Integrand size = 24, antiderivative size = 72 \[ \int \frac {(1-2 x)^{3/2}}{(2+3 x) (3+5 x)} \, dx=-\frac {4}{15} \sqrt {1-2 x}+\frac {14}{3} \sqrt {\frac {7}{3}} \text {arctanh}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )-\frac {22}{5} \sqrt {\frac {11}{5}} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \]
-22/25*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)+14/9*arctanh(1/7*21^( 1/2)*(1-2*x)^(1/2))*21^(1/2)-4/15*(1-2*x)^(1/2)
Time = 0.09 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.92 \[ \int \frac {(1-2 x)^{3/2}}{(2+3 x) (3+5 x)} \, dx=-\frac {2}{225} \left (30 \sqrt {1-2 x}-175 \sqrt {21} \text {arctanh}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )+99 \sqrt {55} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )\right ) \]
(-2*(30*Sqrt[1 - 2*x] - 175*Sqrt[21]*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]] + 99 *Sqrt[55]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]]))/225
Time = 0.17 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.01, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {95, 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.
\(\displaystyle \int \frac {(1-2 x)^{3/2}}{(3 x+2) (5 x+3)} \, dx\) |
\(\Big \downarrow \) 95 |
\(\displaystyle \frac {1}{15} \int -\frac {136 x+9}{\sqrt {1-2 x} (3 x+2) (5 x+3)}dx-\frac {4}{15} \sqrt {1-2 x}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {1}{15} \int \frac {136 x+9}{\sqrt {1-2 x} (3 x+2) (5 x+3)}dx-\frac {4}{15} \sqrt {1-2 x}\) |
\(\Big \downarrow \) 174 |
\(\displaystyle \frac {1}{15} \left (363 \int \frac {1}{\sqrt {1-2 x} (5 x+3)}dx-245 \int \frac {1}{\sqrt {1-2 x} (3 x+2)}dx\right )-\frac {4}{15} \sqrt {1-2 x}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{15} \left (245 \int \frac {1}{\frac {7}{2}-\frac {3}{2} (1-2 x)}d\sqrt {1-2 x}-363 \int \frac {1}{\frac {11}{2}-\frac {5}{2} (1-2 x)}d\sqrt {1-2 x}\right )-\frac {4}{15} \sqrt {1-2 x}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{15} \left (70 \sqrt {\frac {7}{3}} \text {arctanh}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )-66 \sqrt {\frac {11}{5}} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )\right )-\frac {4}{15} \sqrt {1-2 x}\) |
(-4*Sqrt[1 - 2*x])/15 + (70*Sqrt[7/3]*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]] - 6 6*Sqrt[11/5]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/15
3.20.2.3.1 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[((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[(((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 = 47, normalized size of antiderivative = 0.65
method | result | size |
derivativedivides | \(-\frac {22 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{25}+\frac {14 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{9}-\frac {4 \sqrt {1-2 x}}{15}\) | \(47\) |
default | \(-\frac {22 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{25}+\frac {14 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{9}-\frac {4 \sqrt {1-2 x}}{15}\) | \(47\) |
pseudoelliptic | \(-\frac {22 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{25}+\frac {14 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{9}-\frac {4 \sqrt {1-2 x}}{15}\) | \(47\) |
risch | \(\frac {-\frac {4}{15}+\frac {8 x}{15}}{\sqrt {1-2 x}}+\frac {14 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{9}-\frac {22 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{25}\) | \(52\) |
