Integrand size = 24, antiderivative size = 112 \[ \int \frac {1}{(1-2 x)^{3/2} (2+3 x)^3 (3+5 x)} \, dx=-\frac {2525}{3773 \sqrt {1-2 x}}+\frac {3}{14 \sqrt {1-2 x} (2+3 x)^2}+\frac {225}{98 \sqrt {1-2 x} (2+3 x)}+\frac {8025}{343} \sqrt {\frac {3}{7}} \text {arctanh}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )-\frac {250}{11} \sqrt {\frac {5}{11}} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \]
8025/2401*arctanh(1/7*21^(1/2)*(1-2*x)^(1/2))*21^(1/2)-250/121*arctanh(1/1 1*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)-2525/3773/(1-2*x)^(1/2)+3/14/(2+3*x)^2/ (1-2*x)^(1/2)+225/98/(2+3*x)/(1-2*x)^(1/2)
Time = 0.27 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.79 \[ \int \frac {1}{(1-2 x)^{3/2} (2+3 x)^3 (3+5 x)} \, dx=\frac {16067-8625 x-45450 x^2}{7546 \sqrt {1-2 x} (2+3 x)^2}+\frac {8025}{343} \sqrt {\frac {3}{7}} \text {arctanh}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )-\frac {250}{11} \sqrt {\frac {5}{11}} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \]
(16067 - 8625*x - 45450*x^2)/(7546*Sqrt[1 - 2*x]*(2 + 3*x)^2) + (8025*Sqrt [3/7]*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]])/343 - (250*Sqrt[5/11]*ArcTanh[Sqrt [5/11]*Sqrt[1 - 2*x]])/11
Time = 0.21 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.10, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {114, 27, 168, 169, 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.
\(\displaystyle \int \frac {1}{(1-2 x)^{3/2} (3 x+2)^3 (5 x+3)} \, dx\) |
\(\Big \downarrow \) 114 |
\(\displaystyle \frac {1}{14} \int \frac {25 (1-3 x)}{(1-2 x)^{3/2} (3 x+2)^2 (5 x+3)}dx+\frac {3}{14 \sqrt {1-2 x} (3 x+2)^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {25}{14} \int \frac {1-3 x}{(1-2 x)^{3/2} (3 x+2)^2 (5 x+3)}dx+\frac {3}{14 \sqrt {1-2 x} (3 x+2)^2}\) |
\(\Big \downarrow \) 168 |
\(\displaystyle \frac {25}{14} \left (\frac {1}{7} \int \frac {17-135 x}{(1-2 x)^{3/2} (3 x+2) (5 x+3)}dx+\frac {9}{7 \sqrt {1-2 x} (3 x+2)}\right )+\frac {3}{14 \sqrt {1-2 x} (3 x+2)^2}\) |
\(\Big \downarrow \) 169 |
\(\displaystyle \frac {25}{14} \left (\frac {1}{7} \left (-\frac {2}{77} \int -\frac {2521-1515 x}{2 \sqrt {1-2 x} (3 x+2) (5 x+3)}dx-\frac {202}{77 \sqrt {1-2 x}}\right )+\frac {9}{7 \sqrt {1-2 x} (3 x+2)}\right )+\frac {3}{14 \sqrt {1-2 x} (3 x+2)^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {25}{14} \left (\frac {1}{7} \left (\frac {1}{77} \int \frac {2521-1515 x}{\sqrt {1-2 x} (3 x+2) (5 x+3)}dx-\frac {202}{77 \sqrt {1-2 x}}\right )+\frac {9}{7 \sqrt {1-2 x} (3 x+2)}\right )+\frac {3}{14 \sqrt {1-2 x} (3 x+2)^2}\) |
\(\Big \downarrow \) 174 |
\(\displaystyle \frac {25}{14} \left (\frac {1}{7} \left (\frac {1}{77} \left (17150 \int \frac {1}{\sqrt {1-2 x} (5 x+3)}dx-10593 \int \frac {1}{\sqrt {1-2 x} (3 x+2)}dx\right )-\frac {202}{77 \sqrt {1-2 x}}\right )+\frac {9}{7 \sqrt {1-2 x} (3 x+2)}\right )+\frac {3}{14 \sqrt {1-2 x} (3 x+2)^2}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {25}{14} \left (\frac {1}{7} \left (\frac {1}{77} \left (10593 \int \frac {1}{\frac {7}{2}-\frac {3}{2} (1-2 x)}d\sqrt {1-2 