Integrand size = 27, antiderivative size = 137 \[ \int \frac {(5-x) (3+2 x)^4}{\sqrt {2+5 x+3 x^2}} \, dx=\frac {391}{135} (3+2 x)^2 \sqrt {2+5 x+3 x^2}+\frac {53}{60} (3+2 x)^3 \sqrt {2+5 x+3 x^2}-\frac {1}{15} (3+2 x)^4 \sqrt {2+5 x+3 x^2}+\frac {1}{648} (27519+9650 x) \sqrt {2+5 x+3 x^2}+\frac {28051 \text {arctanh}\left (\frac {5+6 x}{2 \sqrt {3} \sqrt {2+5 x+3 x^2}}\right )}{1296 \sqrt {3}} \]
28051/3888*arctanh(1/6*(5+6*x)*3^(1/2)/(3*x^2+5*x+2)^(1/2))*3^(1/2)+391/13 5*(3+2*x)^2*(3*x^2+5*x+2)^(1/2)+53/60*(3+2*x)^3*(3*x^2+5*x+2)^(1/2)-1/15*( 3+2*x)^4*(3*x^2+5*x+2)^(1/2)+1/648*(27519+9650*x)*(3*x^2+5*x+2)^(1/2)
Time = 0.37 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.52 \[ \int \frac {(5-x) (3+2 x)^4}{\sqrt {2+5 x+3 x^2}} \, dx=\frac {-3 \sqrt {2+5 x+3 x^2} \left (-281829-268750 x-93912 x^2-2160 x^3+3456 x^4\right )+140255 \sqrt {3} \text {arctanh}\left (\frac {\sqrt {\frac {2}{3}+\frac {5 x}{3}+x^2}}{1+x}\right )}{9720} \]
(-3*Sqrt[2 + 5*x + 3*x^2]*(-281829 - 268750*x - 93912*x^2 - 2160*x^3 + 345 6*x^4) + 140255*Sqrt[3]*ArcTanh[Sqrt[2/3 + (5*x)/3 + x^2]/(1 + x)])/9720
Time = 0.31 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.11, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {1236, 27, 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.
\(\displaystyle \int \frac {(5-x) (2 x+3)^4}{\sqrt {3 x^2+5 x+2}} \, dx\) |
\(\Big \downarrow \) 1236 |
\(\displaystyle \frac {1}{15} \int \frac {(2 x+3)^3 (318 x+497)}{2 \sqrt {3 x^2+5 x+2}}dx-\frac {1}{15} (2 x+3)^4 \sqrt {3 x^2+5 x+2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{30} \int \frac {(2 x+3)^3 (318 x+497)}{\sqrt {3 x^2+5 x+2}}dx-\frac {1}{15} (2 x+3)^4 \sqrt {3 x^2+5 x+2}\) |
\(\Big \downarrow \) 1236 |
\(\displaystyle \frac {1}{30} \left (\frac {1}{12} \int \frac {3 (2 x+3)^2 (3128 x+3897)}{\sqrt {3 x^2+5 x+2}}dx+\frac {53}{2} \sqrt {3 x^2+5 x+2} (2 x+3)^3\right )-\frac {1}{15} (2 x+3)^4 \sqrt {3 x^2+5 x+2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{30} \left (\frac {1}{4} \int \frac {(2 x+3)^2 (3128 x+3897)}{\sqrt {3 x^2+5 x+2}}dx+\frac {53}{2} \sqrt {3 x^2+5 x+2} (2 x+3)^3\right )-\frac {1}{15} (2 x+3)^4 \sqrt {3 x^2+5 x+2}\) |
\(\Big \downarrow \) 1236 |
\(\displaystyle \frac {1}{30} \left (\frac {1}{4} \left (\frac {1}{9} \int \frac {5 (2 x+3) (9650 x+11347)}{\sqrt {3 x^2+5 x+2}}dx+\frac {3128}{9} \sqrt {3 x^2+5 x+2} (2 x+3)^2\right )+\frac {53}{2} \sqrt {3 x^2+5 x+2} (2 x+3)^3\right )-\frac {1}{15} (2 x+3)^4 \sqrt {3 x^2+5 x+2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{30} \left (\frac {1}{4} \left (\frac {5}{9} \int \frac {(2 x+3) (9650 x+11347)}{\sqrt {3 x^2+5 x+2}}dx+\frac {3128}{9} \sqrt {3 x^2+5 x+2} (2 x+3)^2\right )+\frac {53}{2} \sqrt {3 x^2+5 x+2} (2 x+3)^3\right )-\frac {1}{15} (2 x+3)^4 \sqrt {3 x^2+5 x+2}\) |
\(\Big \downarrow \) 1225 |
