\(\int \frac {(5-x) (3+2 x)^{5/2}}{(2+5 x+3 x^2)^3} \, dx\) [845]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [B] (verification not implemented)
Sympy [A] (verification not implemented)
Maxima [A] (verification not implemented)
Giac [A] (verification not implemented)
Mupad [B] (verification not implemented)
Reduce [B] (verification not implemented)

Optimal result

Integrand size = 27, antiderivative size = 102 \[ \int \frac {(5-x) (3+2 x)^{5/2}}{\left (2+5 x+3 x^2\right )^3} \, dx=-\frac {\sqrt {3+2 x} (533+587 x)}{18 \left (2+5 x+3 x^2\right )^2}+\frac {131 \sqrt {3+2 x} (62+75 x)}{18 \left (2+5 x+3 x^2\right )}+1250 \text {arctanh}\left (\sqrt {3+2 x}\right )-\frac {2905}{3} \sqrt {\frac {5}{3}} \text {arctanh}\left (\sqrt {\frac {3}{5}} \sqrt {3+2 x}\right ) \] Output:

-1/18*(3+2*x)^(1/2)*(533+587*x)/(3*x^2+5*x+2)^2+131*(3+2*x)^(1/2)*(62+75*x 
)/(54*x^2+90*x+36)+1250*arctanh((3+2*x)^(1/2))-2905/9*15^(1/2)*arctanh(1/5 
*15^(1/2)*(3+2*x)^(1/2))
 

Mathematica [A] (verified)

Time = 0.29 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.80 \[ \int \frac {(5-x) (3+2 x)^{5/2}}{\left (2+5 x+3 x^2\right )^3} \, dx=\frac {\sqrt {3+2 x} \left (5237+19891 x+24497 x^2+9825 x^3\right )}{6 \left (2+5 x+3 x^2\right )^2}+1250 \text {arctanh}\left (\sqrt {3+2 x}\right )-\frac {2905}{3} \sqrt {\frac {5}{3}} \text {arctanh}\left (\sqrt {\frac {3}{5}} \sqrt {3+2 x}\right ) \] Input:

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

Output:

(Sqrt[3 + 2*x]*(5237 + 19891*x + 24497*x^2 + 9825*x^3))/(6*(2 + 5*x + 3*x^ 
2)^2) + 1250*ArcTanh[Sqrt[3 + 2*x]] - (2905*Sqrt[5/3]*ArcTanh[Sqrt[3/5]*Sq 
rt[3 + 2*x]])/3
 

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.02, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {1233, 27, 1234, 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.

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

\(\Big \downarrow \) 1233

\(\displaystyle \frac {1}{6} \int -\frac {25 \sqrt {2 x+3} (17 x+36)}{\left (3 x^2+5 x+2\right )^2}dx-\frac {(2 x+3)^{3/2} (139 x+121)}{6 \left (3 x^2+5 x+2\right )^2}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {25}{6} \int \frac {\sqrt {2 x+3} (17 x+36)}{\left (3 x^2+5 x+2\right )^2}dx-\frac {(139 x+121) (2 x+3)^{3/2}}{6 \left (3 x^2+5 x+2\right )^2}\)

\(\Big \downarrow \) 1234

\(\displaystyle -\frac {25}{6} \left (-\int \frac {131 x+281}{\sqrt {2 x+3} \left (3 x^2+5 x+2\right )}dx-\frac {\sqrt {2 x+3} (131 x+112)}{3 x^2+5 x+2}\right )-\frac {(139 x+121) (2 x+3)^{3/2}}{6 \left (3 x^2+5 x+2\right )^2}\)

\(\Big \downarrow \) 1197

\(\displaystyle -\frac {25}{6} \left (-2 \int \frac {131 (2 x+3)+169}{3 (2 x+3)^2-8 (2 x+3)+5}d\sqrt {2 x+3}-\frac {\sqrt {2 x+3} (131 x+112)}{3 x^2+5 x+2}\right )-\frac {(139 x+121) (2 x+3)^{3/2}}{6 \left (3 x^2+5 x+2\right )^2}\)