x}-17150 \int \frac {1}{\frac {11}{2}-\frac {5}{2} (1-2 x)}d\sqrt {1-2 x}\right )-\frac {202}{77 \sqrt {1-2 x}}\right )+\frac {9}{7 \sqrt {1-2 x} (3 x+2)}\right )+\frac {3}{14 \sqrt {1-2 x} (3 x+2)^2}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {25}{14} \left (\frac {1}{7} \left (\frac {1}{77} \left (7062 \sqrt {\frac {3}{7}} \text {arctanh}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )-6860 \sqrt {\frac {5}{11}} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )\right )-\frac {202}{77 \sqrt {1-2 x}}\right )+\frac {9}{7 \sqrt {1-2 x} (3 x+2)}\right )+\frac {3}{14 \sqrt {1-2 x} (3 x+2)^2}\) |
3/(14*Sqrt[1 - 2*x]*(2 + 3*x)^2) + (25*(9/(7*Sqrt[1 - 2*x]*(2 + 3*x)) + (- 202/(77*Sqrt[1 - 2*x]) + (7062*Sqrt[3/7]*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]] - 6860*Sqrt[5/11]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/77)/7))/14
3.22.13.3.1 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*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1 )/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + Simp[1/((m + 1)*(b*c - a*d)*(b*e - a*f)) Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*(m + 1) - b*(d*e*(m + n + 2) + c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && ILtQ[m, -1] && (IntegerQ[n] || IntegersQ[2*n, 2*p] || ILtQ[m + n + p + 3, 0])
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 + 1)*((e + f*x)^(p + 1)/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + S imp[1/((m + 1)*(b*c - a*d)*(b*e - a*f)) Int[(a + b*x)^(m + 1)*(c + d*x)^n *(e + f*x)^p*Simp[(a*d*f*g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a* h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && ILtQ[m, -1]
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 + 1)*((e + f*x)^(p + 1)/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + S imp[1/((m + 1)*(b*c - a*d)*(b*e - a*f)) Int[(a + b*x)^(m + 1)*(c + d*x)^n *(e + f*x)^p*Simp[(a*d*f*g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a* h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && LtQ[m, -1] && 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 = 3.48 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.57
method | result | size |
risch | \(-\frac {45450 x^{2}+8625 x -16067}{7546 \left (2+3 x \right )^{2} \sqrt {1-2 x}}+\frac {8025 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{2401}-\frac {250 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{121}\) | \(64\) |
derivativedivides | \(-\frac {250 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{121}-\frac {486 \left (\frac {77 \left (1-2 x \right )^{\frac {3}{2}}}{18}-\frac {553 \sqrt {1-2 x}}{54}\right )}{343 \left (-4-6 x \right )^{2}}+\frac {8025 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{2401}+\frac {16}{3773 \sqrt {1-2 x}}\) | \(75\) |
default | \(-\frac {250 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{121}-\frac {486 \left (\frac {77 \left (1-2 x \right )^{\frac {3}{2}}}{18}-\frac {553 \sqrt {1-2 x}}{54}\right )}{343 \left (-4-6 x \right )^{2}}+\frac {8025 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{2401}+\frac {16}{3773 \sqrt {1-2 x}}\) | \(75\) |
