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

Optimal. Leaf size=42 \[ \sqrt {b x+c x^2}+\frac {b \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{\sqrt {c}} \]

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Rubi [A]  time = 0.02, antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {664, 620, 206} \begin {gather*} \sqrt {b x+c x^2}+\frac {b \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{\sqrt {c}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[b*x + c*x^2]/x,x]

[Out]

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

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 664

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(
a + b*x + c*x^2)^p)/(e*(m + 2*p + 1)), x] - Dist[(p*(2*c*d - b*e))/(e^2*(m + 2*p + 1)), Int[(d + e*x)^(m + 1)*
(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] && GtQ[p, 0] && (LeQ[-2, m, 0] || EqQ[m + p + 1, 0]) && NeQ[m + 2*p + 1, 0] && IntegerQ[2*p]

Rubi steps

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

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Mathematica [A]  time = 0.09, size = 66, normalized size = 1.57 \begin {gather*} \sqrt {x (b+c x)} \left (\frac {b^{3/2} \sqrt {\frac {c x}{b}+1} \sinh ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{\sqrt {c} \sqrt {x} (b+c x)}+1\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[b*x + c*x^2]/x,x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.14, size = 51, normalized size = 1.21 \begin {gather*} \sqrt {b x+c x^2}-\frac {b \log \left (-2 \sqrt {c} \sqrt {b x+c x^2}+b+2 c x\right )}{2 \sqrt {c}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[Sqrt[b*x + c*x^2]/x,x]

[Out]

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

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fricas [A]  time = 0.42, size = 100, normalized size = 2.38 \begin {gather*} \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}, -\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}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x)^(1/2)/x,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, -(b*sqrt(-c)*arctan(s
qrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) - sqrt(c*x^2 + b*x)*c)/c]

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giac [A]  time = 0.23, size = 48, normalized size = 1.14 \begin {gather*} -\frac {b \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} - b \right |}\right )}{2 \, \sqrt {c}} + \sqrt {c x^{2} + b x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.04, size = 43, normalized size = 1.02 \begin {gather*} \frac {b \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{2 \sqrt {c}}+\sqrt {c \,x^{2}+b x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.13, size = 41, normalized size = 0.98 \begin {gather*} \frac {b \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{2 \, \sqrt {c}} + \sqrt {c x^{2} + b x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.07, size = 42, normalized size = 1.00 \begin {gather*} \sqrt {c\,x^2+b\,x}+\frac {b\,\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x}\right )}{2\,\sqrt {c}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x + c*x^2)^(1/2)/x,x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(x*(b + c*x))/x, x)

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