Integrand size = 27, antiderivative size = 68 \[ \int \frac {(5-x) (3+2 x)^{3/2}}{2+5 x+3 x^2} \, dx=\frac {62}{9} \sqrt {3+2 x}-\frac {2}{9} (3+2 x)^{3/2}+12 \text {arctanh}\left (\sqrt {3+2 x}\right )-\frac {170}{9} \sqrt {\frac {5}{3}} \text {arctanh}\left (\sqrt {\frac {3}{5}} \sqrt {3+2 x}\right ) \] Output:
62/9*(3+2*x)^(1/2)-2/9*(3+2*x)^(3/2)+12*arctanh((3+2*x)^(1/2))-170/27*15^( 1/2)*arctanh(1/5*15^(1/2)*(3+2*x)^(1/2))
Time = 0.11 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.84 \[ \int \frac {(5-x) (3+2 x)^{3/2}}{2+5 x+3 x^2} \, dx=12 \text {arctanh}\left (\sqrt {3+2 x}\right )-\frac {2}{27} \left (6 (-14+x) \sqrt {3+2 x}+85 \sqrt {15} \text {arctanh}\left (\sqrt {\frac {3}{5}} \sqrt {3+2 x}\right )\right ) \] Input:
Integrate[((5 - x)*(3 + 2*x)^(3/2))/(2 + 5*x + 3*x^2),x]
Output:
12*ArcTanh[Sqrt[3 + 2*x]] - (2*(6*(-14 + x)*Sqrt[3 + 2*x] + 85*Sqrt[15]*Ar cTanh[Sqrt[3/5]*Sqrt[3 + 2*x]]))/27
Time = 0.42 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.12, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {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.
\(\displaystyle \int \frac {(5-x) (2 x+3)^{3/2}}{3 x^2+5 x+2} \, dx\) |
\(\Big \downarrow \) 1196 |
\(\displaystyle \frac {1}{3} \int \frac {\sqrt {2 x+3} (31 x+49)}{3 x^2+5 x+2}dx-\frac {2}{9} (2 x+3)^{3/2}\) |
\(\Big \downarrow \) 1196 |
\(\displaystyle \frac {1}{3} \left (\frac {1}{3} \int \frac {263 x+317}{\sqrt {2 x+3} \left (3 x^2+5 x+2\right )}dx+\frac {62}{3} \sqrt {2 x+3}\right )-\frac {2}{9} (2 x+3)^{3/2}\) |
\(\Big \downarrow \) 1197 |
\(\displaystyle \frac {1}{3} \left (\frac {2}{3} \int -\frac {155-263 (2 x+3)}{3 (2 x+3)^2-8 (2 x+3)+5}d\sqrt {2 x+3}+\frac {62}{3} \sqrt {2 x+3}\right )-\frac {2}{9} (2 x+3)^{3/2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{3} \left (\frac {62}{3} \sqrt {2 x+3}-\frac {2}{3} \int \frac {155-263 (2 x+3)}{3 (2 x+3)^2-8 (2 x+3)+5}d\sqrt {2 x+3}\right )-\frac {2}{9} (2 x+3)^{3/2}\) |
\(\Big \downarrow \) 1480 |
\(\displaystyle \frac {1}{3} \left (\frac {2}{3} \left (425 \int \frac {1}{3 (2 x+3)-5}d\sqrt {2 x+3}-162 \int \frac {1}{3 (2 x+3)-3}d\sqrt {2 x+3}\right )+\frac {62}{3} \sqrt {2 x+3}\right )-\frac {2}{9} (2 x+3)^{3/2}\) |
\(\Big \downarrow \) 220 |
\(\displaystyle \frac {1}{3} \left (\frac {2}{3} \left (54 \text {arctanh}\left (\sqrt {2 x+3}\right )-85 \sqrt {\frac {5}{3}} \text {arctanh}\left (\sqrt {\frac {3}{5}} \sqrt {2 x+3}\right )\right )+\frac {62}{3} \sqrt {2 x+3}\right )-\frac {2}{9} (2 x+3)^{3/2}\) |
Input:
Int[((5 - x)*(3 + 2*x)^(3/2))/(2 + 5*x + 3*x^2),x]
Output:
(-2*(3 + 2*x)^(3/2))/9 + ((62*Sqrt[3 + 2*x])/3 + (2*(54*ArcTanh[Sqrt[3 + 2 *x]] - 85*Sqrt[5/3]*ArcTanh[Sqrt[3/5]*Sqrt[3 + 2*x]]))/3)/3
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 = 1.48 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.82
method | result | size |
