Integrand size = 24, antiderivative size = 89 \[ \int \frac {(1-2 x)^{3/2} (2+3 x)^2}{(3+5 x)^2} \, dx=\frac {126}{625} \sqrt {1-2 x}+\frac {42 (1-2 x)^{3/2}}{1375}-\frac {9}{125} (1-2 x)^{5/2}-\frac {(1-2 x)^{5/2}}{275 (3+5 x)}-\frac {126}{625} \sqrt {\frac {11}{5}} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \]
42/1375*(1-2*x)^(3/2)-9/125*(1-2*x)^(5/2)-1/275*(1-2*x)^(5/2)/(3+5*x)-126/ 3125*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)+126/625*(1-2*x)^(1/2)
Time = 0.11 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.71 \[ \int \frac {(1-2 x)^{3/2} (2+3 x)^2}{(3+5 x)^2} \, dx=\frac {\frac {5 \sqrt {1-2 x} \left (298+935 x+160 x^2-900 x^3\right )}{3+5 x}-126 \sqrt {55} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{3125} \]
((5*Sqrt[1 - 2*x]*(298 + 935*x + 160*x^2 - 900*x^3))/(3 + 5*x) - 126*Sqrt[ 55]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/3125
Time = 0.18 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.17, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {100, 27, 90, 60, 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.
\(\displaystyle \int \frac {(1-2 x)^{3/2} (3 x+2)^2}{(5 x+3)^2} \, dx\) |
\(\Big \downarrow \) 100 |
\(\displaystyle \frac {1}{275} \int \frac {45 (1-2 x)^{3/2} (11 x+8)}{5 x+3}dx-\frac {(1-2 x)^{5/2}}{275 (5 x+3)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {9}{55} \int \frac {(1-2 x)^{3/2} (11 x+8)}{5 x+3}dx-\frac {(1-2 x)^{5/2}}{275 (5 x+3)}\) |
\(\Big \downarrow \) 90 |
\(\displaystyle \frac {9}{55} \left (\frac {7}{5} \int \frac {(1-2 x)^{3/2}}{5 x+3}dx-\frac {11}{25} (1-2 x)^{5/2}\right )-\frac {(1-2 x)^{5/2}}{275 (5 x+3)}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {9}{55} \left (\frac {7}{5} \left (\frac {11}{5} \int \frac {\sqrt {1-2 x}}{5 x+3}dx+\frac {2}{15} (1-2 x)^{3/2}\right )-\frac {11}{25} (1-2 x)^{5/2}\right )-\frac {(1-2 x)^{5/2}}{275 (5 x+3)}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {9}{55} \left (\frac {7}{5} \left (\frac {11}{5} \left (\frac {11}{5} \int \frac {1}{\sqrt {1-2 x} (5 x+3)}dx+\frac {2}{5} \sqrt {1-2 x}\right )+\frac {2}{15} (1-2 x)^{3/2}\right )-\frac {11}{25} (1-2 x)^{5/2}\right )-\frac {(1-2 x)^{5/2}}{275 (5 x+3)}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {9}{55} \left (\frac {7}{5} \left (\frac {11}{5} \left (\frac {2}{5} \sqrt {1-2 x}-\frac {11}{5} \int \frac {1}{\frac {11}{2}-\frac {5}{2} (1-2 x)}d\sqrt {1-2 x}\right )+\frac {2}{15} (1-2 x)^{3/2}\right )-\frac {11}{25} (1-2 x)^{5/2}\right )-\frac {(1-2 x)^{5/2}}{275 (5 x+3)}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {9}{55} \left (\frac {7}{5} \left (\frac {11}{5} \left (\frac {2}{5} \sqrt {1-2 x}-\frac {2}{5} \sqrt {\frac {11}{5}} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )\right )+\frac {2}{15} (1-2 x)^{3/2}\right )-\frac {11}{25} (1-2 x)^{5/2}\right )-\frac {(1-2 x)^{5/2}}{275 (5 x+3)}\) |
-1/275*(1 - 2*x)^(5/2)/(3 + 5*x) + (9*((-11*(1 - 2*x)^(5/2))/25 + (7*((2*( 1 - 2*x)^(3/2))/15 + (11*((2*Sqrt[1 - 2*x])/5 - (2*Sqrt[11/5]*ArcTanh[Sqrt [5/11]*Sqrt[1 - 2*x]])/5))/5))/5))/55
