Integrand size = 24, antiderivative size = 97 \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x) (3+5 x)^3} \, dx=-\frac {11 (1-2 x)^{3/2}}{10 (3+5 x)^2}+\frac {803 \sqrt {1-2 x}}{50 (3+5 x)}+98 \sqrt {\frac {7}{3}} \text {arctanh}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )-\frac {2523}{25} \sqrt {\frac {11}{5}} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \]
-11/10*(1-2*x)^(3/2)/(3+5*x)^2-2523/125*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2 ))*55^(1/2)+98/3*arctanh(1/7*21^(1/2)*(1-2*x)^(1/2))*21^(1/2)+803/50*(1-2* x)^(1/2)/(3+5*x)
Time = 0.22 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.84 \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x) (3+5 x)^3} \, dx=98 \sqrt {\frac {7}{3}} \text {arctanh}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )+\frac {1}{250} \left (\frac {55 \sqrt {1-2 x} (214+375 x)}{(3+5 x)^2}-5046 \sqrt {55} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )\right ) \]
98*Sqrt[7/3]*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]] + ((55*Sqrt[1 - 2*x]*(214 + 375*x))/(3 + 5*x)^2 - 5046*Sqrt[55]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/250
Time = 0.20 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.08, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {109, 166, 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.
| \(\displaystyle \int \frac {(1-2 x)^{5/2}}{(3 x+2) (5 x+3)^3} \, dx\) |
| \(\Big \downarrow \) 109 |
| \(\displaystyle -\frac {1}{10} \int \frac {(136-41 x) \sqrt {1-2 x}}{(3 x+2) (5 x+3)^2}dx-\frac {11 (1-2 x)^{3/2}}{10 (5 x+3)^2}\) |
| \(\Big \downarrow \) 166 |
| \(\displaystyle \frac {1}{10} \left (\frac {803 \sqrt {1-2 x}}{5 (5 x+3)}-\frac {1}{5} \int -\frac {4056-2491 x}{\sqrt {1-2 x} (3 x+2) (5 x+3)}dx\right )-\frac {11 (1-2 x)^{3/2}}{10 (5 x+3)^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{10} \left (\frac {1}{5} \int \frac {4056-2491 x}{\sqrt {1-2 x} (3 x+2) (5 x+3)}dx+\frac {803 \sqrt {1-2 x}}{5 (5 x+3)}\right )-\frac {11 (1-2 x)^{3/2}}{10 (5 x+3)^2}\) |
| \(\Big \downarrow \) 174 |
| \(\displaystyle \frac {1}{10} \left (\frac {1}{5} \left (27753 \int \frac {1}{\sqrt {1-2 x} (5 x+3)}dx-17150 \int \frac {1}{\sqrt {1-2 x} (3 x+2)}dx\right )+\frac {803 \sqrt {1-2 x}}{5 (5 x+3)}\right )-\frac {11 (1-2 x)^{3/2}}{10 (5 x+3)^2}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {1}{10} \left (\frac {1}{5} \left (17150 \int \frac {1}{\frac {7}{2}-\frac {3}{2} (1-2 x)}d\sqrt {1-2 x}-27753 \int \frac {1}{\frac {11}{2}-\frac {5}{2} (1-2 x)}d\sqrt {1-2 x}\right )+\frac {803 \sqrt {1-2 x}}{5 (5 x+3)}\right )-\frac {11 (1-2 x)^{3/2}}{10 (5 x+3)^2}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {1}{10} \left (\frac {1}{5} \left (4900 \sqrt {\frac {7}{3}} \text {arctanh}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )-5046 \sqrt {\frac {11}{5}} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )\right )+\frac {803 \sqrt {1-2 x}}{5 (5 x+3)}\right )-\frac {11 (1-2 x)^{3/2}}{10 (5 x+3)^2}\) |
(-11*(1 - 2*x)^(3/2))/(10*(3 + 5*x)^2) + ((803*Sqrt[1 - 2*x])/(5*(3 + 5*x) ) + (4900*Sqrt[7/3]*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]] - 5046*Sqrt[11/5]*Arc Tanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/5)/10
