Integrand size = 24, antiderivative size = 92 \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x)^2 (3+5 x)} \, dx=\frac {26}{15} \sqrt {1-2 x}+\frac {7 (1-2 x)^{3/2}}{3 (2+3 x)}+\frac {140}{3} \sqrt {\frac {7}{3}} \text {arctanh}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )-\frac {242}{5} \sqrt {\frac {11}{5}} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \]
7/3*(1-2*x)^(3/2)/(2+3*x)-242/25*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^( 1/2)+140/9*arctanh(1/7*21^(1/2)*(1-2*x)^(1/2))*21^(1/2)+26/15*(1-2*x)^(1/2 )
Time = 0.16 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.85 \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x)^2 (3+5 x)} \, dx=\frac {1}{225} \left (\frac {15 \sqrt {1-2 x} (87+8 x)}{2+3 x}+3500 \sqrt {21} \text {arctanh}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )-2178 \sqrt {55} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )\right ) \]
((15*Sqrt[1 - 2*x]*(87 + 8*x))/(2 + 3*x) + 3500*Sqrt[21]*ArcTanh[Sqrt[3/7] *Sqrt[1 - 2*x]] - 2178*Sqrt[55]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/225
Time = 0.20 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.01, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {109, 27, 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.
\(\displaystyle \int \frac {(1-2 x)^{5/2}}{(3 x+2)^2 (5 x+3)} \, dx\) |
\(\Big \downarrow \) 109 |
\(\displaystyle \frac {1}{3} \int \frac {3 \sqrt {1-2 x} (13 x+32)}{(3 x+2) (5 x+3)}dx+\frac {7 (1-2 x)^{3/2}}{3 (3 x+2)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {\sqrt {1-2 x} (13 x+32)}{(3 x+2) (5 x+3)}dx+\frac {7 (1-2 x)^{3/2}}{3 (3 x+2)}\) |
\(\Big \downarrow \) 171 |
\(\displaystyle \frac {2}{15} \int \frac {636-271 x}{2 \sqrt {1-2 x} (3 x+2) (5 x+3)}dx+\frac {7 (1-2 x)^{3/2}}{3 (3 x+2)}+\frac {26}{15} \sqrt {1-2 x}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{15} \int \frac {636-271 x}{\sqrt {1-2 x} (3 x+2) (5 x+3)}dx+\frac {7 (1-2 x)^{3/2}}{3 (3 x+2)}+\frac {26}{15} \sqrt {1-2 x}\) |
\(\Big \downarrow \) 174 |
\(\displaystyle \frac {1}{15} \left (3993 \int \frac {1}{\sqrt {1-2 x} (5 x+3)}dx-2450 \int \frac {1}{\sqrt {1-2 x} (3 x+2)}dx\right )+\frac {7 (1-2 x)^{3/2}}{3 (3 x+2)}+\frac {26}{15} \sqrt {1-2 x}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{15} \left (2450 \int \frac {1}{\frac {7}{2}-\frac {3}{2} (1-2 x)}d\sqrt {1-2 x}-3993 \int \frac {1}{\frac {11}{2}-\frac {5}{2} (1-2 x)}d\sqrt {1-2 x}\right )+\frac {7 (1-2 x)^{3/2}}{3 (3 x+2)}+\frac {26}{15} \sqrt {1-2 x}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{15} \left (700 \sqrt {\frac {7}{3}} \text {arctanh}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )-726 \sqrt {\frac {11}{5}} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )\right )+\frac {7 (1-2 x)^{3/2}}{3 (3 x+2)}+\frac {26}{15} \sqrt {1-2 x}\) |
(26*Sqrt[1 - 2*x])/15 + (7*(1 - 2*x)^(3/2))/(3*(2 + 3*x)) + (700*Sqrt[7/3] *ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]] - 726*Sqrt[11/5]*ArcTanh[Sqrt[5/11]*Sqrt [1 - 2*x]])/15
