3.25.96 \(\int \frac {(5-x) (3+2 x)^3}{\sqrt {2+5 x+3 x^2}} \, dx\) [2496]

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

[Out]

19405/3888*arctanh(1/6*(5+6*x)*3^(1/2)/(3*x^2+5*x+2)^(1/2))*3^(1/2)+32/27*(3+2*x)^2*(3*x^2+5*x+2)^(1/2)-1/12*(
3+2*x)^3*(3*x^2+5*x+2)^(1/2)+5/648*(3261+1078*x)*(3*x^2+5*x+2)^(1/2)

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Rubi [A]
time = 0.04, antiderivative size = 112, normalized size of antiderivative = 1.00, 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 = {846, 793, 635, 212} \begin {gather*} -\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}} \end {gather*}

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 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 635

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 793

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 846

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])

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} \text {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*}

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Mathematica [A]
time = 0.28, size = 66, normalized size = 0.59 \begin {gather*} \frac {-3 \sqrt {2+5 x+3 x^2} \left (-21759-11690 x-1128 x^2+432 x^3\right )+19405 \sqrt {3} \tanh ^{-1}\left (\frac {\sqrt {\frac {2}{3}+\frac {5 x}{3}+x^2}}{1+x}\right )}{1944} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*Sqrt[2 + 5*x + 3*x^2]*(-21759 - 11690*x - 1128*x^2 + 432*x^3) + 19405*Sqrt[3]*ArcTanh[Sqrt[2/3 + (5*x)/3 +
x^2]/(1 + x)])/1944

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Maple [A]
time = 0.08, size = 94, normalized size = 0.84

method	result	size
risch	\(-\frac {\left (432 x^{3}-1128 x^{2}-11690 x -21759\right ) \sqrt {3 x^{2}+5 x +2}}{648}+\frac {19405 \ln \left (\frac {\left (\frac {5}{2}+3 x \right ) \sqrt {3}}{3}+\sqrt {3 x^{2}+5 x +2}\right ) \sqrt {3}}{3888}\)	\(60\)
trager	\(\left (-\frac {2}{3} x^{3}+\frac {47}{27} x^{2}+\frac {5845}{324} x +\frac {7253}{216}\right ) \sqrt {3 x^{2}+5 x +2}-\frac {19405 \RootOf \left (\textit {\_Z}^{2}-3\right ) \ln \left (-6 \RootOf \left (\textit {\_Z}^{2}-3\right ) x -5 \RootOf \left (\textit {\_Z}^{2}-3\right )+6 \sqrt {3 x^{2}+5 x +2}\right )}{3888}\)	\(71\)
default	\(-\frac {2 x^{3} \sqrt {3 x^{2}+5 x +2}}{3}+\frac {47 x^{2} \sqrt {3 x^{2}+5 x +2}}{27}+\frac {5845 x \sqrt {3 x^{2}+5 x +2}}{324}+\frac {7253 \sqrt {3 x^{2}+5 x +2}}{216}+\frac {19405 \ln \left (\frac {\left (\frac {5}{2}+3 x \right ) \sqrt {3}}{3}+\sqrt {3 x^{2}+5 x +2}\right ) \sqrt {3}}{3888}\)	\(94\)

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,method=_RETURNVERBOSE)

[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 = 0.52, size = 92, normalized size = 0.82 \begin {gather*} -\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 {gather*}

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 = 3.32, size = 68, normalized size = 0.61 \begin {gather*} -\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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \int \left (- \frac {243 x}{\sqrt {3 x^{2} + 5 x + 2}}\right )\, dx - \int \left (- \frac {126 x^{2}}{\sqrt {3 x^{2} + 5 x + 2}}\right )\, dx - \int \left (- \frac {4 x^{3}}{\sqrt {3 x^{2} + 5 x + 2}}\right )\, dx - \int \frac {8 x^{4}}{\sqrt {3 x^{2} + 5 x + 2}}\, dx - \int \left (- \frac {135}{\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/(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 = 1.82, size = 64, normalized size = 0.57 \begin {gather*} -\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 {gather*}

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))

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} -\int \frac {{\left (2\,x+3\right )}^3\,\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)^3*(x - 5))/(5*x + 3*x^2 + 2)^(1/2),x)

[Out]

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

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