Integrand size = 24, antiderivative size = 141 \[ \int \frac {(1-2 x)^{5/2} (2+3 x)^4}{(3+5 x)^2} \, dx=\frac {2794 \sqrt {1-2 x}}{78125}+\frac {254 (1-2 x)^{3/2}}{46875}-\frac {32 (1-2 x)^{5/2} (2+3 x)^2}{4125}+\frac {39}{275} (1-2 x)^{5/2} (2+3 x)^3-\frac {(1-2 x)^{5/2} (2+3 x)^4}{5 (3+5 x)}-\frac {(1-2 x)^{5/2} (1347116+1110975 x)}{3609375}-\frac {2794 \sqrt {\frac {11}{5}} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{78125} \]
254/46875*(1-2*x)^(3/2)-32/4125*(1-2*x)^(5/2)*(2+3*x)^2+39/275*(1-2*x)^(5/ 2)*(2+3*x)^3-1/5*(1-2*x)^(5/2)*(2+3*x)^4/(3+5*x)-1/3609375*(1-2*x)^(5/2)*( 1347116+1110975*x)-2794/390625*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/ 2)+2794/78125*(1-2*x)^(1/2)
Time = 0.13 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.55 \[ \int \frac {(1-2 x)^{5/2} (2+3 x)^4}{(3+5 x)^2} \, dx=\frac {\frac {5 \sqrt {1-2 x} \left (-15982128+50081215 x+85482115 x^2-214071975 x^3-173598750 x^4+237037500 x^5+212625000 x^6\right )}{3+5 x}-645414 \sqrt {55} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{90234375} \]
((5*Sqrt[1 - 2*x]*(-15982128 + 50081215*x + 85482115*x^2 - 214071975*x^3 - 173598750*x^4 + 237037500*x^5 + 212625000*x^6))/(3 + 5*x) - 645414*Sqrt[5 5]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/90234375
Time = 0.23 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.18, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {108, 170, 25, 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.
| \(\displaystyle \int \frac {(1-2 x)^{5/2} (3 x+2)^4}{(5 x+3)^2} \, dx\) |
| \(\Big \downarrow \) 108 |
| \(\displaystyle \frac {1}{5} \int \frac {(2-39 x) (1-2 x)^{3/2} (3 x+2)^3}{5 x+3}dx-\frac {(1-2 x)^{5/2} (3 x+2)^4}{5 (5 x+3)}\) |
| \(\Big \downarrow \) 170 |
| \(\displaystyle \frac {1}{5} \left (\frac {39}{55} (1-2 x)^{5/2} (3 x+2)^3-\frac {1}{55} \int -\frac {(1-2 x)^{3/2} (3 x+2)^2 (96 x+337)}{5 x+3}dx\right )-\frac {(1-2 x)^{5/2} (3 x+2)^4}{5 (5 x+3)}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{55} \int \frac {(1-2 x)^{3/2} (3 x+2)^2 (96 x+337)}{5 x+3}dx+\frac {39}{55} (1-2 x)^{5/2} (3 x+2)^3\right )-\frac {(1-2 x)^{5/2} (3 x+2)^4}{5 (5 x+3)}\) |
| \(\Big \downarrow \) 170 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{55} \left (-\frac {1}{45} \int -\frac {3 (1-2 x)^{3/2} (3 x+2) (14813 x+9726)}{5 x+3}dx-\frac {32}{15} (3 x+2)^2 (1-2 x)^{5/2}\right )+\frac {39}{55} (1-2 x)^{5/2} (3 x+2)^3\right )-\frac {(1-2 x)^{5/2} (3 x+2)^4}{5 (5 x+3)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{55} \left (\frac {1}{15} \int \frac {(1-2 x)^{3/2} (3 x+2) (14813 x+9726)}{5 x+3}dx-\frac {32}{15} (1-2 x)^{5/2} (3 x+2)^2\right )+\frac {39}{55} (1-2 x)^{5/2} (3 x+2)^3\right )-\frac {(1-2 x)^{5/2} (3 x+2)^4}{5 (5 x+3)}\) |