3.20.96.3.1 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[(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[(b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^n*((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 - 1)*(e + f*x)^p*Simp[b* c*(f*g - e*h)*(m + 1) + (b*g - a*h)*(d*e*n + c*f*(p + 1)) + d*(b*(f*g - e*h )*(m + 1) + f*(b*g - a*h)*(n + p + 1))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, p}, x] && ILtQ[m, -1] && GtQ[n, 0]
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 = 1.08 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.66
| method | result | size |
| risch | \(-\frac {11 \left (750 x^{2}+53 x -214\right )}{50 \left (3+5 x \right )^{2} \sqrt {1-2 x}}+\frac {98 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{3}-\frac {2523 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{125}\) | \(64\) |
| derivativedivides | \(\frac {-165 \left (1-2 x \right )^{\frac {3}{2}}+\frac {8833 \sqrt {1-2 x}}{25}}{\left (-6-10 x \right )^{2}}-\frac {2523 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{125}+\frac {98 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{3}\) | \(66\) |
| default | \(\frac {-165 \left (1-2 x \right )^{\frac {3}{2}}+\frac {8833 \sqrt {1-2 x}}{25}}{\left (-6-10 x \right )^{2}}-\frac {2523 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{125}+\frac {98 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{3}\) | \(66\) |
| pseudoelliptic | \(\frac {24500 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \left (3+5 x \right )^{2} \sqrt {21}-15138 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \left (3+5 x \right )^{2} \sqrt {55}+165 \sqrt {1-2 x}\, \left (375 x +214\right )}{750 \left (3+5 x \right )^{2}}\) | \(75\) |
| trager | \(\frac {11 \left (375 x +214\right ) \sqrt {1-2 x}}{50 \left (3+5 x \right )^{2}}+\frac {49 \operatorname {RootOf}\left (\textit {\_Z}^{2}-21\right ) \ln \left (\frac {-3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-21\right ) x +21 \sqrt {1-2 x}+5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-21\right )}{2+3 x}\right )}{3}+\frac {2523 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) \ln \left (\frac {5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) x +55 \sqrt {1-2 x}-8 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right )}{3+5 x}\right )}{250}\) | \(111\) |
-11/50*(750*x^2+53*x-214)/(3+5*x)^2/(1-2*x)^(1/2)+98/3*arctanh(1/7*21^(1/2 )*(1-2*x)^(1/2))*21^(1/2)-2523/125*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 \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x) (3+5 x)^3} \, dx=\frac {7569 \, \sqrt {11} \sqrt {5} {\left (25 \, x^{2} + 30 \, x + 9\right )} \log \left (\frac {\sqrt {11} \sqrt {5} \sqrt {-2 \, x + 1} + 5 \, x - 8}{5 \, x + 3}\right ) + 12250 \, \sqrt {7} \sqrt {3} {\left (25 \, x^{2} + 30 \, x + 9\right )} \log \left (-\frac {\sqrt {7} \sqrt {3} \sqrt {-2 \, x + 1} - 3 \, x + 5}{3 \, x + 2}\right ) + 165 \, {\left (375 \, x + 214\right )} \sqrt {-2 \, x + 1}}{750 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} \]
1/750*(7569*sqrt(11)*sqrt(5)*(25*x^2 + 30*x + 9)*log((sqrt(11)*sqrt(5)*sqr t(-2*x + 1) + 5*x - 8)/(5*x + 3)) + 12250*sqrt(7)*sqrt(3)*(25*x^2 + 30*x + 9)*log(-(sqrt(7)*sqrt(3)*sqrt(-2*x + 1) - 3*x + 5)/(3*x + 2)) + 165*(375* x + 214)*sqrt(-2*x + 1))/(25*x^2 + 30*x + 9)
Leaf count of result is larger than twice the leaf count of optimal. 167 vs. \(2 (83) = 166\).
