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

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

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Rubi [A]  time = 0.09, antiderivative size = 116, normalized size of antiderivative = 1.00, 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} \begin {gather*} \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} \end {gather*}

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 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 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 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}{\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*}

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Mathematica [A]  time = 0.12, size = 129, normalized size = 1.11 \begin {gather*} \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 )}} \end {gather*}

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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IntegrateAlgebraic [A]  time = 0.25, size = 95, normalized size = 0.82 \begin {gather*} \frac {\sqrt {a x+b \sqrt {x}} \left (8 a^2 x-10 a b \sqrt {x}+15 b^2\right )}{12 a^3}+\frac {5 b^3 \log \left (-2 a^{7/2} \sqrt {a x+b \sqrt {x}}+2 a^4 \sqrt {x}+a^3 b\right )}{8 a^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x/Sqrt[b*Sqrt[x] + a*x],x]

[Out]

(Sqrt[b*Sqrt[x] + a*x]*(15*b^2 - 10*a*b*Sqrt[x] + 8*a^2*x))/(12*a^3) + (5*b^3*Log[a^3*b + 2*a^4*Sqrt[x] - 2*a^
(7/2)*Sqrt[b*Sqrt[x] + a*x]])/(8*a^(7/2))

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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)^(1/2),x, algorithm="fricas")

[Out]

Timed out

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giac [A]  time = 0.31, size = 83, normalized size = 0.72 \begin {gather*} \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 {gather*}

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)

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maple [B]  time = 0.05, size = 181, normalized size = 1.56 \begin {gather*} -\frac {\sqrt {a x +b \sqrt {x}}\, \left (24 a \,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 )-9 a \,b^{3} \ln \left (\frac {2 a \sqrt {x}+b +2 \sqrt {a x +b \sqrt {x}}\, \sqrt {a}}{2 \sqrt {a}}\right )+36 \sqrt {a x +b \sqrt {x}}\, a^{\frac {5}{2}} b \sqrt {x}-48 \sqrt {\left (a \sqrt {x}+b \right ) \sqrt {x}}\, a^{\frac {3}{2}} b^{2}+18 \sqrt {a x +b \sqrt {x}}\, a^{\frac {3}{2}} b^{2}-16 \left (a x +b \sqrt {x}\right )^{\frac {3}{2}} a^{\frac {5}{2}}\right )}{24 \sqrt {\left (a \sqrt {x}+b \right ) \sqrt {x}}\, a^{\frac {9}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x}{\sqrt {a x + b \sqrt {x}}}\,{d x} \end {gather*}

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x}{\sqrt {a\,x+b\,\sqrt {x}}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(a*x + b*x^(1/2))^(1/2),x)

[Out]

int(x/(a*x + b*x^(1/2))^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x}{\sqrt {a x + b \sqrt {x}}}\, dx \end {gather*}

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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