\(\Big \downarrow \) 1480

\(\displaystyle -\frac {25}{6} \left (-2 \left (581 \int \frac {1}{3 (2 x+3)-5}d\sqrt {2 x+3}-450 \int \frac {1}{3 (2 x+3)-3}d\sqrt {2 x+3}\right )-\frac {\sqrt {2 x+3} (131 x+112)}{3 x^2+5 x+2}\right )-\frac {(139 x+121) (2 x+3)^{3/2}}{6 \left (3 x^2+5 x+2\right )^2}\)

\(\Big \downarrow \) 220

\(\displaystyle -\frac {25}{6} \left (-2 \left (150 \text {arctanh}\left (\sqrt {2 x+3}\right )-\frac {581 \text {arctanh}\left (\sqrt {\frac {3}{5}} \sqrt {2 x+3}\right )}{\sqrt {15}}\right )-\frac {\sqrt {2 x+3} (131 x+112)}{3 x^2+5 x+2}\right )-\frac {(139 x+121) (2 x+3)^{3/2}}{6 \left (3 x^2+5 x+2\right )^2}\)

Input:

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

Output:

-1/6*((3 + 2*x)^(3/2)*(121 + 139*x))/(2 + 5*x + 3*x^2)^2 - (25*(-((Sqrt[3 
+ 2*x]*(112 + 131*x))/(2 + 5*x + 3*x^2)) - 2*(150*ArcTanh[Sqrt[3 + 2*x]] - 
 (581*ArcTanh[Sqrt[3/5]*Sqrt[3 + 2*x]])/Sqrt[15])))/6
 

Defintions of rubi rules used

rule 27
Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

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 1233
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c 
_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(-(d + e*x)^(m - 1))*(a + b*x + c*x^2) 
^(p + 1)*((2*a*c*(e*f + d*g) - b*(c*d*f + a*e*g) - (2*c^2*d*f + b^2*e*g - c 
*(b*e*f + b*d*g + 2*a*e*g))*x)/(c*(p + 1)*(b^2 - 4*a*c))), x] - Simp[1/(c*( 
p + 1)*(b^2 - 4*a*c))   Int[(d + e*x)^(m - 2)*(a + b*x + c*x^2)^(p + 1)*Sim 
p[2*c^2*d^2*f*(2*p + 3) + b*e*g*(a*e*(m - 1) + b*d*(p + 2)) - c*(2*a*e*(e*f 
*(m - 1) + d*g*m) + b*d*(d*g*(2*p + 3) - e*f*(m - 2*p - 4))) + e*(b^2*e*g*( 
m + p + 1) + 2*c^2*d*f*(m + 2*p + 2) - c*(2*a*e*g*m + b*(e*f + d*g)*(m + 2* 
p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && LtQ[p, -1] && 
GtQ[m, 1] && ((EqQ[m, 2] && EqQ[p, -3] && RationalQ[a, b, c, d, e, f, g]) | 
|  !ILtQ[m + 2*p + 3, 0])
 

rule 1234
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c 
_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d + e*x)^m*(a + b*x + c*x^2)^(p + 1)*( 
(f*b - 2*a*g + (2*c*f - b*g)*x)/((p + 1)*(b^2 - 4*a*c))), x] + Simp[1/((p + 
 1)*(b^2 - 4*a*c))   Int[(d + e*x)^(m - 1)*(a + b*x + c*x^2)^(p + 1)*Simp[g 
*(2*a*e*m + b*d*(2*p + 3)) - f*(b*e*m + 2*c*d*(2*p + 3)) - e*(2*c*f - b*g)* 
(m + 2*p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && LtQ[p, -1 
] && GtQ[m, 0] && (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 = 1.54 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.78

