Integrand size = 20, antiderivative size = 114 \[ \int \frac {1}{(1+2 x)^2 \left (2+3 x+5 x^2\right )^3} \, dx=-\frac {51516}{329623 (1+2 x)}+\frac {37+20 x}{434 (1+2 x) \left (2+3 x+5 x^2\right )^2}+\frac {6427+5820 x}{47089 (1+2 x) \left (2+3 x+5 x^2\right )}-\frac {1065012 \arctan \left (\frac {3+10 x}{\sqrt {31}}\right )}{2307361 \sqrt {31}}+\frac {384 \log (1+2 x)}{2401}-\frac {192 \log \left (2+3 x+5 x^2\right )}{2401} \]
-51516/329623/(1+2*x)+1/434*(37+20*x)/(1+2*x)/(5*x^2+3*x+2)^2+1/47089*(642 7+5820*x)/(1+2*x)/(5*x^2+3*x+2)+384/2401*ln(1+2*x)-192/2401*ln(5*x^2+3*x+2 )-1065012/71528191*arctan(1/31*(3+10*x)*31^(1/2))*31^(1/2)
Time = 0.05 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.86 \[ \int \frac {1}{(1+2 x)^2 \left (2+3 x+5 x^2\right )^3} \, dx=\frac {4 \left (-\frac {1668296}{1+2 x}-\frac {47089 (-43+270 x)}{8 \left (2+3 x+5 x^2\right )^2}-\frac {217 (-15179+51910 x)}{4 \left (2+3 x+5 x^2\right )}-266253 \sqrt {31} \arctan \left (\frac {3+10 x}{\sqrt {31}}\right )+2859936 \log (1+2 x)-1429968 \log \left (4 \left (2+3 x+5 x^2\right )\right )\right )}{71528191} \]
(4*(-1668296/(1 + 2*x) - (47089*(-43 + 270*x))/(8*(2 + 3*x + 5*x^2)^2) - ( 217*(-15179 + 51910*x))/(4*(2 + 3*x + 5*x^2)) - 266253*Sqrt[31]*ArcTan[(3 + 10*x)/Sqrt[31]] + 2859936*Log[1 + 2*x] - 1429968*Log[4*(2 + 3*x + 5*x^2) ]))/71528191
Time = 0.30 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.09, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {1165, 27, 1235, 27, 1200, 2009}
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.
\(\displaystyle \int \frac {1}{(2 x+1)^2 \left (5 x^2+3 x+2\right )^3} \, dx\) |
\(\Big \downarrow \) 1165 |
\(\displaystyle \frac {1}{434} \int \frac {2 (80 x+191)}{(2 x+1)^2 \left (5 x^2+3 x+2\right )^2}dx+\frac {20 x+37}{434 (2 x+1) \left (5 x^2+3 x+2\right )^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{217} \int \frac {80 x+191}{(2 x+1)^2 \left (5 x^2+3 x+2\right )^2}dx+\frac {20 x+37}{434 (2 x+1) \left (5 x^2+3 x+2\right )^2}\) |
\(\Big \downarrow \) 1235 |
\(\displaystyle \frac {1}{217} \left (\frac {1}{217} \int \frac {6 (3880 x+6233)}{(2 x+1)^2 \left (5 x^2+3 x+2\right )}dx+\frac {5820 x+6427}{217 (2 x+1) \left (5 x^2+3 x+2\right )}\right )+\frac {20 x+37}{434 (2 x+1) \left (5 x^2+3 x+2\right )^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{217} \left (\frac {6}{217} \int \frac {3880 x+6233}{(2 x+1)^2 \left (5 x^2+3 x+2\right )}dx+\frac {5820 x+6427}{217 (2 x+1) \left (5 x^2+3 x+2\right )}\right )+\frac {20 x+37}{434 (2 x+1) \left (5 x^2+3 x+2\right )^2}\) |
\(\Big \downarrow \) 1200 |
