Integrand size = 22, antiderivative size = 81 \[ \int \frac {(1-2 x)^{3/2} (2+3 x)}{(3+5 x)^3} \, dx=-\frac {39}{275} \sqrt {1-2 x}-\frac {(1-2 x)^{5/2}}{110 (3+5 x)^2}-\frac {13 (1-2 x)^{3/2}}{110 (3+5 x)}+\frac {39 \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{25 \sqrt {55}} \]
-1/110*(1-2*x)^(5/2)/(3+5*x)^2-13/110*(1-2*x)^(3/2)/(3+5*x)+39/1375*arctan h(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)-39/275*(1-2*x)^(1/2)
Time = 0.15 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.72 \[ \int \frac {(1-2 x)^{3/2} (2+3 x)}{(3+5 x)^3} \, dx=-\frac {\sqrt {1-2 x} \left (82+205 x+120 x^2\right )}{50 (3+5 x)^2}+\frac {39 \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{25 \sqrt {55}} \]
-1/50*(Sqrt[1 - 2*x]*(82 + 205*x + 120*x^2))/(3 + 5*x)^2 + (39*ArcTanh[Sqr t[5/11]*Sqrt[1 - 2*x]])/(25*Sqrt[55])
Time = 0.17 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.15, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {87, 51, 60, 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.
\(\displaystyle \int \frac {(1-2 x)^{3/2} (3 x+2)}{(5 x+3)^3} \, dx\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {13}{22} \int \frac {(1-2 x)^{3/2}}{(5 x+3)^2}dx-\frac {(1-2 x)^{5/2}}{110 (5 x+3)^2}\) |
\(\Big \downarrow \) 51 |
\(\displaystyle \frac {13}{22} \left (-\frac {3}{5} \int \frac {\sqrt {1-2 x}}{5 x+3}dx-\frac {(1-2 x)^{3/2}}{5 (5 x+3)}\right )-\frac {(1-2 x)^{5/2}}{110 (5 x+3)^2}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {13}{22} \left (-\frac {3}{5} \left (\frac {11}{5} \int \frac {1}{\sqrt {1-2 x} (5 x+3)}dx+\frac {2}{5} \sqrt {1-2 x}\right )-\frac {(1-2 x)^{3/2}}{5 (5 x+3)}\right )-\frac {(1-2 x)^{5/2}}{110 (5 x+3)^2}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {13}{22} \left (-\frac {3}{5} \left (\frac {2}{5} \sqrt {1-2 x}-\frac {11}{5} \int \frac {1}{\frac {11}{2}-\frac {5}{2} (1-2 x)}d\sqrt {1-2 x}\right )-\frac {(1-2 x)^{3/2}}{5 (5 x+3)}\right )-\frac {(1-2 x)^{5/2}}{110 (5 x+3)^2}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {13}{22} \left (-\frac {3}{5} \left (\frac {2}{5} \sqrt {1-2 x}-\frac {2}{5} \sqrt {\frac {11}{5}} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )\right )-\frac {(1-2 x)^{3/2}}{5 (5 x+3)}\right )-\frac {(1-2 x)^{5/2}}{110 (5 x+3)^2}\) |
-1/110*(1 - 2*x)^(5/2)/(3 + 5*x)^2 + (13*(-1/5*(1 - 2*x)^(3/2)/(3 + 5*x) - (3*((2*Sqrt[1 - 2*x])/5 - (2*Sqrt[11/5]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]] )/5))/5))/22
3.20.22.3.1 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] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( b*(m + n + 1))) Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[a, b, c, d, m, n, x]
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 = 1.00 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.63
method | result | size |
risch | \(\frac {240 x^{3}+290 x^{2}-41 x -82}{50 \left (3+5 x \right )^{2} \sqrt {1-2 x}}+\frac {39 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{1375}\) | \(51\) |
pseudoelliptic | \(\frac {78 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \left (3+5 x \right )^{2} \sqrt {55}-55 \sqrt {1-2 x}\, \left (120 x^{2}+205 x +82\right )}{2750 \left (3+5 x \right )^{2}}\) | \(55\) |
derivativedivides | \(-\frac {12 \sqrt {1-2 x}}{125}-\frac {4 \left (-\frac {61 \left (1-2 x \right )^{\frac {3}{2}}}{20}+\frac {693 \sqrt {1-2 x}}{100}\right )}{5 \left (-6-10 x \right )^{2}}+\frac {39 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{1375}\) | \(57\) |
default | \(-\frac {12 \sqrt {1-2 x}}{125}-\frac {4 \left (-\frac {61 \left (1-2 x \right )^{\frac {3}{2}}}{20}+\frac {693 \sqrt {1-2 x}}{100}\right )}{5 \left (-6-10 x \right )^{2}}+\frac {39 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{1375}\) | \(57\) |
trager | \(-\frac {\left (120 x^{2}+205 x +82\right ) \sqrt {1-2 x}}{50 \left (3+5 x \right )^{2}}-\frac {39 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) \ln \left (\frac {5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) x +55 \sqrt {1-2 x}-8 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right )}{3+5 x}\right )}{2750}\) | \(72\) |
