3.2410 \(\int (5-x) \sqrt{2+5 x+3 x^2} \, dx\)

Optimal. Leaf size=80 \[ -\frac{1}{9} \left (3 x^2+5 x+2\right )^{3/2}+\frac{35}{72} (6 x+5) \sqrt{3 x^2+5 x+2}-\frac{35 \tanh ^{-1}\left (\frac{6 x+5}{2 \sqrt{3} \sqrt{3 x^2+5 x+2}}\right )}{144 \sqrt{3}} \]

[Out]

(35*(5 + 6*x)*Sqrt[2 + 5*x + 3*x^2])/72 - (2 + 5*x + 3*x^2)^(3/2)/9 - (35*ArcTanh[(5 + 6*x)/(2*Sqrt[3]*Sqrt[2
+ 5*x + 3*x^2])])/(144*Sqrt[3])

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Rubi [A]  time = 0.0226837, antiderivative size = 80, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {640, 612, 621, 206} \[ -\frac{1}{9} \left (3 x^2+5 x+2\right )^{3/2}+\frac{35}{72} (6 x+5) \sqrt{3 x^2+5 x+2}-\frac{35 \tanh ^{-1}\left (\frac{6 x+5}{2 \sqrt{3} \sqrt{3 x^2+5 x+2}}\right )}{144 \sqrt{3}} \]

Antiderivative was successfully verified.

[In]

Int[(5 - x)*Sqrt[2 + 5*x + 3*x^2],x]

[Out]

(35*(5 + 6*x)*Sqrt[2 + 5*x + 3*x^2])/72 - (2 + 5*x + 3*x^2)^(3/2)/9 - (35*ArcTanh[(5 + 6*x)/(2*Sqrt[3]*Sqrt[2
+ 5*x + 3*x^2])])/(144*Sqrt[3])

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 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 621

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

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])

Rubi steps

\begin{align*} \int (5-x) \sqrt{2+5 x+3 x^2} \, dx &=-\frac{1}{9} \left (2+5 x+3 x^2\right )^{3/2}+\frac{35}{6} \int \sqrt{2+5 x+3 x^2} \, dx\\ &=\frac{35}{72} (5+6 x) \sqrt{2+5 x+3 x^2}-\frac{1}{9} \left (2+5 x+3 x^2\right )^{3/2}-\frac{35}{144} \int \frac{1}{\sqrt{2+5 x+3 x^2}} \, dx\\ &=\frac{35}{72} (5+6 x) \sqrt{2+5 x+3 x^2}-\frac{1}{9} \left (2+5 x+3 x^2\right )^{3/2}-\frac{35}{72} \operatorname{Subst}\left (\int \frac{1}{12-x^2} \, dx,x,\frac{5+6 x}{\sqrt{2+5 x+3 x^2}}\right )\\ &=\frac{35}{72} (5+6 x) \sqrt{2+5 x+3 x^2}-\frac{1}{9} \left (2+5 x+3 x^2\right )^{3/2}-\frac{35 \tanh ^{-1}\left (\frac{5+6 x}{2 \sqrt{3} \sqrt{2+5 x+3 x^2}}\right )}{144 \sqrt{3}}\\ \end{align*}

Mathematica [A]  time = 0.051881, size = 62, normalized size = 0.78 \[ \frac{1}{432} \left (-6 \sqrt{3 x^2+5 x+2} \left (24 x^2-170 x-159\right )-35 \sqrt{3} \tanh ^{-1}\left (\frac{6 x+5}{2 \sqrt{9 x^2+15 x+6}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(5 - x)*Sqrt[2 + 5*x + 3*x^2],x]

[Out]

(-6*Sqrt[2 + 5*x + 3*x^2]*(-159 - 170*x + 24*x^2) - 35*Sqrt[3]*ArcTanh[(5 + 6*x)/(2*Sqrt[6 + 15*x + 9*x^2])])/
432

