Integrand size = 24, antiderivative size = 100 \[ \int \frac {(2+3 x)^4}{(1-2 x)^{5/2} (3+5 x)^2} \, dx=-\frac {3 (40912-24739 x)}{33275 \sqrt {1-2 x}}-\frac {38 (2+3 x)^2}{1815 \sqrt {1-2 x} (3+5 x)}+\frac {7 (2+3 x)^3}{33 (1-2 x)^{3/2} (3+5 x)}-\frac {274 \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{33275 \sqrt {55}} \]
7/33*(2+3*x)^3/(1-2*x)^(3/2)/(3+5*x)-274/1830125*arctanh(1/11*55^(1/2)*(1- 2*x)^(1/2))*55^(1/2)-3/33275*(40912-24739*x)/(1-2*x)^(1/2)-38/1815*(2+3*x) ^2/(3+5*x)/(1-2*x)^(1/2)
Time = 0.11 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.79 \[ \int \frac {(2+3 x)^4}{(1-2 x)^{5/2} (3+5 x)^2} \, dx=\frac {-7263025+20845825 (1-2 x)-4416963 (1-2 x)^2-1617165 (1-2 x)^3}{399300 (-11+5 (1-2 x)) (1-2 x)^{3/2}}-\frac {274 \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{33275 \sqrt {55}} \]
(-7263025 + 20845825*(1 - 2*x) - 4416963*(1 - 2*x)^2 - 1617165*(1 - 2*x)^3 )/(399300*(-11 + 5*(1 - 2*x))*(1 - 2*x)^(3/2)) - (274*ArcTanh[Sqrt[5/11]*S qrt[1 - 2*x]])/(33275*Sqrt[55])
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, 166, 27, 163, 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.
| \(\displaystyle \int \frac {(3 x+2)^4}{(1-2 x)^{5/2} (5 x+3)^2} \, dx\) |
| \(\Big \downarrow \) 109 |
| \(\displaystyle \frac {7 (3 x+2)^3}{33 (1-2 x)^{3/2} (5 x+3)}-\frac {1}{33} \int \frac {(3 x+2)^2 (201 x+113)}{(1-2 x)^{3/2} (5 x+3)^2}dx\) |
| \(\Big \downarrow \) 166 |
| \(\displaystyle \frac {1}{33} \left (-\frac {1}{55} \int \frac {3 (3 x+2) (2249 x+1322)}{(1-2 x)^{3/2} (5 x+3)}dx-\frac {38 (3 x+2)^2}{55 \sqrt {1-2 x} (5 x+3)}\right )+\frac {7 (3 x+2)^3}{33 (1-2 x)^{3/2} (5 x+3)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{33} \left (-\frac {3}{55} \int \frac {(3 x+2) (2249 x+1322)}{(1-2 x)^{3/2} (5 x+3)}dx-\frac {38 (3 x+2)^2}{55 \sqrt {1-2 x} (5 x+3)}\right )+\frac {7 (3 x+2)^3}{33 (1-2 x)^{3/2} (5 x+3)}\) |
| \(\Big \downarrow \) 163 |
| \(\displaystyle \frac {1}{33} \left (-\frac {3}{55} \left (\frac {3 (40912-24739 x)}{55 \sqrt {1-2 x}}-\frac {137}{55} \int \frac {1}{\sqrt {1-2 x} (5 x+3)}dx\right )-\frac {38 (3 x+2)^2}{55 \sqrt {1-2 x} (5 x+3)}\right )+\frac {7 (3 x+2)^3}{33 (1-2 x)^{3/2} (5 x+3)}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {1}{33} \left (-\frac {3}{55} \left (\frac {137}{55} \int \frac {1}{\frac {11}{2}-\frac {5}{2} (1-2 x)}d\sqrt {1-2 x}+\frac {3 (40912-24739 x)}{55 \sqrt {1-2 x}}\right )-\frac {38 (3 x+2)^2}{55 \sqrt {1-2 x} (5 x+3)}\right )+\frac {7 (3 x+2)^3}{33 (1-2 x)^{3/2} (5 x+3)}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {1}{33} \left (-\frac {3}{55} \left (\frac {274 \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{55 \sqrt {55}}+\frac {3 (40912-24739 x)}{55 \sqrt {1-2 x}}\right )-\frac {38 (3 x+2)^2}{55 \sqrt {1-2 x} (5 x+3)}\right )+\frac {7 (3 x+2)^3}{33 (1-2 x)^{3/2} (5 x+3)}\) |
(7*(2 + 3*x)^3)/(33*(1 - 2*x)^(3/2)*(3 + 5*x)) + ((-38*(2 + 3*x)^2)/(55*Sq rt[1 - 2*x]*(3 + 5*x)) - (3*((3*(40912 - 24739*x))/(55*Sqrt[1 - 2*x]) + (2 74*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(55*Sqrt[55])))/55)/33
