Integrand size = 22, antiderivative size = 59 \[ \int \frac {\sqrt {1-2 x} (3+5 x)}{(2+3 x)^2} \, dx=\frac {8}{7} \sqrt {1-2 x}+\frac {(1-2 x)^{3/2}}{21 (2+3 x)}-\frac {8 \text {arctanh}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{\sqrt {21}} \]
1/21*(1-2*x)^(3/2)/(2+3*x)-8/21*arctanh(1/7*21^(1/2)*(1-2*x)^(1/2))*21^(1/ 2)+8/7*(1-2*x)^(1/2)
Time = 0.10 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.93 \[ \int \frac {\sqrt {1-2 x} (3+5 x)}{(2+3 x)^2} \, dx=\frac {7 \sqrt {1-2 x} (7+10 x)-8 \sqrt {21} (2+3 x) \text {arctanh}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{42+63 x} \]
(7*Sqrt[1 - 2*x]*(7 + 10*x) - 8*Sqrt[21]*(2 + 3*x)*ArcTanh[Sqrt[3/7]*Sqrt[ 1 - 2*x]])/(42 + 63*x)
Time = 0.17 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.15, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {87, 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.
\(\displaystyle \int \frac {\sqrt {1-2 x} (5 x+3)}{(3 x+2)^2} \, dx\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {12}{7} \int \frac {\sqrt {1-2 x}}{3 x+2}dx+\frac {(1-2 x)^{3/2}}{21 (3 x+2)}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {12}{7} \left (\frac {7}{3} \int \frac {1}{\sqrt {1-2 x} (3 x+2)}dx+\frac {2}{3} \sqrt {1-2 x}\right )+\frac {(1-2 x)^{3/2}}{21 (3 x+2)}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {12}{7} \left (\frac {2}{3} \sqrt {1-2 x}-\frac {7}{3} \int \frac {1}{\frac {7}{2}-\frac {3}{2} (1-2 x)}d\sqrt {1-2 x}\right )+\frac {(1-2 x)^{3/2}}{21 (3 x+2)}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {12}{7} \left (\frac {2}{3} \sqrt {1-2 x}-\frac {2}{3} \sqrt {\frac {7}{3}} \text {arctanh}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )\right )+\frac {(1-2 x)^{3/2}}{21 (3 x+2)}\) |
(1 - 2*x)^(3/2)/(21*(2 + 3*x)) + (12*((2*Sqrt[1 - 2*x])/3 - (2*Sqrt[7/3]*A rcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]])/3))/7
3.18.98.3.1 Defintions of rubi rules used
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_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)) Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege rQ[p] || !(IntegerQ[n] || !(EqQ[e, 0] || !(EqQ[c, 0] || LtQ[p, n]))))
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 = 2.96 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.76
method | result | size |
derivativedivides | \(\frac {10 \sqrt {1-2 x}}{9}-\frac {2 \sqrt {1-2 x}}{27 \left (-\frac {4}{3}-2 x \right )}-\frac {8 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{21}\) | \(45\) |
default | \(\frac {10 \sqrt {1-2 x}}{9}-\frac {2 \sqrt {1-2 x}}{27 \left (-\frac {4}{3}-2 x \right )}-\frac {8 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{21}\) | \(45\) |
risch | \(-\frac {20 x^{2}+4 x -7}{3 \left (2+3 x \right ) \sqrt {1-2 x}}-\frac {8 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{21}\) | \(46\) |
pseudoelliptic | \(\frac {-8 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \left (2+3 x \right ) \sqrt {21}+7 \sqrt {1-2 x}\, \left (7+10 x \right )}{42+63 x}\) | \(47\) |
