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

Optimal. Leaf size=62 \[ \frac {1}{6} \sqrt {3 x^2+5 x+2} (19-2 x)+\frac {31 \tanh ^{-1}\left (\frac {6 x+5}{2 \sqrt {3} \sqrt {3 x^2+5 x+2}}\right )}{4 \sqrt {3}} \]

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Rubi [A]  time = 0.02, antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {779, 621, 206} \begin {gather*} \frac {1}{6} \sqrt {3 x^2+5 x+2} (19-2 x)+\frac {31 \tanh ^{-1}\left (\frac {6 x+5}{2 \sqrt {3} \sqrt {3 x^2+5 x+2}}\right )}{4 \sqrt {3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((19 - 2*x)*Sqrt[2 + 5*x + 3*x^2])/6 + (31*ArcTanh[(5 + 6*x)/(2*Sqrt[3]*Sqrt[2 + 5*x + 3*x^2])])/(4*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 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 779

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((b
*e*g*(p + 2) - c*(e*f + d*g)*(2*p + 3) - 2*c*e*g*(p + 1)*x)*(a + b*x + c*x^2)^(p + 1))/(2*c^2*(p + 1)*(2*p + 3
)), x] + Dist[(b^2*e*g*(p + 2) - 2*a*c*e*g + c*(2*c*d*f - b*(e*f + d*g))*(2*p + 3))/(2*c^2*(2*p + 3)), Int[(a
+ b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b^2 - 4*a*c, 0] &&  !LeQ[p, -1]

Rubi steps

\begin {align*} \int \frac {(5-x) (3+2 x)}{\sqrt {2+5 x+3 x^2}} \, dx &=\frac {1}{6} (19-2 x) \sqrt {2+5 x+3 x^2}+\frac {31}{4} \int \frac {1}{\sqrt {2+5 x+3 x^2}} \, dx\\ &=\frac {1}{6} (19-2 x) \sqrt {2+5 x+3 x^2}+\frac {31}{2} \operatorname {Subst}\left (\int \frac {1}{12-x^2} \, dx,x,\frac {5+6 x}{\sqrt {2+5 x+3 x^2}}\right )\\ &=\frac {1}{6} (19-2 x) \sqrt {2+5 x+3 x^2}+\frac {31 \tanh ^{-1}\left (\frac {5+6 x}{2 \sqrt {3} \sqrt {2+5 x+3 x^2}}\right )}{4 \sqrt {3}}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 57, normalized size = 0.92 \begin {gather*} \frac {1}{12} \left (31 \sqrt {3} \tanh ^{-1}\left (\frac {6 x+5}{2 \sqrt {9 x^2+15 x+6}}\right )-2 (2 x-19) \sqrt {3 x^2+5 x+2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(-19 + 2*x)*Sqrt[2 + 5*x + 3*x^2] + 31*Sqrt[3]*ArcTanh[(5 + 6*x)/(2*Sqrt[6 + 15*x + 9*x^2])])/12

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IntegrateAlgebraic [A]  time = 0.52, size = 59, normalized size = 0.95 \begin {gather*} \frac {1}{6} \sqrt {3 x^2+5 x+2} (19-2 x)+\frac {31 \tanh ^{-1}\left (\frac {\sqrt {3 x^2+5 x+2}}{\sqrt {3} (x+1)}\right )}{2 \sqrt {3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((19 - 2*x)*Sqrt[2 + 5*x + 3*x^2])/6 + (31*ArcTanh[Sqrt[2 + 5*x + 3*x^2]/(Sqrt[3]*(1 + x))])/(2*Sqrt[3])

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fricas [A]  time = 0.40, size = 58, normalized size = 0.94 \begin {gather*} -\frac {1}{6} \, \sqrt {3 \, x^{2} + 5 \, x + 2} {\left (2 \, x - 19\right )} + \frac {31}{24} \, \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+2*x)/(3*x^2+5*x+2)^(1/2),x, algorithm="fricas")

[Out]

-1/6*sqrt(3*x^2 + 5*x + 2)*(2*x - 19) + 31/24*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 = 0.22, size = 54, normalized size = 0.87 \begin {gather*} -\frac {1}{6} \, \sqrt {3 \, x^{2} + 5 \, x + 2} {\left (2 \, x - 19\right )} - \frac {31}{12} \, \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+2*x)/(3*x^2+5*x+2)^(1/2),x, algorithm="giac")

[Out]

-1/6*sqrt(3*x^2 + 5*x + 2)*(2*x - 19) - 31/12*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 = 0.01, size = 60, normalized size = 0.97 \begin {gather*} -\frac {\sqrt {3 x^{2}+5 x +2}\, x}{3}+\frac {31 \sqrt {3}\, \ln \left (\frac {\left (3 x +\frac {5}{2}\right ) \sqrt {3}}{3}+\sqrt {3 x^{2}+5 x +2}\right )}{12}+\frac {19 \sqrt {3 x^{2}+5 x +2}}{6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(3*x^2+5*x+2)^(1/2)*x+19/6*(3*x^2+5*x+2)^(1/2)+31/12*3^(1/2)*ln(1/3*(3*x+5/2)*3^(1/2)+(3*x^2+5*x+2)^(1/2)
)

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maxima [A]  time = 1.31, size = 58, normalized size = 0.94 \begin {gather*} -\frac {1}{3} \, \sqrt {3 \, x^{2} + 5 \, x + 2} x + \frac {31}{12} \, \sqrt {3} \log \left (2 \, \sqrt {3} \sqrt {3 \, x^{2} + 5 \, x + 2} + 6 \, x + 5\right ) + \frac {19}{6} \, \sqrt {3 \, x^{2} + 5 \, x + 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*sqrt(3*x^2 + 5*x + 2)*x + 31/12*sqrt(3)*log(2*sqrt(3)*sqrt(3*x^2 + 5*x + 2) + 6*x + 5) + 19/6*sqrt(3*x^2
+ 5*x + 2)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} -\int \frac {\left (2\,x+3\right )\,\left (x-5\right )}{\sqrt {3\,x^2+5\,x+2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \int \left (- \frac {7 x}{\sqrt {3 x^{2} + 5 x + 2}}\right )\, dx - \int \frac {2 x^{2}}{\sqrt {3 x^{2} + 5 x + 2}}\, dx - \int \left (- \frac {15}{\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+2*x)/(3*x**2+5*x+2)**(1/2),x)

[Out]

-Integral(-7*x/sqrt(3*x**2 + 5*x + 2), x) - Integral(2*x**2/sqrt(3*x**2 + 5*x + 2), x) - Integral(-15/sqrt(3*x
**2 + 5*x + 2), x)

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