3.22.82.3.1 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^2*d*f*h*(n + 2) + b^2*d*e*g*(m + n + 3) + a*b*(c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b*f*h*(b*c - a*d)* (m + 1)*x)/(b^2*d*(b*c - a*d)*(m + 1)*(m + n + 3)))*(a + b*x)^(m + 1)*(c + d*x)^(n + 1), 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*(b*c - a*d)*(m + 1)*(m + n + 3)) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && ((GeQ[m, -2] && LtQ[m, - 1]) || SumSimplerQ[m, 1]) && NeQ[m, -1] && 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 = 1.12 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.58
| method | result | size |
| risch | \(\frac {1617165 x^{3}-4634229 x^{2}-1790101 x +943584}{99825 \sqrt {1-2 x}\, \left (3+5 x \right ) \left (-1+2 x \right )}-\frac {274 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{1830125}\) | \(58\) |
| derivativedivides | \(-\frac {81 \sqrt {1-2 x}}{100}+\frac {2 \sqrt {1-2 x}}{166375 \left (-\frac {6}{5}-2 x \right )}-\frac {274 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{1830125}+\frac {2401}{1452 \left (1-2 x \right )^{\frac {3}{2}}}-\frac {10633}{2662 \sqrt {1-2 x}}\) | \(63\) |
| default | \(-\frac {81 \sqrt {1-2 x}}{100}+\frac {2 \sqrt {1-2 x}}{166375 \left (-\frac {6}{5}-2 x \right )}-\frac {274 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{1830125}+\frac {2401}{1452 \left (1-2 x \right )^{\frac {3}{2}}}-\frac {10633}{2662 \sqrt {1-2 x}}\) | \(63\) |
| pseudoelliptic | \(\frac {822 \sqrt {1-2 x}\, \operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \left (10 x^{2}+x -3\right ) \sqrt {55}-88944075 x^{3}+254882595 x^{2}+98455555 x -51897120}{\left (1-2 x \right )^{\frac {3}{2}} \left (16471125+27451875 x \right )}\) | \(65\) |
| trager | \(-\frac {\left (1617165 x^{3}-4634229 x^{2}-1790101 x +943584\right ) \sqrt {1-2 x}}{99825 \left (-1+2 x \right )^{2} \left (3+5 x \right )}-\frac {137 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) \ln \left (\frac {-5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) x +55 \sqrt {1-2 x}+8 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right )}{3+5 x}\right )}{1830125}\) | \(84\) |
1/99825*(1617165*x^3-4634229*x^2-1790101*x+943584)/(1-2*x)^(1/2)/(3+5*x)/( -1+2*x)-274/1830125*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)
Time = 0.22 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.89 \[ \int \frac {(2+3 x)^4}{(1-2 x)^{5/2} (3+5 x)^2} \, dx=\frac {411 \, \sqrt {55} {\left (20 \, x^{3} - 8 \, x^{2} - 7 \, x + 3\right )} \log \left (\frac {5 \, x + \sqrt {55} \sqrt {-2 \, x + 1} - 8}{5 \, x + 3}\right ) - 55 \, {\left (1617165 \, x^{3} - 4634229 \, x^{2} - 1790101 \, x + 943584\right )} \sqrt {-2 \, x + 1}}{5490375 \, {\left (20 \, x^{3} - 8 \, x^{2} - 7 \, x + 3\right )}} \]
1/5490375*(411*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*(1617165*x^3 - 4634229*x^2 - 1790101*x + 943584)*sqrt(-2*x + 1))/(20*x^3 - 8*x^2 - 7*x + 3)
