Integrand size = 24, antiderivative size = 93 \[ \int \frac {\sqrt {1-2 x} (3+5 x)^3}{(2+3 x)^2} \, dx=\frac {7}{9} \sqrt {1-2 x} (3+5 x)^2-\frac {\sqrt {1-2 x} (3+5 x)^3}{3 (2+3 x)}-\frac {2}{81} \sqrt {1-2 x} (211+170 x)-\frac {212 \text {arctanh}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{81 \sqrt {21}} \]
-212/1701*arctanh(1/7*21^(1/2)*(1-2*x)^(1/2))*21^(1/2)+7/9*(3+5*x)^2*(1-2* x)^(1/2)-1/3*(3+5*x)^3*(1-2*x)^(1/2)/(2+3*x)-2/81*(211+170*x)*(1-2*x)^(1/2 )
Time = 0.11 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.68 \[ \int \frac {\sqrt {1-2 x} (3+5 x)^3}{(2+3 x)^2} \, dx=\frac {\sqrt {1-2 x} \left (-439-110 x+1725 x^2+1350 x^3\right )}{81 (2+3 x)}-\frac {212 \text {arctanh}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{81 \sqrt {21}} \]
(Sqrt[1 - 2*x]*(-439 - 110*x + 1725*x^2 + 1350*x^3))/(81*(2 + 3*x)) - (212 *ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]])/(81*Sqrt[21])
Time = 0.20 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.11, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {108, 170, 27, 164, 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.
\(\displaystyle \int \frac {\sqrt {1-2 x} (5 x+3)^3}{(3 x+2)^2} \, dx\) |
\(\Big \downarrow \) 108 |
\(\displaystyle \frac {1}{3} \int \frac {(12-35 x) (5 x+3)^2}{\sqrt {1-2 x} (3 x+2)}dx-\frac {\sqrt {1-2 x} (5 x+3)^3}{3 (3 x+2)}\) |
\(\Big \downarrow \) 170 |
\(\displaystyle \frac {1}{3} \left (\frac {7}{3} \sqrt {1-2 x} (5 x+3)^2-\frac {1}{15} \int -\frac {10 (5 x+3) (34 x+5)}{\sqrt {1-2 x} (3 x+2)}dx\right )-\frac {\sqrt {1-2 x} (5 x+3)^3}{3 (3 x+2)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{3} \left (\frac {2}{3} \int \frac {(5 x+3) (34 x+5)}{\sqrt {1-2 x} (3 x+2)}dx+\frac {7}{3} \sqrt {1-2 x} (5 x+3)^2\right )-\frac {\sqrt {1-2 x} (5 x+3)^3}{3 (3 x+2)}\) |
\(\Big \downarrow \) 164 |
\(\displaystyle \frac {1}{3} \left (\frac {2}{3} \left (\frac {53}{9} \int \frac {1}{\sqrt {1-2 x} (3 x+2)}dx-\frac {1}{9} \sqrt {1-2 x} (170 x+211)\right )+\frac {7}{3} \sqrt {1-2 x} (5 x+3)^2\right )-\frac {\sqrt {1-2 x} (5 x+3)^3}{3 (3 x+2)}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{3} \left (\frac {2}{3} \left (-\frac {53}{9} \int \frac {1}{\frac {7}{2}-\frac {3}{2} (1-2 x)}d\sqrt {1-2 x}-\frac {1}{9} \sqrt {1-2 x} (170 x+211)\right )+\frac {7}{3} \sqrt {1-2 x} (5 x+3)^2\right )-\frac {\sqrt {1-2 x} (5 x+3)^3}{3 (3 x+2)}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{3} \left (\frac {2}{3} \left (-\frac {106 \text {arctanh}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{9 \sqrt {21}}-\frac {1}{9} \sqrt {1-2 x} (170 x+211)\right )+\frac {7}{3} \sqrt {1-2 x} (5 x+3)^2\right )-\frac {\sqrt {1-2 x} (5 x+3)^3}{3 (3 x+2)}\) |
