Integrand size = 24, antiderivative size = 102 \[ \int \frac {(1-2 x)^{5/2} (3+5 x)^2}{(2+3 x)^2} \, dx=-\frac {350}{81} \sqrt {1-2 x}-\frac {50}{81} (1-2 x)^{3/2}-\frac {10}{63} (1-2 x)^{5/2}-\frac {25}{63} (1-2 x)^{7/2}-\frac {(1-2 x)^{7/2}}{63 (2+3 x)}+\frac {350}{81} \sqrt {\frac {7}{3}} \text {arctanh}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right ) \]
-50/81*(1-2*x)^(3/2)-10/63*(1-2*x)^(5/2)-25/63*(1-2*x)^(7/2)-1/63*(1-2*x)^ (7/2)/(2+3*x)+350/243*arctanh(1/7*21^(1/2)*(1-2*x)^(1/2))*21^(1/2)-350/81* (1-2*x)^(1/2)
Time = 0.11 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.69 \[ \int \frac {(1-2 x)^{5/2} (3+5 x)^2}{(2+3 x)^2} \, dx=\frac {\sqrt {1-2 x} \left (-6239-4471 x+1002 x^2-5508 x^3+5400 x^4\right )}{567 (2+3 x)}+\frac {350}{81} \sqrt {\frac {7}{3}} \text {arctanh}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right ) \]
(Sqrt[1 - 2*x]*(-6239 - 4471*x + 1002*x^2 - 5508*x^3 + 5400*x^4))/(567*(2 + 3*x)) + (350*Sqrt[7/3]*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]])/81
Time = 0.20 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.16, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {100, 27, 90, 60, 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.
\(\displaystyle \int \frac {(1-2 x)^{5/2} (5 x+3)^2}{(3 x+2)^2} \, dx\) |
\(\Big \downarrow \) 100 |
\(\displaystyle \frac {1}{63} \int \frac {25 (1-2 x)^{5/2} (21 x+11)}{3 x+2}dx-\frac {(1-2 x)^{7/2}}{63 (3 x+2)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {25}{63} \int \frac {(1-2 x)^{5/2} (21 x+11)}{3 x+2}dx-\frac {(1-2 x)^{7/2}}{63 (3 x+2)}\) |
\(\Big \downarrow \) 90 |
\(\displaystyle \frac {25}{63} \left (-3 \int \frac {(1-2 x)^{5/2}}{3 x+2}dx-(1-2 x)^{7/2}\right )-\frac {(1-2 x)^{7/2}}{63 (3 x+2)}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {25}{63} \left (-3 \left (\frac {7}{3} \int \frac {(1-2 x)^{3/2}}{3 x+2}dx+\frac {2}{15} (1-2 x)^{5/2}\right )-(1-2 x)^{7/2}\right )-\frac {(1-2 x)^{7/2}}{63 (3 x+2)}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {25}{63} \left (-3 \left (\frac {7}{3} \left (\frac {7}{3} \int \frac {\sqrt {1-2 x}}{3 x+2}dx+\frac {2}{9} (1-2 x)^{3/2}\right )+\frac {2}{15} (1-2 x)^{5/2}\right )-(1-2 x)^{7/2}\right )-\frac {(1-2 x)^{7/2}}{63 (3 x+2)}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {25}{63} \left (-3 \left (\frac {7}{3} \left (\frac {7}{3} \left (\frac {7}{3} \int \frac {1}{\sqrt {1-2 x} (3 x+2)}dx+\frac {2}{3} \sqrt {1-2 x}\right )+\frac {2}{9} (1-2 x)^{3/2}\right )+\frac {2}{15} (1-2 x)^{5/2}\right )-(1-2 x)^{7/2}\right )-\frac {(1-2 x)^{7/2}}{63 (3 x+2)}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {25}{63} \left (-3 \left (\frac {7}{3} \left (\frac {7}{3} \left (\frac {2}{3} \sqrt {1-2 x}-\frac {7}{3} \int \frac {1}{\frac {7}{2}-\frac {3}{2} (1-2 x)}d\sqrt {1-2 x}\right )+\frac {2}{9} (1-2 x)^{3/2}\right )+\frac {2}{15} (1-2 x)^{5/2}\right )-(1-2 x)^{7/2}\right )-\frac {(1-2 x)^{7/2}}{63 (3 x+2)}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {25}{63} \left (-3 \left (\frac {7}{3} \left (\frac {7}{3} \left (\frac {2}{3} \sqrt {1-2 x}-\frac {2}{3} \sqrt {\frac {7}{3}} \text {arctanh}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )\right )+\frac {2}{9} (1-2 x)^{3/2}\right )+\frac {2}{15} (1-2 x)^{5/2}\right )-(1-2 x)^{7/2}\right )-\frac {(1-2 x)^{7/2}}{63 (3 x+2)}\) |
