Integrand size = 27, antiderivative size = 68 \[ \int \frac {5-x}{(3+2 x)^{5/2} \left (2+5 x+3 x^2\right )} \, dx=-\frac {26}{15 (3+2 x)^{3/2}}-\frac {198}{25 \sqrt {3+2 x}}+12 \text {arctanh}\left (\sqrt {3+2 x}\right )-\frac {102}{25} \sqrt {\frac {3}{5}} \text {arctanh}\left (\sqrt {\frac {3}{5}} \sqrt {3+2 x}\right ) \] Output:
-26/15/(3+2*x)^(3/2)-198/25/(3+2*x)^(1/2)+12*arctanh((3+2*x)^(1/2))-102/12 5*15^(1/2)*arctanh(1/5*15^(1/2)*(3+2*x)^(1/2))
Time = 0.13 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.88 \[ \int \frac {5-x}{(3+2 x)^{5/2} \left (2+5 x+3 x^2\right )} \, dx=-\frac {4 (478+297 x)}{75 (3+2 x)^{3/2}}+12 \text {arctanh}\left (\sqrt {3+2 x}\right )-\frac {102}{25} \sqrt {\frac {3}{5}} \text {arctanh}\left (\sqrt {\frac {3}{5}} \sqrt {3+2 x}\right ) \] Input:
Integrate[(5 - x)/((3 + 2*x)^(5/2)*(2 + 5*x + 3*x^2)),x]
Output:
(-4*(478 + 297*x))/(75*(3 + 2*x)^(3/2)) + 12*ArcTanh[Sqrt[3 + 2*x]] - (102 *Sqrt[3/5]*ArcTanh[Sqrt[3/5]*Sqrt[3 + 2*x]])/25
Time = 0.40 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.12, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {1198, 27, 1198, 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.
\(\displaystyle \int \frac {5-x}{(2 x+3)^{5/2} \left (3 x^2+5 x+2\right )} \, dx\) |
\(\Big \downarrow \) 1198 |
\(\displaystyle \frac {1}{5} \int -\frac {3 (13 x+3)}{(2 x+3)^{3/2} \left (3 x^2+5 x+2\right )}dx-\frac {26}{15 (2 x+3)^{3/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {3}{5} \int \frac {13 x+3}{(2 x+3)^{3/2} \left (3 x^2+5 x+2\right )}dx-\frac {26}{15 (2 x+3)^{3/2}}\) |
\(\Big \downarrow \) 1198 |
\(\displaystyle -\frac {3}{5} \left (\frac {1}{5} \int \frac {99 x+49}{\sqrt {2 x+3} \left (3 x^2+5 x+2\right )}dx+\frac {66}{5 \sqrt {2 x+3}}\right )-\frac {26}{15 (2 x+3)^{3/2}}\) |
\(\Big \downarrow \) 1197 |
\(\displaystyle -\frac {3}{5} \left (\frac {2}{5} \int -\frac {199-99 (2 x+3)}{3 (2 x+3)^2-8 (2 x+3)+5}d\sqrt {2 x+3}+\frac {66}{5 \sqrt {2 x+3}}\right )-\frac {26}{15 (2 x+3)^{3/2}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {3}{5} \left (\frac {66}{5 \sqrt {2 x+3}}-\frac {2}{5} \int \frac {199-99 (2 x+3)}{3 (2 x+3)^2-8 (2 x+3)+5}d\sqrt {2 x+3}\right )-\frac {26}{15 (2 x+3)^{3/2}}\) |
\(\Big \downarrow \) 1480 |
\(\displaystyle -\frac {3}{5} \left (\frac {2}{5} \left (150 \int \frac {1}{3 (2 x+3)-3}d\sqrt {2 x+3}-51 \int \frac {1}{3 (2 x+3)-5}d\sqrt {2 x+3}\right )+\frac {66}{5 \sqrt {2 x+3}}\right )-\frac {26}{15 (2 x+3)^{3/2}}\) |
