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

Optimal. Leaf size=81 \[ -\frac {133 (1-2 x)^{3/2}}{6050 (5 x+3)}-\frac {(1-2 x)^{3/2}}{550 (5 x+3)^2}+\frac {409 \sqrt {1-2 x}}{3025}-\frac {409 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{275 \sqrt {55}} \]

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Rubi [A]  time = 0.02, antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {89, 78, 50, 63, 206} \begin {gather*} -\frac {133 (1-2 x)^{3/2}}{6050 (5 x+3)}-\frac {(1-2 x)^{3/2}}{550 (5 x+3)^2}+\frac {409 \sqrt {1-2 x}}{3025}-\frac {409 \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)^2)/(3 + 5*x)^3,x]

[Out]

(409*Sqrt[1 - 2*x])/3025 - (1 - 2*x)^(3/2)/(550*(3 + 5*x)^2) - (133*(1 - 2*x)^(3/2))/(6050*(3 + 5*x)) - (409*A
rcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(275*Sqrt[55])

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 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 89

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c - a*
d)^2*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d^2*(d*e - c*f)*(n + 1)), x] - Dist[1/(d^2*(d*e - c*f)*(n + 1)), In
t[(c + d*x)^(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*(p + 1)) - 2*a*b*d*(d*e*
(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ
[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))

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)^2}{(3+5 x)^3} \, dx &=-\frac {(1-2 x)^{3/2}}{550 (3+5 x)^2}+\frac {1}{550} \int \frac {\sqrt {1-2 x} (727+990 x)}{(3+5 x)^2} \, dx\\ &=-\frac {(1-2 x)^{3/2}}{550 (3+5 x)^2}-\frac {133 (1-2 x)^{3/2}}{6050 (3+5 x)}+\frac {409 \int \frac {\sqrt {1-2 x}}{3+5 x} \, dx}{1210}\\ &=\frac {409 \sqrt {1-2 x}}{3025}-\frac {(1-2 x)^{3/2}}{550 (3+5 x)^2}-\frac {133 (1-2 x)^{3/2}}{6050 (3+5 x)}+\frac {409}{550} \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx\\ &=\frac {409 \sqrt {1-2 x}}{3025}-\frac {(1-2 x)^{3/2}}{550 (3+5 x)^2}-\frac {133 (1-2 x)^{3/2}}{6050 (3+5 x)}-\frac {409}{550} \operatorname {Subst}\left (\int \frac {1}{\frac {11}{2}-\frac {5 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )\\ &=\frac {409 \sqrt {1-2 x}}{3025}-\frac {(1-2 x)^{3/2}}{550 (3+5 x)^2}-\frac {133 (1-2 x)^{3/2}}{6050 (3+5 x)}-\frac {409 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{275 \sqrt {55}}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 58, normalized size = 0.72 \begin {gather*} \frac {\sqrt {1-2 x} \left (1980 x^2+2245 x+632\right )}{550 (5 x+3)^2}-\frac {409 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{275 \sqrt {55}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[1 - 2*x]*(632 + 2245*x + 1980*x^2))/(550*(3 + 5*x)^2) - (409*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(275*Sqr
t[55])

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IntegrateAlgebraic [A]  time = 0.18, size = 70, normalized size = 0.86 \begin {gather*} \frac {\left (990 (1-2 x)^2-4225 (1-2 x)+4499\right ) \sqrt {1-2 x}}{275 (5 (1-2 x)-11)^2}-\frac {409 \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)^2)/(3 + 5*x)^3,x]

[Out]

((4499 - 4225*(1 - 2*x) + 990*(1 - 2*x)^2)*Sqrt[1 - 2*x])/(275*(-11 + 5*(1 - 2*x))^2) - (409*ArcTanh[Sqrt[5/11
]*Sqrt[1 - 2*x]])/(275*Sqrt[55])

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fricas [A]  time = 1.18, size = 74, normalized size = 0.91 \begin {gather*} \frac {409 \, \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 (1980 \, x^{2} + 2245 \, x + 632\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)^2*(1-2*x)^(1/2)/(3+5*x)^3,x, algorithm="fricas")

[Out]

1/30250*(409*sqrt(55)*(25*x^2 + 30*x + 9)*log((5*x + sqrt(55)*sqrt(-2*x + 1) - 8)/(5*x + 3)) + 55*(1980*x^2 +
2245*x + 632)*sqrt(-2*x + 1))/(25*x^2 + 30*x + 9)

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giac [A]  time = 1.36, size = 77, normalized size = 0.95 \begin {gather*} \frac {409}{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 {18}{125} \, \sqrt {-2 \, x + 1} + \frac {655 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 1463 \, \sqrt {-2 \, x + 1}}{5500 \, {\left (5 \, x + 3\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

409/30250*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 18/125*sqrt(-
2*x + 1) + 1/5500*(655*(-2*x + 1)^(3/2) - 1463*sqrt(-2*x + 1))/(5*x + 3)^2

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maple [A]  time = 0.01, size = 57, normalized size = 0.70 \begin {gather*} -\frac {409 \sqrt {55}\, \arctanh \left (\frac {\sqrt {55}\, \sqrt {-2 x +1}}{11}\right )}{15125}+\frac {18 \sqrt {-2 x +1}}{125}+\frac {\frac {131 \left (-2 x +1\right )^{\frac {3}{2}}}{275}-\frac {133 \sqrt {-2 x +1}}{125}}{\left (-10 x -6\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

18/125*(-2*x+1)^(1/2)+2/5*(131/110*(-2*x+1)^(3/2)-133/50*(-2*x+1)^(1/2))/(-10*x-6)^2-409/15125*arctanh(1/11*55
^(1/2)*(-2*x+1)^(1/2))*55^(1/2)

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maxima [A]  time = 1.35, size = 83, normalized size = 1.02 \begin {gather*} \frac {409}{30250} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) + \frac {18}{125} \, \sqrt {-2 \, x + 1} + \frac {655 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 1463 \, \sqrt {-2 \, x + 1}}{1375 \, {\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)^2*(1-2*x)^(1/2)/(3+5*x)^3,x, algorithm="maxima")

[Out]

409/30250*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 18/125*sqrt(-2*x + 1) +
1/1375*(655*(-2*x + 1)^(3/2) - 1463*sqrt(-2*x + 1))/(25*(2*x - 1)^2 + 220*x + 11)

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mupad [B]  time = 0.06, size = 63, normalized size = 0.78 \begin {gather*} \frac {18\,\sqrt {1-2\,x}}{125}-\frac {409\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )}{15125}-\frac {\frac {133\,\sqrt {1-2\,x}}{3125}-\frac {131\,{\left (1-2\,x\right )}^{3/2}}{6875}}{\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)^2)/(5*x + 3)^3,x)

[Out]

(18*(1 - 2*x)^(1/2))/125 - (409*55^(1/2)*atanh((55^(1/2)*(1 - 2*x)^(1/2))/11))/15125 - ((133*(1 - 2*x)^(1/2))/
3125 - (131*(1 - 2*x)^(3/2))/6875)/((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)**2*(1-2*x)**(1/2)/(3+5*x)**3,x)

[Out]

Timed out

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