Integrand size = 24, antiderivative size = 88 \[ \int \frac {\sqrt {1-2 x} (2+3 x)^3}{(3+5 x)^2} \, dx=-\frac {294}{625} \sqrt {1-2 x}+\frac {21}{125} \sqrt {1-2 x} (2+3 x)^2-\frac {\sqrt {1-2 x} (2+3 x)^3}{5 (3+5 x)}-\frac {196 \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{625 \sqrt {55}} \]
-196/34375*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)-294/625*(1-2*x)^( 1/2)+21/125*(2+3*x)^2*(1-2*x)^(1/2)-1/5*(2+3*x)^3*(1-2*x)^(1/2)/(3+5*x)
Time = 0.11 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.72 \[ \int \frac {\sqrt {1-2 x} (2+3 x)^3}{(3+5 x)^2} \, dx=\frac {\sqrt {1-2 x} \left (-622-90 x+2385 x^2+1350 x^3\right )}{625 (3+5 x)}-\frac {196 \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{625 \sqrt {55}} \]
(Sqrt[1 - 2*x]*(-622 - 90*x + 2385*x^2 + 1350*x^3))/(625*(3 + 5*x)) - (196 *ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(625*Sqrt[55])
Time = 0.20 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.11, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {108, 27, 170, 27, 90, 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.
\(\displaystyle \int \frac {\sqrt {1-2 x} (3 x+2)^3}{(5 x+3)^2} \, dx\) |
\(\Big \downarrow \) 108 |
\(\displaystyle \frac {1}{5} \int \frac {7 (1-3 x) (3 x+2)^2}{\sqrt {1-2 x} (5 x+3)}dx-\frac {\sqrt {1-2 x} (3 x+2)^3}{5 (5 x+3)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {7}{5} \int \frac {(1-3 x) (3 x+2)^2}{\sqrt {1-2 x} (5 x+3)}dx-\frac {\sqrt {1-2 x} (3 x+2)^3}{5 (5 x+3)}\) |
\(\Big \downarrow \) 170 |
\(\displaystyle \frac {7}{5} \left (\frac {3}{25} \sqrt {1-2 x} (3 x+2)^2-\frac {1}{25} \int -\frac {14 (3 x+2)}{\sqrt {1-2 x} (5 x+3)}dx\right )-\frac {\sqrt {1-2 x} (3 x+2)^3}{5 (5 x+3)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {7}{5} \left (\frac {14}{25} \int \frac {3 x+2}{\sqrt {1-2 x} (5 x+3)}dx+\frac {3}{25} \sqrt {1-2 x} (3 x+2)^2\right )-\frac {\sqrt {1-2 x} (3 x+2)^3}{5 (5 x+3)}\) |
\(\Big \downarrow \) 90 |
\(\displaystyle \frac {7}{5} \left (\frac {14}{25} \left (\frac {1}{5} \int \frac {1}{\sqrt {1-2 x} (5 x+3)}dx-\frac {3}{5} \sqrt {1-2 x}\right )+\frac {3}{25} \sqrt {1-2 x} (3 x+2)^2\right )-\frac {\sqrt {1-2 x} (3 x+2)^3}{5 (5 x+3)}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {7}{5} \left (\frac {14}{25} \left (-\frac {1}{5} \int \frac {1}{\frac {11}{2}-\frac {5}{2} (1-2 x)}d\sqrt {1-2 x}-\frac {3}{5} \sqrt {1-2 x}\right )+\frac {3}{25} \sqrt {1-2 x} (3 x+2)^2\right )-\frac {\sqrt {1-2 x} (3 x+2)^3}{5 (5 x+3)}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {7}{5} \left (\frac {14}{25} \left (-\frac {2 \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{5 \sqrt {55}}-\frac {3}{5} \sqrt {1-2 x}\right )+\frac {3}{25} \sqrt {1-2 x} (3 x+2)^2\right )-\frac {\sqrt {1-2 x} (3 x+2)^3}{5 (5 x+3)}\) |
-1/5*(Sqrt[1 - 2*x]*(2 + 3*x)^3)/(3 + 5*x) + (7*((3*Sqrt[1 - 2*x]*(2 + 3*x )^2)/25 + (14*((-3*Sqrt[1 - 2*x])/5 - (2*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]] )/(5*Sqrt[55])))/25))/5
3.19.40.3.1 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] :> 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_))^(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_) )^(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 = 0.98 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.64
method | result | size |
risch | \(-\frac {2700 x^{4}+3420 x^{3}-2565 x^{2}-1154 x +622}{625 \left (3+5 x \right ) \sqrt {1-2 x}}-\frac {196 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{34375}\) | \(56\) |
pseudoelliptic | \(\frac {-196 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \left (3+5 x \right ) \sqrt {55}+55 \sqrt {1-2 x}\, \left (1350 x^{3}+2385 x^{2}-90 x -622\right )}{103125+171875 x}\) | \(57\) |
derivativedivides | \(\frac {27 \left (1-2 x \right )^{\frac {5}{2}}}{250}-\frac {117 \left (1-2 x \right )^{\frac {3}{2}}}{250}+\frac {18 \sqrt {1-2 x}}{625}+\frac {2 \sqrt {1-2 x}}{3125 \left (-\frac {6}{5}-2 x \right )}-\frac {196 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{34375}\) | \(63\) |
