3.2.4 \(\int \frac {x}{\sqrt {b \sqrt {x}+a x}} \, dx\) [104]

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

[Out]

-5/4*b^3*arctanh(a^(1/2)*x^(1/2)/(b*x^(1/2)+a*x)^(1/2))/a^(7/2)+5/4*b^2*(b*x^(1/2)+a*x)^(1/2)/a^3+2/3*x*(b*x^(
1/2)+a*x)^(1/2)/a-5/6*b*x^(1/2)*(b*x^(1/2)+a*x)^(1/2)/a^2

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Rubi [A]
time = 0.07, 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 = {2043, 684, 654, 634, 212} \begin {gather*} -\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} \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 212

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

Rule 634

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 654

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 684

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 2043

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 \text {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) \text {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 ) \text {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 ) \text {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 ) \text {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 = 0.17, size = 95, normalized size = 0.82 \begin {gather*} \frac {\sqrt {b \sqrt {x}+a x} \left (15 b^2-10 a b \sqrt {x}+8 a^2 x\right )}{12 a^3}+\frac {5 b^3 \log \left (a^3 b+2 a^4 \sqrt {x}-2 a^{7/2} \sqrt {b \sqrt {x}+a x}\right )}{8 a^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[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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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(180\) vs. \(2(84)=168\).
time = 0.38, size = 181, normalized size = 1.56

method	result	size
derivativedivides	\(\frac {2 x \sqrt {b \sqrt {x}+a x}}{3 a}-\frac {5 b \left (\frac {\sqrt {x}\, \sqrt {b \sqrt {x}+a x}}{2 a}-\frac {3 b \left (\frac {\sqrt {b \sqrt {x}+a x}}{a}-\frac {b \ln \left (\frac {\frac {b}{2}+a \sqrt {x}}{\sqrt {a}}+\sqrt {b \sqrt {x}+a x}\right )}{2 a^{\frac {3}{2}}}\right )}{4 a}\right )}{3 a}\)	\(99\)
default	\(\frac {\sqrt {b \sqrt {x}+a x}\, \left (16 \left (b \sqrt {x}+a x \right )^{\frac {3}{2}} a^{\frac {5}{2}}-36 \sqrt {b \sqrt {x}+a x}\, \sqrt {x}\, a^{\frac {5}{2}} b -18 \sqrt {b \sqrt {x}+a x}\, a^{\frac {3}{2}} b^{2}+48 \sqrt {\sqrt {x}\, \left (a \sqrt {x}+b \right )}\, a^{\frac {3}{2}} b^{2}-24 a \ln \left (\frac {2 a \sqrt {x}+2 \sqrt {\sqrt {x}\, \left (a \sqrt {x}+b \right )}\, \sqrt {a}+b}{2 \sqrt {a}}\right ) b^{3}+9 \ln \left (\frac {2 a \sqrt {x}+2 \sqrt {b \sqrt {x}+a x}\, \sqrt {a}+b}{2 \sqrt {a}}\right ) a \,b^{3}\right )}{24 a^{\frac {9}{2}} \sqrt {\sqrt {x}\, \left (a \sqrt {x}+b \right )}}\)	\(181\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \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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Fricas [F(-1)] Timed out
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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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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Giac [A]
time = 1.15, 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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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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