3.20.84 \(\int \frac {(1-2 x)^{5/2}}{(3+5 x)^2} \, dx\) [1984]

Optimal. Leaf size=76 \[ -\frac {22}{25} \sqrt {1-2 x}-\frac {2}{15} (1-2 x)^{3/2}-\frac {(1-2 x)^{5/2}}{5 (3+5 x)}+\frac {22}{25} \sqrt {\frac {11}{5}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \]

[Out]

-2/15*(1-2*x)^(3/2)-1/5*(1-2*x)^(5/2)/(3+5*x)+22/125*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)-22/25*(1-2*
x)^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 0.01, antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {43, 52, 65, 212} \begin {gather*} -\frac {(1-2 x)^{5/2}}{5 (5 x+3)}-\frac {2}{15} (1-2 x)^{3/2}-\frac {22}{25} \sqrt {1-2 x}+\frac {22}{25} \sqrt {\frac {11}{5}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 - 2*x)^(5/2)/(3 + 5*x)^2,x]

[Out]

(-22*Sqrt[1 - 2*x])/25 - (2*(1 - 2*x)^(3/2))/15 - (1 - 2*x)^(5/2)/(5*(3 + 5*x)) + (22*Sqrt[11/5]*ArcTanh[Sqrt[
5/11]*Sqrt[1 - 2*x]])/25

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 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*} \int \frac {(1-2 x)^{5/2}}{(3+5 x)^2} \, dx &=-\frac {(1-2 x)^{5/2}}{5 (3+5 x)}-\int \frac {(1-2 x)^{3/2}}{3+5 x} \, dx\\ &=-\frac {2}{15} (1-2 x)^{3/2}-\frac {(1-2 x)^{5/2}}{5 (3+5 x)}-\frac {11}{5} \int \frac {\sqrt {1-2 x}}{3+5 x} \, dx\\ &=-\frac {22}{25} \sqrt {1-2 x}-\frac {2}{15} (1-2 x)^{3/2}-\frac {(1-2 x)^{5/2}}{5 (3+5 x)}-\frac {121}{25} \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx\\ &=-\frac {22}{25} \sqrt {1-2 x}-\frac {2}{15} (1-2 x)^{3/2}-\frac {(1-2 x)^{5/2}}{5 (3+5 x)}+\frac {121}{25} \text {Subst}\left (\int \frac {1}{\frac {11}{2}-\frac {5 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )\\ &=-\frac {22}{25} \sqrt {1-2 x}-\frac {2}{15} (1-2 x)^{3/2}-\frac {(1-2 x)^{5/2}}{5 (3+5 x)}+\frac {22}{25} \sqrt {\frac {11}{5}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.11, size = 58, normalized size = 0.76 \begin {gather*} \frac {1}{375} \left (\frac {5 \sqrt {1-2 x} \left (-243-260 x+40 x^2\right )}{3+5 x}+66 \sqrt {55} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((5*Sqrt[1 - 2*x]*(-243 - 260*x + 40*x^2))/(3 + 5*x) + 66*Sqrt[55]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/375

________________________________________________________________________________________

Maple [A]
time = 0.12, size = 54, normalized size = 0.71

method	result	size
risch	\(-\frac {80 x^{3}-560 x^{2}-226 x +243}{75 \left (3+5 x \right ) \sqrt {1-2 x}}+\frac {22 \arctanh \left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{125}\)	\(51\)
derivativedivides	\(-\frac {4 \left (1-2 x \right )^{\frac {3}{2}}}{75}-\frac {88 \sqrt {1-2 x}}{125}+\frac {242 \sqrt {1-2 x}}{625 \left (-\frac {6}{5}-2 x \right )}+\frac {22 \arctanh \left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{125}\)	\(54\)
default	\(-\frac {4 \left (1-2 x \right )^{\frac {3}{2}}}{75}-\frac {88 \sqrt {1-2 x}}{125}+\frac {242 \sqrt {1-2 x}}{625 \left (-\frac {6}{5}-2 x \right )}+\frac {22 \arctanh \left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{125}\)	\(54\)
trager	\(\frac {\left (40 x^{2}-260 x -243\right ) \sqrt {1-2 x}}{225+375 x}-\frac {11 \RootOf \left (\textit {\_Z}^{2}-55\right ) \ln \left (\frac {5 \RootOf \left (\textit {\_Z}^{2}-55\right ) x +55 \sqrt {1-2 x}-8 \RootOf \left (\textit {\_Z}^{2}-55\right )}{3+5 x}\right )}{125}\)	\(72\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/75*(1-2*x)^(3/2)-88/125*(1-2*x)^(1/2)+242/625*(1-2*x)^(1/2)/(-6/5-2*x)+22/125*arctanh(1/11*55^(1/2)*(1-2*x)
^(1/2))*55^(1/2)

