3.1992 \(\int \frac{(1-2 x)^{5/2} (2+3 x)^3}{(3+5 x)^3} \, dx\)

Optimal. Leaf size=135 \[ -\frac{47 (1-2 x)^{3/2} (3 x+2)^3}{25 (5 x+3)}-\frac{(1-2 x)^{5/2} (3 x+2)^3}{10 (5 x+3)^2}+\frac{954}{875} (1-2 x)^{3/2} (3 x+2)^2+\frac{3 (1-2 x)^{3/2} (2403 x+1618)}{6250}+\frac{5559 \sqrt{1-2 x}}{15625}-\frac{5559 \sqrt{\frac{11}{5}} \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )}{15625} \]

[Out]

(5559*Sqrt[1 - 2*x])/15625 + (954*(1 - 2*x)^(3/2)*(2 + 3*x)^2)/875 - ((1 - 2*x)^(5/2)*(2 + 3*x)^3)/(10*(3 + 5*
x)^2) - (47*(1 - 2*x)^(3/2)*(2 + 3*x)^3)/(25*(3 + 5*x)) + (3*(1 - 2*x)^(3/2)*(1618 + 2403*x))/6250 - (5559*Sqr
t[11/5]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/15625

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Rubi [A]  time = 0.0433697, antiderivative size = 135, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.292, Rules used = {97, 149, 153, 147, 50, 63, 206} \[ -\frac{47 (1-2 x)^{3/2} (3 x+2)^3}{25 (5 x+3)}-\frac{(1-2 x)^{5/2} (3 x+2)^3}{10 (5 x+3)^2}+\frac{954}{875} (1-2 x)^{3/2} (3 x+2)^2+\frac{3 (1-2 x)^{3/2} (2403 x+1618)}{6250}+\frac{5559 \sqrt{1-2 x}}{15625}-\frac{5559 \sqrt{\frac{11}{5}} \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )}{15625} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5559*Sqrt[1 - 2*x])/15625 + (954*(1 - 2*x)^(3/2)*(2 + 3*x)^2)/875 - ((1 - 2*x)^(5/2)*(2 + 3*x)^3)/(10*(3 + 5*
x)^2) - (47*(1 - 2*x)^(3/2)*(2 + 3*x)^3)/(25*(3 + 5*x)) + (3*(1 - 2*x)^(3/2)*(1618 + 2403*x))/6250 - (5559*Sqr
t[11/5]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/15625

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 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 153

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

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 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}{(3+5 x)^3} \, dx &=-\frac{(1-2 x)^{5/2} (2+3 x)^3}{10 (3+5 x)^2}+\frac{1}{10} \int \frac{(-1-33 x) (1-2 x)^{3/2} (2+3 x)^2}{(3+5 x)^2} \, dx\\ &=-\frac{(1-2 x)^{5/2} (2+3 x)^3}{10 (3+5 x)^2}-\frac{47 (1-2 x)^{3/2} (2+3 x)^3}{25 (3+5 x)}+\frac{1}{50} \int \frac{(-33-1908 x) \sqrt{1-2 x} (2+3 x)^2}{3+5 x} \, dx\\ &=\frac{954}{875} (1-2 x)^{3/2} (2+3 x)^2-\frac{(1-2 x)^{5/2} (2+3 x)^3}{10 (3+5 x)^2}-\frac{47 (1-2 x)^{3/2} (2+3 x)^3}{25 (3+5 x)}-\frac{\int \frac{\sqrt{1-2 x} (2+3 x) (2310+16821 x)}{3+5 x} \, dx}{1750}\\ &=\frac{954}{875} (1-2 x)^{3/2} (2+3 x)^2-\frac{(1-2 x)^{5/2} (2+3 x)^3}{10 (3+5 x)^2}-\frac{47 (1-2 x)^{3/2} (2+3 x)^3}{25 (3+5 x)}+\frac{3 (1-2 x)^{3/2} (1618+2403 x)}{6250}+\frac{5559 \int \frac{\sqrt{1-2 x}}{3+5 x} \, dx}{6250}\\ &=\frac{5559 \sqrt{1-2 x}}{15625}+\frac{954}{875} (1-2 x)^{3/2} (2+3 x)^2-\frac{(1-2 x)^{5/2} (2+3 x)^3}{10 (3+5 x)^2}-\frac{47 (1-2 x)^{3/2} (2+3 x)^3}{25 (3+5 x)}+\frac{3 (1-2 x)^{3/2} (1618+2403 x)}{6250}+\frac{61149 \int \frac{1}{\sqrt{1-2 x} (3+5 x)} \, dx}{31250}\\ &=\frac{5559 \sqrt{1-2 x}}{15625}+\frac{954}{875} (1-2 x)^{3/2} (2+3 x)^2-\frac{(1-2 x)^{5/2} (2+3 x)^3}{10 (3+5 x)^2}-\frac{47 (1-2 x)^{3/2} (2+3 x)^3}{25 (3+5 x)}+\frac{3 (1-2 x)^{3/2} (1618+2403 x)}{6250}-\frac{61149 \operatorname{Subst}\left (\int \frac{1}{\frac{11}{2}-\frac{5 x^2}{2}} \, dx,x,\sqrt{1-2 x}\right )}{31250}\\ &=\frac{5559 \sqrt{1-2 x}}{15625}+\frac{954}{875} (1-2 x)^{3/2} (2+3 x)^2-\frac{(1-2 x)^{5/2} (2+3 x)^3}{10 (3+5 x)^2}-\frac{47 (1-2 x)^{3/2} (2+3 x)^3}{25 (3+5 x)}+\frac{3 (1-2 x)^{3/2} (1618+2403 x)}{6250}-\frac{5559 \sqrt{\frac{11}{5}} \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )}{15625}\\ \end{align*}

