Integrand size = 22, antiderivative size = 96 \[ \int \frac {(1-2 x)^{5/2} (2+3 x)}{(3+5 x)^3} \, dx=-\frac {254}{625} \sqrt {1-2 x}-\frac {4}{125} (1-2 x)^{3/2}-\frac {(1-2 x)^{5/2}}{50 (3+5 x)^2}-\frac {671 \sqrt {1-2 x}}{1250 (3+5 x)}+\frac {63}{125} \sqrt {\frac {11}{5}} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \] Output:
-254/625*(1-2*x)^(1/2)-4/125*(1-2*x)^(3/2)-1/50*(1-2*x)^(5/2)/(3+5*x)^2-67 1*(1-2*x)^(1/2)/(3750+6250*x)+63/625*55^(1/2)*arctanh(1/11*55^(1/2)*(1-2*x )^(1/2))
Time = 0.15 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.66 \[ \int \frac {(1-2 x)^{5/2} (2+3 x)}{(3+5 x)^3} \, dx=\frac {\frac {5 \sqrt {1-2 x} \left (-1394-3795 x-2280 x^2+400 x^3\right )}{(3+5 x)^2}+126 \sqrt {55} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{1250} \] Input:
Integrate[((1 - 2*x)^(5/2)*(2 + 3*x))/(3 + 5*x)^3,x]
Output:
((5*Sqrt[1 - 2*x]*(-1394 - 3795*x - 2280*x^2 + 400*x^3))/(3 + 5*x)^2 + 126 *Sqrt[55]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/1250
Time = 0.19 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.10, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {87, 51, 60, 60, 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.
| \(\displaystyle \int \frac {(1-2 x)^{5/2} (3 x+2)}{(5 x+3)^3} \, dx\) |
| \(\Big \downarrow \) 87 |
| \(\displaystyle \frac {63}{110} \int \frac {(1-2 x)^{5/2}}{(5 x+3)^2}dx-\frac {(1-2 x)^{7/2}}{110 (5 x+3)^2}\) |
| \(\Big \downarrow \) 51 |
| \(\displaystyle \frac {63}{110} \left (-\int \frac {(1-2 x)^{3/2}}{5 x+3}dx-\frac {(1-2 x)^{5/2}}{5 (5 x+3)}\right )-\frac {(1-2 x)^{7/2}}{110 (5 x+3)^2}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {63}{110} \left (-\frac {11}{5} \int \frac {\sqrt {1-2 x}}{5 x+3}dx-\frac {(1-2 x)^{5/2}}{5 (5 x+3)}-\frac {2}{15} (1-2 x)^{3/2}\right )-\frac {(1-2 x)^{7/2}}{110 (5 x+3)^2}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {63}{110} \left (-\frac {11}{5} \left (\frac {11}{5} \int \frac {1}{\sqrt {1-2 x} (5 x+3)}dx+\frac {2}{5} \sqrt {1-2 x}\right )-\frac {(1-2 x)^{5/2}}{5 (5 x+3)}-\frac {2}{15} (1-2 x)^{3/2}\right )-\frac {(1-2 x)^{7/2}}{110 (5 x+3)^2}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {63}{110} \left (-\frac {11}{5} \left (\frac {2}{5} \sqrt {1-2 x}-\frac {11}{5} \int \frac {1}{\frac {11}{2}-\frac {5}{2} (1-2 x)}d\sqrt {1-2 x}\right )-\frac {(1-2 x)^{5/2}}{5 (5 x+3)}-\frac {2}{15} (1-2 x)^{3/2}\right )-\frac {(1-2 x)^{7/2}}{110 (5 x+3)^2}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {63}{110} \left (-\frac {11}{5} \left (\frac {2}{5} \sqrt {1-2 x}-\frac {2}{5} \sqrt {\frac {11}{5}} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )\right )-\frac {(1-2 x)^{5/2}}{5 (5 x+3)}-\frac {2}{15} (1-2 x)^{3/2}\right )-\frac {(1-2 x)^{7/2}}{110 (5 x+3)^2}\) |
Input:
Int[((1 - 2*x)^(5/2)*(2 + 3*x))/(3 + 5*x)^3,x]
Output:
-1/110*(1 - 2*x)^(7/2)/(3 + 5*x)^2 + (63*((-2*(1 - 2*x)^(3/2))/15 - (1 - 2 *x)^(5/2)/(5*(3 + 5*x)) - (11*((2*Sqrt[1 - 2*x])/5 - (2*Sqrt[11/5]*ArcTanh [Sqrt[5/11]*Sqrt[1 - 2*x]])/5))/5))/110
