Integrand size = 24, antiderivative size = 113 \[ \int \frac {\sqrt {1-2 x} (2+3 x)^4}{(3+5 x)^2} \, dx=\frac {12}{625} \sqrt {1-2 x} (2+3 x)^2+\frac {27}{175} \sqrt {1-2 x} (2+3 x)^3-\frac {\sqrt {1-2 x} (2+3 x)^4}{5 (3+5 x)}-\frac {3 \sqrt {1-2 x} (1256+375 x)}{3125}-\frac {262 \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{3125 \sqrt {55}} \]
-262/171875*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)+12/625*(2+3*x)^2 *(1-2*x)^(1/2)+27/175*(2+3*x)^3*(1-2*x)^(1/2)-1/5*(2+3*x)^4*(1-2*x)^(1/2)/ (3+5*x)-3/3125*(1256+375*x)*(1-2*x)^(1/2)
Time = 0.12 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.60 \[ \int \frac {\sqrt {1-2 x} (2+3 x)^4}{(3+5 x)^2} \, dx=\frac {\sqrt {1-2 x} \left (-63088-52485 x+206415 x^2+258525 x^3+101250 x^4\right )}{21875 (3+5 x)}-\frac {262 \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{3125 \sqrt {55}} \]
(Sqrt[1 - 2*x]*(-63088 - 52485*x + 206415*x^2 + 258525*x^3 + 101250*x^4))/ (21875*(3 + 5*x)) - (262*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(3125*Sqrt[55] )
Time = 0.22 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.13, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {108, 170, 27, 170, 25, 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.
| \(\displaystyle \int \frac {\sqrt {1-2 x} (3 x+2)^4}{(5 x+3)^2} \, dx\) |
| \(\Big \downarrow \) 108 |
| \(\displaystyle \frac {1}{5} \int \frac {(10-27 x) (3 x+2)^3}{\sqrt {1-2 x} (5 x+3)}dx-\frac {\sqrt {1-2 x} (3 x+2)^4}{5 (5 x+3)}\) |
| \(\Big \downarrow \) 170 |
| \(\displaystyle \frac {1}{5} \left (\frac {27}{35} \sqrt {1-2 x} (3 x+2)^3-\frac {1}{35} \int -\frac {7 (19-12 x) (3 x+2)^2}{\sqrt {1-2 x} (5 x+3)}dx\right )-\frac {\sqrt {1-2 x} (3 x+2)^4}{5 (5 x+3)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{5} \int \frac {(19-12 x) (3 x+2)^2}{\sqrt {1-2 x} (5 x+3)}dx+\frac {27}{35} \sqrt {1-2 x} (3 x+2)^3\right )-\frac {\sqrt {1-2 x} (3 x+2)^4}{5 (5 x+3)}\) |
| \(\Big \downarrow \) 170 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{5} \left (\frac {12}{25} \sqrt {1-2 x} (3 x+2)^2-\frac {1}{25} \int -\frac {(3 x+2) (1125 x+806)}{\sqrt {1-2 x} (5 x+3)}dx\right )+\frac {27}{35} \sqrt {1-2 x} (3 x+2)^3\right )-\frac {\sqrt {1-2 x} (3 x+2)^4}{5 (5 x+3)}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{5} \left (\frac {1}{25} \int \frac {(3 x+2) (1125 x+806)}{\sqrt {1-2 x} (5 x+3)}dx+\frac {12}{25} \sqrt {1-2 x} (3 x+2)^2\right )+\frac {27}{35} \sqrt {1-2 x} (3 x+2)^3\right )-\frac {\sqrt {1-2 x} (3 x+2)^4}{5 (5 x+3)}\) |
| \(\Big \downarrow \) 164 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{5} \left (\frac {1}{25} \left (\frac {131}{5} \int \frac {1}{\sqrt {1-2 x} (5 x+3)}dx-\frac {3}{5} \sqrt {1-2 x} (375 x+1256)\right )+\frac {12}{25} \sqrt {1-2 x} (3 x+2)^2\right )+\frac {27}{35} \sqrt {1-2 x} (3 x+2)^3\right )-\frac {\sqrt {1-2 x} (3 x+2)^4}{5 (5 x+3)}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{5} \left (\frac {1}{25} \left (-\frac {131}{5} \int \frac {1}{\frac {11}{2}-\frac {5}{2} (1-2 x)}d\sqrt {1-2 x}-\frac {3}{5} \sqrt {1-2 x} (375 x+1256)\right )+\frac {12}{25} \sqrt {1-2 x} (3 x+2)^2\right )+\frac {27}{35} \sqrt {1-2 x} (3 x+2)^3\right )-\frac {\sqrt {1-2 x} (3 x+2)^4}{5 (5 x+3)}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{5} \left (\frac {1}{25} \left (-\frac {262 \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{5 \sqrt {55}}-\frac {3}{5} \sqrt {1-2 x} (375 x+1256)\right )+\frac {12}{25} \sqrt {1-2 x} (3 x+2)^2\right )+\frac {27}{35} \sqrt {1-2 x} (3 x+2)^3\right )-\frac {\sqrt {1-2 x} (3 x+2)^4}{5 (5 x+3)}\) |