risch | \(-\frac {4 \left (x -14\right ) \sqrt {2 x +3}}{9}+6 \ln \left (\sqrt {2 x +3}+1\right )-6 \ln \left (\sqrt {2 x +3}-1\right )-\frac {170 \sqrt {15}\, \operatorname {arctanh}\left (\frac {\sqrt {15}\, \sqrt {2 x +3}}{5}\right )}{27}\) | \(56\) |
pseudoelliptic | \(-\frac {170 \sqrt {15}\, \operatorname {arctanh}\left (\frac {\sqrt {15}\, \sqrt {2 x +3}}{5}\right )}{27}-6 \ln \left (\sqrt {2 x +3}-1\right )+6 \ln \left (\sqrt {2 x +3}+1\right )+\frac {4 \left (-x +14\right ) \sqrt {2 x +3}}{9}\) | \(58\) |
derivativedivides | \(-\frac {2 \left (2 x +3\right )^{\frac {3}{2}}}{9}+\frac {62 \sqrt {2 x +3}}{9}-6 \ln \left (\sqrt {2 x +3}-1\right )-\frac {170 \sqrt {15}\, \operatorname {arctanh}\left (\frac {\sqrt {15}\, \sqrt {2 x +3}}{5}\right )}{27}+6 \ln \left (\sqrt {2 x +3}+1\right )\) | \(62\) |
default | \(-\frac {2 \left (2 x +3\right )^{\frac {3}{2}}}{9}+\frac {62 \sqrt {2 x +3}}{9}-6 \ln \left (\sqrt {2 x +3}-1\right )-\frac {170 \sqrt {15}\, \operatorname {arctanh}\left (\frac {\sqrt {15}\, \sqrt {2 x +3}}{5}\right )}{27}+6 \ln \left (\sqrt {2 x +3}+1\right )\) | \(62\) |
trager | \(\left (-\frac {4 x}{9}+\frac {56}{9}\right ) \sqrt {2 x +3}-6 \ln \left (\frac {-2-x +\sqrt {2 x +3}}{x +1}\right )-\frac {85 \operatorname {RootOf}\left (\textit {\_Z}^{2}-15\right ) \ln \left (\frac {3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-15\right ) x +15 \sqrt {2 x +3}+7 \operatorname {RootOf}\left (\textit {\_Z}^{2}-15\right )}{3 x +2}\right )}{27}\) | \(80\) |
Input:
int((5-x)*(2*x+3)^(3/2)/(3*x^2+5*x+2),x,method=_RETURNVERBOSE)
Output:
-4/9*(x-14)*(2*x+3)^(1/2)+6*ln((2*x+3)^(1/2)+1)-6*ln((2*x+3)^(1/2)-1)-170/ 27*15^(1/2)*arctanh(1/5*15^(1/2)*(2*x+3)^(1/2))
Time = 0.08 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.01 \[ \int \frac {(5-x) (3+2 x)^{3/2}}{2+5 x+3 x^2} \, dx=-\frac {4}{9} \, \sqrt {2 \, x + 3} {\left (x - 14\right )} + \frac {85}{9} \, \sqrt {\frac {5}{3}} \log \left (-\frac {3 \, \sqrt {\frac {5}{3}} \sqrt {2 \, x + 3} - 3 \, x - 7}{3 \, x + 2}\right ) + 6 \, \log \left (\sqrt {2 \, x + 3} + 1\right ) - 6 \, \log \left (\sqrt {2 \, x + 3} - 1\right ) \] Input:
integrate((5-x)*(3+2*x)^(3/2)/(3*x^2+5*x+2),x, algorithm="fricas")
Output:
-4/9*sqrt(2*x + 3)*(x - 14) + 85/9*sqrt(5/3)*log(-(3*sqrt(5/3)*sqrt(2*x + 3) - 3*x - 7)/(3*x + 2)) + 6*log(sqrt(2*x + 3) + 1) - 6*log(sqrt(2*x + 3) - 1)
Time = 2.56 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.35 \[ \int \frac {(5-x) (3+2 x)^{3/2}}{2+5 x+3 x^2} \, dx=- \frac {2 \left (2 x + 3\right )^{\frac {3}{2}}}{9} + \frac {62 \sqrt {2 x + 3}}{9} + \frac {85 \sqrt {15} \left (\log {\left (\sqrt {2 x + 3} - \frac {\sqrt {15}}{3} \right )} - \log {\left (\sqrt {2 x + 3} + \frac {\sqrt {15}}{3} \right )}\right )}{27} - 6 \log {\left (\sqrt {2 x + 3} - 1 \right )} + 6 \log {\left (\sqrt {2 x + 3} + 1 \right )} \] Input:
integrate((5-x)*(3+2*x)**(3/2)/(3*x**2+5*x+2),x)
Output:
-2*(2*x + 3)**(3/2)/9 + 62*sqrt(2*x + 3)/9 + 85*sqrt(15)*(log(sqrt(2*x + 3 ) - sqrt(15)/3) - log(sqrt(2*x + 3) + sqrt(15)/3))/27 - 6*log(sqrt(2*x + 3 ) - 1) + 6*log(sqrt(2*x + 3) + 1)
