3.2.14 \(\int \frac {1}{x (b \sqrt {x}+a x)^{3/2}} \, dx\) [114]

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

[Out]

4/b/x^(1/2)/(b*x^(1/2)+a*x)^(1/2)-16/3*(b*x^(1/2)+a*x)^(1/2)/b^2/x+32/3*a*(b*x^(1/2)+a*x)^(1/2)/b^3/x^(1/2)

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Rubi [A]
time = 0.08, antiderivative size = 79, normalized size of antiderivative = 1.00, 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 = {2040, 2041, 2039} \begin {gather*} \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}}} \end {gather*}

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 2039

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] &&
NeQ[n, j] && EqQ[m + n*p + n - j + 1, 0] && (IntegerQ[j] || GtQ[c, 0])

Rule 2040

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 2041

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 \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*}

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Mathematica [A]
time = 0.15, size = 55, normalized size = 0.70 \begin {gather*} -\frac {4 \sqrt {b \sqrt {x}+a x} \left (b^2-4 a b \sqrt {x}-8 a^2 x\right )}{3 b^3 \left (b+a \sqrt {x}\right ) x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 3 vs. order 2.
time = 0.46, size = 524, normalized size = 6.63

method	result	size
derivativedivides	\(-\frac {4}{3 b \sqrt {x}\, \sqrt {b \sqrt {x}+a x}}+\frac {16 a \left (b +2 a \sqrt {x}\right )}{3 b^{3} \sqrt {b \sqrt {x}+a x}}\)	\(46\)
default	\(\frac {\sqrt {b \sqrt {x}+a x}\, \left (24 \left (b \sqrt {x}+a x \right )^{\frac {3}{2}} x^{\frac {5}{2}} a^{\frac {7}{2}}-6 \sqrt {b \sqrt {x}+a x}\, x^{\frac {7}{2}} a^{\frac {9}{2}}-6 x^{\frac {7}{2}} \sqrt {\sqrt {x}\, \left (a \sqrt {x}+b \right )}\, a^{\frac {9}{2}}-3 x^{\frac {7}{2}} \ln \left (\frac {2 a \sqrt {x}+2 \sqrt {b \sqrt {x}+a x}\, \sqrt {a}+b}{2 \sqrt {a}}\right ) a^{4} b +3 x^{\frac {7}{2}} \ln \left (\frac {2 a \sqrt {x}+2 \sqrt {\sqrt {x}\, \left (a \sqrt {x}+b \right )}\, \sqrt {a}+b}{2 \sqrt {a}}\right ) a^{4} b +44 \left (b \sqrt {x}+a x \right )^{\frac {3}{2}} x^{2} a^{\frac {5}{2}} b -12 \sqrt {b \sqrt {x}+a x}\, x^{3} a^{\frac {7}{2}} b -12 x^{3} \sqrt {\sqrt {x}\, \left (a \sqrt {x}+b \right )}\, a^{\frac {7}{2}} b -6 x^{3} \ln \left (\frac {2 a \sqrt {x}+2 \sqrt {b \sqrt {x}+a x}\, \sqrt {a}+b}{2 \sqrt {a}}\right ) a^{3} b^{2}+6 x^{3} \ln \left (\frac {2 a \sqrt {x}+2 \sqrt {\sqrt {x}\, \left (a \sqrt {x}+b \right )}\, \sqrt {a}+b}{2 \sqrt {a}}\right ) a^{3} b^{2}-12 x^{\frac {5}{2}} \left (\sqrt {x}\, \left (a \sqrt {x}+b \right )\right )^{\frac {3}{2}} a^{\frac {7}{2}}+16 \left (b \sqrt {x}+a x \right )^{\frac {3}{2}} x^{\frac {3}{2}} a^{\frac {3}{2}} b^{2}-6 \sqrt {b \sqrt {x}+a x}\, x^{\frac {5}{2}} a^{\frac {5}{2}} b^{2}-6 x^{\frac {5}{2}} \sqrt {\sqrt {x}\, \left (a \sqrt {x}+b \right )}\, a^{\frac {5}{2}} b^{2}-3 x^{\frac {5}{2}} \ln \left (\frac {2 a \sqrt {x}+2 \sqrt {b \sqrt {x}+a x}\, \sqrt {a}+b}{2 \sqrt {a}}\right ) a^{2} b^{3}+3 x^{\frac {5}{2}} \ln \left (\frac {2 a \sqrt {x}+2 \sqrt {\sqrt {x}\, \left (a \sqrt {x}+b \right )}\, \sqrt {a}+b}{2 \sqrt {a}}\right ) a^{2} b^{3}-4 \left (b \sqrt {x}+a x \right )^{\frac {3}{2}} \sqrt {a}\, b^{3} x \right )}{3 \sqrt {\sqrt {x}\, \left (a \sqrt {x}+b \right )}\, b^{4} \sqrt {a}\, \left (a \sqrt {x}+b \right )^{2} x^{\frac {5}{2}}}\)	\(524\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 = 2.94, size = 63, normalized size = 0.80 \begin {gather*} -\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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x \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/(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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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

[Out]

int(1/(x*(a*x + b*x^(1/2))^(3/2)), x)

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