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)/(b*x^(1/2)+a*x)^(1/2)-32/7*(b*x^(1/2)+a*x)^(1/2)/b^2/x^2+192/35*a*(b*x^(1/2)+a*x)^(1/2)/b^3/x^(3/2
)-256/35*a^2*(b*x^(1/2)+a*x)^(1/2)/b^4/x+512/35*a^3*(b*x^(1/2)+a*x)^(1/2)/b^5/x^(1/2)

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Rubi [A]  time = 0.20, antiderivative size = 137, normalized size of antiderivative = 1.00, 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 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 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])

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*}

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Mathematica [A]  time = 0.08, size = 72, normalized size = 0.53 \[ \frac {4 \left (128 a^4 x^2+64 a^3 b x^{3/2}-16 a^2 b^2 x+8 a b^3 \sqrt {x}-5 b^4\right )}{35 b^5 x^{3/2} \sqrt {a x+b \sqrt {x}}} \]

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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fricas [A]  time = 0.75, size = 87, normalized size = 0.64 \[ -\frac {4 \, {\left (64 \, a^{4} b x^{2} - 24 \, a^{2} b^{3} x - 5 \, b^{5} - {\left (128 \, a^{5} x^{2} - 80 \, a^{3} b^{2} x - 13 \, a b^{4}\right )} \sqrt {x}\right )} \sqrt {a x + b \sqrt {x}}}{35 \, {\left (a^{2} b^{5} x^{3} - b^{7} x^{2}\right )}} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (a x + b \sqrt {x}\right )}^{\frac {3}{2}} x^{2}}\,{d x} \]

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)

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maple [C]  time = 0.06, size = 570, normalized size = 4.16 \[ -\frac {\sqrt {a x +b \sqrt {x}}\, \left (-105 a^{6} b \,x^{\frac {11}{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 )+105 a^{6} b \,x^{\frac {11}{2}} \ln \left (\frac {2 a \sqrt {x}+b +2 \sqrt {a x +b \sqrt {x}}\, \sqrt {a}}{2 \sqrt {a}}\right )-210 a^{5} b^{2} x^{5} \ln \left (\frac {2 a \sqrt {x}+b +2 \sqrt {\left (a \sqrt {x}+b \right ) \sqrt {x}}\, \sqrt {a}}{2 \sqrt {a}}\right )+210 a^{5} b^{2} x^{5} \ln \left (\frac {2 a \sqrt {x}+b +2 \sqrt {a x +b \sqrt {x}}\, \sqrt {a}}{2 \sqrt {a}}\right )-105 a^{4} b^{3} x^{\frac {9}{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 )+105 a^{4} b^{3} x^{\frac {9}{2}} \ln \left (\frac {2 a \sqrt {x}+b +2 \sqrt {a x +b \sqrt {x}}\, \sqrt {a}}{2 \sqrt {a}}\right )+210 \sqrt {\left (a \sqrt {x}+b \right ) \sqrt {x}}\, a^{\frac {13}{2}} x^{\frac {11}{2}}+210 \sqrt {a x +b \sqrt {x}}\, a^{\frac {13}{2}} x^{\frac {11}{2}}+420 \sqrt {\left (a \sqrt {x}+b \right ) \sqrt {x}}\, a^{\frac {11}{2}} b \,x^{5}+420 \sqrt {a x +b \sqrt {x}}\, a^{\frac {11}{2}} b \,x^{5}+210 \sqrt {\left (a \sqrt {x}+b \right ) \sqrt {x}}\, a^{\frac {9}{2}} b^{2} x^{\frac {9}{2}}+210 \sqrt {a x +b \sqrt {x}}\, a^{\frac {9}{2}} b^{2} x^{\frac {9}{2}}-560 \left (a x +b \sqrt {x}\right )^{\frac {3}{2}} a^{\frac {11}{2}} x^{\frac {9}{2}}+140 \left (\left (a \sqrt {x}+b \right ) \sqrt {x}\right )^{\frac {3}{2}} a^{\frac {11}{2}} x^{\frac {9}{2}}-932 \left (a x +b \sqrt {x}\right )^{\frac {3}{2}} a^{\frac {9}{2}} b \,x^{4}-256 \left (a x +b \sqrt {x}\right )^{\frac {3}{2}} a^{\frac {7}{2}} b^{2} x^{\frac {7}{2}}+64 \left (a x +b \sqrt {x}\right )^{\frac {3}{2}} a^{\frac {5}{2}} b^{3} x^{3}-32 \left (a x +b \sqrt {x}\right )^{\frac {3}{2}} a^{\frac {3}{2}} b^{4} x^{\frac {5}{2}}+20 \left (a x +b \sqrt {x}\right )^{\frac {3}{2}} \sqrt {a}\, b^{5} x^{2}\right )}{35 \sqrt {\left (a \sqrt {x}+b \right ) \sqrt {x}}\, \left (a \sqrt {x}+b \right )^{2} \sqrt {a}\, b^{6} x^{\frac {9}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/35*(a*x+b*x^(1/2))^(1/2)*(210*x^(11/2)*((a*x^(1/2)+b)*x^(1/2))^(1/2)*a^(13/2)-560*x^(9/2)*(a*x+b*x^(1/2))^(
3/2)*a^(11/2)+210*x^(11/2)*(a*x+b*x^(1/2))^(1/2)*a^(13/2)-105*x^(11/2)*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^6*b+105*x^(11/2)*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^6*b-256*x^(7/2)*(a*x+b*x^(1/2))^(3/2)*a^(7/2)*b^2+420*x^5*((a*x^(1/2)+b)*x^(1/2))^(1/2)*a^(11/2)*b-932
*x^4*(a*x+b*x^(1/2))^(3/2)*a^(9/2)*b+420*x^5*(a*x+b*x^(1/2))^(1/2)*a^(11/2)*b-210*x^5*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^5*b^2+210*x^5*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^5*b^2+140*x^(9/2)*((a*x^(1/2)+b)*x^(1/2))^(3/2)*a^(11/2)+64*x^3*(a*x+b*x^(1/2))^(3/2)*a^(5
/2)*b^3+210*x^(9/2)*((a*x^(1/2)+b)*x^(1/2))^(1/2)*a^(9/2)*b^2+210*x^(9/2)*(a*x+b*x^(1/2))^(1/2)*a^(9/2)*b^2-10
5*x^(9/2)*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^4*b^3+105*x^(9/2)*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^4*b^3-32*x^(5/2)*(a*x+b*x^(1/2))^(3/2)*a^(3/2)*b^4+2
0*(a*x+b*x^(1/2))^(3/2)*a^(1/2)*x^2*b^5)/((a*x^(1/2)+b)*x^(1/2))^(1/2)/b^6/x^(9/2)/a^(1/2)/(a*x^(1/2)+b)^2

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (a x + b \sqrt {x}\right )}^{\frac {3}{2}} x^{2}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{x^2\,{\left (a\,x+b\,\sqrt {x}\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{x^{2} \left (a x + b \sqrt {x}\right )^{\frac {3}{2}}}\, dx \]

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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