Integrand size = 22, antiderivative size = 68 \[ \int \frac {\sqrt {1-2 x} (2+3 x)}{(3+5 x)^3} \, dx=-\frac {(1-2 x)^{3/2}}{110 (3+5 x)^2}-\frac {67 \sqrt {1-2 x}}{550 (3+5 x)}+\frac {67 \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{275 \sqrt {55}} \]
-1/110*(1-2*x)^(3/2)/(3+5*x)^2+67/15125*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2 ))*55^(1/2)-67/550*(1-2*x)^(1/2)/(3+5*x)
Time = 0.14 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.78 \[ \int \frac {\sqrt {1-2 x} (2+3 x)}{(3+5 x)^3} \, dx=-\frac {\sqrt {1-2 x} (206+325 x)}{550 (3+5 x)^2}+\frac {67 \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{275 \sqrt {55}} \]
-1/550*(Sqrt[1 - 2*x]*(206 + 325*x))/(3 + 5*x)^2 + (67*ArcTanh[Sqrt[5/11]* Sqrt[1 - 2*x]])/(275*Sqrt[55])
Time = 0.17 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.07, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {87, 51, 73, 219}
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.
\(\displaystyle \int \frac {\sqrt {1-2 x} (3 x+2)}{(5 x+3)^3} \, dx\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {67}{110} \int \frac {\sqrt {1-2 x}}{(5 x+3)^2}dx-\frac {(1-2 x)^{3/2}}{110 (5 x+3)^2}\) |
\(\Big \downarrow \) 51 |
\(\displaystyle \frac {67}{110} \left (-\frac {1}{5} \int \frac {1}{\sqrt {1-2 x} (5 x+3)}dx-\frac {\sqrt {1-2 x}}{5 (5 x+3)}\right )-\frac {(1-2 x)^{3/2}}{110 (5 x+3)^2}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {67}{110} \left (\frac {1}{5} \int \frac {1}{\frac {11}{2}-\frac {5}{2} (1-2 x)}d\sqrt {1-2 x}-\frac {\sqrt {1-2 x}}{5 (5 x+3)}\right )-\frac {(1-2 x)^{3/2}}{110 (5 x+3)^2}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {67}{110} \left (\frac {2 \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{5 \sqrt {55}}-\frac {\sqrt {1-2 x}}{5 (5 x+3)}\right )-\frac {(1-2 x)^{3/2}}{110 (5 x+3)^2}\) |
-1/110*(1 - 2*x)^(3/2)/(3 + 5*x)^2 + (67*(-1/5*Sqrt[1 - 2*x]/(3 + 5*x) + ( 2*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(5*Sqrt[55])))/110
3.19.51.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + 1))), x] - Simp[d*(n/(b*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n}, x ] && ILtQ[m, -1] && FractionQ[n] && GtQ[n, 0]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)) Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege rQ[p] || !(IntegerQ[n] || !(EqQ[e, 0] || !(EqQ[c, 0] || LtQ[p, n]))))
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Time = 0.98 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.68
method | result | size |
risch | \(\frac {650 x^{2}+87 x -206}{550 \left (3+5 x \right )^{2} \sqrt {1-2 x}}+\frac {67 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{15125}\) | \(46\) |
derivativedivides | \(-\frac {100 \left (-\frac {13 \left (1-2 x \right )^{\frac {3}{2}}}{1100}+\frac {67 \sqrt {1-2 x}}{2500}\right )}{\left (-6-10 x \right )^{2}}+\frac {67 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{15125}\) | \(48\) |
default | \(-\frac {100 \left (-\frac {13 \left (1-2 x \right )^{\frac {3}{2}}}{1100}+\frac {67 \sqrt {1-2 x}}{2500}\right )}{\left (-6-10 x \right )^{2}}+\frac {67 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{15125}\) | \(48\) |
pseudoelliptic | \(\frac {134 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \left (3+5 x \right )^{2} \sqrt {55}-55 \sqrt {1-2 x}\, \left (325 x +206\right )}{30250 \left (3+5 x \right )^{2}}\) | \(50\) |
trager | \(-\frac {\left (325 x +206\right ) \sqrt {1-2 x}}{550 \left (3+5 x \right )^{2}}-\frac {67 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) \ln \left (\frac {5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) x +55 \sqrt {1-2 x}-8 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right )}{3+5 x}\right )}{30250}\) | \(67\) |
1/550*(650*x^2+87*x-206)/(3+5*x)^2/(1-2*x)^(1/2)+67/15125*arctanh(1/11*55^ (1/2)*(1-2*x)^(1/2))*55^(1/2)
Time = 0.22 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.03 \[ \int \frac {\sqrt {1-2 x} (2+3 x)}{(3+5 x)^3} \, dx=\frac {67 \, \sqrt {55} {\left (25 \, x^{2} + 30 \, x + 9\right )} \log \left (\frac {5 \, x - \sqrt {55} \sqrt {-2 \, x + 1} - 8}{5 \, x + 3}\right ) - 55 \, {\left (325 \, x + 206\right )} \sqrt {-2 \, x + 1}}{30250 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} \]
1/30250*(67*sqrt(55)*(25*x^2 + 30*x + 9)*log((5*x - sqrt(55)*sqrt(-2*x + 1 ) - 8)/(5*x + 3)) - 55*(325*x + 206)*sqrt(-2*x + 1))/(25*x^2 + 30*x + 9)
Leaf count of result is larger than twice the leaf count of optimal. 121 vs. \(2 (56) = 112\).
