Integrand size = 27, antiderivative size = 55 \[ \int \frac {5-x}{(3+2 x)^{3/2} \left (2+5 x+3 x^2\right )} \, dx=-\frac {26}{5 \sqrt {3+2 x}}+12 \text {arctanh}\left (\sqrt {3+2 x}\right )-\frac {34}{5} \sqrt {\frac {3}{5}} \text {arctanh}\left (\sqrt {\frac {3}{5}} \sqrt {3+2 x}\right ) \]
12*arctanh((3+2*x)^(1/2))-34/25*arctanh(1/5*15^(1/2)*(3+2*x)^(1/2))*15^(1/ 2)-26/5/(3+2*x)^(1/2)
Time = 0.08 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00 \[ \int \frac {5-x}{(3+2 x)^{3/2} \left (2+5 x+3 x^2\right )} \, dx=-\frac {26}{5 \sqrt {3+2 x}}+12 \text {arctanh}\left (\sqrt {3+2 x}\right )-\frac {34}{5} \sqrt {\frac {3}{5}} \text {arctanh}\left (\sqrt {\frac {3}{5}} \sqrt {3+2 x}\right ) \]
-26/(5*Sqrt[3 + 2*x]) + 12*ArcTanh[Sqrt[3 + 2*x]] - (34*Sqrt[3/5]*ArcTanh[ Sqrt[3/5]*Sqrt[3 + 2*x]])/5
Time = 0.22 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.02, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {1198, 27, 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.
\(\displaystyle \int \frac {5-x}{(2 x+3)^{3/2} \left (3 x^2+5 x+2\right )} \, dx\) |
\(\Big \downarrow \) 1198 |
\(\displaystyle \frac {1}{5} \int -\frac {3 (13 x+3)}{\sqrt {2 x+3} \left (3 x^2+5 x+2\right )}dx-\frac {26}{5 \sqrt {2 x+3}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {3}{5} \int \frac {13 x+3}{\sqrt {2 x+3} \left (3 x^2+5 x+2\right )}dx-\frac {26}{5 \sqrt {2 x+3}}\) |
\(\Big \downarrow \) 1197 |
\(\displaystyle -\frac {6}{5} \int -\frac {33-13 (2 x+3)}{3 (2 x+3)^2-8 (2 x+3)+5}d\sqrt {2 x+3}-\frac {26}{5 \sqrt {2 x+3}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {6}{5} \int \frac {33-13 (2 x+3)}{3 (2 x+3)^2-8 (2 x+3)+5}d\sqrt {2 x+3}-\frac {26}{5 \sqrt {2 x+3}}\) |
\(\Big \downarrow \) 1480 |
\(\displaystyle -\frac {6}{5} \left (30 \int \frac {1}{3 (2 x+3)-3}d\sqrt {2 x+3}-17 \int \frac {1}{3 (2 x+3)-5}d\sqrt {2 x+3}\right )-\frac {26}{5 \sqrt {2 x+3}}\) |
\(\Big \downarrow \) 220 |
\(\displaystyle -\frac {6}{5} \left (\frac {17 \text {arctanh}\left (\sqrt {\frac {3}{5}} \sqrt {2 x+3}\right )}{\sqrt {15}}-10 \text {arctanh}\left (\sqrt {2 x+3}\right )\right )-\frac {26}{5 \sqrt {2 x+3}}\) |
-26/(5*Sqrt[3 + 2*x]) - (6*(-10*ArcTanh[Sqrt[3 + 2*x]] + (17*ArcTanh[Sqrt[ 3/5]*Sqrt[3 + 2*x]])/Sqrt[15]))/5
3.26.55.3.1 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[((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), x_Symbol] :> Simp[(e*f - d*g)*((d + e*x)^(m + 1)/((m + 1)*(c *d^2 - b*d*e + a*e^2))), x] + Simp[1/(c*d^2 - b*d*e + a*e^2) Int[(d + e*x )^(m + 1)*(Simp[c*d*f - f*b*e + a*e*g - c*(e*f - d*g)*x, x]/(a + b*x + c*x^ 2)), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && FractionQ[m] && LtQ[m, -1 ]
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 = 0.36 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.96
method | result | size |
derivativedivides | \(-\frac {26}{5 \sqrt {3+2 x}}-6 \ln \left (\sqrt {3+2 x}-1\right )+6 \ln \left (\sqrt {3+2 x}+1\right )-\frac {34 \,\operatorname {arctanh}\left (\frac {\sqrt {15}\, \sqrt {3+2 x}}{5}\right ) \sqrt {15}}{25}\) | \(53\) |
default | \(-\frac {26}{5 \sqrt {3+2 x}}-6 \ln \left (\sqrt {3+2 x}-1\right )+6 \ln \left (\sqrt {3+2 x}+1\right )-\frac {34 \,\operatorname {arctanh}\left (\frac {\sqrt {15}\, \sqrt {3+2 x}}{5}\right ) \sqrt {15}}{25}\) | \(53\) |
risch | \(-\frac {26}{5 \sqrt {3+2 x}}-6 \ln \left (\sqrt {3+2 x}-1\right )+6 \ln \left (\sqrt {3+2 x}+1\right )-\frac {34 \,\operatorname {arctanh}\left (\frac {\sqrt {15}\, \sqrt {3+2 x}}{5}\right ) \sqrt {15}}{25}\) | \(53\) |
pseudoelliptic | \(-\frac {2 \left (17 \,\operatorname {arctanh}\left (\frac {\sqrt {15}\, \sqrt {3+2 x}}{5}\right ) \sqrt {15}\, \sqrt {3+2 x}+75 \ln \left (\sqrt {3+2 x}-1\right ) \sqrt {3+2 x}-75 \ln \left (\sqrt {3+2 x}+1\right ) \sqrt {3+2 x}+65\right )}{25 \sqrt {3+2 x}}\) | \(75\) |
trager | \(-\frac {26}{5 \sqrt {3+2 x}}-6 \ln \left (\frac {-2-x +\sqrt {3+2 x}}{1+x}\right )-\frac {17 \operatorname {RootOf}\left (\textit {\_Z}^{2}-15\right ) \ln \left (\frac {3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-15\right ) x +15 \sqrt {3+2 x}+7 \operatorname {RootOf}\left (\textit {\_Z}^{2}-15\right )}{2+3 x}\right )}{25}\) | \(76\) |
-26/5/(3+2*x)^(1/2)-6*ln((3+2*x)^(1/2)-1)+6*ln((3+2*x)^(1/2)+1)-34/25*arct anh(1/5*15^(1/2)*(3+2*x)^(1/2))*15^(1/2)
Leaf count of result is larger than twice the leaf count of optimal. 95 vs. \(2 (38) = 76\).
