Integrand size = 27, antiderivative size = 94 \[ \int \frac {(5-x) (3+2 x)^{7/2}}{2+5 x+3 x^2} \, dx=\frac {3278}{81} \sqrt {3+2 x}+\frac {526}{81} (3+2 x)^{3/2}+\frac {62}{45} (3+2 x)^{5/2}-\frac {2}{21} (3+2 x)^{7/2}+12 \text {arctanh}\left (\sqrt {3+2 x}\right )-\frac {4250}{81} \sqrt {\frac {5}{3}} \text {arctanh}\left (\sqrt {\frac {3}{5}} \sqrt {3+2 x}\right ) \]
526/81*(3+2*x)^(3/2)+62/45*(3+2*x)^(5/2)-2/21*(3+2*x)^(7/2)+12*arctanh((3+ 2*x)^(1/2))-4250/243*arctanh(1/5*15^(1/2)*(3+2*x)^(1/2))*15^(1/2)+3278/81* (3+2*x)^(1/2)
Time = 0.10 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.74 \[ \int \frac {(5-x) (3+2 x)^{7/2}}{2+5 x+3 x^2} \, dx=-\frac {8 \sqrt {3+2 x} \left (-24728-8639 x-738 x^2+270 x^3\right )}{2835}+12 \text {arctanh}\left (\sqrt {3+2 x}\right )-\frac {4250}{81} \sqrt {\frac {5}{3}} \text {arctanh}\left (\sqrt {\frac {3}{5}} \sqrt {3+2 x}\right ) \]
(-8*Sqrt[3 + 2*x]*(-24728 - 8639*x - 738*x^2 + 270*x^3))/2835 + 12*ArcTanh [Sqrt[3 + 2*x]] - (4250*Sqrt[5/3]*ArcTanh[Sqrt[3/5]*Sqrt[3 + 2*x]])/81
Time = 0.31 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.19, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {1196, 1196, 1196, 1196, 1197, 25, 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.
\(\displaystyle \int \frac {(5-x) (2 x+3)^{7/2}}{3 x^2+5 x+2} \, dx\) |
\(\Big \downarrow \) 1196 |
\(\displaystyle \frac {1}{3} \int \frac {(2 x+3)^{5/2} (31 x+49)}{3 x^2+5 x+2}dx-\frac {2}{21} (2 x+3)^{7/2}\) |
\(\Big \downarrow \) 1196 |
\(\displaystyle \frac {1}{3} \left (\frac {1}{3} \int \frac {(2 x+3)^{3/2} (263 x+317)}{3 x^2+5 x+2}dx+\frac {62}{15} (2 x+3)^{5/2}\right )-\frac {2}{21} (2 x+3)^{7/2}\) |
\(\Big \downarrow \) 1196 |
\(\displaystyle \frac {1}{3} \left (\frac {1}{3} \left (\frac {1}{3} \int \frac {\sqrt {2 x+3} (1639 x+1801)}{3 x^2+5 x+2}dx+\frac {526}{9} (2 x+3)^{3/2}\right )+\frac {62}{15} (2 x+3)^{5/2}\right )-\frac {2}{21} (2 x+3)^{7/2}\) |
\(\Big \downarrow \) 1196 |
\(\displaystyle \frac {1}{3} \left (\frac {1}{3} \left (\frac {1}{3} \left (\frac {1}{3} \int \frac {9167 x+9653}{\sqrt {2 x+3} \left (3 x^2+5 x+2\right )}dx+\frac {3278}{3} \sqrt {2 x+3}\right )+\frac {526}{9} (2 x+3)^{3/2}\right )+\frac {62}{15} (2 x+3)^{5/2}\right )-\frac {2}{21} (2 x+3)^{7/2}\) |
\(\Big \downarrow \) 1197 |