\(\displaystyle \frac {1}{217} \left (\frac {6}{217} \int \left (\frac {-307520 x-181007}{49 \left (5 x^2+3 x+2\right )}+\frac {123008}{49 (2 x+1)}+\frac {17172}{7 (2 x+1)^2}\right )dx+\frac {5820 x+6427}{217 (2 x+1) \left (5 x^2+3 x+2\right )}\right )+\frac {20 x+37}{434 (2 x+1) \left (5 x^2+3 x+2\right )^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{217} \left (\frac {6}{217} \left (-\frac {177502 \arctan \left (\frac {10 x+3}{\sqrt {31}}\right )}{49 \sqrt {31}}-\frac {30752}{49} \log \left (5 x^2+3 x+2\right )-\frac {8586}{7 (2 x+1)}+\frac {61504}{49} \log (2 x+1)\right )+\frac {5820 x+6427}{217 (2 x+1) \left (5 x^2+3 x+2\right )}\right )+\frac {20 x+37}{434 (2 x+1) \left (5 x^2+3 x+2\right )^2}\) |
(37 + 20*x)/(434*(1 + 2*x)*(2 + 3*x + 5*x^2)^2) + ((6427 + 5820*x)/(217*(1 + 2*x)*(2 + 3*x + 5*x^2)) + (6*(-8586/(7*(1 + 2*x)) - (177502*ArcTan[(3 + 10*x)/Sqrt[31]])/(49*Sqrt[31]) + (61504*Log[1 + 2*x])/49 - (30752*Log[2 + 3*x + 5*x^2])/49))/217)/217
3.23.30.3.1 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[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[(d + e*x)^(m + 1)*(b*c*d - b^2*e + 2*a*c*e + c*(2*c*d - b*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*Simp[b*c*d*e*(2*p - m + 2) + b^2*e^2*(m + p + 2) - 2*c^2*d^2*(2*p + 3) - 2*a*c*e^2*(m + 2*p + 3) - c*e*(2*c*d - b*e)*(m + 2*p + 4)*x, x]*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && LtQ[p, -1] && IntQuadraticQ[a, b, c, d, e, m, p, x]
Int[(((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.))/((a_.) + (b_.)* (x_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*((f + g* x)^n/(a + b*x + c*x^2)), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && In tegersQ[n]
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] )
Time = 21.20 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.68
method | result | size |
default | \(-\frac {32}{343 \left (1+2 x \right )}+\frac {384 \ln \left (1+2 x \right )}{2401}-\frac {25 \left (\frac {72674}{961} x^{3}+\frac {111769}{4805} x^{2}+\frac {613046}{24025} x -\frac {490329}{48050}\right )}{2401 \left (5 x^{2}+3 x +2\right )^{2}}-\frac {192 \ln \left (5 x^{2}+3 x +2\right )}{2401}-\frac {1065012 \arctan \left (\frac {\left (3+10 x \right ) \sqrt {31}}{31}\right ) \sqrt {31}}{71528191}\) | \(77\) |
risch | \(\frac {-\frac {1287900}{329623} x^{4}-\frac {1341780}{329623} x^{3}-\frac {1146799}{329623} x^{2}-\frac {386555}{329623} x -\frac {175969}{659246}}{\left (1+2 x \right ) \left (5 x^{2}+3 x +2\right )^{2}}-\frac {192 \ln \left (100 x^{2}+60 x +40\right )}{2401}-\frac {1065012 \arctan \left (\frac {\left (3+10 x \right ) \sqrt {31}}{31}\right ) \sqrt {31}}{71528191}+\frac {384 \ln \left (1+2 x \right )}{2401}\) | \(80\) |
