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

Optimal. Leaf size=120 \[ -\frac {48 \sqrt {1-2 x} (3 x+2)^3}{25 (5 x+3)}-\frac {(1-2 x)^{3/2} (3 x+2)^3}{10 (5 x+3)^2}+\frac {693}{625} \sqrt {1-2 x} (3 x+2)^2+\frac {63 \sqrt {1-2 x} (125 x+92)}{6250}-\frac {5943 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{3125 \sqrt {55}} \]

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Rubi [A]  time = 0.04, antiderivative size = 120, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {97, 149, 153, 147, 63, 206} \begin {gather*} -\frac {48 \sqrt {1-2 x} (3 x+2)^3}{25 (5 x+3)}-\frac {(1-2 x)^{3/2} (3 x+2)^3}{10 (5 x+3)^2}+\frac {693}{625} \sqrt {1-2 x} (3 x+2)^2+\frac {63 \sqrt {1-2 x} (125 x+92)}{6250}-\frac {5943 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{3125 \sqrt {55}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(693*Sqrt[1 - 2*x]*(2 + 3*x)^2)/625 - ((1 - 2*x)^(3/2)*(2 + 3*x)^3)/(10*(3 + 5*x)^2) - (48*Sqrt[1 - 2*x]*(2 +
3*x)^3)/(25*(3 + 5*x)) + (63*Sqrt[1 - 2*x]*(92 + 125*x))/6250 - (5943*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(3125
*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 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 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)^{3/2} (2+3 x)^3}{(3+5 x)^3} \, dx &=-\frac {(1-2 x)^{3/2} (2+3 x)^3}{10 (3+5 x)^2}+\frac {1}{10} \int \frac {(3-27 x) \sqrt {1-2 x} (2+3 x)^2}{(3+5 x)^2} \, dx\\ &=-\frac {(1-2 x)^{3/2} (2+3 x)^3}{10 (3+5 x)^2}-\frac {48 \sqrt {1-2 x} (2+3 x)^3}{25 (3+5 x)}+\frac {1}{50} \int \frac {(357-1386 x) (2+3 x)^2}{\sqrt {1-2 x} (3+5 x)} \, dx\\ &=\frac {693}{625} \sqrt {1-2 x} (2+3 x)^2-\frac {(1-2 x)^{3/2} (2+3 x)^3}{10 (3+5 x)^2}-\frac {48 \sqrt {1-2 x} (2+3 x)^3}{25 (3+5 x)}-\frac {\int \frac {(2+3 x) (-1218+7875 x)}{\sqrt {1-2 x} (3+5 x)} \, dx}{1250}\\ &=\frac {693}{625} \sqrt {1-2 x} (2+3 x)^2-\frac {(1-2 x)^{3/2} (2+3 x)^3}{10 (3+5 x)^2}-\frac {48 \sqrt {1-2 x} (2+3 x)^3}{25 (3+5 x)}+\frac {63 \sqrt {1-2 x} (92+125 x)}{6250}+\frac {5943 \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx}{6250}\\ &=\frac {693}{625} \sqrt {1-2 x} (2+3 x)^2-\frac {(1-2 x)^{3/2} (2+3 x)^3}{10 (3+5 x)^2}-\frac {48 \sqrt {1-2 x} (2+3 x)^3}{25 (3+5 x)}+\frac {63 \sqrt {1-2 x} (92+125 x)}{6250}-\frac {5943 \operatorname {Subst}\left (\int \frac {1}{\frac {11}{2}-\frac {5 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )}{6250}\\ &=\frac {693}{625} \sqrt {1-2 x} (2+3 x)^2-\frac {(1-2 x)^{3/2} (2+3 x)^3}{10 (3+5 x)^2}-\frac {48 \sqrt {1-2 x} (2+3 x)^3}{25 (3+5 x)}+\frac {63 \sqrt {1-2 x} (92+125 x)}{6250}-\frac {5943 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{3125 \sqrt {55}}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 68, normalized size = 0.57 \begin {gather*} \frac {\sqrt {1-2 x} \left (-27000 x^4-14400 x^3+37530 x^2+36295 x+8644\right )}{6250 (5 x+3)^2}-\frac {5943 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{3125 \sqrt {55}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[1 - 2*x]*(8644 + 36295*x + 37530*x^2 - 14400*x^3 - 27000*x^4))/(6250*(3 + 5*x)^2) - (5943*ArcTanh[Sqrt[5
/11]*Sqrt[1 - 2*x]])/(3125*Sqrt[55])