________________________________________________________________________________________

Maxima [A]
time = 0.51, size = 71, normalized size = 0.93 \begin {gather*} -\frac {4}{75} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - \frac {11}{125} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) - \frac {88}{125} \, \sqrt {-2 \, x + 1} - \frac {121 \, \sqrt {-2 \, x + 1}}{125 \, {\left (5 \, x + 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/75*(-2*x + 1)^(3/2) - 11/125*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 8
8/125*sqrt(-2*x + 1) - 121/125*sqrt(-2*x + 1)/(5*x + 3)

________________________________________________________________________________________

Fricas [A]
time = 1.20, size = 71, normalized size = 0.93 \begin {gather*} \frac {33 \, \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 ) + 5 \, {\left (40 \, x^{2} - 260 \, x - 243\right )} \sqrt {-2 \, x + 1}}{375 \, {\left (5 \, x + 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/375*(33*sqrt(11)*sqrt(5)*(5*x + 3)*log(-(sqrt(11)*sqrt(5)*sqrt(-2*x + 1) - 5*x + 8)/(5*x + 3)) + 5*(40*x^2 -
260*x - 243)*sqrt(-2*x + 1))/(5*x + 3)

________________________________________________________________________________________

Sympy [C] Result contains complex when optimal does not.
time = 3.55, size = 196, normalized size = 2.58 \begin {gather*} \begin {cases} \frac {8 \sqrt {5} i \left (x + \frac {3}{5}\right ) \sqrt {10 x - 5}}{375} - \frac {308 \sqrt {5} i \sqrt {10 x - 5}}{1875} - \frac {22 \sqrt {55} i \operatorname {asin}{\left (\frac {\sqrt {110}}{10 \sqrt {x + \frac {3}{5}}} \right )}}{125} - \frac {121 \sqrt {5} i \sqrt {10 x - 5}}{3125 \left (x + \frac {3}{5}\right )} & \text {for}\: \left |{x + \frac {3}{5}}\right | > \frac {11}{10} \\\frac {8 \sqrt {5} \sqrt {5 - 10 x} \left (x + \frac {3}{5}\right )}{375} - \frac {308 \sqrt {5} \sqrt {5 - 10 x}}{1875} - \frac {121 \sqrt {5} \sqrt {5 - 10 x}}{3125 \left (x + \frac {3}{5}\right )} - \frac {11 \sqrt {55} \log {\left (x + \frac {3}{5} \right )}}{125} + \frac {22 \sqrt {55} \log {\left (\sqrt {\frac {5}{11} - \frac {10 x}{11}} + 1 \right )}}{125} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((8*sqrt(5)*I*(x + 3/5)*sqrt(10*x - 5)/375 - 308*sqrt(5)*I*sqrt(10*x - 5)/1875 - 22*sqrt(55)*I*asin(s
qrt(110)/(10*sqrt(x + 3/5)))/125 - 121*sqrt(5)*I*sqrt(10*x - 5)/(3125*(x + 3/5)), Abs(x + 3/5) > 11/10), (8*sq
rt(5)*sqrt(5 - 10*x)*(x + 3/5)/375 - 308*sqrt(5)*sqrt(5 - 10*x)/1875 - 121*sqrt(5)*sqrt(5 - 10*x)/(3125*(x + 3
/5)) - 11*sqrt(55)*log(x + 3/5)/125 + 22*sqrt(55)*log(sqrt(5/11 - 10*x/11) + 1)/125, True))

________________________________________________________________________________________

Giac [A]
time = 0.60, size = 74, normalized size = 0.97 \begin {gather*} -\frac {4}{75} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - \frac {11}{125} \, \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 {88}{125} \, \sqrt {-2 \, x + 1} - \frac {121 \, \sqrt {-2 \, x + 1}}{125 \, {\left (5 \, x + 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/75*(-2*x + 1)^(3/2) - 11/125*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x
+ 1))) - 88/125*sqrt(-2*x + 1) - 121/125*sqrt(-2*x + 1)/(5*x + 3)

________________________________________________________________________________________

Mupad [B]
time = 1.18, size = 55, normalized size = 0.72 \begin {gather*} -\frac {242\,\sqrt {1-2\,x}}{625\,\left (2\,x+\frac {6}{5}\right )}-\frac {88\,\sqrt {1-2\,x}}{125}-\frac {4\,{\left (1-2\,x\right )}^{3/2}}{75}-\frac {\sqrt {55}\,\mathrm {atan}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}\,1{}\mathrm {i}}{11}\right )\,22{}\mathrm {i}}{125} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1 - 2*x)^(5/2)/(5*x + 3)^2,x)

[Out]

- (55^(1/2)*atan((55^(1/2)*(1 - 2*x)^(1/2)*1i)/11)*22i)/125 - (242*(1 - 2*x)^(1/2))/(625*(2*x + 6/5)) - (88*(1
- 2*x)^(1/2))/125 - (4*(1 - 2*x)^(3/2))/75

________________________________________________________________________________________