-32/343/(1+2*x)+384/2401*ln(1+2*x)-25/2401*(72674/961*x^3+111769/4805*x^2+ 613046/24025*x-490329/48050)/(5*x^2+3*x+2)^2-192/2401*ln(5*x^2+3*x+2)-1065 012/71528191*arctan(1/31*(3+10*x)*31^(1/2))*31^(1/2)
Time = 0.34 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.41 \[ \int \frac {1}{(1+2 x)^2 \left (2+3 x+5 x^2\right )^3} \, dx=-\frac {558948600 \, x^{4} + 582332520 \, x^{3} + 2130024 \, \sqrt {31} {\left (50 \, x^{5} + 85 \, x^{4} + 88 \, x^{3} + 53 \, x^{2} + 20 \, x + 4\right )} \arctan \left (\frac {1}{31} \, \sqrt {31} {\left (10 \, x + 3\right )}\right ) + 497710766 \, x^{2} + 11439744 \, {\left (50 \, x^{5} + 85 \, x^{4} + 88 \, x^{3} + 53 \, x^{2} + 20 \, x + 4\right )} \log \left (5 \, x^{2} + 3 \, x + 2\right ) - 22879488 \, {\left (50 \, x^{5} + 85 \, x^{4} + 88 \, x^{3} + 53 \, x^{2} + 20 \, x + 4\right )} \log \left (2 \, x + 1\right ) + 167764870 \, x + 38185273}{143056382 \, {\left (50 \, x^{5} + 85 \, x^{4} + 88 \, x^{3} + 53 \, x^{2} + 20 \, x + 4\right )}} \]
-1/143056382*(558948600*x^4 + 582332520*x^3 + 2130024*sqrt(31)*(50*x^5 + 8 5*x^4 + 88*x^3 + 53*x^2 + 20*x + 4)*arctan(1/31*sqrt(31)*(10*x + 3)) + 497 710766*x^2 + 11439744*(50*x^5 + 85*x^4 + 88*x^3 + 53*x^2 + 20*x + 4)*log(5 *x^2 + 3*x + 2) - 22879488*(50*x^5 + 85*x^4 + 88*x^3 + 53*x^2 + 20*x + 4)* log(2*x + 1) + 167764870*x + 38185273)/(50*x^5 + 85*x^4 + 88*x^3 + 53*x^2 + 20*x + 4)
Time = 0.13 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.89 \[ \int \frac {1}{(1+2 x)^2 \left (2+3 x+5 x^2\right )^3} \, dx=\frac {- 2575800 x^{4} - 2683560 x^{3} - 2293598 x^{2} - 773110 x - 175969}{32962300 x^{5} + 56035910 x^{4} + 58013648 x^{3} + 34940038 x^{2} + 13184920 x + 2636984} + \frac {384 \log {\left (x + \frac {1}{2} \right )}}{2401} - \frac {192 \log {\left (x^{2} + \frac {3 x}{5} + \frac {2}{5} \right )}}{2401} - \frac {1065012 \sqrt {31} \operatorname {atan}{\left (\frac {10 \sqrt {31} x}{31} + \frac {3 \sqrt {31}}{31} \right )}}{71528191} \]
(-2575800*x**4 - 2683560*x**3 - 2293598*x**2 - 773110*x - 175969)/(3296230 0*x**5 + 56035910*x**4 + 58013648*x**3 + 34940038*x**2 + 13184920*x + 2636 984) + 384*log(x + 1/2)/2401 - 192*log(x**2 + 3*x/5 + 2/5)/2401 - 1065012* sqrt(31)*atan(10*sqrt(31)*x/31 + 3*sqrt(31)/31)/71528191