\(\displaystyle \frac {1}{3} \left (\frac {1}{3} \left (\frac {1}{3} \left (\frac {2}{3} \int -\frac {8195-9167 (2 x+3)}{3 (2 x+3)^2-8 (2 x+3)+5}d\sqrt {2 x+3}+\frac {3278}{3} \sqrt {2 x+3}\right )+\frac {526}{9} (2 x+3)^{3/2}\right )+\frac {62}{15} (2 x+3)^{5/2}\right )-\frac {2}{21} (2 x+3)^{7/2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{3} \left (\frac {1}{3} \left (\frac {1}{3} \left (\frac {3278}{3} \sqrt {2 x+3}-\frac {2}{3} \int \frac {8195-9167 (2 x+3)}{3 (2 x+3)^2-8 (2 x+3)+5}d\sqrt {2 x+3}\right )+\frac {526}{9} (2 x+3)^{3/2}\right )+\frac {62}{15} (2 x+3)^{5/2}\right )-\frac {2}{21} (2 x+3)^{7/2}\) |
\(\Big \downarrow \) 1480 |
\(\displaystyle \frac {1}{3} \left (\frac {1}{3} \left (\frac {1}{3} \left (\frac {2}{3} \left (10625 \int \frac {1}{3 (2 x+3)-5}d\sqrt {2 x+3}-1458 \int \frac {1}{3 (2 x+3)-3}d\sqrt {2 x+3}\right )+\frac {3278}{3} \sqrt {2 x+3}\right )+\frac {526}{9} (2 x+3)^{3/2}\right )+\frac {62}{15} (2 x+3)^{5/2}\right )-\frac {2}{21} (2 x+3)^{7/2}\) |
\(\Big \downarrow \) 220 |
\(\displaystyle \frac {1}{3} \left (\frac {1}{3} \left (\frac {1}{3} \left (\frac {2}{3} \left (486 \text {arctanh}\left (\sqrt {2 x+3}\right )-2125 \sqrt {\frac {5}{3}} \text {arctanh}\left (\sqrt {\frac {3}{5}} \sqrt {2 x+3}\right )\right )+\frac {3278}{3} \sqrt {2 x+3}\right )+\frac {526}{9} (2 x+3)^{3/2}\right )+\frac {62}{15} (2 x+3)^{5/2}\right )-\frac {2}{21} (2 x+3)^{7/2}\) |
(-2*(3 + 2*x)^(7/2))/21 + ((62*(3 + 2*x)^(5/2))/15 + ((526*(3 + 2*x)^(3/2) )/9 + ((3278*Sqrt[3 + 2*x])/3 + (2*(486*ArcTanh[Sqrt[3 + 2*x]] - 2125*Sqrt [5/3]*ArcTanh[Sqrt[3/5]*Sqrt[3 + 2*x]]))/3)/3)/3)/3
3.26.50.3.1 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[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[g*((d + e*x)^m/(c*m)), x] + Simp[1/c Int [(d + e*x)^(m - 1)*(Simp[c*d*f - a*e*g + (g*c*d - b*e*g + c*e*f)*x, x]/(a + b*x + c*x^2)), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && FractionQ[m] & & GtQ[m, 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_)^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 = 68, normalized size of antiderivative = 0.72
method | result | size |
risch | \(-\frac {8 \left (270 x^{3}-738 x^{2}-8639 x -24728\right ) \sqrt {3+2 x}}{2835}+6 \ln \left (\sqrt {3+2 x}+1\right )-\frac {4250 \,\operatorname {arctanh}\left (\frac {\sqrt {15}\, \sqrt {3+2 x}}{5}\right ) \sqrt {15}}{243}-6 \ln \left (\sqrt {3+2 x}-1\right )\) | \(68\) |
pseudoelliptic | \(-\frac {4250 \,\operatorname {arctanh}\left (\frac {\sqrt {15}\, \sqrt {3+2 x}}{5}\right ) \sqrt {15}}{243}-6 \ln \left (\sqrt {3+2 x}-1\right )+6 \ln \left (\sqrt {3+2 x}+1\right )+\frac {8 \left (-270 x^{3}+738 x^{2}+8639 x +24728\right ) \sqrt {3+2 x}}{2835}\) | \(68\) |
