Integrand size = 22, antiderivative size = 96 \[ \int \frac {2+3 x}{(1-2 x)^{5/2} (3+5 x)^3} \, dx=\frac {73}{3993 (1-2 x)^{3/2}}+\frac {365}{14641 \sqrt {1-2 x}}-\frac {1}{110 (1-2 x)^{3/2} (3+5 x)^2}-\frac {73}{1210 (1-2 x)^{3/2} (3+5 x)}-\frac {365 \sqrt {\frac {5}{11}} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{14641} \]
73/3993/(1-2*x)^(3/2)-1/110/(1-2*x)^(3/2)/(3+5*x)^2-73/1210/(1-2*x)^(3/2)/ (3+5*x)-365/161051*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)+365/14641 /(1-2*x)^(1/2)
Time = 0.16 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.68 \[ \int \frac {2+3 x}{(1-2 x)^{5/2} (3+5 x)^3} \, dx=\frac {-\frac {11 \left (-17466-47961 x+36500 x^2+109500 x^3\right )}{2 (1-2 x)^{3/2} (3+5 x)^2}-1095 \sqrt {55} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{483153} \]
((-11*(-17466 - 47961*x + 36500*x^2 + 109500*x^3))/(2*(1 - 2*x)^(3/2)*(3 + 5*x)^2) - 1095*Sqrt[55]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/483153
Time = 0.19 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.16, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {87, 52, 61, 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.
| \(\displaystyle \int \frac {3 x+2}{(1-2 x)^{5/2} (5 x+3)^3} \, dx\) |
| \(\Big \downarrow \) 87 |
| \(\displaystyle \frac {73}{110} \int \frac {1}{(1-2 x)^{5/2} (5 x+3)^2}dx-\frac {1}{110 (1-2 x)^{3/2} (5 x+3)^2}\) |
| \(\Big \downarrow \) 52 |
| \(\displaystyle \frac {73}{110} \left (\frac {5}{11} \int \frac {1}{(1-2 x)^{5/2} (5 x+3)}dx-\frac {1}{11 (1-2 x)^{3/2} (5 x+3)}\right )-\frac {1}{110 (1-2 x)^{3/2} (5 x+3)^2}\) |
| \(\Big \downarrow \) 61 |
| \(\displaystyle \frac {73}{110} \left (\frac {5}{11} \left (\frac {5}{11} \int \frac {1}{(1-2 x)^{3/2} (5 x+3)}dx+\frac {2}{33 (1-2 x)^{3/2}}\right )-\frac {1}{11 (1-2 x)^{3/2} (5 x+3)}\right )-\frac {1}{110 (1-2 x)^{3/2} (5 x+3)^2}\) |
| \(\Big \downarrow \) 61 |
| \(\displaystyle \frac {73}{110} \left (\frac {5}{11} \left (\frac {5}{11} \left (\frac {5}{11} \int \frac {1}{\sqrt {1-2 x} (5 x+3)}dx+\frac {2}{11 \sqrt {1-2 x}}\right )+\frac {2}{33 (1-2 x)^{3/2}}\right )-\frac {1}{11 (1-2 x)^{3/2} (5 x+3)}\right )-\frac {1}{110 (1-2 x)^{3/2} (5 x+3)^2}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {73}{110} \left (\frac {5}{11} \left (\frac {5}{11} \left (\frac {2}{11 \sqrt {1-2 x}}-\frac {5}{11} \int \frac {1}{\frac {11}{2}-\frac {5}{2} (1-2 x)}d\sqrt {1-2 x}\right )+\frac {2}{33 (1-2 x)^{3/2}}\right )-\frac {1}{11 (1-2 x)^{3/2} (5 x+3)}\right )-\frac {1}{110 (1-2 x)^{3/2} (5 x+3)^2}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {73}{110} \left (\frac {5}{11} \left (\frac {5}{11} \left (\frac {2}{11 \sqrt {1-2 x}}-\frac {2}{11} \sqrt {\frac {5}{11}} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )\right )+\frac {2}{33 (1-2 x)^{3/2}}\right )-\frac {1}{11 (1-2 x)^{3/2} (5 x+3)}\right )-\frac {1}{110 (1-2 x)^{3/2} (5 x+3)^2}\) |
-1/110*1/((1 - 2*x)^(3/2)*(3 + 5*x)^2) + (73*(-1/11*1/((1 - 2*x)^(3/2)*(3 + 5*x)) + (5*(2/(33*(1 - 2*x)^(3/2)) + (5*(2/(11*Sqrt[1 - 2*x]) - (2*Sqrt[ 5/11]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/11))/11))/11))/110
3.22.95.3.1 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] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
