Integrand size = 24, antiderivative size = 69 \[ \int \frac {\sqrt {1-2 x}}{(2+3 x) (3+5 x)^2} \, dx=-\frac {\sqrt {1-2 x}}{3+5 x}-2 \sqrt {21} \text {arctanh}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )+\frac {68 \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{\sqrt {55}} \]
-2*arctanh(1/7*21^(1/2)*(1-2*x)^(1/2))*21^(1/2)+68/55*arctanh(1/11*55^(1/2 )*(1-2*x)^(1/2))*55^(1/2)-(1-2*x)^(1/2)/(3+5*x)
Time = 0.17 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.00 \[ \int \frac {\sqrt {1-2 x}}{(2+3 x) (3+5 x)^2} \, dx=-\frac {\sqrt {1-2 x}}{3+5 x}-2 \sqrt {21} \text {arctanh}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )+\frac {68 \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{\sqrt {55}} \]
-(Sqrt[1 - 2*x]/(3 + 5*x)) - 2*Sqrt[21]*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]] + (68*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/Sqrt[55]
Time = 0.18 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {110, 25, 174, 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.
\(\displaystyle \int \frac {\sqrt {1-2 x}}{(3 x+2) (5 x+3)^2} \, dx\) |
\(\Big \downarrow \) 110 |
\(\displaystyle \int -\frac {5-3 x}{\sqrt {1-2 x} (3 x+2) (5 x+3)}dx-\frac {\sqrt {1-2 x}}{5 x+3}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \frac {5-3 x}{\sqrt {1-2 x} (3 x+2) (5 x+3)}dx-\frac {\sqrt {1-2 x}}{5 x+3}\) |
\(\Big \downarrow \) 174 |
\(\displaystyle 21 \int \frac {1}{\sqrt {1-2 x} (3 x+2)}dx-34 \int \frac {1}{\sqrt {1-2 x} (5 x+3)}dx-\frac {\sqrt {1-2 x}}{5 x+3}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle 34 \int \frac {1}{\frac {11}{2}-\frac {5}{2} (1-2 x)}d\sqrt {1-2 x}-21 \int \frac {1}{\frac {7}{2}-\frac {3}{2} (1-2 x)}d\sqrt {1-2 x}-\frac {\sqrt {1-2 x}}{5 x+3}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -2 \sqrt {21} \text {arctanh}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )+\frac {68 \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{\sqrt {55}}-\frac {\sqrt {1-2 x}}{5 x+3}\) |
-(Sqrt[1 - 2*x]/(3 + 5*x)) - 2*Sqrt[21]*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]] + (68*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/Sqrt[55]
3.19.44.3.1 Defintions of rubi rules used
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 + 1)/((m + 1)*(b*e - a*f))), x] - Simp[1/((m + 1)*(b*e - a*f)) Int[(a + b*x)^(m + 1) *(c + d*x)^(n - 1)*(e + f*x)^p*Simp[d*e*n + c*f*(m + p + 2) + d*f*(m + n + p + 2)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && LtQ[m, -1] && Gt Q[n, 0] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p] || IntegersQ[p, m + n])
Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))* ((c_.) + (d_.)*(x_))), x_] :> Simp[(b*g - a*h)/(b*c - a*d) Int[(e + f*x)^ p/(a + b*x), x], x] - Simp[(d*g - c*h)/(b*c - a*d) Int[(e + f*x)^p/(c + d *x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x]
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 = 0.99 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.78
method | result | size |
derivativedivides | \(\frac {2 \sqrt {1-2 x}}{5 \left (-\frac {6}{5}-2 x \right )}+\frac {68 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{55}-2 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}\) | \(54\) |
default | \(\frac {2 \sqrt {1-2 x}}{5 \left (-\frac {6}{5}-2 x \right )}+\frac {68 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{55}-2 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}\) | \(54\) |
risch | \(\frac {-1+2 x}{\left (3+5 x \right ) \sqrt {1-2 x}}-2 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}+\frac {68 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{55}\) | \(58\) |
pseudoelliptic | \(\frac {-110 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \left (3+5 x \right ) \sqrt {21}+68 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \left (3+5 x \right ) \sqrt {55}-55 \sqrt {1-2 x}}{165+275 x}\) | \(65\) |
