Integrand size = 24, antiderivative size = 93 \[ \int \frac {\sqrt {1-2 x}}{(2+3 x) (3+5 x)^3} \, dx=-\frac {\sqrt {1-2 x}}{2 (3+5 x)^2}+\frac {67 \sqrt {1-2 x}}{22 (3+5 x)}+6 \sqrt {21} \text {arctanh}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )-\frac {2243 \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{11 \sqrt {55}} \]
6*arctanh(1/7*21^(1/2)*(1-2*x)^(1/2))*21^(1/2)-2243/605*arctanh(1/11*55^(1 /2)*(1-2*x)^(1/2))*55^(1/2)-1/2*(1-2*x)^(1/2)/(3+5*x)^2+67/22*(1-2*x)^(1/2 )/(3+5*x)
Time = 0.22 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.84 \[ \int \frac {\sqrt {1-2 x}}{(2+3 x) (3+5 x)^3} \, dx=\frac {5 \sqrt {1-2 x} (38+67 x)}{22 (3+5 x)^2}+6 \sqrt {21} \text {arctanh}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )-\frac {2243 \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{11 \sqrt {55}} \]
(5*Sqrt[1 - 2*x]*(38 + 67*x))/(22*(3 + 5*x)^2) + 6*Sqrt[21]*ArcTanh[Sqrt[3 /7]*Sqrt[1 - 2*x]] - (2243*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(11*Sqrt[55] )
Time = 0.20 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.09, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {110, 25, 168, 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.
\(\displaystyle \int \frac {\sqrt {1-2 x}}{(3 x+2) (5 x+3)^3} \, dx\) |
\(\Big \downarrow \) 110 |
\(\displaystyle \frac {1}{2} \int -\frac {8-9 x}{\sqrt {1-2 x} (3 x+2) (5 x+3)^2}dx-\frac {\sqrt {1-2 x}}{2 (5 x+3)^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {1}{2} \int \frac {8-9 x}{\sqrt {1-2 x} (3 x+2) (5 x+3)^2}dx-\frac {\sqrt {1-2 x}}{2 (5 x+3)^2}\) |
\(\Big \downarrow \) 168 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{11} \int \frac {328-201 x}{\sqrt {1-2 x} (3 x+2) (5 x+3)}dx+\frac {67 \sqrt {1-2 x}}{11 (5 x+3)}\right )-\frac {\sqrt {1-2 x}}{2 (5 x+3)^2}\) |
\(\Big \downarrow \) 174 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{11} \left (2243 \int \frac {1}{\sqrt {1-2 x} (5 x+3)}dx-1386 \int \frac {1}{\sqrt {1-2 x} (3 x+2)}dx\right )+\frac {67 \sqrt {1-2 x}}{11 (5 x+3)}\right )-\frac {\sqrt {1-2 x}}{2 (5 x+3)^2}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{11} \left (1386 \int \frac {1}{\frac {7}{2}-\frac {3}{2} (1-2 x)}d\sqrt {1-2 x}-2243 \int \frac {1}{\frac {11}{2}-\frac {5}{2} (1-2 x)}d\sqrt {1-2 x}\right )+\frac {67 \sqrt {1-2 x}}{11 (5 x+3)}\right )-\frac {\sqrt {1-2 x}}{2 (5 x+3)^2}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{11} \left (132 \sqrt {21} \text {arctanh}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )-\frac {4486 \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{\sqrt {55}}\right )+\frac {67 \sqrt {1-2 x}}{11 (5 x+3)}\right )-\frac {\sqrt {1-2 x}}{2 (5 x+3)^2}\) |
-1/2*Sqrt[1 - 2*x]/(3 + 5*x)^2 + ((67*Sqrt[1 - 2*x])/(11*(3 + 5*x)) + (132 *Sqrt[21]*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]] - (4486*ArcTanh[Sqrt[5/11]*Sqrt [1 - 2*x]])/Sqrt[55])/11)/2