3.20.11.3.1 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_))^(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*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(d*f*(n + p + 2)) Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0]
Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^( p_), x_] :> Simp[(b*c - a*d)^2*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d^2*(d *e - c*f)*(n + 1))), x] - Simp[1/(d^2*(d*e - c*f)*(n + 1)) Int[(c + d*x)^ (n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*( p + 1)) - 2*a*b*d*(d*e*(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x , x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] || !SumSimplerQ[p, 1])))
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 = 1.01 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.63
method | result | size |
risch | \(\frac {1800 x^{4}-1220 x^{3}-1710 x^{2}+339 x +298}{625 \left (3+5 x \right ) \sqrt {1-2 x}}-\frac {126 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{3125}\) | \(56\) |
pseudoelliptic | \(\frac {-126 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \left (3+5 x \right ) \sqrt {55}-5 \sqrt {1-2 x}\, \left (900 x^{3}-160 x^{2}-935 x -298\right )}{9375+15625 x}\) | \(57\) |
derivativedivides | \(-\frac {9 \left (1-2 x \right )^{\frac {5}{2}}}{125}+\frac {4 \left (1-2 x \right )^{\frac {3}{2}}}{125}+\frac {128 \sqrt {1-2 x}}{625}+\frac {22 \sqrt {1-2 x}}{3125 \left (-\frac {6}{5}-2 x \right )}-\frac {126 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{3125}\) | \(63\) |
default | \(-\frac {9 \left (1-2 x \right )^{\frac {5}{2}}}{125}+\frac {4 \left (1-2 x \right )^{\frac {3}{2}}}{125}+\frac {128 \sqrt {1-2 x}}{625}+\frac {22 \sqrt {1-2 x}}{3125 \left (-\frac {6}{5}-2 x \right )}-\frac {126 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{3125}\) | \(63\) |
trager | \(-\frac {\left (900 x^{3}-160 x^{2}-935 x -298\right ) \sqrt {1-2 x}}{625 \left (3+5 x \right )}-\frac {63 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) \ln \left (\frac {-5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) x +55 \sqrt {1-2 x}+8 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right )}{3+5 x}\right )}{3125}\) | \(77\) |
1/625*(1800*x^4-1220*x^3-1710*x^2+339*x+298)/(3+5*x)/(1-2*x)^(1/2)-126/312 5*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)
Time = 0.23 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.84 \[ \int \frac {(1-2 x)^{3/2} (2+3 x)^2}{(3+5 x)^2} \, dx=\frac {63 \, \sqrt {11} \sqrt {5} {\left (5 \, x + 3\right )} \log \left (\frac {\sqrt {11} \sqrt {5} \sqrt {-2 \, x + 1} + 5 \, x - 8}{5 \, x + 3}\right ) - 5 \, {\left (900 \, x^{3} - 160 \, x^{2} - 935 \, x - 298\right )} \sqrt {-2 \, x + 1}}{3125 \, {\left (5 \, x + 3\right )}} \]
1/3125*(63*sqrt(11)*sqrt(5)*(5*x + 3)*log((sqrt(11)*sqrt(5)*sqrt(-2*x + 1) + 5*x - 8)/(5*x + 3)) - 5*(900*x^3 - 160*x^2 - 935*x - 298)*sqrt(-2*x + 1 ))/(5*x + 3)
