\(\int \frac {5-x}{\sqrt {3+2 x} (2+5 x+3 x^2)^2} \, dx\) [839]

Optimal result

Integrand size = 27, antiderivative size = 72 \[ \int \frac {5-x}{\sqrt {3+2 x} \left (2+5 x+3 x^2\right )^2} \, dx=-\frac {3 \sqrt {3+2 x} (37+47 x)}{5 \left (2+5 x+3 x^2\right )}-58 \text {arctanh}\left (\sqrt {3+2 x}\right )+\frac {384}{5} \sqrt {\frac {3}{5}} \text {arctanh}\left (\sqrt {\frac {3}{5}} \sqrt {3+2 x}\right ) \] Output:

-3*(3+2*x)^(1/2)*(37+47*x)/(15*x^2+25*x+10)-58*arctanh((3+2*x)^(1/2))+384/ 
25*15^(1/2)*arctanh(1/5*15^(1/2)*(3+2*x)^(1/2))
 

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.00 \[ \int \frac {5-x}{\sqrt {3+2 x} \left (2+5 x+3 x^2\right )^2} \, dx=-\frac {3 \sqrt {3+2 x} (37+47 x)}{5 \left (2+5 x+3 x^2\right )}-58 \text {arctanh}\left (\sqrt {3+2 x}\right )+\frac {384}{5} \sqrt {\frac {3}{5}} \text {arctanh}\left (\sqrt {\frac {3}{5}} \sqrt {3+2 x}\right ) \] Input:

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

Output:

(-3*Sqrt[3 + 2*x]*(37 + 47*x))/(5*(2 + 5*x + 3*x^2)) - 58*ArcTanh[Sqrt[3 + 
 2*x]] + (384*Sqrt[3/5]*ArcTanh[Sqrt[3/5]*Sqrt[3 + 2*x]])/5
 

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.04, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {1235, 1197, 1480, 220}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 1235

\(\displaystyle -\frac {1}{5} \int \frac {141 x+286}{\sqrt {2 x+3} \left (3 x^2+5 x+2\right )}dx-\frac {3 \sqrt {2 x+3} (47 x+37)}{5 \left (3 x^2+5 x+2\right )}\)

\(\Big \downarrow \) 1197

\(\displaystyle -\frac {2}{5} \int \frac {141 (2 x+3)+149}{3 (2 x+3)^2-8 (2 x+3)+5}d\sqrt {2 x+3}-\frac {3 \sqrt {2 x+3} (47 x+37)}{5 \left (3 x^2+5 x+2\right )}\)

\(\Big \downarrow \) 1480

\(\displaystyle -\frac {2}{5} \left (576 \int \frac {1}{3 (2 x+3)-5}d\sqrt {2 x+3}-435 \int \frac {1}{3 (2 x+3)-3}d\sqrt {2 x+3}\right )-\frac {3 \sqrt {2 x+3} (47 x+37)}{5 \left (3 x^2+5 x+2\right )}\)

\(\Big \downarrow \) 220

\(\displaystyle -\frac {2}{5} \left (145 \text {arctanh}\left (\sqrt {2 x+3}\right )-192 \sqrt {\frac {3}{5}} \text {arctanh}\left (\sqrt {\frac {3}{5}} \sqrt {2 x+3}\right )\right )-\frac {3 \sqrt {2 x+3} (47 x+37)}{5 \left (3 x^2+5 x+2\right )}\)

Input:

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

Output:

(-3*Sqrt[3 + 2*x]*(37 + 47*x))/(5*(2 + 5*x + 3*x^2)) - (2*(145*ArcTanh[Sqr 
t[3 + 2*x]] - 192*Sqrt[3/5]*ArcTanh[Sqrt[3/5]*Sqrt[3 + 2*x]]))/5
 

