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

Optimal. Leaf size=77 \[ -\frac {6 b \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {a x+b \sqrt {x}}}\right )}{a^{5/2}}+\frac {6 \sqrt {a x+b \sqrt {x}}}{a^2}-\frac {4 x}{a \sqrt {a x+b \sqrt {x}}} \]

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Rubi [A]  time = 0.07, antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {2018, 668, 640, 620, 206} \begin {gather*} \frac {6 \sqrt {a x+b \sqrt {x}}}{a^2}-\frac {6 b \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {a x+b \sqrt {x}}}\right )}{a^{5/2}}-\frac {4 x}{a \sqrt {a x+b \sqrt {x}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x/(b*Sqrt[x] + a*x)^(3/2),x]

[Out]

(-4*x)/(a*Sqrt[b*Sqrt[x] + a*x]) + (6*Sqrt[b*Sqrt[x] + a*x])/a^2 - (6*b*ArcTanh[(Sqrt[a]*Sqrt[x])/Sqrt[b*Sqrt[
x] + a*x]])/a^(5/2)

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 620

Int[1/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(1 - c*x^2), x], x, x/Sqrt[b*x + c*x^2
]], x] /; FreeQ[{b, c}, x]

Rule 640

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(a + b*x + c*x^2)^(p +
1))/(2*c*(p + 1)), x] + Dist[(2*c*d - b*e)/(2*c), Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}
, x] && NeQ[2*c*d - b*e, 0] && NeQ[p, -1]

Rule 668

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)
*(a + b*x + c*x^2)^(p + 1))/(c*(p + 1)), x] - Dist[(e^2*(m + p))/(c*(p + 1)), Int[(d + e*x)^(m - 2)*(a + b*x +
c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] &
& LtQ[p, -1] && GtQ[m, 1] && IntegerQ[2*p]

Rule 2018

Int[(x_)^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)
/n] - 1)*(a*x^Simplify[j/n] + b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[
n, j] && IntegerQ[Simplify[j/n]] && IntegerQ[Simplify[(m + 1)/n]] && NeQ[n^2, 1]

Rubi steps

\begin {align*} \int \frac {x}{\left (b \sqrt {x}+a x\right )^{3/2}} \, dx &=2 \operatorname {Subst}\left (\int \frac {x^3}{\left (b x+a x^2\right )^{3/2}} \, dx,x,\sqrt {x}\right )\\ &=-\frac {4 x}{a \sqrt {b \sqrt {x}+a x}}+\frac {6 \operatorname {Subst}\left (\int \frac {x}{\sqrt {b x+a x^2}} \, dx,x,\sqrt {x}\right )}{a}\\ &=-\frac {4 x}{a \sqrt {b \sqrt {x}+a x}}+\frac {6 \sqrt {b \sqrt {x}+a x}}{a^2}-\frac {(3 b) \operatorname {Subst}\left (\int \frac {1}{\sqrt {b x+a x^2}} \, dx,x,\sqrt {x}\right )}{a^2}\\ &=-\frac {4 x}{a \sqrt {b \sqrt {x}+a x}}+\frac {6 \sqrt {b \sqrt {x}+a x}}{a^2}-\frac {(6 b) \operatorname {Subst}\left (\int \frac {1}{1-a x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt {b \sqrt {x}+a x}}\right )}{a^2}\\ &=-\frac {4 x}{a \sqrt {b \sqrt {x}+a x}}+\frac {6 \sqrt {b \sqrt {x}+a x}}{a^2}-\frac {6 b \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b \sqrt {x}+a x}}\right )}{a^{5/2}}\\ \end {align*}

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Mathematica [C]  time = 0.04, size = 64, normalized size = 0.83 \begin {gather*} \frac {4 x^{3/2} \sqrt {\frac {a \sqrt {x}}{b}+1} \, _2F_1\left (\frac {3}{2},\frac {5}{2};\frac {7}{2};-\frac {a \sqrt {x}}{b}\right )}{5 b \sqrt {a x+b \sqrt {x}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*Sqrt[1 + (a*Sqrt[x])/b]*x^(3/2)*Hypergeometric2F1[3/2, 5/2, 7/2, -((a*Sqrt[x])/b)])/(5*b*Sqrt[b*Sqrt[x] + a
*x])

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IntegrateAlgebraic [A]  time = 0.39, size = 90, normalized size = 1.17 \begin {gather*} \frac {2 \sqrt {a x+b \sqrt {x}} \left (a \sqrt {x}+3 b\right )}{a^2 \left (a \sqrt {x}+b\right )}+\frac {3 b \log \left (-2 a^{5/2} \sqrt {a x+b \sqrt {x}}+2 a^3 \sqrt {x}+a^2 b\right )}{a^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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giac [A]  time = 0.32, size = 94, normalized size = 1.22 \begin {gather*} \frac {3 \, b \log \left ({\left | -2 \, \sqrt {a} {\left (\sqrt {a} \sqrt {x} - \sqrt {a x + b \sqrt {x}}\right )} - b \right |}\right )}{a^{\frac {5}{2}}} + \frac {4 \, b^{2}}{{\left (a {\left (\sqrt {a} \sqrt {x} - \sqrt {a x + b \sqrt {x}}\right )} + \sqrt {a} b\right )} a^{2}} + \frac {2 \, \sqrt {a x + b \sqrt {x}}}{a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*b*log(abs(-2*sqrt(a)*(sqrt(a)*sqrt(x) - sqrt(a*x + b*sqrt(x))) - b))/a^(5/2) + 4*b^2/((a*(sqrt(a)*sqrt(x) -
sqrt(a*x + b*sqrt(x))) + sqrt(a)*b)*a^2) + 2*sqrt(a*x + b*sqrt(x))/a^2

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maple [B]  time = 0.05, size = 236, normalized size = 3.06 \begin {gather*} \frac {\sqrt {a x +b \sqrt {x}}\, \left (-3 a^{2} b x \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 \,b^{2} \sqrt {x}\, \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 b^{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 \sqrt {\left (a \sqrt {x}+b \right ) \sqrt {x}}\, a^{\frac {5}{2}} x +12 \sqrt {\left (a \sqrt {x}+b \right ) \sqrt {x}}\, a^{\frac {3}{2}} b \sqrt {x}+6 \sqrt {\left (a \sqrt {x}+b \right ) \sqrt {x}}\, \sqrt {a}\, b^{2}-4 \left (\left (a \sqrt {x}+b \right ) \sqrt {x}\right )^{\frac {3}{2}} a^{\frac {3}{2}}\right )}{\sqrt {\left (a \sqrt {x}+b \right ) \sqrt {x}}\, \left (a \sqrt {x}+b \right )^{2} a^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a*x+b*x^(1/2))^(1/2)/a^(5/2)*(6*x*((a*x^(1/2)+b)*x^(1/2))^(1/2)*a^(5/2)-3*x*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^2*b+12*x^(1/2)*((a*x^(1/2)+b)*x^(1/2))^(1/2)*a^(3/2)*b-6*x^(1/2)*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*b^2-4*a^(3/2)*((a*x^(1/2)+b)*x^(1/2))^(3
/2)+6*((a*x^(1/2)+b)*x^(1/2))^(1/2)*a^(1/2)*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))*b^3)/((a*x^(1/2)+b)*x^(1/2))^(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 {x}{{\left (a x + b \sqrt {x}\right )}^{\frac {3}{2}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

[Out]

Integral(x/(a*x + b*sqrt(x))**(3/2), x)

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