3.19.53.3.1 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[((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 = 1.02 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.69
method | result | size |
risch | \(-\frac {5 \left (134 x^{2}+9 x -38\right )}{22 \left (3+5 x \right )^{2} \sqrt {1-2 x}}+6 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}-\frac {2243 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{605}\) | \(64\) |
derivativedivides | \(\frac {-\frac {335 \left (1-2 x \right )^{\frac {3}{2}}}{11}+65 \sqrt {1-2 x}}{\left (-6-10 x \right )^{2}}-\frac {2243 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{605}+6 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}\) | \(66\) |
default | \(\frac {-\frac {335 \left (1-2 x \right )^{\frac {3}{2}}}{11}+65 \sqrt {1-2 x}}{\left (-6-10 x \right )^{2}}-\frac {2243 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{605}+6 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}\) | \(66\) |
pseudoelliptic | \(\frac {7260 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \left (3+5 x \right )^{2} \sqrt {21}-4486 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \left (3+5 x \right )^{2} \sqrt {55}+275 \sqrt {1-2 x}\, \left (67 x +38\right )}{1210 \left (3+5 x \right )^{2}}\) | \(75\) |
trager | \(\frac {5 \left (67 x +38\right ) \sqrt {1-2 x}}{22 \left (3+5 x \right )^{2}}+\frac {2243 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) \ln \left (\frac {5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) x +55 \sqrt {1-2 x}-8 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right )}{3+5 x}\right )}{1210}+3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-21\right ) \ln \left (\frac {-3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-21\right ) x +21 \sqrt {1-2 x}+5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-21\right )}{2+3 x}\right )\) | \(111\) |
-5/22*(134*x^2+9*x-38)/(3+5*x)^2/(1-2*x)^(1/2)+6*arctanh(1/7*21^(1/2)*(1-2 *x)^(1/2))*21^(1/2)-2243/605*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)
Time = 0.23 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.18 \[ \int \frac {\sqrt {1-2 x}}{(2+3 x) (3+5 x)^3} \, dx=\frac {2243 \, \sqrt {55} {\left (25 \, x^{2} + 30 \, x + 9\right )} \log \left (\frac {5 \, x + \sqrt {55} \sqrt {-2 \, x + 1} - 8}{5 \, x + 3}\right ) + 3630 \, \sqrt {21} {\left (25 \, x^{2} + 30 \, x + 9\right )} \log \left (\frac {3 \, x - \sqrt {21} \sqrt {-2 \, x + 1} - 5}{3 \, x + 2}\right ) + 275 \, {\left (67 \, x + 38\right )} \sqrt {-2 \, x + 1}}{1210 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} \]
1/1210*(2243*sqrt(55)*(25*x^2 + 30*x + 9)*log((5*x + sqrt(55)*sqrt(-2*x + 1) - 8)/(5*x + 3)) + 3630*sqrt(21)*(25*x^2 + 30*x + 9)*log((3*x - sqrt(21) *sqrt(-2*x + 1) - 5)/(3*x + 2)) + 275*(67*x + 38)*sqrt(-2*x + 1))/(25*x^2 + 30*x + 9)
Leaf count of result is larger than twice the leaf count of optimal. 165 vs. \(2 (80) = 160\).
