Integrand size = 24, antiderivative size = 80 \[ \int \frac {(2+3 x)^3}{(1-2 x)^{3/2} (3+5 x)^2} \, dx=\frac {7 (2+3 x)^2}{11 \sqrt {1-2 x} (3+5 x)}+\frac {18 \sqrt {1-2 x} (559+935 x)}{3025 (3+5 x)}-\frac {204 \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{3025 \sqrt {55}} \]
-204/166375*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)+7/11*(2+3*x)^2/( 3+5*x)/(1-2*x)^(1/2)+18/3025*(559+935*x)*(1-2*x)^(1/2)/(3+5*x)
Time = 0.12 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.72 \[ \int \frac {(2+3 x)^3}{(1-2 x)^{3/2} (3+5 x)^2} \, dx=\frac {\frac {55 \left (17762+19806 x-16335 x^2\right )}{\sqrt {1-2 x} (3+5 x)}-204 \sqrt {55} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{166375} \]
((55*(17762 + 19806*x - 16335*x^2))/(Sqrt[1 - 2*x]*(3 + 5*x)) - 204*Sqrt[5 5]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/166375
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, 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.
\(\displaystyle \int \frac {(3 x+2)^3}{(1-2 x)^{3/2} (5 x+3)^2} \, dx\) |
\(\Big \downarrow \) 109 |
\(\displaystyle \frac {7 (3 x+2)^2}{11 \sqrt {1-2 x} (5 x+3)}-\frac {1}{11} \int \frac {6 (3 x+2) (17 x+9)}{\sqrt {1-2 x} (5 x+3)^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {7 (3 x+2)^2}{11 \sqrt {1-2 x} (5 x+3)}-\frac {6}{11} \int \frac {(3 x+2) (17 x+9)}{\sqrt {1-2 x} (5 x+3)^2}dx\) |
\(\Big \downarrow \) 163 |
\(\displaystyle \frac {7 (3 x+2)^2}{11 \sqrt {1-2 x} (5 x+3)}-\frac {6}{11} \left (-\frac {17}{275} \int \frac {1}{\sqrt {1-2 x} (5 x+3)}dx-\frac {3 \sqrt {1-2 x} (935 x+559)}{275 (5 x+3)}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {7 (3 x+2)^2}{11 \sqrt {1-2 x} (5 x+3)}-\frac {6}{11} \left (\frac {17}{275} \int \frac {1}{\frac {11}{2}-\frac {5}{2} (1-2 x)}d\sqrt {1-2 x}-\frac {3 \sqrt {1-2 x} (935 x+559)}{275 (5 x+3)}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {7 (3 x+2)^2}{11 \sqrt {1-2 x} (5 x+3)}-\frac {6}{11} \left (\frac {34 \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{275 \sqrt {55}}-\frac {3 \sqrt {1-2 x} (935 x+559)}{275 (5 x+3)}\right )\) |
(7*(2 + 3*x)^2)/(11*Sqrt[1 - 2*x]*(3 + 5*x)) - (6*((-3*Sqrt[1 - 2*x]*(559 + 935*x))/(275*(3 + 5*x)) + (34*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(275*Sq rt[55])))/11
3.22.18.3.1 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_)^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 = 46, normalized size of antiderivative = 0.58
method | result | size |
risch | \(-\frac {16335 x^{2}-19806 x -17762}{3025 \left (3+5 x \right ) \sqrt {1-2 x}}-\frac {204 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{166375}\) | \(46\) |
derivativedivides | \(\frac {27 \sqrt {1-2 x}}{50}+\frac {2 \sqrt {1-2 x}}{15125 \left (-\frac {6}{5}-2 x \right )}-\frac {204 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{166375}+\frac {343}{242 \sqrt {1-2 x}}\) | \(54\) |
default | \(\frac {27 \sqrt {1-2 x}}{50}+\frac {2 \sqrt {1-2 x}}{15125 \left (-\frac {6}{5}-2 x \right )}-\frac {204 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{166375}+\frac {343}{242 \sqrt {1-2 x}}\) | \(54\) |
pseudoelliptic | \(-\frac {1020 \left (\frac {\sqrt {1-2 x}\, \operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \left (3+5 x \right ) \sqrt {55}}{5}+\frac {59895 x^{2}}{68}-\frac {36311 x}{34}-\frac {97691}{102}\right )}{\sqrt {1-2 x}\, \left (499125+831875 x \right )}\) | \(57\) |
