Integrand size = 24, antiderivative size = 80 \[ \int \frac {(2+3 x)^3}{(1-2 x)^{5/2} (3+5 x)^2} \, dx=\frac {7 (2+3 x)^2}{33 (1-2 x)^{3/2} (3+5 x)}-\frac {2 (10309+17112 x)}{19965 \sqrt {1-2 x} (3+5 x)}-\frac {208 \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{6655 \sqrt {55}} \]
7/33*(2+3*x)^2/(1-2*x)^(3/2)/(3+5*x)-208/366025*arctanh(1/11*55^(1/2)*(1-2 *x)^(1/2))*55^(1/2)-2/19965*(10309+17112*x)/(3+5*x)/(1-2*x)^(1/2)
Time = 0.13 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.72 \[ \int \frac {(2+3 x)^3}{(1-2 x)^{5/2} (3+5 x)^2} \, dx=\frac {\frac {55 \left (-3678+57832 x+106563 x^2\right )}{(1-2 x)^{3/2} (3+5 x)}-624 \sqrt {55} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{1098075} \]
((55*(-3678 + 57832*x + 106563*x^2))/((1 - 2*x)^(3/2)*(3 + 5*x)) - 624*Sqr t[55]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/1098075
Time = 0.18 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.06, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {109, 27, 161, 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.
\(\displaystyle \int \frac {(3 x+2)^3}{(1-2 x)^{5/2} (5 x+3)^2} \, dx\) |
\(\Big \downarrow \) 109 |
\(\displaystyle \frac {7 (3 x+2)^2}{33 (1-2 x)^{3/2} (5 x+3)}-\frac {1}{33} \int \frac {2 (3 x+2) (48 x+25)}{(1-2 x)^{3/2} (5 x+3)^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {7 (3 x+2)^2}{33 (1-2 x)^{3/2} (5 x+3)}-\frac {2}{33} \int \frac {(3 x+2) (48 x+25)}{(1-2 x)^{3/2} (5 x+3)^2}dx\) |
\(\Big \downarrow \) 161 |
\(\displaystyle \frac {7 (3 x+2)^2}{33 (1-2 x)^{3/2} (5 x+3)}-\frac {2}{33} \left (\frac {17112 x+10309}{605 \sqrt {1-2 x} (5 x+3)}-\frac {156}{605} \int \frac {1}{\sqrt {1-2 x} (5 x+3)}dx\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {7 (3 x+2)^2}{33 (1-2 x)^{3/2} (5 x+3)}-\frac {2}{33} \left (\frac {156}{605} \int \frac {1}{\frac {11}{2}-\frac {5}{2} (1-2 x)}d\sqrt {1-2 x}+\frac {17112 x+10309}{605 \sqrt {1-2 x} (5 x+3)}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {7 (3 x+2)^2}{33 (1-2 x)^{3/2} (5 x+3)}-\frac {2}{33} \left (\frac {312 \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{605 \sqrt {55}}+\frac {17112 x+10309}{605 \sqrt {1-2 x} (5 x+3)}\right )\) |
(7*(2 + 3*x)^2)/(33*(1 - 2*x)^(3/2)*(3 + 5*x)) - (2*((10309 + 17112*x)/(60 5*Sqrt[1 - 2*x]*(3 + 5*x)) + (312*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(605* Sqrt[55])))/33
3.22.83.3.1 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_)) *((g_.) + (h_.)*(x_)), x_] :> Simp[((b^2*c*d*e*g*(n + 1) + a^2*c*d*f*h*(n + 1) + a*b*(d^2*e*g*(m + 1) + c^2*f*h*(m + 1) - c*d*(f*g + e*h)*(m + n + 2)) + (a^2*d^2*f*h*(n + 1) - a*b*d^2*(f*g + e*h)*(n + 1) + b^2*(c^2*f*h*(m + 1 ) - c*d*(f*g + e*h)*(m + 1) + d^2*e*g*(m + n + 2)))*x)/(b*d*(b*c - a*d)^2*( m + 1)*(n + 1)))*(a + b*x)^(m + 1)*(c + d*x)^(n + 1), x] - Simp[(a^2*d^2*f* h*(2 + 3*n + n^2) + a*b*d*(n + 1)*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b^2*(c^2*f*h*(2 + 3*m + m^2) - c*d*(f*g + e*h)*(m + 1)*(m + n + 3) + d^2*e*g*(6 + m^2 + 5*n + n^2 + m*(2*n + 5))))/(b*d*(b*c - a*d)^2*(m + 1)*( n + 1)) Int[(a + b*x)^(m + 1)*(c + d*x)^(n + 1), x], x] /; FreeQ[{a, b, c , d, e, f, g, h}, x] && LtQ[m, -1] && LtQ[n, -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 = 3.54 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.66
method | result | size |
risch | \(-\frac {106563 x^{2}+57832 x -3678}{19965 \left (-1+2 x \right ) \sqrt {1-2 x}\, \left (3+5 x \right )}-\frac {208 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{366025}\) | \(53\) |
derivativedivides | \(\frac {2 \sqrt {1-2 x}}{33275 \left (-\frac {6}{5}-2 x \right )}-\frac {208 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{366025}+\frac {343}{726 \left (1-2 x \right )^{\frac {3}{2}}}-\frac {1421}{2662 \sqrt {1-2 x}}\) | \(54\) |
default | \(\frac {2 \sqrt {1-2 x}}{33275 \left (-\frac {6}{5}-2 x \right )}-\frac {208 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{366025}+\frac {343}{726 \left (1-2 x \right )^{\frac {3}{2}}}-\frac {1421}{2662 \sqrt {1-2 x}}\) | \(54\) |