default | \(\frac {27 \left (1-2 x \right )^{\frac {5}{2}}}{250}-\frac {117 \left (1-2 x \right )^{\frac {3}{2}}}{250}+\frac {18 \sqrt {1-2 x}}{625}+\frac {2 \sqrt {1-2 x}}{3125 \left (-\frac {6}{5}-2 x \right )}-\frac {196 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{34375}\) | \(63\) |
trager | \(\frac {\sqrt {1-2 x}\, \left (1350 x^{3}+2385 x^{2}-90 x -622\right )}{1875+3125 x}+\frac {98 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) \ln \left (\frac {5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) x +55 \sqrt {1-2 x}-8 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right )}{3+5 x}\right )}{34375}\) | \(77\) |
-1/625*(2700*x^4+3420*x^3-2565*x^2-1154*x+622)/(3+5*x)/(1-2*x)^(1/2)-196/3 4375*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)
Time = 0.23 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.78 \[ \int \frac {\sqrt {1-2 x} (2+3 x)^3}{(3+5 x)^2} \, dx=\frac {98 \, \sqrt {55} {\left (5 \, x + 3\right )} \log \left (\frac {5 \, x + \sqrt {55} \sqrt {-2 \, x + 1} - 8}{5 \, x + 3}\right ) + 55 \, {\left (1350 \, x^{3} + 2385 \, x^{2} - 90 \, x - 622\right )} \sqrt {-2 \, x + 1}}{34375 \, {\left (5 \, x + 3\right )}} \]
1/34375*(98*sqrt(55)*(5*x + 3)*log((5*x + sqrt(55)*sqrt(-2*x + 1) - 8)/(5* x + 3)) + 55*(1350*x^3 + 2385*x^2 - 90*x - 622)*sqrt(-2*x + 1))/(5*x + 3)
Time = 35.81 (sec) , antiderivative size = 197, normalized size of antiderivative = 2.24 \[ \int \frac {\sqrt {1-2 x} (2+3 x)^3}{(3+5 x)^2} \, dx=\frac {27 \left (1 - 2 x\right )^{\frac {5}{2}}}{250} - \frac {117 \left (1 - 2 x\right )^{\frac {3}{2}}}{250} + \frac {18 \sqrt {1 - 2 x}}{625} + \frac {97 \sqrt {55} \left (\log {\left (\sqrt {1 - 2 x} - \frac {\sqrt {55}}{5} \right )} - \log {\left (\sqrt {1 - 2 x} + \frac {\sqrt {55}}{5} \right )}\right )}{34375} - \frac {44 \left (\begin {cases} \frac {\sqrt {55} \left (- \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1 \right )}}{4} + \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1 \right )}}{4} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1\right )} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1\right )}\right )}{605} & \text {for}\: \sqrt {1 - 2 x} > - \frac {\sqrt {55}}{5} \wedge \sqrt {1 - 2 x} < \frac {\sqrt {55}}{5} \end {cases}\right )}{625} \]
27*(1 - 2*x)**(5/2)/250 - 117*(1 - 2*x)**(3/2)/250 + 18*sqrt(1 - 2*x)/625 + 97*sqrt(55)*(log(sqrt(1 - 2*x) - sqrt(55)/5) - log(sqrt(1 - 2*x) + sqrt( 55)/5))/34375 - 44*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) > -s qrt(55)/5) & (sqrt(1 - 2*x) < sqrt(55)/5)))/625
Time = 0.27 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.91 \[ \int \frac {\sqrt {1-2 x} (2+3 x)^3}{(3+5 x)^2} \, dx=\frac {27}{250} \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} - \frac {117}{250} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {98}{34375} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) + \frac {18}{625} \, \sqrt {-2 \, x + 1} - \frac {\sqrt {-2 \, x + 1}}{625 \, {\left (5 \, x + 3\right )}} \]
27/250*(-2*x + 1)^(5/2) - 117/250*(-2*x + 1)^(3/2) + 98/34375*sqrt(55)*log (-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 18/625*sq rt(-2*x + 1) - 1/625*sqrt(-2*x + 1)/(5*x + 3)
Time = 0.27 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.02 \[ \int \frac {\sqrt {1-2 x} (2+3 x)^3}{(3+5 x)^2} \, dx=\frac {27}{250} \, {\left (2 \, x - 1\right )}^{2} \sqrt {-2 \, x + 1} - \frac {117}{250} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {98}{34375} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) + \frac {18}{625} \, \sqrt {-2 \, x + 1} - \frac {\sqrt {-2 \, x + 1}}{625 \, {\left (5 \, x + 3\right )}} \]
27/250*(2*x - 1)^2*sqrt(-2*x + 1) - 117/250*(-2*x + 1)^(3/2) + 98/34375*sq rt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2* x + 1))) + 18/625*sqrt(-2*x + 1) - 1/625*sqrt(-2*x + 1)/(5*x + 3)
Time = 0.08 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.73 \[ \int \frac {\sqrt {1-2 x} (2+3 x)^3}{(3+5 x)^2} \, dx=\frac {18\,\sqrt {1-2\,x}}{625}-\frac {2\,\sqrt {1-2\,x}}{3125\,\left (2\,x+\frac {6}{5}\right )}-\frac {117\,{\left (1-2\,x\right )}^{3/2}}{250}+\frac {27\,{\left (1-2\,x\right )}^{5/2}}{250}+\frac {\sqrt {55}\,\mathrm {atan}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}\,1{}\mathrm {i}}{11}\right )\,196{}\mathrm {i}}{34375} \]