\(\int \frac {(5-x) \sqrt {2+5 x+3 x^2}}{3+2 x} \, dx\) [2411]

Optimal result

Integrand size = 27, antiderivative size = 100 \[ \int \frac {(5-x) \sqrt {2+5 x+3 x^2}}{3+2 x} \, dx=\frac {1}{24} (73-6 x) \sqrt {2+5 x+3 x^2}-\frac {311 \text {arctanh}\left (\frac {5+6 x}{2 \sqrt {3} \sqrt {2+5 x+3 x^2}}\right )}{48 \sqrt {3}}+\frac {13}{8} \sqrt {5} \text {arctanh}\left (\frac {7+8 x}{2 \sqrt {5} \sqrt {2+5 x+3 x^2}}\right ) \]

[Out]

-311/144*arctanh(1/6*(5+6*x)*3^(1/2)/(3*x^2+5*x+2)^(1/2))*3^(1/2)+13/8*arctanh(1/10*(7+8*x)*5^(1/2)/(3*x^2+5*x
+2)^(1/2))*5^(1/2)+1/24*(73-6*x)*(3*x^2+5*x+2)^(1/2)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {828, 857, 635, 212, 738} \[ \int \frac {(5-x) \sqrt {2+5 x+3 x^2}}{3+2 x} \, dx=-\frac {311 \text {arctanh}\left (\frac {6 x+5}{2 \sqrt {3} \sqrt {3 x^2+5 x+2}}\right )}{48 \sqrt {3}}+\frac {13}{8} \sqrt {5} \text {arctanh}\left (\frac {8 x+7}{2 \sqrt {5} \sqrt {3 x^2+5 x+2}}\right )+\frac {1}{24} \sqrt {3 x^2+5 x+2} (73-6 x) \]

[In]

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

[Out]

((73 - 6*x)*Sqrt[2 + 5*x + 3*x^2])/24 - (311*ArcTanh[(5 + 6*x)/(2*Sqrt[3]*Sqrt[2 + 5*x + 3*x^2])])/(48*Sqrt[3]
) + (13*Sqrt[5]*ArcTanh[(7 + 8*x)/(2*Sqrt[5]*Sqrt[2 + 5*x + 3*x^2])])/8

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 738

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

Rule 828

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

Rule 857

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

Rubi steps \begin{align*} \text {integral}& = \frac {1}{24} (73-6 x) \sqrt {2+5 x+3 x^2}-\frac {1}{48} \int \frac {543+622 x}{(3+2 x) \sqrt {2+5 x+3 x^2}} \, dx \\ & = \frac {1}{24} (73-6 x) \sqrt {2+5 x+3 x^2}-\frac {311}{48} \int \frac {1}{\sqrt {2+5 x+3 x^2}} \, dx+\frac {65}{8} \int \frac {1}{(3+2 x) \sqrt {2+5 x+3 x^2}} \, dx \\ & = \frac {1}{24} (73-6 x) \sqrt {2+5 x+3 x^2}-\frac {311}{24} \text {Subst}\left (\int \frac {1}{12-x^2} \, dx,x,\frac {5+6 x}{\sqrt {2+5 x+3 x^2}}\right )-\frac {65}{4} \text {Subst}\left (\int \frac {1}{20-x^2} \, dx,x,\frac {-7-8 x}{\sqrt {2+5 x+3 x^2}}\right ) \\ & = \frac {1}{24} (73-6 x) \sqrt {2+5 x+3 x^2}-\frac {311 \tanh ^{-1}\left (\frac {5+6 x}{2 \sqrt {3} \sqrt {2+5 x+3 x^2}}\right )}{48 \sqrt {3}}+\frac {13}{8} \sqrt {5} \tanh ^{-1}\left (\frac {7+8 x}{2 \sqrt {5} \sqrt {2+5 x+3 x^2}}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.86 \[ \int \frac {(5-x) \sqrt {2+5 x+3 x^2}}{3+2 x} \, dx=\frac {1}{72} \left (-3 (-73+6 x) \sqrt {2+5 x+3 x^2}+234 \sqrt {5} \text {arctanh}\left (\frac {\sqrt {\frac {2}{5}+x+\frac {3 x^2}{5}}}{1+x}\right )-311 \sqrt {3} \text {arctanh}\left (\frac {\sqrt {\frac {2}{3}+\frac {5 x}{3}+x^2}}{1+x}\right )\right ) \]