\(\displaystyle \frac {1}{30} \left (\frac {1}{4} \left (\frac {5}{9} \left (\frac {28051}{6} \int \frac {1}{\sqrt {3 x^2+5 x+2}}dx+\frac {1}{3} \sqrt {3 x^2+5 x+2} (9650 x+27519)\right )+\frac {3128}{9} \sqrt {3 x^2+5 x+2} (2 x+3)^2\right )+\frac {53}{2} \sqrt {3 x^2+5 x+2} (2 x+3)^3\right )-\frac {1}{15} (2 x+3)^4 \sqrt {3 x^2+5 x+2}\) |
\(\Big \downarrow \) 1092 |
\(\displaystyle \frac {1}{30} \left (\frac {1}{4} \left (\frac {5}{9} \left (\frac {28051}{3} \int \frac {1}{12-\frac {(6 x+5)^2}{3 x^2+5 x+2}}d\frac {6 x+5}{\sqrt {3 x^2+5 x+2}}+\frac {1}{3} \sqrt {3 x^2+5 x+2} (9650 x+27519)\right )+\frac {3128}{9} \sqrt {3 x^2+5 x+2} (2 x+3)^2\right )+\frac {53}{2} \sqrt {3 x^2+5 x+2} (2 x+3)^3\right )-\frac {1}{15} (2 x+3)^4 \sqrt {3 x^2+5 x+2}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{30} \left (\frac {1}{4} \left (\frac {5}{9} \left (\frac {28051 \text {arctanh}\left (\frac {6 x+5}{2 \sqrt {3} \sqrt {3 x^2+5 x+2}}\right )}{6 \sqrt {3}}+\frac {1}{3} \sqrt {3 x^2+5 x+2} (9650 x+27519)\right )+\frac {3128}{9} \sqrt {3 x^2+5 x+2} (2 x+3)^2\right )+\frac {53}{2} \sqrt {3 x^2+5 x+2} (2 x+3)^3\right )-\frac {1}{15} (2 x+3)^4 \sqrt {3 x^2+5 x+2}\) |
-1/15*((3 + 2*x)^4*Sqrt[2 + 5*x + 3*x^2]) + ((53*(3 + 2*x)^3*Sqrt[2 + 5*x + 3*x^2])/2 + ((3128*(3 + 2*x)^2*Sqrt[2 + 5*x + 3*x^2])/9 + (5*(((27519 + 9650*x)*Sqrt[2 + 5*x + 3*x^2])/3 + (28051*ArcTanh[(5 + 6*x)/(2*Sqrt[3]*Sqr t[2 + 5*x + 3*x^2])])/(6*Sqrt[3])))/9)/4)/30
3.25.95.3.1 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[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 = 0.37 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.47
method | result | size |
risch | \(-\frac {\left (3456 x^{4}-2160 x^{3}-93912 x^{2}-268750 x -281829\right ) \sqrt {3 x^{2}+5 x +2}}{3240}+\frac {28051 \ln \left (\frac {\left (\frac {5}{2}+3 x \right ) \sqrt {3}}{3}+\sqrt {3 x^{2}+5 x +2}\right ) \sqrt {3}}{3888}\) | \(65\) |
trager | \(\left (-\frac {16}{15} x^{4}+\frac {2}{3} x^{3}+\frac {3913}{135} x^{2}+\frac {26875}{324} x +\frac {93943}{1080}\right ) \sqrt {3 x^{2}+5 x +2}+\frac {28051 \operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right ) \ln \left (6 \operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right ) x +5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right )+6 \sqrt {3 x^{2}+5 x +2}\right )}{3888}\) | \(76\) |
default | \(\frac {28051 \ln \left (\frac {\left (\frac {5}{2}+3 x \right ) \sqrt {3}}{3}+\sqrt {3 x^{2}+5 x +2}\right ) \sqrt {3}}{3888}+\frac {93943 \sqrt {3 x^{2}+5 x +2}}{1080}-\frac {16 x^{4} \sqrt {3 x^{2}+5 x +2}}{15}+\frac {2 x^{3} \sqrt {3 x^{2}+5 x +2}}{3}+\frac {3913 x^{2} \sqrt {3 x^{2}+5 x +2}}{135}+\frac {26875 x \sqrt {3 x^{2}+5 x +2}}{324}\) | \(111\) |
-1/3240*(3456*x^4-2160*x^3-93912*x^2-268750*x-281829)*(3*x^2+5*x+2)^(1/2)+ 28051/3888*ln(1/3*(5/2+3*x)*3^(1/2)+(3*x^2+5*x+2)^(1/2))*3^(1/2)
