}{b}^{2}-140\,{a}^{11/2} \left ( \sqrt{x} \left ( b+a\sqrt{x} \right ) \right ) ^{3/2}{x}^{9/2}-420\,{a}^{11/2}\sqrt{\sqrt{x} \left ( b+a\sqrt{x} \right ) }{x}^{5}b+210\,\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}{a}^{5}{b}^{2}-64\, \left ( b\sqrt{x}+ax \right ) ^{3/2}{a}^{5/2}{x}^{3}{b}^{3}-210\,\sqrt{b\sqrt{x}+ax}{a}^{9/2}{x}^{9/2}{b}^{2}-105\,\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}^{4}{b}^{3}-210\,{a}^{9/2}\sqrt{\sqrt{x} \left ( b+a\sqrt{x} \right ) }{x}^{9/2}{b}^{2}+105\,\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}^{4}{b}^{3}+32\, \left ( b\sqrt{x}+ax \right ) ^{3/2}{a}^{3/2}{x}^{5/2}{b}^{4}-20\, \left ( b\sqrt{x}+ax \right ) ^{3/2}\sqrt{a}{x}^{2}{b}^{5} \right ){\frac{1}{\sqrt{\sqrt{x} \left ( b+a\sqrt{x} \right ) }}}{\frac{1}{\sqrt{a}}}{x}^{-{\frac{9}{2}}} \left ( b+a\sqrt{x} \right ) ^{-2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/35*(b*x^(1/2)+a*x)^(1/2)*(560*(b*x^(1/2)+a*x)^(3/2)*a^(11/2)*x^(9/2)-210*(b*x^(1/2)+a*x)^(1/2)*a^(13/2)*x^(1
1/2)-105*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^(11/2)*a^6*b-210*a^(13/2)*(x^(1/2)*
(b+a*x^(1/2)))^(1/2)*x^(11/2)+105*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^(1
1/2)*a^6*b+256*(b*x^(1/2)+a*x)^(3/2)*a^(7/2)*x^(7/2)*b^2+932*(b*x^(1/2)+a*x)^(3/2)*a^(9/2)*x^4*b-420*(b*x^(1/2
)+a*x)^(1/2)*a^(11/2)*x^5*b-210*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*a^5*b^2-14
0*a^(11/2)*(x^(1/2)*(b+a*x^(1/2)))^(3/2)*x^(9/2)-420*a^(11/2)*(x^(1/2)*(b+a*x^(1/2)))^(1/2)*x^5*b+210*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*a^5*b^2-64*(b*x^(1/2)+a*x)^(3/2)*a^(5/2)*x
^3*b^3-210*(b*x^(1/2)+a*x)^(1/2)*a^(9/2)*x^(9/2)*b^2-105*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^4*b^3-210*a^(9/2)*(x^(1/2)*(b+a*x^(1/2)))^(1/2)*x^(9/2)*b^2+105*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^4*b^3+32*(b*x^(1/2)+a*x)^(3/2)*a^(3/2)*x^(5/2)*b^4-20
*(b*x^(1/2)+a*x)^(3/2)*a^(1/2)*x^2*b^5)/(x^(1/2)*(b+a*x^(1/2)))^(1/2)/b^6/a^(1/2)/x^(9/2)/(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^{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)^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.78855, size = 190, normalized size = 1.39 \begin{align*} -\frac{4 \,{\left (64 \, a^{4} b x^{2} - 24 \, a^{2} b^{3} x - 5 \, b^{5} -{\left (128 \, a^{5} x^{2} - 80 \, a^{3} b^{2} x - 13 \, a b^{4}\right )} \sqrt{x}\right )} \sqrt{a x + b \sqrt{x}}}{35 \,{\left (a^{2} b^{5} x^{3} - b^{7} x^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/35*(64*a^4*b*x^2 - 24*a^2*b^3*x - 5*b^5 - (128*a^5*x^2 - 80*a^3*b^2*x - 13*a*b^4)*sqrt(x))*sqrt(a*x + b*sqr
t(x))/(a^2*b^5*x^3 - b^7*x^2)

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

[Out]

Integral(1/(x**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^{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)^(3/2),x, algorithm="giac")

[Out]

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