Defintions of rubi rules used

rule 220

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

rule 1197

Int[((f_.) + (g_.)*(x_))/(Sqrt[(d_.) + (e_.)*(x_)]*((a_.) + (b_.)*(x_) + (c 
_.)*(x_)^2)), x_Symbol] :> Simp[2   Subst[Int[(e*f - d*g + g*x^2)/(c*d^2 - 
b*d*e + a*e^2 - (2*c*d - b*e)*x^2 + c*x^4), x], x, Sqrt[d + e*x]], x] /; Fr 
eeQ[{a, b, c, d, e, f, g}, x]
 

rule 1235

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c 
_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d + e*x)^(m + 1)*(f*(b*c*d - b^2*e + 2 
*a*c*e) - a*g*(2*c*d - b*e) + c*(f*(2*c*d - b*e) - g*(b*d - 2*a*e))*x)*((a 
+ b*x + c*x^2)^(p + 1)/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2))), x] 
 + Simp[1/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2))   Int[(d + e*x)^m 
*(a + b*x + c*x^2)^(p + 1)*Simp[f*(b*c*d*e*(2*p - m + 2) + b^2*e^2*(p + m + 
 2) - 2*c^2*d^2*(2*p + 3) - 2*a*c*e^2*(m + 2*p + 3)) - g*(a*e*(b*e - 2*c*d* 
m + b*e*m) - b*d*(3*c*d - b*e + 2*c*d*p - b*e*p)) + c*e*(g*(b*d - 2*a*e) - 
f*(2*c*d - b*e))*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, 
 m}, x] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p] 
)
 

rule 1480

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

Maple [A] (verified)

Time = 1.35 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.97

method	result	size
risch	\(-\frac {3 \left (37+47 x \right ) \sqrt {2 x +3}}{5 \left (3 x^{2}+5 x +2\right )}-29 \ln \left (\sqrt {2 x +3}+1\right )+29 \ln \left (\sqrt {2 x +3}-1\right )+\frac {384 \sqrt {15}\, \operatorname {arctanh}\left (\frac {\sqrt {15}\, \sqrt {2 x +3}}{5}\right )}{25}\)	\(70\)
derivativedivides	\(-\frac {34 \sqrt {2 x +3}}{5 \left (2 x +\frac {4}{3}\right )}+\frac {384 \sqrt {15}\, \operatorname {arctanh}\left (\frac {\sqrt {15}\, \sqrt {2 x +3}}{5}\right )}{25}-\frac {6}{\sqrt {2 x +3}-1}+29 \ln \left (\sqrt {2 x +3}-1\right )-\frac {6}{\sqrt {2 x +3}+1}-29 \ln \left (\sqrt {2 x +3}+1\right )\)	\(86\)
default	\(-\frac {34 \sqrt {2 x +3}}{5 \left (2 x +\frac {4}{3}\right )}+\frac {384 \sqrt {15}\, \operatorname {arctanh}\left (\frac {\sqrt {15}\, \sqrt {2 x +3}}{5}\right )}{25}-\frac {6}{\sqrt {2 x +3}-1}+29 \ln \left (\sqrt {2 x +3}-1\right )-\frac {6}{\sqrt {2 x +3}+1}-29 \ln \left (\sqrt {2 x +3}+1\right )\)	\(86\)
trager	\(-\frac {3 \left (37+47 x \right ) \sqrt {2 x +3}}{5 \left (3 x^{2}+5 x +2\right )}+29 \ln \left (\frac {-2-x +\sqrt {2 x +3}}{x +1}\right )-\frac {192 \operatorname {RootOf}\left (\textit {\_Z}^{2}-15\right ) \ln \left (\frac {-3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-15\right ) x +15 \sqrt {2 x +3}-7 \operatorname {RootOf}\left (\textit {\_Z}^{2}-15\right )}{3 x +2}\right )}{25}\)	\(93\)
pseudoelliptic	\(\frac {1152 \sqrt {15}\, \left (x +1\right ) \left (x +\frac {2}{3}\right ) \operatorname {arctanh}\left (\frac {\sqrt {15}\, \sqrt {2 x +3}}{5}\right )+\left (2175 x^{2}+3625 x +1450\right ) \ln \left (\sqrt {2 x +3}-1\right )+\left (-2175 x^{2}-3625 x -1450\right ) \ln \left (\sqrt {2 x +3}+1\right )+\left (-705 x -555\right ) \sqrt {2 x +3}}{75 x^{2}+125 x +50}\)	\(94\)