| \(\Big \downarrow \) 164 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{55} \left (\frac {1}{15} \left (\frac {4191}{25} \int \frac {(1-2 x)^{3/2}}{5 x+3}dx-\frac {1}{875} (1-2 x)^{5/2} (1110975 x+1347116)\right )-\frac {32}{15} (1-2 x)^{5/2} (3 x+2)^2\right )+\frac {39}{55} (1-2 x)^{5/2} (3 x+2)^3\right )-\frac {(1-2 x)^{5/2} (3 x+2)^4}{5 (5 x+3)}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{55} \left (\frac {1}{15} \left (\frac {4191}{25} \left (\frac {11}{5} \int \frac {\sqrt {1-2 x}}{5 x+3}dx+\frac {2}{15} (1-2 x)^{3/2}\right )-\frac {1}{875} (1-2 x)^{5/2} (1110975 x+1347116)\right )-\frac {32}{15} (1-2 x)^{5/2} (3 x+2)^2\right )+\frac {39}{55} (1-2 x)^{5/2} (3 x+2)^3\right )-\frac {(1-2 x)^{5/2} (3 x+2)^4}{5 (5 x+3)}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{55} \left (\frac {1}{15} \left (\frac {4191}{25} \left (\frac {11}{5} \left (\frac {11}{5} \int \frac {1}{\sqrt {1-2 x} (5 x+3)}dx+\frac {2}{5} \sqrt {1-2 x}\right )+\frac {2}{15} (1-2 x)^{3/2}\right )-\frac {1}{875} (1-2 x)^{5/2} (1110975 x+1347116)\right )-\frac {32}{15} (1-2 x)^{5/2} (3 x+2)^2\right )+\frac {39}{55} (1-2 x)^{5/2} (3 x+2)^3\right )-\frac {(1-2 x)^{5/2} (3 x+2)^4}{5 (5 x+3)}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{55} \left (\frac {1}{15} \left (\frac {4191}{25} \left (\frac {11}{5} \left (\frac {2}{5} \sqrt {1-2 x}-\frac {11}{5} \int \frac {1}{\frac {11}{2}-\frac {5}{2} (1-2 x)}d\sqrt {1-2 x}\right )+\frac {2}{15} (1-2 x)^{3/2}\right )-\frac {1}{875} (1-2 x)^{5/2} (1110975 x+1347116)\right )-\frac {32}{15} (1-2 x)^{5/2} (3 x+2)^2\right )+\frac {39}{55} (1-2 x)^{5/2} (3 x+2)^3\right )-\frac {(1-2 x)^{5/2} (3 x+2)^4}{5 (5 x+3)}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{55} \left (\frac {1}{15} \left (\frac {4191}{25} \left (\frac {11}{5} \left (\frac {2}{5} \sqrt {1-2 x}-\frac {2}{5} \sqrt {\frac {11}{5}} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )\right )+\frac {2}{15} (1-2 x)^{3/2}\right )-\frac {1}{875} (1-2 x)^{5/2} (1110975 x+1347116)\right )-\frac {32}{15} (1-2 x)^{5/2} (3 x+2)^2\right )+\frac {39}{55} (1-2 x)^{5/2} (3 x+2)^3\right )-\frac {(1-2 x)^{5/2} (3 x+2)^4}{5 (5 x+3)}\) |
-1/5*((1 - 2*x)^(5/2)*(2 + 3*x)^4)/(3 + 5*x) + ((39*(1 - 2*x)^(5/2)*(2 + 3 *x)^3)/55 + ((-32*(1 - 2*x)^(5/2)*(2 + 3*x)^2)/15 + (-1/875*((1 - 2*x)^(5/ 2)*(1347116 + 1110975*x)) + (4191*((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))/25)/ 15)/55)/5
3.20.80.3.1 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 = 3.22 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.50
| method | result | size |
| risch | \(-\frac {425250000 x^{7}+261450000 x^{6}-584235000 x^{5}-254545200 x^{4}+385036205 x^{3}+14680315 x^{2}-82045471 x +15982128}{18046875 \left (3+5 x \right ) \sqrt {1-2 x}}-\frac {2794 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{390625}\) | \(71\) |
| pseudoelliptic | \(\frac {-645414 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \left (3+5 x \right ) \sqrt {55}+5 \sqrt {1-2 x}\, \left (212625000 x^{6}+237037500 x^{5}-173598750 x^{4}-214071975 x^{3}+85482115 x^{2}+50081215 x -15982128\right )}{270703125+451171875 x}\) | \(72\) |
| derivativedivides | \(-\frac {81 \left (1-2 x \right )^{\frac {11}{2}}}{1100}+\frac {111 \left (1-2 x \right )^{\frac {9}{2}}}{250}-\frac {12393 \left (1-2 x \right )^{\frac {7}{2}}}{17500}+\frac {24 \left (1-2 x \right )^{\frac {5}{2}}}{15625}+\frac {52 \left (1-2 x \right )^{\frac {3}{2}}}{9375}+\frac {2816 \sqrt {1-2 x}}{78125}+\frac {242 \sqrt {1-2 x}}{390625 \left (-\frac {6}{5}-2 x \right )}-\frac {2794 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{390625}\) | \(90\) |
