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

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

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Rubi [A]  time = 0.12, 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 = {2015, 2016, 2014} \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 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 \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 = 0.08, size = 48, normalized size = 0.61 \begin {gather*} \frac {4 \left (8 a^2 x+4 a b \sqrt {x}-b^2\right )}{3 b^3 \sqrt {x} \sqrt {a x+b \sqrt {x}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.30, size = 57, normalized size = 0.72 \begin {gather*} \frac {4 \sqrt {a x+b \sqrt {x}} \left (8 a^2 x+4 a b \sqrt {x}-b^2\right )}{3 b^3 x \left (a \sqrt {x}+b\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[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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fricas [A]  time = 0.82, 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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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}\,{d x} \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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maple [C]  time = 0.07, size = 524, normalized size = 6.63 \begin {gather*} \frac {\sqrt {a x +b \sqrt {x}}\, \left (3 a^{4} b \,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 )-3 a^{4} b \,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 )+6 a^{3} b^{2} x^{3} \ln \left (\frac {2 a \sqrt {x}+b +2 \sqrt {\left (a \sqrt {x}+b \right ) \sqrt {x}}\, \sqrt {a}}{2 \sqrt {a}}\right )-6 a^{3} b^{2} x^{3} \ln \left (\frac {2 a \sqrt {x}+b +2 \sqrt {a x +b \sqrt {x}}\, \sqrt {a}}{2 \sqrt {a}}\right )+3 a^{2} b^{3} x^{\frac {5}{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 )-3 a^{2} b^{3} x^{\frac {5}{2}} \ln \left (\frac {2 a \sqrt {x}+b +2 \sqrt {a x +b \sqrt {x}}\, \sqrt {a}}{2 \sqrt {a}}\right )-6 \sqrt {a x +b \sqrt {x}}\, a^{\frac {9}{2}} x^{\frac {7}{2}}-6 \sqrt {\left (a \sqrt {x}+b \right ) \sqrt {x}}\, a^{\frac {9}{2}} x^{\frac {7}{2}}-12 \sqrt {a x +b \sqrt {x}}\, a^{\frac {7}{2}} b \,x^{3}-12 \sqrt {\left (a \sqrt {x}+b \right ) \sqrt {x}}\, a^{\frac {7}{2}} b \,x^{3}-6 \sqrt {a x +b \sqrt {x}}\, a^{\frac {5}{2}} b^{2} x^{\frac {5}{2}}-6 \sqrt {\left (a \sqrt {x}+b \right ) \sqrt {x}}\, a^{\frac {5}{2}} b^{2} x^{\frac {5}{2}}+24 \left (a x +b \sqrt {x}\right )^{\frac {3}{2}} a^{\frac {7}{2}} x^{\frac {5}{2}}-12 \left (\left (a \sqrt {x}+b \right ) \sqrt {x}\right )^{\frac {3}{2}} a^{\frac {7}{2}} x^{\frac {5}{2}}+44 \left (a x +b \sqrt {x}\right )^{\frac {3}{2}} a^{\frac {5}{2}} b \,x^{2}+16 \left (a x +b \sqrt {x}\right )^{\frac {3}{2}} a^{\frac {3}{2}} b^{2} x^{\frac {3}{2}}-4 \left (a x +b \sqrt {x}\right )^{\frac {3}{2}} \sqrt {a}\, b^{3} x \right )}{3 \sqrt {\left (a \sqrt {x}+b \right ) \sqrt {x}}\, \left (a \sqrt {x}+b \right )^{2} \sqrt {a}\, b^{4} x^{\frac {5}{2}}} \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]

1/3*(a*x+b*x^(1/2))^(1/2)*(24*(a*x+b*x^(1/2))^(3/2)*a^(7/2)*x^(5/2)-6*(a*x+b*x^(1/2))^(1/2)*a^(9/2)*x^(7/2)+3*
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^4*b-6*a^(9/2)*x^(7/2)*((a*x^
(1/2)+b)*x^(1/2))^(1/2)-3*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^4*b+44*(a*
x+b*x^(1/2))^(3/2)*a^(5/2)*x^2*b-12*(a*x+b*x^(1/2))^(1/2)*a^(7/2)*x^3*b+6*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^3*a^3*b^2-12*a^(7/2)*x^3*((a*x^(1/2)+b)*x^(1/2))^(1/2)*b-12*a^(7/2)*x^(5
/2)*((a*x^(1/2)+b)*x^(1/2))^(3/2)-6*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^3*a^3*b^
2+16*(a*x+b*x^(1/2))^(3/2)*a^(3/2)*x^(3/2)*b^2-6*(a*x+b*x^(1/2))^(1/2)*a^(5/2)*x^(5/2)*b^2+3*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^(5/2)*a^2*b^3-6*a^(5/2)*x^(5/2)*((a*x^(1/2)+b)*x^(1/2
))^(1/2)*b^2-3*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^(5/2)*a^2*b^3-4*(a*x+b*x^(1/2
))^(3/2)*a^(1/2)*x*b^3)/((a*x^(1/2)+b)*x^(1/2))^(1/2)/b^4/a^(1/2)/x^(5/2)/(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}\,{d x} \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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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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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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