Integrand size = 24, antiderivative size = 100 \[ \int \frac {(2+3 x)^4}{(1-2 x)^{3/2} (3+5 x)^2} \, dx=-\frac {36 \sqrt {1-2 x} (2+3 x)^2}{605 (3+5 x)}+\frac {7 (2+3 x)^3}{11 \sqrt {1-2 x} (3+5 x)}+\frac {27 \sqrt {1-2 x} (792+265 x)}{3025}-\frac {54 \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{3025 \sqrt {55}} \]
-54/166375*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)+7/11*(2+3*x)^3/(3 +5*x)/(1-2*x)^(1/2)-36/605*(2+3*x)^2*(1-2*x)^(1/2)/(3+5*x)+27/3025*(792+26 5*x)*(1-2*x)^(1/2)
Time = 0.12 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.63 \[ \int \frac {(2+3 x)^4}{(1-2 x)^{3/2} (3+5 x)^2} \, dx=\frac {-\frac {55 \left (-78832-68661 x+114345 x^2+16335 x^3\right )}{\sqrt {1-2 x} (3+5 x)}-54 \sqrt {55} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{166375} \]
((-55*(-78832 - 68661*x + 114345*x^2 + 16335*x^3))/(Sqrt[1 - 2*x]*(3 + 5*x )) - 54*Sqrt[55]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/166375
Time = 0.21 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.10, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {109, 27, 166, 27, 164, 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.
| \(\displaystyle \int \frac {(3 x+2)^4}{(1-2 x)^{3/2} (5 x+3)^2} \, dx\) |
| \(\Big \downarrow \) 109 |
| \(\displaystyle \frac {7 (3 x+2)^3}{11 \sqrt {1-2 x} (5 x+3)}-\frac {1}{11} \int \frac {9 (3 x+2)^2 (23 x+13)}{\sqrt {1-2 x} (5 x+3)^2}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {7 (3 x+2)^3}{11 \sqrt {1-2 x} (5 x+3)}-\frac {9}{11} \int \frac {(3 x+2)^2 (23 x+13)}{\sqrt {1-2 x} (5 x+3)^2}dx\) |
| \(\Big \downarrow \) 166 |
| \(\displaystyle \frac {7 (3 x+2)^3}{11 \sqrt {1-2 x} (5 x+3)}-\frac {9}{11} \left (\frac {1}{55} \int \frac {3 (3 x+2) (265 x+158)}{\sqrt {1-2 x} (5 x+3)}dx+\frac {4 \sqrt {1-2 x} (3 x+2)^2}{55 (5 x+3)}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {7 (3 x+2)^3}{11 \sqrt {1-2 x} (5 x+3)}-\frac {9}{11} \left (\frac {3}{55} \int \frac {(3 x+2) (265 x+158)}{\sqrt {1-2 x} (5 x+3)}dx+\frac {4 \sqrt {1-2 x} (3 x+2)^2}{55 (5 x+3)}\right )\) |
| \(\Big \downarrow \) 164 |
| \(\displaystyle \frac {7 (3 x+2)^3}{11 \sqrt {1-2 x} (5 x+3)}-\frac {9}{11} \left (\frac {3}{55} \left (-\frac {1}{5} \int \frac {1}{\sqrt {1-2 x} (5 x+3)}dx-\frac {1}{5} \sqrt {1-2 x} (265 x+792)\right )+\frac {4 \sqrt {1-2 x} (3 x+2)^2}{55 (5 x+3)}\right )\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {7 (3 x+2)^3}{11 \sqrt {1-2 x} (5 x+3)}-\frac {9}{11} \left (\frac {3}{55} \left (\frac {1}{5} \int \frac {1}{\frac {11}{2}-\frac {5}{2} (1-2 x)}d\sqrt {1-2 x}-\frac {1}{5} \sqrt {1-2 x} (265 x+792)\right )+\frac {4 \sqrt {1-2 x} (3 x+2)^2}{55 (5 x+3)}\right )\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {7 (3 x+2)^3}{11 \sqrt {1-2 x} (5 x+3)}-\frac {9}{11} \left (\frac {3}{55} \left (\frac {2 \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{5 \sqrt {55}}-\frac {1}{5} \sqrt {1-2 x} (265 x+792)\right )+\frac {4 \sqrt {1-2 x} (3 x+2)^2}{55 (5 x+3)}\right )\) |
(7*(2 + 3*x)^3)/(11*Sqrt[1 - 2*x]*(3 + 5*x)) - (9*((4*Sqrt[1 - 2*x]*(2 + 3 *x)^2)/(55*(3 + 5*x)) + (3*(-1/5*(Sqrt[1 - 2*x]*(792 + 265*x)) + (2*ArcTan h[Sqrt[5/11]*Sqrt[1 - 2*x]])/(5*Sqrt[55])))/55))/11
