Integrand size = 24, antiderivative size = 121 \[ \int \frac {(1-2 x)^{5/2} (3+5 x)^3}{(2+3 x)^2} \, dx=\frac {1540}{729} \sqrt {1-2 x}+\frac {220}{729} (1-2 x)^{3/2}+\frac {55}{81} (1-2 x)^{5/2} (3+5 x)^2-\frac {(1-2 x)^{5/2} (3+5 x)^3}{3 (2+3 x)}-\frac {22}{567} (1-2 x)^{5/2} (69+100 x)-\frac {1540}{729} \sqrt {\frac {7}{3}} \text {arctanh}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right ) \]
220/729*(1-2*x)^(3/2)+55/81*(1-2*x)^(5/2)*(3+5*x)^2-1/3*(1-2*x)^(5/2)*(3+5 *x)^3/(2+3*x)-22/567*(1-2*x)^(5/2)*(69+100*x)-1540/2187*arctanh(1/7*21^(1/ 2)*(1-2*x)^(1/2))*21^(1/2)+1540/729*(1-2*x)^(1/2)
Time = 0.11 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.60 \[ \int \frac {(1-2 x)^{5/2} (3+5 x)^3}{(2+3 x)^2} \, dx=\frac {\frac {3 \sqrt {1-2 x} \left (13759+65558 x+25275 x^2-159714 x^3-17100 x^4+189000 x^5\right )}{2+3 x}-10780 \sqrt {21} \text {arctanh}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{15309} \]
((3*Sqrt[1 - 2*x]*(13759 + 65558*x + 25275*x^2 - 159714*x^3 - 17100*x^4 + 189000*x^5))/(2 + 3*x) - 10780*Sqrt[21]*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]])/ 15309
Time = 0.22 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.12, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {108, 27, 170, 27, 164, 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.
| \(\displaystyle \int \frac {(1-2 x)^{5/2} (5 x+3)^3}{(3 x+2)^2} \, dx\) |
| \(\Big \downarrow \) 108 |
| \(\displaystyle \frac {1}{3} \int -\frac {55 (1-2 x)^{3/2} x (5 x+3)^2}{3 x+2}dx-\frac {(1-2 x)^{5/2} (5 x+3)^3}{3 (3 x+2)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {55}{3} \int \frac {(1-2 x)^{3/2} x (5 x+3)^2}{3 x+2}dx-\frac {(1-2 x)^{5/2} (5 x+3)^3}{3 (3 x+2)}\) |
| \(\Big \downarrow \) 170 |
| \(\displaystyle -\frac {55}{3} \left (-\frac {1}{27} \int \frac {2 (1-2 x)^{3/2} (5 x+3) (12 x+5)}{3 x+2}dx-\frac {1}{27} (5 x+3)^2 (1-2 x)^{5/2}\right )-\frac {(1-2 x)^{5/2} (5 x+3)^3}{3 (3 x+2)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {55}{3} \left (-\frac {2}{27} \int \frac {(1-2 x)^{3/2} (5 x+3) (12 x+5)}{3 x+2}dx-\frac {1}{27} (5 x+3)^2 (1-2 x)^{5/2}\right )-\frac {(1-2 x)^{5/2} (5 x+3)^3}{3 (3 x+2)}\) |
| \(\Big \downarrow \) 164 |
| \(\displaystyle -\frac {55}{3} \left (-\frac {2}{27} \left (\int \frac {(1-2 x)^{3/2}}{3 x+2}dx-\frac {1}{35} (1-2 x)^{5/2} (100 x+69)\right )-\frac {1}{27} (5 x+3)^2 (1-2 x)^{5/2}\right )-\frac {(1-2 x)^{5/2} (5 x+3)^3}{3 (3 x+2)}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle -\frac {55}{3} \left (-\frac {2}{27} \left (\frac {7}{3} \int \frac {\sqrt {1-2 x}}{3 x+2}dx-\frac {1}{35} (100 x+69) (1-2 x)^{5/2}+\frac {2}{9} (1-2 x)^{3/2}\right )-\frac {1}{27} (5 x+3)^2 (1-2 x)^{5/2}\right )-\frac {(1-2 x)^{5/2} (5 x+3)^3}{3 (3 x+2)}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle -\frac {55}{3} \left (-\frac {2}{27} \left (\frac {7}{3} \left (\frac {7}{3} \int \frac {1}{\sqrt {1-2 x} (3 x+2)}dx+\frac {2}{3} \sqrt {1-2 x}\right )-\frac {1}{35} (100 x+69) (1-2 x)^{5/2}+\frac {2}{9} (1-2 x)^{3/2}\right )-\frac {1}{27} (5 x+3)^2 (1-2 x)^{5/2}\right )-\frac {(1-2 x)^{5/2} (5 x+3)^3}{3 (3 x+2)}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle -\frac {55}{3} \left (-\frac {2}{27} \left (\frac {7}{3} \left (\frac {2}{3} \sqrt {1-2 x}-\frac {7}{3} \int \frac {1}{\frac {7}{2}-\frac {3}{2} (1-2 x)}d\sqrt {1-2 x}\right )-\frac {1}{35} (100 x+69) (1-2 x)^{5/2}+\frac {2}{9} (1-2 x)^{3/2}\right )-\frac {1}{27} (5 x+3)^2 (1-2 x)^{5/2}\right )-\frac {(1-2 x)^{5/2} (5 x+3)^3}{3 (3 x+2)}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle -\frac {55}{3} \left (-\frac {2}{27} \left (\frac {7}{3} \left (\frac {2}{3} \sqrt {1-2 x}-\frac {2}{3} \sqrt {\frac {7}{3}} \text {arctanh}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )\right )-\frac {1}{35} (100 x+69) (1-2 x)^{5/2}+\frac {2}{9} (1-2 x)^{3/2}\right )-\frac {1}{27} (5 x+3)^2 (1-2 x)^{5/2}\right )-\frac {(1-2 x)^{5/2} (5 x+3)^3}{3 (3 x+2)}\) |
-1/3*((1 - 2*x)^(5/2)*(3 + 5*x)^3)/(2 + 3*x) - (55*(-1/27*((1 - 2*x)^(5/2) *(3 + 5*x)^2) - (2*((2*(1 - 2*x)^(3/2))/9 - ((1 - 2*x)^(5/2)*(69 + 100*x)) /35 + (7*((2*Sqrt[1 - 2*x])/3 - (2*Sqrt[7/3]*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2* x]])/3))/3))/27))/3
3.20.61.3.1 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_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Simp[(a + b*x)^(m + 1)*(c + d*x)^n*((e + f*x)^p/(b*(m + 1))) , x] - Simp[1/(b*(m + 1)) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f* x)^(p - 1)*Simp[d*e*n + c*f*p + d*f*(n + p)*x, x], x], x] /; FreeQ[{a, b, c , d, e, f}, x] && LtQ[m, -1] && GtQ[n, 0] && GtQ[p, 0] && (IntegersQ[2*m, 2 *n, 2*p] || IntegersQ[m, n + p] || IntegersQ[p, m + n])
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_ ))*((g_.) + (h_.)*(x_)), x_] :> Simp[(-(a*d*f*h*(n + 2) + b*c*f*h*(m + 2) - b*d*(f*g + e*h)*(m + n + 3) - b*d*f*h*(m + n + 2)*x))*(a + b*x)^(m + 1)*(( c + d*x)^(n + 1)/(b^2*d^2*(m + n + 2)*(m + n + 3))), x] + Simp[(a^2*d^2*f*h *(n + 1)*(n + 2) + a*b*d*(n + 1)*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b^2*(c^2*f*h*(m + 1)*(m + 2) - c*d*(f*g + e*h)*(m + 1)*(m + n + 3) + d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2*d^2*(m + n + 2)*(m + n + 3)) Int[( a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && NeQ[m + n + 2, 0] && NeQ[m + n + 3, 0]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_)*((g_.) + (h_.)*(x_)), x_] :> Simp[h*(a + b*x)^m*(c + d*x)^(n + 1)*(( e + f*x)^(p + 1)/(d*f*(m + n + p + 2))), x] + Simp[1/(d*f*(m + n + p + 2)) Int[(a + b*x)^(m - 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*g*(m + n + p + 2 ) - h*(b*c*e*m + a*(d*e*(n + 1) + c*f*(p + 1))) + (b*d*f*g*(m + n + p + 2) + h*(a*d*f*m - b*(d*e*(m + n + 1) + c*f*(m + p + 1))))*x, x], x], x] /; Fre eQ[{a, b, c, d, e, f, g, h, n, p}, x] && GtQ[m, 0] && NeQ[m + n + p + 2, 0] && IntegerQ[m]
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 = 66, normalized size of antiderivative = 0.55
| method | result | size |
| risch | \(-\frac {378000 x^{6}-223200 x^{5}-302328 x^{4}+210264 x^{3}+105841 x^{2}-38040 x -13759}{5103 \left (2+3 x \right ) \sqrt {1-2 x}}-\frac {1540 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{2187}\) | \(66\) |