trager | \(-\frac {4 \sqrt {1-2 x}}{15}+\frac {7 \operatorname {RootOf}\left (\textit {\_Z}^{2}-21\right ) \ln \left (\frac {-3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-21\right ) x +21 \sqrt {1-2 x}+5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-21\right )}{2+3 x}\right )}{9}+\frac {11 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) \ln \left (\frac {5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) x +55 \sqrt {1-2 x}-8 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right )}{3+5 x}\right )}{25}\) | \(99\) |
-22/25*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)+14/9*arctanh(1/7*21^( 1/2)*(1-2*x)^(1/2))*21^(1/2)-4/15*(1-2*x)^(1/2)
Time = 0.24 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.15 \[ \int \frac {(1-2 x)^{3/2}}{(2+3 x) (3+5 x)} \, dx=\frac {11}{25} \, \sqrt {11} \sqrt {5} \log \left (\frac {\sqrt {11} \sqrt {5} \sqrt {-2 \, x + 1} + 5 \, x - 8}{5 \, x + 3}\right ) + \frac {7}{9} \, \sqrt {7} \sqrt {3} \log \left (-\frac {\sqrt {7} \sqrt {3} \sqrt {-2 \, x + 1} - 3 \, x + 5}{3 \, x + 2}\right ) - \frac {4}{15} \, \sqrt {-2 \, x + 1} \]
11/25*sqrt(11)*sqrt(5)*log((sqrt(11)*sqrt(5)*sqrt(-2*x + 1) + 5*x - 8)/(5* x + 3)) + 7/9*sqrt(7)*sqrt(3)*log(-(sqrt(7)*sqrt(3)*sqrt(-2*x + 1) - 3*x + 5)/(3*x + 2)) - 4/15*sqrt(-2*x + 1)
Time = 1.27 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.32 \[ \int \frac {(1-2 x)^{3/2}}{(2+3 x) (3+5 x)} \, dx=- \frac {4 \sqrt {1 - 2 x}}{15} - \frac {7 \sqrt {21} \left (\log {\left (\sqrt {1 - 2 x} - \frac {\sqrt {21}}{3} \right )} - \log {\left (\sqrt {1 - 2 x} + \frac {\sqrt {21}}{3} \right )}\right )}{9} + \frac {11 \sqrt {55} \left (\log {\left (\sqrt {1 - 2 x} - \frac {\sqrt {55}}{5} \right )} - \log {\left (\sqrt {1 - 2 x} + \frac {\sqrt {55}}{5} \right )}\right )}{25} \]
-4*sqrt(1 - 2*x)/15 - 7*sqrt(21)*(log(sqrt(1 - 2*x) - sqrt(21)/3) - log(sq rt(1 - 2*x) + sqrt(21)/3))/9 + 11*sqrt(55)*(log(sqrt(1 - 2*x) - sqrt(55)/5 ) - log(sqrt(1 - 2*x) + sqrt(55)/5))/25
Time = 0.29 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.14 \[ \int \frac {(1-2 x)^{3/2}}{(2+3 x) (3+5 x)} \, dx=\frac {11}{25} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) - \frac {7}{9} \, \sqrt {21} \log \left (-\frac {\sqrt {21} - 3 \, \sqrt {-2 \, x + 1}}{\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}}\right ) - \frac {4}{15} \, \sqrt {-2 \, x + 1} \]
11/25*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 7/9*sqrt(21)*log(-(sqrt(21) - 3*sqrt(-2*x + 1))/(sqrt(21) + 3*sqr t(-2*x + 1))) - 4/15*sqrt(-2*x + 1)
Time = 0.29 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.22 \[ \int \frac {(1-2 x)^{3/2}}{(2+3 x) (3+5 x)} \, dx=\frac {11}{25} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) - \frac {7}{9} \, \sqrt {21} \log \left (\frac {{\left | -2 \, \sqrt {21} + 6 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}\right )}}\right ) - \frac {4}{15} \, \sqrt {-2 \, x + 1} \]
11/25*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5* sqrt(-2*x + 1))) - 7/9*sqrt(21)*log(1/2*abs(-2*sqrt(21) + 6*sqrt(-2*x + 1) )/(sqrt(21) + 3*sqrt(-2*x + 1))) - 4/15*sqrt(-2*x + 1)
Time = 0.11 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.64 \[ \int \frac {(1-2 x)^{3/2}}{(2+3 x) (3+5 x)} \, dx=\frac {14\,\sqrt {21}\,\mathrm {atanh}\left (\frac {\sqrt {21}\,\sqrt {1-2\,x}}{7}\right )}{9}-\frac {22\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )}{25}-\frac {4\,\sqrt {1-2\,x}}{15} \]