Integrand size = 24, antiderivative size = 97 \[ \int \frac {\sqrt {1-2 x}}{(2+3 x)^2 (3+5 x)^2} \, dx=-\frac {10 \sqrt {1-2 x}}{3+5 x}+\frac {\sqrt {1-2 x}}{(2+3 x) (3+5 x)}-138 \sqrt {\frac {3}{7}} \text {arctanh}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )+134 \sqrt {\frac {5}{11}} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \]
-138/7*arctanh(1/7*21^(1/2)*(1-2*x)^(1/2))*21^(1/2)+134/11*arctanh(1/11*55 ^(1/2)*(1-2*x)^(1/2))*55^(1/2)-10*(1-2*x)^(1/2)/(3+5*x)+(1-2*x)^(1/2)/(2+3 *x)/(3+5*x)
Time = 0.23 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.86 \[ \int \frac {\sqrt {1-2 x}}{(2+3 x)^2 (3+5 x)^2} \, dx=-\frac {\sqrt {1-2 x} (19+30 x)}{6+19 x+15 x^2}-138 \sqrt {\frac {3}{7}} \text {arctanh}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )+134 \sqrt {\frac {5}{11}} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \]
-((Sqrt[1 - 2*x]*(19 + 30*x))/(6 + 19*x + 15*x^2)) - 138*Sqrt[3/7]*ArcTanh [Sqrt[3/7]*Sqrt[1 - 2*x]] + 134*Sqrt[5/11]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x ]]
Time = 0.20 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {110, 25, 168, 27, 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.
| \(\displaystyle \int \frac {\sqrt {1-2 x}}{(3 x+2)^2 (5 x+3)^2} \, dx\) |
| \(\Big \downarrow \) 110 |
| \(\displaystyle \frac {\sqrt {1-2 x}}{(3 x+2) (5 x+3)}-\int -\frac {13-15 x}{\sqrt {1-2 x} (3 x+2) (5 x+3)^2}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \int \frac {13-15 x}{\sqrt {1-2 x} (3 x+2) (5 x+3)^2}dx+\frac {\sqrt {1-2 x}}{(3 x+2) (5 x+3)}\) |
| \(\Big \downarrow \) 168 |
| \(\displaystyle -\frac {1}{11} \int \frac {11 (49-30 x)}{\sqrt {1-2 x} (3 x+2) (5 x+3)}dx-\frac {10 \sqrt {1-2 x}}{5 x+3}+\frac {\sqrt {1-2 x}}{(3 x+2) (5 x+3)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\int \frac {49-30 x}{\sqrt {1-2 x} (3 x+2) (5 x+3)}dx-\frac {10 \sqrt {1-2 x}}{5 x+3}+\frac {\sqrt {1-2 x}}{(3 x+2) (5 x+3)}\) |
| \(\Big \downarrow \) 174 |
| \(\displaystyle 207 \int \frac {1}{\sqrt {1-2 x} (3 x+2)}dx-335 \int \frac {1}{\sqrt {1-2 x} (5 x+3)}dx-\frac {10 \sqrt {1-2 x}}{5 x+3}+\frac {\sqrt {1-2 x}}{(3 x+2) (5 x+3)}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle 335 \int \frac {1}{\frac {11}{2}-\frac {5}{2} (1-2 x)}d\sqrt {1-2 x}-207 \int \frac {1}{\frac {7}{2}-\frac {3}{2} (1-2 x)}d\sqrt {1-2 x}-\frac {10 \sqrt {1-2 x}}{5 x+3}+\frac {\sqrt {1-2 x}}{(3 x+2) (5 x+3)}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle -138 \sqrt {\frac {3}{7}} \text {arctanh}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )+134 \sqrt {\frac {5}{11}} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )-\frac {10 \sqrt {1-2 x}}{5 x+3}+\frac {\sqrt {1-2 x}}{(3 x+2) (5 x+3)}\) |
(-10*Sqrt[1 - 2*x])/(3 + 5*x) + Sqrt[1 - 2*x]/((2 + 3*x)*(3 + 5*x)) - 138* Sqrt[3/7]*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]] + 134*Sqrt[5/11]*ArcTanh[Sqrt[5 /11]*Sqrt[1 - 2*x]]
3.19.45.3.1 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 + 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[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_)*((g_.) + (h_.)*(x_)), x_] :> Simp[(b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + S imp[1/((m + 1)*(b*c - a*d)*(b*e - a*f)) Int[(a + b*x)^(m + 1)*(c + d*x)^n *(e + f*x)^p*Simp[(a*d*f*g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a* h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && ILtQ[m, -1]
