Integrand size = 24, antiderivative size = 100 \[ \int \frac {(2+3 x)^4}{\sqrt {1-2 x} (3+5 x)^3} \, dx=-\frac {\sqrt {1-2 x} (2+3 x)^3}{110 (3+5 x)^2}-\frac {84 \sqrt {1-2 x} (2+3 x)^2}{3025 (3+5 x)}-\frac {63 \sqrt {1-2 x} (352+75 x)}{30250}-\frac {2667 \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{15125 \sqrt {55}} \]
-2667/831875*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)-1/110*(2+3*x)^3 *(1-2*x)^(1/2)/(3+5*x)^2-84/3025*(2+3*x)^2*(1-2*x)^(1/2)/(3+5*x)-63/30250* (352+75*x)*(1-2*x)^(1/2)
Time = 0.15 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.63 \[ \int \frac {(2+3 x)^4}{\sqrt {1-2 x} (3+5 x)^3} \, dx=\frac {-\frac {55 \sqrt {1-2 x} \left (211864+764745 x+784080 x^2+163350 x^3\right )}{(3+5 x)^2}-5334 \sqrt {55} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{1663750} \]
((-55*Sqrt[1 - 2*x]*(211864 + 764745*x + 784080*x^2 + 163350*x^3))/(3 + 5* x)^2 - 5334*Sqrt[55]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/1663750
Time = 0.20 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.10, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {109, 27, 166, 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.
| \(\displaystyle \int \frac {(3 x+2)^4}{\sqrt {1-2 x} (5 x+3)^3} \, dx\) |
| \(\Big \downarrow \) 109 |
| \(\displaystyle -\frac {1}{110} \int -\frac {21 (3 x+2)^2 (9 x+7)}{\sqrt {1-2 x} (5 x+3)^2}dx-\frac {\sqrt {1-2 x} (3 x+2)^3}{110 (5 x+3)^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {21}{110} \int \frac {(3 x+2)^2 (9 x+7)}{\sqrt {1-2 x} (5 x+3)^2}dx-\frac {\sqrt {1-2 x} (3 x+2)^3}{110 (5 x+3)^2}\) |
| \(\Big \downarrow \) 166 |
| \(\displaystyle \frac {21}{110} \left (\frac {1}{55} \int \frac {(3 x+2) (225 x+262)}{\sqrt {1-2 x} (5 x+3)}dx-\frac {8 \sqrt {1-2 x} (3 x+2)^2}{55 (5 x+3)}\right )-\frac {\sqrt {1-2 x} (3 x+2)^3}{110 (5 x+3)^2}\) |
| \(\Big \downarrow \) 164 |
| \(\displaystyle \frac {21}{110} \left (\frac {1}{55} \left (\frac {127}{5} \int \frac {1}{\sqrt {1-2 x} (5 x+3)}dx-\frac {3}{5} \sqrt {1-2 x} (75 x+352)\right )-\frac {8 \sqrt {1-2 x} (3 x+2)^2}{55 (5 x+3)}\right )-\frac {\sqrt {1-2 x} (3 x+2)^3}{110 (5 x+3)^2}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {21}{110} \left (\frac {1}{55} \left (-\frac {127}{5} \int \frac {1}{\frac {11}{2}-\frac {5}{2} (1-2 x)}d\sqrt {1-2 x}-\frac {3}{5} \sqrt {1-2 x} (75 x+352)\right )-\frac {8 \sqrt {1-2 x} (3 x+2)^2}{55 (5 x+3)}\right )-\frac {\sqrt {1-2 x} (3 x+2)^3}{110 (5 x+3)^2}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {21}{110} \left (\frac {1}{55} \left (-\frac {254 \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{5 \sqrt {55}}-\frac {3}{5} \sqrt {1-2 x} (75 x+352)\right )-\frac {8 \sqrt {1-2 x} (3 x+2)^2}{55 (5 x+3)}\right )-\frac {\sqrt {1-2 x} (3 x+2)^3}{110 (5 x+3)^2}\) |
-1/110*(Sqrt[1 - 2*x]*(2 + 3*x)^3)/(3 + 5*x)^2 + (21*((-8*Sqrt[1 - 2*x]*(2 + 3*x)^2)/(55*(3 + 5*x)) + ((-3*Sqrt[1 - 2*x]*(352 + 75*x))/5 - (254*ArcT anh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(5*Sqrt[55]))/55))/110
3.21.60.3.1 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 = 1.07 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.56