Time = 38.11 (sec) , antiderivative size = 372, normalized size of antiderivative = 3.84 \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x) (3+5 x)^3} \, dx=- \frac {49 \sqrt {21} \left (\log {\left (\sqrt {1 - 2 x} - \frac {\sqrt {21}}{3} \right )} - \log {\left (\sqrt {1 - 2 x} + \frac {\sqrt {21}}{3} \right )}\right )}{3} + \frac {1299 \sqrt {55} \left (\log {\left (\sqrt {1 - 2 x} - \frac {\sqrt {55}}{5} \right )} - \log {\left (\sqrt {1 - 2 x} + \frac {\sqrt {55}}{5} \right )}\right )}{125} + \frac {18876 \left (\begin {cases} \frac {\sqrt {55} \left (- \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1 \right )}}{4} + \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1 \right )}}{4} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1\right )} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1\right )}\right )}{605} & \text {for}\: \sqrt {1 - 2 x} > - \frac {\sqrt {55}}{5} \wedge \sqrt {1 - 2 x} < \frac {\sqrt {55}}{5} \end {cases}\right )}{25} + \frac {10648 \left (\begin {cases} \frac {\sqrt {55} \cdot \left (\frac {3 \log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1 \right )}}{16} - \frac {3 \log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1 \right )}}{16} + \frac {3}{16 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1\right )} + \frac {1}{16 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1\right )^{2}} + \frac {3}{16 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1\right )} - \frac {1}{16 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1\right )^{2}}\right )}{6655} & \text {for}\: \sqrt {1 - 2 x} > - \frac {\sqrt {55}}{5} \wedge \sqrt {1 - 2 x} < \frac {\sqrt {55}}{5} \end {cases}\right )}{25} \]
-49*sqrt(21)*(log(sqrt(1 - 2*x) - sqrt(21)/3) - log(sqrt(1 - 2*x) + sqrt(2 1)/3))/3 + 1299*sqrt(55)*(log(sqrt(1 - 2*x) - sqrt(55)/5) - log(sqrt(1 - 2 *x) + sqrt(55)/5))/125 + 18876*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)*sq rt(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)))/25 + 10648*Piecewi se((sqrt(55)*(3*log(sqrt(55)*sqrt(1 - 2*x)/11 - 1)/16 - 3*log(sqrt(55)*sqr t(1 - 2*x)/11 + 1)/16 + 3/(16*(sqrt(55)*sqrt(1 - 2*x)/11 + 1)) + 1/(16*(sq rt(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)))/25
Time = 0.29 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.13 \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x) (3+5 x)^3} \, dx=\frac {2523}{250} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) - \frac {49}{3} \, \sqrt {21} \log \left (-\frac {\sqrt {21} - 3 \, \sqrt {-2 \, x + 1}}{\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}}\right ) - \frac {11 \, {\left (375 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 803 \, \sqrt {-2 \, x + 1}\right )}}{25 \, {\left (25 \, {\left (2 \, x - 1\right )}^{2} + 220 \, x + 11\right )}} \]
2523/250*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2 *x + 1))) - 49/3*sqrt(21)*log(-(sqrt(21) - 3*sqrt(-2*x + 1))/(sqrt(21) + 3 *sqrt(-2*x + 1))) - 11/25*(375*(-2*x + 1)^(3/2) - 803*sqrt(-2*x + 1))/(25* (2*x - 1)^2 + 220*x + 11)
Time = 0.29 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.10 \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x) (3+5 x)^3} \, dx=\frac {2523}{250} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) - \frac {49}{3} \, \sqrt {21} \log \left (\frac {{\left | -2 \, \sqrt {21} + 6 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}\right )}}\right ) - \frac {11 \, {\left (375 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 803 \, \sqrt {-2 \, x + 1}\right )}}{100 \, {\left (5 \, x + 3\right )}^{2}} \]
2523/250*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 49/3*sqrt(21)*log(1/2*abs(-2*sqrt(21) + 6*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) - 11/100*(375*(-2*x + 1)^(3/2) - 803* sqrt(-2*x + 1))/(5*x + 3)^2
Time = 1.36 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.73 \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x) (3+5 x)^3} \, dx=\frac {98\,\sqrt {21}\,\mathrm {atanh}\left (\frac {\sqrt {21}\,\sqrt {1-2\,x}}{7}\right )}{3}-\frac {2523\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )}{125}+\frac {\frac {8833\,\sqrt {1-2\,x}}{625}-\frac {33\,{\left (1-2\,x\right )}^{3/2}}{5}}{\frac {44\,x}{5}+{\left (2\,x-1\right )}^2+\frac {11}{25}} \]