Time = 75.60 (sec) , antiderivative size = 330, normalized size of antiderivative = 4.85 \[ \int \frac {\sqrt {1-2 x} (2+3 x)}{(3+5 x)^3} \, dx=- \frac {6 \sqrt {55} \left (\log {\left (\sqrt {1 - 2 x} - \frac {\sqrt {55}}{5} \right )} - \log {\left (\sqrt {1 - 2 x} + \frac {\sqrt {55}}{5} \right )}\right )}{1375} - \frac {124 \left (\begin {cases} \frac {\sqrt {55} \left (- \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1 \right )}}{4} + \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1 \right )}}{4} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1\right )} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1\right )}\right )}{605} & \text {for}\: \sqrt {1 - 2 x} > - \frac {\sqrt {55}}{5} \wedge \sqrt {1 - 2 x} < \frac {\sqrt {55}}{5} \end {cases}\right )}{25} + \frac {88 \left (\begin {cases} \frac {\sqrt {55} \cdot \left (\frac {3 \log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1 \right )}}{16} - \frac {3 \log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1 \right )}}{16} + \frac {3}{16 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1\right )} + \frac {1}{16 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1\right )^{2}} + \frac {3}{16 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1\right )} - \frac {1}{16 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1\right )^{2}}\right )}{6655} & \text {for}\: \sqrt {1 - 2 x} > - \frac {\sqrt {55}}{5} \wedge \sqrt {1 - 2 x} < \frac {\sqrt {55}}{5} \end {cases}\right )}{25} \]
-6*sqrt(55)*(log(sqrt(1 - 2*x) - sqrt(55)/5) - log(sqrt(1 - 2*x) + sqrt(55 )/5))/1375 - 124*Piecewise((sqrt(55)*(-log(sqrt(55)*sqrt(1 - 2*x)/11 - 1)/ 4 + log(sqrt(55)*sqrt(1 - 2*x)/11 + 1)/4 - 1/(4*(sqrt(55)*sqrt(1 - 2*x)/11 + 1)) - 1/(4*(sqrt(55)*sqrt(1 - 2*x)/11 - 1)))/605, (sqrt(1 - 2*x) > -sqr t(55)/5) & (sqrt(1 - 2*x) < sqrt(55)/5)))/25 + 88*Piecewise((sqrt(55)*(3*l og(sqrt(55)*sqrt(1 - 2*x)/11 - 1)/16 - 3*log(sqrt(55)*sqrt(1 - 2*x)/11 + 1 )/16 + 3/(16*(sqrt(55)*sqrt(1 - 2*x)/11 + 1)) + 1/(16*(sqrt(55)*sqrt(1 - 2 *x)/11 + 1)**2) + 3/(16*(sqrt(55)*sqrt(1 - 2*x)/11 - 1)) - 1/(16*(sqrt(55) *sqrt(1 - 2*x)/11 - 1)**2))/6655, (sqrt(1 - 2*x) > -sqrt(55)/5) & (sqrt(1 - 2*x) < sqrt(55)/5)))/25
Time = 0.34 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.09 \[ \int \frac {\sqrt {1-2 x} (2+3 x)}{(3+5 x)^3} \, dx=-\frac {67}{30250} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) + \frac {325 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 737 \, \sqrt {-2 \, x + 1}}{275 \, {\left (25 \, {\left (2 \, x - 1\right )}^{2} + 220 \, x + 11\right )}} \]
-67/30250*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(- 2*x + 1))) + 1/275*(325*(-2*x + 1)^(3/2) - 737*sqrt(-2*x + 1))/(25*(2*x - 1)^2 + 220*x + 11)
Time = 0.29 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.00 \[ \int \frac {\sqrt {1-2 x} (2+3 x)}{(3+5 x)^3} \, dx=-\frac {67}{30250} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) + \frac {325 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 737 \, \sqrt {-2 \, x + 1}}{1100 \, {\left (5 \, x + 3\right )}^{2}} \]
-67/30250*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 1/1100*(325*(-2*x + 1)^(3/2) - 737*sqrt(-2*x + 1))/ (5*x + 3)^2
Time = 1.38 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.79 \[ \int \frac {\sqrt {1-2 x} (2+3 x)}{(3+5 x)^3} \, dx=\frac {67\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )}{15125}-\frac {\frac {67\,\sqrt {1-2\,x}}{625}-\frac {13\,{\left (1-2\,x\right )}^{3/2}}{275}}{\frac {44\,x}{5}+{\left (2\,x-1\right )}^2+\frac {11}{25}} \]