| default | \(-\frac {81 \left (1-2 x \right )^{\frac {11}{2}}}{1100}+\frac {111 \left (1-2 x \right )^{\frac {9}{2}}}{250}-\frac {12393 \left (1-2 x \right )^{\frac {7}{2}}}{17500}+\frac {24 \left (1-2 x \right )^{\frac {5}{2}}}{15625}+\frac {52 \left (1-2 x \right )^{\frac {3}{2}}}{9375}+\frac {2816 \sqrt {1-2 x}}{78125}+\frac {242 \sqrt {1-2 x}}{390625 \left (-\frac {6}{5}-2 x \right )}-\frac {2794 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{390625}\) | \(90\) |
| trager | \(\frac {\sqrt {1-2 x}\, \left (212625000 x^{6}+237037500 x^{5}-173598750 x^{4}-214071975 x^{3}+85482115 x^{2}+50081215 x -15982128\right )}{54140625+90234375 x}-\frac {1397 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) \ln \left (\frac {-5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) x +55 \sqrt {1-2 x}+8 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right )}{3+5 x}\right )}{390625}\) | \(92\) |
-1/18046875*(425250000*x^7+261450000*x^6-584235000*x^5-254545200*x^4+38503 6205*x^3+14680315*x^2-82045471*x+15982128)/(3+5*x)/(1-2*x)^(1/2)-2794/3906 25*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)
Time = 0.22 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.64 \[ \int \frac {(1-2 x)^{5/2} (2+3 x)^4}{(3+5 x)^2} \, dx=\frac {322707 \, \sqrt {11} \sqrt {5} {\left (5 \, x + 3\right )} \log \left (\frac {\sqrt {11} \sqrt {5} \sqrt {-2 \, x + 1} + 5 \, x - 8}{5 \, x + 3}\right ) + 5 \, {\left (212625000 \, x^{6} + 237037500 \, x^{5} - 173598750 \, x^{4} - 214071975 \, x^{3} + 85482115 \, x^{2} + 50081215 \, x - 15982128\right )} \sqrt {-2 \, x + 1}}{90234375 \, {\left (5 \, x + 3\right )}} \]
1/90234375*(322707*sqrt(11)*sqrt(5)*(5*x + 3)*log((sqrt(11)*sqrt(5)*sqrt(- 2*x + 1) + 5*x - 8)/(5*x + 3)) + 5*(212625000*x^6 + 237037500*x^5 - 173598 750*x^4 - 214071975*x^3 + 85482115*x^2 + 50081215*x - 15982128)*sqrt(-2*x + 1))/(5*x + 3)
Time = 41.38 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.65 \[ \int \frac {(1-2 x)^{5/2} (2+3 x)^4}{(3+5 x)^2} \, dx=- \frac {81 \left (1 - 2 x\right )^{\frac {11}{2}}}{1100} + \frac {111 \left (1 - 2 x\right )^{\frac {9}{2}}}{250} - \frac {12393 \left (1 - 2 x\right )^{\frac {7}{2}}}{17500} + \frac {24 \left (1 - 2 x\right )^{\frac {5}{2}}}{15625} + \frac {52 \left (1 - 2 x\right )^{\frac {3}{2}}}{9375} + \frac {2816 \sqrt {1 - 2 x}}{78125} + \frac {1386 \sqrt {55} \left (\log {\left (\sqrt {1 - 2 x} - \frac {\sqrt {55}}{5} \right )} - \log {\left (\sqrt {1 - 2 x} + \frac {\sqrt {55}}{5} \right )}\right )}{390625} - \frac {5324 \left (\begin {cases} \frac {\sqrt {55} \left (- \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1 \right )}}{4} + \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1 \right )}}{4} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1\right )} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1\right )}\right )}{605} & \text {for}\: \sqrt {1 - 2 x} > - \frac {\sqrt {55}}{5} \wedge \sqrt {1 - 2 x} < \frac {\sqrt {55}}{5} \end {cases}\right )}{78125} \]
