Integrand size = 27, antiderivative size = 107 \[ \int \frac {(5-x) (3+2 x)^3}{\sqrt {2+5 x+3 x^2}} \, dx=\frac {32}{27} (3+2 x)^2 \sqrt {2+5 x+3 x^2}-\frac {1}{12} (3+2 x)^3 \sqrt {2+5 x+3 x^2}+\frac {5}{648} (3261+1078 x) \sqrt {2+5 x+3 x^2}+\frac {19405 \text {arctanh}\left (\frac {\sqrt {3} (1+x)}{\sqrt {2+5 x+3 x^2}}\right )}{648 \sqrt {3}} \] Output:
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)+5/6 48*(3261+1078*x)*(3*x^2+5*x+2)^(1/2)+19405/1944*arctanh(3^(1/2)*(1+x)/(3*x ^2+5*x+2)^(1/2))*3^(1/2)
Time = 0.45 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.62 \[ \int \frac {(5-x) (3+2 x)^3}{\sqrt {2+5 x+3 x^2}} \, dx=\frac {-3 \sqrt {2+5 x+3 x^2} \left (-21759-11690 x-1128 x^2+432 x^3\right )+19405 \sqrt {3} \text {arctanh}\left (\frac {\sqrt {\frac {2}{3}+\frac {5 x}{3}+x^2}}{1+x}\right )}{1944} \] Input:
Integrate[((5 - x)*(3 + 2*x)^3)/Sqrt[2 + 5*x + 3*x^2],x]
Output:
(-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
Time = 0.44 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.14, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {1236, 27, 1236, 27, 1225, 1092, 219}
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.
\(\displaystyle \int \frac {(5-x) (2 x+3)^3}{\sqrt {3 x^2+5 x+2}} \, dx\) |
\(\Big \downarrow \) 1236 |
\(\displaystyle \frac {1}{12} \int \frac {(2 x+3)^2 (256 x+399)}{2 \sqrt {3 x^2+5 x+2}}dx-\frac {1}{12} (2 x+3)^3 \sqrt {3 x^2+5 x+2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{24} \int \frac {(2 x+3)^2 (256 x+399)}{\sqrt {3 x^2+5 x+2}}dx-\frac {1}{12} (2 x+3)^3 \sqrt {3 x^2+5 x+2}\) |
\(\Big \downarrow \) 1236 |
\(\displaystyle \frac {1}{24} \left (\frac {1}{9} \int \frac {5 (2 x+3) (1078 x+1361)}{\sqrt {3 x^2+5 x+2}}dx+\frac {256}{9} \sqrt {3 x^2+5 x+2} (2 x+3)^2\right )-\frac {1}{12} (2 x+3)^3 \sqrt {3 x^2+5 x+2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{24} \left (\frac {5}{9} \int \frac {(2 x+3) (1078 x+1361)}{\sqrt {3 x^2+5 x+2}}dx+\frac {256}{9} \sqrt {3 x^2+5 x+2} (2 x+3)^2\right )-\frac {1}{12} (2 x+3)^3 \sqrt {3 x^2+5 x+2}\) |
\(\Big \downarrow \) 1225 |
\(\displaystyle \frac {1}{24} \left (\frac {5}{9} \left (\frac {3881}{6} \int \frac {1}{\sqrt {3 x^2+5 x+2}}dx+\frac {1}{3} \sqrt {3 x^2+5 x+2} (1078 x+3261)\right )+\frac {256}{9} \sqrt {3 x^2+5 x+2} (2 x+3)^2\right )-\frac {1}{12} (2 x+3)^3 \sqrt {3 x^2+5 x+2}\) |
\(\Big \downarrow \) 1092 |
\(\displaystyle \frac {1}{24} \left (\frac {5}{9} \left (\frac {3881}{3} \int \frac {1}{12-\frac {(6 x+5)^2}{3 x^2+5 x+2}}d\frac {6 x+5}{\sqrt {3 x^2+5 x+2}}+\frac {1}{3} \sqrt {3 x^2+5 x+2} (1078 x+3261)\right )+\frac {256}{9} \sqrt {3 x^2+5 x+2} (2 x+3)^2\right )-\frac {1}{12} (2 x+3)^3 \sqrt {3 x^2+5 x+2}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{24} \left (\frac {5}{9} \left (\frac {3881 \text {arctanh}\left (\frac {6 x+5}{2 \sqrt {3} \sqrt {3 x^2+5 x+2}}\right )}{6 \sqrt {3}}+\frac {1}{3} \sqrt {3 x^2+5 x+2} (1078 x+3261)\right )+\frac {256}{9} \sqrt {3 x^2+5 x+2} (2 x+3)^2\right )-\frac {1}{12} (2 x+3)^3 \sqrt {3 x^2+5 x+2}\) |
