Integrand size = 24, antiderivative size = 108 \[ \int \frac {(1-2 x)^{3/2} (2+3 x)^3}{(3+5 x)^2} \, dx=\frac {192 \sqrt {1-2 x}}{3125}+\frac {27}{175} (1-2 x)^{3/2} (2+3 x)^2-\frac {(1-2 x)^{3/2} (2+3 x)^3}{5 (3+5 x)}-\frac {6}{625} (1-2 x)^{3/2} (29+9 x)-\frac {192 \sqrt {\frac {11}{5}} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{3125} \]
27/175*(1-2*x)^(3/2)*(2+3*x)^2-1/5*(1-2*x)^(3/2)*(2+3*x)^3/(3+5*x)-6/625*( 1-2*x)^(3/2)*(29+9*x)-192/15625*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1 /2)+192/3125*(1-2*x)^(1/2)
Time = 0.11 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.63 \[ \int \frac {(1-2 x)^{3/2} (2+3 x)^3}{(3+5 x)^2} \, dx=\frac {-\frac {5 \sqrt {1-2 x} \left (8738-27640 x-57165 x^2+62100 x^3+67500 x^4\right )}{3+5 x}-1344 \sqrt {55} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{109375} \]
((-5*Sqrt[1 - 2*x]*(8738 - 27640*x - 57165*x^2 + 62100*x^3 + 67500*x^4))/( 3 + 5*x) - 1344*Sqrt[55]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/109375
Time = 0.21 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.14, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {108, 27, 170, 27, 164, 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.
| \(\displaystyle \int \frac {(1-2 x)^{3/2} (3 x+2)^3}{(5 x+3)^2} \, dx\) |
| \(\Big \downarrow \) 108 |
| \(\displaystyle \frac {1}{5} \int \frac {3 (1-9 x) \sqrt {1-2 x} (3 x+2)^2}{5 x+3}dx-\frac {(1-2 x)^{3/2} (3 x+2)^3}{5 (5 x+3)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {3}{5} \int \frac {(1-9 x) \sqrt {1-2 x} (3 x+2)^2}{5 x+3}dx-\frac {(1-2 x)^{3/2} (3 x+2)^3}{5 (5 x+3)}\) |
| \(\Big \downarrow \) 170 |
| \(\displaystyle \frac {3}{5} \left (\frac {9}{35} (1-2 x)^{3/2} (3 x+2)^2-\frac {1}{35} \int -\frac {14 \sqrt {1-2 x} (3 x+2) (3 x+5)}{5 x+3}dx\right )-\frac {(1-2 x)^{3/2} (3 x+2)^3}{5 (5 x+3)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {3}{5} \left (\frac {2}{5} \int \frac {\sqrt {1-2 x} (3 x+2) (3 x+5)}{5 x+3}dx+\frac {9}{35} (1-2 x)^{3/2} (3 x+2)^2\right )-\frac {(1-2 x)^{3/2} (3 x+2)^3}{5 (5 x+3)}\) |
| \(\Big \downarrow \) 164 |
| \(\displaystyle \frac {3}{5} \left (\frac {2}{5} \left (\frac {16}{25} \int \frac {\sqrt {1-2 x}}{5 x+3}dx-\frac {1}{25} (1-2 x)^{3/2} (9 x+29)\right )+\frac {9}{35} (1-2 x)^{3/2} (3 x+2)^2\right )-\frac {(1-2 x)^{3/2} (3 x+2)^3}{5 (5 x+3)}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {3}{5} \left (\frac {2}{5} \left (\frac {16}{25} \left (\frac {11}{5} \int \frac {1}{\sqrt {1-2 x} (5 x+3)}dx+\frac {2}{5} \sqrt {1-2 x}\right )-\frac {1}{25} (1-2 x)^{3/2} (9 x+29)\right )+\frac {9}{35} (1-2 x)^{3/2} (3 x+2)^2\right )-\frac {(1-2 x)^{3/2} (3 x+2)^3}{5 (5 x+3)}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {3}{5} \left (\frac {2}{5} \left (\frac {16}{25} \left (\frac {2}{5} \sqrt {1-2 x}-\frac {11}{5} \int \frac {1}{\frac {11}{2}-\frac {5}{2} (1-2 x)}d\sqrt {1-2 x}\right )-\frac {1}{25} (1-2 x)^{3/2} (9 x+29)\right )+\frac {9}{35} (1-2 x)^{3/2} (3 x+2)^2\right )-\frac {(1-2 x)^{3/2} (3 x+2)^3}{5 (5 x+3)}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {3}{5} \left (\frac {2}{5} \left (\frac {16}{25} \left (\frac {2}{5} \sqrt {1-2 x}-\frac {2}{5} \sqrt {\frac {11}{5}} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )\right )-\frac {1}{25} (1-2 x)^{3/2} (9 x+29)\right )+\frac {9}{35} (1-2 x)^{3/2} (3 x+2)^2\right )-\frac {(1-2 x)^{3/2} (3 x+2)^3}{5 (5 x+3)}\) |
