Integrand size = 27, antiderivative size = 110 \[ \int (5-x) (3+2 x)^2 \sqrt {2+5 x+3 x^2} \, dx=\frac {2267 (5+6 x) \sqrt {2+5 x+3 x^2}}{2592}-\frac {1}{15} (3+2 x)^2 \left (2+5 x+3 x^2\right )^{3/2}+\frac {(7969+3006 x) \left (2+5 x+3 x^2\right )^{3/2}}{1620}-\frac {2267 \text {arctanh}\left (\frac {5+6 x}{2 \sqrt {3} \sqrt {2+5 x+3 x^2}}\right )}{5184 \sqrt {3}} \]
-1/15*(3+2*x)^2*(3*x^2+5*x+2)^(3/2)+1/1620*(7969+3006*x)*(3*x^2+5*x+2)^(3/ 2)-2267/15552*arctanh(1/6*(5+6*x)*3^(1/2)/(3*x^2+5*x+2)^(1/2))*3^(1/2)+226 7/2592*(5+6*x)*(3*x^2+5*x+2)^(1/2)
Time = 0.28 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.65 \[ \int (5-x) (3+2 x)^2 \sqrt {2+5 x+3 x^2} \, dx=\frac {-3 \sqrt {2+5 x+3 x^2} \left (-168627-375250 x-229416 x^2-23760 x^3+10368 x^4\right )-11335 \sqrt {3} \text {arctanh}\left (\frac {\sqrt {\frac {2}{3}+\frac {5 x}{3}+x^2}}{1+x}\right )}{38880} \]
(-3*Sqrt[2 + 5*x + 3*x^2]*(-168627 - 375250*x - 229416*x^2 - 23760*x^3 + 1 0368*x^4) - 11335*Sqrt[3]*ArcTanh[Sqrt[2/3 + (5*x)/3 + x^2]/(1 + x)])/3888 0
Time = 0.26 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.09, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {1236, 27, 1225, 1087, 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.
\(\displaystyle \int (5-x) (2 x+3)^2 \sqrt {3 x^2+5 x+2} \, dx\) |
\(\Big \downarrow \) 1236 |
\(\displaystyle \frac {1}{15} \int \frac {1}{2} (2 x+3) (334 x+511) \sqrt {3 x^2+5 x+2}dx-\frac {1}{15} (2 x+3)^2 \left (3 x^2+5 x+2\right )^{3/2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{30} \int (2 x+3) (334 x+511) \sqrt {3 x^2+5 x+2}dx-\frac {1}{15} (2 x+3)^2 \left (3 x^2+5 x+2\right )^{3/2}\) |
\(\Big \downarrow \) 1225 |
\(\displaystyle \frac {1}{30} \left (\frac {11335}{36} \int \sqrt {3 x^2+5 x+2}dx+\frac {1}{54} (3006 x+7969) \left (3 x^2+5 x+2\right )^{3/2}\right )-\frac {1}{15} (2 x+3)^2 \left (3 x^2+5 x+2\right )^{3/2}\) |
\(\Big \downarrow \) 1087 |
\(\displaystyle \frac {1}{30} \left (\frac {11335}{36} \left (\frac {1}{12} (6 x+5) \sqrt {3 x^2+5 x+2}-\frac {1}{24} \int \frac {1}{\sqrt {3 x^2+5 x+2}}dx\right )+\frac {1}{54} (3006 x+7969) \left (3 x^2+5 x+2\right )^{3/2}\right )-\frac {1}{15} (2 x+3)^2 \left (3 x^2+5 x+2\right )^{3/2}\) |
\(\Big \downarrow \) 1092 |
\(\displaystyle \frac {1}{30} \left (\frac {11335}{36} \left (\frac {1}{12} (6 x+5) \sqrt {3 x^2+5 x+2}-\frac {1}{12} \int \frac {1}{12-\frac {(6 x+5)^2}{3 x^2+5 x+2}}d\frac {6 x+5}{\sqrt {3 x^2+5 x+2}}\right )+\frac {1}{54} (3006 x+7969) \left (3 x^2+5 x+2\right )^{3/2}\right )-\frac {1}{15} (2 x+3)^2 \left (3 x^2+5 x+2\right )^{3/2}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{30} \left (\frac {11335}{36} \left (\frac {1}{12} (6 x+5) \sqrt {3 x^2+5 x+2}-\frac {\text {arctanh}\left (\frac {6 x+5}{2 \sqrt {3} \sqrt {3 x^2+5 x+2}}\right )}{24 \sqrt {3}}\right )+\frac {1}{54} (3006 x+7969) \left (3 x^2+5 x+2\right )^{3/2}\right )-\frac {1}{15} (2 x+3)^2 \left (3 x^2+5 x+2\right )^{3/2}\) |