Time = 38.30 (sec) , antiderivative size = 367, normalized size of antiderivative = 3.95 \[ \int \frac {\sqrt {1-2 x}}{(2+3 x) (3+5 x)^3} \, dx=- 3 \sqrt {21} \left (\log {\left (\sqrt {1 - 2 x} - \frac {\sqrt {21}}{3} \right )} - \log {\left (\sqrt {1 - 2 x} + \frac {\sqrt {21}}{3} \right )}\right ) + \frac {21 \sqrt {55} \left (\log {\left (\sqrt {1 - 2 x} - \frac {\sqrt {55}}{5} \right )} - \log {\left (\sqrt {1 - 2 x} + \frac {\sqrt {55}}{5} \right )}\right )}{11} + 140 \left (\begin {cases} \frac {\sqrt {55} \left (- \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1 \right )}}{4} + \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1 \right )}}{4} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1\right )} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1\right )}\right )}{605} & \text {for}\: \sqrt {1 - 2 x} > - \frac {\sqrt {55}}{5} \wedge \sqrt {1 - 2 x} < \frac {\sqrt {55}}{5} \end {cases}\right ) + 88 \left (\begin {cases} \frac {\sqrt {55} \cdot \left (\frac {3 \log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1 \right )}}{16} - \frac {3 \log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1 \right )}}{16} + \frac {3}{16 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1\right )} + \frac {1}{16 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1\right )^{2}} + \frac {3}{16 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1\right )} - \frac {1}{16 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1\right )^{2}}\right )}{6655} & \text {for}\: \sqrt {1 - 2 x} > - \frac {\sqrt {55}}{5} \wedge \sqrt {1 - 2 x} < \frac {\sqrt {55}}{5} \end {cases}\right ) \]
-3*sqrt(21)*(log(sqrt(1 - 2*x) - sqrt(21)/3) - log(sqrt(1 - 2*x) + sqrt(21 )/3)) + 21*sqrt(55)*(log(sqrt(1 - 2*x) - sqrt(55)/5) - log(sqrt(1 - 2*x) + sqrt(55)/5))/11 + 140*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)/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))) + 88*Piecewise((sqrt(55)*( 3*log(sqrt(55)*sqrt(1 - 2*x)/11 - 1)/16 - 3*log(sqrt(55)*sqrt(1 - 2*x)/11 + 1)/16 + 3/(16*(sqrt(55)*sqrt(1 - 2*x)/11 + 1)) + 1/(16*(sqrt(55)*sqrt(1 - 2*x)/11 + 1)**2) + 3/(16*(sqrt(55)*sqrt(1 - 2*x)/11 - 1)) - 1/(16*(sqrt( 55)*sqrt(1 - 2*x)/11 - 1)**2))/6655, (sqrt(1 - 2*x) > -sqrt(55)/5) & (sqrt (1 - 2*x) < sqrt(55)/5)))
Time = 0.31 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.18 \[ \int \frac {\sqrt {1-2 x}}{(2+3 x) (3+5 x)^3} \, dx=\frac {2243}{1210} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) - 3 \, \sqrt {21} \log \left (-\frac {\sqrt {21} - 3 \, \sqrt {-2 \, x + 1}}{\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}}\right ) - \frac {5 \, {\left (67 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 143 \, \sqrt {-2 \, x + 1}\right )}}{11 \, {\left (25 \, {\left (2 \, x - 1\right )}^{2} + 220 \, x + 11\right )}} \]
2243/1210*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(- 2*x + 1))) - 3*sqrt(21)*log(-(sqrt(21) - 3*sqrt(-2*x + 1))/(sqrt(21) + 3*s qrt(-2*x + 1))) - 5/11*(67*(-2*x + 1)^(3/2) - 143*sqrt(-2*x + 1))/(25*(2*x - 1)^2 + 220*x + 11)
Time = 0.29 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.15 \[ \int \frac {\sqrt {1-2 x}}{(2+3 x) (3+5 x)^3} \, dx=\frac {2243}{1210} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) - 3 \, \sqrt {21} \log \left (\frac {{\left | -2 \, \sqrt {21} + 6 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}\right )}}\right ) - \frac {5 \, {\left (67 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 143 \, \sqrt {-2 \, x + 1}\right )}}{44 \, {\left (5 \, x + 3\right )}^{2}} \]
2243/1210*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 3*sqrt(21)*log(1/2*abs(-2*sqrt(21) + 6*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) - 5/44*(67*(-2*x + 1)^(3/2) - 143*sqrt( -2*x + 1))/(5*x + 3)^2
Time = 0.10 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.76 \[ \int \frac {\sqrt {1-2 x}}{(2+3 x) (3+5 x)^3} \, dx=6\,\sqrt {21}\,\mathrm {atanh}\left (\frac {\sqrt {21}\,\sqrt {1-2\,x}}{7}\right )-\frac {2243\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )}{605}+\frac {\frac {13\,\sqrt {1-2\,x}}{5}-\frac {67\,{\left (1-2\,x\right )}^{3/2}}{55}}{\frac {44\,x}{5}+{\left (2\,x-1\right )}^2+\frac {11}{25}} \]