Integrand size = 24, antiderivative size = 80 \[ \int \frac {(2+3 x)^3}{(1-2 x)^{3/2} (3+5 x)^3} \, dx=\frac {7 (2+3 x)^2}{11 \sqrt {1-2 x} (3+5 x)^2}-\frac {\sqrt {1-2 x} (15676+24825 x)}{66550 (3+5 x)^2}-\frac {7143 \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{33275 \sqrt {55}} \]
-7143/1830125*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)+7/11*(2+3*x)^2 /(3+5*x)^2/(1-2*x)^(1/2)-1/66550*(15676+24825*x)*(1-2*x)^(1/2)/(3+5*x)^2
Time = 0.15 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.72 \[ \int \frac {(2+3 x)^3}{(1-2 x)^{3/2} (3+5 x)^3} \, dx=\frac {\frac {55 \left (153724+514727 x+430800 x^2\right )}{\sqrt {1-2 x} (3+5 x)^2}-14286 \sqrt {55} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{3660250} \]
((55*(153724 + 514727*x + 430800*x^2))/(Sqrt[1 - 2*x]*(3 + 5*x)^2) - 14286 *Sqrt[55]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/3660250
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, 25, 162, 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.
\(\displaystyle \int \frac {(3 x+2)^3}{(1-2 x)^{3/2} (5 x+3)^3} \, dx\) |
\(\Big \downarrow \) 109 |
\(\displaystyle \frac {7 (3 x+2)^2}{11 \sqrt {1-2 x} (5 x+3)^2}-\frac {1}{11} \int -\frac {(3 x+2) (3 x+16)}{\sqrt {1-2 x} (5 x+3)^3}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{11} \int \frac {(3 x+2) (3 x+16)}{\sqrt {1-2 x} (5 x+3)^3}dx+\frac {7 (3 x+2)^2}{11 \sqrt {1-2 x} (5 x+3)^2}\) |
\(\Big \downarrow \) 162 |
\(\displaystyle \frac {1}{11} \left (\frac {7143 \int \frac {1}{\sqrt {1-2 x} (5 x+3)}dx}{6050}-\frac {\sqrt {1-2 x} (24825 x+15676)}{6050 (5 x+3)^2}\right )+\frac {7 (3 x+2)^2}{11 \sqrt {1-2 x} (5 x+3)^2}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{11} \left (-\frac {7143 \int \frac {1}{\frac {11}{2}-\frac {5}{2} (1-2 x)}d\sqrt {1-2 x}}{6050}-\frac {\sqrt {1-2 x} (24825 x+15676)}{6050 (5 x+3)^2}\right )+\frac {7 (3 x+2)^2}{11 \sqrt {1-2 x} (5 x+3)^2}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{11} \left (-\frac {7143 \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{3025 \sqrt {55}}-\frac {\sqrt {1-2 x} (24825 x+15676)}{6050 (5 x+3)^2}\right )+\frac {7 (3 x+2)^2}{11 \sqrt {1-2 x} (5 x+3)^2}\) |
(7*(2 + 3*x)^2)/(11*Sqrt[1 - 2*x]*(3 + 5*x)^2) + (-1/6050*(Sqrt[1 - 2*x]*( 15676 + 24825*x))/(3 + 5*x)^2 - (7143*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/( 3025*Sqrt[55]))/11
3.22.28.3.1 Defintions of rubi rules used
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^3*c*e*g*(m + 2) - a^3*d*f*h*(n + 2) - a^2*b*(c*f*h*m - d*(f*g + e*h)*(m + n + 3)) - a*b^2*(c*(f*g + e*h) + d*e *g*(2*m + n + 4)) + b*(a^2*d*f*h*(m - n) - a*b*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(n + 1)) + b^2*(c*(f*g + e*h)*(m + 1) - d*e*g*(m + n + 2)))*x)/(b^2*(b *c - a*d)^2*(m + 1)*(m + 2)))*(a + b*x)^(m + 1)*(c + d*x)^(n + 1), x] + Sim p[(f*(h/b^2) - (d*(m + n + 3)*(a^2*d*f*h*(m - n) - a*b*(2*c*f*h*(m + 1) - d *(f*g + e*h)*(n + 1)) + b^2*(c*(f*g + e*h)*(m + 1) - d*e*g*(m + n + 2))))/( b^2*(b*c - a*d)^2*(m + 1)*(m + 2))) Int[(a + b*x)^(m + 2)*(c + d*x)^n, x] , x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && (LtQ[m, -2] || (EqQ[m + n + 3, 0] && !LtQ[n, -2]))
