3.104 \(\int \frac{x}{\sqrt{b \sqrt{x}+a x}} \, dx\)

Optimal. Leaf size=116 \[ \frac{5 b^2 \sqrt{a x+b \sqrt{x}}}{4 a^3}-\frac{5 b^3 \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{x}}{\sqrt{a x+b \sqrt{x}}}\right )}{4 a^{7/2}}-\frac{5 b \sqrt{x} \sqrt{a x+b \sqrt{x}}}{6 a^2}+\frac{2 x \sqrt{a x+b \sqrt{x}}}{3 a} \]

[Out]

(5*b^2*Sqrt[b*Sqrt[x] + a*x])/(4*a^3) - (5*b*Sqrt[x]*Sqrt[b*Sqrt[x] + a*x])/(6*a^2) + (2*x*Sqrt[b*Sqrt[x] + a*
x])/(3*a) - (5*b^3*ArcTanh[(Sqrt[a]*Sqrt[x])/Sqrt[b*Sqrt[x] + a*x]])/(4*a^(7/2))

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Rubi [A]  time = 0.0868686, antiderivative size = 116, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.294, Rules used = {2018, 670, 640, 620, 206} \[ \frac{5 b^2 \sqrt{a x+b \sqrt{x}}}{4 a^3}-\frac{5 b^3 \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{x}}{\sqrt{a x+b \sqrt{x}}}\right )}{4 a^{7/2}}-\frac{5 b \sqrt{x} \sqrt{a x+b \sqrt{x}}}{6 a^2}+\frac{2 x \sqrt{a x+b \sqrt{x}}}{3 a} \]

Antiderivative was successfully verified.

[In]

Int[x/Sqrt[b*Sqrt[x] + a*x],x]

[Out]

(5*b^2*Sqrt[b*Sqrt[x] + a*x])/(4*a^3) - (5*b*Sqrt[x]*Sqrt[b*Sqrt[x] + a*x])/(6*a^2) + (2*x*Sqrt[b*Sqrt[x] + a*
x])/(3*a) - (5*b^3*ArcTanh[(Sqrt[a]*Sqrt[x])/Sqrt[b*Sqrt[x] + a*x]])/(4*a^(7/2))

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 670

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*(m + 2*p + 1)), x] + Dist[((m + p)*(2*c*d - b*e))/(c*(m + 2*p + 1)), Int[(d + e
*x)^(m - 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{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] && GtQ[m, 1] && NeQ[m + 2*p + 1, 0] && 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}{\sqrt{b \sqrt{x}+a x}} \, dx &=2 \operatorname{Subst}\left (\int \frac{x^3}{\sqrt{b x+a x^2}} \, dx,x,\sqrt{x}\right )\\ &=\frac{2 x \sqrt{b \sqrt{x}+a x}}{3 a}-\frac{(5 b) \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{b x+a x^2}} \, dx,x,\sqrt{x}\right )}{3 a}\\ &=-\frac{5 b \sqrt{x} \sqrt{b \sqrt{x}+a x}}{6 a^2}+\frac{2 x \sqrt{b \sqrt{x}+a x}}{3 a}+\frac{\left (5 b^2\right ) \operatorname{Subst}\left (\int \frac{x}{\sqrt{b x+a x^2}} \, dx,x,\sqrt{x}\right )}{4 a^2}\\ &=\frac{5 b^2 \sqrt{b \sqrt{x}+a x}}{4 a^3}-\frac{5 b \sqrt{x} \sqrt{b \sqrt{x}+a x}}{6 a^2}+\frac{2 x \sqrt{b \sqrt{x}+a x}}{3 a}-\frac{\left (5 b^3\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{b x+a x^2}} \, dx,x,\sqrt{x}\right )}{8 a^3}\\ &=\frac{5 b^2 \sqrt{b \sqrt{x}+a x}}{4 a^3}-\frac{5 b \sqrt{x} \sqrt{b \sqrt{x}+a x}}{6 a^2}+\frac{2 x \sqrt{b \sqrt{x}+a x}}{3 a}-\frac{\left (5 b^3\right ) \operatorname{Subst}\left (\int \frac{1}{1-a x^2} \, dx,x,\frac{\sqrt{x}}{\sqrt{b \sqrt{x}+a x}}\right )}{4 a^3}\\ &=\frac{5 b^2 \sqrt{b \sqrt{x}+a x}}{4 a^3}-\frac{5 b \sqrt{x} \sqrt{b \sqrt{x}+a x}}{6 a^2}+\frac{2 x \sqrt{b \sqrt{x}+a x}}{3 a}-\frac{5 b^3 \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{x}}{\sqrt{b \sqrt{x}+a x}}\right )}{4 a^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.106052, size = 129, normalized size = 1.11 \[ \frac{5 b^4 \left (\frac{a \sqrt{x}}{b}+1\right ) \left (\frac{16 a^3 x^{3/2}}{15 b^3}-\frac{4 a^2 x}{3 b^2}+\frac{2 a \sqrt{x}}{b}-\frac{2 \sqrt{a} \sqrt [4]{x} \sinh ^{-1}\left (\frac{\sqrt{a} \sqrt [4]{x}}{\sqrt{b}}\right )}{\sqrt{b} \sqrt{\frac{a \sqrt{x}}{b}+1}}\right )}{8 a^4 \sqrt{\sqrt{x} \left (a \sqrt{x}+b\right )}} \]

