3.2496 \(\int \frac{(5-x) (3+2 x)^3}{\sqrt{2+5 x+3 x^2}} \, dx\)

Optimal. Leaf size=112 \[ -\frac{1}{12} \sqrt{3 x^2+5 x+2} (2 x+3)^3+\frac{32}{27} \sqrt{3 x^2+5 x+2} (2 x+3)^2+\frac{5}{648} (1078 x+3261) \sqrt{3 x^2+5 x+2}+\frac{19405 \tanh ^{-1}\left (\frac{6 x+5}{2 \sqrt{3} \sqrt{3 x^2+5 x+2}}\right )}{1296 \sqrt{3}} \]

[Out]

(32*(3 + 2*x)^2*Sqrt[2 + 5*x + 3*x^2])/27 - ((3 + 2*x)^3*Sqrt[2 + 5*x + 3*x^2])/12 + (5*(3261 + 1078*x)*Sqrt[2
+ 5*x + 3*x^2])/648 + (19405*ArcTanh[(5 + 6*x)/(2*Sqrt[3]*Sqrt[2 + 5*x + 3*x^2])])/(1296*Sqrt[3])

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Rubi [A]  time = 0.0639784, antiderivative size = 112, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.148, Rules used = {832, 779, 621, 206} \[ -\frac{1}{12} \sqrt{3 x^2+5 x+2} (2 x+3)^3+\frac{32}{27} \sqrt{3 x^2+5 x+2} (2 x+3)^2+\frac{5}{648} (1078 x+3261) \sqrt{3 x^2+5 x+2}+\frac{19405 \tanh ^{-1}\left (\frac{6 x+5}{2 \sqrt{3} \sqrt{3 x^2+5 x+2}}\right )}{1296 \sqrt{3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(32*(3 + 2*x)^2*Sqrt[2 + 5*x + 3*x^2])/27 - ((3 + 2*x)^3*Sqrt[2 + 5*x + 3*x^2])/12 + (5*(3261 + 1078*x)*Sqrt[2
+ 5*x + 3*x^2])/648 + (19405*ArcTanh[(5 + 6*x)/(2*Sqrt[3]*Sqrt[2 + 5*x + 3*x^2])])/(1296*Sqrt[3])

Rule 832

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

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 \frac{(5-x) (3+2 x)^3}{\sqrt{2+5 x+3 x^2}} \, dx &=-\frac{1}{12} (3+2 x)^3 \sqrt{2+5 x+3 x^2}+\frac{1}{12} \int \frac{(3+2 x)^2 \left (\frac{399}{2}+128 x\right )}{\sqrt{2+5 x+3 x^2}} \, dx\\ &=\frac{32}{27} (3+2 x)^2 \sqrt{2+5 x+3 x^2}-\frac{1}{12} (3+2 x)^3 \sqrt{2+5 x+3 x^2}+\frac{1}{108} \int \frac{(3+2 x) \left (\frac{6805}{2}+2695 x\right )}{\sqrt{2+5 x+3 x^2}} \, dx\\ &=\frac{32}{27} (3+2 x)^2 \sqrt{2+5 x+3 x^2}-\frac{1}{12} (3+2 x)^3 \sqrt{2+5 x+3 x^2}+\frac{5}{648} (3261+1078 x) \sqrt{2+5 x+3 x^2}+\frac{19405 \int \frac{1}{\sqrt{2+5 x+3 x^2}} \, dx}{1296}\\ &=\frac{32}{27} (3+2 x)^2 \sqrt{2+5 x+3 x^2}-\frac{1}{12} (3+2 x)^3 \sqrt{2+5 x+3 x^2}+\frac{5}{648} (3261+1078 x) \sqrt{2+5 x+3 x^2}+\frac{19405}{648} \operatorname{Subst}\left (\int \frac{1}{12-x^2} \, dx,x,\frac{5+6 x}{\sqrt{2+5 x+3 x^2}}\right )\\ &=\frac{32}{27} (3+2 x)^2 \sqrt{2+5 x+3 x^2}-\frac{1}{12} (3+2 x)^3 \sqrt{2+5 x+3 x^2}+\frac{5}{648} (3261+1078 x) \sqrt{2+5 x+3 x^2}+\frac{19405 \tanh ^{-1}\left (\frac{5+6 x}{2 \sqrt{3} \sqrt{2+5 x+3 x^2}}\right )}{1296 \sqrt{3}}\\ \end{align*}

Mathematica [A]  time = 0.0371025, size = 67, normalized size = 0.6 \[ \frac{19405 \sqrt{3} \tanh ^{-1}\left (\frac{6 x+5}{2 \sqrt{9 x^2+15 x+6}}\right )-6 \sqrt{3 x^2+5 x+2} \left (432 x^3-1128 x^2-11690 x-21759\right )}{3888} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-6*Sqrt[2 + 5*x + 3*x^2]*(-21759 - 11690*x - 1128*x^2 + 432*x^3) + 19405*Sqrt[3]*ArcTanh[(5 + 6*x)/(2*Sqrt[6
+ 15*x + 9*x^2])])/3888

