Integrand size = 27, antiderivative size = 72 \[ \int \frac {(5-x) (3+2 x)^{3/2}}{\left (2+5 x+3 x^2\right )^2} \, dx=-\frac {\sqrt {3+2 x} (121+139 x)}{3 \left (2+5 x+3 x^2\right )}-106 \text {arctanh}\left (\sqrt {3+2 x}\right )+\frac {248}{3} \sqrt {\frac {5}{3}} \text {arctanh}\left (\sqrt {\frac {3}{5}} \sqrt {3+2 x}\right ) \]
-106*arctanh((3+2*x)^(1/2))+248/9*arctanh(1/5*15^(1/2)*(3+2*x)^(1/2))*15^( 1/2)-1/3*(121+139*x)*(3+2*x)^(1/2)/(3*x^2+5*x+2)
Time = 0.16 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.97 \[ \int \frac {(5-x) (3+2 x)^{3/2}}{\left (2+5 x+3 x^2\right )^2} \, dx=-\frac {2}{9} \left (\frac {3 \sqrt {3+2 x} (121+139 x)}{4+10 x+6 x^2}+477 \text {arctanh}\left (\sqrt {3+2 x}\right )-124 \sqrt {15} \text {arctanh}\left (\sqrt {\frac {3}{5}} \sqrt {3+2 x}\right )\right ) \]
(-2*((3*Sqrt[3 + 2*x]*(121 + 139*x))/(4 + 10*x + 6*x^2) + 477*ArcTanh[Sqrt [3 + 2*x]] - 124*Sqrt[15]*ArcTanh[Sqrt[3/5]*Sqrt[3 + 2*x]]))/9
Time = 0.22 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.04, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {1233, 25, 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.
\(\displaystyle \int \frac {(5-x) (2 x+3)^{3/2}}{\left (3 x^2+5 x+2\right )^2} \, dx\) |
\(\Big \downarrow \) 1233 |
\(\displaystyle \frac {1}{3} \int -\frac {143 x+302}{\sqrt {2 x+3} \left (3 x^2+5 x+2\right )}dx-\frac {\sqrt {2 x+3} (139 x+121)}{3 \left (3 x^2+5 x+2\right )}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {1}{3} \int \frac {143 x+302}{\sqrt {2 x+3} \left (3 x^2+5 x+2\right )}dx-\frac {\sqrt {2 x+3} (139 x+121)}{3 \left (3 x^2+5 x+2\right )}\) |
\(\Big \downarrow \) 1197 |
\(\displaystyle -\frac {2}{3} \int \frac {143 (2 x+3)+175}{3 (2 x+3)^2-8 (2 x+3)+5}d\sqrt {2 x+3}-\frac {\sqrt {2 x+3} (139 x+121)}{3 \left (3 x^2+5 x+2\right )}\) |
\(\Big \downarrow \) 1480 |
\(\displaystyle -\frac {2}{3} \left (620 \int \frac {1}{3 (2 x+3)-5}d\sqrt {2 x+3}-477 \int \frac {1}{3 (2 x+3)-3}d\sqrt {2 x+3}\right )-\frac {\sqrt {2 x+3} (139 x+121)}{3 \left (3 x^2+5 x+2\right )}\) |
\(\Big \downarrow \) 220 |
\(\displaystyle -\frac {2}{3} \left (159 \text {arctanh}\left (\sqrt {2 x+3}\right )-124 \sqrt {\frac {5}{3}} \text {arctanh}\left (\sqrt {\frac {3}{5}} \sqrt {2 x+3}\right )\right )-\frac {\sqrt {2 x+3} (139 x+121)}{3 \left (3 x^2+5 x+2\right )}\) |
-1/3*(Sqrt[3 + 2*x]*(121 + 139*x))/(2 + 5*x + 3*x^2) - (2*(159*ArcTanh[Sqr t[3 + 2*x]] - 124*Sqrt[5/3]*ArcTanh[Sqrt[3/5]*Sqrt[3 + 2*x]]))/3