Input:
Int[((5 - x)*(3 + 2*x)^3)/Sqrt[2 + 5*x + 3*x^2],x]
Output:
-1/12*((3 + 2*x)^3*Sqrt[2 + 5*x + 3*x^2]) + ((256*(3 + 2*x)^2*Sqrt[2 + 5*x + 3*x^2])/9 + (5*(((3261 + 1078*x)*Sqrt[2 + 5*x + 3*x^2])/3 + (3881*ArcTa nh[(5 + 6*x)/(2*Sqrt[3]*Sqrt[2 + 5*x + 3*x^2])])/(6*Sqrt[3])))/9)/24
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[2 Subst[I nt[1/(4*c - x^2), x], x, (b + 2*c*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a , b, c}, x]
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] + Simp[(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] && !LeQ[p, -1]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)^(p_.), x_Symbol] :> Simp[g*(d + e*x)^m*((a + b*x + c*x^2)^(p + 1)/(c*(m + 2*p + 2))), x] + Simp[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] && GtQ[m, 0] && NeQ[m + 2*p + 2, 0] && (Intege rQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p]) && !(IGtQ[m, 0] && EqQ[f, 0])
Time = 1.57 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.56
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 | \(\left (-\frac {2}{3} x^{3}+\frac {47}{27} x^{2}+\frac {5845}{324} x +\frac {7253}{216}\right ) \sqrt {3 x^{2}+5 x +2}+\frac {19405 \operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right ) \ln \left (6 \operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right ) x +5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right )+6 \sqrt {3 x^{2}+5 x +2}\right )}{3888}\) | \(71\) |
default | \(\frac {19405 \ln \left (\frac {\left (\frac {5}{2}+3 x \right ) \sqrt {3}}{3}+\sqrt {3 x^{2}+5 x +2}\right ) \sqrt {3}}{3888}+\frac {7253 \sqrt {3 x^{2}+5 x +2}}{216}+\frac {5845 x \sqrt {3 x^{2}+5 x +2}}{324}+\frac {47 x^{2} \sqrt {3 x^{2}+5 x +2}}{27}-\frac {2 x^{3} \sqrt {3 x^{2}+5 x +2}}{3}\) | \(94\) |
Input:
int((5-x)*(2*x+3)^3/(3*x^2+5*x+2)^(1/2),x,method=_RETURNVERBOSE)
Output:
-1/648*(432*x^3-1128*x^2-11690*x-21759)*(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)
Time = 0.09 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.64 \[ \int \frac {(5-x) (3+2 x)^3}{\sqrt {2+5 x+3 x^2}} \, dx=-\frac {1}{648} \, {\left (432 \, x^{3} - 1128 \, x^{2} - 11690 \, x - 21759\right )} \sqrt {3 \, x^{2} + 5 \, x + 2} + \frac {19405}{7776} \, \sqrt {3} \log \left (4 \, \sqrt {3} \sqrt {3 \, x^{2} + 5 \, x + 2} {\left (6 \, x + 5\right )} + 72 \, x^{2} + 120 \, x + 49\right ) \] Input:
integrate((5-x)*(3+2*x)^3/(3*x^2+5*x+2)^(1/2),x, algorithm="fricas")
Output:
-1/648*(432*x^3 - 1128*x^2 - 11690*x - 21759)*sqrt(3*x^2 + 5*x + 2) + 1940 5/7776*sqrt(3)*log(4*sqrt(3)*sqrt(3*x^2 + 5*x + 2)*(6*x + 5) + 72*x^2 + 12 0*x + 49)
