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

Optimal. Leaf size=68 \[ -\frac {(1-2 x)^{3/2}}{10 (5 x+3)^2}+\frac {3 \sqrt {1-2 x}}{50 (5 x+3)}-\frac {3 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{25 \sqrt {55}} \]

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Rubi [A]  time = 0.01, antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {47, 63, 206} \begin {gather*} -\frac {(1-2 x)^{3/2}}{10 (5 x+3)^2}+\frac {3 \sqrt {1-2 x}}{50 (5 x+3)}-\frac {3 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{25 \sqrt {55}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 - 2*x)^(3/2)/(3 + 5*x)^3,x]

[Out]

-(1 - 2*x)^(3/2)/(10*(3 + 5*x)^2) + (3*Sqrt[1 - 2*x])/(50*(3 + 5*x)) - (3*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(
25*Sqrt[55])

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 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)^{3/2}}{(3+5 x)^3} \, dx &=-\frac {(1-2 x)^{3/2}}{10 (3+5 x)^2}-\frac {3}{10} \int \frac {\sqrt {1-2 x}}{(3+5 x)^2} \, dx\\ &=-\frac {(1-2 x)^{3/2}}{10 (3+5 x)^2}+\frac {3 \sqrt {1-2 x}}{50 (3+5 x)}+\frac {3}{50} \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx\\ &=-\frac {(1-2 x)^{3/2}}{10 (3+5 x)^2}+\frac {3 \sqrt {1-2 x}}{50 (3+5 x)}-\frac {3}{50} \operatorname {Subst}\left (\int \frac {1}{\frac {11}{2}-\frac {5 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )\\ &=-\frac {(1-2 x)^{3/2}}{10 (3+5 x)^2}+\frac {3 \sqrt {1-2 x}}{50 (3+5 x)}-\frac {3 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{25 \sqrt {55}}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 74, normalized size = 1.09 \begin {gather*} \frac {55 \left (-50 x^2+17 x+4\right )+6 \sqrt {55} \sqrt {2 x-1} (5 x+3)^2 \tan ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {2 x-1}\right )}{2750 \sqrt {1-2 x} (5 x+3)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(55*(4 + 17*x - 50*x^2) + 6*Sqrt[55]*Sqrt[-1 + 2*x]*(3 + 5*x)^2*ArcTan[Sqrt[5/11]*Sqrt[-1 + 2*x]])/(2750*Sqrt[
1 - 2*x]*(3 + 5*x)^2)

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IntegrateAlgebraic [A]  time = 0.18, size = 61, normalized size = 0.90 \begin {gather*} -\frac {\sqrt {1-2 x} (25 (1-2 x)-33)}{25 (5 (1-2 x)-11)^2}-\frac {3 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{25 \sqrt {55}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/25*((-33 + 25*(1 - 2*x))*Sqrt[1 - 2*x])/(-11 + 5*(1 - 2*x))^2 - (3*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(25*S
qrt[55])

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fricas [A]  time = 1.39, size = 69, normalized size = 1.01 \begin {gather*} \frac {3 \, \sqrt {55} {\left (25 \, x^{2} + 30 \, x + 9\right )} \log \left (\frac {5 \, x + \sqrt {55} \sqrt {-2 \, x + 1} - 8}{5 \, x + 3}\right ) + 55 \, {\left (25 \, x + 4\right )} \sqrt {-2 \, x + 1}}{2750 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2750*(3*sqrt(55)*(25*x^2 + 30*x + 9)*log((5*x + sqrt(55)*sqrt(-2*x + 1) - 8)/(5*x + 3)) + 55*(25*x + 4)*sqrt
(-2*x + 1))/(25*x^2 + 30*x + 9)

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giac [A]  time = 1.17, size = 68, normalized size = 1.00 \begin {gather*} \frac {3}{2750} \, \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 {25 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 33 \, \sqrt {-2 \, x + 1}}{100 \, {\left (5 \, x + 3\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/2750*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 1/100*(25*(-2*x
+ 1)^(3/2) - 33*sqrt(-2*x + 1))/(5*x + 3)^2