Time = 0.37 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.73 \[ \int \frac {5-x}{(3+2 x)^{3/2} \left (2+5 x+3 x^2\right )} \, dx=\frac {17 \, \sqrt {5} \sqrt {3} {\left (2 \, x + 3\right )} \log \left (-\frac {\sqrt {5} \sqrt {3} \sqrt {2 \, x + 3} - 3 \, x - 7}{3 \, x + 2}\right ) + 150 \, {\left (2 \, x + 3\right )} \log \left (\sqrt {2 \, x + 3} + 1\right ) - 150 \, {\left (2 \, x + 3\right )} \log \left (\sqrt {2 \, x + 3} - 1\right ) - 130 \, \sqrt {2 \, x + 3}}{25 \, {\left (2 \, x + 3\right )}} \]
1/25*(17*sqrt(5)*sqrt(3)*(2*x + 3)*log(-(sqrt(5)*sqrt(3)*sqrt(2*x + 3) - 3 *x - 7)/(3*x + 2)) + 150*(2*x + 3)*log(sqrt(2*x + 3) + 1) - 150*(2*x + 3)* log(sqrt(2*x + 3) - 1) - 130*sqrt(2*x + 3))/(2*x + 3)
Time = 3.41 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.45 \[ \int \frac {5-x}{(3+2 x)^{3/2} \left (2+5 x+3 x^2\right )} \, dx=\frac {17 \sqrt {15} \left (\log {\left (\sqrt {2 x + 3} - \frac {\sqrt {15}}{3} \right )} - \log {\left (\sqrt {2 x + 3} + \frac {\sqrt {15}}{3} \right )}\right )}{25} - 6 \log {\left (\sqrt {2 x + 3} - 1 \right )} + 6 \log {\left (\sqrt {2 x + 3} + 1 \right )} - \frac {26}{5 \sqrt {2 x + 3}} \]
17*sqrt(15)*(log(sqrt(2*x + 3) - sqrt(15)/3) - log(sqrt(2*x + 3) + sqrt(15 )/3))/25 - 6*log(sqrt(2*x + 3) - 1) + 6*log(sqrt(2*x + 3) + 1) - 26/(5*sqr t(2*x + 3))
Time = 0.27 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.27 \[ \int \frac {5-x}{(3+2 x)^{3/2} \left (2+5 x+3 x^2\right )} \, dx=\frac {17}{25} \, \sqrt {15} \log \left (-\frac {\sqrt {15} - 3 \, \sqrt {2 \, x + 3}}{\sqrt {15} + 3 \, \sqrt {2 \, x + 3}}\right ) - \frac {26}{5 \, \sqrt {2 \, x + 3}} + 6 \, \log \left (\sqrt {2 \, x + 3} + 1\right ) - 6 \, \log \left (\sqrt {2 \, x + 3} - 1\right ) \]
17/25*sqrt(15)*log(-(sqrt(15) - 3*sqrt(2*x + 3))/(sqrt(15) + 3*sqrt(2*x + 3))) - 26/5/sqrt(2*x + 3) + 6*log(sqrt(2*x + 3) + 1) - 6*log(sqrt(2*x + 3) - 1)
Time = 0.27 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.35 \[ \int \frac {5-x}{(3+2 x)^{3/2} \left (2+5 x+3 x^2\right )} \, dx=\frac {17}{25} \, \sqrt {15} \log \left (\frac {{\left | -2 \, \sqrt {15} + 6 \, \sqrt {2 \, x + 3} \right |}}{2 \, {\left (\sqrt {15} + 3 \, \sqrt {2 \, x + 3}\right )}}\right ) - \frac {26}{5 \, \sqrt {2 \, x + 3}} + 6 \, \log \left (\sqrt {2 \, x + 3} + 1\right ) - 6 \, \log \left ({\left | \sqrt {2 \, x + 3} - 1 \right |}\right ) \]
17/25*sqrt(15)*log(1/2*abs(-2*sqrt(15) + 6*sqrt(2*x + 3))/(sqrt(15) + 3*sq rt(2*x + 3))) - 26/5/sqrt(2*x + 3) + 6*log(sqrt(2*x + 3) + 1) - 6*log(abs( sqrt(2*x + 3) - 1))
Time = 11.13 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.69 \[ \int \frac {5-x}{(3+2 x)^{3/2} \left (2+5 x+3 x^2\right )} \, dx=12\,\mathrm {atanh}\left (\sqrt {2\,x+3}\right )-\frac {34\,\sqrt {15}\,\mathrm {atanh}\left (\frac {\sqrt {15}\,\sqrt {2\,x+3}}{5}\right )}{25}-\frac {26}{5\,\sqrt {2\,x+3}} \]