\(\Big \downarrow \) 220 |
\(\displaystyle -\frac {3}{5} \left (\frac {2}{5} \left (17 \sqrt {\frac {3}{5}} \text {arctanh}\left (\sqrt {\frac {3}{5}} \sqrt {2 x+3}\right )-50 \text {arctanh}\left (\sqrt {2 x+3}\right )\right )+\frac {66}{5 \sqrt {2 x+3}}\right )-\frac {26}{15 (2 x+3)^{3/2}}\) |
Input:
Int[(5 - x)/((3 + 2*x)^(5/2)*(2 + 5*x + 3*x^2)),x]
Output:
-26/(15*(3 + 2*x)^(3/2)) - (3*(66/(5*Sqrt[3 + 2*x]) + (2*(-50*ArcTanh[Sqrt [3 + 2*x]] + 17*Sqrt[3/5]*ArcTanh[Sqrt[3/5]*Sqrt[3 + 2*x]]))/5))/5
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 = 1.54 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.91
method | result | size |
derivativedivides | \(-\frac {102 \sqrt {15}\, \operatorname {arctanh}\left (\frac {\sqrt {15}\, \sqrt {2 x +3}}{5}\right )}{125}-\frac {26}{15 \left (2 x +3\right )^{\frac {3}{2}}}-\frac {198}{25 \sqrt {2 x +3}}+6 \ln \left (\sqrt {2 x +3}+1\right )-6 \ln \left (\sqrt {2 x +3}-1\right )\) | \(62\) |
default | \(-\frac {102 \sqrt {15}\, \operatorname {arctanh}\left (\frac {\sqrt {15}\, \sqrt {2 x +3}}{5}\right )}{125}-\frac {26}{15 \left (2 x +3\right )^{\frac {3}{2}}}-\frac {198}{25 \sqrt {2 x +3}}+6 \ln \left (\sqrt {2 x +3}+1\right )-6 \ln \left (\sqrt {2 x +3}-1\right )\) | \(62\) |
trager | \(-\frac {4 \left (297 x +478\right )}{75 \left (2 x +3\right )^{\frac {3}{2}}}+\frac {51 \operatorname {RootOf}\left (\textit {\_Z}^{2}-15\right ) \ln \left (\frac {-3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-15\right ) x +15 \sqrt {2 x +3}-7 \operatorname {RootOf}\left (\textit {\_Z}^{2}-15\right )}{3 x +2}\right )}{125}+6 \ln \left (\frac {\sqrt {2 x +3}+2+x}{x +1}\right )\) | \(79\) |
pseudoelliptic | \(-\frac {204 \left (\sqrt {15}\, \sqrt {2 x +3}\, \left (x +\frac {3}{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {15}\, \sqrt {2 x +3}}{5}\right )+\frac {125 \sqrt {2 x +3}\, \left (x +\frac {3}{2}\right ) \ln \left (\sqrt {2 x +3}-1\right )}{17}-\frac {125 \sqrt {2 x +3}\, \left (x +\frac {3}{2}\right ) \ln \left (\sqrt {2 x +3}+1\right )}{17}+\frac {165 x}{17}+\frac {2390}{153}\right )}{125 \left (2 x +3\right )^{\frac {3}{2}}}\) | \(86\) |
Input:
int((5-x)/(2*x+3)^(5/2)/(3*x^2+5*x+2),x,method=_RETURNVERBOSE)
Output:
-102/125*15^(1/2)*arctanh(1/5*15^(1/2)*(2*x+3)^(1/2))-26/15/(2*x+3)^(3/2)- 198/25/(2*x+3)^(1/2)+6*ln((2*x+3)^(1/2)+1)-6*ln((2*x+3)^(1/2)-1)
Leaf count of result is larger than twice the leaf count of optimal. 115 vs. \(2 (47) = 94\).
