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

Optimal. Leaf size=68 \[ -\frac {(1-2 x)^{3/2}}{110 (5 x+3)^2}-\frac {67 \sqrt {1-2 x}}{550 (5 x+3)}+\frac {67 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{275 \sqrt {55}} \]

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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 = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {78, 47, 63, 206} \begin {gather*} -\frac {(1-2 x)^{3/2}}{110 (5 x+3)^2}-\frac {67 \sqrt {1-2 x}}{550 (5 x+3)}+\frac {67 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{275 \sqrt {55}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(1 - 2*x)^(3/2)/(110*(3 + 5*x)^2) - (67*Sqrt[1 - 2*x])/(550*(3 + 5*x)) + (67*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]
])/(275*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 78

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] || LtQ
[p, n]))))

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 {\sqrt {1-2 x} (2+3 x)}{(3+5 x)^3} \, dx &=-\frac {(1-2 x)^{3/2}}{110 (3+5 x)^2}+\frac {67}{110} \int \frac {\sqrt {1-2 x}}{(3+5 x)^2} \, dx\\ &=-\frac {(1-2 x)^{3/2}}{110 (3+5 x)^2}-\frac {67 \sqrt {1-2 x}}{550 (3+5 x)}-\frac {67}{550} \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx\\ &=-\frac {(1-2 x)^{3/2}}{110 (3+5 x)^2}-\frac {67 \sqrt {1-2 x}}{550 (3+5 x)}+\frac {67}{550} \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}}{110 (3+5 x)^2}-\frac {67 \sqrt {1-2 x}}{550 (3+5 x)}+\frac {67 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{275 \sqrt {55}}\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 74, normalized size = 1.09 \begin {gather*} \frac {55 \left (650 x^2+87 x-206\right )-134 \sqrt {55} \sqrt {2 x-1} (5 x+3)^2 \tan ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {2 x-1}\right )}{30250 \sqrt {1-2 x} (5 x+3)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(55*(-206 + 87*x + 650*x^2) - 134*Sqrt[55]*Sqrt[-1 + 2*x]*(3 + 5*x)^2*ArcTan[Sqrt[5/11]*Sqrt[-1 + 2*x]])/(3025
0*Sqrt[1 - 2*x]*(3 + 5*x)^2)

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IntegrateAlgebraic [A]  time = 0.17, size = 61, normalized size = 0.90 \begin {gather*} \frac {\sqrt {1-2 x} (325 (1-2 x)-737)}{275 (5 (1-2 x)-11)^2}+\frac {67 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{275 \sqrt {55}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-737 + 325*(1 - 2*x))*Sqrt[1 - 2*x])/(275*(-11 + 5*(1 - 2*x))^2) + (67*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(2
75*Sqrt[55])

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fricas [A]  time = 1.78, size = 70, normalized size = 1.03 \begin {gather*} \frac {67 \, \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 (325 \, x + 206\right )} \sqrt {-2 \, x + 1}}{30250 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/30250*(67*sqrt(55)*(25*x^2 + 30*x + 9)*log((5*x - sqrt(55)*sqrt(-2*x + 1) - 8)/(5*x + 3)) - 55*(325*x + 206)
*sqrt(-2*x + 1))/(25*x^2 + 30*x + 9)

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giac [A]  time = 1.31, size = 68, normalized size = 1.00 \begin {gather*} -\frac {67}{30250} \, \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 {325 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 737 \, \sqrt {-2 \, x + 1}}{1100 \, {\left (5 \, x + 3\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-67/30250*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 1/1100*(325*(
-2*x + 1)^(3/2) - 737*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 {67 \sqrt {55}\, \arctanh \left (\frac {\sqrt {55}\, \sqrt {-2 x +1}}{11}\right )}{15125}-\frac {100 \left (-\frac {13 \left (-2 x +1\right )^{\frac {3}{2}}}{1100}+\frac {67 \sqrt {-2 x +1}}{2500}\right )}{\left (-10 x -6\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-100*(-13/1100*(-2*x+1)^(3/2)+67/2500*(-2*x+1)^(1/2))/(-10*x-6)^2+67/15125*arctanh(1/11*55^(1/2)*(-2*x+1)^(1/2
))*55^(1/2)

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maxima [A]  time = 1.34, size = 74, normalized size = 1.09 \begin {gather*} -\frac {67}{30250} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) + \frac {325 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 737 \, \sqrt {-2 \, x + 1}}{275 \, {\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((2+3*x)*(1-2*x)^(1/2)/(3+5*x)^3,x, algorithm="maxima")

[Out]

-67/30250*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 1/275*(325*(-2*x + 1)^(
3/2) - 737*sqrt(-2*x + 1))/(25*(2*x - 1)^2 + 220*x + 11)

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mupad [B]  time = 1.18, size = 54, normalized size = 0.79 \begin {gather*} \frac {67\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )}{15125}-\frac {\frac {67\,\sqrt {1-2\,x}}{625}-\frac {13\,{\left (1-2\,x\right )}^{3/2}}{275}}{\frac {44\,x}{5}+{\left (2\,x-1\right )}^2+\frac {11}{25}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(67*55^(1/2)*atanh((55^(1/2)*(1 - 2*x)^(1/2))/11))/15125 - ((67*(1 - 2*x)^(1/2))/625 - (13*(1 - 2*x)^(3/2))/27
5)/((44*x)/5 + (2*x - 1)^2 + 11/25)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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