3.1993 \(\int \frac{(1-2 x)^{5/2} (2+3 x)^2}{(3+5 x)^3} \, dx\)

Optimal. Leaf size=109 \[ -\frac{129 (1-2 x)^{7/2}}{6050 (5 x+3)}-\frac{(1-2 x)^{7/2}}{550 (5 x+3)^2}+\frac{1533 (1-2 x)^{5/2}}{75625}+\frac{511 (1-2 x)^{3/2}}{6875}+\frac{1533 \sqrt{1-2 x}}{3125}-\frac{1533 \sqrt{\frac{11}{5}} \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )}{3125} \]

[Out]

(1533*Sqrt[1 - 2*x])/3125 + (511*(1 - 2*x)^(3/2))/6875 + (1533*(1 - 2*x)^(5/2))/75625 - (1 - 2*x)^(7/2)/(550*(
3 + 5*x)^2) - (129*(1 - 2*x)^(7/2))/(6050*(3 + 5*x)) - (1533*Sqrt[11/5]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/312
5

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Rubi [A]  time = 0.0317155, antiderivative size = 109, normalized size of antiderivative = 1., number of steps used = 7, 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} \[ -\frac{129 (1-2 x)^{7/2}}{6050 (5 x+3)}-\frac{(1-2 x)^{7/2}}{550 (5 x+3)^2}+\frac{1533 (1-2 x)^{5/2}}{75625}+\frac{511 (1-2 x)^{3/2}}{6875}+\frac{1533 \sqrt{1-2 x}}{3125}-\frac{1533 \sqrt{\frac{11}{5}} \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )}{3125} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(1533*Sqrt[1 - 2*x])/3125 + (511*(1 - 2*x)^(3/2))/6875 + (1533*(1 - 2*x)^(5/2))/75625 - (1 - 2*x)^(7/2)/(550*(
3 + 5*x)^2) - (129*(1 - 2*x)^(7/2))/(6050*(3 + 5*x)) - (1533*Sqrt[11/5]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/312
5

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 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 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 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)^{5/2} (2+3 x)^2}{(3+5 x)^3} \, dx &=-\frac{(1-2 x)^{7/2}}{550 (3+5 x)^2}+\frac{1}{550} \int \frac{(1-2 x)^{5/2} (723+990 x)}{(3+5 x)^2} \, dx\\ &=-\frac{(1-2 x)^{7/2}}{550 (3+5 x)^2}-\frac{129 (1-2 x)^{7/2}}{6050 (3+5 x)}+\frac{1533 \int \frac{(1-2 x)^{5/2}}{3+5 x} \, dx}{6050}\\ &=\frac{1533 (1-2 x)^{5/2}}{75625}-\frac{(1-2 x)^{7/2}}{550 (3+5 x)^2}-\frac{129 (1-2 x)^{7/2}}{6050 (3+5 x)}+\frac{1533 \int \frac{(1-2 x)^{3/2}}{3+5 x} \, dx}{2750}\\ &=\frac{511 (1-2 x)^{3/2}}{6875}+\frac{1533 (1-2 x)^{5/2}}{75625}-\frac{(1-2 x)^{7/2}}{550 (3+5 x)^2}-\frac{129 (1-2 x)^{7/2}}{6050 (3+5 x)}+\frac{1533 \int \frac{\sqrt{1-2 x}}{3+5 x} \, dx}{1250}\\ &=\frac{1533 \sqrt{1-2 x}}{3125}+\frac{511 (1-2 x)^{3/2}}{6875}+\frac{1533 (1-2 x)^{5/2}}{75625}-\frac{(1-2 x)^{7/2}}{550 (3+5 x)^2}-\frac{129 (1-2 x)^{7/2}}{6050 (3+5 x)}+\frac{16863 \int \frac{1}{\sqrt{1-2 x} (3+5 x)} \, dx}{6250}\\ &=\frac{1533 \sqrt{1-2 x}}{3125}+\frac{511 (1-2 x)^{3/2}}{6875}+\frac{1533 (1-2 x)^{5/2}}{75625}-\frac{(1-2 x)^{7/2}}{550 (3+5 x)^2}-\frac{129 (1-2 x)^{7/2}}{6050 (3+5 x)}-\frac{16863 \operatorname{Subst}\left (\int \frac{1}{\frac{11}{2}-\frac{5 x^2}{2}} \, dx,x,\sqrt{1-2 x}\right )}{6250}\\ &=\frac{1533 \sqrt{1-2 x}}{3125}+\frac{511 (1-2 x)^{3/2}}{6875}+\frac{1533 (1-2 x)^{5/2}}{75625}-\frac{(1-2 x)^{7/2}}{550 (3+5 x)^2}-\frac{129 (1-2 x)^{7/2}}{6050 (3+5 x)}-\frac{1533 \sqrt{\frac{11}{5}} \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )}{3125}\\ \end{align*}