1/50*(240*x^3+290*x^2-41*x-82)/(3+5*x)^2/(1-2*x)^(1/2)+39/1375*arctanh(1/1 1*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)
Time = 0.23 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.93 \[ \int \frac {(1-2 x)^{3/2} (2+3 x)}{(3+5 x)^3} \, dx=\frac {39 \, \sqrt {55} {\left (25 \, x^{2} + 30 \, x + 9\right )} \log \left (\frac {5 \, x - \sqrt {55} \sqrt {-2 \, x + 1} - 8}{5 \, x + 3}\right ) - 55 \, {\left (120 \, x^{2} + 205 \, x + 82\right )} \sqrt {-2 \, x + 1}}{2750 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} \]
1/2750*(39*sqrt(55)*(25*x^2 + 30*x + 9)*log((5*x - sqrt(55)*sqrt(-2*x + 1) - 8)/(5*x + 3)) - 55*(120*x^2 + 205*x + 82)*sqrt(-2*x + 1))/(25*x^2 + 30* x + 9)
Time = 79.70 (sec) , antiderivative size = 342, normalized size of antiderivative = 4.22 \[ \int \frac {(1-2 x)^{3/2} (2+3 x)}{(3+5 x)^3} \, dx=- \frac {12 \sqrt {1 - 2 x}}{125} - \frac {128 \sqrt {55} \left (\log {\left (\sqrt {1 - 2 x} - \frac {\sqrt {55}}{5} \right )} - \log {\left (\sqrt {1 - 2 x} + \frac {\sqrt {55}}{5} \right )}\right )}{6875} - \frac {1276 \left (\begin {cases} \frac {\sqrt {55} \left (- \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1 \right )}}{4} + \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1 \right )}}{4} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1\right )} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1\right )}\right )}{605} & \text {for}\: \sqrt {1 - 2 x} > - \frac {\sqrt {55}}{5} \wedge \sqrt {1 - 2 x} < \frac {\sqrt {55}}{5} \end {cases}\right )}{125} + \frac {968 \left (\begin {cases} \frac {\sqrt {55} \cdot \left (\frac {3 \log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1 \right )}}{16} - \frac {3 \log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1 \right )}}{16} + \frac {3}{16 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1\right )} + \frac {1}{16 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1\right )^{2}} + \frac {3}{16 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1\right )} - \frac {1}{16 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1\right )^{2}}\right )}{6655} & \text {for}\: \sqrt {1 - 2 x} > - \frac {\sqrt {55}}{5} \wedge \sqrt {1 - 2 x} < \frac {\sqrt {55}}{5} \end {cases}\right )}{125} \]
-12*sqrt(1 - 2*x)/125 - 128*sqrt(55)*(log(sqrt(1 - 2*x) - sqrt(55)/5) - lo g(sqrt(1 - 2*x) + sqrt(55)/5))/6875 - 1276*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) > -sqrt(55)/5) & (sqrt(1 - 2*x) < sqrt(55)/5)))/125 + 968*Piecewise((sqrt(55)*(3*log(sqrt(55)*sqrt(1 - 2*x)/11 - 1)/16 - 3*log(s qrt(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)))/125
Time = 0.28 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.02 \[ \int \frac {(1-2 x)^{3/2} (2+3 x)}{(3+5 x)^3} \, dx=-\frac {39}{2750} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) - \frac {12}{125} \, \sqrt {-2 \, x + 1} + \frac {305 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 693 \, \sqrt {-2 \, x + 1}}{125 \, {\left (25 \, {\left (2 \, x - 1\right )}^{2} + 220 \, x + 11\right )}} \]
-39/2750*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2 *x + 1))) - 12/125*sqrt(-2*x + 1) + 1/125*(305*(-2*x + 1)^(3/2) - 693*sqrt (-2*x + 1))/(25*(2*x - 1)^2 + 220*x + 11)
Time = 0.28 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.95 \[ \int \frac {(1-2 x)^{3/2} (2+3 x)}{(3+5 x)^3} \, dx=-\frac {39}{2750} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) - \frac {12}{125} \, \sqrt {-2 \, x + 1} + \frac {305 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 693 \, \sqrt {-2 \, x + 1}}{500 \, {\left (5 \, x + 3\right )}^{2}} \]
-39/2750*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 12/125*sqrt(-2*x + 1) + 1/500*(305*(-2*x + 1)^(3/2) - 693*sqrt(-2*x + 1))/(5*x + 3)^2
Time = 1.54 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.78 \[ \int \frac {(1-2 x)^{3/2} (2+3 x)}{(3+5 x)^3} \, dx=\frac {39\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )}{1375}-\frac {12\,\sqrt {1-2\,x}}{125}-\frac {\frac {693\,\sqrt {1-2\,x}}{3125}-\frac {61\,{\left (1-2\,x\right )}^{3/2}}{625}}{\frac {44\,x}{5}+{\left (2\,x-1\right )}^2+\frac {11}{25}} \]