Mathematica [A]  time = 0.047895, size = 73, normalized size = 0.54 \[ \frac{\frac{5 \sqrt{1-2 x} \left (1350000 x^5-27000 x^4-1506900 x^3+1651030 x^2+2637795 x+770444\right )}{(5 x+3)^2}-77826 \sqrt{55} \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )}{1093750} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((5*Sqrt[1 - 2*x]*(770444 + 2637795*x + 1651030*x^2 - 1506900*x^3 - 27000*x^4 + 1350000*x^5))/(3 + 5*x)^2 - 77
826*Sqrt[55]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/1093750

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Maple [A]  time = 0.009, size = 84, normalized size = 0.6 \begin{align*} -{\frac{27}{875} \left ( 1-2\,x \right ) ^{{\frac{7}{2}}}}+{\frac{54}{3125} \left ( 1-2\,x \right ) ^{{\frac{5}{2}}}}+{\frac{186}{3125} \left ( 1-2\,x \right ) ^{{\frac{3}{2}}}}+{\frac{46}{125}\sqrt{1-2\,x}}+{\frac{22}{125\, \left ( -10\,x-6 \right ) ^{2}} \left ({\frac{189}{50} \left ( 1-2\,x \right ) ^{{\frac{3}{2}}}}-{\frac{2101}{250}\sqrt{1-2\,x}} \right ) }-{\frac{5559\,\sqrt{55}}{78125}{\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/(3+5*x)^3,x)

[Out]

-27/875*(1-2*x)^(7/2)+54/3125*(1-2*x)^(5/2)+186/3125*(1-2*x)^(3/2)+46/125*(1-2*x)^(1/2)+22/125*(189/50*(1-2*x)
^(3/2)-2101/250*(1-2*x)^(1/2))/(-10*x-6)^2-5559/78125*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)

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Maxima [A]  time = 3.12695, size = 149, normalized size = 1.1 \begin{align*} -\frac{27}{875} \,{\left (-2 \, x + 1\right )}^{\frac{7}{2}} + \frac{54}{3125} \,{\left (-2 \, x + 1\right )}^{\frac{5}{2}} + \frac{186}{3125} \,{\left (-2 \, x + 1\right )}^{\frac{3}{2}} + \frac{5559}{156250} \, \sqrt{55} \log \left (-\frac{\sqrt{55} - 5 \, \sqrt{-2 \, x + 1}}{\sqrt{55} + 5 \, \sqrt{-2 \, x + 1}}\right ) + \frac{46}{125} \, \sqrt{-2 \, x + 1} + \frac{11 \,{\left (945 \,{\left (-2 \, x + 1\right )}^{\frac{3}{2}} - 2101 \, \sqrt{-2 \, x + 1}\right )}}{15625 \,{\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)^3/(3+5*x)^3,x, algorithm="maxima")

[Out]

-27/875*(-2*x + 1)^(7/2) + 54/3125*(-2*x + 1)^(5/2) + 186/3125*(-2*x + 1)^(3/2) + 5559/156250*sqrt(55)*log(-(s
qrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 46/125*sqrt(-2*x + 1) + 11/15625*(945*(-2*x + 1)^
(3/2) - 2101*sqrt(-2*x + 1))/(25*(2*x - 1)^2 + 220*x + 11)

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Fricas [A]  time = 1.3585, size = 309, normalized size = 2.29 \begin{align*} \frac{38913 \, \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 (1350000 \, x^{5} - 27000 \, x^{4} - 1506900 \, x^{3} + 1651030 \, x^{2} + 2637795 \, x + 770444\right )} \sqrt{-2 \, x + 1}}{1093750 \,{\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)^3/(3+5*x)^3,x, algorithm="fricas")

[Out]

1/1093750*(38913*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*(1350000*x^5 - 27000*x^4 - 1506900*x^3 + 1651030*x^2 + 2637795*x + 770444)*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)**3/(3+5*x)**3,x)

[Out]

Timed out

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Giac [A]  time = 2.48108, size = 159, normalized size = 1.18 \begin{align*} \frac{27}{875} \,{\left (2 \, x - 1\right )}^{3} \sqrt{-2 \, x + 1} + \frac{54}{3125} \,{\left (2 \, x - 1\right )}^{2} \sqrt{-2 \, x + 1} + \frac{186}{3125} \,{\left (-2 \, x + 1\right )}^{\frac{3}{2}} + \frac{5559}{156250} \, \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{46}{125} \, \sqrt{-2 \, x + 1} + \frac{11 \,{\left (945 \,{\left (-2 \, x + 1\right )}^{\frac{3}{2}} - 2101 \, \sqrt{-2 \, x + 1}\right )}}{62500 \,{\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)^3/(3+5*x)^3,x, algorithm="giac")

[Out]

27/875*(2*x - 1)^3*sqrt(-2*x + 1) + 54/3125*(2*x - 1)^2*sqrt(-2*x + 1) + 186/3125*(-2*x + 1)^(3/2) + 5559/1562
50*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 46/125*sqrt(-2*x + 1
) + 11/62500*(945*(-2*x + 1)^(3/2) - 2101*sqrt(-2*x + 1))/(5*x + 3)^2