Integrand size = 27, antiderivative size = 66 \[ \int \frac {(5-x) \sqrt {3+2 x}}{\left (2+5 x+3 x^2\right )^2} \, dx=-\frac {\sqrt {3+2 x} (29+35 x)}{2+5 x+3 x^2}-82 \text {arctanh}\left (\sqrt {3+2 x}\right )+\frac {316 \text {arctanh}\left (\sqrt {\frac {3}{5}} \sqrt {3+2 x}\right )}{\sqrt {15}} \]
-82*arctanh((3+2*x)^(1/2))+316/15*arctanh(1/5*15^(1/2)*(3+2*x)^(1/2))*15^( 1/2)-(29+35*x)*(3+2*x)^(1/2)/(3*x^2+5*x+2)
Time = 0.15 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.00 \[ \int \frac {(5-x) \sqrt {3+2 x}}{\left (2+5 x+3 x^2\right )^2} \, dx=-\frac {\sqrt {3+2 x} (29+35 x)}{2+5 x+3 x^2}-82 \text {arctanh}\left (\sqrt {3+2 x}\right )+\frac {316 \text {arctanh}\left (\sqrt {\frac {3}{5}} \sqrt {3+2 x}\right )}{\sqrt {15}} \]
-((Sqrt[3 + 2*x]*(29 + 35*x))/(2 + 5*x + 3*x^2)) - 82*ArcTanh[Sqrt[3 + 2*x ]] + (316*ArcTanh[Sqrt[3/5]*Sqrt[3 + 2*x]])/Sqrt[15]
Time = 0.22 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.05, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {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.
\(\displaystyle \int \frac {(5-x) \sqrt {2 x+3}}{\left (3 x^2+5 x+2\right )^2} \, dx\) |
\(\Big \downarrow \) 1234 |
\(\displaystyle -\int \frac {35 x+76}{\sqrt {2 x+3} \left (3 x^2+5 x+2\right )}dx-\frac {\sqrt {2 x+3} (35 x+29)}{3 x^2+5 x+2}\) |
\(\Big \downarrow \) 1197 |
\(\displaystyle -2 \int \frac {35 (2 x+3)+47}{3 (2 x+3)^2-8 (2 x+3)+5}d\sqrt {2 x+3}-\frac {\sqrt {2 x+3} (35 x+29)}{3 x^2+5 x+2}\) |
\(\Big \downarrow \) 1480 |
\(\displaystyle -2 \left (158 \int \frac {1}{3 (2 x+3)-5}d\sqrt {2 x+3}-123 \int \frac {1}{3 (2 x+3)-3}d\sqrt {2 x+3}\right )-\frac {\sqrt {2 x+3} (35 x+29)}{3 x^2+5 x+2}\) |
\(\Big \downarrow \) 220 |
\(\displaystyle -2 \left (41 \text {arctanh}\left (\sqrt {2 x+3}\right )-\frac {158 \text {arctanh}\left (\sqrt {\frac {3}{5}} \sqrt {2 x+3}\right )}{\sqrt {15}}\right )-\frac {\sqrt {2 x+3} (35 x+29)}{3 x^2+5 x+2}\) |
-((Sqrt[3 + 2*x]*(29 + 35*x))/(2 + 5*x + 3*x^2)) - 2*(41*ArcTanh[Sqrt[3 + 2*x]] - (158*ArcTanh[Sqrt[3/5]*Sqrt[3 + 2*x]])/Sqrt[15])
3.26.61.3.1 Defintions of rubi rules used
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])
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]
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])
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]
Time = 0.37 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.06
method | result | size |
risch | \(-\frac {\left (29+35 x \right ) \sqrt {3+2 x}}{3 x^{2}+5 x +2}-41 \ln \left (\sqrt {3+2 x}+1\right )+\frac {316 \,\operatorname {arctanh}\left (\frac {\sqrt {15}\, \sqrt {3+2 x}}{5}\right ) \sqrt {15}}{15}+41 \ln \left (\sqrt {3+2 x}-1\right )\) | \(70\) |
