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

Optimal. Leaf size=60 \[ \frac {(b+2 c x) \sqrt {b x+c x^2}}{4 c}-\frac {b^2 \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{4 c^{3/2}} \]

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Rubi [A]  time = 0.02, antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {612, 620, 206} \begin {gather*} \frac {(b+2 c x) \sqrt {b x+c x^2}}{4 c}-\frac {b^2 \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{4 c^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[b*x + c*x^2],x]

[Out]

((b + 2*c*x)*Sqrt[b*x + c*x^2])/(4*c) - (b^2*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(4*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 612

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b + 2*c*x)*(a + b*x + c*x^2)^p)/(2*c*(2*p +
1)), x] - Dist[(p*(b^2 - 4*a*c))/(2*c*(2*p + 1)), Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x]
&& NeQ[b^2 - 4*a*c, 0] && GtQ[p, 0] && IntegerQ[4*p]

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]

Rubi steps

\begin {align*} \int \sqrt {b x+c x^2} \, dx &=\frac {(b+2 c x) \sqrt {b x+c x^2}}{4 c}-\frac {b^2 \int \frac {1}{\sqrt {b x+c x^2}} \, dx}{8 c}\\ &=\frac {(b+2 c x) \sqrt {b x+c x^2}}{4 c}-\frac {b^2 \operatorname {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x}{\sqrt {b x+c x^2}}\right )}{4 c}\\ &=\frac {(b+2 c x) \sqrt {b x+c x^2}}{4 c}-\frac {b^2 \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{4 c^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 74, normalized size = 1.23 \begin {gather*} \frac {\sqrt {x (b+c x)} \left (\sqrt {c} (b+2 c x)-\frac {b^{3/2} \sinh ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{\sqrt {x} \sqrt {\frac {c x}{b}+1}}\right )}{4 c^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[b*x + c*x^2],x]

[Out]

(Sqrt[x*(b + c*x)]*(Sqrt[c]*(b + 2*c*x) - (b^(3/2)*ArcSinh[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/(Sqrt[x]*Sqrt[1 + (c*x)
/b])))/(4*c^(3/2))

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IntegrateAlgebraic [A]  time = 0.00, size = 70, normalized size = 1.17 \begin {gather*} \frac {b^2 \log \left (-2 c^{3/2} \sqrt {b x+c x^2}+b c+2 c^2 x\right )}{8 c^{3/2}}+\frac {(b+2 c x) \sqrt {b x+c x^2}}{4 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((b + 2*c*x)*Sqrt[b*x + c*x^2])/(4*c) + (b^2*Log[b*c + 2*c^2*x - 2*c^(3/2)*Sqrt[b*x + c*x^2]])/(8*c^(3/2))

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fricas [A]  time = 0.42, size = 121, normalized size = 2.02 \begin {gather*} \left [\frac {b^{2} \sqrt {c} \log \left (2 \, c x + b - 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) + 2 \, {\left (2 \, c^{2} x + b c\right )} \sqrt {c x^{2} + b x}}{8 \, c^{2}}, \frac {b^{2} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) + {\left (2 \, c^{2} x + b c\right )} \sqrt {c x^{2} + b x}}{4 \, c^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*(b^2*sqrt(c)*log(2*c*x + b - 2*sqrt(c*x^2 + b*x)*sqrt(c)) + 2*(2*c^2*x + b*c)*sqrt(c*x^2 + b*x))/c^2, 1/4
*(b^2*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) + (2*c^2*x + b*c)*sqrt(c*x^2 + b*x))/c^2]

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giac [A]  time = 0.17, size = 61, normalized size = 1.02 \begin {gather*} \frac {1}{4} \, \sqrt {c x^{2} + b x} {\left (2 \, x + \frac {b}{c}\right )} + \frac {b^{2} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} - b \right |}\right )}{8 \, c^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*sqrt(c*x^2 + b*x)*(2*x + b/c) + 1/8*b^2*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) - b))/c^(3/2)

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maple [A]  time = 0.04, size = 56, normalized size = 0.93 \begin {gather*} -\frac {b^{2} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {3}{2}}}+\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x}}{4 c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*b^2/c^(3/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x)^(1/2))+1/4*(2*c*x+b)*(c*x^2+b*x)^(1/2)/c

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maxima [A]  time = 1.37, size = 63, normalized size = 1.05 \begin {gather*} \frac {1}{2} \, \sqrt {c x^{2} + b x} x - \frac {b^{2} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{8 \, c^{\frac {3}{2}}} + \frac {\sqrt {c x^{2} + b x} b}{4 \, c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(c*x^2 + b*x)*x - 1/8*b^2*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(3/2) + 1/4*sqrt(c*x^2 + b*x)
*b/c

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mupad [B]  time = 0.22, size = 55, normalized size = 0.92 \begin {gather*} \sqrt {c\,x^2+b\,x}\,\left (\frac {x}{2}+\frac {b}{4\,c}\right )-\frac {b^2\,\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x}\right )}{8\,c^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x + c*x^2)^(1/2),x)

[Out]

(b*x + c*x^2)^(1/2)*(x/2 + b/(4*c)) - (b^2*log((b/2 + c*x)/c^(1/2) + (b*x + c*x^2)^(1/2)))/(8*c^(3/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt {b x + c x^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(b*x + c*x**2), x)

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