Integrand size = 24, antiderivative size = 121 \[ \int \frac {(1-2 x)^{5/2} (2+3 x)^3}{(3+5 x)^2} \, dx=\frac {2068 \sqrt {1-2 x}}{15625}+\frac {188 (1-2 x)^{3/2}}{9375}+\frac {11}{75} (1-2 x)^{5/2} (2+3 x)^2-\frac {(1-2 x)^{5/2} (2+3 x)^3}{5 (3+5 x)}-\frac {2 (1-2 x)^{5/2} (6191+2850 x)}{65625}-\frac {2068 \sqrt {\frac {11}{5}} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{15625} \]
188/9375*(1-2*x)^(3/2)+11/75*(1-2*x)^(5/2)*(2+3*x)^2-1/5*(1-2*x)^(5/2)*(2+ 3*x)^3/(3+5*x)-2/65625*(1-2*x)^(5/2)*(6191+2850*x)-2068/78125*arctanh(1/11 *55^(1/2)*(1-2*x)^(1/2))*55^(1/2)+2068/15625*(1-2*x)^(1/2)
Time = 0.11 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.60 \[ \int \frac {(1-2 x)^{5/2} (2+3 x)^3}{(3+5 x)^2} \, dx=\frac {\frac {5 \sqrt {1-2 x} \left (16794+680930 x+152105 x^2-1858950 x^3+427500 x^4+1575000 x^5\right )}{3+5 x}-43428 \sqrt {55} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{1640625} \]
((5*Sqrt[1 - 2*x]*(16794 + 680930*x + 152105*x^2 - 1858950*x^3 + 427500*x^ 4 + 1575000*x^5))/(3 + 5*x) - 43428*Sqrt[55]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2 *x]])/1640625
Time = 0.21 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.17, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {108, 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.
| \(\displaystyle \int \frac {(1-2 x)^{5/2} (3 x+2)^3}{(5 x+3)^2} \, dx\) |
| \(\Big \downarrow \) 108 |
| \(\displaystyle \frac {1}{5} \int -\frac {(1-2 x)^{3/2} (3 x+2)^2 (33 x+1)}{5 x+3}dx-\frac {(1-2 x)^{5/2} (3 x+2)^3}{5 (5 x+3)}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {1}{5} \int \frac {(1-2 x)^{3/2} (3 x+2)^2 (33 x+1)}{5 x+3}dx-\frac {(1-2 x)^{5/2} (3 x+2)^3}{5 (5 x+3)}\) |
| \(\Big \downarrow \) 170 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{45} \int \frac {6 (1-2 x)^{3/2} (3 x+2) (38 x+51)}{5 x+3}dx+\frac {11}{15} (3 x+2)^2 (1-2 x)^{5/2}\right )-\frac {(1-2 x)^{5/2} (3 x+2)^3}{5 (5 x+3)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \left (\frac {2}{15} \int \frac {(1-2 x)^{3/2} (3 x+2) (38 x+51)}{5 x+3}dx+\frac {11}{15} (3 x+2)^2 (1-2 x)^{5/2}\right )-\frac {(1-2 x)^{5/2} (3 x+2)^3}{5 (5 x+3)}\) |
| \(\Big \downarrow \) 164 |
| \(\displaystyle \frac {1}{5} \left (\frac {2}{15} \left (\frac {141}{25} \int \frac {(1-2 x)^{3/2}}{5 x+3}dx-\frac {1}{875} (1-2 x)^{5/2} (2850 x+6191)\right )+\frac {11}{15} (3 x+2)^2 (1-2 x)^{5/2}\right )-\frac {(1-2 x)^{5/2} (3 x+2)^3}{5 (5 x+3)}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {1}{5} \left (\frac {2}{15} \left (\frac {141}{25} \left (\frac {11}{5} \int \frac {\sqrt {1-2 x}}{5 x+3}dx+\frac {2}{15} (1-2 x)^{3/2}\right )-\frac {1}{875} (1-2 x)^{5/2} (2850 x+6191)\right )+\frac {11}{15} (3 x+2)^2 (1-2 x)^{5/2}\right )-\frac {(1-2 x)^{5/2} (3 x+2)^3}{5 (5 x+3)}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {1}{5} \left (\frac {2}{15} \left (\frac {141}{25} \left (\frac {11}{5} \left (\frac {11}{5} \int \frac {1}{\sqrt {1-2 x} (5 x+3)}dx+\frac {2}{5} \sqrt {1-2 x}\right )+\frac {2}{15} (1-2 x)^{3/2}\right )-\frac {1}{875} (1-2 x)^{5/2} (2850 x+6191)\right )+\frac {11}{15} (3 x+2)^2 (1-2 x)^{5/2}\right )-\frac {(1-2 x)^{5/2} (3 x+2)^3}{5 (5 x+3)}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {1}{5} \left (\frac {2}{15} \left (\frac {141}{25} \left (\frac {11}{5} \left (\frac {2}{5} \sqrt {1-2 x}-\frac {11}{5} \int \frac {1}{\frac {11}{2}-\frac {5}{2} (1-2 x)}d\sqrt {1-2 x}\right )+\frac {2}{15} (1-2 x)^{3/2}\right )-\frac {1}{875} (1-2 x)^{5/2} (2850 x+6191)\right )+\frac {11}{15} (3 x+2)^2 (1-2 x)^{5/2}\right )-\frac {(1-2 x)^{5/2} (3 x+2)^3}{5 (5 x+3)}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {1}{5} \left (\frac {2}{15} \left (\frac {141}{25} \left (\frac {11}{5} \left (\frac {2}{5} \sqrt {1-2 x}-\frac {2}{5} \sqrt {\frac {11}{5}} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )\right )+\frac {2}{15} (1-2 x)^{3/2}\right )-\frac {1}{875} (1-2 x)^{5/2} (2850 x+6191)\right )+\frac {11}{15} (3 x+2)^2 (1-2 x)^{5/2}\right )-\frac {(1-2 x)^{5/2} (3 x+2)^3}{5 (5 x+3)}\) |