trager | \(\frac {\sqrt {1-2 x}\, \left (7+10 x \right )}{9 x +6}+\frac {4 \operatorname {RootOf}\left (\textit {\_Z}^{2}-21\right ) \ln \left (\frac {3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-21\right ) x -5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-21\right )+21 \sqrt {1-2 x}}{2+3 x}\right )}{21}\) | \(67\) |
10/9*(1-2*x)^(1/2)-2/27*(1-2*x)^(1/2)/(-4/3-2*x)-8/21*arctanh(1/7*21^(1/2) *(1-2*x)^(1/2))*21^(1/2)
Time = 0.23 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.00 \[ \int \frac {\sqrt {1-2 x} (3+5 x)}{(2+3 x)^2} \, dx=\frac {4 \, \sqrt {21} {\left (3 \, x + 2\right )} \log \left (\frac {3 \, x + \sqrt {21} \sqrt {-2 \, x + 1} - 5}{3 \, x + 2}\right ) + 7 \, {\left (10 \, x + 7\right )} \sqrt {-2 \, x + 1}}{21 \, {\left (3 \, x + 2\right )}} \]
1/21*(4*sqrt(21)*(3*x + 2)*log((3*x + sqrt(21)*sqrt(-2*x + 1) - 5)/(3*x + 2)) + 7*(10*x + 7)*sqrt(-2*x + 1))/(3*x + 2)
Time = 28.83 (sec) , antiderivative size = 173, normalized size of antiderivative = 2.93 \[ \int \frac {\sqrt {1-2 x} (3+5 x)}{(2+3 x)^2} \, dx=\frac {10 \sqrt {1 - 2 x}}{9} + \frac {37 \sqrt {21} \left (\log {\left (\sqrt {1 - 2 x} - \frac {\sqrt {21}}{3} \right )} - \log {\left (\sqrt {1 - 2 x} + \frac {\sqrt {21}}{3} \right )}\right )}{189} + \frac {28 \left (\begin {cases} \frac {\sqrt {21} \left (- \frac {\log {\left (\frac {\sqrt {21} \sqrt {1 - 2 x}}{7} - 1 \right )}}{4} + \frac {\log {\left (\frac {\sqrt {21} \sqrt {1 - 2 x}}{7} + 1 \right )}}{4} - \frac {1}{4 \left (\frac {\sqrt {21} \sqrt {1 - 2 x}}{7} + 1\right )} - \frac {1}{4 \left (\frac {\sqrt {21} \sqrt {1 - 2 x}}{7} - 1\right )}\right )}{147} & \text {for}\: \sqrt {1 - 2 x} > - \frac {\sqrt {21}}{3} \wedge \sqrt {1 - 2 x} < \frac {\sqrt {21}}{3} \end {cases}\right )}{9} \]
10*sqrt(1 - 2*x)/9 + 37*sqrt(21)*(log(sqrt(1 - 2*x) - sqrt(21)/3) - log(sq rt(1 - 2*x) + sqrt(21)/3))/189 + 28*Piecewise((sqrt(21)*(-log(sqrt(21)*sqr t(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)))/9
Time = 0.28 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.05 \[ \int \frac {\sqrt {1-2 x} (3+5 x)}{(2+3 x)^2} \, dx=\frac {4}{21} \, \sqrt {21} \log \left (-\frac {\sqrt {21} - 3 \, \sqrt {-2 \, x + 1}}{\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}}\right ) + \frac {10}{9} \, \sqrt {-2 \, x + 1} + \frac {\sqrt {-2 \, x + 1}}{9 \, {\left (3 \, x + 2\right )}} \]
4/21*sqrt(21)*log(-(sqrt(21) - 3*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) + 10/9*sqrt(-2*x + 1) + 1/9*sqrt(-2*x + 1)/(3*x + 2)
Time = 0.29 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.10 \[ \int \frac {\sqrt {1-2 x} (3+5 x)}{(2+3 x)^2} \, dx=\frac {4}{21} \, \sqrt {21} \log \left (\frac {{\left | -2 \, \sqrt {21} + 6 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}\right )}}\right ) + \frac {10}{9} \, \sqrt {-2 \, x + 1} + \frac {\sqrt {-2 \, x + 1}}{9 \, {\left (3 \, x + 2\right )}} \]
4/21*sqrt(21)*log(1/2*abs(-2*sqrt(21) + 6*sqrt(-2*x + 1))/(sqrt(21) + 3*sq rt(-2*x + 1))) + 10/9*sqrt(-2*x + 1) + 1/9*sqrt(-2*x + 1)/(3*x + 2)
Time = 1.44 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.75 \[ \int \frac {\sqrt {1-2 x} (3+5 x)}{(2+3 x)^2} \, dx=\frac {2\,\sqrt {1-2\,x}}{27\,\left (2\,x+\frac {4}{3}\right )}-\frac {8\,\sqrt {21}\,\mathrm {atanh}\left (\frac {\sqrt {21}\,\sqrt {1-2\,x}}{7}\right )}{21}+\frac {10\,\sqrt {1-2\,x}}{9} \]