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + 1))), x] - Simp[d*(n/(b*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n}, x ] && ILtQ[m, -1] && FractionQ[n] && GtQ[n, 0]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( b*(m + n + 1))) Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[a, b, c, d, m, n, 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)^(-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 = 0.21 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.58
| method | result | size |
| risch | \(-\frac {800 x^{4}-4960 x^{3}-5310 x^{2}+1007 x +1394}{250 \left (3+5 x \right )^{2} \sqrt {1-2 x}}+\frac {63 \sqrt {55}\, \operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right )}{625}\) | \(56\) |
| pseudoelliptic | \(\frac {126 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \left (3+5 x \right )^{2} \sqrt {55}+5 \sqrt {1-2 x}\, \left (400 x^{3}-2280 x^{2}-3795 x -1394\right )}{1250 \left (3+5 x \right )^{2}}\) | \(60\) |
| derivativedivides | \(-\frac {4 \left (1-2 x \right )^{\frac {3}{2}}}{125}-\frac {256 \sqrt {1-2 x}}{625}-\frac {44 \left (-\frac {57 \left (1-2 x \right )^{\frac {3}{2}}}{20}+\frac {649 \sqrt {1-2 x}}{100}\right )}{25 \left (-6-10 x \right )^{2}}+\frac {63 \sqrt {55}\, \operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right )}{625}\) | \(66\) |
| default | \(-\frac {4 \left (1-2 x \right )^{\frac {3}{2}}}{125}-\frac {256 \sqrt {1-2 x}}{625}-\frac {44 \left (-\frac {57 \left (1-2 x \right )^{\frac {3}{2}}}{20}+\frac {649 \sqrt {1-2 x}}{100}\right )}{25 \left (-6-10 x \right )^{2}}+\frac {63 \sqrt {55}\, \operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right )}{625}\) | \(66\) |
| trager | \(\frac {\left (400 x^{3}-2280 x^{2}-3795 x -1394\right ) \sqrt {1-2 x}}{250 \left (3+5 x \right )^{2}}+\frac {63 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) \ln \left (\frac {-5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) x +55 \sqrt {1-2 x}+8 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right )}{3+5 x}\right )}{1250}\) | \(77\) |
Input:
int((1-2*x)^(5/2)*(2+3*x)/(3+5*x)^3,x,method=_RETURNVERBOSE)
Output:
-1/250*(800*x^4-4960*x^3-5310*x^2+1007*x+1394)/(3+5*x)^2/(1-2*x)^(1/2)+63/ 625*55^(1/2)*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))
Time = 0.07 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.82 \[ \int \frac {(1-2 x)^{5/2} (2+3 x)}{(3+5 x)^3} \, dx=\frac {63 \, \sqrt {\frac {11}{5}} {\left (25 \, x^{2} + 30 \, x + 9\right )} \log \left (\frac {5 \, x - 5 \, \sqrt {\frac {11}{5}} \sqrt {-2 \, x + 1} - 8}{5 \, x + 3}\right ) + {\left (400 \, x^{3} - 2280 \, x^{2} - 3795 \, x - 1394\right )} \sqrt {-2 \, x + 1}}{250 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} \] Input:
integrate((1-2*x)^(5/2)*(2+3*x)/(3+5*x)^3,x, algorithm="fricas")
Output:
1/250*(63*sqrt(11/5)*(25*x^2 + 30*x + 9)*log((5*x - 5*sqrt(11/5)*sqrt(-2*x + 1) - 8)/(5*x + 3)) + (400*x^3 - 2280*x^2 - 3795*x - 1394)*sqrt(-2*x + 1 ))/(25*x^2 + 30*x + 9)