method result size
risch \(\frac {\left (9825 x^{3}+24497 x^{2}+19891 x +5237\right ) \sqrt {2 x +3}}{6 \left (3 x^{2}+5 x +2\right )^{2}}+625 \ln \left (\sqrt {2 x +3}+1\right )-625 \ln \left (\sqrt {2 x +3}-1\right )-\frac {2905 \sqrt {15}\, \operatorname {arctanh}\left (\frac {\sqrt {15}\, \sqrt {2 x +3}}{5}\right )}{9}\) \(80\)
trager \(\frac {\left (9825 x^{3}+24497 x^{2}+19891 x +5237\right ) \sqrt {2 x +3}}{6 \left (3 x^{2}+5 x +2\right )^{2}}+625 \ln \left (\frac {\sqrt {2 x +3}+2+x}{x +1}\right )-\frac {2905 \operatorname {RootOf}\left (\textit {\_Z}^{2}-15\right ) \ln \left (\frac {3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-15\right ) x +15 \sqrt {2 x +3}+7 \operatorname {RootOf}\left (\textit {\_Z}^{2}-15\right )}{3 x +2}\right )}{18}\) \(101\)
derivativedivides \(\frac {3}{\left (\sqrt {2 x +3}-1\right )^{2}}+\frac {80}{\sqrt {2 x +3}-1}-625 \ln \left (\sqrt {2 x +3}-1\right )-\frac {3}{\left (\sqrt {2 x +3}+1\right )^{2}}+\frac {80}{\sqrt {2 x +3}+1}+625 \ln \left (\sqrt {2 x +3}+1\right )+\frac {1835 \left (2 x +3\right )^{\frac {3}{2}}-\frac {10025 \sqrt {2 x +3}}{3}}{\left (6 x +4\right )^{2}}-\frac {2905 \sqrt {15}\, \operatorname {arctanh}\left (\frac {\sqrt {15}\, \sqrt {2 x +3}}{5}\right )}{9}\) \(124\)
default \(\frac {3}{\left (\sqrt {2 x +3}-1\right )^{2}}+\frac {80}{\sqrt {2 x +3}-1}-625 \ln \left (\sqrt {2 x +3}-1\right )-\frac {3}{\left (\sqrt {2 x +3}+1\right )^{2}}+\frac {80}{\sqrt {2 x +3}+1}+625 \ln \left (\sqrt {2 x +3}+1\right )+\frac {1835 \left (2 x +3\right )^{\frac {3}{2}}-\frac {10025 \sqrt {2 x +3}}{3}}{\left (6 x +4\right )^{2}}-\frac {2905 \sqrt {15}\, \operatorname {arctanh}\left (\frac {\sqrt {15}\, \sqrt {2 x +3}}{5}\right )}{9}\) \(124\)
pseudoelliptic \(\frac {-11620 \sqrt {15}\, \left (x +1\right )^{2} \left (x +\frac {2}{3}\right )^{2} \operatorname {arctanh}\left (\frac {\sqrt {15}\, \sqrt {2 x +3}}{5}\right )-22500 \left (x +1\right )^{2} \left (x +\frac {2}{3}\right )^{2} \ln \left (\sqrt {2 x +3}-1\right )+22500 \left (x +1\right )^{2} \left (x +\frac {2}{3}\right )^{2} \ln \left (\sqrt {2 x +3}+1\right )+6550 \left (x^{3}+\frac {187}{75} x^{2}+\frac {19891}{9825} x +\frac {5237}{9825}\right ) \sqrt {2 x +3}}{\left (\sqrt {2 x +3}-1\right )^{2} \left (\sqrt {2 x +3}+1\right )^{2} \left (3 x +2\right )^{2}}\) \(127\)

Input:

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

Output:

1/6*(9825*x^3+24497*x^2+19891*x+5237)/(3*x^2+5*x+2)^2*(2*x+3)^(1/2)+625*ln 
((2*x+3)^(1/2)+1)-625*ln((2*x+3)^(1/2)-1)-2905/9*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. 164 vs. \(2 (81) = 162\).