Input:

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

Output:

-3/5*(37+47*x)/(3*x^2+5*x+2)*(2*x+3)^(1/2)-29*ln((2*x+3)^(1/2)+1)+29*ln((2 
*x+3)^(1/2)-1)+384/25*15^(1/2)*arctanh(1/5*15^(1/2)*(2*x+3)^(1/2))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 114 vs. \(2 (55) = 110\).

Time = 0.09 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.58 \[ \int \frac {5-x}{\sqrt {3+2 x} \left (2+5 x+3 x^2\right )^2} \, dx=\frac {192 \, \sqrt {\frac {3}{5}} {\left (3 \, x^{2} + 5 \, x + 2\right )} \log \left (\frac {5 \, \sqrt {\frac {3}{5}} \sqrt {2 \, x + 3} + 3 \, x + 7}{3 \, x + 2}\right ) - 145 \, {\left (3 \, x^{2} + 5 \, x + 2\right )} \log \left (\sqrt {2 \, x + 3} + 1\right ) + 145 \, {\left (3 \, x^{2} + 5 \, x + 2\right )} \log \left (\sqrt {2 \, x + 3} - 1\right ) - 3 \, {\left (47 \, x + 37\right )} \sqrt {2 \, x + 3}}{5 \, {\left (3 \, x^{2} + 5 \, x + 2\right )}} \] Input:

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

Output:

1/5*(192*sqrt(3/5)*(3*x^2 + 5*x + 2)*log((5*sqrt(3/5)*sqrt(2*x + 3) + 3*x 
+ 7)/(3*x + 2)) - 145*(3*x^2 + 5*x + 2)*log(sqrt(2*x + 3) + 1) + 145*(3*x^ 
2 + 5*x + 2)*log(sqrt(2*x + 3) - 1) - 3*(47*x + 37)*sqrt(2*x + 3))/(3*x^2 
+ 5*x + 2)
 

Sympy [A] (verification not implemented)

Time = 39.22 (sec) , antiderivative size = 209, normalized size of antiderivative = 2.90 \[ \int \frac {5-x}{\sqrt {3+2 x} \left (2+5 x+3 x^2\right )^2} \, dx=- 7 \sqrt {15} \left (\log {\left (\sqrt {2 x + 3} - \frac {\sqrt {15}}{3} \right )} - \log {\left (\sqrt {2 x + 3} + \frac {\sqrt {15}}{3} \right )}\right ) + 204 \left (\begin {cases} \frac {\sqrt {15} \left (- \frac {\log {\left (\frac {\sqrt {15} \sqrt {2 x + 3}}{5} - 1 \right )}}{4} + \frac {\log {\left (\frac {\sqrt {15} \sqrt {2 x + 3}}{5} + 1 \right )}}{4} - \frac {1}{4 \left (\frac {\sqrt {15} \sqrt {2 x + 3}}{5} + 1\right )} - \frac {1}{4 \left (\frac {\sqrt {15} \sqrt {2 x + 3}}{5} - 1\right )}\right )}{75} & \text {for}\: \sqrt {2 x + 3} > - \frac {\sqrt {15}}{3} \wedge \sqrt {2 x + 3} < \frac {\sqrt {15}}{3} \end {cases}\right ) + 29 \log {\left (\sqrt {2 x + 3} - 1 \right )} - 29 \log {\left (\sqrt {2 x + 3} + 1 \right )} - \frac {6}{\sqrt {2 x + 3} + 1} - \frac {6}{\sqrt {2 x + 3} - 1} \] Input:

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

Output:

-7*sqrt(15)*(log(sqrt(2*x + 3) - sqrt(15)/3) - log(sqrt(2*x + 3) + sqrt(15 
)/3)) + 204*Piecewise((sqrt(15)*(-log(sqrt(15)*sqrt(2*x + 3)/5 - 1)/4 + lo 
g(sqrt(15)*sqrt(2*x + 3)/5 + 1)/4 - 1/(4*(sqrt(15)*sqrt(2*x + 3)/5 + 1)) - 
 1/(4*(sqrt(15)*sqrt(2*x + 3)/5 - 1)))/75, (sqrt(2*x + 3) > -sqrt(15)/3) & 
 (sqrt(2*x + 3) < sqrt(15)/3))) + 29*log(sqrt(2*x + 3) - 1) - 29*log(sqrt( 
2*x + 3) + 1) - 6/(sqrt(2*x + 3) + 1) - 6/(sqrt(2*x + 3) - 1)
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.36 \[ \int \frac {5-x}{\sqrt {3+2 x} \left (2+5 x+3 x^2\right )^2} \, dx=-\frac {192}{25} \, \sqrt {15} \log \left (-\frac {\sqrt {15} - 3 \, \sqrt {2 \, x + 3}}{\sqrt {15} + 3 \, \sqrt {2 \, x + 3}}\right ) - \frac {6 \, {\left (47 \, {\left (2 \, x + 3\right )}^{\frac {3}{2}} - 67 \, \sqrt {2 \, x + 3}\right )}}{5 \, {\left (3 \, {\left (2 \, x + 3\right )}^{2} - 16 \, x - 19\right )}} - 29 \, \log \left (\sqrt {2 \, x + 3} + 1\right ) + 29 \, \log \left (\sqrt {2 \, x + 3} - 1\right ) \] Input:

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

Output:

-192/25*sqrt(15)*log(-(sqrt(15) - 3*sqrt(2*x + 3))/(sqrt(15) + 3*sqrt(2*x 
+ 3))) - 6/5*(47*(2*x + 3)^(3/2) - 67*sqrt(2*x + 3))/(3*(2*x + 3)^2 - 16*x 
 - 19) - 29*log(sqrt(2*x + 3) + 1) + 29*log(sqrt(2*x + 3) - 1)
 

Giac [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.42 \[ \int \frac {5-x}{\sqrt {3+2 x} \left (2+5 x+3 x^2\right )^2} \, dx=-\frac {192}{25} \, \sqrt {15} \log \left (\frac {{\left | -2 \, \sqrt {15} + 6 \, \sqrt {2 \, x + 3} \right |}}{2 \, {\left (\sqrt {15} + 3 \, \sqrt {2 \, x + 3}\right )}}\right ) - \frac {6 \, {\left (47 \, {\left (2 \, x + 3\right )}^{\frac {3}{2}} - 67 \, \sqrt {2 \, x + 3}\right )}}{5 \, {\left (3 \, {\left (2 \, x + 3\right )}^{2} - 16 \, x - 19\right )}} - 29 \, \log \left (\sqrt {2 \, x + 3} + 1\right ) + 29 \, \log \left ({\left | \sqrt {2 \, x + 3} - 1 \right |}\right ) \] Input:

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

Output:

-192/25*sqrt(15)*log(1/2*abs(-2*sqrt(15) + 6*sqrt(2*x + 3))/(sqrt(15) + 3* 
sqrt(2*x + 3))) - 6/5*(47*(2*x + 3)^(3/2) - 67*sqrt(2*x + 3))/(3*(2*x + 3) 
^2 - 16*x - 19) - 29*log(sqrt(2*x + 3) + 1) + 29*log(abs(sqrt(2*x + 3) - 1 
))
 

Mupad [B] (verification not implemented)