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 = 3.05 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.70
| method | result | size |
| risch | \(\frac {\left (-1+2 x \right ) \left (19+30 x \right )}{\left (15 x^{2}+19 x +6\right ) \sqrt {1-2 x}}-\frac {138 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{7}+\frac {134 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{11}\) | \(68\) |
| derivativedivides | \(\frac {2 \sqrt {1-2 x}}{-\frac {6}{5}-2 x}+\frac {134 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{11}+\frac {2 \sqrt {1-2 x}}{-\frac {4}{3}-2 x}-\frac {138 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{7}\) | \(70\) |
| default | \(\frac {2 \sqrt {1-2 x}}{-\frac {6}{5}-2 x}+\frac {134 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{11}+\frac {2 \sqrt {1-2 x}}{-\frac {4}{3}-2 x}-\frac {138 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{7}\) | \(70\) |
| pseudoelliptic | \(\frac {-1518 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \left (15 x^{2}+19 x +6\right ) \sqrt {21}+938 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \left (15 x^{2}+19 x +6\right ) \sqrt {55}-77 \sqrt {1-2 x}\, \left (19+30 x \right )}{1155 x^{2}+1463 x +462}\) | \(85\) |
| trager | \(-\frac {\left (19+30 x \right ) \sqrt {1-2 x}}{15 x^{2}+19 x +6}+\frac {67 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) \ln \left (\frac {-5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) x +55 \sqrt {1-2 x}+8 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right )}{3+5 x}\right )}{11}-\frac {69 \operatorname {RootOf}\left (\textit {\_Z}^{2}-21\right ) \ln \left (\frac {-3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-21\right ) x +21 \sqrt {1-2 x}+5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-21\right )}{2+3 x}\right )}{7}\) | \(116\) |
(-1+2*x)*(19+30*x)/(15*x^2+19*x+6)/(1-2*x)^(1/2)-138/7*arctanh(1/7*21^(1/2 )*(1-2*x)^(1/2))*21^(1/2)+134/11*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^( 1/2)
Time = 0.23 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.26 \[ \int \frac {\sqrt {1-2 x}}{(2+3 x)^2 (3+5 x)^2} \, dx=\frac {469 \, \sqrt {11} \sqrt {5} {\left (15 \, x^{2} + 19 \, x + 6\right )} \log \left (-\frac {\sqrt {11} \sqrt {5} \sqrt {-2 \, x + 1} - 5 \, x + 8}{5 \, x + 3}\right ) + 759 \, \sqrt {7} \sqrt {3} {\left (15 \, x^{2} + 19 \, x + 6\right )} \log \left (\frac {\sqrt {7} \sqrt {3} \sqrt {-2 \, x + 1} + 3 \, x - 5}{3 \, x + 2}\right ) - 77 \, {\left (30 \, x + 19\right )} \sqrt {-2 \, x + 1}}{77 \, {\left (15 \, x^{2} + 19 \, x + 6\right )}} \]
1/77*(469*sqrt(11)*sqrt(5)*(15*x^2 + 19*x + 6)*log(-(sqrt(11)*sqrt(5)*sqrt (-2*x + 1) - 5*x + 8)/(5*x + 3)) + 759*sqrt(7)*sqrt(3)*(15*x^2 + 19*x + 6) *log((sqrt(7)*sqrt(3)*sqrt(-2*x + 1) + 3*x - 5)/(3*x + 2)) - 77*(30*x + 19 )*sqrt(-2*x + 1))/(15*x^2 + 19*x + 6)