Antiderivative was successfully verified.

[In]

Integrate[x/Sqrt[b*Sqrt[x] + a*x],x]

[Out]

(5*b^4*(1 + (a*Sqrt[x])/b)*((2*a*Sqrt[x])/b - (4*a^2*x)/(3*b^2) + (16*a^3*x^(3/2))/(15*b^3) - (2*Sqrt[a]*x^(1/
4)*ArcSinh[(Sqrt[a]*x^(1/4))/Sqrt[b]])/(Sqrt[b]*Sqrt[1 + (a*Sqrt[x])/b])))/(8*a^4*Sqrt[(b + a*Sqrt[x])*Sqrt[x]
])

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Maple [B]  time = 0.007, size = 181, normalized size = 1.6 \begin{align*}{\frac{1}{24}\sqrt{b\sqrt{x}+ax} \left ( 16\, \left ( b\sqrt{x}+ax \right ) ^{3/2}{a}^{5/2}-36\,\sqrt{b\sqrt{x}+ax}{a}^{5/2}\sqrt{x}b-18\,\sqrt{b\sqrt{x}+ax}{a}^{3/2}{b}^{2}+48\,{a}^{3/2}\sqrt{\sqrt{x} \left ( b+a\sqrt{x} \right ) }{b}^{2}-24\,a\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}+9\,\ln \left ( 1/2\,{\frac{2\,a\sqrt{x}+2\,\sqrt{b\sqrt{x}+ax}\sqrt{a}+b}{\sqrt{a}}} \right ) a{b}^{3} \right ){a}^{-{\frac{9}{2}}}{\frac{1}{\sqrt{\sqrt{x} \left ( b+a\sqrt{x} \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(b*x^(1/2)+a*x)^(1/2)/a^(9/2)*(16*(b*x^(1/2)+a*x)^(3/2)*a^(5/2)-36*(b*x^(1/2)+a*x)^(1/2)*a^(5/2)*x^(1/2)*
b-18*(b*x^(1/2)+a*x)^(1/2)*a^(3/2)*b^2+48*a^(3/2)*(x^(1/2)*(b+a*x^(1/2)))^(1/2)*b^2-24*a*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+9*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*b^3)/(x^(1/2)*(b+a*x^(1/2)))^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{\sqrt{a x + b \sqrt{x}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{\sqrt{a x + b \sqrt{x}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(b*x**(1/2)+a*x)**(1/2),x)

[Out]

Integral(x/sqrt(a*x + b*sqrt(x)), x)

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Giac [A]  time = 1.36586, size = 112, normalized size = 0.97 \begin{align*} \frac{1}{12} \, \sqrt{a x + b \sqrt{x}}{\left (2 \, \sqrt{x}{\left (\frac{4 \, \sqrt{x}}{a} - \frac{5 \, b}{a^{2}}\right )} + \frac{15 \, b^{2}}{a^{3}}\right )} + \frac{5 \, b^{3} \log \left ({\left | -2 \, \sqrt{a}{\left (\sqrt{a} \sqrt{x} - \sqrt{a x + b \sqrt{x}}\right )} - b \right |}\right )}{8 \, a^{\frac{7}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*sqrt(a*x + b*sqrt(x))*(2*sqrt(x)*(4*sqrt(x)/a - 5*b/a^2) + 15*b^2/a^3) + 5/8*b^3*log(abs(-2*sqrt(a)*(sqrt
(a)*sqrt(x) - sqrt(a*x + b*sqrt(x))) - b))/a^(7/2)