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

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

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Rubi [A]  time = 0.16, antiderivative size = 107, normalized size of antiderivative = 1.00, 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} \begin {gather*} -\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}}} \end {gather*}

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 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^{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*}

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Mathematica [A]  time = 0.05, size = 57, normalized size = 0.53 \begin {gather*} -\frac {4 \left (16 a^3 x^{3/2}+8 a^2 b x-2 a b^2 \sqrt {x}+b^3\right )}{5 b^4 x \sqrt {a x+b \sqrt {x}}} \end {gather*}

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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IntegrateAlgebraic [A]  time = 0.31, size = 70, normalized size = 0.65 \begin {gather*} -\frac {4 \sqrt {a x+b \sqrt {x}} \left (16 a^3 x^{3/2}+8 a^2 b x-2 a b^2 \sqrt {x}+b^3\right )}{5 b^4 x^{3/2} \left (a \sqrt {x}+b\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.65, size = 79, normalized size = 0.74 \begin {gather*} \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 {gather*}

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

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)

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maple [C]  time = 0.06, size = 548, normalized size = 5.12 \begin {gather*} \frac {2 \sqrt {a x +b \sqrt {x}}\, \left (-5 a^{5} b \,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 )+5 a^{5} b \,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 )-10 a^{4} b^{2} x^{4} \ln \left (\frac {2 a \sqrt {x}+b +2 \sqrt {\left (a \sqrt {x}+b \right ) \sqrt {x}}\, \sqrt {a}}{2 \sqrt {a}}\right )+10 a^{4} b^{2} x^{4} \ln \left (\frac {2 a \sqrt {x}+b +2 \sqrt {a x +b \sqrt {x}}\, \sqrt {a}}{2 \sqrt {a}}\right )-5 a^{3} b^{3} x^{\frac {7}{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 )+5 a^{3} b^{3} x^{\frac {7}{2}} \ln \left (\frac {2 a \sqrt {x}+b +2 \sqrt {a x +b \sqrt {x}}\, \sqrt {a}}{2 \sqrt {a}}\right )+10 \sqrt {a x +b \sqrt {x}}\, a^{\frac {11}{2}} x^{\frac {9}{2}}+10 \sqrt {\left (a \sqrt {x}+b \right ) \sqrt {x}}\, a^{\frac {11}{2}} x^{\frac {9}{2}}+20 \sqrt {a x +b \sqrt {x}}\, a^{\frac {9}{2}} b \,x^{4}+20 \sqrt {\left (a \sqrt {x}+b \right ) \sqrt {x}}\, a^{\frac {9}{2}} b \,x^{4}+10 \sqrt {a x +b \sqrt {x}}\, a^{\frac {7}{2}} b^{2} x^{\frac {7}{2}}+10 \sqrt {\left (a \sqrt {x}+b \right ) \sqrt {x}}\, a^{\frac {7}{2}} b^{2} x^{\frac {7}{2}}-30 \left (a x +b \sqrt {x}\right )^{\frac {3}{2}} a^{\frac {9}{2}} x^{\frac {7}{2}}+10 \left (\left (a \sqrt {x}+b \right ) \sqrt {x}\right )^{\frac {3}{2}} a^{\frac {9}{2}} x^{\frac {7}{2}}-52 \left (a x +b \sqrt {x}\right )^{\frac {3}{2}} a^{\frac {7}{2}} b \,x^{3}-16 \left (a x +b \sqrt {x}\right )^{\frac {3}{2}} a^{\frac {5}{2}} b^{2} x^{\frac {5}{2}}+4 \left (a x +b \sqrt {x}\right )^{\frac {3}{2}} a^{\frac {3}{2}} b^{3} x^{2}-2 \left (a x +b \sqrt {x}\right )^{\frac {3}{2}} \sqrt {a}\, b^{4} x^{\frac {3}{2}}\right )}{5 \sqrt {\left (a \sqrt {x}+b \right ) \sqrt {x}}\, \left (a \sqrt {x}+b \right )^{2} \sqrt {a}\, b^{5} x^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*(a*x+b*x^(1/2))^(1/2)*(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))*x^(9/2)*a^5*b-30*
(a*x+b*x^(1/2))^(3/2)*a^(9/2)*x^(7/2)+10*(a*x+b*x^(1/2))^(1/2)*a^(11/2)*x^(9/2)-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))*x^(9/2)*a^5*b+10*a^(11/2)*x^(9/2)*((a*x^(1/2)+b)*x^(1/2))^(1/2)+10
*ln(1/2*(2*a*x^(1/2)+b+2*(a*x+b*x^(1/2))^(1/2)*a^(1/2))/a^(1/2))*x^4*a^4*b^2-16*(a*x+b*x^(1/2))^(3/2)*a^(5/2)*
x^(5/2)*b^2-52*(a*x+b*x^(1/2))^(3/2)*a^(7/2)*x^3*b+20*(a*x+b*x^(1/2))^(1/2)*a^(9/2)*x^4*b-10*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))*x^4*a^4*b^2+20*a^(9/2)*x^4*((a*x^(1/2)+b)*x^(1/2))^(1/2
)*b+10*a^(9/2)*x^(7/2)*((a*x^(1/2)+b)*x^(1/2))^(3/2)+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))*x^(7/2)*a^3*b^3+4*(a*x+b*x^(1/2))^(3/2)*a^(3/2)*x^2*b^3+10*(a*x+b*x^(1/2))^(1/2)*a^(7/2)*x^(7/2)*b^2-
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))*x^(7/2)*a^3*b^3+10*a^(7/2)*x^(7/2)*(
(a*x^(1/2)+b)*x^(1/2))^(1/2)*b^2-2*(a*x+b*x^(1/2))^(3/2)*a^(1/2)*x^(3/2)*b^4)/((a*x^(1/2)+b)*x^(1/2))^(1/2)/b^
5/x^(7/2)/a^(1/2)/(a*x^(1/2)+b)^2

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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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