3.1995 \(\int \frac {(1-2 x)^{5/2}}{(3+5 x)^3} \, dx\)

Optimal. Leaf size=83 \[ -\frac {(1-2 x)^{5/2}}{10 (5 x+3)^2}+\frac {(1-2 x)^{3/2}}{10 (5 x+3)}+\frac {3}{25} \sqrt {1-2 x}-\frac {3}{25} \sqrt {\frac {11}{5}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \]

[Out]

-1/10*(1-2*x)^(5/2)/(3+5*x)^2+1/10*(1-2*x)^(3/2)/(3+5*x)-3/125*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)+3
/25*(1-2*x)^(1/2)

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Rubi [A]  time = 0.02, antiderivative size = 83, 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 = {47, 50, 63, 206} \[ -\frac {(1-2 x)^{5/2}}{10 (5 x+3)^2}+\frac {(1-2 x)^{3/2}}{10 (5 x+3)}+\frac {3}{25} \sqrt {1-2 x}-\frac {3}{25} \sqrt {\frac {11}{5}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(1 - 2*x)^(5/2)/(3 + 5*x)^3,x]

[Out]

(3*Sqrt[1 - 2*x])/25 - (1 - 2*x)^(5/2)/(10*(3 + 5*x)^2) + (1 - 2*x)^(3/2)/(10*(3 + 5*x)) - (3*Sqrt[11/5]*ArcTa
nh[Sqrt[5/11]*Sqrt[1 - 2*x]])/25

Rule 47

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},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 50

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 63

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 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 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)^3} \, dx &=-\frac {(1-2 x)^{5/2}}{10 (3+5 x)^2}-\frac {1}{2} \int \frac {(1-2 x)^{3/2}}{(3+5 x)^2} \, dx\\ &=-\frac {(1-2 x)^{5/2}}{10 (3+5 x)^2}+\frac {(1-2 x)^{3/2}}{10 (3+5 x)}+\frac {3}{10} \int \frac {\sqrt {1-2 x}}{3+5 x} \, dx\\ &=\frac {3}{25} \sqrt {1-2 x}-\frac {(1-2 x)^{5/2}}{10 (3+5 x)^2}+\frac {(1-2 x)^{3/2}}{10 (3+5 x)}+\frac {33}{50} \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx\\ &=\frac {3}{25} \sqrt {1-2 x}-\frac {(1-2 x)^{5/2}}{10 (3+5 x)^2}+\frac {(1-2 x)^{3/2}}{10 (3+5 x)}-\frac {33}{50} \operatorname {Subst}\left (\int \frac {1}{\frac {11}{2}-\frac {5 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )\\ &=\frac {3}{25} \sqrt {1-2 x}-\frac {(1-2 x)^{5/2}}{10 (3+5 x)^2}+\frac {(1-2 x)^{3/2}}{10 (3+5 x)}-\frac {3}{25} \sqrt {\frac {11}{5}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 30, normalized size = 0.36 \[ -\frac {8 (1-2 x)^{7/2} \, _2F_1\left (3,\frac {7}{2};\frac {9}{2};-\frac {5}{11} (2 x-1)\right )}{9317} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-8*(1 - 2*x)^(7/2)*Hypergeometric2F1[3, 7/2, 9/2, (-5*(-1 + 2*x))/11])/9317

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fricas [A]  time = 0.98, size = 80, normalized size = 0.96 \[ \frac {3 \, \sqrt {11} \sqrt {5} {\left (25 \, x^{2} + 30 \, x + 9\right )} \log \left (\frac {\sqrt {11} \sqrt {5} \sqrt {-2 \, x + 1} + 5 \, x - 8}{5 \, x + 3}\right ) + 5 \, {\left (80 \, x^{2} + 195 \, x + 64\right )} \sqrt {-2 \, x + 1}}{250 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/250*(3*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*(
80*x^2 + 195*x + 64)*sqrt(-2*x + 1))/(25*x^2 + 30*x + 9)

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giac [A]  time = 1.25, size = 77, normalized size = 0.93 \[ \frac {3}{250} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) + \frac {8}{125} \, \sqrt {-2 \, x + 1} - \frac {11 \, {\left (45 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 77 \, \sqrt {-2 \, x + 1}\right )}}{500 \, {\left (5 \, x + 3\right )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/250*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 8/125*sqrt(-2*x +
1) - 11/500*(45*(-2*x + 1)^(3/2) - 77*sqrt(-2*x + 1))/(5*x + 3)^2

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maple [A]  time = 0.01, size = 57, normalized size = 0.69 \[ -\frac {3 \sqrt {55}\, \arctanh \left (\frac {\sqrt {55}\, \sqrt {-2 x +1}}{11}\right )}{125}+\frac {8 \sqrt {-2 x +1}}{125}+\frac {-\frac {99 \left (-2 x +1\right )^{\frac {3}{2}}}{25}+\frac {847 \sqrt {-2 x +1}}{125}}{\left (-10 x -6\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

8/125*(-2*x+1)^(1/2)+88/5*(-9/40*(-2*x+1)^(3/2)+77/200*(-2*x+1)^(1/2))/(-10*x-6)^2-3/125*arctanh(1/11*55^(1/2)
*(-2*x+1)^(1/2))*55^(1/2)

