Integrand size = 24, antiderivative size = 68 \[ \int \frac {(2+3 x)^2}{(1-2 x)^{3/2} (3+5 x)^2} \, dx=\frac {49}{22 \sqrt {1-2 x} (3+5 x)}-\frac {1227 \sqrt {1-2 x}}{1210 (3+5 x)}-\frac {138 \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{605 \sqrt {55}} \]
-138/33275*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)+49/22/(3+5*x)/(1- 2*x)^(1/2)-1227/1210*(1-2*x)^(1/2)/(3+5*x)
Time = 0.13 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.78 \[ \int \frac {(2+3 x)^2}{(1-2 x)^{3/2} (3+5 x)^2} \, dx=\frac {734+1227 x}{605 \sqrt {1-2 x} (3+5 x)}-\frac {138 \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{605 \sqrt {55}} \]
(734 + 1227*x)/(605*Sqrt[1 - 2*x]*(3 + 5*x)) - (138*ArcTanh[Sqrt[5/11]*Sqr t[1 - 2*x]])/(605*Sqrt[55])
Time = 0.18 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.07, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {100, 27, 87, 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.
\(\displaystyle \int \frac {(3 x+2)^2}{(1-2 x)^{3/2} (5 x+3)^2} \, dx\) |
\(\Big \downarrow \) 100 |
\(\displaystyle \frac {49}{22 \sqrt {1-2 x} (5 x+3)}-\frac {1}{22} \int -\frac {3 (62-33 x)}{\sqrt {1-2 x} (5 x+3)^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {3}{22} \int \frac {62-33 x}{\sqrt {1-2 x} (5 x+3)^2}dx+\frac {49}{22 \sqrt {1-2 x} (5 x+3)}\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {3}{22} \left (\frac {46}{55} \int \frac {1}{\sqrt {1-2 x} (5 x+3)}dx-\frac {409 \sqrt {1-2 x}}{55 (5 x+3)}\right )+\frac {49}{22 \sqrt {1-2 x} (5 x+3)}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {3}{22} \left (-\frac {46}{55} \int \frac {1}{\frac {11}{2}-\frac {5}{2} (1-2 x)}d\sqrt {1-2 x}-\frac {409 \sqrt {1-2 x}}{55 (5 x+3)}\right )+\frac {49}{22 \sqrt {1-2 x} (5 x+3)}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {3}{22} \left (-\frac {92 \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{55 \sqrt {55}}-\frac {409 \sqrt {1-2 x}}{55 (5 x+3)}\right )+\frac {49}{22 \sqrt {1-2 x} (5 x+3)}\) |
49/(22*Sqrt[1 - 2*x]*(3 + 5*x)) + (3*((-409*Sqrt[1 - 2*x])/(55*(3 + 5*x)) - (92*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(55*Sqrt[55])))/22
3.22.19.3.1 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_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)) Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege rQ[p] || !(IntegerQ[n] || !(EqQ[e, 0] || !(EqQ[c, 0] || LtQ[p, n]))))
Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^( p_), x_] :> Simp[(b*c - a*d)^2*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d^2*(d *e - c*f)*(n + 1))), x] - Simp[1/(d^2*(d*e - c*f)*(n + 1)) Int[(c + d*x)^ (n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*( p + 1)) - 2*a*b*d*(d*e*(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x , x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] || !SumSimplerQ[p, 1])))
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 = 1.08 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.60
method | result | size |
risch | \(\frac {1227 x +734}{605 \left (3+5 x \right ) \sqrt {1-2 x}}-\frac {138 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{33275}\) | \(41\) |
derivativedivides | \(\frac {2 \sqrt {1-2 x}}{3025 \left (-\frac {6}{5}-2 x \right )}-\frac {138 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{33275}+\frac {49}{121 \sqrt {1-2 x}}\) | \(45\) |
default | \(\frac {2 \sqrt {1-2 x}}{3025 \left (-\frac {6}{5}-2 x \right )}-\frac {138 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{33275}+\frac {49}{121 \sqrt {1-2 x}}\) | \(45\) |