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Maple [A]  time = 0.007, size = 94, normalized size = 0.8 \begin{align*} -{\frac{2\,{x}^{3}}{3}\sqrt{3\,{x}^{2}+5\,x+2}}+{\frac{47\,{x}^{2}}{27}\sqrt{3\,{x}^{2}+5\,x+2}}+{\frac{5845\,x}{324}\sqrt{3\,{x}^{2}+5\,x+2}}+{\frac{7253}{216}\sqrt{3\,{x}^{2}+5\,x+2}}+{\frac{19405\,\sqrt{3}}{3888}\ln \left ({\frac{\sqrt{3}}{3} \left ({\frac{5}{2}}+3\,x \right ) }+\sqrt{3\,{x}^{2}+5\,x+2} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*x^3*(3*x^2+5*x+2)^(1/2)+47/27*x^2*(3*x^2+5*x+2)^(1/2)+5845/324*x*(3*x^2+5*x+2)^(1/2)+7253/216*(3*x^2+5*x+
2)^(1/2)+19405/3888*ln(1/3*(5/2+3*x)*3^(1/2)+(3*x^2+5*x+2)^(1/2))*3^(1/2)

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Maxima [A]  time = 1.59942, size = 124, normalized size = 1.11 \begin{align*} -\frac{2}{3} \, \sqrt{3 \, x^{2} + 5 \, x + 2} x^{3} + \frac{47}{27} \, \sqrt{3 \, x^{2} + 5 \, x + 2} x^{2} + \frac{5845}{324} \, \sqrt{3 \, x^{2} + 5 \, x + 2} x + \frac{19405}{3888} \, \sqrt{3} \log \left (2 \, \sqrt{3} \sqrt{3 \, x^{2} + 5 \, x + 2} + 6 \, x + 5\right ) + \frac{7253}{216} \, \sqrt{3 \, x^{2} + 5 \, x + 2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*sqrt(3*x^2 + 5*x + 2)*x^3 + 47/27*sqrt(3*x^2 + 5*x + 2)*x^2 + 5845/324*sqrt(3*x^2 + 5*x + 2)*x + 19405/38
88*sqrt(3)*log(2*sqrt(3)*sqrt(3*x^2 + 5*x + 2) + 6*x + 5) + 7253/216*sqrt(3*x^2 + 5*x + 2)

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Fricas [A]  time = 2.0625, size = 215, normalized size = 1.92 \begin{align*} -\frac{1}{648} \,{\left (432 \, x^{3} - 1128 \, x^{2} - 11690 \, x - 21759\right )} \sqrt{3 \, x^{2} + 5 \, x + 2} + \frac{19405}{7776} \, \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+2*x)^3/(3*x^2+5*x+2)^(1/2),x, algorithm="fricas")

[Out]

-1/648*(432*x^3 - 1128*x^2 - 11690*x - 21759)*sqrt(3*x^2 + 5*x + 2) + 19405/7776*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 - \frac{243 x}{\sqrt{3 x^{2} + 5 x + 2}}\, dx - \int - \frac{126 x^{2}}{\sqrt{3 x^{2} + 5 x + 2}}\, dx - \int - \frac{4 x^{3}}{\sqrt{3 x^{2} + 5 x + 2}}\, dx - \int \frac{8 x^{4}}{\sqrt{3 x^{2} + 5 x + 2}}\, dx - \int - \frac{135}{\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+2*x)**3/(3*x**2+5*x+2)**(1/2),x)

[Out]

-Integral(-243*x/sqrt(3*x**2 + 5*x + 2), x) - Integral(-126*x**2/sqrt(3*x**2 + 5*x + 2), x) - Integral(-4*x**3
/sqrt(3*x**2 + 5*x + 2), x) - Integral(8*x**4/sqrt(3*x**2 + 5*x + 2), x) - Integral(-135/sqrt(3*x**2 + 5*x + 2
), x)

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Giac [A]  time = 1.14358, size = 86, normalized size = 0.77 \begin{align*} -\frac{1}{648} \,{\left (2 \,{\left (12 \,{\left (18 \, x - 47\right )} x - 5845\right )} x - 21759\right )} \sqrt{3 \, x^{2} + 5 \, x + 2} - \frac{19405}{3888} \, \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+2*x)^3/(3*x^2+5*x+2)^(1/2),x, algorithm="giac")

[Out]

-1/648*(2*(12*(18*x - 47)*x - 5845)*x - 21759)*sqrt(3*x^2 + 5*x + 2) - 19405/3888*sqrt(3)*log(abs(-2*sqrt(3)*(
sqrt(3)*x - sqrt(3*x^2 + 5*x + 2)) - 5))