Integrand size = 24, antiderivative size = 77 \[ \int \frac {(1-2 x)^{3/2}}{(2+3 x) (3+5 x)^2} \, dx=-\frac {11 \sqrt {1-2 x}}{5 (3+5 x)}-14 \sqrt {\frac {7}{3}} \text {arctanh}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )+\frac {72}{5} \sqrt {\frac {11}{5}} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \]
72/25*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)-14/3*arctanh(1/7*21^(1 /2)*(1-2*x)^(1/2))*21^(1/2)-11/5*(1-2*x)^(1/2)/(3+5*x)
Time = 0.23 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.97 \[ \int \frac {(1-2 x)^{3/2}}{(2+3 x) (3+5 x)^2} \, dx=-\frac {11 \sqrt {1-2 x}}{15+25 x}-14 \sqrt {\frac {7}{3}} \text {arctanh}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )+\frac {72}{5} \sqrt {\frac {11}{5}} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \]
(-11*Sqrt[1 - 2*x])/(15 + 25*x) - 14*Sqrt[7/3]*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]] + (72*Sqrt[11/5]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/5
Time = 0.18 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.04, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {109, 174, 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.
\(\displaystyle \int \frac {(1-2 x)^{3/2}}{(3 x+2) (5 x+3)^2} \, dx\) |
\(\Big \downarrow \) 109 |
\(\displaystyle -\frac {1}{5} \int \frac {57-37 x}{\sqrt {1-2 x} (3 x+2) (5 x+3)}dx-\frac {11 \sqrt {1-2 x}}{5 (5 x+3)}\) |
\(\Big \downarrow \) 174 |
\(\displaystyle \frac {1}{5} \left (245 \int \frac {1}{\sqrt {1-2 x} (3 x+2)}dx-396 \int \frac {1}{\sqrt {1-2 x} (5 x+3)}dx\right )-\frac {11 \sqrt {1-2 x}}{5 (5 x+3)}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{5} \left (396 \int \frac {1}{\frac {11}{2}-\frac {5}{2} (1-2 x)}d\sqrt {1-2 x}-245 \int \frac {1}{\frac {7}{2}-\frac {3}{2} (1-2 x)}d\sqrt {1-2 x}\right )-\frac {11 \sqrt {1-2 x}}{5 (5 x+3)}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{5} \left (72 \sqrt {\frac {11}{5}} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )-70 \sqrt {\frac {7}{3}} \text {arctanh}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )\right )-\frac {11 \sqrt {1-2 x}}{5 (5 x+3)}\) |
(-11*Sqrt[1 - 2*x])/(5*(3 + 5*x)) + (-70*Sqrt[7/3]*ArcTanh[Sqrt[3/7]*Sqrt[ 1 - 2*x]] + 72*Sqrt[11/5]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/5
3.20.14.3.1 Defintions of rubi rules used
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[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))* ((c_.) + (d_.)*(x_))), x_] :> Simp[(b*g - a*h)/(b*c - a*d) Int[(e + f*x)^ p/(a + b*x), x], x] - Simp[(d*g - c*h)/(b*c - a*d) Int[(e + f*x)^p/(c + d *x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x]
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 = 1.02 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.70
method | result | size |
derivativedivides | \(\frac {22 \sqrt {1-2 x}}{25 \left (-\frac {6}{5}-2 x \right )}+\frac {72 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{25}-\frac {14 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{3}\) | \(54\) |
default | \(\frac {22 \sqrt {1-2 x}}{25 \left (-\frac {6}{5}-2 x \right )}+\frac {72 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{25}-\frac {14 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{3}\) | \(54\) |
risch | \(\frac {-\frac {11}{5}+\frac {22 x}{5}}{\left (3+5 x \right ) \sqrt {1-2 x}}-\frac {14 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{3}+\frac {72 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{25}\) | \(59\) |
pseudoelliptic | \(\frac {-350 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \left (3+5 x \right ) \sqrt {21}+216 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \left (3+5 x \right ) \sqrt {55}-165 \sqrt {1-2 x}}{225+375 x}\) | \(65\) |