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)^(-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 = 1.12 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.60
| method | result | size |
| risch | \(\frac {109500 x^{3}+36500 x^{2}-47961 x -17466}{87846 \left (3+5 x \right )^{2} \sqrt {1-2 x}\, \left (-1+2 x \right )}-\frac {365 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{161051}\) | \(58\) |
| derivativedivides | \(\frac {\frac {175 \left (1-2 x \right )^{\frac {3}{2}}}{1331}-\frac {395 \sqrt {1-2 x}}{1331}}{\left (-6-10 x \right )^{2}}-\frac {365 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{161051}+\frac {28}{3993 \left (1-2 x \right )^{\frac {3}{2}}}+\frac {288}{14641 \sqrt {1-2 x}}\) | \(66\) |
| default | \(\frac {\frac {175 \left (1-2 x \right )^{\frac {3}{2}}}{1331}-\frac {395 \sqrt {1-2 x}}{1331}}{\left (-6-10 x \right )^{2}}-\frac {365 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{161051}+\frac {28}{3993 \left (1-2 x \right )^{\frac {3}{2}}}+\frac {288}{14641 \sqrt {1-2 x}}\) | \(66\) |
| pseudoelliptic | \(\frac {\frac {365 \sqrt {1-2 x}\, \operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \left (-1+2 x \right ) \left (3+5 x \right )^{2} \sqrt {55}}{161051}-\frac {18250 x^{3}}{14641}-\frac {18250 x^{2}}{43923}+\frac {15987 x}{29282}+\frac {2911}{14641}}{\left (1-2 x \right )^{\frac {3}{2}} \left (3+5 x \right )^{2}}\) | \(69\) |
| trager | \(-\frac {\left (109500 x^{3}+36500 x^{2}-47961 x -17466\right ) \sqrt {1-2 x}}{87846 \left (10 x^{2}+x -3\right )^{2}}+\frac {365 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) \ln \left (\frac {5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) x +55 \sqrt {1-2 x}-8 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right )}{3+5 x}\right )}{322102}\) | \(80\) |
1/87846*(109500*x^3+36500*x^2-47961*x-17466)/(3+5*x)^2/(1-2*x)^(1/2)/(-1+2 *x)-365/161051*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)
Time = 0.23 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.09 \[ \int \frac {2+3 x}{(1-2 x)^{5/2} (3+5 x)^3} \, dx=\frac {1095 \, \sqrt {11} \sqrt {5} {\left (100 \, x^{4} + 20 \, x^{3} - 59 \, x^{2} - 6 \, x + 9\right )} \log \left (\frac {\sqrt {11} \sqrt {5} \sqrt {-2 \, x + 1} + 5 \, x - 8}{5 \, x + 3}\right ) - 11 \, {\left (109500 \, x^{3} + 36500 \, x^{2} - 47961 \, x - 17466\right )} \sqrt {-2 \, x + 1}}{966306 \, {\left (100 \, x^{4} + 20 \, x^{3} - 59 \, x^{2} - 6 \, x + 9\right )}} \]
1/966306*(1095*sqrt(11)*sqrt(5)*(100*x^4 + 20*x^3 - 59*x^2 - 6*x + 9)*log( (sqrt(11)*sqrt(5)*sqrt(-2*x + 1) + 5*x - 8)/(5*x + 3)) - 11*(109500*x^3 + 36500*x^2 - 47961*x - 17466)*sqrt(-2*x + 1))/(100*x^4 + 20*x^3 - 59*x^2 - 6*x + 9)
Time = 86.93 (sec) , antiderivative size = 354, normalized size of antiderivative = 3.69 \[ \int \frac {2+3 x}{(1-2 x)^{5/2} (3+5 x)^3} \, dx=\frac {144 \sqrt {55} \left (\log {\left (\sqrt {1 - 2 x} - \frac {\sqrt {55}}{5} \right )} - \log {\left (\sqrt {1 - 2 x} + \frac {\sqrt {55}}{5} \right )}\right )}{161051} - \frac {740 \left (\begin {cases} \frac {\sqrt {55} \left (- \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1 \right )}}{4} + \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1 \right )}}{4} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1\right )} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1\right )}\right )}{605} & \text {for}\: \sqrt {1 - 2 x} > - \frac {\sqrt {55}}{5} \wedge \sqrt {1 - 2 x} < \frac {\sqrt {55}}{5} \end {cases}\right )}{1331} + \frac {40 \left (\begin {cases} \frac {\sqrt {55} \cdot \left (\frac {3 \log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1 \right )}}{16} - \frac {3 \log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1 \right )}}{16} + \frac {3}{16 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1\right )} + \frac {1}{16 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1\right )^{2}} + \frac {3}{16 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1\right )} - \frac {1}{16 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1\right )^{2}}\right )}{6655} & \text {for}\: \sqrt {1 - 2 x} > - \frac {\sqrt {55}}{5} \wedge \sqrt {1 - 2 x} < \frac {\sqrt {55}}{5} \end {cases}\right )}{121} + \frac {288}{14641 \sqrt {1 - 2 x}} + \frac {28}{3993 \left (1 - 2 x\right )^{\frac {3}{2}}} \]
144*sqrt(55)*(log(sqrt(1 - 2*x) - sqrt(55)/5) - log(sqrt(1 - 2*x) + sqrt(5 5)/5))/161051 - 740*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)))/1331 + 40*Piecewise((sqrt(55) *(3*log(sqrt(55)*sqrt(1 - 2*x)/11 - 1)/16 - 3*log(sqrt(55)*sqrt(1 - 2*x)/1 1 + 1)/16 + 3/(16*(sqrt(55)*sqrt(1 - 2*x)/11 + 1)) + 1/(16*(sqrt(55)*sqrt( 1 - 2*x)/11 + 1)**2) + 3/(16*(sqrt(55)*sqrt(1 - 2*x)/11 - 1)) - 1/(16*(sqr t(55)*sqrt(1 - 2*x)/11 - 1)**2))/6655, (sqrt(1 - 2*x) > -sqrt(55)/5) & (sq rt(1 - 2*x) < sqrt(55)/5)))/121 + 288/(14641*sqrt(1 - 2*x)) + 28/(3993*(1 - 2*x)**(3/2))
Time = 0.30 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.96 \[ \int \frac {2+3 x}{(1-2 x)^{5/2} (3+5 x)^3} \, dx=\frac {365}{322102} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) - \frac {27375 \, {\left (2 \, x - 1\right )}^{3} + 100375 \, {\left (2 \, x - 1\right )}^{2} + 141328 \, x - 107932}{43923 \, {\left (25 \, {\left (-2 \, x + 1\right )}^{\frac {7}{2}} - 110 \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} + 121 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}}\right )}} \]
365/322102*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt( -2*x + 1))) - 1/43923*(27375*(2*x - 1)^3 + 100375*(2*x - 1)^2 + 141328*x - 107932)/(25*(-2*x + 1)^(7/2) - 110*(-2*x + 1)^(5/2) + 121*(-2*x + 1)^(3/2 ))
Time = 0.29 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.93 \[ \int \frac {2+3 x}{(1-2 x)^{5/2} (3+5 x)^3} \, dx=\frac {365}{322102} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) + \frac {4 \, {\left (432 \, x - 293\right )}}{43923 \, {\left (2 \, x - 1\right )} \sqrt {-2 \, x + 1}} + \frac {5 \, {\left (35 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 79 \, \sqrt {-2 \, x + 1}\right )}}{5324 \, {\left (5 \, x + 3\right )}^{2}} \]
365/322102*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 4/43923*(432*x - 293)/((2*x - 1)*sqrt(-2*x + 1)) + 5/5324*(35*(-2*x + 1)^(3/2) - 79*sqrt(-2*x + 1))/(5*x + 3)^2
Time = 0.08 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.75 \[ \int \frac {2+3 x}{(1-2 x)^{5/2} (3+5 x)^3} \, dx=-\frac {365\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )}{161051}-\frac {\frac {1168\,x}{9075}+\frac {365\,{\left (2\,x-1\right )}^2}{3993}+\frac {365\,{\left (2\,x-1\right )}^3}{14641}-\frac {892}{9075}}{\frac {121\,{\left (1-2\,x\right )}^{3/2}}{25}-\frac {22\,{\left (1-2\,x\right )}^{5/2}}{5}+{\left (1-2\,x\right )}^{7/2}} \]