3.1983 \(\int \frac{(1-2 x)^{5/2} (2+3 x)}{(3+5 x)^2} \, dx\)

Optimal. Leaf size=89 \[ -\frac{(1-2 x)^{7/2}}{55 (5 x+3)}+\frac{56 (1-2 x)^{5/2}}{1375}+\frac{56}{375} (1-2 x)^{3/2}+\frac{616}{625} \sqrt{1-2 x}-\frac{616}{625} \sqrt{\frac{11}{5}} \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right ) \]

[Out]

(616*Sqrt[1 - 2*x])/625 + (56*(1 - 2*x)^(3/2))/375 + (56*(1 - 2*x)^(5/2))/1375 - (1 - 2*x)^(7/2)/(55*(3 + 5*x)
) - (616*Sqrt[11/5]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/625

________________________________________________________________________________________

Rubi [A]  time = 0.0247181, antiderivative size = 89, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {78, 50, 63, 206} \[ -\frac{(1-2 x)^{7/2}}{55 (5 x+3)}+\frac{56 (1-2 x)^{5/2}}{1375}+\frac{56}{375} (1-2 x)^{3/2}+\frac{616}{625} \sqrt{1-2 x}-\frac{616}{625} \sqrt{\frac{11}{5}} \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(616*Sqrt[1 - 2*x])/625 + (56*(1 - 2*x)^(3/2))/375 + (56*(1 - 2*x)^(5/2))/1375 - (1 - 2*x)^(7/2)/(55*(3 + 5*x)
) - (616*Sqrt[11/5]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/625

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)}{(3+5 x)^2} \, dx &=-\frac{(1-2 x)^{7/2}}{55 (3+5 x)}+\frac{28}{55} \int \frac{(1-2 x)^{5/2}}{3+5 x} \, dx\\ &=\frac{56 (1-2 x)^{5/2}}{1375}-\frac{(1-2 x)^{7/2}}{55 (3+5 x)}+\frac{28}{25} \int \frac{(1-2 x)^{3/2}}{3+5 x} \, dx\\ &=\frac{56}{375} (1-2 x)^{3/2}+\frac{56 (1-2 x)^{5/2}}{1375}-\frac{(1-2 x)^{7/2}}{55 (3+5 x)}+\frac{308}{125} \int \frac{\sqrt{1-2 x}}{3+5 x} \, dx\\ &=\frac{616}{625} \sqrt{1-2 x}+\frac{56}{375} (1-2 x)^{3/2}+\frac{56 (1-2 x)^{5/2}}{1375}-\frac{(1-2 x)^{7/2}}{55 (3+5 x)}+\frac{3388}{625} \int \frac{1}{\sqrt{1-2 x} (3+5 x)} \, dx\\ &=\frac{616}{625} \sqrt{1-2 x}+\frac{56}{375} (1-2 x)^{3/2}+\frac{56 (1-2 x)^{5/2}}{1375}-\frac{(1-2 x)^{7/2}}{55 (3+5 x)}-\frac{3388}{625} \operatorname{Subst}\left (\int \frac{1}{\frac{11}{2}-\frac{5 x^2}{2}} \, dx,x,\sqrt{1-2 x}\right )\\ &=\frac{616}{625} \sqrt{1-2 x}+\frac{56}{375} (1-2 x)^{3/2}+\frac{56 (1-2 x)^{5/2}}{1375}-\frac{(1-2 x)^{7/2}}{55 (3+5 x)}-\frac{616}{625} \sqrt{\frac{11}{5}} \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )\\ \end{align*}

Mathematica [A]  time = 0.0353224, size = 63, normalized size = 0.71 \[ \frac{\frac{5 \sqrt{1-2 x} \left (1800 x^3-3820 x^2+8630 x+6579\right )}{5 x+3}-1848 \sqrt{55} \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )}{9375} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((5*Sqrt[1 - 2*x]*(6579 + 8630*x - 3820*x^2 + 1800*x^3))/(3 + 5*x) - 1848*Sqrt[55]*ArcTanh[Sqrt[5/11]*Sqrt[1 -
2*x]])/9375

