Integrand size = 24, antiderivative size = 92 \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x) (3+5 x)^2} \, dx=-\frac {58}{75} \sqrt {1-2 x}-\frac {11 (1-2 x)^{3/2}}{5 (3+5 x)}-\frac {98}{3} \sqrt {\frac {7}{3}} \text {arctanh}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )+\frac {836}{25} \sqrt {\frac {11}{5}} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \]
-11/5*(1-2*x)^(3/2)/(3+5*x)+836/125*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*5 5^(1/2)-98/9*arctanh(1/7*21^(1/2)*(1-2*x)^(1/2))*21^(1/2)-58/75*(1-2*x)^(1 /2)
Time = 0.19 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.90 \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x) (3+5 x)^2} \, dx=-\frac {98}{3} \sqrt {\frac {7}{3}} \text {arctanh}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )+\frac {1}{375} \left (\frac {5 \sqrt {1-2 x} (-339+40 x)}{3+5 x}+2508 \sqrt {55} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )\right ) \]
(-98*Sqrt[7/3]*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]])/3 + ((5*Sqrt[1 - 2*x]*(-3 39 + 40*x))/(3 + 5*x) + 2508*Sqrt[55]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/3 75
Time = 0.20 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.07, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {109, 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.
| \(\displaystyle \int \frac {(1-2 x)^{5/2}}{(3 x+2) (5 x+3)^2} \, dx\) |
| \(\Big \downarrow \) 109 |
| \(\displaystyle -\frac {1}{5} \int \frac {\sqrt {1-2 x} (29 x+101)}{(3 x+2) (5 x+3)}dx-\frac {11 (1-2 x)^{3/2}}{5 (5 x+3)}\) |
| \(\Big \downarrow \) 171 |
| \(\displaystyle \frac {1}{5} \left (-\frac {2}{15} \int \frac {1863-1493 x}{2 \sqrt {1-2 x} (3 x+2) (5 x+3)}dx-\frac {58}{15} \sqrt {1-2 x}\right )-\frac {11 (1-2 x)^{3/2}}{5 (5 x+3)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \left (-\frac {1}{15} \int \frac {1863-1493 x}{\sqrt {1-2 x} (3 x+2) (5 x+3)}dx-\frac {58}{15} \sqrt {1-2 x}\right )-\frac {11 (1-2 x)^{3/2}}{5 (5 x+3)}\) |
| \(\Big \downarrow \) 174 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{15} \left (8575 \int \frac {1}{\sqrt {1-2 x} (3 x+2)}dx-13794 \int \frac {1}{\sqrt {1-2 x} (5 x+3)}dx\right )-\frac {58}{15} \sqrt {1-2 x}\right )-\frac {11 (1-2 x)^{3/2}}{5 (5 x+3)}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{15} \left (13794 \int \frac {1}{\frac {11}{2}-\frac {5}{2} (1-2 x)}d\sqrt {1-2 x}-8575 \int \frac {1}{\frac {7}{2}-\frac {3}{2} (1-2 x)}d\sqrt {1-2 x}\right )-\frac {58}{15} \sqrt {1-2 x}\right )-\frac {11 (1-2 x)^{3/2}}{5 (5 x+3)}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{15} \left (2508 \sqrt {\frac {11}{5}} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )-2450 \sqrt {\frac {7}{3}} \text {arctanh}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )\right )-\frac {58}{15} \sqrt {1-2 x}\right )-\frac {11 (1-2 x)^{3/2}}{5 (5 x+3)}\) |
(-11*(1 - 2*x)^(3/2))/(5*(3 + 5*x)) + ((-58*Sqrt[1 - 2*x])/15 + (-2450*Sqr t[7/3]*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]] + 2508*Sqrt[11/5]*ArcTanh[Sqrt[5/1 1]*Sqrt[1 - 2*x]])/15)/5
3.20.85.3.1 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 = 1.07 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.68
| method | result | size |
| derivativedivides | \(\frac {8 \sqrt {1-2 x}}{75}+\frac {242 \sqrt {1-2 x}}{125 \left (-\frac {6}{5}-2 x \right )}+\frac {836 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{125}-\frac {98 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{9}\) | \(63\) |
| default | \(\frac {8 \sqrt {1-2 x}}{75}+\frac {242 \sqrt {1-2 x}}{125 \left (-\frac {6}{5}-2 x \right )}+\frac {836 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{125}-\frac {98 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{9}\) | \(63\) |
| risch | \(-\frac {80 x^{2}-718 x +339}{75 \left (3+5 x \right ) \sqrt {1-2 x}}-\frac {98 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{9}+\frac {836 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{125}\) | \(64\) |
| pseudoelliptic | \(\frac {-12250 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \left (3+5 x \right ) \sqrt {21}+7524 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \left (3+5 x \right ) \sqrt {55}+15 \sqrt {1-2 x}\, \left (-339+40 x \right )}{3375+5625 x}\) | \(70\) |