Time = 11.54 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.92 \[ \int \frac {5-x}{\sqrt {3+2 x} \left (2+5 x+3 x^2\right )^2} \, dx=\frac {384\,\sqrt {15}\,\mathrm {atanh}\left (\frac {\sqrt {15}\,\sqrt {2\,x+3}}{5}\right )}{25}-\frac {\frac {134\,\sqrt {2\,x+3}}{5}-\frac {94\,{\left (2\,x+3\right )}^{3/2}}{5}}{\frac {16\,x}{3}-{\left (2\,x+3\right )}^2+\frac {19}{3}}-58\,\mathrm {atanh}\left (\sqrt {2\,x+3}\right ) \] Input:

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

Output:

(384*15^(1/2)*atanh((15^(1/2)*(2*x + 3)^(1/2))/5))/25 - ((134*(2*x + 3)^(1 
/2))/5 - (94*(2*x + 3)^(3/2))/5)/((16*x)/3 - (2*x + 3)^2 + 19/3) - 58*atan 
h((2*x + 3)^(1/2))
 

Reduce [B] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 215, normalized size of antiderivative = 2.99 \[ \int \frac {5-x}{\sqrt {3+2 x} \left (2+5 x+3 x^2\right )^2} \, dx=\frac {-705 \sqrt {2 x +3}\, x -555 \sqrt {2 x +3}-576 \sqrt {15}\, \mathrm {log}\left (3 \sqrt {2 x +3}-\sqrt {15}\right ) x^{2}-960 \sqrt {15}\, \mathrm {log}\left (3 \sqrt {2 x +3}-\sqrt {15}\right ) x -384 \sqrt {15}\, \mathrm {log}\left (3 \sqrt {2 x +3}-\sqrt {15}\right )+576 \sqrt {15}\, \mathrm {log}\left (3 \sqrt {2 x +3}+\sqrt {15}\right ) x^{2}+960 \sqrt {15}\, \mathrm {log}\left (3 \sqrt {2 x +3}+\sqrt {15}\right ) x +384 \sqrt {15}\, \mathrm {log}\left (3 \sqrt {2 x +3}+\sqrt {15}\right )+2175 \,\mathrm {log}\left (\sqrt {2 x +3}-1\right ) x^{2}+3625 \,\mathrm {log}\left (\sqrt {2 x +3}-1\right ) x +1450 \,\mathrm {log}\left (\sqrt {2 x +3}-1\right )-2175 \,\mathrm {log}\left (\sqrt {2 x +3}+1\right ) x^{2}-3625 \,\mathrm {log}\left (\sqrt {2 x +3}+1\right ) x -1450 \,\mathrm {log}\left (\sqrt {2 x +3}+1\right )}{75 x^{2}+125 x +50} \] Input:

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

Output:

( - 705*sqrt(2*x + 3)*x - 555*sqrt(2*x + 3) - 576*sqrt(15)*log(3*sqrt(2*x 
+ 3) - sqrt(15))*x**2 - 960*sqrt(15)*log(3*sqrt(2*x + 3) - sqrt(15))*x - 3 
84*sqrt(15)*log(3*sqrt(2*x + 3) - sqrt(15)) + 576*sqrt(15)*log(3*sqrt(2*x 
+ 3) + sqrt(15))*x**2 + 960*sqrt(15)*log(3*sqrt(2*x + 3) + sqrt(15))*x + 3 
84*sqrt(15)*log(3*sqrt(2*x + 3) + sqrt(15)) + 2175*log(sqrt(2*x + 3) - 1)* 
x**2 + 3625*log(sqrt(2*x + 3) - 1)*x + 1450*log(sqrt(2*x + 3) - 1) - 2175* 
log(sqrt(2*x + 3) + 1)*x**2 - 3625*log(sqrt(2*x + 3) + 1)*x - 1450*log(sqr 
t(2*x + 3) + 1))/(25*(3*x**2 + 5*x + 2))