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maple [A]  time = 0.01, size = 48, normalized size = 0.71 \begin {gather*} -\frac {3 \sqrt {55}\, \arctanh \left (\frac {\sqrt {55}\, \sqrt {-2 x +1}}{11}\right )}{1375}+\frac {-\left (-2 x +1\right )^{\frac {3}{2}}+\frac {33 \sqrt {-2 x +1}}{25}}{\left (-10 x -6\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

200*(-1/200*(-2*x+1)^(3/2)+33/5000*(-2*x+1)^(1/2))/(-10*x-6)^2-3/1375*arctanh(1/11*55^(1/2)*(-2*x+1)^(1/2))*55
^(1/2)

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maxima [A]  time = 1.25, size = 74, normalized size = 1.09 \begin {gather*} \frac {3}{2750} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) - \frac {25 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 33 \, \sqrt {-2 \, x + 1}}{25 \, {\left (25 \, {\left (2 \, x - 1\right )}^{2} + 220 \, x + 11\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/2750*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 1/25*(25*(-2*x + 1)^(3/2)
- 33*sqrt(-2*x + 1))/(25*(2*x - 1)^2 + 220*x + 11)

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mupad [B]  time = 0.06, size = 53, normalized size = 0.78 \begin {gather*} \frac {\frac {33\,\sqrt {1-2\,x}}{625}-\frac {{\left (1-2\,x\right )}^{3/2}}{25}}{\frac {44\,x}{5}+{\left (2\,x-1\right )}^2+\frac {11}{25}}-\frac {3\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )}{1375} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1 - 2*x)^(3/2)/(5*x + 3)^3,x)

[Out]

((33*(1 - 2*x)^(1/2))/625 - (1 - 2*x)^(3/2)/25)/((44*x)/5 + (2*x - 1)^2 + 11/25) - (3*55^(1/2)*atanh((55^(1/2)
*(1 - 2*x)^(1/2))/11))/1375

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sympy [A]  time = 2.34, size = 235, normalized size = 3.46 \begin {gather*} \begin {cases} - \frac {3 \sqrt {55} \operatorname {acosh}{\left (\frac {\sqrt {110}}{10 \sqrt {x + \frac {3}{5}}} \right )}}{1375} - \frac {\sqrt {2}}{50 \sqrt {-1 + \frac {11}{10 \left (x + \frac {3}{5}\right )}} \sqrt {x + \frac {3}{5}}} + \frac {77 \sqrt {2}}{2500 \sqrt {-1 + \frac {11}{10 \left (x + \frac {3}{5}\right )}} \left (x + \frac {3}{5}\right )^{\frac {3}{2}}} - \frac {121 \sqrt {2}}{12500 \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 )}}{1375} + \frac {\sqrt {2} i}{50 \sqrt {1 - \frac {11}{10 \left (x + \frac {3}{5}\right )}} \sqrt {x + \frac {3}{5}}} - \frac {77 \sqrt {2} i}{2500 \sqrt {1 - \frac {11}{10 \left (x + \frac {3}{5}\right )}} \left (x + \frac {3}{5}\right )^{\frac {3}{2}}} + \frac {121 \sqrt {2} i}{12500 \sqrt {1 - \frac {11}{10 \left (x + \frac {3}{5}\right )}} \left (x + \frac {3}{5}\right )^{\frac {5}{2}}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-3*sqrt(55)*acosh(sqrt(110)/(10*sqrt(x + 3/5)))/1375 - sqrt(2)/(50*sqrt(-1 + 11/(10*(x + 3/5)))*sqr
t(x + 3/5)) + 77*sqrt(2)/(2500*sqrt(-1 + 11/(10*(x + 3/5)))*(x + 3/5)**(3/2)) - 121*sqrt(2)/(12500*sqrt(-1 + 1
1/(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)
))/1375 + sqrt(2)*I/(50*sqrt(1 - 11/(10*(x + 3/5)))*sqrt(x + 3/5)) - 77*sqrt(2)*I/(2500*sqrt(1 - 11/(10*(x + 3
/5)))*(x + 3/5)**(3/2)) + 121*sqrt(2)*I/(12500*sqrt(1 - 11/(10*(x + 3/5)))*(x + 3/5)**(5/2)), True))

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