Time = 0.26 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.53 \[ \int \frac {(5-x) (3+2 x)^4}{\sqrt {2+5 x+3 x^2}} \, dx=-\frac {1}{3240} \, {\left (3456 \, x^{4} - 2160 \, x^{3} - 93912 \, x^{2} - 268750 \, x - 281829\right )} \sqrt {3 \, x^{2} + 5 \, x + 2} + \frac {28051}{7776} \, \sqrt {3} \log \left (4 \, \sqrt {3} \sqrt {3 \, x^{2} + 5 \, x + 2} {\left (6 \, x + 5\right )} + 72 \, x^{2} + 120 \, x + 49\right ) \]
-1/3240*(3456*x^4 - 2160*x^3 - 93912*x^2 - 268750*x - 281829)*sqrt(3*x^2 + 5*x + 2) + 28051/7776*sqrt(3)*log(4*sqrt(3)*sqrt(3*x^2 + 5*x + 2)*(6*x + 5) + 72*x^2 + 120*x + 49)
Time = 0.55 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.55 \[ \int \frac {(5-x) (3+2 x)^4}{\sqrt {2+5 x+3 x^2}} \, dx=\sqrt {3 x^{2} + 5 x + 2} \left (- \frac {16 x^{4}}{15} + \frac {2 x^{3}}{3} + \frac {3913 x^{2}}{135} + \frac {26875 x}{324} + \frac {93943}{1080}\right ) + \frac {28051 \sqrt {3} \log {\left (6 x + 2 \sqrt {3} \sqrt {3 x^{2} + 5 x + 2} + 5 \right )}}{3888} \]
sqrt(3*x**2 + 5*x + 2)*(-16*x**4/15 + 2*x**3/3 + 3913*x**2/135 + 26875*x/3 24 + 93943/1080) + 28051*sqrt(3)*log(6*x + 2*sqrt(3)*sqrt(3*x**2 + 5*x + 2 ) + 5)/3888
Time = 0.27 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.80 \[ \int \frac {(5-x) (3+2 x)^4}{\sqrt {2+5 x+3 x^2}} \, dx=-\frac {16}{15} \, \sqrt {3 \, x^{2} + 5 \, x + 2} x^{4} + \frac {2}{3} \, \sqrt {3 \, x^{2} + 5 \, x + 2} x^{3} + \frac {3913}{135} \, \sqrt {3 \, x^{2} + 5 \, x + 2} x^{2} + \frac {26875}{324} \, \sqrt {3 \, x^{2} + 5 \, x + 2} x + \frac {28051}{3888} \, \sqrt {3} \log \left (2 \, \sqrt {3} \sqrt {3 \, x^{2} + 5 \, x + 2} + 6 \, x + 5\right ) + \frac {93943}{1080} \, \sqrt {3 \, x^{2} + 5 \, x + 2} \]
-16/15*sqrt(3*x^2 + 5*x + 2)*x^4 + 2/3*sqrt(3*x^2 + 5*x + 2)*x^3 + 3913/13 5*sqrt(3*x^2 + 5*x + 2)*x^2 + 26875/324*sqrt(3*x^2 + 5*x + 2)*x + 28051/38 88*sqrt(3)*log(2*sqrt(3)*sqrt(3*x^2 + 5*x + 2) + 6*x + 5) + 93943/1080*sqr t(3*x^2 + 5*x + 2)
Time = 0.28 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.50 \[ \int \frac {(5-x) (3+2 x)^4}{\sqrt {2+5 x+3 x^2}} \, dx=-\frac {1}{3240} \, {\left (2 \, {\left (12 \, {\left (18 \, {\left (8 \, x - 5\right )} x - 3913\right )} x - 134375\right )} x - 281829\right )} \sqrt {3 \, x^{2} + 5 \, x + 2} - \frac {28051}{3888} \, \sqrt {3} \log \left ({\left | -2 \, \sqrt {3} {\left (\sqrt {3} x - \sqrt {3 \, x^{2} + 5 \, x + 2}\right )} - 5 \right |}\right ) \]
-1/3240*(2*(12*(18*(8*x - 5)*x - 3913)*x - 134375)*x - 281829)*sqrt(3*x^2 + 5*x + 2) - 28051/3888*sqrt(3)*log(abs(-2*sqrt(3)*(sqrt(3)*x - sqrt(3*x^2 + 5*x + 2)) - 5))
Timed out. \[ \int \frac {(5-x) (3+2 x)^4}{\sqrt {2+5 x+3 x^2}} \, dx=-\int \frac {{\left (2\,x+3\right )}^4\,\left (x-5\right )}{\sqrt {3\,x^2+5\,x+2}} \,d x \]