Time = 28.06 (sec) , antiderivative size = 321, normalized size of antiderivative = 3.31 \[ \int \frac {\sqrt {1-2 x}}{(2+3 x)^2 (3+5 x)^2} \, dx=\frac {68 \sqrt {21} \left (\log {\left (\sqrt {1 - 2 x} - \frac {\sqrt {21}}{3} \right )} - \log {\left (\sqrt {1 - 2 x} + \frac {\sqrt {21}}{3} \right )}\right )}{7} - \frac {68 \sqrt {55} \left (\log {\left (\sqrt {1 - 2 x} - \frac {\sqrt {55}}{5} \right )} - \log {\left (\sqrt {1 - 2 x} + \frac {\sqrt {55}}{5} \right )}\right )}{11} - 84 \left (\begin {cases} \frac {\sqrt {21} \left (- \frac {\log {\left (\frac {\sqrt {21} \sqrt {1 - 2 x}}{7} - 1 \right )}}{4} + \frac {\log {\left (\frac {\sqrt {21} \sqrt {1 - 2 x}}{7} + 1 \right )}}{4} - \frac {1}{4 \left (\frac {\sqrt {21} \sqrt {1 - 2 x}}{7} + 1\right )} - \frac {1}{4 \left (\frac {\sqrt {21} \sqrt {1 - 2 x}}{7} - 1\right )}\right )}{147} & \text {for}\: \sqrt {1 - 2 x} > - \frac {\sqrt {21}}{3} \wedge \sqrt {1 - 2 x} < \frac {\sqrt {21}}{3} \end {cases}\right ) - 220 \left (\begin {cases} \frac {\sqrt {55} \left (- \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1 \right )}}{4} + \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1 \right )}}{4} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1\right )} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1\right )}\right )}{605} & \text {for}\: \sqrt {1 - 2 x} > - \frac {\sqrt {55}}{5} \wedge \sqrt {1 - 2 x} < \frac {\sqrt {55}}{5} \end {cases}\right ) \]
68*sqrt(21)*(log(sqrt(1 - 2*x) - sqrt(21)/3) - log(sqrt(1 - 2*x) + sqrt(21 )/3))/7 - 68*sqrt(55)*(log(sqrt(1 - 2*x) - sqrt(55)/5) - log(sqrt(1 - 2*x) + sqrt(55)/5))/11 - 84*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))) - 220*Piecewise((sqrt(55)*(-l og(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) > -sqrt(55)/5) & (sqrt(1 - 2*x) < sqrt(55)/5)) )
Time = 0.27 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.13 \[ \int \frac {\sqrt {1-2 x}}{(2+3 x)^2 (3+5 x)^2} \, dx=-\frac {67}{11} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) + \frac {69}{7} \, \sqrt {21} \log \left (-\frac {\sqrt {21} - 3 \, \sqrt {-2 \, x + 1}}{\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}}\right ) + \frac {4 \, {\left (15 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 34 \, \sqrt {-2 \, x + 1}\right )}}{15 \, {\left (2 \, x - 1\right )}^{2} + 136 \, x + 9} \]
-67/11*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 69/7*sqrt(21)*log(-(sqrt(21) - 3*sqrt(-2*x + 1))/(sqrt(21) + 3*s qrt(-2*x + 1))) + 4*(15*(-2*x + 1)^(3/2) - 34*sqrt(-2*x + 1))/(15*(2*x - 1 )^2 + 136*x + 9)
Time = 0.30 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.20 \[ \int \frac {\sqrt {1-2 x}}{(2+3 x)^2 (3+5 x)^2} \, dx=-\frac {67}{11} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) + \frac {69}{7} \, \sqrt {21} \log \left (\frac {{\left | -2 \, \sqrt {21} + 6 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}\right )}}\right ) + \frac {4 \, {\left (15 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 34 \, \sqrt {-2 \, x + 1}\right )}}{15 \, {\left (2 \, x - 1\right )}^{2} + 136 \, x + 9} \]
-67/11*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5 *sqrt(-2*x + 1))) + 69/7*sqrt(21)*log(1/2*abs(-2*sqrt(21) + 6*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) + 4*(15*(-2*x + 1)^(3/2) - 34*sqrt(-2*x + 1))/(15*(2*x - 1)^2 + 136*x + 9)
Time = 0.11 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.74 \[ \int \frac {\sqrt {1-2 x}}{(2+3 x)^2 (3+5 x)^2} \, dx=\frac {134\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )}{11}-\frac {138\,\sqrt {21}\,\mathrm {atanh}\left (\frac {\sqrt {21}\,\sqrt {1-2\,x}}{7}\right )}{7}-\frac {\frac {136\,\sqrt {1-2\,x}}{15}-4\,{\left (1-2\,x\right )}^{3/2}}{\frac {136\,x}{15}+{\left (2\,x-1\right )}^2+\frac {3}{5}} \]