pseudoelliptic | \(\frac {624 \sqrt {1-2 x}\, \operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \left (10 x^{2}+x -3\right ) \sqrt {55}+5860965 x^{2}+3180760 x -202290}{\left (1-2 x \right )^{\frac {3}{2}} \left (3294225+5490375 x \right )}\) | \(60\) |
trager | \(\frac {\left (106563 x^{2}+57832 x -3678\right ) \sqrt {1-2 x}}{19965 \left (-1+2 x \right )^{2} \left (3+5 x \right )}-\frac {104 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) \ln \left (\frac {-5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) x +55 \sqrt {1-2 x}+8 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right )}{3+5 x}\right )}{366025}\) | \(79\) |
-1/19965*(106563*x^2+57832*x-3678)/(-1+2*x)/(1-2*x)^(1/2)/(3+5*x)-208/3660 25*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)
Time = 0.23 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.05 \[ \int \frac {(2+3 x)^3}{(1-2 x)^{5/2} (3+5 x)^2} \, dx=\frac {312 \, \sqrt {55} {\left (20 \, x^{3} - 8 \, x^{2} - 7 \, x + 3\right )} \log \left (\frac {5 \, x + \sqrt {55} \sqrt {-2 \, x + 1} - 8}{5 \, x + 3}\right ) + 55 \, {\left (106563 \, x^{2} + 57832 \, x - 3678\right )} \sqrt {-2 \, x + 1}}{1098075 \, {\left (20 \, x^{3} - 8 \, x^{2} - 7 \, x + 3\right )}} \]
1/1098075*(312*sqrt(55)*(20*x^3 - 8*x^2 - 7*x + 3)*log((5*x + sqrt(55)*sqr t(-2*x + 1) - 8)/(5*x + 3)) + 55*(106563*x^2 + 57832*x - 3678)*sqrt(-2*x + 1))/(20*x^3 - 8*x^2 - 7*x + 3)
Time = 54.38 (sec) , antiderivative size = 185, normalized size of antiderivative = 2.31 \[ \int \frac {(2+3 x)^3}{(1-2 x)^{5/2} (3+5 x)^2} \, dx=\frac {103 \sqrt {55} \left (\log {\left (\sqrt {1 - 2 x} - \frac {\sqrt {55}}{5} \right )} - \log {\left (\sqrt {1 - 2 x} + \frac {\sqrt {55}}{5} \right )}\right )}{366025} - \frac {4 \left (\begin {cases} \frac {\sqrt {55} \left (- \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1 \right )}}{4} + \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1 \right )}}{4} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1\right )} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1\right )}\right )}{605} & \text {for}\: \sqrt {1 - 2 x} > - \frac {\sqrt {55}}{5} \wedge \sqrt {1 - 2 x} < \frac {\sqrt {55}}{5} \end {cases}\right )}{605} - \frac {1421}{2662 \sqrt {1 - 2 x}} + \frac {343}{726 \left (1 - 2 x\right )^{\frac {3}{2}}} \]
103*sqrt(55)*(log(sqrt(1 - 2*x) - sqrt(55)/5) - log(sqrt(1 - 2*x) + sqrt(5 5)/5))/366025 - 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)/1 1 + 1)) - 1/(4*(sqrt(55)*sqrt(1 - 2*x)/11 - 1)))/605, (sqrt(1 - 2*x) > -sq rt(55)/5) & (sqrt(1 - 2*x) < sqrt(55)/5)))/605 - 1421/(2662*sqrt(1 - 2*x)) + 343/(726*(1 - 2*x)**(3/2))
Time = 0.27 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.92 \[ \int \frac {(2+3 x)^3}{(1-2 x)^{5/2} (3+5 x)^2} \, dx=\frac {104}{366025} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) - \frac {106563 \, {\left (2 \, x - 1\right )}^{2} + 657580 \, x - 121275}{39930 \, {\left (5 \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} - 11 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}}\right )}} \]
104/366025*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt( -2*x + 1))) - 1/39930*(106563*(2*x - 1)^2 + 657580*x - 121275)/(5*(-2*x + 1)^(5/2) - 11*(-2*x + 1)^(3/2))
Time = 0.29 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.96 \[ \int \frac {(2+3 x)^3}{(1-2 x)^{5/2} (3+5 x)^2} \, dx=\frac {104}{366025} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) - \frac {49 \, {\left (87 \, x - 5\right )}}{3993 \, {\left (2 \, x - 1\right )} \sqrt {-2 \, x + 1}} - \frac {\sqrt {-2 \, x + 1}}{6655 \, {\left (5 \, x + 3\right )}} \]
104/366025*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 49/3993*(87*x - 5)/((2*x - 1)*sqrt(-2*x + 1)) - 1/ 6655*sqrt(-2*x + 1)/(5*x + 3)
Time = 1.44 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.69 \[ \int \frac {(2+3 x)^3}{(1-2 x)^{5/2} (3+5 x)^2} \, dx=\frac {\frac {5978\,x}{1815}+\frac {35521\,{\left (2\,x-1\right )}^2}{66550}-\frac {147}{242}}{\frac {11\,{\left (1-2\,x\right )}^{3/2}}{5}-{\left (1-2\,x\right )}^{5/2}}-\frac {208\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )}{366025} \]