derivativedivides | \(-\frac {2 \left (3+2 x \right )^{\frac {7}{2}}}{21}+\frac {62 \left (3+2 x \right )^{\frac {5}{2}}}{45}+\frac {526 \left (3+2 x \right )^{\frac {3}{2}}}{81}+\frac {3278 \sqrt {3+2 x}}{81}-\frac {4250 \,\operatorname {arctanh}\left (\frac {\sqrt {15}\, \sqrt {3+2 x}}{5}\right ) \sqrt {15}}{243}-6 \ln \left (\sqrt {3+2 x}-1\right )+6 \ln \left (\sqrt {3+2 x}+1\right )\) | \(80\) |
default | \(-\frac {2 \left (3+2 x \right )^{\frac {7}{2}}}{21}+\frac {62 \left (3+2 x \right )^{\frac {5}{2}}}{45}+\frac {526 \left (3+2 x \right )^{\frac {3}{2}}}{81}+\frac {3278 \sqrt {3+2 x}}{81}-\frac {4250 \,\operatorname {arctanh}\left (\frac {\sqrt {15}\, \sqrt {3+2 x}}{5}\right ) \sqrt {15}}{243}-6 \ln \left (\sqrt {3+2 x}-1\right )+6 \ln \left (\sqrt {3+2 x}+1\right )\) | \(80\) |
trager | \(\left (-\frac {16}{21} x^{3}+\frac {656}{315} x^{2}+\frac {69112}{2835} x +\frac {197824}{2835}\right ) \sqrt {3+2 x}+\frac {2125 \operatorname {RootOf}\left (\textit {\_Z}^{2}-15\right ) \ln \left (\frac {-3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-15\right ) x +15 \sqrt {3+2 x}-7 \operatorname {RootOf}\left (\textit {\_Z}^{2}-15\right )}{2+3 x}\right )}{243}+6 \ln \left (\frac {\sqrt {3+2 x}+2+x}{1+x}\right )\) | \(88\) |
-8/2835*(270*x^3-738*x^2-8639*x-24728)*(3+2*x)^(1/2)+6*ln((3+2*x)^(1/2)+1) -4250/243*arctanh(1/5*15^(1/2)*(3+2*x)^(1/2))*15^(1/2)-6*ln((3+2*x)^(1/2)- 1)
Time = 0.26 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.91 \[ \int \frac {(5-x) (3+2 x)^{7/2}}{2+5 x+3 x^2} \, dx=\frac {2125}{243} \, \sqrt {5} \sqrt {3} \log \left (-\frac {\sqrt {5} \sqrt {3} \sqrt {2 \, x + 3} - 3 \, x - 7}{3 \, x + 2}\right ) - \frac {8}{2835} \, {\left (270 \, x^{3} - 738 \, x^{2} - 8639 \, x - 24728\right )} \sqrt {2 \, x + 3} + 6 \, \log \left (\sqrt {2 \, x + 3} + 1\right ) - 6 \, \log \left (\sqrt {2 \, x + 3} - 1\right ) \]
2125/243*sqrt(5)*sqrt(3)*log(-(sqrt(5)*sqrt(3)*sqrt(2*x + 3) - 3*x - 7)/(3 *x + 2)) - 8/2835*(270*x^3 - 738*x^2 - 8639*x - 24728)*sqrt(2*x + 3) + 6*l og(sqrt(2*x + 3) + 1) - 6*log(sqrt(2*x + 3) - 1)
Time = 3.03 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.23 \[ \int \frac {(5-x) (3+2 x)^{7/2}}{2+5 x+3 x^2} \, dx=- \frac {2 \left (2 x + 3\right )^{\frac {7}{2}}}{21} + \frac {62 \left (2 x + 3\right )^{\frac {5}{2}}}{45} + \frac {526 \left (2 x + 3\right )^{\frac {3}{2}}}{81} + \frac {3278 \sqrt {2 x + 3}}{81} + \frac {2125 \sqrt {15} \left (\log {\left (\sqrt {2 x + 3} - \frac {\sqrt {15}}{3} \right )} - \log {\left (\sqrt {2 x + 3} + \frac {\sqrt {15}}{3} \right )}\right )}{243} - 6 \log {\left (\sqrt {2 x + 3} - 1 \right )} + 6 \log {\left (\sqrt {2 x + 3} + 1 \right )} \]