Time = 0.14 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.16 \[ \int \frac {(5-x) (3+2 x)^{3/2}}{2+5 x+3 x^2} \, dx=-\frac {2}{9} \, {\left (2 \, x + 3\right )}^{\frac {3}{2}} + \frac {85}{27} \, \sqrt {15} \log \left (-\frac {\sqrt {15} - 3 \, \sqrt {2 \, x + 3}}{\sqrt {15} + 3 \, \sqrt {2 \, x + 3}}\right ) + \frac {62}{9} \, \sqrt {2 \, x + 3} + 6 \, \log \left (\sqrt {2 \, x + 3} + 1\right ) - 6 \, \log \left (\sqrt {2 \, x + 3} - 1\right ) \] Input:
integrate((5-x)*(3+2*x)^(3/2)/(3*x^2+5*x+2),x, algorithm="maxima")
Output:
-2/9*(2*x + 3)^(3/2) + 85/27*sqrt(15)*log(-(sqrt(15) - 3*sqrt(2*x + 3))/(s qrt(15) + 3*sqrt(2*x + 3))) + 62/9*sqrt(2*x + 3) + 6*log(sqrt(2*x + 3) + 1 ) - 6*log(sqrt(2*x + 3) - 1)
Time = 0.21 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.22 \[ \int \frac {(5-x) (3+2 x)^{3/2}}{2+5 x+3 x^2} \, dx=-\frac {2}{9} \, {\left (2 \, x + 3\right )}^{\frac {3}{2}} + \frac {85}{27} \, \sqrt {15} \log \left (\frac {{\left | -2 \, \sqrt {15} + 6 \, \sqrt {2 \, x + 3} \right |}}{2 \, {\left (\sqrt {15} + 3 \, \sqrt {2 \, x + 3}\right )}}\right ) + \frac {62}{9} \, \sqrt {2 \, x + 3} + 6 \, \log \left (\sqrt {2 \, x + 3} + 1\right ) - 6 \, \log \left ({\left | \sqrt {2 \, x + 3} - 1 \right |}\right ) \] Input:
integrate((5-x)*(3+2*x)^(3/2)/(3*x^2+5*x+2),x, algorithm="giac")
Output:
-2/9*(2*x + 3)^(3/2) + 85/27*sqrt(15)*log(1/2*abs(-2*sqrt(15) + 6*sqrt(2*x + 3))/(sqrt(15) + 3*sqrt(2*x + 3))) + 62/9*sqrt(2*x + 3) + 6*log(sqrt(2*x + 3) + 1) - 6*log(abs(sqrt(2*x + 3) - 1))
Time = 0.08 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.78 \[ \int \frac {(5-x) (3+2 x)^{3/2}}{2+5 x+3 x^2} \, dx=\frac {62\,\sqrt {2\,x+3}}{9}-\frac {2\,{\left (2\,x+3\right )}^{3/2}}{9}-\mathrm {atan}\left (\sqrt {2\,x+3}\,1{}\mathrm {i}\right )\,12{}\mathrm {i}+\frac {\sqrt {15}\,\mathrm {atan}\left (\frac {\sqrt {15}\,\sqrt {2\,x+3}\,1{}\mathrm {i}}{5}\right )\,170{}\mathrm {i}}{27} \] Input:
int(-((2*x + 3)^(3/2)*(x - 5))/(5*x + 3*x^2 + 2),x)
Output:
(15^(1/2)*atan((15^(1/2)*(2*x + 3)^(1/2)*1i)/5)*170i)/27 - atan((2*x + 3)^ (1/2)*1i)*12i + (62*(2*x + 3)^(1/2))/9 - (2*(2*x + 3)^(3/2))/9
Time = 0.25 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.09 \[ \int \frac {(5-x) (3+2 x)^{3/2}}{2+5 x+3 x^2} \, dx=-\frac {4 \sqrt {2 x +3}\, x}{9}+\frac {56 \sqrt {2 x +3}}{9}+\frac {85 \sqrt {15}\, \mathrm {log}\left (3 \sqrt {2 x +3}-\sqrt {15}\right )}{27}-\frac {85 \sqrt {15}\, \mathrm {log}\left (3 \sqrt {2 x +3}+\sqrt {15}\right )}{27}-6 \,\mathrm {log}\left (\sqrt {2 x +3}-1\right )+6 \,\mathrm {log}\left (\sqrt {2 x +3}+1\right ) \] Input:
int((5-x)*(3+2*x)^(3/2)/(3*x^2+5*x+2),x)
Output:
( - 12*sqrt(2*x + 3)*x + 168*sqrt(2*x + 3) + 85*sqrt(15)*log(3*sqrt(2*x + 3) - sqrt(15)) - 85*sqrt(15)*log(3*sqrt(2*x + 3) + sqrt(15)) - 162*log(sqr t(2*x + 3) - 1) + 162*log(sqrt(2*x + 3) + 1))/27