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IntegrateAlgebraic [A]  time = 0.19, size = 88, normalized size = 0.73 \begin {gather*} -\frac {\sqrt {1-2 x} \left (3375 (1-2 x)^4-17100 (1-2 x)^3+12285 (1-2 x)^2+49525 (1-2 x)-65373\right )}{3125 (5 (1-2 x)-11)^2}-\frac {5943 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{3125 \sqrt {55}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3125*((-65373 + 49525*(1 - 2*x) + 12285*(1 - 2*x)^2 - 17100*(1 - 2*x)^3 + 3375*(1 - 2*x)^4)*Sqrt[1 - 2*x])/
(-11 + 5*(1 - 2*x))^2 - (5943*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(3125*Sqrt[55])

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fricas [A]  time = 1.74, size = 84, normalized size = 0.70 \begin {gather*} \frac {5943 \, \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 (27000 \, x^{4} + 14400 \, x^{3} - 37530 \, x^{2} - 36295 \, x - 8644\right )} \sqrt {-2 \, x + 1}}{343750 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/343750*(5943*sqrt(55)*(25*x^2 + 30*x + 9)*log((5*x + sqrt(55)*sqrt(-2*x + 1) - 8)/(5*x + 3)) - 55*(27000*x^4
+ 14400*x^3 - 37530*x^2 - 36295*x - 8644)*sqrt(-2*x + 1))/(25*x^2 + 30*x + 9)

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giac [A]  time = 1.04, size = 102, normalized size = 0.85 \begin {gather*} -\frac {27}{625} \, {\left (2 \, x - 1\right )}^{2} \sqrt {-2 \, x + 1} + \frac {18}{625} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {5943}{343750} \, \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 {558}{3125} \, \sqrt {-2 \, x + 1} + \frac {193 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 429 \, \sqrt {-2 \, x + 1}}{2500 \, {\left (5 \, x + 3\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-27/625*(2*x - 1)^2*sqrt(-2*x + 1) + 18/625*(-2*x + 1)^(3/2) + 5943/343750*sqrt(55)*log(1/2*abs(-2*sqrt(55) +
10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 558/3125*sqrt(-2*x + 1) + 1/2500*(193*(-2*x + 1)^(3/2) - 4
29*sqrt(-2*x + 1))/(5*x + 3)^2

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maple [A]  time = 0.01, size = 75, normalized size = 0.62 \begin {gather*} -\frac {5943 \sqrt {55}\, \arctanh \left (\frac {\sqrt {55}\, \sqrt {-2 x +1}}{11}\right )}{171875}-\frac {27 \left (-2 x +1\right )^{\frac {5}{2}}}{625}+\frac {18 \left (-2 x +1\right )^{\frac {3}{2}}}{625}+\frac {558 \sqrt {-2 x +1}}{3125}+\frac {\frac {193 \left (-2 x +1\right )^{\frac {3}{2}}}{625}-\frac {429 \sqrt {-2 x +1}}{625}}{\left (-10 x -6\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-27/625*(-2*x+1)^(5/2)+18/625*(-2*x+1)^(3/2)+558/3125*(-2*x+1)^(1/2)+2/125*(193/10*(-2*x+1)^(3/2)-429/10*(-2*x
+1)^(1/2))/(-10*x-6)^2-5943/171875*arctanh(1/11*55^(1/2)*(-2*x+1)^(1/2))*55^(1/2)

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maxima [A]  time = 1.24, size = 101, normalized size = 0.84 \begin {gather*} -\frac {27}{625} \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} + \frac {18}{625} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {5943}{343750} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) + \frac {558}{3125} \, \sqrt {-2 \, x + 1} + \frac {193 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 429 \, \sqrt {-2 \, x + 1}}{625 \, {\left (25 \, {\left (2 \, x - 1\right )}^{2} + 220 \, x + 11\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-27/625*(-2*x + 1)^(5/2) + 18/625*(-2*x + 1)^(3/2) + 5943/343750*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(
sqrt(55) + 5*sqrt(-2*x + 1))) + 558/3125*sqrt(-2*x + 1) + 1/625*(193*(-2*x + 1)^(3/2) - 429*sqrt(-2*x + 1))/(2
5*(2*x - 1)^2 + 220*x + 11)

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mupad [B]  time = 0.05, size = 83, normalized size = 0.69 \begin {gather*} \frac {558\,\sqrt {1-2\,x}}{3125}+\frac {18\,{\left (1-2\,x\right )}^{3/2}}{625}-\frac {27\,{\left (1-2\,x\right )}^{5/2}}{625}-\frac {\frac {429\,\sqrt {1-2\,x}}{15625}-\frac {193\,{\left (1-2\,x\right )}^{3/2}}{15625}}{\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 )\,5943{}\mathrm {i}}{171875} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(55^(1/2)*atan((55^(1/2)*(1 - 2*x)^(1/2)*1i)/11)*5943i)/171875 + (558*(1 - 2*x)^(1/2))/3125 + (18*(1 - 2*x)^(3
/2))/625 - (27*(1 - 2*x)^(5/2))/625 - ((429*(1 - 2*x)^(1/2))/15625 - (193*(1 - 2*x)^(3/2))/15625)/((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 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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