3.26.60.3.1 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 - 1))*(a + b*x + c*x^2) ^(p + 1)*((2*a*c*(e*f + d*g) - b*(c*d*f + a*e*g) - (2*c^2*d*f + b^2*e*g - c *(b*e*f + b*d*g + 2*a*e*g))*x)/(c*(p + 1)*(b^2 - 4*a*c))), x] - Simp[1/(c*( p + 1)*(b^2 - 4*a*c)) Int[(d + e*x)^(m - 2)*(a + b*x + c*x^2)^(p + 1)*Sim p[2*c^2*d^2*f*(2*p + 3) + b*e*g*(a*e*(m - 1) + b*d*(p + 2)) - c*(2*a*e*(e*f *(m - 1) + d*g*m) + b*d*(d*g*(2*p + 3) - e*f*(m - 2*p - 4))) + e*(b^2*e*g*( m + p + 1) + 2*c^2*d*f*(m + 2*p + 2) - c*(2*a*e*g*m + b*(e*f + d*g)*(m + 2* p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && LtQ[p, -1] && GtQ[m, 1] && ((EqQ[m, 2] && EqQ[p, -3] && RationalQ[a, b, c, d, e, f, g]) | | !ILtQ[m + 2*p + 3, 0])
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 = 0.38 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.97
method | result | size |
risch | \(-\frac {\left (121+139 x \right ) \sqrt {3+2 x}}{3 \left (3 x^{2}+5 x +2\right )}-53 \ln \left (\sqrt {3+2 x}+1\right )+\frac {248 \,\operatorname {arctanh}\left (\frac {\sqrt {15}\, \sqrt {3+2 x}}{5}\right ) \sqrt {15}}{9}+53 \ln \left (\sqrt {3+2 x}-1\right )\) | \(70\) |
derivativedivides | \(-\frac {6}{\sqrt {3+2 x}+1}-53 \ln \left (\sqrt {3+2 x}+1\right )-\frac {170 \sqrt {3+2 x}}{9 \left (\frac {4}{3}+2 x \right )}+\frac {248 \,\operatorname {arctanh}\left (\frac {\sqrt {15}\, \sqrt {3+2 x}}{5}\right ) \sqrt {15}}{9}-\frac {6}{\sqrt {3+2 x}-1}+53 \ln \left (\sqrt {3+2 x}-1\right )\) | \(86\) |
default | \(-\frac {6}{\sqrt {3+2 x}+1}-53 \ln \left (\sqrt {3+2 x}+1\right )-\frac {170 \sqrt {3+2 x}}{9 \left (\frac {4}{3}+2 x \right )}+\frac {248 \,\operatorname {arctanh}\left (\frac {\sqrt {15}\, \sqrt {3+2 x}}{5}\right ) \sqrt {15}}{9}-\frac {6}{\sqrt {3+2 x}-1}+53 \ln \left (\sqrt {3+2 x}-1\right )\) | \(86\) |
trager | \(-\frac {\left (121+139 x \right ) \sqrt {3+2 x}}{3 \left (3 x^{2}+5 x +2\right )}-53 \ln \left (\frac {\sqrt {3+2 x}+2+x}{1+x}\right )-\frac {124 \operatorname {RootOf}\left (\textit {\_Z}^{2}-15\right ) \ln \left (\frac {-3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-15\right ) x +15 \sqrt {3+2 x}-7 \operatorname {RootOf}\left (\textit {\_Z}^{2}-15\right )}{2+3 x}\right )}{9}\) | \(91\) |
pseudoelliptic | \(\frac {744 \sqrt {15}\, \left (x +\frac {2}{3}\right ) \left (1+x \right ) \operatorname {arctanh}\left (\frac {\sqrt {15}\, \sqrt {3+2 x}}{5}\right )+\left (1431 x^{2}+2385 x +954\right ) \ln \left (\sqrt {3+2 x}-1\right )+\left (-1431 x^{2}-2385 x -954\right ) \ln \left (\sqrt {3+2 x}+1\right )+\left (-417 x -363\right ) \sqrt {3+2 x}}{27 x^{2}+45 x +18}\) | \(94\) |
-1/3*(121+139*x)*(3+2*x)^(1/2)/(3*x^2+5*x+2)-53*ln((3+2*x)^(1/2)+1)+248/9* arctanh(1/5*15^(1/2)*(3+2*x)^(1/2))*15^(1/2)+53*ln((3+2*x)^(1/2)-1)
Leaf count of result is larger than twice the leaf count of optimal. 119 vs. \(2 (55) = 110\).
