3.2062 \(\int \frac {(2+3 x)^2}{\sqrt {1-2 x} (3+5 x)^3} \, dx\)

Optimal. Leaf size=68 \[ -\frac {27 \sqrt {1-2 x}}{1210 (5 x+3)}-\frac {\sqrt {1-2 x}}{550 (5 x+3)^2}-\frac {2313 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{3025 \sqrt {55}} \]

[Out]

-2313/166375*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)-1/550*(1-2*x)^(1/2)/(3+5*x)^2-27/1210*(1-2*x)^(1/2)
/(3+5*x)

________________________________________________________________________________________

Rubi [A]  time = 0.02, antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {89, 78, 63, 206} \[ -\frac {27 \sqrt {1-2 x}}{1210 (5 x+3)}-\frac {\sqrt {1-2 x}}{550 (5 x+3)^2}-\frac {2313 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{3025 \sqrt {55}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-Sqrt[1 - 2*x]/(550*(3 + 5*x)^2) - (27*Sqrt[1 - 2*x])/(1210*(3 + 5*x)) - (2313*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x
]])/(3025*Sqrt[55])

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 {(2+3 x)^2}{\sqrt {1-2 x} (3+5 x)^3} \, dx &=-\frac {\sqrt {1-2 x}}{550 (3+5 x)^2}+\frac {1}{550} \int \frac {729+990 x}{\sqrt {1-2 x} (3+5 x)^2} \, dx\\ &=-\frac {\sqrt {1-2 x}}{550 (3+5 x)^2}-\frac {27 \sqrt {1-2 x}}{1210 (3+5 x)}+\frac {2313 \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx}{6050}\\ &=-\frac {\sqrt {1-2 x}}{550 (3+5 x)^2}-\frac {27 \sqrt {1-2 x}}{1210 (3+5 x)}-\frac {2313 \operatorname {Subst}\left (\int \frac {1}{\frac {11}{2}-\frac {5 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )}{6050}\\ &=-\frac {\sqrt {1-2 x}}{550 (3+5 x)^2}-\frac {27 \sqrt {1-2 x}}{1210 (3+5 x)}-\frac {2313 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{3025 \sqrt {55}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.05, size = 53, normalized size = 0.78 \[ \frac {-\frac {55 \sqrt {1-2 x} (675 x+416)}{(5 x+3)^2}-4626 \sqrt {55} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{332750} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-55*Sqrt[1 - 2*x]*(416 + 675*x))/(3 + 5*x)^2 - 4626*Sqrt[55]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/332750

________________________________________________________________________________________

fricas [A]  time = 1.00, size = 69, normalized size = 1.01 \[ \frac {2313 \, \sqrt {55} {\left (25 \, x^{2} + 30 \, x + 9\right )} \log \left (\frac {5 \, x + \sqrt {55} \sqrt {-2 \, x + 1} - 8}{5 \, x + 3}\right ) - 55 \, {\left (675 \, x + 416\right )} \sqrt {-2 \, x + 1}}{332750 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/332750*(2313*sqrt(55)*(25*x^2 + 30*x + 9)*log((5*x + sqrt(55)*sqrt(-2*x + 1) - 8)/(5*x + 3)) - 55*(675*x + 4
16)*sqrt(-2*x + 1))/(25*x^2 + 30*x + 9)

________________________________________________________________________________________

giac [A]  time = 1.23, size = 68, normalized size = 1.00 \[ \frac {2313}{332750} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) + \frac {675 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 1507 \, \sqrt {-2 \, x + 1}}{12100 \, {\left (5 \, x + 3\right )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2313/332750*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 1/12100*(67
5*(-2*x + 1)^(3/2) - 1507*sqrt(-2*x + 1))/(5*x + 3)^2

________________________________________________________________________________________

maple [A]  time = 0.01, size = 48, normalized size = 0.71 \[ -\frac {2313 \sqrt {55}\, \arctanh \left (\frac {\sqrt {55}\, \sqrt {-2 x +1}}{11}\right )}{166375}+\frac {\frac {27 \left (-2 x +1\right )^{\frac {3}{2}}}{121}-\frac {137 \sqrt {-2 x +1}}{275}}{\left (-10 x -6\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

50*(27/6050*(-2*x+1)^(3/2)-137/13750*(-2*x+1)^(1/2))/(-10*x-6)^2-2313/166375*arctanh(1/11*55^(1/2)*(-2*x+1)^(1
/2))*55^(1/2)

________________________________________________________________________________________

maxima [A]  time = 1.24, size = 74, normalized size = 1.09 \[ \frac {2313}{332750} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) + \frac {675 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 1507 \, \sqrt {-2 \, x + 1}}{3025 \, {\left (25 \, {\left (2 \, x - 1\right )}^{2} + 220 \, x + 11\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2313/332750*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 1/3025*(675*(-2*x + 1
)^(3/2) - 1507*sqrt(-2*x + 1))/(25*(2*x - 1)^2 + 220*x + 11)

________________________________________________________________________________________

mupad [B]  time = 1.23, size = 54, normalized size = 0.79 \[ -\frac {2313\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )}{166375}-\frac {\frac {137\,\sqrt {1-2\,x}}{6875}-\frac {27\,{\left (1-2\,x\right )}^{3/2}}{3025}}{\frac {44\,x}{5}+{\left (2\,x-1\right )}^2+\frac {11}{25}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- (2313*55^(1/2)*atanh((55^(1/2)*(1 - 2*x)^(1/2))/11))/166375 - ((137*(1 - 2*x)^(1/2))/6875 - (27*(1 - 2*x)^(3
/2))/3025)/((44*x)/5 + (2*x - 1)^2 + 11/25)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________