-1/15*((3 + 2*x)^2*(2 + 5*x + 3*x^2)^(3/2)) + (((7969 + 3006*x)*(2 + 5*x + 3*x^2)^(3/2))/54 + (11335*(((5 + 6*x)*Sqrt[2 + 5*x + 3*x^2])/12 - ArcTanh [(5 + 6*x)/(2*Sqrt[3]*Sqrt[2 + 5*x + 3*x^2])]/(24*Sqrt[3])))/36)/30
3.25.8.3.1 Defintions of rubi rules used
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[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b + 2*c*x) *((a + b*x + c*x^2)^p/(2*c*(2*p + 1))), x] - Simp[p*((b^2 - 4*a*c)/(2*c*(2* p + 1))) Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[3*p])
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 = 0.33 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.59
method | result | size |
risch | \(-\frac {\left (10368 x^{4}-23760 x^{3}-229416 x^{2}-375250 x -168627\right ) \sqrt {3 x^{2}+5 x +2}}{12960}-\frac {2267 \ln \left (\frac {\left (\frac {5}{2}+3 x \right ) \sqrt {3}}{3}+\sqrt {3 x^{2}+5 x +2}\right ) \sqrt {3}}{15552}\) | \(65\) |
trager | \(\left (-\frac {4}{5} x^{4}+\frac {11}{6} x^{3}+\frac {9559}{540} x^{2}+\frac {37525}{1296} x +\frac {56209}{4320}\right ) \sqrt {3 x^{2}+5 x +2}-\frac {2267 \operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right ) \ln \left (6 \operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right ) x +5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right )+6 \sqrt {3 x^{2}+5 x +2}\right )}{15552}\) | \(76\) |
default | \(\frac {2267 \left (5+6 x \right ) \sqrt {3 x^{2}+5 x +2}}{2592}-\frac {2267 \ln \left (\frac {\left (\frac {5}{2}+3 x \right ) \sqrt {3}}{3}+\sqrt {3 x^{2}+5 x +2}\right ) \sqrt {3}}{15552}+\frac {6997 \left (3 x^{2}+5 x +2\right )^{\frac {3}{2}}}{1620}-\frac {4 x^{2} \left (3 x^{2}+5 x +2\right )^{\frac {3}{2}}}{15}+\frac {19 x \left (3 x^{2}+5 x +2\right )^{\frac {3}{2}}}{18}\) | \(96\) |
-1/12960*(10368*x^4-23760*x^3-229416*x^2-375250*x-168627)*(3*x^2+5*x+2)^(1 /2)-2267/15552*ln(1/3*(5/2+3*x)*3^(1/2)+(3*x^2+5*x+2)^(1/2))*3^(1/2)
Time = 0.28 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.66 \[ \int (5-x) (3+2 x)^2 \sqrt {2+5 x+3 x^2} \, dx=-\frac {1}{12960} \, {\left (10368 \, x^{4} - 23760 \, x^{3} - 229416 \, x^{2} - 375250 \, x - 168627\right )} \sqrt {3 \, x^{2} + 5 \, x + 2} + \frac {2267}{31104} \, \sqrt {3} \log \left (-4 \, \sqrt {3} \sqrt {3 \, x^{2} + 5 \, x + 2} {\left (6 \, x + 5\right )} + 72 \, x^{2} + 120 \, x + 49\right ) \]
-1/12960*(10368*x^4 - 23760*x^3 - 229416*x^2 - 375250*x - 168627)*sqrt(3*x ^2 + 5*x + 2) + 2267/31104*sqrt(3)*log(-4*sqrt(3)*sqrt(3*x^2 + 5*x + 2)*(6 *x + 5) + 72*x^2 + 120*x + 49)