-81*(1 - 2*x)**(11/2)/1100 + 111*(1 - 2*x)**(9/2)/250 - 12393*(1 - 2*x)**( 7/2)/17500 + 24*(1 - 2*x)**(5/2)/15625 + 52*(1 - 2*x)**(3/2)/9375 + 2816*s qrt(1 - 2*x)/78125 + 1386*sqrt(55)*(log(sqrt(1 - 2*x) - sqrt(55)/5) - log( sqrt(1 - 2*x) + sqrt(55)/5))/390625 - 5324*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)))/78125
Time = 0.29 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.76 \[ \int \frac {(1-2 x)^{5/2} (2+3 x)^4}{(3+5 x)^2} \, dx=-\frac {81}{1100} \, {\left (-2 \, x + 1\right )}^{\frac {11}{2}} + \frac {111}{250} \, {\left (-2 \, x + 1\right )}^{\frac {9}{2}} - \frac {12393}{17500} \, {\left (-2 \, x + 1\right )}^{\frac {7}{2}} + \frac {24}{15625} \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} + \frac {52}{9375} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {1397}{390625} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) + \frac {2816}{78125} \, \sqrt {-2 \, x + 1} - \frac {121 \, \sqrt {-2 \, x + 1}}{78125 \, {\left (5 \, x + 3\right )}} \]
-81/1100*(-2*x + 1)^(11/2) + 111/250*(-2*x + 1)^(9/2) - 12393/17500*(-2*x + 1)^(7/2) + 24/15625*(-2*x + 1)^(5/2) + 52/9375*(-2*x + 1)^(3/2) + 1397/3 90625*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 2816/78125*sqrt(-2*x + 1) - 121/78125*sqrt(-2*x + 1)/(5*x + 3)
Time = 0.29 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.98 \[ \int \frac {(1-2 x)^{5/2} (2+3 x)^4}{(3+5 x)^2} \, dx=\frac {81}{1100} \, {\left (2 \, x - 1\right )}^{5} \sqrt {-2 \, x + 1} + \frac {111}{250} \, {\left (2 \, x - 1\right )}^{4} \sqrt {-2 \, x + 1} + \frac {12393}{17500} \, {\left (2 \, x - 1\right )}^{3} \sqrt {-2 \, x + 1} + \frac {24}{15625} \, {\left (2 \, x - 1\right )}^{2} \sqrt {-2 \, x + 1} + \frac {52}{9375} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {1397}{390625} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) + \frac {2816}{78125} \, \sqrt {-2 \, x + 1} - \frac {121 \, \sqrt {-2 \, x + 1}}{78125 \, {\left (5 \, x + 3\right )}} \]
81/1100*(2*x - 1)^5*sqrt(-2*x + 1) + 111/250*(2*x - 1)^4*sqrt(-2*x + 1) + 12393/17500*(2*x - 1)^3*sqrt(-2*x + 1) + 24/15625*(2*x - 1)^2*sqrt(-2*x + 1) + 52/9375*(-2*x + 1)^(3/2) + 1397/390625*sqrt(55)*log(1/2*abs(-2*sqrt(5 5) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 2816/78125*sqrt(- 2*x + 1) - 121/78125*sqrt(-2*x + 1)/(5*x + 3)
Time = 1.48 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.65 \[ \int \frac {(1-2 x)^{5/2} (2+3 x)^4}{(3+5 x)^2} \, dx=\frac {2816\,\sqrt {1-2\,x}}{78125}-\frac {242\,\sqrt {1-2\,x}}{390625\,\left (2\,x+\frac {6}{5}\right )}+\frac {52\,{\left (1-2\,x\right )}^{3/2}}{9375}+\frac {24\,{\left (1-2\,x\right )}^{5/2}}{15625}-\frac {12393\,{\left (1-2\,x\right )}^{7/2}}{17500}+\frac {111\,{\left (1-2\,x\right )}^{9/2}}{250}-\frac {81\,{\left (1-2\,x\right )}^{11/2}}{1100}+\frac {\sqrt {55}\,\mathrm {atan}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}\,1{}\mathrm {i}}{11}\right )\,2794{}\mathrm {i}}{390625} \]