3.1849 \(\int \frac {\sqrt {1-2 x} (2+3 x)^3}{(3+5 x)^3} \, dx\)

Optimal. Leaf size=100 \[ -\frac {\sqrt {1-2 x} (3 x+2)^3}{10 (5 x+3)^2}-\frac {49 \sqrt {1-2 x} (3 x+2)^2}{275 (5 x+3)}+\frac {21 \sqrt {1-2 x} (75 x+44)}{2750}-\frac {1267 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{1375 \sqrt {55}} \]

[Out]

-1267/75625*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)-1/10*(2+3*x)^3*(1-2*x)^(1/2)/(3+5*x)^2-49/275*(2+3*x
)^2*(1-2*x)^(1/2)/(3+5*x)+21/2750*(44+75*x)*(1-2*x)^(1/2)

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Rubi [A]  time = 0.03, antiderivative size = 100, 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 = {97, 149, 147, 63, 206} \[ -\frac {\sqrt {1-2 x} (3 x+2)^3}{10 (5 x+3)^2}-\frac {49 \sqrt {1-2 x} (3 x+2)^2}{275 (5 x+3)}+\frac {21 \sqrt {1-2 x} (75 x+44)}{2750}-\frac {1267 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{1375 \sqrt {55}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[1 - 2*x]*(2 + 3*x)^3)/(10*(3 + 5*x)^2) - (49*Sqrt[1 - 2*x]*(2 + 3*x)^2)/(275*(3 + 5*x)) + (21*Sqrt[1 -
2*x]*(44 + 75*x))/2750 - (1267*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(1375*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 97

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

Rule 147

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol]
:> -Simp[((a*d*f*h*(n + 2) + b*c*f*h*(m + 2) - b*d*(f*g + e*h)*(m + n + 3) - b*d*f*h*(m + n + 2)*x)*(a + b*x)^
(m + 1)*(c + d*x)^(n + 1))/(b^2*d^2*(m + n + 2)*(m + n + 3)), x] + Dist[(a^2*d^2*f*h*(n + 1)*(n + 2) + a*b*d*(
n + 1)*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b^2*(c^2*f*h*(m + 1)*(m + 2) - c*d*(f*g + e*h)*(m + 1)*
(m + n + 3) + d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2*d^2*(m + n + 2)*(m + n + 3)), Int[(a + b*x)^m*(c + d*x)^n
, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && NeQ[m + n + 2, 0] && NeQ[m + n + 3, 0]

Rule 149

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[((b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^(p + 1))/(b*(b*e - a*f)*(m + 1)), x] - Dist[1
/(b*(b*e - a*f)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p*Simp[b*c*(f*g - e*h)*(m + 1) + (
b*g - a*h)*(d*e*n + c*f*(p + 1)) + d*(b*(f*g - e*h)*(m + 1) + f*(b*g - a*h)*(n + p + 1))*x, x], x], x] /; Free
Q[{a, b, c, d, e, f, g, h, p}, x] && LtQ[m, -1] && GtQ[n, 0] && IntegerQ[m]

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}{(3+5 x)^3} \, dx &=-\frac {\sqrt {1-2 x} (2+3 x)^3}{10 (3+5 x)^2}+\frac {1}{10} \int \frac {(7-21 x) (2+3 x)^2}{\sqrt {1-2 x} (3+5 x)^2} \, dx\\ &=-\frac {\sqrt {1-2 x} (2+3 x)^3}{10 (3+5 x)^2}-\frac {49 \sqrt {1-2 x} (2+3 x)^2}{275 (3+5 x)}+\frac {1}{550} \int \frac {(322-1575 x) (2+3 x)}{\sqrt {1-2 x} (3+5 x)} \, dx\\ &=-\frac {\sqrt {1-2 x} (2+3 x)^3}{10 (3+5 x)^2}-\frac {49 \sqrt {1-2 x} (2+3 x)^2}{275 (3+5 x)}+\frac {21 \sqrt {1-2 x} (44+75 x)}{2750}+\frac {1267 \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx}{2750}\\ &=-\frac {\sqrt {1-2 x} (2+3 x)^3}{10 (3+5 x)^2}-\frac {49 \sqrt {1-2 x} (2+3 x)^2}{275 (3+5 x)}+\frac {21 \sqrt {1-2 x} (44+75 x)}{2750}-\frac {1267 \operatorname {Subst}\left (\int \frac {1}{\frac {11}{2}-\frac {5 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )}{2750}\\ &=-\frac {\sqrt {1-2 x} (2+3 x)^3}{10 (3+5 x)^2}-\frac {49 \sqrt {1-2 x} (2+3 x)^2}{275 (3+5 x)}+\frac {21 \sqrt {1-2 x} (44+75 x)}{2750}-\frac {1267 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{1375 \sqrt {55}}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 63, normalized size = 0.63 \[ \frac {\frac {55 \sqrt {1-2 x} \left (9900 x^3+12870 x^2+4555 x+236\right )}{(5 x+3)^2}-2534 \sqrt {55} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{151250} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((55*Sqrt[1 - 2*x]*(236 + 4555*x + 12870*x^2 + 9900*x^3))/(3 + 5*x)^2 - 2534*Sqrt[55]*ArcTanh[Sqrt[5/11]*Sqrt[
1 - 2*x]])/151250

