Integrand size = 17, antiderivative size = 83 \[ \int \frac {1}{(1-2 x)^{3/2} (3+5 x)^3} \, dx=\frac {15}{1331 \sqrt {1-2 x}}-\frac {1}{22 \sqrt {1-2 x} (3+5 x)^2}-\frac {5}{242 \sqrt {1-2 x} (3+5 x)}-\frac {15 \sqrt {\frac {5}{11}} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{1331} \]
-15/14641*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)+15/1331/(1-2*x)^(1 /2)-1/22/(3+5*x)^2/(1-2*x)^(1/2)-5/242/(3+5*x)/(1-2*x)^(1/2)
Time = 0.15 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.70 \[ \int \frac {1}{(1-2 x)^{3/2} (3+5 x)^3} \, dx=\frac {\frac {11 \left (-16+625 x+750 x^2\right )}{\sqrt {1-2 x} (3+5 x)^2}-30 \sqrt {55} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{29282} \]
((11*(-16 + 625*x + 750*x^2))/(Sqrt[1 - 2*x]*(3 + 5*x)^2) - 30*Sqrt[55]*Ar cTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/29282
Time = 0.17 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.12, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {52, 52, 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.
\(\displaystyle \int \frac {1}{(1-2 x)^{3/2} (5 x+3)^3} \, dx\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {5}{22} \int \frac {1}{(1-2 x)^{3/2} (5 x+3)^2}dx-\frac {1}{22 \sqrt {1-2 x} (5 x+3)^2}\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {5}{22} \left (\frac {3}{11} \int \frac {1}{(1-2 x)^{3/2} (5 x+3)}dx-\frac {1}{11 \sqrt {1-2 x} (5 x+3)}\right )-\frac {1}{22 \sqrt {1-2 x} (5 x+3)^2}\) |
\(\Big \downarrow \) 61 |
\(\displaystyle \frac {5}{22} \left (\frac {3}{11} \left (\frac {5}{11} \int \frac {1}{\sqrt {1-2 x} (5 x+3)}dx+\frac {2}{11 \sqrt {1-2 x}}\right )-\frac {1}{11 \sqrt {1-2 x} (5 x+3)}\right )-\frac {1}{22 \sqrt {1-2 x} (5 x+3)^2}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {5}{22} \left (\frac {3}{11} \left (\frac {2}{11 \sqrt {1-2 x}}-\frac {5}{11} \int \frac {1}{\frac {11}{2}-\frac {5}{2} (1-2 x)}d\sqrt {1-2 x}\right )-\frac {1}{11 \sqrt {1-2 x} (5 x+3)}\right )-\frac {1}{22 \sqrt {1-2 x} (5 x+3)^2}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {5}{22} \left (\frac {3}{11} \left (\frac {2}{11 \sqrt {1-2 x}}-\frac {2}{11} \sqrt {\frac {5}{11}} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )\right )-\frac {1}{11 \sqrt {1-2 x} (5 x+3)}\right )-\frac {1}{22 \sqrt {1-2 x} (5 x+3)^2}\) |
-1/22*1/(Sqrt[1 - 2*x]*(3 + 5*x)^2) + (5*(-1/11*1/(Sqrt[1 - 2*x]*(3 + 5*x) ) + (3*(2/(11*Sqrt[1 - 2*x]) - (2*Sqrt[5/11]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2 *x]])/11))/11))/22
3.22.31.3.1 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_)^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 = 46, normalized size of antiderivative = 0.55
method | result | size |
risch | \(\frac {750 x^{2}+625 x -16}{2662 \left (3+5 x \right )^{2} \sqrt {1-2 x}}-\frac {15 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{14641}\) | \(46\) |
pseudoelliptic | \(-\frac {375 \left (\sqrt {55}\, \left (x +\frac {3}{5}\right )^{2} \sqrt {1-2 x}\, \operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right )-11 x^{2}-\frac {55 x}{6}+\frac {88}{375}\right )}{14641 \sqrt {1-2 x}\, \left (3+5 x \right )^{2}}\) | \(56\) |
derivativedivides | \(\frac {\frac {175 \left (1-2 x \right )^{\frac {3}{2}}}{1331}-\frac {45 \sqrt {1-2 x}}{121}}{\left (-6-10 x \right )^{2}}-\frac {15 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{14641}+\frac {8}{1331 \sqrt {1-2 x}}\) | \(57\) |
default | \(\frac {\frac {175 \left (1-2 x \right )^{\frac {3}{2}}}{1331}-\frac {45 \sqrt {1-2 x}}{121}}{\left (-6-10 x \right )^{2}}-\frac {15 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{14641}+\frac {8}{1331 \sqrt {1-2 x}}\) | \(57\) |
trager | \(-\frac {\left (750 x^{2}+625 x -16\right ) \sqrt {1-2 x}}{2662 \left (3+5 x \right )^{2} \left (-1+2 x \right )}+\frac {15 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) \ln \left (\frac {5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) x +55 \sqrt {1-2 x}-8 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right )}{3+5 x}\right )}{29282}\) | \(79\) |
1/2662*(750*x^2+625*x-16)/(3+5*x)^2/(1-2*x)^(1/2)-15/14641*arctanh(1/11*55 ^(1/2)*(1-2*x)^(1/2))*55^(1/2)
Time = 0.22 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.08 \[ \int \frac {1}{(1-2 x)^{3/2} (3+5 x)^3} \, dx=\frac {15 \, \sqrt {11} \sqrt {5} {\left (50 \, x^{3} + 35 \, x^{2} - 12 \, x - 9\right )} \log \left (\frac {\sqrt {11} \sqrt {5} \sqrt {-2 \, x + 1} + 5 \, x - 8}{5 \, x + 3}\right ) - 11 \, {\left (750 \, x^{2} + 625 \, x - 16\right )} \sqrt {-2 \, x + 1}}{29282 \, {\left (50 \, x^{3} + 35 \, x^{2} - 12 \, x - 9\right )}} \]
1/29282*(15*sqrt(11)*sqrt(5)*(50*x^3 + 35*x^2 - 12*x - 9)*log((sqrt(11)*sq rt(5)*sqrt(-2*x + 1) + 5*x - 8)/(5*x + 3)) - 11*(750*x^2 + 625*x - 16)*sqr t(-2*x + 1))/(50*x^3 + 35*x^2 - 12*x - 9)
Result contains complex when optimal does not.
