\(\int \frac {(1-2 x)^{5/2} (2+3 x)}{(3+5 x)^3} \, dx\) [1994]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 22, antiderivative size = 96 \[ \int \frac {(1-2 x)^{5/2} (2+3 x)}{(3+5 x)^3} \, dx=-\frac {63}{125} \sqrt {1-2 x}-\frac {21}{275} (1-2 x)^{3/2}-\frac {(1-2 x)^{7/2}}{110 (3+5 x)^2}-\frac {63 (1-2 x)^{5/2}}{550 (3+5 x)}+\frac {63}{125} \sqrt {\frac {11}{5}} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \]

[Out]

-21/275*(1-2*x)^(3/2)-1/110*(1-2*x)^(7/2)/(3+5*x)^2-63/550*(1-2*x)^(5/2)/(3+5*x)+63/625*arctanh(1/11*55^(1/2)*
(1-2*x)^(1/2))*55^(1/2)-63/125*(1-2*x)^(1/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {79, 43, 52, 65, 212} \[ \int \frac {(1-2 x)^{5/2} (2+3 x)}{(3+5 x)^3} \, dx=\frac {63}{125} \sqrt {\frac {11}{5}} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )-\frac {(1-2 x)^{7/2}}{110 (5 x+3)^2}-\frac {63 (1-2 x)^{5/2}}{550 (5 x+3)}-\frac {21}{275} (1-2 x)^{3/2}-\frac {63}{125} \sqrt {1-2 x} \]

[In]

Int[((1 - 2*x)^(5/2)*(2 + 3*x))/(3 + 5*x)^3,x]

[Out]

(-63*Sqrt[1 - 2*x])/125 - (21*(1 - 2*x)^(3/2))/275 - (1 - 2*x)^(7/2)/(110*(3 + 5*x)^2) - (63*(1 - 2*x)^(5/2))/
(550*(3 + 5*x)) + (63*Sqrt[11/5]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/125

Rule 43

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] - Dist[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] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && GtQ[n, 0]

Rule 52

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] + Dist[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] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[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] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 79

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(-(b*e - a*f
))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Dist[(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] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || L
tQ[p, n]))))

Rule 212

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] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps \begin{align*} \text {integral}& = -\frac {(1-2 x)^{7/2}}{110 (3+5 x)^2}+\frac {63}{110} \int \frac {(1-2 x)^{5/2}}{(3+5 x)^2} \, dx \\ & = -\frac {(1-2 x)^{7/2}}{110 (3+5 x)^2}-\frac {63 (1-2 x)^{5/2}}{550 (3+5 x)}-\frac {63}{110} \int \frac {(1-2 x)^{3/2}}{3+5 x} \, dx \\ & = -\frac {21}{275} (1-2 x)^{3/2}-\frac {(1-2 x)^{7/2}}{110 (3+5 x)^2}-\frac {63 (1-2 x)^{5/2}}{550 (3+5 x)}-\frac {63}{50} \int \frac {\sqrt {1-2 x}}{3+5 x} \, dx \\ & = -\frac {63}{125} \sqrt {1-2 x}-\frac {21}{275} (1-2 x)^{3/2}-\frac {(1-2 x)^{7/2}}{110 (3+5 x)^2}-\frac {63 (1-2 x)^{5/2}}{550 (3+5 x)}-\frac {693}{250} \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx \\ & = -\frac {63}{125} \sqrt {1-2 x}-\frac {21}{275} (1-2 x)^{3/2}-\frac {(1-2 x)^{7/2}}{110 (3+5 x)^2}-\frac {63 (1-2 x)^{5/2}}{550 (3+5 x)}+\frac {693}{250} \text {Subst}\left (\int \frac {1}{\frac {11}{2}-\frac {5 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right ) \\ & = -\frac {63}{125} \sqrt {1-2 x}-\frac {21}{275} (1-2 x)^{3/2}-\frac {(1-2 x)^{7/2}}{110 (3+5 x)^2}-\frac {63 (1-2 x)^{5/2}}{550 (3+5 x)}+\frac {63}{125} \sqrt {\frac {11}{5}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.66 \[ \int \frac {(1-2 x)^{5/2} (2+3 x)}{(3+5 x)^3} \, dx=\frac {\frac {5 \sqrt {1-2 x} \left (-1394-3795 x-2280 x^2+400 x^3\right )}{(3+5 x)^2}+126 \sqrt {55} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{1250} \]