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fricas [A]  time = 0.70, size = 79, normalized size = 0.79 \[ \frac {1267 \, \sqrt {55} {\left (25 \, x^{2} + 30 \, x + 9\right )} \log \left (\frac {5 \, x + \sqrt {55} \sqrt {-2 \, x + 1} - 8}{5 \, x + 3}\right ) + 55 \, {\left (9900 \, x^{3} + 12870 \, x^{2} + 4555 \, x + 236\right )} \sqrt {-2 \, x + 1}}{151250 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/151250*(1267*sqrt(55)*(25*x^2 + 30*x + 9)*log((5*x + sqrt(55)*sqrt(-2*x + 1) - 8)/(5*x + 3)) + 55*(9900*x^3
+ 12870*x^2 + 4555*x + 236)*sqrt(-2*x + 1))/(25*x^2 + 30*x + 9)

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giac [A]  time = 1.37, size = 86, normalized size = 0.86 \[ -\frac {9}{125} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {1267}{151250} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) + \frac {54}{625} \, \sqrt {-2 \, x + 1} + \frac {985 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 2189 \, \sqrt {-2 \, x + 1}}{27500 \, {\left (5 \, x + 3\right )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-9/125*(-2*x + 1)^(3/2) + 1267/151250*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt
(-2*x + 1))) + 54/625*sqrt(-2*x + 1) + 1/27500*(985*(-2*x + 1)^(3/2) - 2189*sqrt(-2*x + 1))/(5*x + 3)^2

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maple [A]  time = 0.01, size = 66, normalized size = 0.66 \[ -\frac {1267 \sqrt {55}\, \arctanh \left (\frac {\sqrt {55}\, \sqrt {-2 x +1}}{11}\right )}{75625}-\frac {9 \left (-2 x +1\right )^{\frac {3}{2}}}{125}+\frac {54 \sqrt {-2 x +1}}{625}+\frac {\frac {197 \left (-2 x +1\right )^{\frac {3}{2}}}{1375}-\frac {199 \sqrt {-2 x +1}}{625}}{\left (-10 x -6\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-9/125*(-2*x+1)^(3/2)+54/625*(-2*x+1)^(1/2)+2/25*(197/110*(-2*x+1)^(3/2)-199/50*(-2*x+1)^(1/2))/(-10*x-6)^2-12
67/75625*arctanh(1/11*55^(1/2)*(-2*x+1)^(1/2))*55^(1/2)

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maxima [A]  time = 1.09, size = 92, normalized size = 0.92 \[ -\frac {9}{125} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {1267}{151250} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) + \frac {54}{625} \, \sqrt {-2 \, x + 1} + \frac {985 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 2189 \, \sqrt {-2 \, x + 1}}{6875 \, {\left (25 \, {\left (2 \, x - 1\right )}^{2} + 220 \, x + 11\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-9/125*(-2*x + 1)^(3/2) + 1267/151250*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1)
)) + 54/625*sqrt(-2*x + 1) + 1/6875*(985*(-2*x + 1)^(3/2) - 2189*sqrt(-2*x + 1))/(25*(2*x - 1)^2 + 220*x + 11)

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mupad [B]  time = 0.07, size = 74, normalized size = 0.74 \[ \frac {54\,\sqrt {1-2\,x}}{625}-\frac {9\,{\left (1-2\,x\right )}^{3/2}}{125}-\frac {\frac {199\,\sqrt {1-2\,x}}{15625}-\frac {197\,{\left (1-2\,x\right )}^{3/2}}{34375}}{\frac {44\,x}{5}+{\left (2\,x-1\right )}^2+\frac {11}{25}}+\frac {\sqrt {55}\,\mathrm {atan}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}\,1{}\mathrm {i}}{11}\right )\,1267{}\mathrm {i}}{75625} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(55^(1/2)*atan((55^(1/2)*(1 - 2*x)^(1/2)*1i)/11)*1267i)/75625 + (54*(1 - 2*x)^(1/2))/625 - (9*(1 - 2*x)^(3/2))
/125 - ((199*(1 - 2*x)^(1/2))/15625 - (197*(1 - 2*x)^(3/2))/34375)/((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 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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