| pseudoelliptic | \(\frac {-10780 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \left (2+3 x \right ) \sqrt {21}+3 \sqrt {1-2 x}\, \left (189000 x^{5}-17100 x^{4}-159714 x^{3}+25275 x^{2}+65558 x +13759\right )}{30618+45927 x}\) | \(67\) |
| derivativedivides | \(\frac {125 \left (1-2 x \right )^{\frac {9}{2}}}{162}-\frac {725 \left (1-2 x \right )^{\frac {7}{2}}}{378}+\frac {2 \left (1-2 x \right )^{\frac {5}{2}}}{27}+\frac {214 \left (1-2 x \right )^{\frac {3}{2}}}{729}+\frac {1526 \sqrt {1-2 x}}{729}-\frac {98 \sqrt {1-2 x}}{2187 \left (-\frac {4}{3}-2 x \right )}-\frac {1540 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{2187}\) | \(81\) |
| default | \(\frac {125 \left (1-2 x \right )^{\frac {9}{2}}}{162}-\frac {725 \left (1-2 x \right )^{\frac {7}{2}}}{378}+\frac {2 \left (1-2 x \right )^{\frac {5}{2}}}{27}+\frac {214 \left (1-2 x \right )^{\frac {3}{2}}}{729}+\frac {1526 \sqrt {1-2 x}}{729}-\frac {98 \sqrt {1-2 x}}{2187 \left (-\frac {4}{3}-2 x \right )}-\frac {1540 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{2187}\) | \(81\) |
| trager | \(\frac {\sqrt {1-2 x}\, \left (189000 x^{5}-17100 x^{4}-159714 x^{3}+25275 x^{2}+65558 x +13759\right )}{10206+15309 x}-\frac {770 \operatorname {RootOf}\left (\textit {\_Z}^{2}-21\right ) \ln \left (\frac {-3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-21\right ) x +21 \sqrt {1-2 x}+5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-21\right )}{2+3 x}\right )}{2187}\) | \(87\) |
-1/5103*(378000*x^6-223200*x^5-302328*x^4+210264*x^3+105841*x^2-38040*x-13 759)/(2+3*x)/(1-2*x)^(1/2)-1540/2187*arctanh(1/7*21^(1/2)*(1-2*x)^(1/2))*2 1^(1/2)
Time = 0.23 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.70 \[ \int \frac {(1-2 x)^{5/2} (3+5 x)^3}{(2+3 x)^2} \, dx=\frac {5390 \, \sqrt {7} \sqrt {3} {\left (3 \, x + 2\right )} \log \left (\frac {\sqrt {7} \sqrt {3} \sqrt {-2 \, x + 1} + 3 \, x - 5}{3 \, x + 2}\right ) + 3 \, {\left (189000 \, x^{5} - 17100 \, x^{4} - 159714 \, x^{3} + 25275 \, x^{2} + 65558 \, x + 13759\right )} \sqrt {-2 \, x + 1}}{15309 \, {\left (3 \, x + 2\right )}} \]
1/15309*(5390*sqrt(7)*sqrt(3)*(3*x + 2)*log((sqrt(7)*sqrt(3)*sqrt(-2*x + 1 ) + 3*x - 5)/(3*x + 2)) + 3*(189000*x^5 - 17100*x^4 - 159714*x^3 + 25275*x ^2 + 65558*x + 13759)*sqrt(-2*x + 1))/(3*x + 2)
Time = 35.31 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.83 \[ \int \frac {(1-2 x)^{5/2} (3+5 x)^3}{(2+3 x)^2} \, dx=\frac {125 \left (1 - 2 x\right )^{\frac {9}{2}}}{162} - \frac {725 \left (1 - 2 x\right )^{\frac {7}{2}}}{378} + \frac {2 \left (1 - 2 x\right )^{\frac {5}{2}}}{27} + \frac {214 \left (1 - 2 x\right )^{\frac {3}{2}}}{729} + \frac {1526 \sqrt {1 - 2 x}}{729} + \frac {259 \sqrt {21} \left (\log {\left (\sqrt {1 - 2 x} - \frac {\sqrt {21}}{3} \right )} - \log {\left (\sqrt {1 - 2 x} + \frac {\sqrt {21}}{3} \right )}\right )}{729} + \frac {1372 \left (\begin {cases} \frac {\sqrt {21} \left (- \frac {\log {\left (\frac {\sqrt {21} \sqrt {1 - 2 x}}{7} - 1 \right )}}{4} + \frac {\log {\left (\frac {\sqrt {21} \sqrt {1 - 2 x}}{7} + 1 \right )}}{4} - \frac {1}{4 \left (\frac {\sqrt {21} \sqrt {1 - 2 x}}{7} + 1\right )} - \frac {1}{4 \left (\frac {\sqrt {21} \sqrt {1 - 2 x}}{7} - 1\right )}\right )}{147} & \text {for}\: \sqrt {1 - 2 x} > - \frac {\sqrt {21}}{3} \wedge \sqrt {1 - 2 x} < \frac {\sqrt {21}}{3} \end {cases}\right )}{729} \]