pseudoelliptic | \(\frac {\frac {16067}{7546}+\frac {8025 \sqrt {1-2 x}\, \operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \left (2+3 x \right )^{2} \sqrt {21}}{2401}-\frac {250 \sqrt {1-2 x}\, \operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \left (2+3 x \right )^{2} \sqrt {55}}{121}-\frac {22725 x^{2}}{3773}-\frac {8625 x}{7546}}{\left (2+3 x \right )^{2} \sqrt {1-2 x}}\) | \(91\) |
trager | \(\frac {\left (45450 x^{2}+8625 x -16067\right ) \sqrt {1-2 x}}{7546 \left (2+3 x \right )^{2} \left (-1+2 x \right )}-\frac {75 \operatorname {RootOf}\left (\textit {\_Z}^{2}-240429\right ) \ln \left (\frac {3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-240429\right ) x -5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-240429\right )+2247 \sqrt {1-2 x}}{2+3 x}\right )}{4802}+\frac {125 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) \ln \left (\frac {5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) x +55 \sqrt {1-2 x}-8 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right )}{3+5 x}\right )}{121}\) | \(123\) |
-1/7546*(45450*x^2+8625*x-16067)/(2+3*x)^2/(1-2*x)^(1/2)+8025/2401*arctanh (1/7*21^(1/2)*(1-2*x)^(1/2))*21^(1/2)-250/121*arctanh(1/11*55^(1/2)*(1-2*x )^(1/2))*55^(1/2)
Time = 0.23 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.27 \[ \int \frac {1}{(1-2 x)^{3/2} (2+3 x)^3 (3+5 x)} \, dx=\frac {600250 \, \sqrt {11} \sqrt {5} {\left (18 \, x^{3} + 15 \, x^{2} - 4 \, x - 4\right )} \log \left (\frac {\sqrt {11} \sqrt {5} \sqrt {-2 \, x + 1} + 5 \, x - 8}{5 \, x + 3}\right ) + 971025 \, \sqrt {7} \sqrt {3} {\left (18 \, x^{3} + 15 \, x^{2} - 4 \, x - 4\right )} \log \left (-\frac {\sqrt {7} \sqrt {3} \sqrt {-2 \, x + 1} - 3 \, x + 5}{3 \, x + 2}\right ) + 77 \, {\left (45450 \, x^{2} + 8625 \, x - 16067\right )} \sqrt {-2 \, x + 1}}{581042 \, {\left (18 \, x^{3} + 15 \, x^{2} - 4 \, x - 4\right )}} \]
1/581042*(600250*sqrt(11)*sqrt(5)*(18*x^3 + 15*x^2 - 4*x - 4)*log((sqrt(11 )*sqrt(5)*sqrt(-2*x + 1) + 5*x - 8)/(5*x + 3)) + 971025*sqrt(7)*sqrt(3)*(1 8*x^3 + 15*x^2 - 4*x - 4)*log(-(sqrt(7)*sqrt(3)*sqrt(-2*x + 1) - 3*x + 5)/ (3*x + 2)) + 77*(45450*x^2 + 8625*x - 16067)*sqrt(-2*x + 1))/(18*x^3 + 15* x^2 - 4*x - 4)
Result contains complex when optimal does not.
Time = 9.03 (sec) , antiderivative size = 1875, normalized size of antiderivative = 16.74 \[ \int \frac {1}{(1-2 x)^{3/2} (2+3 x)^3 (3+5 x)} \, dx=\text {Too large to display} \]
-1555848000*sqrt(55)*I*(x - 1/2)**(11/2)*atan(sqrt(110)*sqrt(x - 1/2)/11)/ (753030432*(x - 1/2)**(11/2) + 3514142016*(x - 1/2)**(9/2) + 6149748528*(x - 1/2)**(7/2) + 4783137744*(x - 1/2)**(5/2) + 1395081842*(x - 1/2)**(3/2) ) + 2516896800*sqrt(21)*I*(x - 1/2)**(11/2)*atan(sqrt(42)*sqrt(x - 1/2)/7) /(753030432*(x - 1/2)**(11/2) + 3514142016*(x - 1/2)**(9/2) + 6149748528*( x - 1/2)**(7/2) + 4783137744*(x - 1/2)**(5/2) + 1395081842*(x - 1/2)**(3/2 )) - 1258448400*sqrt(21)*I*pi*(x - 1/2)**(11/2)/(753030432*(x - 1/2)**(11/ 2) + 3514142016*(x - 1/2)**(9/2) + 6149748528*(x - 1/2)**(7/2) + 478313774 4*(x - 1/2)**(5/2) + 