Integrand size = 22, antiderivative size = 56 \[ \int \frac {\sqrt {1-2 x} (2+3 x)}{3+5 x} \, dx=\frac {2}{25} \sqrt {1-2 x}-\frac {1}{5} (1-2 x)^{3/2}-\frac {2}{25} \sqrt {\frac {11}{5}} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \] Output:
2/25*(1-2*x)^(1/2)-1/5*(1-2*x)^(3/2)-2/125*55^(1/2)*arctanh(1/11*55^(1/2)* (1-2*x)^(1/2))
Time = 0.04 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.82 \[ \int \frac {\sqrt {1-2 x} (2+3 x)}{3+5 x} \, dx=\frac {1}{125} \left (5 \sqrt {1-2 x} (-3+10 x)-2 \sqrt {55} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )\right ) \] Input:
Integrate[(Sqrt[1 - 2*x]*(2 + 3*x))/(3 + 5*x),x]
Output:
(5*Sqrt[1 - 2*x]*(-3 + 10*x) - 2*Sqrt[55]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x] ])/125
Time = 0.17 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.09, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {90, 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.
\(\displaystyle \int \frac {\sqrt {1-2 x} (3 x+2)}{5 x+3} \, dx\) |
\(\Big \downarrow \) 90 |
\(\displaystyle \frac {1}{5} \int \frac {\sqrt {1-2 x}}{5 x+3}dx-\frac {1}{5} (1-2 x)^{3/2}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {1}{5} \left (\frac {11}{5} \int \frac {1}{\sqrt {1-2 x} (5 x+3)}dx+\frac {2}{5} \sqrt {1-2 x}\right )-\frac {1}{5} (1-2 x)^{3/2}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{5} \left (\frac {2}{5} \sqrt {1-2 x}-\frac {11}{5} \int \frac {1}{\frac {11}{2}-\frac {5}{2} (1-2 x)}d\sqrt {1-2 x}\right )-\frac {1}{5} (1-2 x)^{3/2}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{5} \left (\frac {2}{5} \sqrt {1-2 x}-\frac {2}{5} \sqrt {\frac {11}{5}} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )\right )-\frac {1}{5} (1-2 x)^{3/2}\) |
Input:
Int[(Sqrt[1 - 2*x]*(2 + 3*x))/(3 + 5*x),x]
Output:
-1/5*(1 - 2*x)^(3/2) + ((2*Sqrt[1 - 2*x])/5 - (2*Sqrt[11/5]*ArcTanh[Sqrt[5 /11]*Sqrt[1 - 2*x]])/5)/5
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*(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_)^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 = 0.18 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.61
method | result | size |
pseudoelliptic | \(-\frac {2 \sqrt {55}\, \operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right )}{125}+\frac {\sqrt {1-2 x}\, \left (10 x -3\right )}{25}\) | \(34\) |
derivativedivides | \(\frac {2 \sqrt {1-2 x}}{25}-\frac {\left (1-2 x \right )^{\frac {3}{2}}}{5}-\frac {2 \sqrt {55}\, \operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right )}{125}\) | \(38\) |
default | \(\frac {2 \sqrt {1-2 x}}{25}-\frac {\left (1-2 x \right )^{\frac {3}{2}}}{5}-\frac {2 \sqrt {55}\, \operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right )}{125}\) | \(38\) |
risch | \(-\frac {\left (10 x -3\right ) \left (-1+2 x \right )}{25 \sqrt {1-2 x}}-\frac {2 \sqrt {55}\, \operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right )}{125}\) | \(39\) |
trager | \(\left (\frac {2 x}{5}-\frac {3}{25}\right ) \sqrt {1-2 x}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) \ln \left (\frac {5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) x +55 \sqrt {1-2 x}-8 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right )}{3+5 x}\right )}{125}\) | \(59\) |
Input:
int((1-2*x)^(1/2)*(2+3*x)/(3+5*x),x,method=_RETURNVERBOSE)
Output:
-2/125*55^(1/2)*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))+1/25*(1-2*x)^(1/2)*(1 0*x-3)
Time = 0.07 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.82 \[ \int \frac {\sqrt {1-2 x} (2+3 x)}{3+5 x} \, dx=\frac {1}{25} \, {\left (10 \, x - 3\right )} \sqrt {-2 \, x + 1} + \frac {1}{25} \, \sqrt {\frac {11}{5}} \log \left (\frac {5 \, x + 5 \, \sqrt {\frac {11}{5}} \sqrt {-2 \, x + 1} - 8}{5 \, x + 3}\right ) \] Input:
integrate((1-2*x)^(1/2)*(2+3*x)/(3+5*x),x, algorithm="fricas")
Output:
1/25*(10*x - 3)*sqrt(-2*x + 1) + 1/25*sqrt(11/5)*log((5*x + 5*sqrt(11/5)*s qrt(-2*x + 1) - 8)/(5*x + 3))
Time = 2.11 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.09 \[ \int \frac {\sqrt {1-2 x} (2+3 x)}{3+5 x} \, dx=- \frac {\left (1 - 2 x\right )^{\frac {3}{2}}}{5} + \frac {2 \sqrt {1 - 2 x}}{25} + \frac {\sqrt {55} \left (\log {\left (\sqrt {1 - 2 x} - \frac {\sqrt {55}}{5} \right )} - \log {\left (\sqrt {1 - 2 x} + \frac {\sqrt {55}}{5} \right )}\right )}{125} \] Input:
integrate((1-2*x)**(1/2)*(2+3*x)/(3+5*x),x)
Output:
-(1 - 2*x)**(3/2)/5 + 2*sqrt(1 - 2*x)/25 + sqrt(55)*(log(sqrt(1 - 2*x) - s qrt(55)/5) - log(sqrt(1 - 2*x) + sqrt(55)/5))/125
Time = 0.12 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.98 \[ \int \frac {\sqrt {1-2 x} (2+3 x)}{3+5 x} \, dx=-\frac {1}{5} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {1}{125} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) + \frac {2}{25} \, \sqrt {-2 \, x + 1} \] Input:
integrate((1-2*x)^(1/2)*(2+3*x)/(3+5*x),x, algorithm="maxima")
Output:
-1/5*(-2*x + 1)^(3/2) + 1/125*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/ (sqrt(55) + 5*sqrt(-2*x + 1))) + 2/25*sqrt(-2*x + 1)
Time = 0.12 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.04 \[ \int \frac {\sqrt {1-2 x} (2+3 x)}{3+5 x} \, dx=-\frac {1}{5} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {1}{125} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) + \frac {2}{25} \, \sqrt {-2 \, x + 1} \] Input:
integrate((1-2*x)^(1/2)*(2+3*x)/(3+5*x),x, algorithm="giac")
Output:
-1/5*(-2*x + 1)^(3/2) + 1/125*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(- 2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 2/25*sqrt(-2*x + 1)
Time = 0.04 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.66 \[ \int \frac {\sqrt {1-2 x} (2+3 x)}{3+5 x} \, dx=\frac {2\,\sqrt {1-2\,x}}{25}-\frac {2\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )}{125}-\frac {{\left (1-2\,x\right )}^{3/2}}{5} \] Input:
int(((1 - 2*x)^(1/2)*(3*x + 2))/(5*x + 3),x)
Output:
(2*(1 - 2*x)^(1/2))/25 - (2*55^(1/2)*atanh((55^(1/2)*(1 - 2*x)^(1/2))/11)) /125 - (1 - 2*x)^(3/2)/5
Time = 0.16 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.93 \[ \int \frac {\sqrt {1-2 x} (2+3 x)}{3+5 x} \, dx=\frac {2 \sqrt {-2 x +1}\, x}{5}-\frac {3 \sqrt {-2 x +1}}{25}+\frac {\sqrt {55}\, \mathrm {log}\left (5 \sqrt {-2 x +1}-\sqrt {55}\right )}{125}-\frac {\sqrt {55}\, \mathrm {log}\left (5 \sqrt {-2 x +1}+\sqrt {55}\right )}{125} \] Input:
int((1-2*x)^(1/2)*(2+3*x)/(3+5*x),x)
Output:
(50*sqrt( - 2*x + 1)*x - 15*sqrt( - 2*x + 1) + sqrt(55)*log(5*sqrt( - 2*x + 1) - sqrt(55)) - sqrt(55)*log(5*sqrt( - 2*x + 1) + sqrt(55)))/125