trager | \(\frac {\left (16335 x^{2}-19806 x -17762\right ) \sqrt {1-2 x}}{30250 x^{2}+3025 x -9075}-\frac {102 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) \ln \left (\frac {-5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) x +55 \sqrt {1-2 x}+8 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right )}{3+5 x}\right )}{166375}\) | \(75\) |
-1/3025*(16335*x^2-19806*x-17762)/(3+5*x)/(1-2*x)^(1/2)-204/166375*arctanh (1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)
Time = 0.22 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.88 \[ \int \frac {(2+3 x)^3}{(1-2 x)^{3/2} (3+5 x)^2} \, dx=\frac {102 \, \sqrt {55} {\left (10 \, x^{2} + x - 3\right )} \log \left (\frac {5 \, x + \sqrt {55} \sqrt {-2 \, x + 1} - 8}{5 \, x + 3}\right ) + 55 \, {\left (16335 \, x^{2} - 19806 \, x - 17762\right )} \sqrt {-2 \, x + 1}}{166375 \, {\left (10 \, x^{2} + x - 3\right )}} \]
1/166375*(102*sqrt(55)*(10*x^2 + x - 3)*log((5*x + sqrt(55)*sqrt(-2*x + 1) - 8)/(5*x + 3)) + 55*(16335*x^2 - 19806*x - 17762)*sqrt(-2*x + 1))/(10*x^ 2 + x - 3)
Time = 54.80 (sec) , antiderivative size = 185, normalized size of antiderivative = 2.31 \[ \int \frac {(2+3 x)^3}{(1-2 x)^{3/2} (3+5 x)^2} \, dx=\frac {27 \sqrt {1 - 2 x}}{50} + \frac {101 \sqrt {55} \left (\log {\left (\sqrt {1 - 2 x} - \frac {\sqrt {55}}{5} \right )} - \log {\left (\sqrt {1 - 2 x} + \frac {\sqrt {55}}{5} \right )}\right )}{166375} - \frac {4 \left (\begin {cases} \frac {\sqrt {55} \left (- \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1 \right )}}{4} + \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1 \right )}}{4} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1\right )} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1\right )}\right )}{605} & \text {for}\: \sqrt {1 - 2 x} > - \frac {\sqrt {55}}{5} \wedge \sqrt {1 - 2 x} < \frac {\sqrt {55}}{5} \end {cases}\right )}{275} + \frac {343}{242 \sqrt {1 - 2 x}} \]
27*sqrt(1 - 2*x)/50 + 101*sqrt(55)*(log(sqrt(1 - 2*x) - sqrt(55)/5) - log( sqrt(1 - 2*x) + sqrt(55)/5))/166375 - 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*(sq rt(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)))/275 + 343 /(242*sqrt(1 - 2*x))
Time = 0.28 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.92 \[ \int \frac {(2+3 x)^3}{(1-2 x)^{3/2} (3+5 x)^2} \, dx=\frac {102}{166375} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) + \frac {27}{50} \, \sqrt {-2 \, x + 1} - \frac {42879 \, x + 25723}{3025 \, {\left (5 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 11 \, \sqrt {-2 \, x + 1}\right )}} \]
102/166375*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt( -2*x + 1))) + 27/50*sqrt(-2*x + 1) - 1/3025*(42879*x + 25723)/(5*(-2*x + 1 )^(3/2) - 11*sqrt(-2*x + 1))
Time = 0.38 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.96 \[ \int \frac {(2+3 x)^3}{(1-2 x)^{3/2} (3+5 x)^2} \, dx=\frac {102}{166375} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) + \frac {27}{50} \, \sqrt {-2 \, x + 1} - \frac {42879 \, x + 25723}{3025 \, {\left (5 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 11 \, \sqrt {-2 \, x + 1}\right )}} \]
102/166375*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 27/50*sqrt(-2*x + 1) - 1/3025*(42879*x + 25723)/(5 *(-2*x + 1)^(3/2) - 11*sqrt(-2*x + 1))
Time = 0.06 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.69 \[ \int \frac {(2+3 x)^3}{(1-2 x)^{3/2} (3+5 x)^2} \, dx=\frac {\frac {42879\,x}{15125}+\frac {25723}{15125}}{\frac {11\,\sqrt {1-2\,x}}{5}-{\left (1-2\,x\right )}^{3/2}}-\frac {204\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )}{166375}+\frac {27\,\sqrt {1-2\,x}}{50} \]