Time = 85.83 (sec) , antiderivative size = 354, normalized size of antiderivative = 3.69 \[ \int \frac {(1-2 x)^{5/2} (2+3 x)}{(3+5 x)^3} \, dx=- \frac {4 \left (1 - 2 x\right )^{\frac {3}{2}}}{125} - \frac {256 \sqrt {1 - 2 x}}{625} - \frac {186 \sqrt {55} \left (\log {\left (\sqrt {1 - 2 x} - \frac {\sqrt {55}}{5} \right )} - \log {\left (\sqrt {1 - 2 x} + \frac {\sqrt {55}}{5} \right )}\right )}{3125} - \frac {13068 \left (\begin {cases} \frac {\sqrt {55} \left (- \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1 \right )}}{4} + \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1 \right )}}{4} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1\right )} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1\right )}\right )}{605} & \text {for}\: \sqrt {1 - 2 x} > - \frac {\sqrt {55}}{5} \wedge \sqrt {1 - 2 x} < \frac {\sqrt {55}}{5} \end {cases}\right )}{625} + \frac {10648 \left (\begin {cases} \frac {\sqrt {55} \cdot \left (\frac {3 \log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1 \right )}}{16} - \frac {3 \log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1 \right )}}{16} + \frac {3}{16 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1\right )} + \frac {1}{16 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1\right )^{2}} + \frac {3}{16 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1\right )} - \frac {1}{16 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1\right )^{2}}\right )}{6655} & \text {for}\: \sqrt {1 - 2 x} > - \frac {\sqrt {55}}{5} \wedge \sqrt {1 - 2 x} < \frac {\sqrt {55}}{5} \end {cases}\right )}{625} \] Input:
integrate((1-2*x)**(5/2)*(2+3*x)/(3+5*x)**3,x)
Output:
-4*(1 - 2*x)**(3/2)/125 - 256*sqrt(1 - 2*x)/625 - 186*sqrt(55)*(log(sqrt(1 - 2*x) - sqrt(55)/5) - log(sqrt(1 - 2*x) + sqrt(55)/5))/3125 - 13068*Piec ewise((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(5 5)*sqrt(1 - 2*x)/11 - 1)))/605, (sqrt(1 - 2*x) > -sqrt(55)/5) & (sqrt(1 - 2*x) < sqrt(55)/5)))/625 + 10648*Piecewise((sqrt(55)*(3*log(sqrt(55)*sqrt( 1 - 2*x)/11 - 1)/16 - 3*log(sqrt(55)*sqrt(1 - 2*x)/11 + 1)/16 + 3/(16*(sqr t(55)*sqrt(1 - 2*x)/11 + 1)) + 1/(16*(sqrt(55)*sqrt(1 - 2*x)/11 + 1)**2) + 3/(16*(sqrt(55)*sqrt(1 - 2*x)/11 - 1)) - 1/(16*(sqrt(55)*sqrt(1 - 2*x)/11 - 1)**2))/6655, (sqrt(1 - 2*x) > -sqrt(55)/5) & (sqrt(1 - 2*x) < sqrt(55) /5)))/625
Time = 0.11 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.96 \[ \int \frac {(1-2 x)^{5/2} (2+3 x)}{(3+5 x)^3} \, dx=-\frac {4}{125} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - \frac {63}{1250} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) - \frac {256}{625} \, \sqrt {-2 \, x + 1} + \frac {11 \, {\left (285 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 649 \, \sqrt {-2 \, x + 1}\right )}}{625 \, {\left (25 \, {\left (2 \, x - 1\right )}^{2} + 220 \, x + 11\right )}} \] Input:
integrate((1-2*x)^(5/2)*(2+3*x)/(3+5*x)^3,x, algorithm="maxima")
Output:
-4/125*(-2*x + 1)^(3/2) - 63/1250*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 256/625*sqrt(-2*x + 1) + 11/625*(285* (-2*x + 1)^(3/2) - 649*sqrt(-2*x + 1))/(25*(2*x - 1)^2 + 220*x + 11)
Time = 0.13 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.90 \[ \int \frac {(1-2 x)^{5/2} (2+3 x)}{(3+5 x)^3} \, dx=-\frac {4}{125} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - \frac {63}{1250} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) - \frac {256}{625} \, \sqrt {-2 \, x + 1} + \frac {11 \, {\left (285 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 649 \, \sqrt {-2 \, x + 1}\right )}}{2500 \, {\left (5 \, x + 3\right )}^{2}} \] Input:
integrate((1-2*x)^(5/2)*(2+3*x)/(3+5*x)^3,x, algorithm="giac")
Output:
-4/125*(-2*x + 1)^(3/2) - 63/1250*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sq rt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 256/625*sqrt(-2*x + 1) + 11 /2500*(285*(-2*x + 1)^(3/2) - 649*sqrt(-2*x + 1))/(5*x + 3)^2
Time = 0.04 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.77 \[ \int \frac {(1-2 x)^{5/2} (2+3 x)}{(3+5 x)^3} \, dx=-\frac {256\,\sqrt {1-2\,x}}{625}-\frac {4\,{\left (1-2\,x\right )}^{3/2}}{125}-\frac {\frac {7139\,\sqrt {1-2\,x}}{15625}-\frac {627\,{\left (1-2\,x\right )}^{3/2}}{3125}}{\frac {44\,x}{5}+{\left (2\,x-1\right )}^2+\frac {11}{25}}-\frac {\sqrt {55}\,\mathrm {atan}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}\,1{}\mathrm {i}}{11}\right )\,63{}\mathrm {i}}{625} \] Input:
int(((1 - 2*x)^(5/2)*(3*x + 2))/(5*x + 3)^3,x)
Output:
- (55^(1/2)*atan((55^(1/2)*(1 - 2*x)^(1/2)*1i)/11)*63i)/625 - (256*(1 - 2* x)^(1/2))/625 - (4*(1 - 2*x)^(3/2))/125 - ((7139*(1 - 2*x)^(1/2))/15625 - (627*(1 - 2*x)^(3/2))/3125)/((44*x)/5 + (2*x - 1)^2 + 11/25)
Time = 0.17 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.70 \[ \int \frac {(1-2 x)^{5/2} (2+3 x)}{(3+5 x)^3} \, dx=\frac {2000 \sqrt {-2 x +1}\, x^{3}-11400 \sqrt {-2 x +1}\, x^{2}-18975 \sqrt {-2 x +1}\, x -6970 \sqrt {-2 x +1}-1575 \sqrt {55}\, \mathrm {log}\left (5 \sqrt {-2 x +1}-\sqrt {55}\right ) x^{2}-1890 \sqrt {55}\, \mathrm {log}\left (5 \sqrt {-2 x +1}-\sqrt {55}\right ) x -567 \sqrt {55}\, \mathrm {log}\left (5 \sqrt {-2 x +1}-\sqrt {55}\right )+1575 \sqrt {55}\, \mathrm {log}\left (5 \sqrt {-2 x +1}+\sqrt {55}\right ) x^{2}+1890 \sqrt {55}\, \mathrm {log}\left (5 \sqrt {-2 x +1}+\sqrt {55}\right ) x +567 \sqrt {55}\, \mathrm {log}\left (5 \sqrt {-2 x +1}+\sqrt {55}\right )}{31250 x^{2}+37500 x +11250} \] Input:
int((1-2*x)^(5/2)*(2+3*x)/(3+5*x)^3,x)
Output:
(2000*sqrt( - 2*x + 1)*x**3 - 11400*sqrt( - 2*x + 1)*x**2 - 18975*sqrt( - 2*x + 1)*x - 6970*sqrt( - 2*x + 1) - 1575*sqrt(55)*log(5*sqrt( - 2*x + 1) - sqrt(55))*x**2 - 1890*sqrt(55)*log(5*sqrt( - 2*x + 1) - sqrt(55))*x - 56 7*sqrt(55)*log(5*sqrt( - 2*x + 1) - sqrt(55)) + 1575*sqrt(55)*log(5*sqrt( - 2*x + 1) + sqrt(55))*x**2 + 1890*sqrt(55)*log(5*sqrt( - 2*x + 1) + sqrt( 55))*x + 567*sqrt(55)*log(5*sqrt( - 2*x + 1) + sqrt(55)))/(1250*(25*x**2 + 30*x + 9))