pseudoelliptic | \(-\frac {690 \left (\sqrt {55}\, \left (x +\frac {3}{5}\right ) \sqrt {1-2 x}\, \operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right )-\frac {4499 x}{46}-\frac {4037}{69}\right )}{\sqrt {1-2 x}\, \left (99825+166375 x \right )}\) | \(49\) |
trager | \(-\frac {\left (1227 x +734\right ) \sqrt {1-2 x}}{605 \left (10 x^{2}+x -3\right )}+\frac {69 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) \ln \left (\frac {5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) x +55 \sqrt {1-2 x}-8 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right )}{3+5 x}\right )}{33275}\) | \(70\) |
1/605*(1227*x+734)/(3+5*x)/(1-2*x)^(1/2)-138/33275*arctanh(1/11*55^(1/2)*( 1-2*x)^(1/2))*55^(1/2)
Time = 0.22 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.96 \[ \int \frac {(2+3 x)^2}{(1-2 x)^{3/2} (3+5 x)^2} \, dx=\frac {69 \, \sqrt {55} {\left (10 \, x^{2} + x - 3\right )} \log \left (\frac {5 \, x + \sqrt {55} \sqrt {-2 \, x + 1} - 8}{5 \, x + 3}\right ) - 55 \, {\left (1227 \, x + 734\right )} \sqrt {-2 \, x + 1}}{33275 \, {\left (10 \, x^{2} + x - 3\right )}} \]
1/33275*(69*sqrt(55)*(10*x^2 + x - 3)*log((5*x + sqrt(55)*sqrt(-2*x + 1) - 8)/(5*x + 3)) - 55*(1227*x + 734)*sqrt(-2*x + 1))/(10*x^2 + x - 3)
Leaf count of result is larger than twice the leaf count of optimal. 121 vs. \(2 (56) = 112\).
Time = 41.13 (sec) , antiderivative size = 173, normalized size of antiderivative = 2.54 \[ \int \frac {(2+3 x)^2}{(1-2 x)^{3/2} (3+5 x)^2} \, dx=\frac {68 \sqrt {55} \left (\log {\left (\sqrt {1 - 2 x} - \frac {\sqrt {55}}{5} \right )} - \log {\left (\sqrt {1 - 2 x} + \frac {\sqrt {55}}{5} \right )}\right )}{33275} - \frac {4 \left (\begin {cases} \frac {\sqrt {55} \left (- \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1 \right )}}{4} + \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1 \right )}}{4} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1\right )} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1\right )}\right )}{605} & \text {for}\: \sqrt {1 - 2 x} > - \frac {\sqrt {55}}{5} \wedge \sqrt {1 - 2 x} < \frac {\sqrt {55}}{5} \end {cases}\right )}{55} + \frac {49}{121 \sqrt {1 - 2 x}} \]
68*sqrt(55)*(log(sqrt(1 - 2*x) - sqrt(55)/5) - log(sqrt(1 - 2*x) + sqrt(55 )/5))/33275 - 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)))/55 + 49/(121*sqrt(1 - 2*x))
Time = 0.28 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.96 \[ \int \frac {(2+3 x)^2}{(1-2 x)^{3/2} (3+5 x)^2} \, dx=\frac {69}{33275} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) - \frac {2 \, {\left (1227 \, x + 734\right )}}{605 \, {\left (5 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 11 \, \sqrt {-2 \, x + 1}\right )}} \]
69/33275*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2 *x + 1))) - 2/605*(1227*x + 734)/(5*(-2*x + 1)^(3/2) - 11*sqrt(-2*x + 1))
Time = 0.38 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.00 \[ \int \frac {(2+3 x)^2}{(1-2 x)^{3/2} (3+5 x)^2} \, dx=\frac {69}{33275} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) - \frac {2 \, {\left (1227 \, x + 734\right )}}{605 \, {\left (5 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 11 \, \sqrt {-2 \, x + 1}\right )}} \]
69/33275*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 2/605*(1227*x + 734)/(5*(-2*x + 1)^(3/2) - 11*sqrt(- 2*x + 1))
Time = 1.49 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.68 \[ \int \frac {(2+3 x)^2}{(1-2 x)^{3/2} (3+5 x)^2} \, dx=\frac {\frac {2454\,x}{3025}+\frac {1468}{3025}}{\frac {11\,\sqrt {1-2\,x}}{5}-{\left (1-2\,x\right )}^{3/2}}-\frac {138\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )}{33275} \]