[In]

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

[Out]

(-3*(-73 + 6*x)*Sqrt[2 + 5*x + 3*x^2] + 234*Sqrt[5]*ArcTanh[Sqrt[2/5 + x + (3*x^2)/5]/(1 + x)] - 311*Sqrt[3]*A
rcTanh[Sqrt[2/3 + (5*x)/3 + x^2]/(1 + x)])/72

Maple [A] (verified)

Time = 0.33 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.80

method	result	size
risch	\(-\frac {\left (-73+6 x \right ) \sqrt {3 x^{2}+5 x +2}}{24}-\frac {311 \ln \left (\frac {\left (\frac {5}{2}+3 x \right ) \sqrt {3}}{3}+\sqrt {3 x^{2}+5 x +2}\right ) \sqrt {3}}{144}-\frac {13 \sqrt {5}\, \operatorname {arctanh}\left (\frac {2 \left (-\frac {7}{2}-4 x \right ) \sqrt {5}}{5 \sqrt {12 \left (x +\frac {3}{2}\right )^{2}-16 x -19}}\right )}{8}\)	\(80\)
trager	\(\left (\frac {73}{24}-\frac {x}{4}\right ) \sqrt {3 x^{2}+5 x +2}-\frac {13 \operatorname {RootOf}\left (\textit {\_Z}^{2}-5\right ) \ln \left (\frac {-8 \operatorname {RootOf}\left (\textit {\_Z}^{2}-5\right ) x +10 \sqrt {3 x^{2}+5 x +2}-7 \operatorname {RootOf}\left (\textit {\_Z}^{2}-5\right )}{3+2 x}\right )}{8}-\frac {311 \operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right ) \ln \left (6 \operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right ) x +5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right )+6 \sqrt {3 x^{2}+5 x +2}\right )}{144}\)	\(110\)
default	\(-\frac {\left (5+6 x \right ) \sqrt {3 x^{2}+5 x +2}}{24}+\frac {\ln \left (\frac {\left (\frac {5}{2}+3 x \right ) \sqrt {3}}{3}+\sqrt {3 x^{2}+5 x +2}\right ) \sqrt {3}}{144}+\frac {13 \sqrt {12 \left (x +\frac {3}{2}\right )^{2}-16 x -19}}{8}-\frac {13 \ln \left (\frac {\left (\frac {5}{2}+3 x \right ) \sqrt {3}}{3}+\sqrt {3 \left (x +\frac {3}{2}\right )^{2}-4 x -\frac {19}{4}}\right ) \sqrt {3}}{6}-\frac {13 \sqrt {5}\, \operatorname {arctanh}\left (\frac {2 \left (-\frac {7}{2}-4 x \right ) \sqrt {5}}{5 \sqrt {12 \left (x +\frac {3}{2}\right )^{2}-16 x -19}}\right )}{8}\)	\(127\)

[In]

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

[Out]

-1/24*(-73+6*x)*(3*x^2+5*x+2)^(1/2)-311/144*ln(1/3*(5/2+3*x)*3^(1/2)+(3*x^2+5*x+2)^(1/2))*3^(1/2)-13/8*5^(1/2)
*arctanh(2/5*(-7/2-4*x)*5^(1/2)/(12*(x+3/2)^2-16*x-19)^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.09 \[ \int \frac {(5-x) \sqrt {2+5 x+3 x^2}}{3+2 x} \, dx=-\frac {1}{24} \, \sqrt {3 \, x^{2} + 5 \, x + 2} {\left (6 \, x - 73\right )} + \frac {311}{288} \, \sqrt {3} \log \left (-4 \, \sqrt {3} \sqrt {3 \, x^{2} + 5 \, x + 2} {\left (6 \, x + 5\right )} + 72 \, x^{2} + 120 \, x + 49\right ) + \frac {13}{16} \, \sqrt {5} \log \left (\frac {4 \, \sqrt {5} \sqrt {3 \, x^{2} + 5 \, x + 2} {\left (8 \, x + 7\right )} + 124 \, x^{2} + 212 \, x + 89}{4 \, x^{2} + 12 \, x + 9}\right ) \]