Time = 0.09 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.69 \[ \int \frac {5-x}{(3+2 x)^{5/2} \left (2+5 x+3 x^2\right )} \, dx=\frac {153 \, \sqrt {\frac {3}{5}} {\left (4 \, x^{2} + 12 \, x + 9\right )} \log \left (-\frac {5 \, \sqrt {\frac {3}{5}} \sqrt {2 \, x + 3} - 3 \, x - 7}{3 \, x + 2}\right ) + 450 \, {\left (4 \, x^{2} + 12 \, x + 9\right )} \log \left (\sqrt {2 \, x + 3} + 1\right ) - 450 \, {\left (4 \, x^{2} + 12 \, x + 9\right )} \log \left (\sqrt {2 \, x + 3} - 1\right ) - 4 \, {\left (297 \, x + 478\right )} \sqrt {2 \, x + 3}}{75 \, {\left (4 \, x^{2} + 12 \, x + 9\right )}} \] Input:
integrate((5-x)/(3+2*x)^(5/2)/(3*x^2+5*x+2),x, algorithm="fricas")
Output:
1/75*(153*sqrt(3/5)*(4*x^2 + 12*x + 9)*log(-(5*sqrt(3/5)*sqrt(2*x + 3) - 3 *x - 7)/(3*x + 2)) + 450*(4*x^2 + 12*x + 9)*log(sqrt(2*x + 3) + 1) - 450*( 4*x^2 + 12*x + 9)*log(sqrt(2*x + 3) - 1) - 4*(297*x + 478)*sqrt(2*x + 3))/ (4*x^2 + 12*x + 9)
Time = 3.57 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.35 \[ \int \frac {5-x}{(3+2 x)^{5/2} \left (2+5 x+3 x^2\right )} \, dx=\frac {51 \sqrt {15} \left (\log {\left (\sqrt {2 x + 3} - \frac {\sqrt {15}}{3} \right )} - \log {\left (\sqrt {2 x + 3} + \frac {\sqrt {15}}{3} \right )}\right )}{125} - 6 \log {\left (\sqrt {2 x + 3} - 1 \right )} + 6 \log {\left (\sqrt {2 x + 3} + 1 \right )} - \frac {198}{25 \sqrt {2 x + 3}} - \frac {26}{15 \left (2 x + 3\right )^{\frac {3}{2}}} \] Input:
integrate((5-x)/(3+2*x)**(5/2)/(3*x**2+5*x+2),x)
Output:
51*sqrt(15)*(log(sqrt(2*x + 3) - sqrt(15)/3) - log(sqrt(2*x + 3) + sqrt(15 )/3))/125 - 6*log(sqrt(2*x + 3) - 1) + 6*log(sqrt(2*x + 3) + 1) - 198/(25* sqrt(2*x + 3)) - 26/(15*(2*x + 3)**(3/2))
Time = 0.11 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.10 \[ \int \frac {5-x}{(3+2 x)^{5/2} \left (2+5 x+3 x^2\right )} \, dx=\frac {51}{125} \, \sqrt {15} \log \left (-\frac {\sqrt {15} - 3 \, \sqrt {2 \, x + 3}}{\sqrt {15} + 3 \, \sqrt {2 \, x + 3}}\right ) - \frac {4 \, {\left (297 \, x + 478\right )}}{75 \, {\left (2 \, x + 3\right )}^{\frac {3}{2}}} + 6 \, \log \left (\sqrt {2 \, x + 3} + 1\right ) - 6 \, \log \left (\sqrt {2 \, x + 3} - 1\right ) \] Input:
integrate((5-x)/(3+2*x)^(5/2)/(3*x^2+5*x+2),x, algorithm="maxima")
Output:
51/125*sqrt(15)*log(-(sqrt(15) - 3*sqrt(2*x + 3))/(sqrt(15) + 3*sqrt(2*x + 3))) - 4/75*(297*x + 478)/(2*x + 3)^(3/2) + 6*log(sqrt(2*x + 3) + 1) - 6* log(sqrt(2*x + 3) - 1)