________________________________________________________________________________________

Maple [A]  time = 0.01, size = 63, normalized size = 0.7 \begin{align*}{\frac{6}{125} \left ( 1-2\,x \right ) ^{{\frac{5}{2}}}}+{\frac{62}{375} \left ( 1-2\,x \right ) ^{{\frac{3}{2}}}}+{\frac{638}{625}\sqrt{1-2\,x}}+{\frac{242}{3125}\sqrt{1-2\,x} \left ( -2\,x-{\frac{6}{5}} \right ) ^{-1}}-{\frac{616\,\sqrt{55}}{3125}{\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)/(3+5*x)^2,x)

[Out]

6/125*(1-2*x)^(5/2)+62/375*(1-2*x)^(3/2)+638/625*(1-2*x)^(1/2)+242/3125*(1-2*x)^(1/2)/(-2*x-6/5)-616/3125*arct
anh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)

________________________________________________________________________________________

Maxima [A]  time = 2.60725, size = 108, normalized size = 1.21 \begin{align*} \frac{6}{125} \,{\left (-2 \, x + 1\right )}^{\frac{5}{2}} + \frac{62}{375} \,{\left (-2 \, x + 1\right )}^{\frac{3}{2}} + \frac{308}{3125} \, \sqrt{55} \log \left (-\frac{\sqrt{55} - 5 \, \sqrt{-2 \, x + 1}}{\sqrt{55} + 5 \, \sqrt{-2 \, x + 1}}\right ) + \frac{638}{625} \, \sqrt{-2 \, x + 1} - \frac{121 \, \sqrt{-2 \, x + 1}}{625 \,{\left (5 \, x + 3\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

6/125*(-2*x + 1)^(5/2) + 62/375*(-2*x + 1)^(3/2) + 308/3125*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(
55) + 5*sqrt(-2*x + 1))) + 638/625*sqrt(-2*x + 1) - 121/625*sqrt(-2*x + 1)/(5*x + 3)

________________________________________________________________________________________

Fricas [A]  time = 1.36627, size = 225, normalized size = 2.53 \begin{align*} \frac{924 \, \sqrt{11} \sqrt{5}{\left (5 \, x + 3\right )} \log \left (\frac{\sqrt{11} \sqrt{5} \sqrt{-2 \, x + 1} + 5 \, x - 8}{5 \, x + 3}\right ) + 5 \,{\left (1800 \, x^{3} - 3820 \, x^{2} + 8630 \, x + 6579\right )} \sqrt{-2 \, x + 1}}{9375 \,{\left (5 \, x + 3\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/9375*(924*sqrt(11)*sqrt(5)*(5*x + 3)*log((sqrt(11)*sqrt(5)*sqrt(-2*x + 1) + 5*x - 8)/(5*x + 3)) + 5*(1800*x^
3 - 3820*x^2 + 8630*x + 6579)*sqrt(-2*x + 1))/(5*x + 3)

________________________________________________________________________________________

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)/(3+5*x)**2,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.60035, size = 122, normalized size = 1.37 \begin{align*} \frac{6}{125} \,{\left (2 \, x - 1\right )}^{2} \sqrt{-2 \, x + 1} + \frac{62}{375} \,{\left (-2 \, x + 1\right )}^{\frac{3}{2}} + \frac{308}{3125} \, \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{638}{625} \, \sqrt{-2 \, x + 1} - \frac{121 \, \sqrt{-2 \, x + 1}}{625 \,{\left (5 \, x + 3\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

6/125*(2*x - 1)^2*sqrt(-2*x + 1) + 62/375*(-2*x + 1)^(3/2) + 308/3125*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sq
rt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 638/625*sqrt(-2*x + 1) - 121/625*sqrt(-2*x + 1)/(5*x + 3)