Time = 0.11 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.61 \[ \int \frac {(5-x) (3+2 x)^{5/2}}{\left (2+5 x+3 x^2\right )^3} \, dx=\frac {2905 \, \sqrt {\frac {5}{3}} {\left (9 \, x^{4} + 30 \, x^{3} + 37 \, x^{2} + 20 \, x + 4\right )} \log \left (-\frac {3 \, \sqrt {\frac {5}{3}} \sqrt {2 \, x + 3} - 3 \, x - 7}{3 \, x + 2}\right ) + 3750 \, {\left (9 \, x^{4} + 30 \, x^{3} + 37 \, x^{2} + 20 \, x + 4\right )} \log \left (\sqrt {2 \, x + 3} + 1\right ) - 3750 \, {\left (9 \, x^{4} + 30 \, x^{3} + 37 \, x^{2} + 20 \, x + 4\right )} \log \left (\sqrt {2 \, x + 3} - 1\right ) + {\left (9825 \, x^{3} + 24497 \, x^{2} + 19891 \, x + 5237\right )} \sqrt {2 \, x + 3}}{6 \, {\left (9 \, x^{4} + 30 \, x^{3} + 37 \, x^{2} + 20 \, x + 4\right )}} \] Input:

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

Output:

1/6*(2905*sqrt(5/3)*(9*x^4 + 30*x^3 + 37*x^2 + 20*x + 4)*log(-(3*sqrt(5/3) 
*sqrt(2*x + 3) - 3*x - 7)/(3*x + 2)) + 3750*(9*x^4 + 30*x^3 + 37*x^2 + 20* 
x + 4)*log(sqrt(2*x + 3) + 1) - 3750*(9*x^4 + 30*x^3 + 37*x^2 + 20*x + 4)* 
log(sqrt(2*x + 3) - 1) + (9825*x^3 + 24497*x^2 + 19891*x + 5237)*sqrt(2*x 
+ 3))/(9*x^4 + 30*x^3 + 37*x^2 + 20*x + 4)
 

Sympy [A] (verification not implemented)

Time = 102.99 (sec) , antiderivative size = 406, normalized size of antiderivative = 3.98 \[ \int \frac {(5-x) (3+2 x)^{5/2}}{\left (2+5 x+3 x^2\right )^3} \, dx=141 \sqrt {15} \left (\log {\left (\sqrt {2 x + 3} - \frac {\sqrt {15}}{3} \right )} - \log {\left (\sqrt {2 x + 3} + \frac {\sqrt {15}}{3} \right )}\right ) - \frac {15800 \left (\begin {cases} \frac {\sqrt {15} \left (- \frac {\log {\left (\frac {\sqrt {15} \sqrt {2 x + 3}}{5} - 1 \right )}}{4} + \frac {\log {\left (\frac {\sqrt {15} \sqrt {2 x + 3}}{5} + 1 \right )}}{4} - \frac {1}{4 \left (\frac {\sqrt {15} \sqrt {2 x + 3}}{5} + 1\right )} - \frac {1}{4 \left (\frac {\sqrt {15} \sqrt {2 x + 3}}{5} - 1\right )}\right )}{75} & \text {for}\: \sqrt {2 x + 3} > - \frac {\sqrt {15}}{3} \wedge \sqrt {2 x + 3} < \frac {\sqrt {15}}{3} \end {cases}\right )}{3} + \frac {17000 \left (\begin {cases} \frac {\sqrt {15} \cdot \left (\frac {3 \log {\left (\frac {\sqrt {15} \sqrt {2 x + 3}}{5} - 1 \right )}}{16} - \frac {3 \log {\left (\frac {\sqrt {15} \sqrt {2 x + 3}}{5} + 1 \right )}}{16} + \frac {3}{16 \left (\frac {\sqrt {15} \sqrt {2 x + 3}}{5} + 1\right )} + \frac {1}{16 \left (\frac {\sqrt {15} \sqrt {2 x + 3}}{5} + 1\right )^{2}} + \frac {3}{16 \left (\frac {\sqrt {15} \sqrt {2 x + 3}}{5} - 1\right )} - \frac {1}{16 \left (\frac {\sqrt {15} \sqrt {2 x + 3}}{5} - 1\right )^{2}}\right )}{375} & \text {for}\: \sqrt {2 x + 3} > - \frac {\sqrt {15}}{3} \wedge \sqrt {2 x + 3} < \frac {\sqrt {15}}{3} \end {cases}\right )}{3} - 625 \log {\left (\sqrt {2 x + 3} - 1 \right )} + 625 \log {\left (\sqrt {2 x + 3} + 1 \right )} + \frac {80}{\sqrt {2 x + 3} + 1} - \frac {3}{\left (\sqrt {2 x + 3} + 1\right )^{2}} + \frac {80}{\sqrt {2 x + 3} - 1} + \frac {3}{\left (\sqrt {2 x + 3} - 1\right )^{2}} \] Input:

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

Output:

141*sqrt(15)*(log(sqrt(2*x + 3) - sqrt(15)/3) - log(sqrt(2*x + 3) + sqrt(1 
5)/3)) - 15800*Piecewise((sqrt(15)*(-log(sqrt(15)*sqrt(2*x + 3)/5 - 1)/4 + 
 log(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)))/3 + 17000*Piecewise((sqrt(15)*(3*log(sq 
rt(15)*sqrt(2*x + 3)/5 - 1)/16 - 3*log(sqrt(15)*sqrt(2*x + 3)/5 + 1)/16 + 
3/(16*(sqrt(15)*sqrt(2*x + 3)/5 + 1)) + 1/(16*(sqrt(15)*sqrt(2*x + 3)/5 + 
1)**2) + 3/(16*(sqrt(15)*sqrt(2*x + 3)/5 - 1)) - 1/(16*(sqrt(15)*sqrt(2*x 
+ 3)/5 - 1)**2))/375, (sqrt(2*x + 3) > -sqrt(15)/3) & (sqrt(2*x + 3) < sqr 
t(15)/3)))/3 - 625*log(sqrt(2*x + 3) - 1) + 625*log(sqrt(2*x + 3) + 1) + 8 
0/(sqrt(2*x + 3) + 1) - 3/(sqrt(2*x + 3) + 1)**2 + 80/(sqrt(2*x + 3) - 1) 
+ 3/(sqrt(2*x + 3) - 1)**2
 

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.31 \[ \int \frac {(5-x) (3+2 x)^{5/2}}{\left (2+5 x+3 x^2\right )^3} \, dx=\frac {2905}{18} \, \sqrt {15} \log \left (-\frac {\sqrt {15} - 3 \, \sqrt {2 \, x + 3}}{\sqrt {15} + 3 \, \sqrt {2 \, x + 3}}\right ) + \frac {9825 \, {\left (2 \, x + 3\right )}^{\frac {7}{2}} - 39431 \, {\left (2 \, x + 3\right )}^{\frac {5}{2}} + 50875 \, {\left (2 \, x + 3\right )}^{\frac {3}{2}} - 21125 \, \sqrt {2 \, x + 3}}{3 \, {\left (9 \, {\left (2 \, x + 3\right )}^{4} - 48 \, {\left (2 \, x + 3\right )}^{3} + 94 \, {\left (2 \, x + 3\right )}^{2} - 160 \, x - 215\right )}} + 625 \, \log \left (\sqrt {2 \, x + 3} + 1\right ) - 625 \, \log \left (\sqrt {2 \, x + 3} - 1\right ) \] Input:

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

Output:

2905/18*sqrt(15)*log(-(sqrt(15) - 3*sqrt(2*x + 3))/(sqrt(15) + 3*sqrt(2*x 
+ 3))) + 1/3*(9825*(2*x + 3)^(7/2) - 39431*(2*x + 3)^(5/2) + 50875*(2*x + 
3)^(3/2) - 21125*sqrt(2*x + 3))/(9*(2*x + 3)^4 - 48*(2*x + 3)^3 + 94*(2*x 
+ 3)^2 - 160*x - 215) + 625*log(sqrt(2*x + 3) + 1) - 625*log(sqrt(2*x + 3) 
 - 1)
 