| method | result | size |
| risch | \(\frac {326700 x^{4}+1404810 x^{3}+745410 x^{2}-341017 x -211864}{30250 \left (3+5 x \right )^{2} \sqrt {1-2 x}}-\frac {2667 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{831875}\) | \(56\) |
| pseudoelliptic | \(\frac {-5334 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \left (3+5 x \right )^{2} \sqrt {55}-55 \sqrt {1-2 x}\, \left (163350 x^{3}+784080 x^{2}+764745 x +211864\right )}{1663750 \left (3+5 x \right )^{2}}\) | \(60\) |
| derivativedivides | \(\frac {27 \left (1-2 x \right )^{\frac {3}{2}}}{250}-\frac {1107 \sqrt {1-2 x}}{1250}+\frac {\frac {267 \left (1-2 x \right )^{\frac {3}{2}}}{15125}-\frac {269 \sqrt {1-2 x}}{6875}}{\left (-6-10 x \right )^{2}}-\frac {2667 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{831875}\) | \(66\) |
| default | \(\frac {27 \left (1-2 x \right )^{\frac {3}{2}}}{250}-\frac {1107 \sqrt {1-2 x}}{1250}+\frac {\frac {267 \left (1-2 x \right )^{\frac {3}{2}}}{15125}-\frac {269 \sqrt {1-2 x}}{6875}}{\left (-6-10 x \right )^{2}}-\frac {2667 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{831875}\) | \(66\) |
| trager | \(-\frac {\left (163350 x^{3}+784080 x^{2}+764745 x +211864\right ) \sqrt {1-2 x}}{30250 \left (3+5 x \right )^{2}}+\frac {2667 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) \ln \left (\frac {5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) x +55 \sqrt {1-2 x}-8 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right )}{3+5 x}\right )}{1663750}\) | \(77\) |
1/30250*(326700*x^4+1404810*x^3+745410*x^2-341017*x-211864)/(3+5*x)^2/(1-2 *x)^(1/2)-2667/831875*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)
Time = 0.22 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.79 \[ \int \frac {(2+3 x)^4}{\sqrt {1-2 x} (3+5 x)^3} \, dx=\frac {2667 \, \sqrt {55} {\left (25 \, x^{2} + 30 \, x + 9\right )} \log \left (\frac {5 \, x + \sqrt {55} \sqrt {-2 \, x + 1} - 8}{5 \, x + 3}\right ) - 55 \, {\left (163350 \, x^{3} + 784080 \, x^{2} + 764745 \, x + 211864\right )} \sqrt {-2 \, x + 1}}{1663750 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} \]
1/1663750*(2667*sqrt(55)*(25*x^2 + 30*x + 9)*log((5*x + sqrt(55)*sqrt(-2*x + 1) - 8)/(5*x + 3)) - 55*(163350*x^3 + 784080*x^2 + 764745*x + 211864)*s qrt(-2*x + 1))/(25*x^2 + 30*x + 9)
Time = 153.79 (sec) , antiderivative size = 354, normalized size of antiderivative = 3.54 \[ \int \frac {(2+3 x)^4}{\sqrt {1-2 x} (3+5 x)^3} \, dx=\frac {27 \left (1 - 2 x\right )^{\frac {3}{2}}}{250} - \frac {1107 \sqrt {1 - 2 x}}{1250} + \frac {54 \sqrt {55} \left (\log {\left (\sqrt {1 - 2 x} - \frac {\sqrt {55}}{5} \right )} - \log {\left (\sqrt {1 - 2 x} + \frac {\sqrt {55}}{5} \right )}\right )}{34375} - \frac {48 \left (\begin {cases} \frac {\sqrt {55} \left (- \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1 \right )}}{4} + \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1 \right )}}{4} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1\right )} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1\right )}\right )}{605} & \text {for}\: \sqrt {1 - 2 x} > - \frac {\sqrt {55}}{5} \wedge \sqrt {1 - 2 x} < \frac {\sqrt {55}}{5} \end {cases}\right )}{625} + \frac {8 \left (\begin {cases} \frac {\sqrt {55} \cdot \left (\frac {3 \log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1 \right )}}{16} - \frac {3 \log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1 \right )}}{16} + \frac {3}{16 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1\right )} + \frac {1}{16 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1\right )^{2}} + \frac {3}{16 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1\right )} - \frac {1}{16 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1\right )^{2}}\right )}{6655} & \text {for}\: \sqrt {1 - 2 x} > - \frac {\sqrt {55}}{5} \wedge \sqrt {1 - 2 x} < \frac {\sqrt {55}}{5} \end {cases}\right )}{625} \]
27*(1 - 2*x)**(3/2)/250 - 1107*sqrt(1 - 2*x)/1250 + 54*sqrt(55)*(log(sqrt( 1 - 2*x) - sqrt(55)/5) - log(sqrt(1 - 2*x) + sqrt(55)/5))/34375 - 48*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)))/625 + 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/(1 6*(sqrt(55)*sqrt(1 - 2*x)/11 - 1)) - 1/(16*(sqrt(55)*sqrt(1 - 2*x)/11 - 1) **2))/6655, (sqrt(1 - 2*x) > -sqrt(55)/5) & (sqrt(1 - 2*x) < sqrt(55)/5))) /625
Time = 0.28 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.92 \[ \int \frac {(2+3 x)^4}{\sqrt {1-2 x} (3+5 x)^3} \, dx=\frac {27}{250} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {2667}{1663750} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) - \frac {1107}{1250} \, \sqrt {-2 \, x + 1} + \frac {1335 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 2959 \, \sqrt {-2 \, x + 1}}{75625 \, {\left (25 \, {\left (2 \, x - 1\right )}^{2} + 220 \, x + 11\right )}} \]
27/250*(-2*x + 1)^(3/2) + 2667/1663750*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2 *x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 1107/1250*sqrt(-2*x + 1) + 1/756 25*(1335*(-2*x + 1)^(3/2) - 2959*sqrt(-2*x + 1))/(25*(2*x - 1)^2 + 220*x + 11)
Time = 0.29 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.86 \[ \int \frac {(2+3 x)^4}{\sqrt {1-2 x} (3+5 x)^3} \, dx=\frac {27}{250} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {2667}{1663750} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) - \frac {1107}{1250} \, \sqrt {-2 \, x + 1} + \frac {1335 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 2959 \, \sqrt {-2 \, x + 1}}{302500 \, {\left (5 \, x + 3\right )}^{2}} \]
27/250*(-2*x + 1)^(3/2) + 2667/1663750*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 1107/1250*sqrt(-2*x + 1) + 1/302500*(1335*(-2*x + 1)^(3/2) - 2959*sqrt(-2*x + 1))/(5*x + 3)^2
Time = 1.34 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.74 \[ \int \frac {(2+3 x)^4}{\sqrt {1-2 x} (3+5 x)^3} \, dx=\frac {27\,{\left (1-2\,x\right )}^{3/2}}{250}-\frac {1107\,\sqrt {1-2\,x}}{1250}-\frac {\frac {269\,\sqrt {1-2\,x}}{171875}-\frac {267\,{\left (1-2\,x\right )}^{3/2}}{378125}}{\frac {44\,x}{5}+{\left (2\,x-1\right )}^2+\frac {11}{25}}+\frac {\sqrt {55}\,\mathrm {atan}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}\,1{}\mathrm {i}}{11}\right )\,2667{}\mathrm {i}}{831875} \]