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

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

[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)

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Rubi [A]  time = 0.0742663, antiderivative size = 77, normalized size of antiderivative = 1., 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} \[ \frac{6 \sqrt{a x+b \sqrt{x}}}{a^2}-\frac{6 b \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{x}}{\sqrt{a x+b \sqrt{x}}}\right )}{a^{5/2}}-\frac{4 x}{a \sqrt{a x+b \sqrt{x}}} \]

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 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]

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 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 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 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])

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*}

Mathematica [C]  time = 0.0378811, size = 64, normalized size = 0.83 \[ \frac{4 x^{3/2} \sqrt{\frac{a \sqrt{x}}{b}+1} \, _2F_1\left (\frac{3}{2},\frac{5}{2};\frac{7}{2};-\frac{a \sqrt{x}}{b}\right )}{5 b \sqrt{a x+b \sqrt{x}}} \]

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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Maple [B]  time = 0.006, size = 237, normalized size = 3.1 \begin{align*} -{\sqrt{b\sqrt{x}+ax} \left ( 3\,\ln \left ( 1/2\,{\frac{2\,\sqrt{\sqrt{x} \left ( b+a\sqrt{x} \right ) }\sqrt{a}+2\,a\sqrt{x}+b}{\sqrt{a}}} \right ) x{a}^{2}b-6\,{a}^{5/2}x\sqrt{\sqrt{x} \left ( b+a\sqrt{x} \right ) }+6\,\ln \left ( 1/2\,{\frac{2\,\sqrt{\sqrt{x} \left ( b+a\sqrt{x} \right ) }\sqrt{a}+2\,a\sqrt{x}+b}{\sqrt{a}}} \right ) \sqrt{x}a{b}^{2}-12\,{a}^{3/2}\sqrt{x}\sqrt{\sqrt{x} \left ( b+a\sqrt{x} \right ) }b+4\,{a}^{3/2} \left ( \sqrt{x} \left ( b+a\sqrt{x} \right ) \right ) ^{3/2}+3\,\ln \left ( 1/2\,{\frac{2\,\sqrt{\sqrt{x} \left ( b+a\sqrt{x} \right ) }\sqrt{a}+2\,a\sqrt{x}+b}{\sqrt{a}}} \right ){b}^{3}-6\,\sqrt{a}\sqrt{\sqrt{x} \left ( b+a\sqrt{x} \right ) }{b}^{2} \right ){a}^{-{\frac{5}{2}}}{\frac{1}{\sqrt{\sqrt{x} \left ( b+a\sqrt{x} \right ) }}} \left ( b+a\sqrt{x} \right ) ^{-2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(b*x^(1/2)+a*x)^(1/2)/a^(5/2)*(3*ln(1/2*(2*(x^(1/2)*(b+a*x^(1/2)))^(1/2)*a^(1/2)+2*a*x^(1/2)+b)/a^(1/2))*x*a^
2*b-6*a^(5/2)*x*(x^(1/2)*(b+a*x^(1/2)))^(1/2)+6*ln(1/2*(2*(x^(1/2)*(b+a*x^(1/2)))^(1/2)*a^(1/2)+2*a*x^(1/2)+b)
/a^(1/2))*x^(1/2)*a*b^2-12*a^(3/2)*x^(1/2)*(x^(1/2)*(b+a*x^(1/2)))^(1/2)*b+4*a^(3/2)*(x^(1/2)*(b+a*x^(1/2)))^(
3/2)+3*ln(1/2*(2*(x^(1/2)*(b+a*x^(1/2)))^(1/2)*a^(1/2)+2*a*x^(1/2)+b)/a^(1/2))*b^3-6*a^(1/2)*(x^(1/2)*(b+a*x^(
1/2)))^(1/2)*b^2)/(x^(1/2)*(b+a*x^(1/2)))^(1/2)/(b+a*x^(1/2))^2

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

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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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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

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

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]

Exception raised: TypeError