Time = 0.53 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.69 \[ \int (5-x) (3+2 x)^2 \sqrt {2+5 x+3 x^2} \, dx=\sqrt {3 x^{2} + 5 x + 2} \left (- \frac {4 x^{4}}{5} + \frac {11 x^{3}}{6} + \frac {9559 x^{2}}{540} + \frac {37525 x}{1296} + \frac {56209}{4320}\right ) - \frac {2267 \sqrt {3} \log {\left (6 x + 2 \sqrt {3} \sqrt {3 x^{2} + 5 x + 2} + 5 \right )}}{15552} \]
sqrt(3*x**2 + 5*x + 2)*(-4*x**4/5 + 11*x**3/6 + 9559*x**2/540 + 37525*x/12 96 + 56209/4320) - 2267*sqrt(3)*log(6*x + 2*sqrt(3)*sqrt(3*x**2 + 5*x + 2) + 5)/15552
Time = 0.28 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.95 \[ \int (5-x) (3+2 x)^2 \sqrt {2+5 x+3 x^2} \, dx=-\frac {4}{15} \, {\left (3 \, x^{2} + 5 \, x + 2\right )}^{\frac {3}{2}} x^{2} + \frac {19}{18} \, {\left (3 \, x^{2} + 5 \, x + 2\right )}^{\frac {3}{2}} x + \frac {6997}{1620} \, {\left (3 \, x^{2} + 5 \, x + 2\right )}^{\frac {3}{2}} + \frac {2267}{432} \, \sqrt {3 \, x^{2} + 5 \, x + 2} x - \frac {2267}{15552} \, \sqrt {3} \log \left (2 \, \sqrt {3} \sqrt {3 \, x^{2} + 5 \, x + 2} + 6 \, x + 5\right ) + \frac {11335}{2592} \, \sqrt {3 \, x^{2} + 5 \, x + 2} \]
-4/15*(3*x^2 + 5*x + 2)^(3/2)*x^2 + 19/18*(3*x^2 + 5*x + 2)^(3/2)*x + 6997 /1620*(3*x^2 + 5*x + 2)^(3/2) + 2267/432*sqrt(3*x^2 + 5*x + 2)*x - 2267/15 552*sqrt(3)*log(2*sqrt(3)*sqrt(3*x^2 + 5*x + 2) + 6*x + 5) + 11335/2592*sq rt(3*x^2 + 5*x + 2)
Time = 0.28 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.63 \[ \int (5-x) (3+2 x)^2 \sqrt {2+5 x+3 x^2} \, dx=-\frac {1}{12960} \, {\left (2 \, {\left (12 \, {\left (18 \, {\left (24 \, x - 55\right )} x - 9559\right )} x - 187625\right )} x - 168627\right )} \sqrt {3 \, x^{2} + 5 \, x + 2} + \frac {2267}{15552} \, \sqrt {3} \log \left ({\left | -2 \, \sqrt {3} {\left (\sqrt {3} x - \sqrt {3 \, x^{2} + 5 \, x + 2}\right )} - 5 \right |}\right ) \]
-1/12960*(2*(12*(18*(24*x - 55)*x - 9559)*x - 187625)*x - 168627)*sqrt(3*x ^2 + 5*x + 2) + 2267/15552*sqrt(3)*log(abs(-2*sqrt(3)*(sqrt(3)*x - sqrt(3* x^2 + 5*x + 2)) - 5))
Time = 12.55 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.24 \[ \int (5-x) (3+2 x)^2 \sqrt {2+5 x+3 x^2} \, dx=\frac {386\,\left (\frac {x}{2}+\frac {5}{12}\right )\,\sqrt {3\,x^2+5\,x+2}}{9}-\frac {193\,\sqrt {3}\,\ln \left (\sqrt {3\,x^2+5\,x+2}+\frac {\sqrt {3}\,\left (3\,x+\frac {5}{2}\right )}{3}\right )}{324}-\frac {4\,x^2\,{\left (3\,x^2+5\,x+2\right )}^{3/2}}{15}+\frac {6997\,\sqrt {3\,x^2+5\,x+2}\,\left (72\,x^2+30\,x-27\right )}{38880}+\frac {19\,x\,{\left (3\,x^2+5\,x+2\right )}^{3/2}}{18}+\frac {6997\,\sqrt {3}\,\ln \left (2\,\sqrt {3\,x^2+5\,x+2}+\frac {\sqrt {3}\,\left (6\,x+5\right )}{3}\right )}{15552} \]
(386*(x/2 + 5/12)*(5*x + 3*x^2 + 2)^(1/2))/9 - (193*3^(1/2)*log((5*x + 3*x ^2 + 2)^(1/2) + (3^(1/2)*(3*x + 5/2))/3))/324 - (4*x^2*(5*x + 3*x^2 + 2)^( 3/2))/15 + (6997*(5*x + 3*x^2 + 2)^(1/2)*(30*x + 72*x^2 - 27))/38880 + (19 *x*(5*x + 3*x^2 + 2)^(3/2))/18 + (6997*3^(1/2)*log(2*(5*x + 3*x^2 + 2)^(1/ 2) + (3^(1/2)*(6*x + 5))/3))/15552