Time = 3.28 (sec) , antiderivative size = 231, normalized size of antiderivative = 2.78 \[ \int \frac {1}{(1-2 x)^{3/2} (3+5 x)^3} \, dx=\begin {cases} - \frac {15 \sqrt {55} \operatorname {acosh}{\left (\frac {\sqrt {110}}{10 \sqrt {x + \frac {3}{5}}} \right )}}{14641} + \frac {15 \sqrt {2}}{2662 \sqrt {-1 + \frac {11}{10 \left (x + \frac {3}{5}\right )}} \sqrt {x + \frac {3}{5}}} - \frac {\sqrt {2}}{484 \sqrt {-1 + \frac {11}{10 \left (x + \frac {3}{5}\right )}} \left (x + \frac {3}{5}\right )^{\frac {3}{2}}} - \frac {\sqrt {2}}{1100 \sqrt {-1 + \frac {11}{10 \left (x + \frac {3}{5}\right )}} \left (x + \frac {3}{5}\right )^{\frac {5}{2}}} & \text {for}\: \frac {1}{\left |{x + \frac {3}{5}}\right |} > \frac {10}{11} \\\frac {15 \sqrt {55} i \operatorname {asin}{\left (\frac {\sqrt {110}}{10 \sqrt {x + \frac {3}{5}}} \right )}}{14641} - \frac {15 \sqrt {2} i}{2662 \sqrt {1 - \frac {11}{10 \left (x + \frac {3}{5}\right )}} \sqrt {x + \frac {3}{5}}} + \frac {\sqrt {2} i}{484 \sqrt {1 - \frac {11}{10 \left (x + \frac {3}{5}\right )}} \left (x + \frac {3}{5}\right )^{\frac {3}{2}}} + \frac {\sqrt {2} i}{1100 \sqrt {1 - \frac {11}{10 \left (x + \frac {3}{5}\right )}} \left (x + \frac {3}{5}\right )^{\frac {5}{2}}} & \text {otherwise} \end {cases} \]
Piecewise((-15*sqrt(55)*acosh(sqrt(110)/(10*sqrt(x + 3/5)))/14641 + 15*sqr t(2)/(2662*sqrt(-1 + 11/(10*(x + 3/5)))*sqrt(x + 3/5)) - sqrt(2)/(484*sqrt (-1 + 11/(10*(x + 3/5)))*(x + 3/5)**(3/2)) - sqrt(2)/(1100*sqrt(-1 + 11/(1 0*(x + 3/5)))*(x + 3/5)**(5/2)), 1/Abs(x + 3/5) > 10/11), (15*sqrt(55)*I*a sin(sqrt(110)/(10*sqrt(x + 3/5)))/14641 - 15*sqrt(2)*I/(2662*sqrt(1 - 11/( 10*(x + 3/5)))*sqrt(x + 3/5)) + sqrt(2)*I/(484*sqrt(1 - 11/(10*(x + 3/5))) *(x + 3/5)**(3/2)) + sqrt(2)*I/(1100*sqrt(1 - 11/(10*(x + 3/5)))*(x + 3/5) **(5/2)), True))
Time = 0.28 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.00 \[ \int \frac {1}{(1-2 x)^{3/2} (3+5 x)^3} \, dx=\frac {15}{29282} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) + \frac {375 \, {\left (2 \, x - 1\right )}^{2} + 2750 \, x - 407}{1331 \, {\left (25 \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} - 110 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + 121 \, \sqrt {-2 \, x + 1}\right )}} \]
15/29282*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2 *x + 1))) + 1/1331*(375*(2*x - 1)^2 + 2750*x - 407)/(25*(-2*x + 1)^(5/2) - 110*(-2*x + 1)^(3/2) + 121*sqrt(-2*x + 1))
Time = 0.30 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.93 \[ \int \frac {1}{(1-2 x)^{3/2} (3+5 x)^3} \, dx=\frac {15}{29282} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) + \frac {8}{1331 \, \sqrt {-2 \, x + 1}} + \frac {5 \, {\left (35 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 99 \, \sqrt {-2 \, x + 1}\right )}}{5324 \, {\left (5 \, x + 3\right )}^{2}} \]
15/29282*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 8/1331/sqrt(-2*x + 1) + 5/5324*(35*(-2*x + 1)^(3/2) - 99*sqrt(-2*x + 1))/(5*x + 3)^2
Time = 0.12 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.75 \[ \int \frac {1}{(1-2 x)^{3/2} (3+5 x)^3} \, dx=\frac {\frac {10\,x}{121}+\frac {15\,{\left (2\,x-1\right )}^2}{1331}-\frac {37}{3025}}{\frac {121\,\sqrt {1-2\,x}}{25}-\frac {22\,{\left (1-2\,x\right )}^{3/2}}{5}+{\left (1-2\,x\right )}^{5/2}}-\frac {15\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )}{14641} \]