[In]

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

[Out]

-1/24*sqrt(3*x^2 + 5*x + 2)*(6*x - 73) + 311/288*sqrt(3)*log(-4*sqrt(3)*sqrt(3*x^2 + 5*x + 2)*(6*x + 5) + 72*x
^2 + 120*x + 49) + 13/16*sqrt(5)*log((4*sqrt(5)*sqrt(3*x^2 + 5*x + 2)*(8*x + 7) + 124*x^2 + 212*x + 89)/(4*x^2
+ 12*x + 9))

Sympy [F]

\[ \int \frac {(5-x) \sqrt {2+5 x+3 x^2}}{3+2 x} \, dx=- \int \left (- \frac {5 \sqrt {3 x^{2} + 5 x + 2}}{2 x + 3}\right )\, dx - \int \frac {x \sqrt {3 x^{2} + 5 x + 2}}{2 x + 3}\, dx \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.99 \[ \int \frac {(5-x) \sqrt {2+5 x+3 x^2}}{3+2 x} \, dx=-\frac {1}{4} \, \sqrt {3 \, x^{2} + 5 \, x + 2} x - \frac {311}{144} \, \sqrt {3} \log \left (\sqrt {3} \sqrt {3 \, x^{2} + 5 \, x + 2} + 3 \, x + \frac {5}{2}\right ) - \frac {13}{8} \, \sqrt {5} \log \left (\frac {\sqrt {5} \sqrt {3 \, x^{2} + 5 \, x + 2}}{{\left | 2 \, x + 3 \right |}} + \frac {5}{2 \, {\left | 2 \, x + 3 \right |}} - 2\right ) + \frac {73}{24} \, \sqrt {3 \, x^{2} + 5 \, x + 2} \]

[In]

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

[Out]

-1/4*sqrt(3*x^2 + 5*x + 2)*x - 311/144*sqrt(3)*log(sqrt(3)*sqrt(3*x^2 + 5*x + 2) + 3*x + 5/2) - 13/8*sqrt(5)*l
og(sqrt(5)*sqrt(3*x^2 + 5*x + 2)/abs(2*x + 3) + 5/2/abs(2*x + 3) - 2) + 73/24*sqrt(3*x^2 + 5*x + 2)

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.26 \[ \int \frac {(5-x) \sqrt {2+5 x+3 x^2}}{3+2 x} \, dx=-\frac {1}{24} \, \sqrt {3 \, x^{2} + 5 \, x + 2} {\left (6 \, x - 73\right )} + \frac {13}{8} \, \sqrt {5} \log \left (\frac {{\left | -4 \, \sqrt {3} x - 2 \, \sqrt {5} - 6 \, \sqrt {3} + 4 \, \sqrt {3 \, x^{2} + 5 \, x + 2} \right |}}{{\left | -4 \, \sqrt {3} x + 2 \, \sqrt {5} - 6 \, \sqrt {3} + 4 \, \sqrt {3 \, x^{2} + 5 \, x + 2} \right |}}\right ) + \frac {311}{144} \, \sqrt {3} \log \left ({\left | -6 \, \sqrt {3} x - 5 \, \sqrt {3} + 6 \, \sqrt {3 \, x^{2} + 5 \, x + 2} \right |}\right ) \]

[In]

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

[Out]

-1/24*sqrt(3*x^2 + 5*x + 2)*(6*x - 73) + 13/8*sqrt(5)*log(abs(-4*sqrt(3)*x - 2*sqrt(5) - 6*sqrt(3) + 4*sqrt(3*
x^2 + 5*x + 2))/abs(-4*sqrt(3)*x + 2*sqrt(5) - 6*sqrt(3) + 4*sqrt(3*x^2 + 5*x + 2))) + 311/144*sqrt(3)*log(abs
(-6*sqrt(3)*x - 5*sqrt(3) + 6*sqrt(3*x^2 + 5*x + 2)))

Mupad [F(-1)]

Timed out. \[ \int \frac {(5-x) \sqrt {2+5 x+3 x^2}}{3+2 x} \, dx=-\int \frac {\left (x-5\right )\,\sqrt {3\,x^2+5\,x+2}}{2\,x+3} \,d x \]

[In]

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

[Out]

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