-1/5*(Sqrt[1 - 2*x]*(2 + 3*x)^4)/(3 + 5*x) + ((27*Sqrt[1 - 2*x]*(2 + 3*x)^ 3)/35 + ((12*Sqrt[1 - 2*x]*(2 + 3*x)^2)/25 + ((-3*Sqrt[1 - 2*x]*(1256 + 37 5*x))/5 - (262*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(5*Sqrt[55]))/25)/5)/5
3.19.39.3.1 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[(a + b*x)^(m + 1)*(c + d*x)^n*((e + f*x)^p/(b*(m + 1))) , x] - Simp[1/(b*(m + 1)) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f* x)^(p - 1)*Simp[d*e*n + c*f*p + d*f*(n + p)*x, x], x], x] /; FreeQ[{a, b, c , d, e, f}, x] && LtQ[m, -1] && GtQ[n, 0] && GtQ[p, 0] && (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[h*(a + b*x)^m*(c + d*x)^(n + 1)*(( e + f*x)^(p + 1)/(d*f*(m + n + p + 2))), x] + Simp[1/(d*f*(m + n + p + 2)) Int[(a + b*x)^(m - 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*g*(m + n + p + 2 ) - h*(b*c*e*m + a*(d*e*(n + 1) + c*f*(p + 1))) + (b*d*f*g*(m + n + p + 2) + h*(a*d*f*m - b*(d*e*(m + n + 1) + c*f*(m + p + 1))))*x, x], x], x] /; Fre eQ[{a, b, c, d, e, f, g, h, n, p}, x] && GtQ[m, 0] && NeQ[m + n + p + 2, 0] && IntegerQ[m]
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 = 0.98 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.54
| method | result | size |
| risch | \(-\frac {202500 x^{5}+415800 x^{4}+154305 x^{3}-311385 x^{2}-73691 x +63088}{21875 \left (3+5 x \right ) \sqrt {1-2 x}}-\frac {262 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{171875}\) | \(61\) |
| pseudoelliptic | \(\frac {-1834 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \left (3+5 x \right ) \sqrt {55}+55 \sqrt {1-2 x}\, \left (101250 x^{4}+258525 x^{3}+206415 x^{2}-52485 x -63088\right )}{3609375+6015625 x}\) | \(62\) |
| derivativedivides | \(-\frac {81 \left (1-2 x \right )^{\frac {7}{2}}}{700}+\frac {999 \left (1-2 x \right )^{\frac {5}{2}}}{1250}-\frac {4131 \left (1-2 x \right )^{\frac {3}{2}}}{2500}+\frac {24 \sqrt {1-2 x}}{3125}+\frac {2 \sqrt {1-2 x}}{15625 \left (-\frac {6}{5}-2 x \right )}-\frac {262 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{171875}\) | \(72\) |
| default | \(-\frac {81 \left (1-2 x \right )^{\frac {7}{2}}}{700}+\frac {999 \left (1-2 x \right )^{\frac {5}{2}}}{1250}-\frac {4131 \left (1-2 x \right )^{\frac {3}{2}}}{2500}+\frac {24 \sqrt {1-2 x}}{3125}+\frac {2 \sqrt {1-2 x}}{15625 \left (-\frac {6}{5}-2 x \right )}-\frac {262 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{171875}\) | \(72\) |
| trager | \(\frac {\sqrt {1-2 x}\, \left (101250 x^{4}+258525 x^{3}+206415 x^{2}-52485 x -63088\right )}{65625+109375 x}+\frac {131 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) \ln \left (\frac {5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) x +55 \sqrt {1-2 x}-8 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right )}{3+5 x}\right )}{171875}\) | \(82\) |
-1/21875*(202500*x^5+415800*x^4+154305*x^3-311385*x^2-73691*x+63088)/(3+5* x)/(1-2*x)^(1/2)-262/171875*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)
Time = 0.22 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.65 \[ \int \frac {\sqrt {1-2 x} (2+3 x)^4}{(3+5 x)^2} \, dx=\frac {917 \, \sqrt {55} {\left (5 \, x + 3\right )} \log \left (\frac {5 \, x + \sqrt {55} \sqrt {-2 \, x + 1} - 8}{5 \, x + 3}\right ) + 55 \, {\left (101250 \, x^{4} + 258525 \, x^{3} + 206415 \, x^{2} - 52485 \, x - 63088\right )} \sqrt {-2 \, x + 1}}{1203125 \, {\left (5 \, x + 3\right )}} \]