3.20.74.3.1 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[(b*c - a*d)*(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*((e + f *x)^(p + 1)/(b*(b*e - a*f)*(m + 1))), x] + Simp[1/(b*(b*e - a*f)*(m + 1)) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 2)*(e + f*x)^p*Simp[a*d*(d*e*(n - 1) + c*f*(p + 1)) + b*c*(d*e*(m - n + 2) - c*f*(m + p + 2)) + d*(a*d*f*(n + p) + b*(d*e*(m + 1) - c*f*(m + n + p + 1)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && LtQ[m, -1] && GtQ[n, 1] && (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[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 = 1.06 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.68
method | result | size |
derivativedivides | \(\frac {8 \sqrt {1-2 x}}{45}-\frac {242 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{25}-\frac {98 \sqrt {1-2 x}}{27 \left (-\frac {4}{3}-2 x \right )}+\frac {140 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{9}\) | \(63\) |
default | \(\frac {8 \sqrt {1-2 x}}{45}-\frac {242 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{25}-\frac {98 \sqrt {1-2 x}}{27 \left (-\frac {4}{3}-2 x \right )}+\frac {140 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{9}\) | \(63\) |
risch | \(-\frac {16 x^{2}+166 x -87}{15 \left (2+3 x \right ) \sqrt {1-2 x}}+\frac {140 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{9}-\frac {242 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{25}\) | \(64\) |
pseudoelliptic | \(\frac {3500 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \left (2+3 x \right ) \sqrt {21}-2178 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \left (2+3 x \right ) \sqrt {55}+15 \sqrt {1-2 x}\, \left (87+8 x \right )}{450+675 x}\) | \(70\) |
trager | \(\frac {\sqrt {1-2 x}\, \left (87+8 x \right )}{30+45 x}+\frac {70 \operatorname {RootOf}\left (\textit {\_Z}^{2}-21\right ) \ln \left (\frac {-3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-21\right ) x +21 \sqrt {1-2 x}+5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-21\right )}{2+3 x}\right )}{9}-\frac {121 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) \ln \left (\frac {-5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) x +55 \sqrt {1-2 x}+8 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right )}{3+5 x}\right )}{25}\) | \(111\) |
8/45*(1-2*x)^(1/2)-242/25*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)-98 /27*(1-2*x)^(1/2)/(-4/3-2*x)+140/9*arctanh(1/7*21^(1/2)*(1-2*x)^(1/2))*21^ (1/2)
Time = 0.23 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.16 \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x)^2 (3+5 x)} \, dx=\frac {1089 \, \sqrt {11} \sqrt {5} {\left (3 \, x + 2\right )} \log \left (\frac {\sqrt {11} \sqrt {5} \sqrt {-2 \, x + 1} + 5 \, x - 8}{5 \, x + 3}\right ) + 1750 \, \sqrt {7} \sqrt {3} {\left (3 \, x + 2\right )} \log \left (-\frac {\sqrt {7} \sqrt {3} \sqrt {-2 \, x + 1} - 3 \, x + 5}{3 \, x + 2}\right ) + 15 \, {\left (8 \, x + 87\right )} \sqrt {-2 \, x + 1}}{225 \, {\left (3 \, x + 2\right )}} \]
1/225*(1089*sqrt(11)*sqrt(5)*(3*x + 2)*log((sqrt(11)*sqrt(5)*sqrt(-2*x + 1 ) + 5*x - 8)/(5*x + 3)) + 1750*sqrt(7)*sqrt(3)*(3*x + 2)*log(-(sqrt(7)*sqr t(3)*sqrt(-2*x + 1) - 3*x + 5)/(3*x + 2)) + 15*(8*x + 87)*sqrt(-2*x + 1))/ (3*x + 2)