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maxima [A]  time = 1.42, size = 83, normalized size = 1.00 \[ \frac {3}{250} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) + \frac {8}{125} \, \sqrt {-2 \, x + 1} - \frac {11 \, {\left (45 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 77 \, \sqrt {-2 \, x + 1}\right )}}{125 \, {\left (25 \, {\left (2 \, x - 1\right )}^{2} + 220 \, x + 11\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/250*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 8/125*sqrt(-2*x + 1) - 11/1
25*(45*(-2*x + 1)^(3/2) - 77*sqrt(-2*x + 1))/(25*(2*x - 1)^2 + 220*x + 11)

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mupad [B]  time = 1.18, size = 62, normalized size = 0.75 \[ \frac {8\,\sqrt {1-2\,x}}{125}-\frac {3\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )}{125}+\frac {\frac {847\,\sqrt {1-2\,x}}{3125}-\frac {99\,{\left (1-2\,x\right )}^{3/2}}{625}}{\frac {44\,x}{5}+{\left (2\,x-1\right )}^2+\frac {11}{25}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1 - 2*x)^(5/2)/(5*x + 3)^3,x)

[Out]

(8*(1 - 2*x)^(1/2))/125 - (3*55^(1/2)*atanh((55^(1/2)*(1 - 2*x)^(1/2))/11))/125 + ((847*(1 - 2*x)^(1/2))/3125
- (99*(1 - 2*x)^(3/2))/625)/((44*x)/5 + (2*x - 1)^2 + 11/25)

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sympy [B]  time = 2.97, size = 298, normalized size = 3.59 \[ \begin {cases} - \frac {3 \sqrt {55} \operatorname {acosh}{\left (\frac {\sqrt {110}}{10 \sqrt {x + \frac {3}{5}}} \right )}}{125} - \frac {8 \sqrt {2} \sqrt {x + \frac {3}{5}}}{125 \sqrt {-1 + \frac {11}{10 \left (x + \frac {3}{5}\right )}}} - \frac {11 \sqrt {2}}{1250 \sqrt {-1 + \frac {11}{10 \left (x + \frac {3}{5}\right )}} \sqrt {x + \frac {3}{5}}} + \frac {1331 \sqrt {2}}{12500 \sqrt {-1 + \frac {11}{10 \left (x + \frac {3}{5}\right )}} \left (x + \frac {3}{5}\right )^{\frac {3}{2}}} - \frac {1331 \sqrt {2}}{62500 \sqrt {-1 + \frac {11}{10 \left (x + \frac {3}{5}\right )}} \left (x + \frac {3}{5}\right )^{\frac {5}{2}}} & \text {for}\: \frac {11}{10 \left |{x + \frac {3}{5}}\right |} > 1 \\\frac {3 \sqrt {55} i \operatorname {asin}{\left (\frac {\sqrt {110}}{10 \sqrt {x + \frac {3}{5}}} \right )}}{125} + \frac {8 \sqrt {2} i \sqrt {x + \frac {3}{5}}}{125 \sqrt {1 - \frac {11}{10 \left (x + \frac {3}{5}\right )}}} + \frac {11 \sqrt {2} i}{1250 \sqrt {1 - \frac {11}{10 \left (x + \frac {3}{5}\right )}} \sqrt {x + \frac {3}{5}}} - \frac {1331 \sqrt {2} i}{12500 \sqrt {1 - \frac {11}{10 \left (x + \frac {3}{5}\right )}} \left (x + \frac {3}{5}\right )^{\frac {3}{2}}} + \frac {1331 \sqrt {2} i}{62500 \sqrt {1 - \frac {11}{10 \left (x + \frac {3}{5}\right )}} \left (x + \frac {3}{5}\right )^{\frac {5}{2}}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-3*sqrt(55)*acosh(sqrt(110)/(10*sqrt(x + 3/5)))/125 - 8*sqrt(2)*sqrt(x + 3/5)/(125*sqrt(-1 + 11/(10
*(x + 3/5)))) - 11*sqrt(2)/(1250*sqrt(-1 + 11/(10*(x + 3/5)))*sqrt(x + 3/5)) + 1331*sqrt(2)/(12500*sqrt(-1 + 1
1/(10*(x + 3/5)))*(x + 3/5)**(3/2)) - 1331*sqrt(2)/(62500*sqrt(-1 + 11/(10*(x + 3/5)))*(x + 3/5)**(5/2)), 11/(
10*Abs(x + 3/5)) > 1), (3*sqrt(55)*I*asin(sqrt(110)/(10*sqrt(x + 3/5)))/125 + 8*sqrt(2)*I*sqrt(x + 3/5)/(125*s
qrt(1 - 11/(10*(x + 3/5)))) + 11*sqrt(2)*I/(1250*sqrt(1 - 11/(10*(x + 3/5)))*sqrt(x + 3/5)) - 1331*sqrt(2)*I/(
12500*sqrt(1 - 11/(10*(x + 3/5)))*(x + 3/5)**(3/2)) + 1331*sqrt(2)*I/(62500*sqrt(1 - 11/(10*(x + 3/5)))*(x + 3
/5)**(5/2)), True))

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