Time = 0.29 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.76 \[ \int \frac {1}{(1+2 x)^2 \left (2+3 x+5 x^2\right )^3} \, dx=-\frac {1065012}{71528191} \, \sqrt {31} \arctan \left (\frac {1}{31} \, \sqrt {31} {\left (10 \, x + 3\right )}\right ) - \frac {2575800 \, x^{4} + 2683560 \, x^{3} + 2293598 \, x^{2} + 773110 \, x + 175969}{659246 \, {\left (50 \, x^{5} + 85 \, x^{4} + 88 \, x^{3} + 53 \, x^{2} + 20 \, x + 4\right )}} - \frac {192}{2401} \, \log \left (5 \, x^{2} + 3 \, x + 2\right ) + \frac {384}{2401} \, \log \left (2 \, x + 1\right ) \]
-1065012/71528191*sqrt(31)*arctan(1/31*sqrt(31)*(10*x + 3)) - 1/659246*(25 75800*x^4 + 2683560*x^3 + 2293598*x^2 + 773110*x + 175969)/(50*x^5 + 85*x^ 4 + 88*x^3 + 53*x^2 + 20*x + 4) - 192/2401*log(5*x^2 + 3*x + 2) + 384/2401 *log(2*x + 1)
Time = 0.27 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.95 \[ \int \frac {1}{(1+2 x)^2 \left (2+3 x+5 x^2\right )^3} \, dx=-\frac {1065012}{71528191} \, \sqrt {31} \arctan \left (-\frac {1}{31} \, \sqrt {31} {\left (\frac {7}{2 \, x + 1} - 2\right )}\right ) - \frac {32}{343 \, {\left (2 \, x + 1\right )}} + \frac {4 \, {\left (\frac {1178375}{2 \, x + 1} - \frac {2320190}{{\left (2 \, x + 1\right )}^{2}} + \frac {87843}{{\left (2 \, x + 1\right )}^{3}} - 1304250\right )}}{2307361 \, {\left (\frac {4}{2 \, x + 1} - \frac {7}{{\left (2 \, x + 1\right )}^{2}} - 5\right )}^{2}} - \frac {192}{2401} \, \log \left (-\frac {4}{2 \, x + 1} + \frac {7}{{\left (2 \, x + 1\right )}^{2}} + 5\right ) \]
-1065012/71528191*sqrt(31)*arctan(-1/31*sqrt(31)*(7/(2*x + 1) - 2)) - 32/3 43/(2*x + 1) + 4/2307361*(1178375/(2*x + 1) - 2320190/(2*x + 1)^2 + 87843/ (2*x + 1)^3 - 1304250)/(4/(2*x + 1) - 7/(2*x + 1)^2 - 5)^2 - 192/2401*log( -4/(2*x + 1) + 7/(2*x + 1)^2 + 5)
Time = 0.12 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.82 \[ \int \frac {1}{(1+2 x)^2 \left (2+3 x+5 x^2\right )^3} \, dx=\frac {384\,\ln \left (x+\frac {1}{2}\right )}{2401}-\frac {\frac {25758\,x^4}{329623}+\frac {134178\,x^3}{1648115}+\frac {1146799\,x^2}{16481150}+\frac {77311\,x}{3296230}+\frac {175969}{32962300}}{x^5+\frac {17\,x^4}{10}+\frac {44\,x^3}{25}+\frac {53\,x^2}{50}+\frac {2\,x}{5}+\frac {2}{25}}+\ln \left (x+\frac {3}{10}-\frac {\sqrt {31}\,1{}\mathrm {i}}{10}\right )\,\left (-\frac {192}{2401}+\frac {\sqrt {31}\,532506{}\mathrm {i}}{71528191}\right )-\ln \left (x+\frac {3}{10}+\frac {\sqrt {31}\,1{}\mathrm {i}}{10}\right )\,\left (\frac {192}{2401}+\frac {\sqrt {31}\,532506{}\mathrm {i}}{71528191}\right ) \]
(384*log(x + 1/2))/2401 - ((77311*x)/3296230 + (1146799*x^2)/16481150 + (1 34178*x^3)/1648115 + (25758*x^4)/329623 + 175969/32962300)/((2*x)/5 + (53* x^2)/50 + (44*x^3)/25 + (17*x^4)/10 + x^5 + 2/25) + log(x - (31^(1/2)*1i)/ 10 + 3/10)*((31^(1/2)*532506i)/71528191 - 192/2401) - log(x + (31^(1/2)*1i )/10 + 3/10)*((31^(1/2)*532506i)/71528191 + 192/2401)