3.18.9 \(\int \frac {\sqrt {1-2 x} (2+3 x)^4}{(3+5 x)^3} \, dx\)

Optimal. Leaf size=120 \[ -\frac {\sqrt {1-2 x} (3 x+2)^4}{10 (5 x+3)^2}-\frac {131 \sqrt {1-2 x} (3 x+2)^3}{550 (5 x+3)}+\frac {1428 \sqrt {1-2 x} (3 x+2)^2}{6875}-\frac {21 (704-375 x) \sqrt {1-2 x}}{68750}-\frac {12803 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{34375 \sqrt {55}} \]

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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 {\sqrt {1-2 x} (3 x+2)^4}{10 (5 x+3)^2}-\frac {131 \sqrt {1-2 x} (3 x+2)^3}{550 (5 x+3)}+\frac {1428 \sqrt {1-2 x} (3 x+2)^2}{6875}-\frac {21 (704-375 x) \sqrt {1-2 x}}{68750}-\frac {12803 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{34375 \sqrt {55}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(Sqrt[1 - 2*x]*(2 + 3*x)^4)/(3 + 5*x)^3,x]

[Out]

(-21*(704 - 375*x)*Sqrt[1 - 2*x])/68750 + (1428*Sqrt[1 - 2*x]*(2 + 3*x)^2)/6875 - (Sqrt[1 - 2*x]*(2 + 3*x)^4)/
(10*(3 + 5*x)^2) - (131*Sqrt[1 - 2*x]*(2 + 3*x)^3)/(550*(3 + 5*x)) - (12803*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])
/(34375*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 {\sqrt {1-2 x} (2+3 x)^4}{(3+5 x)^3} \, dx &=-\frac {\sqrt {1-2 x} (2+3 x)^4}{10 (3+5 x)^2}+\frac {1}{10} \int \frac {(10-27 x) (2+3 x)^3}{\sqrt {1-2 x} (3+5 x)^2} \, dx\\ &=-\frac {\sqrt {1-2 x} (2+3 x)^4}{10 (3+5 x)^2}-\frac {131 \sqrt {1-2 x} (2+3 x)^3}{550 (3+5 x)}+\frac {1}{550} \int \frac {(847-2856 x) (2+3 x)^2}{\sqrt {1-2 x} (3+5 x)} \, dx\\ &=\frac {1428 \sqrt {1-2 x} (2+3 x)^2}{6875}-\frac {\sqrt {1-2 x} (2+3 x)^4}{10 (3+5 x)^2}-\frac {131 \sqrt {1-2 x} (2+3 x)^3}{550 (3+5 x)}-\frac {\int \frac {(2+3 x) (-8078+7875 x)}{\sqrt {1-2 x} (3+5 x)} \, dx}{13750}\\ &=-\frac {21 (704-375 x) \sqrt {1-2 x}}{68750}+\frac {1428 \sqrt {1-2 x} (2+3 x)^2}{6875}-\frac {\sqrt {1-2 x} (2+3 x)^4}{10 (3+5 x)^2}-\frac {131 \sqrt {1-2 x} (2+3 x)^3}{550 (3+5 x)}+\frac {12803 \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx}{68750}\\ &=-\frac {21 (704-375 x) \sqrt {1-2 x}}{68750}+\frac {1428 \sqrt {1-2 x} (2+3 x)^2}{6875}-\frac {\sqrt {1-2 x} (2+3 x)^4}{10 (3+5 x)^2}-\frac {131 \sqrt {1-2 x} (2+3 x)^3}{550 (3+5 x)}-\frac {12803 \operatorname {Subst}\left (\int \frac {1}{\frac {11}{2}-\frac {5 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )}{68750}\\ &=-\frac {21 (704-375 x) \sqrt {1-2 x}}{68750}+\frac {1428 \sqrt {1-2 x} (2+3 x)^2}{6875}-\frac {\sqrt {1-2 x} (2+3 x)^4}{10 (3+5 x)^2}-\frac {131 \sqrt {1-2 x} (2+3 x)^3}{550 (3+5 x)}-\frac {12803 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{34375 \sqrt {55}}\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 68, normalized size = 0.57 \begin {gather*} \frac {\frac {55 \sqrt {1-2 x} \left (445500 x^4+1103850 x^3+506880 x^2-200305 x-121976\right )}{(5 x+3)^2}-25606 \sqrt {55} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{3781250} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(Sqrt[1 - 2*x]*(2 + 3*x)^4)/(3 + 5*x)^3,x]

[Out]

((55*Sqrt[1 - 2*x]*(-121976 - 200305*x + 506880*x^2 + 1103850*x^3 + 445500*x^4))/(3 + 5*x)^2 - 25606*Sqrt[55]*
ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/3781250

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IntegrateAlgebraic [A]  time = 0.19, size = 88, normalized size = 0.73 \begin {gather*} \frac {\left (111375 (1-2 x)^4-997425 (1-2 x)^3+2830905 (1-2 x)^2-2714425 (1-2 x)+281666\right ) \sqrt {1-2 x}}{68750 (5 (1-2 x)-11)^2}-\frac {12803 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{34375 \sqrt {55}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(Sqrt[1 - 2*x]*(2 + 3*x)^4)/(3 + 5*x)^3,x]

[Out]