Giac [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.18 \[ \int \frac {(5-x) (3+2 x)^{5/2}}{\left (2+5 x+3 x^2\right )^3} \, dx=\frac {2905}{18} \, \sqrt {15} \log \left (\frac {{\left | -2 \, \sqrt {15} + 6 \, \sqrt {2 \, x + 3} \right |}}{2 \, {\left (\sqrt {15} + 3 \, \sqrt {2 \, x + 3}\right )}}\right ) + \frac {9825 \, {\left (2 \, x + 3\right )}^{\frac {7}{2}} - 39431 \, {\left (2 \, x + 3\right )}^{\frac {5}{2}} + 50875 \, {\left (2 \, x + 3\right )}^{\frac {3}{2}} - 21125 \, \sqrt {2 \, x + 3}}{3 \, {\left (3 \, {\left (2 \, x + 3\right )}^{2} - 16 \, x - 19\right )}^{2}} + 625 \, \log \left (\sqrt {2 \, x + 3} + 1\right ) - 625 \, \log \left ({\left | \sqrt {2 \, x + 3} - 1 \right |}\right ) \] Input:

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

Output:

2905/18*sqrt(15)*log(1/2*abs(-2*sqrt(15) + 6*sqrt(2*x + 3))/(sqrt(15) + 3* 
sqrt(2*x + 3))) + 1/3*(9825*(2*x + 3)^(7/2) - 39431*(2*x + 3)^(5/2) + 5087 
5*(2*x + 3)^(3/2) - 21125*sqrt(2*x + 3))/(3*(2*x + 3)^2 - 16*x - 19)^2 + 6 
25*log(sqrt(2*x + 3) + 1) - 625*log(abs(sqrt(2*x + 3) - 1))
 

Mupad [B] (verification not implemented)

Time = 11.38 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.99 \[ \int \frac {(5-x) (3+2 x)^{5/2}}{\left (2+5 x+3 x^2\right )^3} \, dx=1250\,\mathrm {atanh}\left (\sqrt {2\,x+3}\right )+\frac {\frac {21125\,\sqrt {2\,x+3}}{27}-\frac {50875\,{\left (2\,x+3\right )}^{3/2}}{27}+\frac {39431\,{\left (2\,x+3\right )}^{5/2}}{27}-\frac {3275\,{\left (2\,x+3\right )}^{7/2}}{9}}{\frac {160\,x}{9}-\frac {94\,{\left (2\,x+3\right )}^2}{9}+\frac {16\,{\left (2\,x+3\right )}^3}{3}-{\left (2\,x+3\right )}^4+\frac {215}{9}}-\frac {2905\,\sqrt {15}\,\mathrm {atanh}\left (\frac {\sqrt {15}\,\sqrt {2\,x+3}}{5}\right )}{9} \] Input:

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

Output:

1250*atanh((2*x + 3)^(1/2)) + ((21125*(2*x + 3)^(1/2))/27 - (50875*(2*x + 
3)^(3/2))/27 + (39431*(2*x + 3)^(5/2))/27 - (3275*(2*x + 3)^(7/2))/9)/((16 
0*x)/9 - (94*(2*x + 3)^2)/9 + (16*(2*x + 3)^3)/3 - (2*x + 3)^4 + 215/9) - 
(2905*15^(1/2)*atanh((15^(1/2)*(2*x + 3)^(1/2))/5))/9
 

