Integrand size = 24, antiderivative size = 81 \[ \int \frac {(2+3 x)^2}{(1-2 x)^{5/2} (3+5 x)^2} \, dx=\frac {142}{6655 \sqrt {1-2 x}}+\frac {49}{66 (1-2 x)^{3/2} (3+5 x)}-\frac {1231}{3630 \sqrt {1-2 x} (3+5 x)}-\frac {142 \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{1331 \sqrt {55}} \]
49/66/(1-2*x)^(3/2)/(3+5*x)-142/73205*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2)) *55^(1/2)+142/6655/(1-2*x)^(1/2)-1231/3630/(3+5*x)/(1-2*x)^(1/2)
Time = 0.13 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.72 \[ \int \frac {(2+3 x)^2}{(1-2 x)^{5/2} (3+5 x)^2} \, dx=\frac {-\frac {55 \left (-1866-2623 x+852 x^2\right )}{(1-2 x)^{3/2} (3+5 x)}-426 \sqrt {55} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{219615} \]
((-55*(-1866 - 2623*x + 852*x^2))/((1 - 2*x)^(3/2)*(3 + 5*x)) - 426*Sqrt[5 5]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/219615
Time = 0.18 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.15, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {100, 25, 87, 61, 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.
\(\displaystyle \int \frac {(3 x+2)^2}{(1-2 x)^{5/2} (5 x+3)^2} \, dx\) |
\(\Big \downarrow \) 100 |
\(\displaystyle \frac {49}{66 (1-2 x)^{3/2} (5 x+3)}-\frac {1}{66} \int -\frac {68-297 x}{(1-2 x)^{3/2} (5 x+3)^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{66} \int \frac {68-297 x}{(1-2 x)^{3/2} (5 x+3)^2}dx+\frac {49}{66 (1-2 x)^{3/2} (5 x+3)}\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {1}{66} \left (\frac {426}{55} \int \frac {1}{(1-2 x)^{3/2} (5 x+3)}dx-\frac {1231}{55 \sqrt {1-2 x} (5 x+3)}\right )+\frac {49}{66 (1-2 x)^{3/2} (5 x+3)}\) |
\(\Big \downarrow \) 61 |
\(\displaystyle \frac {1}{66} \left (\frac {426}{55} \left (\frac {5}{11} \int \frac {1}{\sqrt {1-2 x} (5 x+3)}dx+\frac {2}{11 \sqrt {1-2 x}}\right )-\frac {1231}{55 \sqrt {1-2 x} (5 x+3)}\right )+\frac {49}{66 (1-2 x)^{3/2} (5 x+3)}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{66} \left (\frac {426}{55} \left (\frac {2}{11 \sqrt {1-2 x}}-\frac {5}{11} \int \frac {1}{\frac {11}{2}-\frac {5}{2} (1-2 x)}d\sqrt {1-2 x}\right )-\frac {1231}{55 \sqrt {1-2 x} (5 x+3)}\right )+\frac {49}{66 (1-2 x)^{3/2} (5 x+3)}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{66} \left (\frac {426}{55} \left (\frac {2}{11 \sqrt {1-2 x}}-\frac {2}{11} \sqrt {\frac {5}{11}} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )\right )-\frac {1231}{55 \sqrt {1-2 x} (5 x+3)}\right )+\frac {49}{66 (1-2 x)^{3/2} (5 x+3)}\) |
49/(66*(1 - 2*x)^(3/2)*(3 + 5*x)) + (-1231/(55*Sqrt[1 - 2*x]*(3 + 5*x)) + (426*(2/(11*Sqrt[1 - 2*x]) - (2*Sqrt[5/11]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x ]])/11))/55)/66
3.22.84.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( m + n + 2)/((b*c - a*d)*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[m, -1] && !(LtQ[n, -1] && (EqQ[a, 0 ] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d , m, n, 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_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)) Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege rQ[p] || !(IntegerQ[n] || !(EqQ[e, 0] || !(EqQ[c, 0] || LtQ[p, n]))))
Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^( p_), x_] :> Simp[(b*c - a*d)^2*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d^2*(d *e - c*f)*(n + 1))), x] - Simp[1/(d^2*(d*e - c*f)*(n + 1)) Int[(c + d*x)^ (n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*( p + 1)) - 2*a*b*d*(d*e*(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x , x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] || !SumSimplerQ[p, 1])))
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 = 3.51 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.65
method | result | size |
risch | \(\frac {852 x^{2}-2623 x -1866}{3993 \left (3+5 x \right ) \sqrt {1-2 x}\, \left (-1+2 x \right )}-\frac {142 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{73205}\) | \(53\) |
derivativedivides | \(\frac {2 \sqrt {1-2 x}}{6655 \left (-\frac {6}{5}-2 x \right )}-\frac {142 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{73205}+\frac {49}{363 \left (1-2 x \right )^{\frac {3}{2}}}+\frac {28}{1331 \sqrt {1-2 x}}\) | \(54\) |