-1/5*((1 - 2*x)^(5/2)*(2 + 3*x)^3)/(3 + 5*x) + ((11*(1 - 2*x)^(5/2)*(2 + 3 *x)^2)/15 + (2*(-1/875*((1 - 2*x)^(5/2)*(6191 + 2850*x)) + (141*((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)/5
3.20.81.3.1 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 = 1.05 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.55
| method | result | size |
| risch | \(-\frac {3150000 x^{6}-720000 x^{5}-4145400 x^{4}+2163160 x^{3}+1209755 x^{2}-647342 x -16794}{328125 \left (3+5 x \right ) \sqrt {1-2 x}}-\frac {2068 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{78125}\) | \(66\) |
| pseudoelliptic | \(\frac {-43428 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \left (3+5 x \right ) \sqrt {55}+5 \sqrt {1-2 x}\, \left (1575000 x^{5}+427500 x^{4}-1858950 x^{3}+152105 x^{2}+680930 x +16794\right )}{4921875+8203125 x}\) | \(67\) |
| derivativedivides | \(\frac {3 \left (1-2 x \right )^{\frac {9}{2}}}{50}-\frac {351 \left (1-2 x \right )^{\frac {7}{2}}}{1750}+\frac {18 \left (1-2 x \right )^{\frac {5}{2}}}{3125}+\frac {194 \left (1-2 x \right )^{\frac {3}{2}}}{9375}+\frac {418 \sqrt {1-2 x}}{3125}+\frac {242 \sqrt {1-2 x}}{78125 \left (-\frac {6}{5}-2 x \right )}-\frac {2068 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{78125}\) | \(81\) |
| default | \(\frac {3 \left (1-2 x \right )^{\frac {9}{2}}}{50}-\frac {351 \left (1-2 x \right )^{\frac {7}{2}}}{1750}+\frac {18 \left (1-2 x \right )^{\frac {5}{2}}}{3125}+\frac {194 \left (1-2 x \right )^{\frac {3}{2}}}{9375}+\frac {418 \sqrt {1-2 x}}{3125}+\frac {242 \sqrt {1-2 x}}{78125 \left (-\frac {6}{5}-2 x \right )}-\frac {2068 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{78125}\) | \(81\) |
| trager | \(\frac {\sqrt {1-2 x}\, \left (1575000 x^{5}+427500 x^{4}-1858950 x^{3}+152105 x^{2}+680930 x +16794\right )}{984375+1640625 x}+\frac {1034 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) \ln \left (\frac {5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) x +55 \sqrt {1-2 x}-8 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right )}{3+5 x}\right )}{78125}\) | \(87\) |
-1/328125*(3150000*x^6-720000*x^5-4145400*x^4+2163160*x^3+1209755*x^2-6473 42*x-16794)/(3+5*x)/(1-2*x)^(1/2)-2068/78125*arctanh(1/11*55^(1/2)*(1-2*x) ^(1/2))*55^(1/2)
Time = 0.23 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.70 \[ \int \frac {(1-2 x)^{5/2} (2+3 x)^3}{(3+5 x)^2} \, dx=\frac {21714 \, \sqrt {11} \sqrt {5} {\left (5 \, x + 3\right )} \log \left (\frac {\sqrt {11} \sqrt {5} \sqrt {-2 \, x + 1} + 5 \, x - 8}{5 \, x + 3}\right ) + 5 \, {\left (1575000 \, x^{5} + 427500 \, x^{4} - 1858950 \, x^{3} + 152105 \, x^{2} + 680930 \, x + 16794\right )} \sqrt {-2 \, x + 1}}{1640625 \, {\left (5 \, x + 3\right )}} \]
1/1640625*(21714*sqrt(11)*sqrt(5)*(5*x + 3)*log((sqrt(11)*sqrt(5)*sqrt(-2* x + 1) + 5*x - 8)/(5*x + 3)) + 5*(1575000*x^5 + 427500*x^4 - 1858950*x^3 + 152105*x^2 + 680930*x + 16794)*sqrt(-2*x + 1))/(5*x + 3)
