Integrand size = 24, antiderivative size = 85 \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x) (3+5 x)} \, dx=-\frac {272}{225} \sqrt {1-2 x}-\frac {4}{45} (1-2 x)^{3/2}+\frac {98}{9} \sqrt {\frac {7}{3}} \text {arctanh}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )-\frac {242}{25} \sqrt {\frac {11}{5}} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \]
-4/45*(1-2*x)^(3/2)-242/125*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)+ 98/27*arctanh(1/7*21^(1/2)*(1-2*x)^(1/2))*21^(1/2)-272/225*(1-2*x)^(1/2)
Time = 0.10 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.84 \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x) (3+5 x)} \, dx=\frac {60 \sqrt {1-2 x} (-73+10 x)+12250 \sqrt {21} \text {arctanh}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )-6534 \sqrt {55} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{3375} \]
(60*Sqrt[1 - 2*x]*(-73 + 10*x) + 12250*Sqrt[21]*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]] - 6534*Sqrt[55]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/3375
Time = 0.19 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.07, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {95, 25, 171, 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.
\(\displaystyle \int \frac {(1-2 x)^{5/2}}{(3 x+2) (5 x+3)} \, dx\) |
\(\Big \downarrow \) 95 |
\(\displaystyle \frac {1}{15} \int -\frac {\sqrt {1-2 x} (136 x+9)}{(3 x+2) (5 x+3)}dx-\frac {4}{45} (1-2 x)^{3/2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {1}{15} \int \frac {\sqrt {1-2 x} (136 x+9)}{(3 x+2) (5 x+3)}dx-\frac {4}{45} (1-2 x)^{3/2}\) |
\(\Big \downarrow \) 171 |
\(\displaystyle \frac {1}{15} \left (-\frac {2}{15} \int \frac {6938 x+1767}{2 \sqrt {1-2 x} (3 x+2) (5 x+3)}dx-\frac {272}{15} \sqrt {1-2 x}\right )-\frac {4}{45} (1-2 x)^{3/2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{15} \left (-\frac {1}{15} \int \frac {6938 x+1767}{\sqrt {1-2 x} (3 x+2) (5 x+3)}dx-\frac {272}{15} \sqrt {1-2 x}\right )-\frac {4}{45} (1-2 x)^{3/2}\) |
\(\Big \downarrow \) 174 |
\(\displaystyle \frac {1}{15} \left (\frac {1}{15} \left (11979 \int \frac {1}{\sqrt {1-2 x} (5 x+3)}dx-8575 \int \frac {1}{\sqrt {1-2 x} (3 x+2)}dx\right )-\frac {272}{15} \sqrt {1-2 x}\right )-\frac {4}{45} (1-2 x)^{3/2}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{15} \left (\frac {1}{15} \left (8575 \int \frac {1}{\frac {7}{2}-\frac {3}{2} (1-2 x)}d\sqrt {1-2 x}-11979 \int \frac {1}{\frac {11}{2}-\frac {5}{2} (1-2 x)}d\sqrt {1-2 x}\right )-\frac {272}{15} \sqrt {1-2 x}\right )-\frac {4}{45} (1-2 x)^{3/2}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{15} \left (\frac {1}{15} \left (2450 \sqrt {\frac {7}{3}} \text {arctanh}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )-2178 \sqrt {\frac {11}{5}} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )\right )-\frac {272}{15} \sqrt {1-2 x}\right )-\frac {4}{45} (1-2 x)^{3/2}\) |
(-4*(1 - 2*x)^(3/2))/45 + ((-272*Sqrt[1 - 2*x])/15 + (2450*Sqrt[7/3]*ArcTa nh[Sqrt[3/7]*Sqrt[1 - 2*x]] - 2178*Sqrt[11/5]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/15)/15
3.20.73.3.1 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[((e_.) + (f_.)*(x_))^(p_)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_] :> Simp[f*((e + f*x)^(p - 1)/(b*d*(p - 1))), x] + Simp[1/(b*d) Int[(b *d*e^2 - a*c*f^2 + f*(2*b*d*e - b*c*f - a*d*f)*x)*((e + f*x)^(p - 2)/((a + b*x)*(c + d*x))), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && GtQ[p, 1]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_)*((g_.) + (h_.)*(x_)), x_] :> Simp[h*(a + b*x)^m*(c + d*x)^(n + 1)*(( e + f*x)^(p + 1)/(d*f*(m + n + p + 2))), x] + Simp[1/(d*f*(m + n + p + 2)) Int[(a + b*x)^(m - 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*g*(m + n + p + 2 ) - h*(b*c*e*m + a*(d*e*(n + 1) + c*f*(p + 1))) + (b*d*f*g*(m + n + p + 2) + h*(a*d*f*m - b*(d*e*(m + n + 1) + c*f*(m + p + 1))))*x, x], x], x] /; Fre eQ[{a, b, c, d, e, f, g, h, n, p}, x] && GtQ[m, 0] && NeQ[m + n + p + 2, 0] && IntegersQ[2*m, 2*n, 2*p]
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 = 1.04 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.61
method | result | size |
pseudoelliptic | \(-\frac {242 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{125}+\frac {98 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{27}+\frac {4 \sqrt {1-2 x}\, \left (10 x -73\right )}{225}\) | \(52\) |
derivativedivides | \(-\frac {4 \left (1-2 x \right )^{\frac {3}{2}}}{45}-\frac {242 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{125}+\frac {98 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{27}-\frac {272 \sqrt {1-2 x}}{225}\) | \(56\) |