-1/3*(Sqrt[1 - 2*x]*(3 + 5*x)^3)/(2 + 3*x) + ((7*Sqrt[1 - 2*x]*(3 + 5*x)^2 )/3 + (2*(-1/9*(Sqrt[1 - 2*x]*(211 + 170*x)) - (106*ArcTanh[Sqrt[3/7]*Sqrt [1 - 2*x]])/(9*Sqrt[21])))/3)/3
3.19.21.3.1 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_))^(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 = 0.97 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.60
method | result | size |
risch | \(-\frac {2700 x^{4}+2100 x^{3}-1945 x^{2}-768 x +439}{81 \left (2+3 x \right ) \sqrt {1-2 x}}-\frac {212 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{1701}\) | \(56\) |
pseudoelliptic | \(\frac {-212 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \left (2+3 x \right ) \sqrt {21}+21 \sqrt {1-2 x}\, \left (1350 x^{3}+1725 x^{2}-110 x -439\right )}{3402+5103 x}\) | \(57\) |
derivativedivides | \(\frac {25 \left (1-2 x \right )^{\frac {5}{2}}}{18}-\frac {725 \left (1-2 x \right )^{\frac {3}{2}}}{162}+\frac {10 \sqrt {1-2 x}}{27}-\frac {2 \sqrt {1-2 x}}{243 \left (-\frac {4}{3}-2 x \right )}-\frac {212 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{1701}\) | \(63\) |
default | \(\frac {25 \left (1-2 x \right )^{\frac {5}{2}}}{18}-\frac {725 \left (1-2 x \right )^{\frac {3}{2}}}{162}+\frac {10 \sqrt {1-2 x}}{27}-\frac {2 \sqrt {1-2 x}}{243 \left (-\frac {4}{3}-2 x \right )}-\frac {212 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{1701}\) | \(63\) |
trager | \(\frac {\sqrt {1-2 x}\, \left (1350 x^{3}+1725 x^{2}-110 x -439\right )}{162+243 x}+\frac {106 \operatorname {RootOf}\left (\textit {\_Z}^{2}-21\right ) \ln \left (\frac {3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-21\right ) x -5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-21\right )+21 \sqrt {1-2 x}}{2+3 x}\right )}{1701}\) | \(77\) |
-1/81*(2700*x^4+2100*x^3-1945*x^2-768*x+439)/(2+3*x)/(1-2*x)^(1/2)-212/170 1*arctanh(1/7*21^(1/2)*(1-2*x)^(1/2))*21^(1/2)
Time = 0.23 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.74 \[ \int \frac {\sqrt {1-2 x} (3+5 x)^3}{(2+3 x)^2} \, dx=\frac {106 \, \sqrt {21} {\left (3 \, x + 2\right )} \log \left (\frac {3 \, x + \sqrt {21} \sqrt {-2 \, x + 1} - 5}{3 \, x + 2}\right ) + 21 \, {\left (1350 \, x^{3} + 1725 \, x^{2} - 110 \, x - 439\right )} \sqrt {-2 \, x + 1}}{1701 \, {\left (3 \, x + 2\right )}} \]
1/1701*(106*sqrt(21)*(3*x + 2)*log((3*x + sqrt(21)*sqrt(-2*x + 1) - 5)/(3* x + 2)) + 21*(1350*x^3 + 1725*x^2 - 110*x - 439)*sqrt(-2*x + 1))/(3*x + 2)
Time = 36.65 (sec) , antiderivative size = 197, normalized size of antiderivative = 2.12 \[ \int \frac {\sqrt {1-2 x} (3+5 x)^3}{(2+3 x)^2} \, dx=\frac {25 \left (1 - 2 x\right )^{\frac {5}{2}}}{18} - \frac {725 \left (1 - 2 x\right )^{\frac {3}{2}}}{162} + \frac {10 \sqrt {1 - 2 x}}{27} + \frac {107 \sqrt {21} \left (\log {\left (\sqrt {1 - 2 x} - \frac {\sqrt {21}}{3} \right )} - \log {\left (\sqrt {1 - 2 x} + \frac {\sqrt {21}}{3} \right )}\right )}{1701} + \frac {28 \left (\begin {cases} \frac {\sqrt {21} \left (- \frac {\log {\left (\frac {\sqrt {21} \sqrt {1 - 2 x}}{7} - 1 \right )}}{4} + \frac {\log {\left (\frac {\sqrt {21} \sqrt {1 - 2 x}}{7} + 1 \right )}}{4} - \frac {1}{4 \left (\frac {\sqrt {21} \sqrt {1 - 2 x}}{7} + 1\right )} - \frac {1}{4 \left (\frac {\sqrt {21} \sqrt {1 - 2 x}}{7} - 1\right )}\right )}{147} & \text {for}\: \sqrt {1 - 2 x} > - \frac {\sqrt {21}}{3} \wedge \sqrt {1 - 2 x} < \frac {\sqrt {21}}{3} \end {cases}\right )}{81} \]
25*(1 - 2*x)**(5/2)/18 - 725*(1 - 2*x)**(3/2)/162 + 10*sqrt(1 - 2*x)/27 + 107*sqrt(21)*(log(sqrt(1 - 2*x) - sqrt(21)/3) - log(sqrt(1 - 2*x) + sqrt(2 1)/3))/1701 + 28*Piecewise((sqrt(21)*(-log(sqrt(21)*sqrt(1 - 2*x)/7 - 1)/4 + log(sqrt(21)*sqrt(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)))/81
Time = 0.28 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.86 \[ \int \frac {\sqrt {1-2 x} (3+5 x)^3}{(2+3 x)^2} \, dx=\frac {25}{18} \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} - \frac {725}{162} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {106}{1701} \, \sqrt {21} \log \left (-\frac {\sqrt {21} - 3 \, \sqrt {-2 \, x + 1}}{\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}}\right ) + \frac {10}{27} \, \sqrt {-2 \, x + 1} + \frac {\sqrt {-2 \, x + 1}}{81 \, {\left (3 \, x + 2\right )}} \]
25/18*(-2*x + 1)^(5/2) - 725/162*(-2*x + 1)^(3/2) + 106/1701*sqrt(21)*log( -(sqrt(21) - 3*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) + 10/27*sqrt (-2*x + 1) + 1/81*sqrt(-2*x + 1)/(3*x + 2)
Time = 0.30 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.97 \[ \int \frac {\sqrt {1-2 x} (3+5 x)^3}{(2+3 x)^2} \, dx=\frac {25}{18} \, {\left (2 \, x - 1\right )}^{2} \sqrt {-2 \, x + 1} - \frac {725}{162} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {106}{1701} \, \sqrt {21} \log \left (\frac {{\left | -2 \, \sqrt {21} + 6 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}\right )}}\right ) + \frac {10}{27} \, \sqrt {-2 \, x + 1} + \frac {\sqrt {-2 \, x + 1}}{81 \, {\left (3 \, x + 2\right )}} \]
25/18*(2*x - 1)^2*sqrt(-2*x + 1) - 725/162*(-2*x + 1)^(3/2) + 106/1701*sqr t(21)*log(1/2*abs(-2*sqrt(21) + 6*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) + 10/27*sqrt(-2*x + 1) + 1/81*sqrt(-2*x + 1)/(3*x + 2)
Time = 0.05 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.69 \[ \int \frac {\sqrt {1-2 x} (3+5 x)^3}{(2+3 x)^2} \, dx=\frac {2\,\sqrt {1-2\,x}}{243\,\left (2\,x+\frac {4}{3}\right )}+\frac {10\,\sqrt {1-2\,x}}{27}-\frac {725\,{\left (1-2\,x\right )}^{3/2}}{162}+\frac {25\,{\left (1-2\,x\right )}^{5/2}}{18}+\frac {\sqrt {21}\,\mathrm {atan}\left (\frac {\sqrt {21}\,\sqrt {1-2\,x}\,1{}\mathrm {i}}{7}\right )\,212{}\mathrm {i}}{1701} \]