1395081842*(x - 1/2)**(3/2)) + 777924000*sqrt(55)*I*p i*(x - 1/2)**(11/2)/(753030432*(x - 1/2)**(11/2) + 3514142016*(x - 1/2)**( 9/2) + 6149748528*(x - 1/2)**(7/2) + 4783137744*(x - 1/2)**(5/2) + 1395081 842*(x - 1/2)**(3/2)) - 7260624000*sqrt(55)*I*(x - 1/2)**(9/2)*atan(sqrt(1 10)*sqrt(x - 1/2)/11)/(753030432*(x - 1/2)**(11/2) + 3514142016*(x - 1/2)* *(9/2) + 6149748528*(x - 1/2)**(7/2) + 4783137744*(x - 1/2)**(5/2) + 13950 81842*(x - 1/2)**(3/2)) + 11745518400*sqrt(21)*I*(x - 1/2)**(9/2)*atan(sqr t(42)*sqrt(x - 1/2)/7)/(753030432*(x - 1/2)**(11/2) + 3514142016*(x - 1/2) **(9/2) + 6149748528*(x - 1/2)**(7/2) + 4783137744*(x - 1/2)**(5/2) + 1395 081842*(x - 1/2)**(3/2)) - 5872759200*sqrt(21)*I*pi*(x - 1/2)**(9/2)/(7530 30432*(x - 1/2)**(11/2) + 3514142016*(x - 1/2)**(9/2) + 6149748528*(x - 1/ 2)**(7/2) + 4783137744*(x - 1/2)**(5/2) + 1395081842*(x - 1/2)**(3/2)) ...
Time = 0.27 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.06 \[ \int \frac {1}{(1-2 x)^{3/2} (2+3 x)^3 (3+5 x)} \, dx=\frac {125}{121} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) - \frac {8025}{4802} \, \sqrt {21} \log \left (-\frac {\sqrt {21} - 3 \, \sqrt {-2 \, x + 1}}{\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}}\right ) - \frac {22725 \, {\left (2 \, x - 1\right )}^{2} + 108150 \, x - 54859}{3773 \, {\left (9 \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} - 42 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + 49 \, \sqrt {-2 \, x + 1}\right )}} \]
125/121*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2* x + 1))) - 8025/4802*sqrt(21)*log(-(sqrt(21) - 3*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) - 1/3773*(22725*(2*x - 1)^2 + 108150*x - 54859)/(9*( -2*x + 1)^(5/2) - 42*(-2*x + 1)^(3/2) + 49*sqrt(-2*x + 1))
Time = 0.29 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.04 \[ \int \frac {1}{(1-2 x)^{3/2} (2+3 x)^3 (3+5 x)} \, dx=\frac {125}{121} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) - \frac {8025}{4802} \, \sqrt {21} \log \left (\frac {{\left | -2 \, \sqrt {21} + 6 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}\right )}}\right ) + \frac {16}{3773 \, \sqrt {-2 \, x + 1}} - \frac {9 \, {\left (33 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 79 \, \sqrt {-2 \, x + 1}\right )}}{196 \, {\left (3 \, x + 2\right )}^{2}} \]
125/121*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 8025/4802*sqrt(21)*log(1/2*abs(-2*sqrt(21) + 6*sqrt(- 2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) + 16/3773/sqrt(-2*x + 1) - 9/196* (33*(-2*x + 1)^(3/2) - 79*sqrt(-2*x + 1))/(3*x + 2)^2
Time = 0.11 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.72 \[ \int \frac {1}{(1-2 x)^{3/2} (2+3 x)^3 (3+5 x)} \, dx=\frac {8025\,\sqrt {21}\,\mathrm {atanh}\left (\frac {\sqrt {21}\,\sqrt {1-2\,x}}{7}\right )}{2401}-\frac {250\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )}{121}-\frac {\frac {5150\,x}{1617}+\frac {2525\,{\left (2\,x-1\right )}^2}{3773}-\frac {7837}{4851}}{\frac {49\,\sqrt {1-2\,x}}{9}-\frac {14\,{\left (1-2\,x\right )}^{3/2}}{3}+{\left (1-2\,x\right )}^{5/2}} \]