Time = 67.57 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.97 \[ \int \frac {(2+3 x)^4}{(1-2 x)^{5/2} (3+5 x)^2} \, dx=- \frac {81 \sqrt {1 - 2 x}}{100} + \frac {136 \sqrt {55} \left (\log {\left (\sqrt {1 - 2 x} - \frac {\sqrt {55}}{5} \right )} - \log {\left (\sqrt {1 - 2 x} + \frac {\sqrt {55}}{5} \right )}\right )}{1830125} - \frac {4 \left (\begin {cases} \frac {\sqrt {55} \left (- \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1 \right )}}{4} + \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1 \right )}}{4} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1\right )} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1\right )}\right )}{605} & \text {for}\: \sqrt {1 - 2 x} > - \frac {\sqrt {55}}{5} \wedge \sqrt {1 - 2 x} < \frac {\sqrt {55}}{5} \end {cases}\right )}{3025} - \frac {10633}{2662 \sqrt {1 - 2 x}} + \frac {2401}{1452 \left (1 - 2 x\right )^{\frac {3}{2}}} \]
-81*sqrt(1 - 2*x)/100 + 136*sqrt(55)*(log(sqrt(1 - 2*x) - sqrt(55)/5) - lo g(sqrt(1 - 2*x) + sqrt(55)/5))/1830125 - 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)/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 - 10633/(2662*sqrt(1 - 2*x)) + 2401/(1452*(1 - 2*x)**(3/2))
Time = 0.29 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.83 \[ \int \frac {(2+3 x)^4}{(1-2 x)^{5/2} (3+5 x)^2} \, dx=\frac {137}{1830125} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) - \frac {81}{100} \, \sqrt {-2 \, x + 1} - \frac {3987363 \, {\left (2 \, x - 1\right )}^{2} + 20845825 \, x - 6791400}{199650 \, {\left (5 \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} - 11 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}}\right )}} \]
137/1830125*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt (-2*x + 1))) - 81/100*sqrt(-2*x + 1) - 1/199650*(3987363*(2*x - 1)^2 + 208 45825*x - 6791400)/(5*(-2*x + 1)^(5/2) - 11*(-2*x + 1)^(3/2))
Time = 0.29 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.86 \[ \int \frac {(2+3 x)^4}{(1-2 x)^{5/2} (3+5 x)^2} \, dx=\frac {137}{1830125} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) - \frac {81}{100} \, \sqrt {-2 \, x + 1} - \frac {343 \, {\left (372 \, x - 109\right )}}{15972 \, {\left (2 \, x - 1\right )} \sqrt {-2 \, x + 1}} - \frac {\sqrt {-2 \, x + 1}}{33275 \, {\left (5 \, x + 3\right )}} \]
137/1830125*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55 ) + 5*sqrt(-2*x + 1))) - 81/100*sqrt(-2*x + 1) - 343/15972*(372*x - 109)/( (2*x - 1)*sqrt(-2*x + 1)) - 1/33275*sqrt(-2*x + 1)/(5*x + 3)
Time = 0.07 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.64 \[ \int \frac {(2+3 x)^4}{(1-2 x)^{5/2} (3+5 x)^2} \, dx=\frac {\frac {75803\,x}{3630}+\frac {1329121\,{\left (2\,x-1\right )}^2}{332750}-\frac {4116}{605}}{\frac {11\,{\left (1-2\,x\right )}^{3/2}}{5}-{\left (1-2\,x\right )}^{5/2}}-\frac {274\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )}{1830125}-\frac {81\,\sqrt {1-2\,x}}{100} \]