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Maple [A]  time = 0.004, size = 64, normalized size = 0.8 \begin{align*}{\frac{175+210\,x}{72}\sqrt{3\,{x}^{2}+5\,x+2}}-{\frac{35\,\sqrt{3}}{432}\ln \left ({\frac{\sqrt{3}}{3} \left ({\frac{5}{2}}+3\,x \right ) }+\sqrt{3\,{x}^{2}+5\,x+2} \right ) }-{\frac{1}{9} \left ( 3\,{x}^{2}+5\,x+2 \right ) ^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5-x)*(3*x^2+5*x+2)^(1/2),x)

[Out]

35/72*(5+6*x)*(3*x^2+5*x+2)^(1/2)-35/432*ln(1/3*(5/2+3*x)*3^(1/2)+(3*x^2+5*x+2)^(1/2))*3^(1/2)-1/9*(3*x^2+5*x+
2)^(3/2)

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Maxima [A]  time = 1.49398, size = 97, normalized size = 1.21 \begin{align*} -\frac{1}{9} \,{\left (3 \, x^{2} + 5 \, x + 2\right )}^{\frac{3}{2}} + \frac{35}{12} \, \sqrt{3 \, x^{2} + 5 \, x + 2} x - \frac{35}{432} \, \sqrt{3} \log \left (2 \, \sqrt{3} \sqrt{3 \, x^{2} + 5 \, x + 2} + 6 \, x + 5\right ) + \frac{175}{72} \, \sqrt{3 \, x^{2} + 5 \, x + 2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5-x)*(3*x^2+5*x+2)^(1/2),x, algorithm="maxima")

[Out]

-1/9*(3*x^2 + 5*x + 2)^(3/2) + 35/12*sqrt(3*x^2 + 5*x + 2)*x - 35/432*sqrt(3)*log(2*sqrt(3)*sqrt(3*x^2 + 5*x +
2) + 6*x + 5) + 175/72*sqrt(3*x^2 + 5*x + 2)

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Fricas [A]  time = 1.36908, size = 188, normalized size = 2.35 \begin{align*} -\frac{1}{72} \,{\left (24 \, x^{2} - 170 \, x - 159\right )} \sqrt{3 \, x^{2} + 5 \, x + 2} + \frac{35}{864} \, \sqrt{3} \log \left (-4 \, \sqrt{3} \sqrt{3 \, x^{2} + 5 \, x + 2}{\left (6 \, x + 5\right )} + 72 \, x^{2} + 120 \, x + 49\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5-x)*(3*x^2+5*x+2)^(1/2),x, algorithm="fricas")

[Out]

-1/72*(24*x^2 - 170*x - 159)*sqrt(3*x^2 + 5*x + 2) + 35/864*sqrt(3)*log(-4*sqrt(3)*sqrt(3*x^2 + 5*x + 2)*(6*x
+ 5) + 72*x^2 + 120*x + 49)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \int x \sqrt{3 x^{2} + 5 x + 2}\, dx - \int - 5 \sqrt{3 x^{2} + 5 x + 2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5-x)*(3*x**2+5*x+2)**(1/2),x)

[Out]

-Integral(x*sqrt(3*x**2 + 5*x + 2), x) - Integral(-5*sqrt(3*x**2 + 5*x + 2), x)

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Giac [A]  time = 1.18067, size = 80, normalized size = 1. \begin{align*} -\frac{1}{72} \,{\left (2 \,{\left (12 \, x - 85\right )} x - 159\right )} \sqrt{3 \, x^{2} + 5 \, x + 2} + \frac{35}{432} \, \sqrt{3} \log \left ({\left | -2 \, \sqrt{3}{\left (\sqrt{3} x - \sqrt{3 \, x^{2} + 5 \, x + 2}\right )} - 5 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5-x)*(3*x^2+5*x+2)^(1/2),x, algorithm="giac")

[Out]

-1/72*(2*(12*x - 85)*x - 159)*sqrt(3*x^2 + 5*x + 2) + 35/432*sqrt(3)*log(abs(-2*sqrt(3)*(sqrt(3)*x - sqrt(3*x^
2 + 5*x + 2)) - 5))