Time = 34.11 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.83 \[ \int \frac {(1-2 x)^{5/2} (2+3 x)^3}{(3+5 x)^2} \, dx=\frac {3 \left (1 - 2 x\right )^{\frac {9}{2}}}{50} - \frac {351 \left (1 - 2 x\right )^{\frac {7}{2}}}{1750} + \frac {18 \left (1 - 2 x\right )^{\frac {5}{2}}}{3125} + \frac {194 \left (1 - 2 x\right )^{\frac {3}{2}}}{9375} + \frac {418 \sqrt {1 - 2 x}}{3125} + \frac {1023 \sqrt {55} \left (\log {\left (\sqrt {1 - 2 x} - \frac {\sqrt {55}}{5} \right )} - \log {\left (\sqrt {1 - 2 x} + \frac {\sqrt {55}}{5} \right )}\right )}{78125} - \frac {5324 \left (\begin {cases} \frac {\sqrt {55} \left (- \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1 \right )}}{4} + \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1 \right )}}{4} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1\right )} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1\right )}\right )}{605} & \text {for}\: \sqrt {1 - 2 x} > - \frac {\sqrt {55}}{5} \wedge \sqrt {1 - 2 x} < \frac {\sqrt {55}}{5} \end {cases}\right )}{15625} \]
3*(1 - 2*x)**(9/2)/50 - 351*(1 - 2*x)**(7/2)/1750 + 18*(1 - 2*x)**(5/2)/31 25 + 194*(1 - 2*x)**(3/2)/9375 + 418*sqrt(1 - 2*x)/3125 + 1023*sqrt(55)*(l og(sqrt(1 - 2*x) - sqrt(55)/5) - log(sqrt(1 - 2*x) + sqrt(55)/5))/78125 - 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)))/15625
Time = 0.28 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.81 \[ \int \frac {(1-2 x)^{5/2} (2+3 x)^3}{(3+5 x)^2} \, dx=\frac {3}{50} \, {\left (-2 \, x + 1\right )}^{\frac {9}{2}} - \frac {351}{1750} \, {\left (-2 \, x + 1\right )}^{\frac {7}{2}} + \frac {18}{3125} \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} + \frac {194}{9375} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {1034}{78125} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) + \frac {418}{3125} \, \sqrt {-2 \, x + 1} - \frac {121 \, \sqrt {-2 \, x + 1}}{15625 \, {\left (5 \, x + 3\right )}} \]
3/50*(-2*x + 1)^(9/2) - 351/1750*(-2*x + 1)^(7/2) + 18/3125*(-2*x + 1)^(5/ 2) + 194/9375*(-2*x + 1)^(3/2) + 1034/78125*sqrt(55)*log(-(sqrt(55) - 5*sq rt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 418/3125*sqrt(-2*x + 1) - 1 21/15625*sqrt(-2*x + 1)/(5*x + 3)
Time = 0.27 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.01 \[ \int \frac {(1-2 x)^{5/2} (2+3 x)^3}{(3+5 x)^2} \, dx=\frac {3}{50} \, {\left (2 \, x - 1\right )}^{4} \sqrt {-2 \, x + 1} + \frac {351}{1750} \, {\left (2 \, x - 1\right )}^{3} \sqrt {-2 \, x + 1} + \frac {18}{3125} \, {\left (2 \, x - 1\right )}^{2} \sqrt {-2 \, x + 1} + \frac {194}{9375} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {1034}{78125} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) + \frac {418}{3125} \, \sqrt {-2 \, x + 1} - \frac {121 \, \sqrt {-2 \, x + 1}}{15625 \, {\left (5 \, x + 3\right )}} \]
3/50*(2*x - 1)^4*sqrt(-2*x + 1) + 351/1750*(2*x - 1)^3*sqrt(-2*x + 1) + 18 /3125*(2*x - 1)^2*sqrt(-2*x + 1) + 194/9375*(-2*x + 1)^(3/2) + 1034/78125* sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(- 2*x + 1))) + 418/3125*sqrt(-2*x + 1) - 121/15625*sqrt(-2*x + 1)/(5*x + 3)
Time = 0.07 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.68 \[ \int \frac {(1-2 x)^{5/2} (2+3 x)^3}{(3+5 x)^2} \, dx=\frac {418\,\sqrt {1-2\,x}}{3125}-\frac {242\,\sqrt {1-2\,x}}{78125\,\left (2\,x+\frac {6}{5}\right )}+\frac {194\,{\left (1-2\,x\right )}^{3/2}}{9375}+\frac {18\,{\left (1-2\,x\right )}^{5/2}}{3125}-\frac {351\,{\left (1-2\,x\right )}^{7/2}}{1750}+\frac {3\,{\left (1-2\,x\right )}^{9/2}}{50}+\frac {\sqrt {55}\,\mathrm {atan}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}\,1{}\mathrm {i}}{11}\right )\,2068{}\mathrm {i}}{78125} \]