[In]

Integrate[((1 - 2*x)^(5/2)*(2 + 3*x))/(3 + 5*x)^3,x]

[Out]

((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

Maple [A] (verified)

Time = 3.29 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.58

method result size
risch \(-\frac {800 x^{4}-4960 x^{3}-5310 x^{2}+1007 x +1394}{250 \left (3+5 x \right )^{2} \sqrt {1-2 x}}+\frac {63 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{625}\) \(56\)
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 \(-\frac {4 \left (1-2 x \right )^{\frac {3}{2}}}{125}-\frac {256 \sqrt {1-2 x}}{625}-\frac {44 \left (-\frac {57 \left (1-2 x \right )^{\frac {3}{2}}}{20}+\frac {649 \sqrt {1-2 x}}{100}\right )}{25 \left (-6-10 x \right )^{2}}+\frac {63 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{625}\) \(66\)
default \(-\frac {4 \left (1-2 x \right )^{\frac {3}{2}}}{125}-\frac {256 \sqrt {1-2 x}}{625}-\frac {44 \left (-\frac {57 \left (1-2 x \right )^{\frac {3}{2}}}{20}+\frac {649 \sqrt {1-2 x}}{100}\right )}{25 \left (-6-10 x \right )^{2}}+\frac {63 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{625}\) \(66\)
trager \(\frac {\left (400 x^{3}-2280 x^{2}-3795 x -1394\right ) \sqrt {1-2 x}}{250 \left (3+5 x \right )^{2}}+\frac {63 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) \ln \left (\frac {-5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) x +55 \sqrt {1-2 x}+8 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right )}{3+5 x}\right )}{1250}\) \(77\)

[In]

int((1-2*x)^(5/2)*(2+3*x)/(3+5*x)^3,x,method=_RETURNVERBOSE)

[Out]

-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*arctanh(1/11*55^(1/2)*(1-2*x)^(1
/2))*55^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.90 \[ \int \frac {(1-2 x)^{5/2} (2+3 x)}{(3+5 x)^3} \, dx=\frac {63 \, \sqrt {11} \sqrt {5} {\left (25 \, x^{2} + 30 \, x + 9\right )} \log \left (-\frac {\sqrt {11} \sqrt {5} \sqrt {-2 \, x + 1} - 5 \, x + 8}{5 \, x + 3}\right ) + 5 \, {\left (400 \, x^{3} - 2280 \, x^{2} - 3795 \, x - 1394\right )} \sqrt {-2 \, x + 1}}{1250 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} \]

[In]

integrate((1-2*x)^(5/2)*(2+3*x)/(3+5*x)^3,x, algorithm="fricas")

[Out]

1/1250*(63*sqrt(11)*sqrt(5)*(25*x^2 + 30*x + 9)*log(-(sqrt(11)*sqrt(5)*sqrt(-2*x + 1) - 5*x + 8)/(5*x + 3)) +
5*(400*x^3 - 2280*x^2 - 3795*x - 1394)*sqrt(-2*x + 1))/(25*x^2 + 30*x + 9)

Sympy [A] (verification not implemented)