-1/63*(1 - 2*x)^(7/2)/(2 + 3*x) + (25*(-(1 - 2*x)^(7/2) - 3*((2*(1 - 2*x)^ (5/2))/15 + (7*((2*(1 - 2*x)^(3/2))/9 + (7*((2*Sqrt[1 - 2*x])/3 - (2*Sqrt[ 7/3]*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]])/3))/3))/3)))/63
3.20.48.3.1 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 = 1.02 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.60
method | result | size |
risch | \(-\frac {10800 x^{5}-16416 x^{4}+7512 x^{3}-9944 x^{2}-8007 x +6239}{567 \left (2+3 x \right ) \sqrt {1-2 x}}+\frac {350 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{243}\) | \(61\) |
pseudoelliptic | \(\frac {2450 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \left (2+3 x \right ) \sqrt {21}+3 \sqrt {1-2 x}\, \left (5400 x^{4}-5508 x^{3}+1002 x^{2}-4471 x -6239\right )}{3402+5103 x}\) | \(62\) |
derivativedivides | \(-\frac {25 \left (1-2 x \right )^{\frac {7}{2}}}{63}-\frac {4 \left (1-2 x \right )^{\frac {5}{2}}}{27}-\frac {16 \left (1-2 x \right )^{\frac {3}{2}}}{27}-\frac {1036 \sqrt {1-2 x}}{243}+\frac {98 \sqrt {1-2 x}}{729 \left (-\frac {4}{3}-2 x \right )}+\frac {350 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{243}\) | \(72\) |
default | \(-\frac {25 \left (1-2 x \right )^{\frac {7}{2}}}{63}-\frac {4 \left (1-2 x \right )^{\frac {5}{2}}}{27}-\frac {16 \left (1-2 x \right )^{\frac {3}{2}}}{27}-\frac {1036 \sqrt {1-2 x}}{243}+\frac {98 \sqrt {1-2 x}}{729 \left (-\frac {4}{3}-2 x \right )}+\frac {350 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{243}\) | \(72\) |
trager | \(\frac {\sqrt {1-2 x}\, \left (5400 x^{4}-5508 x^{3}+1002 x^{2}-4471 x -6239\right )}{1134+1701 x}+\frac {175 \operatorname {RootOf}\left (\textit {\_Z}^{2}-21\right ) \ln \left (\frac {-3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-21\right ) x +21 \sqrt {1-2 x}+5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-21\right )}{2+3 x}\right )}{243}\) | \(82\) |
-1/567*(10800*x^5-16416*x^4+7512*x^3-9944*x^2-8007*x+6239)/(2+3*x)/(1-2*x) ^(1/2)+350/243*arctanh(1/7*21^(1/2)*(1-2*x)^(1/2))*21^(1/2)
Time = 0.24 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.79 \[ \int \frac {(1-2 x)^{5/2} (3+5 x)^2}{(2+3 x)^2} \, dx=\frac {1225 \, \sqrt {7} \sqrt {3} {\left (3 \, x + 2\right )} \log \left (-\frac {\sqrt {7} \sqrt {3} \sqrt {-2 \, x + 1} - 3 \, x + 5}{3 \, x + 2}\right ) + 3 \, {\left (5400 \, x^{4} - 5508 \, x^{3} + 1002 \, x^{2} - 4471 \, x - 6239\right )} \sqrt {-2 \, x + 1}}{1701 \, {\left (3 \, x + 2\right )}} \]
1/1701*(1225*sqrt(7)*sqrt(3)*(3*x + 2)*log(-(sqrt(7)*sqrt(3)*sqrt(-2*x + 1 ) - 3*x + 5)/(3*x + 2)) + 3*(5400*x^4 - 5508*x^3 + 1002*x^2 - 4471*x - 623 9)*sqrt(-2*x + 1))/(3*x + 2)