Time = 0.28 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.65 \[ \int \frac {(5-x) (3+2 x)^{3/2}}{\left (2+5 x+3 x^2\right )^2} \, dx=\frac {124 \, \sqrt {5} \sqrt {3} {\left (3 \, x^{2} + 5 \, x + 2\right )} \log \left (\frac {\sqrt {5} \sqrt {3} \sqrt {2 \, x + 3} + 3 \, x + 7}{3 \, x + 2}\right ) - 477 \, {\left (3 \, x^{2} + 5 \, x + 2\right )} \log \left (\sqrt {2 \, x + 3} + 1\right ) + 477 \, {\left (3 \, x^{2} + 5 \, x + 2\right )} \log \left (\sqrt {2 \, x + 3} - 1\right ) - 3 \, {\left (139 \, x + 121\right )} \sqrt {2 \, x + 3}}{9 \, {\left (3 \, x^{2} + 5 \, x + 2\right )}} \]
1/9*(124*sqrt(5)*sqrt(3)*(3*x^2 + 5*x + 2)*log((sqrt(5)*sqrt(3)*sqrt(2*x + 3) + 3*x + 7)/(3*x + 2)) - 477*(3*x^2 + 5*x + 2)*log(sqrt(2*x + 3) + 1) + 477*(3*x^2 + 5*x + 2)*log(sqrt(2*x + 3) - 1) - 3*(139*x + 121)*sqrt(2*x + 3))/(3*x^2 + 5*x + 2)
Time = 40.67 (sec) , antiderivative size = 212, normalized size of antiderivative = 2.94 \[ \int \frac {(5-x) (3+2 x)^{3/2}}{\left (2+5 x+3 x^2\right )^2} \, dx=- \frac {107 \sqrt {15} \left (\log {\left (\sqrt {2 x + 3} - \frac {\sqrt {15}}{3} \right )} - \log {\left (\sqrt {2 x + 3} + \frac {\sqrt {15}}{3} \right )}\right )}{9} + \frac {1700 \left (\begin {cases} \frac {\sqrt {15} \left (- \frac {\log {\left (\frac {\sqrt {15} \sqrt {2 x + 3}}{5} - 1 \right )}}{4} + \frac {\log {\left (\frac {\sqrt {15} \sqrt {2 x + 3}}{5} + 1 \right )}}{4} - \frac {1}{4 \left (\frac {\sqrt {15} \sqrt {2 x + 3}}{5} + 1\right )} - \frac {1}{4 \left (\frac {\sqrt {15} \sqrt {2 x + 3}}{5} - 1\right )}\right )}{75} & \text {for}\: \sqrt {2 x + 3} > - \frac {\sqrt {15}}{3} \wedge \sqrt {2 x + 3} < \frac {\sqrt {15}}{3} \end {cases}\right )}{3} + 53 \log {\left (\sqrt {2 x + 3} - 1 \right )} - 53 \log {\left (\sqrt {2 x + 3} + 1 \right )} - \frac {6}{\sqrt {2 x + 3} + 1} - \frac {6}{\sqrt {2 x + 3} - 1} \]
-107*sqrt(15)*(log(sqrt(2*x + 3) - sqrt(15)/3) - log(sqrt(2*x + 3) + sqrt( 15)/3))/9 + 1700*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)))/3 + 53*log(sqrt(2*x + 3) - 1) - 53*lo g(sqrt(2*x + 3) + 1) - 6/(sqrt(2*x + 3) + 1) - 6/(sqrt(2*x + 3) - 1)