Time = 78.55 (sec) , antiderivative size = 354, normalized size of antiderivative = 3.69 \[ \int \frac {(1-2 x)^{5/2} (2+3 x)}{(3+5 x)^3} \, dx=- \frac {4 \left (1 - 2 x\right )^{\frac {3}{2}}}{125} - \frac {256 \sqrt {1 - 2 x}}{625} - \frac {186 \sqrt {55} \left (\log {\left (\sqrt {1 - 2 x} - \frac {\sqrt {55}}{5} \right )} - \log {\left (\sqrt {1 - 2 x} + \frac {\sqrt {55}}{5} \right )}\right )}{3125} - \frac {13068 \left (\begin {cases} \frac {\sqrt {55} \left (- \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1 \right )}}{4} + \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1 \right )}}{4} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1\right )} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1\right )}\right )}{605} & \text {for}\: \sqrt {1 - 2 x} > - \frac {\sqrt {55}}{5} \wedge \sqrt {1 - 2 x} < \frac {\sqrt {55}}{5} \end {cases}\right )}{625} + \frac {10648 \left (\begin {cases} \frac {\sqrt {55} \cdot \left (\frac {3 \log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1 \right )}}{16} - \frac {3 \log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1 \right )}}{16} + \frac {3}{16 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1\right )} + \frac {1}{16 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1\right )^{2}} + \frac {3}{16 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1\right )} - \frac {1}{16 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1\right )^{2}}\right )}{6655} & \text {for}\: \sqrt {1 - 2 x} > - \frac {\sqrt {55}}{5} \wedge \sqrt {1 - 2 x} < \frac {\sqrt {55}}{5} \end {cases}\right )}{625} \]

[In]

integrate((1-2*x)**(5/2)*(2+3*x)/(3+5*x)**3,x)

[Out]

-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*Piecewise((sqrt(55)*(-log(sqrt(55)*sqrt(1 - 2*x)/11 - 1)/4 + log(sqrt(55)*sq
rt(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, (s
qrt(1 - 2*x) > -sqrt(55)/5) & (sqrt(1 - 2*x) < sqrt(55)/5)))/625 + 10648*Piecewise((sqrt(55)*(3*log(sqrt(55)*s
qrt(1 - 2*x)/11 - 1)/16 - 3*log(sqrt(55)*sqrt(1 - 2*x)/11 + 1)/16 + 3/(16*(sqrt(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

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.96 \[ \int \frac {(1-2 x)^{5/2} (2+3 x)}{(3+5 x)^3} \, dx=-\frac {4}{125} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - \frac {63}{1250} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) - \frac {256}{625} \, \sqrt {-2 \, x + 1} + \frac {11 \, {\left (285 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 649 \, \sqrt {-2 \, x + 1}\right )}}{625 \, {\left (25 \, {\left (2 \, x - 1\right )}^{2} + 220 \, x + 11\right )}} \]

[In]

integrate((1-2*x)^(5/2)*(2+3*x)/(3+5*x)^3,x, algorithm="maxima")

[Out]

-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)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.90 \[ \int \frac {(1-2 x)^{5/2} (2+3 x)}{(3+5 x)^3} \, dx=-\frac {4}{125} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - \frac {63}{1250} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) - \frac {256}{625} \, \sqrt {-2 \, x + 1} + \frac {11 \, {\left (285 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 649 \, \sqrt {-2 \, x + 1}\right )}}{2500 \, {\left (5 \, x + 3\right )}^{2}} \]

[In]

integrate((1-2*x)^(5/2)*(2+3*x)/(3+5*x)^3,x, algorithm="giac")

[Out]

-4/125*(-2*x + 1)^(3/2) - 63/1250*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-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

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.77 \[ \int \frac {(1-2 x)^{5/2} (2+3 x)}{(3+5 x)^3} \, dx=-\frac {256\,\sqrt {1-2\,x}}{625}-\frac {4\,{\left (1-2\,x\right )}^{3/2}}{125}-\frac {\frac {7139\,\sqrt {1-2\,x}}{15625}-\frac {627\,{\left (1-2\,x\right )}^{3/2}}{3125}}{\frac {44\,x}{5}+{\left (2\,x-1\right )}^2+\frac {11}{25}}-\frac {\sqrt {55}\,\mathrm {atan}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}\,1{}\mathrm {i}}{11}\right )\,63{}\mathrm {i}}{625} \]

[In]

int(((1 - 2*x)^(5/2)*(3*x + 2))/(5*x + 3)^3,x)

[Out]

- (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)