Time = 31.61 (sec) , antiderivative size = 211, normalized size of antiderivative = 2.07 \[ \int \frac {(1-2 x)^{5/2} (3+5 x)^2}{(2+3 x)^2} \, dx=- \frac {25 \left (1 - 2 x\right )^{\frac {7}{2}}}{63} - \frac {4 \left (1 - 2 x\right )^{\frac {5}{2}}}{27} - \frac {16 \left (1 - 2 x\right )^{\frac {3}{2}}}{27} - \frac {1036 \sqrt {1 - 2 x}}{243} - \frac {532 \sqrt {21} \left (\log {\left (\sqrt {1 - 2 x} - \frac {\sqrt {21}}{3} \right )} - \log {\left (\sqrt {1 - 2 x} + \frac {\sqrt {21}}{3} \right )}\right )}{729} - \frac {1372 \left (\begin {cases} \frac {\sqrt {21} \left (- \frac {\log {\left (\frac {\sqrt {21} \sqrt {1 - 2 x}}{7} - 1 \right )}}{4} + \frac {\log {\left (\frac {\sqrt {21} \sqrt {1 - 2 x}}{7} + 1 \right )}}{4} - \frac {1}{4 \left (\frac {\sqrt {21} \sqrt {1 - 2 x}}{7} + 1\right )} - \frac {1}{4 \left (\frac {\sqrt {21} \sqrt {1 - 2 x}}{7} - 1\right )}\right )}{147} & \text {for}\: \sqrt {1 - 2 x} > - \frac {\sqrt {21}}{3} \wedge \sqrt {1 - 2 x} < \frac {\sqrt {21}}{3} \end {cases}\right )}{243} \]
-25*(1 - 2*x)**(7/2)/63 - 4*(1 - 2*x)**(5/2)/27 - 16*(1 - 2*x)**(3/2)/27 - 1036*sqrt(1 - 2*x)/243 - 532*sqrt(21)*(log(sqrt(1 - 2*x) - sqrt(21)/3) - log(sqrt(1 - 2*x) + sqrt(21)/3))/729 - 1372*Piecewise((sqrt(21)*(-log(sqrt (21)*sqrt(1 - 2*x)/7 - 1)/4 + log(sqrt(21)*sqrt(1 - 2*x)/7 + 1)/4 - 1/(4*( sqrt(21)*sqrt(1 - 2*x)/7 + 1)) - 1/(4*(sqrt(21)*sqrt(1 - 2*x)/7 - 1)))/147 , (sqrt(1 - 2*x) > -sqrt(21)/3) & (sqrt(1 - 2*x) < sqrt(21)/3)))/243
Time = 0.30 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.87 \[ \int \frac {(1-2 x)^{5/2} (3+5 x)^2}{(2+3 x)^2} \, dx=-\frac {25}{63} \, {\left (-2 \, x + 1\right )}^{\frac {7}{2}} - \frac {4}{27} \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} - \frac {16}{27} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - \frac {175}{243} \, \sqrt {21} \log \left (-\frac {\sqrt {21} - 3 \, \sqrt {-2 \, x + 1}}{\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}}\right ) - \frac {1036}{243} \, \sqrt {-2 \, x + 1} - \frac {49 \, \sqrt {-2 \, x + 1}}{243 \, {\left (3 \, x + 2\right )}} \]
-25/63*(-2*x + 1)^(7/2) - 4/27*(-2*x + 1)^(5/2) - 16/27*(-2*x + 1)^(3/2) - 175/243*sqrt(21)*log(-(sqrt(21) - 3*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2 *x + 1))) - 1036/243*sqrt(-2*x + 1) - 49/243*sqrt(-2*x + 1)/(3*x + 2)
Time = 0.29 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.04 \[ \int \frac {(1-2 x)^{5/2} (3+5 x)^2}{(2+3 x)^2} \, dx=\frac {25}{63} \, {\left (2 \, x - 1\right )}^{3} \sqrt {-2 \, x + 1} - \frac {4}{27} \, {\left (2 \, x - 1\right )}^{2} \sqrt {-2 \, x + 1} - \frac {16}{27} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - \frac {175}{243} \, \sqrt {21} \log \left (\frac {{\left | -2 \, \sqrt {21} + 6 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}\right )}}\right ) - \frac {1036}{243} \, \sqrt {-2 \, x + 1} - \frac {49 \, \sqrt {-2 \, x + 1}}{243 \, {\left (3 \, x + 2\right )}} \]
25/63*(2*x - 1)^3*sqrt(-2*x + 1) - 4/27*(2*x - 1)^2*sqrt(-2*x + 1) - 16/27 *(-2*x + 1)^(3/2) - 175/243*sqrt(21)*log(1/2*abs(-2*sqrt(21) + 6*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) - 1036/243*sqrt(-2*x + 1) - 49/243*s qrt(-2*x + 1)/(3*x + 2)
Time = 1.60 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.72 \[ \int \frac {(1-2 x)^{5/2} (3+5 x)^2}{(2+3 x)^2} \, dx=-\frac {98\,\sqrt {1-2\,x}}{729\,\left (2\,x+\frac {4}{3}\right )}-\frac {1036\,\sqrt {1-2\,x}}{243}-\frac {16\,{\left (1-2\,x\right )}^{3/2}}{27}-\frac {4\,{\left (1-2\,x\right )}^{5/2}}{27}-\frac {25\,{\left (1-2\,x\right )}^{7/2}}{63}-\frac {\sqrt {21}\,\mathrm {atan}\left (\frac {\sqrt {21}\,\sqrt {1-2\,x}\,1{}\mathrm {i}}{7}\right )\,350{}\mathrm {i}}{243} \]