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

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

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Rubi [A]  time = 0.02, antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {640, 612, 621, 206} \begin {gather*} -\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}} \end {gather*}

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 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 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 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 (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*}

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Mathematica [A]  time = 0.05, size = 62, normalized size = 0.78 \begin {gather*} \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 ) \end {gather*}

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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IntegrateAlgebraic [A]  time = 0.30, size = 64, normalized size = 0.80 \begin {gather*} \frac {1}{72} \left (-24 x^2+170 x+159\right ) \sqrt {3 x^2+5 x+2}-\frac {35 \tanh ^{-1}\left (\frac {\sqrt {3 x^2+5 x+2}}{\sqrt {3} (x+1)}\right )}{72 \sqrt {3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((159 + 170*x - 24*x^2)*Sqrt[2 + 5*x + 3*x^2])/72 - (35*ArcTanh[Sqrt[2 + 5*x + 3*x^2]/(Sqrt[3]*(1 + x))])/(72*
Sqrt[3])

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fricas [A]  time = 0.38, size = 63, normalized size = 0.79 \begin {gather*} -\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 {gather*}

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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giac [A]  time = 0.18, size = 59, normalized size = 0.74 \begin {gather*} -\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 {gather*}

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))

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maple [A]  time = 0.05, size = 64, normalized size = 0.80 \begin {gather*} -\frac {35 \sqrt {3}\, \ln \left (\frac {\left (3 x +\frac {5}{2}\right ) \sqrt {3}}{3}+\sqrt {3 x^{2}+5 x +2}\right )}{432}+\frac {35 \left (6 x +5\right ) \sqrt {3 x^{2}+5 x +2}}{72}-\frac {\left (3 x^{2}+5 x +2\right )^{\frac {3}{2}}}{9} \end {gather*}

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

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maxima [A]  time = 1.35, size = 72, normalized size = 0.90 \begin {gather*} -\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 {gather*}

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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mupad [B]  time = 2.66, size = 104, normalized size = 1.30 \begin {gather*} 5\,\left (\frac {x}{2}+\frac {5}{12}\right )\,\sqrt {3\,x^2+5\,x+2}-\frac {5\,\sqrt {3}\,\ln \left (\sqrt {3\,x^2+5\,x+2}+\frac {\sqrt {3}\,\left (3\,x+\frac {5}{2}\right )}{3}\right )}{72}-\frac {\sqrt {3\,x^2+5\,x+2}\,\left (72\,x^2+30\,x-27\right )}{216}-\frac {5\,\sqrt {3}\,\ln \left (2\,\sqrt {3\,x^2+5\,x+2}+\frac {\sqrt {3}\,\left (6\,x+5\right )}{3}\right )}{432} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5*(x/2 + 5/12)*(5*x + 3*x^2 + 2)^(1/2) - (5*3^(1/2)*log((5*x + 3*x^2 + 2)^(1/2) + (3^(1/2)*(3*x + 5/2))/3))/72
- ((5*x + 3*x^2 + 2)^(1/2)*(30*x + 72*x^2 - 27))/216 - (5*3^(1/2)*log(2*(5*x + 3*x^2 + 2)^(1/2) + (3^(1/2)*(6
*x + 5))/3))/432

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \int x \sqrt {3 x^{2} + 5 x + 2}\, dx - \int \left (- 5 \sqrt {3 x^{2} + 5 x + 2}\right )\, dx \end {gather*}

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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