trager | \(-\frac {11 \sqrt {1-2 x}}{5 \left (3+5 x \right )}-\frac {7 \operatorname {RootOf}\left (\textit {\_Z}^{2}-21\right ) \ln \left (\frac {-3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-21\right ) x +21 \sqrt {1-2 x}+5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-21\right )}{2+3 x}\right )}{3}-\frac {36 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) \ln \left (\frac {5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) x +55 \sqrt {1-2 x}-8 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right )}{3+5 x}\right )}{25}\) | \(106\) |
22/25*(1-2*x)^(1/2)/(-6/5-2*x)+72/25*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))* 55^(1/2)-14/3*arctanh(1/7*21^(1/2)*(1-2*x)^(1/2))*21^(1/2)
Time = 0.24 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.32 \[ \int \frac {(1-2 x)^{3/2}}{(2+3 x) (3+5 x)^2} \, dx=\frac {108 \, \sqrt {11} \sqrt {5} {\left (5 \, x + 3\right )} \log \left (-\frac {\sqrt {11} \sqrt {5} \sqrt {-2 \, x + 1} - 5 \, x + 8}{5 \, x + 3}\right ) + 175 \, \sqrt {7} \sqrt {3} {\left (5 \, x + 3\right )} \log \left (\frac {\sqrt {7} \sqrt {3} \sqrt {-2 \, x + 1} + 3 \, x - 5}{3 \, x + 2}\right ) - 165 \, \sqrt {-2 \, x + 1}}{75 \, {\left (5 \, x + 3\right )}} \]
1/75*(108*sqrt(11)*sqrt(5)*(5*x + 3)*log(-(sqrt(11)*sqrt(5)*sqrt(-2*x + 1) - 5*x + 8)/(5*x + 3)) + 175*sqrt(7)*sqrt(3)*(5*x + 3)*log((sqrt(7)*sqrt(3 )*sqrt(-2*x + 1) + 3*x - 5)/(3*x + 2)) - 165*sqrt(-2*x + 1))/(5*x + 3)
Time = 14.83 (sec) , antiderivative size = 204, normalized size of antiderivative = 2.65 \[ \int \frac {(1-2 x)^{3/2}}{(2+3 x) (3+5 x)^2} \, dx=\frac {7 \sqrt {21} \left (\log {\left (\sqrt {1 - 2 x} - \frac {\sqrt {21}}{3} \right )} - \log {\left (\sqrt {1 - 2 x} + \frac {\sqrt {21}}{3} \right )}\right )}{3} - \frac {37 \sqrt {55} \left (\log {\left (\sqrt {1 - 2 x} - \frac {\sqrt {55}}{5} \right )} - \log {\left (\sqrt {1 - 2 x} + \frac {\sqrt {55}}{5} \right )}\right )}{25} - \frac {484 \left (\begin {cases} \frac {\sqrt {55} \left (- \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1 \right )}}{4} + \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1 \right )}}{4} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1\right )} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1\right )}\right )}{605} & \text {for}\: \sqrt {1 - 2 x} > - \frac {\sqrt {55}}{5} \wedge \sqrt {1 - 2 x} < \frac {\sqrt {55}}{5} \end {cases}\right )}{5} \]
7*sqrt(21)*(log(sqrt(1 - 2*x) - sqrt(21)/3) - log(sqrt(1 - 2*x) + sqrt(21) /3))/3 - 37*sqrt(55)*(log(sqrt(1 - 2*x) - sqrt(55)/5) - log(sqrt(1 - 2*x) + sqrt(55)/5))/25 - 484*Piecewise((sqrt(55)*(-log(sqrt(55)*sqrt(1 - 2*x)/1 1 - 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)))/5
Time = 0.29 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.16 \[ \int \frac {(1-2 x)^{3/2}}{(2+3 x) (3+5 x)^2} \, dx=-\frac {36}{25} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) + \frac {7}{3} \, \sqrt {21} \log \left (-\frac {\sqrt {21} - 3 \, \sqrt {-2 \, x + 1}}{\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}}\right ) - \frac {11 \, \sqrt {-2 \, x + 1}}{5 \, {\left (5 \, x + 3\right )}} \]
-36/25*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 7/3*sqrt(21)*log(-(sqrt(21) - 3*sqrt(-2*x + 1))/(sqrt(21) + 3*sq rt(-2*x + 1))) - 11/5*sqrt(-2*x + 1)/(5*x + 3)
Time = 0.29 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.23 \[ \int \frac {(1-2 x)^{3/2}}{(2+3 x) (3+5 x)^2} \, dx=-\frac {36}{25} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) + \frac {7}{3} \, \sqrt {21} \log \left (\frac {{\left | -2 \, \sqrt {21} + 6 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}\right )}}\right ) - \frac {11 \, \sqrt {-2 \, x + 1}}{5 \, {\left (5 \, x + 3\right )}} \]
-36/25*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5 *sqrt(-2*x + 1))) + 7/3*sqrt(21)*log(1/2*abs(-2*sqrt(21) + 6*sqrt(-2*x + 1 ))/(sqrt(21) + 3*sqrt(-2*x + 1))) - 11/5*sqrt(-2*x + 1)/(5*x + 3)
Time = 0.11 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.69 \[ \int \frac {(1-2 x)^{3/2}}{(2+3 x) (3+5 x)^2} \, dx=\frac {72\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )}{25}-\frac {14\,\sqrt {21}\,\mathrm {atanh}\left (\frac {\sqrt {21}\,\sqrt {1-2\,x}}{7}\right )}{3}-\frac {22\,\sqrt {1-2\,x}}{25\,\left (2\,x+\frac {6}{5}\right )} \]