((281666 - 2714425*(1 - 2*x) + 2830905*(1 - 2*x)^2 - 997425*(1 - 2*x)^3 + 111375*(1 - 2*x)^4)*Sqrt[1 - 2*x])/(
68750*(-11 + 5*(1 - 2*x))^2) - (12803*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(34375*Sqrt[55])

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fricas [A]  time = 1.50, size = 84, normalized size = 0.70 \begin {gather*} \frac {12803 \, \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 (445500 \, x^{4} + 1103850 \, x^{3} + 506880 \, x^{2} - 200305 \, x - 121976\right )} \sqrt {-2 \, x + 1}}{3781250 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2+3*x)^4*(1-2*x)^(1/2)/(3+5*x)^3,x, algorithm="fricas")

[Out]

1/3781250*(12803*sqrt(55)*(25*x^2 + 30*x + 9)*log((5*x + sqrt(55)*sqrt(-2*x + 1) - 8)/(5*x + 3)) + 55*(445500*
x^4 + 1103850*x^3 + 506880*x^2 - 200305*x - 121976)*sqrt(-2*x + 1))/(25*x^2 + 30*x + 9)

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giac [A]  time = 1.23, size = 102, normalized size = 0.85 \begin {gather*} \frac {81}{1250} \, {\left (2 \, x - 1\right )}^{2} \sqrt {-2 \, x + 1} - \frac {369}{1250} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {12803}{3781250} \, \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 {108}{3125} \, \sqrt {-2 \, x + 1} + \frac {263 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 583 \, \sqrt {-2 \, x + 1}}{27500 \, {\left (5 \, x + 3\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2+3*x)^4*(1-2*x)^(1/2)/(3+5*x)^3,x, algorithm="giac")

[Out]

81/1250*(2*x - 1)^2*sqrt(-2*x + 1) - 369/1250*(-2*x + 1)^(3/2) + 12803/3781250*sqrt(55)*log(1/2*abs(-2*sqrt(55
) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 108/3125*sqrt(-2*x + 1) + 1/27500*(263*(-2*x + 1)^(3/2
) - 583*sqrt(-2*x + 1))/(5*x + 3)^2

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maple [A]  time = 0.02, size = 75, normalized size = 0.62 \begin {gather*} -\frac {12803 \sqrt {55}\, \arctanh \left (\frac {\sqrt {55}\, \sqrt {-2 x +1}}{11}\right )}{1890625}+\frac {81 \left (-2 x +1\right )^{\frac {5}{2}}}{1250}-\frac {369 \left (-2 x +1\right )^{\frac {3}{2}}}{1250}+\frac {108 \sqrt {-2 x +1}}{3125}+\frac {\frac {263 \left (-2 x +1\right )^{\frac {3}{2}}}{6875}-\frac {53 \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((3*x+2)^4*(-2*x+1)^(1/2)/(5*x+3)^3,x)

[Out]

81/1250*(-2*x+1)^(5/2)-369/1250*(-2*x+1)^(3/2)+108/3125*(-2*x+1)^(1/2)+4/125*(263/220*(-2*x+1)^(3/2)-53/20*(-2
*x+1)^(1/2))/(-6-10*x)^2-12803/1890625*arctanh(1/11*55^(1/2)*(-2*x+1)^(1/2))*55^(1/2)

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maxima [A]  time = 1.36, size = 101, normalized size = 0.84 \begin {gather*} \frac {81}{1250} \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} - \frac {369}{1250} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {12803}{3781250} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) + \frac {108}{3125} \, \sqrt {-2 \, x + 1} + \frac {263 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 583 \, \sqrt {-2 \, x + 1}}{6875 \, {\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((2+3*x)^4*(1-2*x)^(1/2)/(3+5*x)^3,x, algorithm="maxima")

[Out]

81/1250*(-2*x + 1)^(5/2) - 369/1250*(-2*x + 1)^(3/2) + 12803/3781250*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1
))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 108/3125*sqrt(-2*x + 1) + 1/6875*(263*(-2*x + 1)^(3/2) - 583*sqrt(-2*x + 1
))/(25*(2*x - 1)^2 + 220*x + 11)

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mupad [B]  time = 0.06, size = 83, normalized size = 0.69 \begin {gather*} \frac {108\,\sqrt {1-2\,x}}{3125}-\frac {369\,{\left (1-2\,x\right )}^{3/2}}{1250}+\frac {81\,{\left (1-2\,x\right )}^{5/2}}{1250}-\frac {\frac {53\,\sqrt {1-2\,x}}{15625}-\frac {263\,{\left (1-2\,x\right )}^{3/2}}{171875}}{\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 )\,12803{}\mathrm {i}}{1890625} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((1 - 2*x)^(1/2)*(3*x + 2)^4)/(5*x + 3)^3,x)

[Out]

(55^(1/2)*atan((55^(1/2)*(1 - 2*x)^(1/2)*1i)/11)*12803i)/1890625 + (108*(1 - 2*x)^(1/2))/3125 - (369*(1 - 2*x)
^(3/2))/1250 + (81*(1 - 2*x)^(5/2))/1250 - ((53*(1 - 2*x)^(1/2))/15625 - (263*(1 - 2*x)^(3/2))/171875)/((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((2+3*x)**4*(1-2*x)**(1/2)/(3+5*x)**3,x)

[Out]

Timed out

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