default | \(-\frac {4 \left (1-2 x \right )^{\frac {3}{2}}}{45}-\frac {242 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{125}+\frac {98 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{27}-\frac {272 \sqrt {1-2 x}}{225}\) | \(56\) |
risch | \(-\frac {4 \left (10 x -73\right ) \left (-1+2 x \right )}{225 \sqrt {1-2 x}}+\frac {98 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{27}-\frac {242 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{125}\) | \(57\) |
trager | \(\left (\frac {8 x}{45}-\frac {292}{225}\right ) \sqrt {1-2 x}-\frac {49 \operatorname {RootOf}\left (\textit {\_Z}^{2}-21\right ) \ln \left (\frac {3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-21\right ) x -5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-21\right )+21 \sqrt {1-2 x}}{2+3 x}\right )}{27}+\frac {121 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) \ln \left (\frac {5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) x +55 \sqrt {1-2 x}-8 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right )}{3+5 x}\right )}{125}\) | \(103\) |
-242/125*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)+98/27*arctanh(1/7*2 1^(1/2)*(1-2*x)^(1/2))*21^(1/2)+4/225*(1-2*x)^(1/2)*(10*x-73)
Time = 0.24 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.04 \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x) (3+5 x)} \, dx=\frac {121}{125} \, \sqrt {11} \sqrt {5} \log \left (\frac {\sqrt {11} \sqrt {5} \sqrt {-2 \, x + 1} + 5 \, x - 8}{5 \, x + 3}\right ) + \frac {49}{27} \, \sqrt {7} \sqrt {3} \log \left (-\frac {\sqrt {7} \sqrt {3} \sqrt {-2 \, x + 1} - 3 \, x + 5}{3 \, x + 2}\right ) + \frac {4}{225} \, {\left (10 \, x - 73\right )} \sqrt {-2 \, x + 1} \]
121/125*sqrt(11)*sqrt(5)*log((sqrt(11)*sqrt(5)*sqrt(-2*x + 1) + 5*x - 8)/( 5*x + 3)) + 49/27*sqrt(7)*sqrt(3)*log(-(sqrt(7)*sqrt(3)*sqrt(-2*x + 1) - 3 *x + 5)/(3*x + 2)) + 4/225*(10*x - 73)*sqrt(-2*x + 1)
Time = 1.31 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.26 \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x) (3+5 x)} \, dx=- \frac {4 \left (1 - 2 x\right )^{\frac {3}{2}}}{45} - \frac {272 \sqrt {1 - 2 x}}{225} - \frac {49 \sqrt {21} \left (\log {\left (\sqrt {1 - 2 x} - \frac {\sqrt {21}}{3} \right )} - \log {\left (\sqrt {1 - 2 x} + \frac {\sqrt {21}}{3} \right )}\right )}{27} + \frac {121 \sqrt {55} \left (\log {\left (\sqrt {1 - 2 x} - \frac {\sqrt {55}}{5} \right )} - \log {\left (\sqrt {1 - 2 x} + \frac {\sqrt {55}}{5} \right )}\right )}{125} \]
-4*(1 - 2*x)**(3/2)/45 - 272*sqrt(1 - 2*x)/225 - 49*sqrt(21)*(log(sqrt(1 - 2*x) - sqrt(21)/3) - log(sqrt(1 - 2*x) + sqrt(21)/3))/27 + 121*sqrt(55)*( log(sqrt(1 - 2*x) - sqrt(55)/5) - log(sqrt(1 - 2*x) + sqrt(55)/5))/125
Time = 0.29 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.07 \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x) (3+5 x)} \, dx=-\frac {4}{45} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {121}{125} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) - \frac {49}{27} \, \sqrt {21} \log \left (-\frac {\sqrt {21} - 3 \, \sqrt {-2 \, x + 1}}{\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}}\right ) - \frac {272}{225} \, \sqrt {-2 \, x + 1} \]
-4/45*(-2*x + 1)^(3/2) + 121/125*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1 ))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 49/27*sqrt(21)*log(-(sqrt(21) - 3*sqrt (-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) - 272/225*sqrt(-2*x + 1)
Time = 0.28 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.14 \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x) (3+5 x)} \, dx=-\frac {4}{45} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {121}{125} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) - \frac {49}{27} \, \sqrt {21} \log \left (\frac {{\left | -2 \, \sqrt {21} + 6 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}\right )}}\right ) - \frac {272}{225} \, \sqrt {-2 \, x + 1} \]
-4/45*(-2*x + 1)^(3/2) + 121/125*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqr t(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 49/27*sqrt(21)*log(1/2*abs(- 2*sqrt(21) + 6*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) - 272/225*sq rt(-2*x + 1)
Time = 1.56 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.69 \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x) (3+5 x)} \, dx=-\frac {272\,\sqrt {1-2\,x}}{225}-\frac {4\,{\left (1-2\,x\right )}^{3/2}}{45}-\frac {\sqrt {21}\,\mathrm {atan}\left (\frac {\sqrt {21}\,\sqrt {1-2\,x}\,1{}\mathrm {i}}{7}\right )\,98{}\mathrm {i}}{27}+\frac {\sqrt {55}\,\mathrm {atan}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}\,1{}\mathrm {i}}{11}\right )\,242{}\mathrm {i}}{125} \]