1/1203125*(917*sqrt(55)*(5*x + 3)*log((5*x + sqrt(55)*sqrt(-2*x + 1) - 8)/ (5*x + 3)) + 55*(101250*x^4 + 258525*x^3 + 206415*x^2 - 52485*x - 63088)*s qrt(-2*x + 1))/(5*x + 3)
Time = 43.11 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.85 \[ \int \frac {\sqrt {1-2 x} (2+3 x)^4}{(3+5 x)^2} \, dx=- \frac {81 \left (1 - 2 x\right )^{\frac {7}{2}}}{700} + \frac {999 \left (1 - 2 x\right )^{\frac {5}{2}}}{1250} - \frac {4131 \left (1 - 2 x\right )^{\frac {3}{2}}}{2500} + \frac {24 \sqrt {1 - 2 x}}{3125} + \frac {26 \sqrt {55} \left (\log {\left (\sqrt {1 - 2 x} - \frac {\sqrt {55}}{5} \right )} - \log {\left (\sqrt {1 - 2 x} + \frac {\sqrt {55}}{5} \right )}\right )}{34375} - \frac {44 \left (\begin {cases} \frac {\sqrt {55} \left (- \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1 \right )}}{4} + \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1 \right )}}{4} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1\right )} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1\right )}\right )}{605} & \text {for}\: \sqrt {1 - 2 x} > - \frac {\sqrt {55}}{5} \wedge \sqrt {1 - 2 x} < \frac {\sqrt {55}}{5} \end {cases}\right )}{3125} \]
-81*(1 - 2*x)**(7/2)/700 + 999*(1 - 2*x)**(5/2)/1250 - 4131*(1 - 2*x)**(3/ 2)/2500 + 24*sqrt(1 - 2*x)/3125 + 26*sqrt(55)*(log(sqrt(1 - 2*x) - sqrt(55 )/5) - log(sqrt(1 - 2*x) + sqrt(55)/5))/34375 - 44*Piecewise((sqrt(55)*(-l og(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)) )/3125
Time = 0.28 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.79 \[ \int \frac {\sqrt {1-2 x} (2+3 x)^4}{(3+5 x)^2} \, dx=-\frac {81}{700} \, {\left (-2 \, x + 1\right )}^{\frac {7}{2}} + \frac {999}{1250} \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} - \frac {4131}{2500} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {131}{171875} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) + \frac {24}{3125} \, \sqrt {-2 \, x + 1} - \frac {\sqrt {-2 \, x + 1}}{3125 \, {\left (5 \, x + 3\right )}} \]
-81/700*(-2*x + 1)^(7/2) + 999/1250*(-2*x + 1)^(5/2) - 4131/2500*(-2*x + 1 )^(3/2) + 131/171875*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 24/3125*sqrt(-2*x + 1) - 1/3125*sqrt(-2*x + 1)/(5* x + 3)
Time = 0.29 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.94 \[ \int \frac {\sqrt {1-2 x} (2+3 x)^4}{(3+5 x)^2} \, dx=\frac {81}{700} \, {\left (2 \, x - 1\right )}^{3} \sqrt {-2 \, x + 1} + \frac {999}{1250} \, {\left (2 \, x - 1\right )}^{2} \sqrt {-2 \, x + 1} - \frac {4131}{2500} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {131}{171875} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) + \frac {24}{3125} \, \sqrt {-2 \, x + 1} - \frac {\sqrt {-2 \, x + 1}}{3125 \, {\left (5 \, x + 3\right )}} \]
81/700*(2*x - 1)^3*sqrt(-2*x + 1) + 999/1250*(2*x - 1)^2*sqrt(-2*x + 1) - 4131/2500*(-2*x + 1)^(3/2) + 131/171875*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 24/3125*sqrt(-2*x + 1 ) - 1/3125*sqrt(-2*x + 1)/(5*x + 3)
Time = 1.39 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.65 \[ \int \frac {\sqrt {1-2 x} (2+3 x)^4}{(3+5 x)^2} \, dx=\frac {24\,\sqrt {1-2\,x}}{3125}-\frac {2\,\sqrt {1-2\,x}}{15625\,\left (2\,x+\frac {6}{5}\right )}-\frac {4131\,{\left (1-2\,x\right )}^{3/2}}{2500}+\frac {999\,{\left (1-2\,x\right )}^{5/2}}{1250}-\frac {81\,{\left (1-2\,x\right )}^{7/2}}{700}+\frac {\sqrt {55}\,\mathrm {atan}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}\,1{}\mathrm {i}}{11}\right )\,262{}\mathrm {i}}{171875} \]