3.22.17.3.1 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[(-(a*d*f*h*(n + 2) + b*c*f*h*(m + 2) - b*d*(f*g + e*h)*(m + n + 3) - b*d*f*h*(m + n + 2)*x))*(a + b*x)^(m + 1)*(( c + d*x)^(n + 1)/(b^2*d^2*(m + n + 2)*(m + n + 3))), x] + Simp[(a^2*d^2*f*h *(n + 1)*(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*(m + 1)*(m + 2) - c*d*(f*g + e*h)*(m + 1)*(m + n + 3) + d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2*d^2*(m + n + 2)*(m + n + 3)) Int[( a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && NeQ[m + n + 2, 0] && NeQ[m + n + 3, 0]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_)*((g_.) + (h_.)*(x_)), x_] :> Simp[(b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^n*((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 - 1)*(e + f*x)^p*Simp[b* c*(f*g - e*h)*(m + 1) + (b*g - a*h)*(d*e*n + c*f*(p + 1)) + d*(b*(f*g - e*h )*(m + 1) + f*(b*g - a*h)*(n + p + 1))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, p}, x] && ILtQ[m, -1] && GtQ[n, 0]
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 = 1.09 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.51
| method | result | size |
| risch | \(-\frac {16335 x^{3}+114345 x^{2}-68661 x -78832}{3025 \left (3+5 x \right ) \sqrt {1-2 x}}-\frac {54 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{166375}\) | \(51\) |
| pseudoelliptic | \(-\frac {898425 \left (\frac {2 \sqrt {55}\, \left (x +\frac {3}{5}\right ) \sqrt {1-2 x}\, \operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right )}{6655}+x^{3}+7 x^{2}-\frac {2543 x}{605}-\frac {78832}{16335}\right )}{\sqrt {1-2 x}\, \left (499125+831875 x \right )}\) | \(58\) |
| derivativedivides | \(-\frac {27 \left (1-2 x \right )^{\frac {3}{2}}}{100}+\frac {999 \sqrt {1-2 x}}{250}+\frac {2 \sqrt {1-2 x}}{75625 \left (-\frac {6}{5}-2 x \right )}-\frac {54 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{166375}+\frac {2401}{484 \sqrt {1-2 x}}\) | \(63\) |
| default | \(-\frac {27 \left (1-2 x \right )^{\frac {3}{2}}}{100}+\frac {999 \sqrt {1-2 x}}{250}+\frac {2 \sqrt {1-2 x}}{75625 \left (-\frac {6}{5}-2 x \right )}-\frac {54 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{166375}+\frac {2401}{484 \sqrt {1-2 x}}\) | \(63\) |
| trager | \(\frac {\left (16335 x^{3}+114345 x^{2}-68661 x -78832\right ) \sqrt {1-2 x}}{30250 x^{2}+3025 x -9075}+\frac {27 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) \ln \left (\frac {5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) x +55 \sqrt {1-2 x}-8 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right )}{3+5 x}\right )}{166375}\) | \(80\) |
-1/3025*(16335*x^3+114345*x^2-68661*x-78832)/(3+5*x)/(1-2*x)^(1/2)-54/1663 75*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)
Time = 0.22 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.75 \[ \int \frac {(2+3 x)^4}{(1-2 x)^{3/2} (3+5 x)^2} \, dx=\frac {27 \, \sqrt {55} {\left (10 \, x^{2} + x - 3\right )} \log \left (\frac {5 \, x + \sqrt {55} \sqrt {-2 \, x + 1} - 8}{5 \, x + 3}\right ) + 55 \, {\left (16335 \, x^{3} + 114345 \, x^{2} - 68661 \, x - 78832\right )} \sqrt {-2 \, x + 1}}{166375 \, {\left (10 \, x^{2} + x - 3\right )}} \]