125*(1 - 2*x)**(9/2)/162 - 725*(1 - 2*x)**(7/2)/378 + 2*(1 - 2*x)**(5/2)/2 7 + 214*(1 - 2*x)**(3/2)/729 + 1526*sqrt(1 - 2*x)/729 + 259*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)*s qrt(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)))/729
Time = 0.29 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.81 \[ \int \frac {(1-2 x)^{5/2} (3+5 x)^3}{(2+3 x)^2} \, dx=\frac {125}{162} \, {\left (-2 \, x + 1\right )}^{\frac {9}{2}} - \frac {725}{378} \, {\left (-2 \, x + 1\right )}^{\frac {7}{2}} + \frac {2}{27} \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} + \frac {214}{729} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {770}{2187} \, \sqrt {21} \log \left (-\frac {\sqrt {21} - 3 \, \sqrt {-2 \, x + 1}}{\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}}\right ) + \frac {1526}{729} \, \sqrt {-2 \, x + 1} + \frac {49 \, \sqrt {-2 \, x + 1}}{729 \, {\left (3 \, x + 2\right )}} \]
125/162*(-2*x + 1)^(9/2) - 725/378*(-2*x + 1)^(7/2) + 2/27*(-2*x + 1)^(5/2 ) + 214/729*(-2*x + 1)^(3/2) + 770/2187*sqrt(21)*log(-(sqrt(21) - 3*sqrt(- 2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) + 1526/729*sqrt(-2*x + 1) + 49/72 9*sqrt(-2*x + 1)/(3*x + 2)
Time = 0.29 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.01 \[ \int \frac {(1-2 x)^{5/2} (3+5 x)^3}{(2+3 x)^2} \, dx=\frac {125}{162} \, {\left (2 \, x - 1\right )}^{4} \sqrt {-2 \, x + 1} + \frac {725}{378} \, {\left (2 \, x - 1\right )}^{3} \sqrt {-2 \, x + 1} + \frac {2}{27} \, {\left (2 \, x - 1\right )}^{2} \sqrt {-2 \, x + 1} + \frac {214}{729} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {770}{2187} \, \sqrt {21} \log \left (\frac {{\left | -2 \, \sqrt {21} + 6 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}\right )}}\right ) + \frac {1526}{729} \, \sqrt {-2 \, x + 1} + \frac {49 \, \sqrt {-2 \, x + 1}}{729 \, {\left (3 \, x + 2\right )}} \]
125/162*(2*x - 1)^4*sqrt(-2*x + 1) + 725/378*(2*x - 1)^3*sqrt(-2*x + 1) + 2/27*(2*x - 1)^2*sqrt(-2*x + 1) + 214/729*(-2*x + 1)^(3/2) + 770/2187*sqrt (21)*log(1/2*abs(-2*sqrt(21) + 6*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) + 1526/729*sqrt(-2*x + 1) + 49/729*sqrt(-2*x + 1)/(3*x + 2)
Time = 1.44 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.68 \[ \int \frac {(1-2 x)^{5/2} (3+5 x)^3}{(2+3 x)^2} \, dx=\frac {98\,\sqrt {1-2\,x}}{2187\,\left (2\,x+\frac {4}{3}\right )}+\frac {1526\,\sqrt {1-2\,x}}{729}+\frac {214\,{\left (1-2\,x\right )}^{3/2}}{729}+\frac {2\,{\left (1-2\,x\right )}^{5/2}}{27}-\frac {725\,{\left (1-2\,x\right )}^{7/2}}{378}+\frac {125\,{\left (1-2\,x\right )}^{9/2}}{162}+\frac {\sqrt {21}\,\mathrm {atan}\left (\frac {\sqrt {21}\,\sqrt {1-2\,x}\,1{}\mathrm {i}}{7}\right )\,1540{}\mathrm {i}}{2187} \]