Reduce [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 383, normalized size of antiderivative = 3.75 \[ \int \frac {(5-x) (3+2 x)^{5/2}}{\left (2+5 x+3 x^2\right )^3} \, dx=\frac {29475 \sqrt {2 x +3}\, x^{3}+73491 \sqrt {2 x +3}\, x^{2}+59673 \sqrt {2 x +3}\, x +15711 \sqrt {2 x +3}+26145 \sqrt {15}\, \mathrm {log}\left (3 \sqrt {2 x +3}-\sqrt {15}\right ) x^{4}+87150 \sqrt {15}\, \mathrm {log}\left (3 \sqrt {2 x +3}-\sqrt {15}\right ) x^{3}+107485 \sqrt {15}\, \mathrm {log}\left (3 \sqrt {2 x +3}-\sqrt {15}\right ) x^{2}+58100 \sqrt {15}\, \mathrm {log}\left (3 \sqrt {2 x +3}-\sqrt {15}\right ) x +11620 \sqrt {15}\, \mathrm {log}\left (3 \sqrt {2 x +3}-\sqrt {15}\right )-26145 \sqrt {15}\, \mathrm {log}\left (3 \sqrt {2 x +3}+\sqrt {15}\right ) x^{4}-87150 \sqrt {15}\, \mathrm {log}\left (3 \sqrt {2 x +3}+\sqrt {15}\right ) x^{3}-107485 \sqrt {15}\, \mathrm {log}\left (3 \sqrt {2 x +3}+\sqrt {15}\right ) x^{2}-58100 \sqrt {15}\, \mathrm {log}\left (3 \sqrt {2 x +3}+\sqrt {15}\right ) x -11620 \sqrt {15}\, \mathrm {log}\left (3 \sqrt {2 x +3}+\sqrt {15}\right )-101250 \,\mathrm {log}\left (\sqrt {2 x +3}-1\right ) x^{4}-337500 \,\mathrm {log}\left (\sqrt {2 x +3}-1\right ) x^{3}-416250 \,\mathrm {log}\left (\sqrt {2 x +3}-1\right ) x^{2}-225000 \,\mathrm {log}\left (\sqrt {2 x +3}-1\right ) x -45000 \,\mathrm {log}\left (\sqrt {2 x +3}-1\right )+101250 \,\mathrm {log}\left (\sqrt {2 x +3}+1\right ) x^{4}+337500 \,\mathrm {log}\left (\sqrt {2 x +3}+1\right ) x^{3}+416250 \,\mathrm {log}\left (\sqrt {2 x +3}+1\right ) x^{2}+225000 \,\mathrm {log}\left (\sqrt {2 x +3}+1\right ) x +45000 \,\mathrm {log}\left (\sqrt {2 x +3}+1\right )}{162 x^{4}+540 x^{3}+666 x^{2}+360 x +72} \] Input:

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

Output:

(29475*sqrt(2*x + 3)*x**3 + 73491*sqrt(2*x + 3)*x**2 + 59673*sqrt(2*x + 3) 
*x + 15711*sqrt(2*x + 3) + 26145*sqrt(15)*log(3*sqrt(2*x + 3) - sqrt(15))* 
x**4 + 87150*sqrt(15)*log(3*sqrt(2*x + 3) - sqrt(15))*x**3 + 107485*sqrt(1 
5)*log(3*sqrt(2*x + 3) - sqrt(15))*x**2 + 58100*sqrt(15)*log(3*sqrt(2*x + 
3) - sqrt(15))*x + 11620*sqrt(15)*log(3*sqrt(2*x + 3) - sqrt(15)) - 26145* 
sqrt(15)*log(3*sqrt(2*x + 3) + sqrt(15))*x**4 - 87150*sqrt(15)*log(3*sqrt( 
2*x + 3) + sqrt(15))*x**3 - 107485*sqrt(15)*log(3*sqrt(2*x + 3) + sqrt(15) 
)*x**2 - 58100*sqrt(15)*log(3*sqrt(2*x + 3) + sqrt(15))*x - 11620*sqrt(15) 
*log(3*sqrt(2*x + 3) + sqrt(15)) - 101250*log(sqrt(2*x + 3) - 1)*x**4 - 33 
7500*log(sqrt(2*x + 3) - 1)*x**3 - 416250*log(sqrt(2*x + 3) - 1)*x**2 - 22 
5000*log(sqrt(2*x + 3) - 1)*x - 45000*log(sqrt(2*x + 3) - 1) + 101250*log( 
sqrt(2*x + 3) + 1)*x**4 + 337500*log(sqrt(2*x + 3) + 1)*x**3 + 416250*log( 
sqrt(2*x + 3) + 1)*x**2 + 225000*log(sqrt(2*x + 3) + 1)*x + 45000*log(sqrt 
(2*x + 3) + 1))/(18*(9*x**4 + 30*x**3 + 37*x**2 + 20*x + 4))