-2*(2*x + 3)**(7/2)/21 + 62*(2*x + 3)**(5/2)/45 + 526*(2*x + 3)**(3/2)/81 + 3278*sqrt(2*x + 3)/81 + 2125*sqrt(15)*(log(sqrt(2*x + 3) - sqrt(15)/3) - log(sqrt(2*x + 3) + sqrt(15)/3))/243 - 6*log(sqrt(2*x + 3) - 1) + 6*log(s qrt(2*x + 3) + 1)
Time = 0.27 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.03 \[ \int \frac {(5-x) (3+2 x)^{7/2}}{2+5 x+3 x^2} \, dx=-\frac {2}{21} \, {\left (2 \, x + 3\right )}^{\frac {7}{2}} + \frac {62}{45} \, {\left (2 \, x + 3\right )}^{\frac {5}{2}} + \frac {526}{81} \, {\left (2 \, x + 3\right )}^{\frac {3}{2}} + \frac {2125}{243} \, \sqrt {15} \log \left (-\frac {\sqrt {15} - 3 \, \sqrt {2 \, x + 3}}{\sqrt {15} + 3 \, \sqrt {2 \, x + 3}}\right ) + \frac {3278}{81} \, \sqrt {2 \, x + 3} + 6 \, \log \left (\sqrt {2 \, x + 3} + 1\right ) - 6 \, \log \left (\sqrt {2 \, x + 3} - 1\right ) \]
-2/21*(2*x + 3)^(7/2) + 62/45*(2*x + 3)^(5/2) + 526/81*(2*x + 3)^(3/2) + 2 125/243*sqrt(15)*log(-(sqrt(15) - 3*sqrt(2*x + 3))/(sqrt(15) + 3*sqrt(2*x + 3))) + 3278/81*sqrt(2*x + 3) + 6*log(sqrt(2*x + 3) + 1) - 6*log(sqrt(2*x + 3) - 1)
Time = 0.29 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.07 \[ \int \frac {(5-x) (3+2 x)^{7/2}}{2+5 x+3 x^2} \, dx=-\frac {2}{21} \, {\left (2 \, x + 3\right )}^{\frac {7}{2}} + \frac {62}{45} \, {\left (2 \, x + 3\right )}^{\frac {5}{2}} + \frac {526}{81} \, {\left (2 \, x + 3\right )}^{\frac {3}{2}} + \frac {2125}{243} \, \sqrt {15} \log \left (\frac {{\left | -2 \, \sqrt {15} + 6 \, \sqrt {2 \, x + 3} \right |}}{2 \, {\left (\sqrt {15} + 3 \, \sqrt {2 \, x + 3}\right )}}\right ) + \frac {3278}{81} \, \sqrt {2 \, x + 3} + 6 \, \log \left (\sqrt {2 \, x + 3} + 1\right ) - 6 \, \log \left ({\left | \sqrt {2 \, x + 3} - 1 \right |}\right ) \]
-2/21*(2*x + 3)^(7/2) + 62/45*(2*x + 3)^(5/2) + 526/81*(2*x + 3)^(3/2) + 2 125/243*sqrt(15)*log(1/2*abs(-2*sqrt(15) + 6*sqrt(2*x + 3))/(sqrt(15) + 3* sqrt(2*x + 3))) + 3278/81*sqrt(2*x + 3) + 6*log(sqrt(2*x + 3) + 1) - 6*log (abs(sqrt(2*x + 3) - 1))
Time = 11.12 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.76 \[ \int \frac {(5-x) (3+2 x)^{7/2}}{2+5 x+3 x^2} \, dx=\frac {3278\,\sqrt {2\,x+3}}{81}+\frac {526\,{\left (2\,x+3\right )}^{3/2}}{81}+\frac {62\,{\left (2\,x+3\right )}^{5/2}}{45}-\frac {2\,{\left (2\,x+3\right )}^{7/2}}{21}-\mathrm {atan}\left (\sqrt {2\,x+3}\,1{}\mathrm {i}\right )\,12{}\mathrm {i}+\frac {\sqrt {15}\,\mathrm {atan}\left (\frac {\sqrt {15}\,\sqrt {2\,x+3}\,1{}\mathrm {i}}{5}\right )\,4250{}\mathrm {i}}{243} \]