Time = 14.26 (sec) , antiderivative size = 216, normalized size of antiderivative = 2.35 \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x)^2 (3+5 x)} \, dx=\frac {8 \sqrt {1 - 2 x}}{45} - \frac {203 \sqrt {21} \left (\log {\left (\sqrt {1 - 2 x} - \frac {\sqrt {21}}{3} \right )} - \log {\left (\sqrt {1 - 2 x} + \frac {\sqrt {21}}{3} \right )}\right )}{27} + \frac {121 \sqrt {55} \left (\log {\left (\sqrt {1 - 2 x} - \frac {\sqrt {55}}{5} \right )} - \log {\left (\sqrt {1 - 2 x} + \frac {\sqrt {55}}{5} \right )}\right )}{25} + \frac {1372 \left (\begin {cases} \frac {\sqrt {21} \left (- \frac {\log {\left (\frac {\sqrt {21} \sqrt {1 - 2 x}}{7} - 1 \right )}}{4} + \frac {\log {\left (\frac {\sqrt {21} \sqrt {1 - 2 x}}{7} + 1 \right )}}{4} - \frac {1}{4 \left (\frac {\sqrt {21} \sqrt {1 - 2 x}}{7} + 1\right )} - \frac {1}{4 \left (\frac {\sqrt {21} \sqrt {1 - 2 x}}{7} - 1\right )}\right )}{147} & \text {for}\: \sqrt {1 - 2 x} > - \frac {\sqrt {21}}{3} \wedge \sqrt {1 - 2 x} < \frac {\sqrt {21}}{3} \end {cases}\right )}{9} \]
8*sqrt(1 - 2*x)/45 - 203*sqrt(21)*(log(sqrt(1 - 2*x) - sqrt(21)/3) - log(s qrt(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))/25 + 1372*Piecewise((sqrt(21)*(-lo g(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)))/9
Time = 0.29 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.07 \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x)^2 (3+5 x)} \, dx=\frac {121}{25} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) - \frac {70}{9} \, \sqrt {21} \log \left (-\frac {\sqrt {21} - 3 \, \sqrt {-2 \, x + 1}}{\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}}\right ) + \frac {8}{45} \, \sqrt {-2 \, x + 1} + \frac {49 \, \sqrt {-2 \, x + 1}}{9 \, {\left (3 \, x + 2\right )}} \]
121/25*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 70/9*sqrt(21)*log(-(sqrt(21) - 3*sqrt(-2*x + 1))/(sqrt(21) + 3*s qrt(-2*x + 1))) + 8/45*sqrt(-2*x + 1) + 49/9*sqrt(-2*x + 1)/(3*x + 2)
Time = 0.28 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.13 \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x)^2 (3+5 x)} \, dx=\frac {121}{25} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) - \frac {70}{9} \, \sqrt {21} \log \left (\frac {{\left | -2 \, \sqrt {21} + 6 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}\right )}}\right ) + \frac {8}{45} \, \sqrt {-2 \, x + 1} + \frac {49 \, \sqrt {-2 \, x + 1}}{9 \, {\left (3 \, x + 2\right )}} \]
121/25*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5 *sqrt(-2*x + 1))) - 70/9*sqrt(21)*log(1/2*abs(-2*sqrt(21) + 6*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) + 8/45*sqrt(-2*x + 1) + 49/9*sqrt(-2*x + 1)/(3*x + 2)
Time = 0.11 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.72 \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x)^2 (3+5 x)} \, dx=\frac {98\,\sqrt {1-2\,x}}{27\,\left (2\,x+\frac {4}{3}\right )}+\frac {8\,\sqrt {1-2\,x}}{45}-\frac {\sqrt {21}\,\mathrm {atan}\left (\frac {\sqrt {21}\,\sqrt {1-2\,x}\,1{}\mathrm {i}}{7}\right )\,140{}\mathrm {i}}{9}+\frac {\sqrt {55}\,\mathrm {atan}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}\,1{}\mathrm {i}}{11}\right )\,242{}\mathrm {i}}{25} \]