default | \(\frac {2 \sqrt {1-2 x}}{6655 \left (-\frac {6}{5}-2 x \right )}-\frac {142 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{73205}+\frac {49}{363 \left (1-2 x \right )^{\frac {3}{2}}}+\frac {28}{1331 \sqrt {1-2 x}}\) | \(54\) |
pseudoelliptic | \(\frac {426 \sqrt {1-2 x}\, \operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \left (10 x^{2}+x -3\right ) \sqrt {55}-46860 x^{2}+144265 x +102630}{\left (1-2 x \right )^{\frac {3}{2}} \left (658845+1098075 x \right )}\) | \(60\) |
trager | \(-\frac {\left (852 x^{2}-2623 x -1866\right ) \sqrt {1-2 x}}{3993 \left (-1+2 x \right )^{2} \left (3+5 x \right )}+\frac {71 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) \ln \left (\frac {5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) x +55 \sqrt {1-2 x}-8 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right )}{3+5 x}\right )}{73205}\) | \(79\) |
1/3993*(852*x^2-2623*x-1866)/(3+5*x)/(1-2*x)^(1/2)/(-1+2*x)-142/73205*arct anh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)
Time = 0.23 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.04 \[ \int \frac {(2+3 x)^2}{(1-2 x)^{5/2} (3+5 x)^2} \, dx=\frac {213 \, \sqrt {55} {\left (20 \, x^{3} - 8 \, x^{2} - 7 \, x + 3\right )} \log \left (\frac {5 \, x + \sqrt {55} \sqrt {-2 \, x + 1} - 8}{5 \, x + 3}\right ) - 55 \, {\left (852 \, x^{2} - 2623 \, x - 1866\right )} \sqrt {-2 \, x + 1}}{219615 \, {\left (20 \, x^{3} - 8 \, x^{2} - 7 \, x + 3\right )}} \]
1/219615*(213*sqrt(55)*(20*x^3 - 8*x^2 - 7*x + 3)*log((5*x + sqrt(55)*sqrt (-2*x + 1) - 8)/(5*x + 3)) - 55*(852*x^2 - 2623*x - 1866)*sqrt(-2*x + 1))/ (20*x^3 - 8*x^2 - 7*x + 3)
Time = 40.88 (sec) , antiderivative size = 185, normalized size of antiderivative = 2.28 \[ \int \frac {(2+3 x)^2}{(1-2 x)^{5/2} (3+5 x)^2} \, dx=\frac {14 \sqrt {55} \left (\log {\left (\sqrt {1 - 2 x} - \frac {\sqrt {55}}{5} \right )} - \log {\left (\sqrt {1 - 2 x} + \frac {\sqrt {55}}{5} \right )}\right )}{14641} - \frac {4 \left (\begin {cases} \frac {\sqrt {55} \left (- \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1 \right )}}{4} + \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1 \right )}}{4} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1\right )} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1\right )}\right )}{605} & \text {for}\: \sqrt {1 - 2 x} > - \frac {\sqrt {55}}{5} \wedge \sqrt {1 - 2 x} < \frac {\sqrt {55}}{5} \end {cases}\right )}{121} + \frac {28}{1331 \sqrt {1 - 2 x}} + \frac {49}{363 \left (1 - 2 x\right )^{\frac {3}{2}}} \]
14*sqrt(55)*(log(sqrt(1 - 2*x) - sqrt(55)/5) - log(sqrt(1 - 2*x) + sqrt(55 )/5))/14641 - 4*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)))/121 + 28/(1331*sqrt(1 - 2*x)) + 4 9/(363*(1 - 2*x)**(3/2))
Time = 0.28 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.91 \[ \int \frac {(2+3 x)^2}{(1-2 x)^{5/2} (3+5 x)^2} \, dx=\frac {71}{73205} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) + \frac {2 \, {\left (213 \, {\left (2 \, x - 1\right )}^{2} - 1771 \, x - 2079\right )}}{3993 \, {\left (5 \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} - 11 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}}\right )}} \]
71/73205*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2 *x + 1))) + 2/3993*(213*(2*x - 1)^2 - 1771*x - 2079)/(5*(-2*x + 1)^(5/2) - 11*(-2*x + 1)^(3/2))
Time = 0.29 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.95 \[ \int \frac {(2+3 x)^2}{(1-2 x)^{5/2} (3+5 x)^2} \, dx=\frac {71}{73205} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) + \frac {7 \, {\left (24 \, x - 89\right )}}{3993 \, {\left (2 \, x - 1\right )} \sqrt {-2 \, x + 1}} - \frac {\sqrt {-2 \, x + 1}}{1331 \, {\left (5 \, x + 3\right )}} \]
71/73205*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 7/3993*(24*x - 89)/((2*x - 1)*sqrt(-2*x + 1)) - 1/13 31*sqrt(-2*x + 1)/(5*x + 3)
Time = 1.60 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.68 \[ \int \frac {(2+3 x)^2}{(1-2 x)^{5/2} (3+5 x)^2} \, dx=\frac {\frac {322\,x}{1815}-\frac {142\,{\left (2\,x-1\right )}^2}{6655}+\frac {126}{605}}{\frac {11\,{\left (1-2\,x\right )}^{3/2}}{5}-{\left (1-2\,x\right )}^{5/2}}-\frac {142\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )}{73205} \]