Time = 0.26 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.36 \[ \int \frac {(5-x) (3+2 x)^{3/2}}{\left (2+5 x+3 x^2\right )^2} \, dx=-\frac {124}{9} \, \sqrt {15} \log \left (-\frac {\sqrt {15} - 3 \, \sqrt {2 \, x + 3}}{\sqrt {15} + 3 \, \sqrt {2 \, x + 3}}\right ) - \frac {2 \, {\left (139 \, {\left (2 \, x + 3\right )}^{\frac {3}{2}} - 175 \, \sqrt {2 \, x + 3}\right )}}{3 \, {\left (3 \, {\left (2 \, x + 3\right )}^{2} - 16 \, x - 19\right )}} - 53 \, \log \left (\sqrt {2 \, x + 3} + 1\right ) + 53 \, \log \left (\sqrt {2 \, x + 3} - 1\right ) \]
-124/9*sqrt(15)*log(-(sqrt(15) - 3*sqrt(2*x + 3))/(sqrt(15) + 3*sqrt(2*x + 3))) - 2/3*(139*(2*x + 3)^(3/2) - 175*sqrt(2*x + 3))/(3*(2*x + 3)^2 - 16* x - 19) - 53*log(sqrt(2*x + 3) + 1) + 53*log(sqrt(2*x + 3) - 1)
Time = 0.29 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.42 \[ \int \frac {(5-x) (3+2 x)^{3/2}}{\left (2+5 x+3 x^2\right )^2} \, dx=-\frac {124}{9} \, \sqrt {15} \log \left (\frac {{\left | -2 \, \sqrt {15} + 6 \, \sqrt {2 \, x + 3} \right |}}{2 \, {\left (\sqrt {15} + 3 \, \sqrt {2 \, x + 3}\right )}}\right ) - \frac {2 \, {\left (139 \, {\left (2 \, x + 3\right )}^{\frac {3}{2}} - 175 \, \sqrt {2 \, x + 3}\right )}}{3 \, {\left (3 \, {\left (2 \, x + 3\right )}^{2} - 16 \, x - 19\right )}} - 53 \, \log \left (\sqrt {2 \, x + 3} + 1\right ) + 53 \, \log \left ({\left | \sqrt {2 \, x + 3} - 1 \right |}\right ) \]
-124/9*sqrt(15)*log(1/2*abs(-2*sqrt(15) + 6*sqrt(2*x + 3))/(sqrt(15) + 3*s qrt(2*x + 3))) - 2/3*(139*(2*x + 3)^(3/2) - 175*sqrt(2*x + 3))/(3*(2*x + 3 )^2 - 16*x - 19) - 53*log(sqrt(2*x + 3) + 1) + 53*log(abs(sqrt(2*x + 3) - 1))
Time = 11.13 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.92 \[ \int \frac {(5-x) (3+2 x)^{3/2}}{\left (2+5 x+3 x^2\right )^2} \, dx=\frac {248\,\sqrt {15}\,\mathrm {atanh}\left (\frac {\sqrt {15}\,\sqrt {2\,x+3}}{5}\right )}{9}-\frac {\frac {350\,\sqrt {2\,x+3}}{9}-\frac {278\,{\left (2\,x+3\right )}^{3/2}}{9}}{\frac {16\,x}{3}-{\left (2\,x+3\right )}^2+\frac {19}{3}}-106\,\mathrm {atanh}\left (\sqrt {2\,x+3}\right ) \]