-1/5*((1 - 2*x)^(3/2)*(2 + 3*x)^3)/(3 + 5*x) + (3*((9*(1 - 2*x)^(3/2)*(2 + 3*x)^2)/35 + (2*(-1/25*((1 - 2*x)^(3/2)*(29 + 9*x)) + (16*((2*Sqrt[1 - 2* x])/5 - (2*Sqrt[11/5]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/5))/25))/5))/5
3.20.10.3.1 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 = 1.00 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.56
| method | result | size |
| risch | \(\frac {135000 x^{5}+56700 x^{4}-176430 x^{3}+1885 x^{2}+45116 x -8738}{21875 \left (3+5 x \right ) \sqrt {1-2 x}}-\frac {192 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{15625}\) | \(61\) |
| pseudoelliptic | \(\frac {-1344 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \left (3+5 x \right ) \sqrt {55}-5 \sqrt {1-2 x}\, \left (67500 x^{4}+62100 x^{3}-57165 x^{2}-27640 x +8738\right )}{328125+546875 x}\) | \(62\) |
| derivativedivides | \(\frac {27 \left (1-2 x \right )^{\frac {7}{2}}}{350}-\frac {351 \left (1-2 x \right )^{\frac {5}{2}}}{1250}+\frac {6 \left (1-2 x \right )^{\frac {3}{2}}}{625}+\frac {194 \sqrt {1-2 x}}{3125}+\frac {22 \sqrt {1-2 x}}{15625 \left (-\frac {6}{5}-2 x \right )}-\frac {192 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{15625}\) | \(72\) |
| default | \(\frac {27 \left (1-2 x \right )^{\frac {7}{2}}}{350}-\frac {351 \left (1-2 x \right )^{\frac {5}{2}}}{1250}+\frac {6 \left (1-2 x \right )^{\frac {3}{2}}}{625}+\frac {194 \sqrt {1-2 x}}{3125}+\frac {22 \sqrt {1-2 x}}{15625 \left (-\frac {6}{5}-2 x \right )}-\frac {192 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{15625}\) | \(72\) |
| trager | \(-\frac {\left (67500 x^{4}+62100 x^{3}-57165 x^{2}-27640 x +8738\right ) \sqrt {1-2 x}}{21875 \left (3+5 x \right )}+\frac {96 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) \ln \left (\frac {5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) x +55 \sqrt {1-2 x}-8 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right )}{3+5 x}\right )}{15625}\) | \(82\) |
1/21875*(135000*x^5+56700*x^4-176430*x^3+1885*x^2+45116*x-8738)/(3+5*x)/(1 -2*x)^(1/2)-192/15625*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)
Time = 0.22 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.74 \[ \int \frac {(1-2 x)^{3/2} (2+3 x)^3}{(3+5 x)^2} \, dx=\frac {672 \, \sqrt {11} \sqrt {5} {\left (5 \, x + 3\right )} \log \left (\frac {\sqrt {11} \sqrt {5} \sqrt {-2 \, x + 1} + 5 \, x - 8}{5 \, x + 3}\right ) - 5 \, {\left (67500 \, x^{4} + 62100 \, x^{3} - 57165 \, x^{2} - 27640 \, x + 8738\right )} \sqrt {-2 \, x + 1}}{109375 \, {\left (5 \, x + 3\right )}} \]
1/109375*(672*sqrt(11)*sqrt(5)*(5*x + 3)*log((sqrt(11)*sqrt(5)*sqrt(-2*x + 1) + 5*x - 8)/(5*x + 3)) - 5*(67500*x^4 + 62100*x^3 - 57165*x^2 - 27640*x + 8738)*sqrt(-2*x + 1))/(5*x + 3)