trager | \(-\frac {\sqrt {1-2 x}}{3+5 x}-\frac {34 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) \ln \left (\frac {5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) x +55 \sqrt {1-2 x}-8 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right )}{3+5 x}\right )}{55}-\operatorname {RootOf}\left (\textit {\_Z}^{2}-21\right ) \ln \left (\frac {-3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-21\right ) x +21 \sqrt {1-2 x}+5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-21\right )}{2+3 x}\right )\) | \(106\) |
2/5*(1-2*x)^(1/2)/(-6/5-2*x)+68/55*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55 ^(1/2)-2*arctanh(1/7*21^(1/2)*(1-2*x)^(1/2))*21^(1/2)
Time = 0.23 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.30 \[ \int \frac {\sqrt {1-2 x}}{(2+3 x) (3+5 x)^2} \, dx=\frac {34 \, \sqrt {55} {\left (5 \, x + 3\right )} \log \left (\frac {5 \, x - \sqrt {55} \sqrt {-2 \, x + 1} - 8}{5 \, x + 3}\right ) + 55 \, \sqrt {21} {\left (5 \, x + 3\right )} \log \left (\frac {3 \, x + \sqrt {21} \sqrt {-2 \, x + 1} - 5}{3 \, x + 2}\right ) - 55 \, \sqrt {-2 \, x + 1}}{55 \, {\left (5 \, x + 3\right )}} \]
1/55*(34*sqrt(55)*(5*x + 3)*log((5*x - sqrt(55)*sqrt(-2*x + 1) - 8)/(5*x + 3)) + 55*sqrt(21)*(5*x + 3)*log((3*x + sqrt(21)*sqrt(-2*x + 1) - 5)/(3*x + 2)) - 55*sqrt(-2*x + 1))/(5*x + 3)
Time = 14.88 (sec) , antiderivative size = 199, normalized size of antiderivative = 2.88 \[ \int \frac {\sqrt {1-2 x}}{(2+3 x) (3+5 x)^2} \, dx=\sqrt {21} \left (\log {\left (\sqrt {1 - 2 x} - \frac {\sqrt {21}}{3} \right )} - \log {\left (\sqrt {1 - 2 x} + \frac {\sqrt {21}}{3} \right )}\right ) - \frac {7 \sqrt {55} \left (\log {\left (\sqrt {1 - 2 x} - \frac {\sqrt {55}}{5} \right )} - \log {\left (\sqrt {1 - 2 x} + \frac {\sqrt {55}}{5} \right )}\right )}{11} - 44 \left (\begin {cases} \frac {\sqrt {55} \left (- \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1 \right )}}{4} + \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1 \right )}}{4} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1\right )} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1\right )}\right )}{605} & \text {for}\: \sqrt {1 - 2 x} > - \frac {\sqrt {55}}{5} \wedge \sqrt {1 - 2 x} < \frac {\sqrt {55}}{5} \end {cases}\right ) \]
sqrt(21)*(log(sqrt(1 - 2*x) - sqrt(21)/3) - log(sqrt(1 - 2*x) + sqrt(21)/3 )) - 7*sqrt(55)*(log(sqrt(1 - 2*x) - sqrt(55)/5) - log(sqrt(1 - 2*x) + sqr t(55)/5))/11 - 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)/1 1 + 1)) - 1/(4*(sqrt(55)*sqrt(1 - 2*x)/11 - 1)))/605, (sqrt(1 - 2*x) > -sq rt(55)/5) & (sqrt(1 - 2*x) < sqrt(55)/5)))
Time = 0.29 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.28 \[ \int \frac {\sqrt {1-2 x}}{(2+3 x) (3+5 x)^2} \, dx=-\frac {34}{55} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) + \sqrt {21} \log \left (-\frac {\sqrt {21} - 3 \, \sqrt {-2 \, x + 1}}{\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}}\right ) - \frac {\sqrt {-2 \, x + 1}}{5 \, x + 3} \]
-34/55*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + sqrt(21)*log(-(sqrt(21) - 3*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(- 2*x + 1))) - sqrt(-2*x + 1)/(5*x + 3)
Time = 0.28 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.36 \[ \int \frac {\sqrt {1-2 x}}{(2+3 x) (3+5 x)^2} \, dx=-\frac {34}{55} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) + \sqrt {21} \log \left (\frac {{\left | -2 \, \sqrt {21} + 6 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}\right )}}\right ) - \frac {\sqrt {-2 \, x + 1}}{5 \, x + 3} \]
-34/55*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5 *sqrt(-2*x + 1))) + sqrt(21)*log(1/2*abs(-2*sqrt(21) + 6*sqrt(-2*x + 1))/( sqrt(21) + 3*sqrt(-2*x + 1))) - sqrt(-2*x + 1)/(5*x + 3)
Time = 1.49 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.77 \[ \int \frac {\sqrt {1-2 x}}{(2+3 x) (3+5 x)^2} \, dx=\frac {68\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )}{55}-2\,\sqrt {21}\,\mathrm {atanh}\left (\frac {\sqrt {21}\,\sqrt {1-2\,x}}{7}\right )-\frac {2\,\sqrt {1-2\,x}}{5\,\left (2\,x+\frac {6}{5}\right )} \]