| trager | \(\frac {\sqrt {1-2 x}\, \left (-339+40 x \right )}{225+375 x}+\frac {49 \operatorname {RootOf}\left (\textit {\_Z}^{2}-21\right ) \ln \left (\frac {3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-21\right ) x -5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-21\right )+21 \sqrt {1-2 x}}{2+3 x}\right )}{9}-\frac {418 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) \ln \left (\frac {5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) x +55 \sqrt {1-2 x}-8 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right )}{3+5 x}\right )}{125}\) | \(111\) |
8/75*(1-2*x)^(1/2)+242/125*(1-2*x)^(1/2)/(-6/5-2*x)+836/125*arctanh(1/11*5 5^(1/2)*(1-2*x)^(1/2))*55^(1/2)-98/9*arctanh(1/7*21^(1/2)*(1-2*x)^(1/2))*2 1^(1/2)
Time = 0.22 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.16 \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x) (3+5 x)^2} \, dx=\frac {3762 \, \sqrt {11} \sqrt {5} {\left (5 \, x + 3\right )} \log \left (-\frac {\sqrt {11} \sqrt {5} \sqrt {-2 \, x + 1} - 5 \, x + 8}{5 \, x + 3}\right ) + 6125 \, \sqrt {7} \sqrt {3} {\left (5 \, x + 3\right )} \log \left (\frac {\sqrt {7} \sqrt {3} \sqrt {-2 \, x + 1} + 3 \, x - 5}{3 \, x + 2}\right ) + 15 \, {\left (40 \, x - 339\right )} \sqrt {-2 \, x + 1}}{1125 \, {\left (5 \, x + 3\right )}} \]
1/1125*(3762*sqrt(11)*sqrt(5)*(5*x + 3)*log(-(sqrt(11)*sqrt(5)*sqrt(-2*x + 1) - 5*x + 8)/(5*x + 3)) + 6125*sqrt(7)*sqrt(3)*(5*x + 3)*log((sqrt(7)*sq rt(3)*sqrt(-2*x + 1) + 3*x - 5)/(3*x + 2)) + 15*(40*x - 339)*sqrt(-2*x + 1 ))/(5*x + 3)
Time = 14.33 (sec) , antiderivative size = 216, normalized size of antiderivative = 2.35 \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x) (3+5 x)^2} \, dx=\frac {8 \sqrt {1 - 2 x}}{75} + \frac {49 \sqrt {21} \left (\log {\left (\sqrt {1 - 2 x} - \frac {\sqrt {21}}{3} \right )} - \log {\left (\sqrt {1 - 2 x} + \frac {\sqrt {21}}{3} \right )}\right )}{9} - \frac {429 \sqrt {55} \left (\log {\left (\sqrt {1 - 2 x} - \frac {\sqrt {55}}{5} \right )} - \log {\left (\sqrt {1 - 2 x} + \frac {\sqrt {55}}{5} \right )}\right )}{125} - \frac {5324 \left (\begin {cases} \frac {\sqrt {55} \left (- \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1 \right )}}{4} + \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1 \right )}}{4} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1\right )} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1\right )}\right )}{605} & \text {for}\: \sqrt {1 - 2 x} > - \frac {\sqrt {55}}{5} \wedge \sqrt {1 - 2 x} < \frac {\sqrt {55}}{5} \end {cases}\right )}{25} \]
8*sqrt(1 - 2*x)/75 + 49*sqrt(21)*(log(sqrt(1 - 2*x) - sqrt(21)/3) - log(sq rt(1 - 2*x) + sqrt(21)/3))/9 - 429*sqrt(55)*(log(sqrt(1 - 2*x) - sqrt(55)/ 5) - log(sqrt(1 - 2*x) + sqrt(55)/5))/125 - 5324*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)))/ 25
Time = 0.28 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.07 \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x) (3+5 x)^2} \, dx=-\frac {418}{125} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) + \frac {49}{9} \, \sqrt {21} \log \left (-\frac {\sqrt {21} - 3 \, \sqrt {-2 \, x + 1}}{\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}}\right ) + \frac {8}{75} \, \sqrt {-2 \, x + 1} - \frac {121 \, \sqrt {-2 \, x + 1}}{25 \, {\left (5 \, x + 3\right )}} \]
-418/125*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2 *x + 1))) + 49/9*sqrt(21)*log(-(sqrt(21) - 3*sqrt(-2*x + 1))/(sqrt(21) + 3 *sqrt(-2*x + 1))) + 8/75*sqrt(-2*x + 1) - 121/25*sqrt(-2*x + 1)/(5*x + 3)
Time = 0.32 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.13 \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x) (3+5 x)^2} \, dx=-\frac {418}{125} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) + \frac {49}{9} \, \sqrt {21} \log \left (\frac {{\left | -2 \, \sqrt {21} + 6 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}\right )}}\right ) + \frac {8}{75} \, \sqrt {-2 \, x + 1} - \frac {121 \, \sqrt {-2 \, x + 1}}{25 \, {\left (5 \, x + 3\right )}} \]
-418/125*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 49/9*sqrt(21)*log(1/2*abs(-2*sqrt(21) + 6*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) + 8/75*sqrt(-2*x + 1) - 121/25*sqrt(- 2*x + 1)/(5*x + 3)
Time = 0.12 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.72 \[ \int \frac {(1-2 x)^{5/2}}{(2+3 x) (3+5 x)^2} \, dx=\frac {8\,\sqrt {1-2\,x}}{75}-\frac {242\,\sqrt {1-2\,x}}{125\,\left (2\,x+\frac {6}{5}\right )}+\frac {\sqrt {21}\,\mathrm {atan}\left (\frac {\sqrt {21}\,\sqrt {1-2\,x}\,1{}\mathrm {i}}{7}\right )\,98{}\mathrm {i}}{9}-\frac {\sqrt {55}\,\mathrm {atan}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}\,1{}\mathrm {i}}{11}\right )\,836{}\mathrm {i}}{125} \]