3.45 \(\int \frac {x}{\sqrt {b x+c x^2}} \, dx\)

Optimal. Leaf size=47 \[ \frac {\sqrt {b x+c x^2}}{c}-\frac {b \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{c^{3/2}} \]

[Out]

-b*arctanh(x*c^(1/2)/(c*x^2+b*x)^(1/2))/c^(3/2)+(c*x^2+b*x)^(1/2)/c

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Rubi [A]  time = 0.02, antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {640, 620, 206} \[ \frac {\sqrt {b x+c x^2}}{c}-\frac {b \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{c^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[x/Sqrt[b*x + c*x^2],x]

[Out]

Sqrt[b*x + c*x^2]/c - (b*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/c^(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 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]

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 71, normalized size = 1.51 \[ \frac {\sqrt {c} x (b+c x)-b^{3/2} \sqrt {x} \sqrt {\frac {c x}{b}+1} \sinh ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{c^{3/2} \sqrt {x (b+c x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[x/Sqrt[b*x + c*x^2],x]

[Out]

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

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fricas [A]  time = 0.66, size = 98, normalized size = 2.09 \[ \left [\frac {b \sqrt {c} \log \left (2 \, c x + b - 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) + 2 \, \sqrt {c x^{2} + b x} c}{2 \, c^{2}}, \frac {b \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) + \sqrt {c x^{2} + b x} c}{c^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(b*sqrt(c)*log(2*c*x + b - 2*sqrt(c*x^2 + b*x)*sqrt(c)) + 2*sqrt(c*x^2 + b*x)*c)/c^2, (b*sqrt(-c)*arctan(
sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) + sqrt(c*x^2 + b*x)*c)/c^2]

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giac [A]  time = 0.23, size = 52, normalized size = 1.11 \[ \frac {b \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} - b \right |}\right )}{2 \, c^{\frac {3}{2}}} + \frac {\sqrt {c x^{2} + b x}}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*b*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) - b))/c^(3/2) + sqrt(c*x^2 + b*x)/c

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maple [A]  time = 0.04, size = 47, normalized size = 1.00 \[ -\frac {b \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{2 c^{\frac {3}{2}}}+\frac {\sqrt {c \,x^{2}+b x}}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(c*x^2+b*x)^(1/2)/c-1/2*b/c^(3/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x)^(1/2))

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maxima [A]  time = 1.37, size = 45, normalized size = 0.96 \[ -\frac {b \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{2 \, c^{\frac {3}{2}}} + \frac {\sqrt {c x^{2} + b x}}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*b*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(3/2) + sqrt(c*x^2 + b*x)/c

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mupad [B]  time = 0.20, size = 46, normalized size = 0.98 \[ \frac {\sqrt {c\,x^2+b\,x}}{c}-\frac {b\,\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x}\right )}{2\,c^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(b*x + c*x^2)^(1/2),x)

[Out]

(b*x + c*x^2)^(1/2)/c - (b*log((b/2 + c*x)/c^(1/2) + (b*x + c*x^2)^(1/2)))/(2*c^(3/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x}{\sqrt {x \left (b + c x\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x/sqrt(x*(b + c*x)), x)

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