derivativedivides | \(-\frac {6}{\sqrt {3+2 x}-1}+41 \ln \left (\sqrt {3+2 x}-1\right )-\frac {34 \sqrt {3+2 x}}{3 \left (\frac {4}{3}+2 x \right )}+\frac {316 \,\operatorname {arctanh}\left (\frac {\sqrt {15}\, \sqrt {3+2 x}}{5}\right ) \sqrt {15}}{15}-\frac {6}{\sqrt {3+2 x}+1}-41 \ln \left (\sqrt {3+2 x}+1\right )\) | \(86\) |
default | \(-\frac {6}{\sqrt {3+2 x}-1}+41 \ln \left (\sqrt {3+2 x}-1\right )-\frac {34 \sqrt {3+2 x}}{3 \left (\frac {4}{3}+2 x \right )}+\frac {316 \,\operatorname {arctanh}\left (\frac {\sqrt {15}\, \sqrt {3+2 x}}{5}\right ) \sqrt {15}}{15}-\frac {6}{\sqrt {3+2 x}+1}-41 \ln \left (\sqrt {3+2 x}+1\right )\) | \(86\) |
trager | \(-\frac {\left (29+35 x \right ) \sqrt {3+2 x}}{3 x^{2}+5 x +2}-\frac {158 \operatorname {RootOf}\left (\textit {\_Z}^{2}-15\right ) \ln \left (\frac {-3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-15\right ) x +15 \sqrt {3+2 x}-7 \operatorname {RootOf}\left (\textit {\_Z}^{2}-15\right )}{2+3 x}\right )}{15}+41 \ln \left (\frac {-2-x +\sqrt {3+2 x}}{1+x}\right )\) | \(93\) |
pseudoelliptic | \(\frac {948 \sqrt {15}\, \left (x +\frac {2}{3}\right ) \left (1+x \right ) \operatorname {arctanh}\left (\frac {\sqrt {15}\, \sqrt {3+2 x}}{5}\right )+\left (1845 x^{2}+3075 x +1230\right ) \ln \left (\sqrt {3+2 x}-1\right )+\left (-1845 x^{2}-3075 x -1230\right ) \ln \left (\sqrt {3+2 x}+1\right )+\left (-525 x -435\right ) \sqrt {3+2 x}}{45 x^{2}+75 x +30}\) | \(94\) |
-(29+35*x)*(3+2*x)^(1/2)/(3*x^2+5*x+2)-41*ln((3+2*x)^(1/2)+1)+316/15*arcta nh(1/5*15^(1/2)*(3+2*x)^(1/2))*15^(1/2)+41*ln((3+2*x)^(1/2)-1)
Leaf count of result is larger than twice the leaf count of optimal. 113 vs. \(2 (55) = 110\).
Time = 0.28 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.71 \[ \int \frac {(5-x) \sqrt {3+2 x}}{\left (2+5 x+3 x^2\right )^2} \, dx=\frac {158 \, \sqrt {15} {\left (3 \, x^{2} + 5 \, x + 2\right )} \log \left (\frac {\sqrt {15} \sqrt {2 \, x + 3} + 3 \, x + 7}{3 \, x + 2}\right ) - 615 \, {\left (3 \, x^{2} + 5 \, x + 2\right )} \log \left (\sqrt {2 \, x + 3} + 1\right ) + 615 \, {\left (3 \, x^{2} + 5 \, x + 2\right )} \log \left (\sqrt {2 \, x + 3} - 1\right ) - 15 \, {\left (35 \, x + 29\right )} \sqrt {2 \, x + 3}}{15 \, {\left (3 \, x^{2} + 5 \, x + 2\right )}} \]
1/15*(158*sqrt(15)*(3*x^2 + 5*x + 2)*log((sqrt(15)*sqrt(2*x + 3) + 3*x + 7 )/(3*x + 2)) - 615*(3*x^2 + 5*x + 2)*log(sqrt(2*x + 3) + 1) + 615*(3*x^2 + 5*x + 2)*log(sqrt(2*x + 3) - 1) - 15*(35*x + 29)*sqrt(2*x + 3))/(3*x^2 + 5*x + 2)
Time = 40.69 (sec) , antiderivative size = 211, normalized size of antiderivative = 3.20 \[ \int \frac {(5-x) \sqrt {3+2 x}}{\left (2+5 x+3 x^2\right )^2} \, dx=- \frac {47 \sqrt {15} \left (\log {\left (\sqrt {2 x + 3} - \frac {\sqrt {15}}{3} \right )} - \log {\left (\sqrt {2 x + 3} + \frac {\sqrt {15}}{3} \right )}\right )}{5} + 340 \left (\begin {cases} \frac {\sqrt {15} \left (- \frac {\log {\left (\frac {\sqrt {15} \sqrt {2 x + 3}}{5} - 1 \right )}}{4} + \frac {\log {\left (\frac {\sqrt {15} \sqrt {2 x + 3}}{5} + 1 \right )}}{4} - \frac {1}{4 \left (\frac {\sqrt {15} \sqrt {2 x + 3}}{5} + 1\right )} - \frac {1}{4 \left (\frac {\sqrt {15} \sqrt {2 x + 3}}{5} - 1\right )}\right )}{75} & \text {for}\: \sqrt {2 x + 3} > - \frac {\sqrt {15}}{3} \wedge \sqrt {2 x + 3} < \frac {\sqrt {15}}{3} \end {cases}\right ) + 41 \log {\left (\sqrt {2 x + 3} - 1 \right )} - 41 \log {\left (\sqrt {2 x + 3} + 1 \right )} - \frac {6}{\sqrt {2 x + 3} + 1} - \frac {6}{\sqrt {2 x + 3} - 1} \]
-47*sqrt(15)*(log(sqrt(2*x + 3) - sqrt(15)/3) - log(sqrt(2*x + 3) + sqrt(1 5)/3))/5 + 340*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))) + 41*log(sqrt(2*x + 3) - 1) - 41*log(sq rt(2*x + 3) + 1) - 6/(sqrt(2*x + 3) + 1) - 6/(sqrt(2*x + 3) - 1)
Time = 0.27 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.48 \[ \int \frac {(5-x) \sqrt {3+2 x}}{\left (2+5 x+3 x^2\right )^2} \, dx=-\frac {158}{15} \, \sqrt {15} \log \left (-\frac {\sqrt {15} - 3 \, \sqrt {2 \, x + 3}}{\sqrt {15} + 3 \, \sqrt {2 \, x + 3}}\right ) - \frac {2 \, {\left (35 \, {\left (2 \, x + 3\right )}^{\frac {3}{2}} - 47 \, \sqrt {2 \, x + 3}\right )}}{3 \, {\left (2 \, x + 3\right )}^{2} - 16 \, x - 19} - 41 \, \log \left (\sqrt {2 \, x + 3} + 1\right ) + 41 \, \log \left (\sqrt {2 \, x + 3} - 1\right ) \]
-158/15*sqrt(15)*log(-(sqrt(15) - 3*sqrt(2*x + 3))/(sqrt(15) + 3*sqrt(2*x + 3))) - 2*(35*(2*x + 3)^(3/2) - 47*sqrt(2*x + 3))/(3*(2*x + 3)^2 - 16*x - 19) - 41*log(sqrt(2*x + 3) + 1) + 41*log(sqrt(2*x + 3) - 1)
Time = 0.29 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.55 \[ \int \frac {(5-x) \sqrt {3+2 x}}{\left (2+5 x+3 x^2\right )^2} \, dx=-\frac {158}{15} \, \sqrt {15} \log \left (\frac {{\left | -2 \, \sqrt {15} + 6 \, \sqrt {2 \, x + 3} \right |}}{2 \, {\left (\sqrt {15} + 3 \, \sqrt {2 \, x + 3}\right )}}\right ) - \frac {2 \, {\left (35 \, {\left (2 \, x + 3\right )}^{\frac {3}{2}} - 47 \, \sqrt {2 \, x + 3}\right )}}{3 \, {\left (2 \, x + 3\right )}^{2} - 16 \, x - 19} - 41 \, \log \left (\sqrt {2 \, x + 3} + 1\right ) + 41 \, \log \left ({\left | \sqrt {2 \, x + 3} - 1 \right |}\right ) \]
-158/15*sqrt(15)*log(1/2*abs(-2*sqrt(15) + 6*sqrt(2*x + 3))/(sqrt(15) + 3* sqrt(2*x + 3))) - 2*(35*(2*x + 3)^(3/2) - 47*sqrt(2*x + 3))/(3*(2*x + 3)^2 - 16*x - 19) - 41*log(sqrt(2*x + 3) + 1) + 41*log(abs(sqrt(2*x + 3) - 1))
Time = 0.07 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.00 \[ \int \frac {(5-x) \sqrt {3+2 x}}{\left (2+5 x+3 x^2\right )^2} \, dx=\frac {316\,\sqrt {15}\,\mathrm {atanh}\left (\frac {\sqrt {15}\,\sqrt {2\,x+3}}{5}\right )}{15}-\frac {\frac {94\,\sqrt {2\,x+3}}{3}-\frac {70\,{\left (2\,x+3\right )}^{3/2}}{3}}{\frac {16\,x}{3}-{\left (2\,x+3\right )}^2+\frac {19}{3}}-82\,\mathrm {atanh}\left (\sqrt {2\,x+3}\right ) \]