1/166375*(27*sqrt(55)*(10*x^2 + x - 3)*log((5*x + sqrt(55)*sqrt(-2*x + 1) - 8)/(5*x + 3)) + 55*(16335*x^3 + 114345*x^2 - 68661*x - 78832)*sqrt(-2*x + 1))/(10*x^2 + x - 3)
Time = 68.02 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.97 \[ \int \frac {(2+3 x)^4}{(1-2 x)^{3/2} (3+5 x)^2} \, dx=- \frac {27 \left (1 - 2 x\right )^{\frac {3}{2}}}{100} + \frac {999 \sqrt {1 - 2 x}}{250} + \frac {134 \sqrt {55} \left (\log {\left (\sqrt {1 - 2 x} - \frac {\sqrt {55}}{5} \right )} - \log {\left (\sqrt {1 - 2 x} + \frac {\sqrt {55}}{5} \right )}\right )}{831875} - \frac {4 \left (\begin {cases} \frac {\sqrt {55} \left (- \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1 \right )}}{4} + \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1 \right )}}{4} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1\right )} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1\right )}\right )}{605} & \text {for}\: \sqrt {1 - 2 x} > - \frac {\sqrt {55}}{5} \wedge \sqrt {1 - 2 x} < \frac {\sqrt {55}}{5} \end {cases}\right )}{1375} + \frac {2401}{484 \sqrt {1 - 2 x}} \]
-27*(1 - 2*x)**(3/2)/100 + 999*sqrt(1 - 2*x)/250 + 134*sqrt(55)*(log(sqrt( 1 - 2*x) - sqrt(55)/5) - log(sqrt(1 - 2*x) + sqrt(55)/5))/831875 - 4*Piece wise((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)))/1375 + 2401/(484*sqrt(1 - 2*x))
Time = 0.27 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.83 \[ \int \frac {(2+3 x)^4}{(1-2 x)^{3/2} (3+5 x)^2} \, dx=-\frac {27}{100} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {27}{166375} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) + \frac {999}{250} \, \sqrt {-2 \, x + 1} - \frac {1500633 \, x + 900371}{30250 \, {\left (5 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 11 \, \sqrt {-2 \, x + 1}\right )}} \]
-27/100*(-2*x + 1)^(3/2) + 27/166375*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 999/250*sqrt(-2*x + 1) - 1/30250*( 1500633*x + 900371)/(5*(-2*x + 1)^(3/2) - 11*sqrt(-2*x + 1))
Time = 0.47 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.86 \[ \int \frac {(2+3 x)^4}{(1-2 x)^{3/2} (3+5 x)^2} \, dx=-\frac {27}{100} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {27}{166375} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) + \frac {999}{250} \, \sqrt {-2 \, x + 1} - \frac {1500633 \, x + 900371}{30250 \, {\left (5 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 11 \, \sqrt {-2 \, x + 1}\right )}} \]
-27/100*(-2*x + 1)^(3/2) + 27/166375*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10 *sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 999/250*sqrt(-2*x + 1) - 1/30250*(1500633*x + 900371)/(5*(-2*x + 1)^(3/2) - 11*sqrt(-2*x + 1))
Time = 0.07 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.66 \[ \int \frac {(2+3 x)^4}{(1-2 x)^{3/2} (3+5 x)^2} \, dx=\frac {\frac {1500633\,x}{151250}+\frac {900371}{151250}}{\frac {11\,\sqrt {1-2\,x}}{5}-{\left (1-2\,x\right )}^{3/2}}+\frac {999\,\sqrt {1-2\,x}}{250}-\frac {27\,{\left (1-2\,x\right )}^{3/2}}{100}+\frac {\sqrt {55}\,\mathrm {atan}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}\,1{}\mathrm {i}}{11}\right )\,54{}\mathrm {i}}{166375} \]