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

Optimal. Leaf size=60 \[ \frac {4 \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {a x+b \sqrt {x}}}\right )}{a^{3/2}}-\frac {4 \sqrt {x}}{a \sqrt {a x+b \sqrt {x}}} \]

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Rubi [A]  time = 0.07, antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {2018, 652, 620, 206} \begin {gather*} \frac {4 \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {a x+b \sqrt {x}}}\right )}{a^{3/2}}-\frac {4 \sqrt {x}}{a \sqrt {a x+b \sqrt {x}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 652

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

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

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Mathematica [A]  time = 0.11, size = 79, normalized size = 1.32 \begin {gather*} \frac {4 \sqrt [4]{x} \left (\sqrt {b} \sqrt {\frac {a \sqrt {x}}{b}+1} \sinh ^{-1}\left (\frac {\sqrt {a} \sqrt [4]{x}}{\sqrt {b}}\right )-\sqrt {a} \sqrt [4]{x}\right )}{a^{3/2} \sqrt {a x+b \sqrt {x}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*x^(1/4)*(-(Sqrt[a]*x^(1/4)) + Sqrt[b]*Sqrt[1 + (a*Sqrt[x])/b]*ArcSinh[(Sqrt[a]*x^(1/4))/Sqrt[b]]))/(a^(3/2)
*Sqrt[b*Sqrt[x] + a*x])

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IntegrateAlgebraic [A]  time = 0.30, size = 76, normalized size = 1.27 \begin {gather*} -\frac {2 \log \left (-2 a^{3/2} \sqrt {a x+b \sqrt {x}}+2 a^2 \sqrt {x}+a b\right )}{a^{3/2}}-\frac {4 \sqrt {a x+b \sqrt {x}}}{a \left (a \sqrt {x}+b\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

[Out]

Timed out

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Unab
le to divide, perhaps due to rounding error%%%{%%%{1,[1]%%%},[2,2]%%%}+%%%{%%{[-2,0]:[1,0,%%%{-1,[1]%%%}]%%},[
1,3]%%%}+%%%{1,[0,4]%%%} / %%%{%%%{1,[2]%%%},[2,0]%%%}+%%%{%%{[%%%{-2,[1]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[1,1]
%%%}+%%%{%%%{1,[1]%%%},[0,2]%%%} Error: Bad Argument Value

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maple [B]  time = 0.05, size = 240, normalized size = 4.00 \begin {gather*} -\frac {2 \sqrt {a x +b \sqrt {x}}\, \left (-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 )-2 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 )-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 )+2 \sqrt {\left (a \sqrt {x}+b \right ) \sqrt {x}}\, a^{\frac {5}{2}} x +4 \sqrt {\left (a \sqrt {x}+b \right ) \sqrt {x}}\, a^{\frac {3}{2}} b \sqrt {x}+2 \sqrt {\left (a \sqrt {x}+b \right ) \sqrt {x}}\, \sqrt {a}\, b^{2}-2 \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 {3}{2}} b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(a*x+b*x^(1/2))^(1/2)/a^(3/2)*(2*((a*x^(1/2)+b)*x^(1/2))^(1/2)*a^(5/2)*x-a^2*b*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))+4*((a*x^(1/2)+b)*x^(1/2))^(1/2)*a^(3/2)*b*x^(1/2)-2*a*b^2*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))-2*((a*x^(1/2)+b)*x^(1/2))^(3/2)*a^(3
/2)+2*((a*x^(1/2)+b)*x^(1/2))^(1/2)*a^(1/2)*b^2-b^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)))/((a*x^(1/2)+b)*x^(1/2))^(1/2)/b/(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 {\sqrt {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^(1/2)/(b*x^(1/2)+a*x)^(3/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {\sqrt {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^(1/2)/(a*x + b*x^(1/2))^(3/2),x)

[Out]

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

[Out]

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

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