Mathematica [A]  time = 0.0449123, size = 68, normalized size = 0.62 \[ \frac{\frac{5 \sqrt{1-2 x} \left (18000 x^4-25400 x^3+51980 x^2+98595 x+32504\right )}{(5 x+3)^2}-3066 \sqrt{55} \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )}{31250} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((5*Sqrt[1 - 2*x]*(32504 + 98595*x + 51980*x^2 - 25400*x^3 + 18000*x^4))/(3 + 5*x)^2 - 3066*Sqrt[55]*ArcTanh[S
qrt[5/11]*Sqrt[1 - 2*x]])/31250

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Maple [A]  time = 0.009, size = 75, normalized size = 0.7 \begin{align*}{\frac{18}{625} \left ( 1-2\,x \right ) ^{{\frac{5}{2}}}}+{\frac{58}{625} \left ( 1-2\,x \right ) ^{{\frac{3}{2}}}}+{\frac{1658}{3125}\sqrt{1-2\,x}}+{\frac{22}{125\, \left ( -10\,x-6 \right ) ^{2}} \left ({\frac{123}{10} \left ( 1-2\,x \right ) ^{{\frac{3}{2}}}}-{\frac{55}{2}\sqrt{1-2\,x}} \right ) }-{\frac{1533\,\sqrt{55}}{15625}{\it Artanh} \left ({\frac{\sqrt{55}}{11}\sqrt{1-2\,x}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

18/625*(1-2*x)^(5/2)+58/625*(1-2*x)^(3/2)+1658/3125*(1-2*x)^(1/2)+22/125*(123/10*(1-2*x)^(3/2)-55/2*(1-2*x)^(1
/2))/(-10*x-6)^2-1533/15625*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)

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Maxima [A]  time = 1.70067, size = 136, normalized size = 1.25 \begin{align*} \frac{18}{625} \,{\left (-2 \, x + 1\right )}^{\frac{5}{2}} + \frac{58}{625} \,{\left (-2 \, x + 1\right )}^{\frac{3}{2}} + \frac{1533}{31250} \, \sqrt{55} \log \left (-\frac{\sqrt{55} - 5 \, \sqrt{-2 \, x + 1}}{\sqrt{55} + 5 \, \sqrt{-2 \, x + 1}}\right ) + \frac{1658}{3125} \, \sqrt{-2 \, x + 1} + \frac{11 \,{\left (123 \,{\left (-2 \, x + 1\right )}^{\frac{3}{2}} - 275 \, \sqrt{-2 \, x + 1}\right )}}{625 \,{\left (25 \,{\left (2 \, x - 1\right )}^{2} + 220 \, x + 11\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

18/625*(-2*x + 1)^(5/2) + 58/625*(-2*x + 1)^(3/2) + 1533/31250*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sq
rt(55) + 5*sqrt(-2*x + 1))) + 1658/3125*sqrt(-2*x + 1) + 11/625*(123*(-2*x + 1)^(3/2) - 275*sqrt(-2*x + 1))/(2
5*(2*x - 1)^2 + 220*x + 11)

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Fricas [A]  time = 1.45958, size = 277, normalized size = 2.54 \begin{align*} \frac{1533 \, \sqrt{11} \sqrt{5}{\left (25 \, x^{2} + 30 \, x + 9\right )} \log \left (\frac{\sqrt{11} \sqrt{5} \sqrt{-2 \, x + 1} + 5 \, x - 8}{5 \, x + 3}\right ) + 5 \,{\left (18000 \, x^{4} - 25400 \, x^{3} + 51980 \, x^{2} + 98595 \, x + 32504\right )} \sqrt{-2 \, x + 1}}{31250 \,{\left (25 \, x^{2} + 30 \, x + 9\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/31250*(1533*sqrt(11)*sqrt(5)*(25*x^2 + 30*x + 9)*log((sqrt(11)*sqrt(5)*sqrt(-2*x + 1) + 5*x - 8)/(5*x + 3))
+ 5*(18000*x^4 - 25400*x^3 + 51980*x^2 + 98595*x + 32504)*sqrt(-2*x + 1))/(25*x^2 + 30*x + 9)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.79801, size = 138, normalized size = 1.27 \begin{align*} \frac{18}{625} \,{\left (2 \, x - 1\right )}^{2} \sqrt{-2 \, x + 1} + \frac{58}{625} \,{\left (-2 \, x + 1\right )}^{\frac{3}{2}} + \frac{1533}{31250} \, \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{1658}{3125} \, \sqrt{-2 \, x + 1} + \frac{11 \,{\left (123 \,{\left (-2 \, x + 1\right )}^{\frac{3}{2}} - 275 \, \sqrt{-2 \, x + 1}\right )}}{2500 \,{\left (5 \, x + 3\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

18/625*(2*x - 1)^2*sqrt(-2*x + 1) + 58/625*(-2*x + 1)^(3/2) + 1533/31250*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10
*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 1658/3125*sqrt(-2*x + 1) + 11/2500*(123*(-2*x + 1)^(3/2) - 2
75*sqrt(-2*x + 1))/(5*x + 3)^2