Time = 31.47 (sec) , antiderivative size = 197, normalized size of antiderivative = 2.21 \[ \int \frac {(1-2 x)^{3/2} (2+3 x)^2}{(3+5 x)^2} \, dx=- \frac {9 \left (1 - 2 x\right )^{\frac {5}{2}}}{125} + \frac {4 \left (1 - 2 x\right )^{\frac {3}{2}}}{125} + \frac {128 \sqrt {1 - 2 x}}{625} + \frac {62 \sqrt {55} \left (\log {\left (\sqrt {1 - 2 x} - \frac {\sqrt {55}}{5} \right )} - \log {\left (\sqrt {1 - 2 x} + \frac {\sqrt {55}}{5} \right )}\right )}{3125} - \frac {484 \left (\begin {cases} \frac {\sqrt {55} \left (- \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1 \right )}}{4} + \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1 \right )}}{4} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1\right )} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1\right )}\right )}{605} & \text {for}\: \sqrt {1 - 2 x} > - \frac {\sqrt {55}}{5} \wedge \sqrt {1 - 2 x} < \frac {\sqrt {55}}{5} \end {cases}\right )}{625} \]
-9*(1 - 2*x)**(5/2)/125 + 4*(1 - 2*x)**(3/2)/125 + 128*sqrt(1 - 2*x)/625 + 62*sqrt(55)*(log(sqrt(1 - 2*x) - sqrt(55)/5) - log(sqrt(1 - 2*x) + sqrt(5 5)/5))/3125 - 484*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)/1 1 + 1)) - 1/(4*(sqrt(55)*sqrt(1 - 2*x)/11 - 1)))/605, (sqrt(1 - 2*x) > -sq rt(55)/5) & (sqrt(1 - 2*x) < sqrt(55)/5)))/625
Time = 0.28 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.90 \[ \int \frac {(1-2 x)^{3/2} (2+3 x)^2}{(3+5 x)^2} \, dx=-\frac {9}{125} \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} + \frac {4}{125} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {63}{3125} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) + \frac {128}{625} \, \sqrt {-2 \, x + 1} - \frac {11 \, \sqrt {-2 \, x + 1}}{625 \, {\left (5 \, x + 3\right )}} \]
-9/125*(-2*x + 1)^(5/2) + 4/125*(-2*x + 1)^(3/2) + 63/3125*sqrt(55)*log(-( sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 128/625*sqrt (-2*x + 1) - 11/625*sqrt(-2*x + 1)/(5*x + 3)
Time = 0.28 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.01 \[ \int \frac {(1-2 x)^{3/2} (2+3 x)^2}{(3+5 x)^2} \, dx=-\frac {9}{125} \, {\left (2 \, x - 1\right )}^{2} \sqrt {-2 \, x + 1} + \frac {4}{125} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {63}{3125} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) + \frac {128}{625} \, \sqrt {-2 \, x + 1} - \frac {11 \, \sqrt {-2 \, x + 1}}{625 \, {\left (5 \, x + 3\right )}} \]
-9/125*(2*x - 1)^2*sqrt(-2*x + 1) + 4/125*(-2*x + 1)^(3/2) + 63/3125*sqrt( 55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 128/625*sqrt(-2*x + 1) - 11/625*sqrt(-2*x + 1)/(5*x + 3)
Time = 1.49 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.72 \[ \int \frac {(1-2 x)^{3/2} (2+3 x)^2}{(3+5 x)^2} \, dx=\frac {128\,\sqrt {1-2\,x}}{625}-\frac {22\,\sqrt {1-2\,x}}{3125\,\left (2\,x+\frac {6}{5}\right )}+\frac {4\,{\left (1-2\,x\right )}^{3/2}}{125}-\frac {9\,{\left (1-2\,x\right )}^{5/2}}{125}+\frac {\sqrt {55}\,\mathrm {atan}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}\,1{}\mathrm {i}}{11}\right )\,126{}\mathrm {i}}{3125} \]