Time = 34.90 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.94 \[ \int \frac {(1-2 x)^{3/2} (2+3 x)^3}{(3+5 x)^2} \, dx=\frac {27 \left (1 - 2 x\right )^{\frac {7}{2}}}{350} - \frac {351 \left (1 - 2 x\right )^{\frac {5}{2}}}{1250} + \frac {6 \left (1 - 2 x\right )^{\frac {3}{2}}}{625} + \frac {194 \sqrt {1 - 2 x}}{3125} + \frac {19 \sqrt {55} \left (\log {\left (\sqrt {1 - 2 x} - \frac {\sqrt {55}}{5} \right )} - \log {\left (\sqrt {1 - 2 x} + \frac {\sqrt {55}}{5} \right )}\right )}{3125} - \frac {484 \left (\begin {cases} \frac {\sqrt {55} \left (- \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1 \right )}}{4} + \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1 \right )}}{4} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1\right )} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1\right )}\right )}{605} & \text {for}\: \sqrt {1 - 2 x} > - \frac {\sqrt {55}}{5} \wedge \sqrt {1 - 2 x} < \frac {\sqrt {55}}{5} \end {cases}\right )}{3125} \]
27*(1 - 2*x)**(7/2)/350 - 351*(1 - 2*x)**(5/2)/1250 + 6*(1 - 2*x)**(3/2)/6 25 + 194*sqrt(1 - 2*x)/3125 + 19*sqrt(55)*(log(sqrt(1 - 2*x) - sqrt(55)/5) - log(sqrt(1 - 2*x) + sqrt(55)/5))/3125 - 484*Piecewise((sqrt(55)*(-log(s qrt(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)))/31 25
Time = 0.29 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.82 \[ \int \frac {(1-2 x)^{3/2} (2+3 x)^3}{(3+5 x)^2} \, dx=\frac {27}{350} \, {\left (-2 \, x + 1\right )}^{\frac {7}{2}} - \frac {351}{1250} \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} + \frac {6}{625} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {96}{15625} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) + \frac {194}{3125} \, \sqrt {-2 \, x + 1} - \frac {11 \, \sqrt {-2 \, x + 1}}{3125 \, {\left (5 \, x + 3\right )}} \]
27/350*(-2*x + 1)^(7/2) - 351/1250*(-2*x + 1)^(5/2) + 6/625*(-2*x + 1)^(3/ 2) + 96/15625*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sq rt(-2*x + 1))) + 194/3125*sqrt(-2*x + 1) - 11/3125*sqrt(-2*x + 1)/(5*x + 3 )
Time = 0.28 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.98 \[ \int \frac {(1-2 x)^{3/2} (2+3 x)^3}{(3+5 x)^2} \, dx=-\frac {27}{350} \, {\left (2 \, x - 1\right )}^{3} \sqrt {-2 \, x + 1} - \frac {351}{1250} \, {\left (2 \, x - 1\right )}^{2} \sqrt {-2 \, x + 1} + \frac {6}{625} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {96}{15625} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) + \frac {194}{3125} \, \sqrt {-2 \, x + 1} - \frac {11 \, \sqrt {-2 \, x + 1}}{3125 \, {\left (5 \, x + 3\right )}} \]
-27/350*(2*x - 1)^3*sqrt(-2*x + 1) - 351/1250*(2*x - 1)^2*sqrt(-2*x + 1) + 6/625*(-2*x + 1)^(3/2) + 96/15625*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*s qrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 194/3125*sqrt(-2*x + 1) - 11/3125*sqrt(-2*x + 1)/(5*x + 3)
Time = 0.06 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.68 \[ \int \frac {(1-2 x)^{3/2} (2+3 x)^3}{(3+5 x)^2} \, dx=\frac {194\,\sqrt {1-2\,x}}{3125}-\frac {22\,\sqrt {1-2\,x}}{15625\,\left (2\,x+\frac {6}{5}\right )}+\frac {6\,{\left (1-2\,x\right )}^{3/2}}{625}-\frac {351\,{\left (1-2\,x\right )}^{5/2}}{1250}+\frac {27\,{\left (1-2\,x\right )}^{7/2}}{350}+\frac {\sqrt {55}\,\mathrm {atan}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}\,1{}\mathrm {i}}{11}\right )\,192{}\mathrm {i}}{15625} \]