Time = 0.22 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.16 \[ \int \frac {5-x}{(3+2 x)^{5/2} \left (2+5 x+3 x^2\right )} \, dx=\frac {51}{125} \, \sqrt {15} \log \left (\frac {{\left | -2 \, \sqrt {15} + 6 \, \sqrt {2 \, x + 3} \right |}}{2 \, {\left (\sqrt {15} + 3 \, \sqrt {2 \, x + 3}\right )}}\right ) - \frac {4 \, {\left (297 \, x + 478\right )}}{75 \, {\left (2 \, x + 3\right )}^{\frac {3}{2}}} + 6 \, \log \left (\sqrt {2 \, x + 3} + 1\right ) - 6 \, \log \left ({\left | \sqrt {2 \, x + 3} - 1 \right |}\right ) \] Input:
integrate((5-x)/(3+2*x)^(5/2)/(3*x^2+5*x+2),x, algorithm="giac")
Output:
51/125*sqrt(15)*log(1/2*abs(-2*sqrt(15) + 6*sqrt(2*x + 3))/(sqrt(15) + 3*s qrt(2*x + 3))) - 4/75*(297*x + 478)/(2*x + 3)^(3/2) + 6*log(sqrt(2*x + 3) + 1) - 6*log(abs(sqrt(2*x + 3) - 1))
Time = 0.08 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.63 \[ \int \frac {5-x}{(3+2 x)^{5/2} \left (2+5 x+3 x^2\right )} \, dx=12\,\mathrm {atanh}\left (\sqrt {2\,x+3}\right )-\frac {\frac {396\,x}{25}+\frac {1912}{75}}{{\left (2\,x+3\right )}^{3/2}}-\frac {102\,\sqrt {15}\,\mathrm {atanh}\left (\frac {\sqrt {15}\,\sqrt {2\,x+3}}{5}\right )}{125} \] Input:
int(-(x - 5)/((2*x + 3)^(5/2)*(5*x + 3*x^2 + 2)),x)
Output:
12*atanh((2*x + 3)^(1/2)) - ((396*x)/25 + 1912/75)/(2*x + 3)^(3/2) - (102* 15^(1/2)*atanh((15^(1/2)*(2*x + 3)^(1/2))/5))/125
Time = 0.24 (sec) , antiderivative size = 186, normalized size of antiderivative = 2.74 \[ \int \frac {5-x}{(3+2 x)^{5/2} \left (2+5 x+3 x^2\right )} \, dx=\frac {306 \sqrt {2 x +3}\, \sqrt {15}\, \mathrm {log}\left (3 \sqrt {2 x +3}-\sqrt {15}\right ) x +459 \sqrt {2 x +3}\, \sqrt {15}\, \mathrm {log}\left (3 \sqrt {2 x +3}-\sqrt {15}\right )-306 \sqrt {2 x +3}\, \sqrt {15}\, \mathrm {log}\left (3 \sqrt {2 x +3}+\sqrt {15}\right ) x -459 \sqrt {2 x +3}\, \sqrt {15}\, \mathrm {log}\left (3 \sqrt {2 x +3}+\sqrt {15}\right )-4500 \sqrt {2 x +3}\, \mathrm {log}\left (\sqrt {2 x +3}-1\right ) x -6750 \sqrt {2 x +3}\, \mathrm {log}\left (\sqrt {2 x +3}-1\right )+4500 \sqrt {2 x +3}\, \mathrm {log}\left (\sqrt {2 x +3}+1\right ) x +6750 \sqrt {2 x +3}\, \mathrm {log}\left (\sqrt {2 x +3}+1\right )-5940 x -9560}{375 \sqrt {2 x +3}\, \left (2 x +3\right )} \] Input:
int((5-x)/(3+2*x)^(5/2)/(3*x^2+5*x+2),x)
Output:
(306*sqrt(2*x + 3)*sqrt(15)*log(3*sqrt(2*x + 3) - sqrt(15))*x + 459*sqrt(2 *x + 3)*sqrt(15)*log(3*sqrt(2*x + 3) - sqrt(15)) - 306*sqrt(2*x + 3)*sqrt( 15)*log(3*sqrt(2*x + 3) + sqrt(15))*x - 459*sqrt(2*x + 3)*sqrt(15)*log(3*s qrt(2*x + 3) + sqrt(15)) - 4500*sqrt(2*x + 3)*log(sqrt(2*x + 3) - 1)*x - 6 750*sqrt(2*x + 3)*log(sqrt(2*x + 3) - 1) + 4500*sqrt(2*x + 3)*log(sqrt(2*x + 3) + 1)*x + 6750*sqrt(2*x + 3)*log(sqrt(2*x + 3) + 1) - 5940*x - 9560)/ (375*sqrt(2*x + 3)*(2*x + 3))