Time = 0.55 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.65 \[ \int \frac {(5-x) (3+2 x)^3}{\sqrt {2+5 x+3 x^2}} \, dx=\sqrt {3 x^{2} + 5 x + 2} \left (- \frac {2 x^{3}}{3} + \frac {47 x^{2}}{27} + \frac {5845 x}{324} + \frac {7253}{216}\right ) + \frac {19405 \sqrt {3} \log {\left (6 x + 2 \sqrt {3} \sqrt {3 x^{2} + 5 x + 2} + 5 \right )}}{3888} \] Input:
integrate((5-x)*(3+2*x)**3/(3*x**2+5*x+2)**(1/2),x)
Output:
sqrt(3*x**2 + 5*x + 2)*(-2*x**3/3 + 47*x**2/27 + 5845*x/324 + 7253/216) + 19405*sqrt(3)*log(6*x + 2*sqrt(3)*sqrt(3*x**2 + 5*x + 2) + 5)/3888
Time = 0.11 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.86 \[ \int \frac {(5-x) (3+2 x)^3}{\sqrt {2+5 x+3 x^2}} \, dx=-\frac {2}{3} \, \sqrt {3 \, x^{2} + 5 \, x + 2} x^{3} + \frac {47}{27} \, \sqrt {3 \, x^{2} + 5 \, x + 2} x^{2} + \frac {5845}{324} \, \sqrt {3 \, x^{2} + 5 \, x + 2} x + \frac {19405}{3888} \, \sqrt {3} \log \left (2 \, \sqrt {3} \sqrt {3 \, x^{2} + 5 \, x + 2} + 6 \, x + 5\right ) + \frac {7253}{216} \, \sqrt {3 \, x^{2} + 5 \, x + 2} \] Input:
integrate((5-x)*(3+2*x)^3/(3*x^2+5*x+2)^(1/2),x, algorithm="maxima")
Output:
-2/3*sqrt(3*x^2 + 5*x + 2)*x^3 + 47/27*sqrt(3*x^2 + 5*x + 2)*x^2 + 5845/32 4*sqrt(3*x^2 + 5*x + 2)*x + 19405/3888*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)
Time = 0.36 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.60 \[ \int \frac {(5-x) (3+2 x)^3}{\sqrt {2+5 x+3 x^2}} \, dx=-\frac {1}{648} \, {\left (2 \, {\left (12 \, {\left (18 \, x - 47\right )} x - 5845\right )} x - 21759\right )} \sqrt {3 \, x^{2} + 5 \, x + 2} - \frac {19405}{3888} \, \sqrt {3} \log \left ({\left | -2 \, \sqrt {3} {\left (\sqrt {3} x - \sqrt {3 \, x^{2} + 5 \, x + 2}\right )} - 5 \right |}\right ) \] Input:
integrate((5-x)*(3+2*x)^3/(3*x^2+5*x+2)^(1/2),x, algorithm="giac")
Output:
-1/648*(2*(12*(18*x - 47)*x - 5845)*x - 21759)*sqrt(3*x^2 + 5*x + 2) - 194 05/3888*sqrt(3)*log(abs(-2*sqrt(3)*(sqrt(3)*x - sqrt(3*x^2 + 5*x + 2)) - 5 ))
Timed out. \[ \int \frac {(5-x) (3+2 x)^3}{\sqrt {2+5 x+3 x^2}} \, dx=-\int \frac {{\left (2\,x+3\right )}^3\,\left (x-5\right )}{\sqrt {3\,x^2+5\,x+2}} \,d x \] Input:
int(-((2*x + 3)^3*(x - 5))/(5*x + 3*x^2 + 2)^(1/2),x)
Output:
-int(((2*x + 3)^3*(x - 5))/(5*x + 3*x^2 + 2)^(1/2), x)
Time = 0.19 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.79 \[ \int \frac {(5-x) (3+2 x)^3}{\sqrt {2+5 x+3 x^2}} \, dx=-\frac {2 \sqrt {3 x^{2}+5 x +2}\, x^{3}}{3}+\frac {47 \sqrt {3 x^{2}+5 x +2}\, x^{2}}{27}+\frac {5845 \sqrt {3 x^{2}+5 x +2}\, x}{324}+\frac {7253 \sqrt {3 x^{2}+5 x +2}}{216}+\frac {19405 \sqrt {3}\, \mathrm {log}\left (2 \sqrt {3 x^{2}+5 x +2}\, \sqrt {3}+6 x +5\right )}{3888} \] Input:
int((5-x)*(3+2*x)^3/(3*x^2+5*x+2)^(1/2),x)
Output:
( - 2592*sqrt(3*x**2 + 5*x + 2)*x**3 + 6768*sqrt(3*x**2 + 5*x + 2)*x**2 + 70140*sqrt(3*x**2 + 5*x + 2)*x + 130554*sqrt(3*x**2 + 5*x + 2) + 19405*sqr t(3)*log(2*sqrt(3*x**2 + 5*x + 2)*sqrt(3) + 6*x + 5))/3888