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 = 3.44 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.58
method | result | size |
risch | \(\frac {430800 x^{2}+514727 x +153724}{66550 \left (3+5 x \right )^{2} \sqrt {1-2 x}}-\frac {7143 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{1830125}\) | \(46\) |
pseudoelliptic | \(-\frac {7143 \left (\sqrt {55}\, \left (x +\frac {3}{5}\right )^{2} \sqrt {1-2 x}\, \operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right )-\frac {157960 x^{2}}{2381}-\frac {5661997 x}{71430}-\frac {845482}{35715}\right )}{73205 \sqrt {1-2 x}\, \left (3+5 x \right )^{2}}\) | \(56\) |
derivativedivides | \(\frac {\frac {41 \left (1-2 x \right )^{\frac {3}{2}}}{1331}-\frac {207 \sqrt {1-2 x}}{3025}}{\left (-6-10 x \right )^{2}}-\frac {7143 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{1830125}+\frac {343}{1331 \sqrt {1-2 x}}\) | \(57\) |
default | \(\frac {\frac {41 \left (1-2 x \right )^{\frac {3}{2}}}{1331}-\frac {207 \sqrt {1-2 x}}{3025}}{\left (-6-10 x \right )^{2}}-\frac {7143 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{1830125}+\frac {343}{1331 \sqrt {1-2 x}}\) | \(57\) |
trager | \(-\frac {\left (430800 x^{2}+514727 x +153724\right ) \sqrt {1-2 x}}{66550 \left (3+5 x \right )^{2} \left (-1+2 x \right )}+\frac {7143 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) \ln \left (\frac {5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) x +55 \sqrt {1-2 x}-8 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right )}{3+5 x}\right )}{3660250}\) | \(79\) |
1/66550*(430800*x^2+514727*x+153724)/(3+5*x)^2/(1-2*x)^(1/2)-7143/1830125* arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)
Time = 0.24 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.05 \[ \int \frac {(2+3 x)^3}{(1-2 x)^{3/2} (3+5 x)^3} \, dx=\frac {7143 \, \sqrt {55} {\left (50 \, x^{3} + 35 \, x^{2} - 12 \, x - 9\right )} \log \left (\frac {5 \, x + \sqrt {55} \sqrt {-2 \, x + 1} - 8}{5 \, x + 3}\right ) - 55 \, {\left (430800 \, x^{2} + 514727 \, x + 153724\right )} \sqrt {-2 \, x + 1}}{3660250 \, {\left (50 \, x^{3} + 35 \, x^{2} - 12 \, x - 9\right )}} \]
1/3660250*(7143*sqrt(55)*(50*x^3 + 35*x^2 - 12*x - 9)*log((5*x + sqrt(55)* sqrt(-2*x + 1) - 8)/(5*x + 3)) - 55*(430800*x^2 + 514727*x + 153724)*sqrt( -2*x + 1))/(50*x^3 + 35*x^2 - 12*x - 9)
Time = 145.65 (sec) , antiderivative size = 342, normalized size of antiderivative = 4.28 \[ \int \frac {(2+3 x)^3}{(1-2 x)^{3/2} (3+5 x)^3} \, dx=\frac {3469 \sqrt {55} \left (\log {\left (\sqrt {1 - 2 x} - \frac {\sqrt {55}}{5} \right )} - \log {\left (\sqrt {1 - 2 x} + \frac {\sqrt {55}}{5} \right )}\right )}{1830125} - \frac {404 \left (\begin {cases} \frac {\sqrt {55} \left (- \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1 \right )}}{4} + \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1 \right )}}{4} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1\right )} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1\right )}\right )}{605} & \text {for}\: \sqrt {1 - 2 x} > - \frac {\sqrt {55}}{5} \wedge \sqrt {1 - 2 x} < \frac {\sqrt {55}}{5} \end {cases}\right )}{3025} + \frac {8 \left (\begin {cases} \frac {\sqrt {55} \cdot \left (\frac {3 \log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1 \right )}}{16} - \frac {3 \log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1 \right )}}{16} + \frac {3}{16 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1\right )} + \frac {1}{16 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1\right )^{2}} + \frac {3}{16 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1\right )} - \frac {1}{16 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1\right )^{2}}\right )}{6655} & \text {for}\: \sqrt {1 - 2 x} > - \frac {\sqrt {55}}{5} \wedge \sqrt {1 - 2 x} < \frac {\sqrt {55}}{5} \end {cases}\right )}{275} + \frac {343}{1331 \sqrt {1 - 2 x}} \]
3469*sqrt(55)*(log(sqrt(1 - 2*x) - sqrt(55)/5) - log(sqrt(1 - 2*x) + sqrt( 55)/5))/1830125 - 404*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)))/3025 + 8*Piecewise((sqrt(55 )*(3*log(sqrt(55)*sqrt(1 - 2*x)/11 - 1)/16 - 3*log(sqrt(55)*sqrt(1 - 2*x)/ 11 + 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*(sq rt(55)*sqrt(1 - 2*x)/11 - 1)**2))/6655, (sqrt(1 - 2*x) > -sqrt(55)/5) & (s qrt(1 - 2*x) < sqrt(55)/5)))/275 + 343/(1331*sqrt(1 - 2*x))
Time = 0.27 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.04 \[ \int \frac {(2+3 x)^3}{(1-2 x)^{3/2} (3+5 x)^3} \, dx=\frac {7143}{3660250} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) + \frac {2 \, {\left (107700 \, {\left (2 \, x - 1\right )}^{2} + 945527 \, x + 46024\right )}}{33275 \, {\left (25 \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} - 110 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + 121 \, \sqrt {-2 \, x + 1}\right )}} \]
7143/3660250*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqr t(-2*x + 1))) + 2/33275*(107700*(2*x - 1)^2 + 945527*x + 46024)/(25*(-2*x + 1)^(5/2) - 110*(-2*x + 1)^(3/2) + 121*sqrt(-2*x + 1))
Time = 0.28 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.96 \[ \int \frac {(2+3 x)^3}{(1-2 x)^{3/2} (3+5 x)^3} \, dx=\frac {7143}{3660250} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) + \frac {343}{1331 \, \sqrt {-2 \, x + 1}} + \frac {1025 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 2277 \, \sqrt {-2 \, x + 1}}{133100 \, {\left (5 \, x + 3\right )}^{2}} \]
7143/3660250*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(5 5) + 5*sqrt(-2*x + 1))) + 343/1331/sqrt(-2*x + 1) + 1/133100*(1025*(-2*x + 1)^(3/2) - 2277*sqrt(-2*x + 1))/(5*x + 3)^2
Time = 1.37 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.78 \[ \int \frac {(2+3 x)^3}{(1-2 x)^{3/2} (3+5 x)^3} \, dx=\frac {\frac {171914\,x}{75625}+\frac {8616\,{\left (2\,x-1\right )}^2}{33275}+\frac {8368}{75625}}{\frac {121\,\sqrt {1-2\,x}}{25}-\frac {22\,{\left (1-2\,x\right )}^{3/2}}{5}+{\left (1-2\,x\right )}^{5/2}}-\frac {7143\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )}{1830125} \]