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

Optimal. Leaf size=101 \[ \frac {2873}{73205 \sqrt {1-2 x}}-\frac {2873}{39930 \sqrt {1-2 x} (5 x+3)}-\frac {614}{1815 \sqrt {1-2 x} (5 x+3)^2}+\frac {49}{66 (1-2 x)^{3/2} (5 x+3)^2}-\frac {2873 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{14641 \sqrt {55}} \]

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Rubi [A]  time = 0.03, antiderivative size = 108, normalized size of antiderivative = 1.07, number of steps used = 6, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {89, 78, 51, 63, 206} \begin {gather*} -\frac {2873 \sqrt {1-2 x}}{29282 (5 x+3)}+\frac {2873}{19965 \sqrt {1-2 x} (5 x+3)}-\frac {614}{1815 \sqrt {1-2 x} (5 x+3)^2}+\frac {49}{66 (1-2 x)^{3/2} (5 x+3)^2}-\frac {2873 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{14641 \sqrt {55}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

49/(66*(1 - 2*x)^(3/2)*(3 + 5*x)^2) - 614/(1815*Sqrt[1 - 2*x]*(3 + 5*x)^2) + 2873/(19965*Sqrt[1 - 2*x]*(3 + 5*
x)) - (2873*Sqrt[1 - 2*x])/(29282*(3 + 5*x)) - (2873*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(14641*Sqrt[55])

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && 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 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 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 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 {(2+3 x)^2}{(1-2 x)^{5/2} (3+5 x)^3} \, dx &=\frac {49}{66 (1-2 x)^{3/2} (3+5 x)^2}-\frac {1}{66} \int \frac {-313+297 x}{(1-2 x)^{3/2} (3+5 x)^3} \, dx\\ &=\frac {49}{66 (1-2 x)^{3/2} (3+5 x)^2}-\frac {614}{1815 \sqrt {1-2 x} (3+5 x)^2}+\frac {2873 \int \frac {1}{(1-2 x)^{3/2} (3+5 x)^2} \, dx}{3630}\\ &=\frac {49}{66 (1-2 x)^{3/2} (3+5 x)^2}-\frac {614}{1815 \sqrt {1-2 x} (3+5 x)^2}+\frac {2873}{19965 \sqrt {1-2 x} (3+5 x)}+\frac {2873 \int \frac {1}{\sqrt {1-2 x} (3+5 x)^2} \, dx}{2662}\\ &=\frac {49}{66 (1-2 x)^{3/2} (3+5 x)^2}-\frac {614}{1815 \sqrt {1-2 x} (3+5 x)^2}+\frac {2873}{19965 \sqrt {1-2 x} (3+5 x)}-\frac {2873 \sqrt {1-2 x}}{29282 (3+5 x)}+\frac {2873 \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx}{29282}\\ &=\frac {49}{66 (1-2 x)^{3/2} (3+5 x)^2}-\frac {614}{1815 \sqrt {1-2 x} (3+5 x)^2}+\frac {2873}{19965 \sqrt {1-2 x} (3+5 x)}-\frac {2873 \sqrt {1-2 x}}{29282 (3+5 x)}-\frac {2873 \operatorname {Subst}\left (\int \frac {1}{\frac {11}{2}-\frac {5 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )}{29282}\\ &=\frac {49}{66 (1-2 x)^{3/2} (3+5 x)^2}-\frac {614}{1815 \sqrt {1-2 x} (3+5 x)^2}+\frac {2873}{19965 \sqrt {1-2 x} (3+5 x)}-\frac {2873 \sqrt {1-2 x}}{29282 (3+5 x)}-\frac {2873 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{14641 \sqrt {55}}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 59, normalized size = 0.58 \begin {gather*} -\frac {11492 (2 x-1) (5 x+3)^2 \, _2F_1\left (-\frac {1}{2},2;\frac {1}{2};-\frac {5}{11} (2 x-1)\right )-121 (2456 x+1467)}{439230 (1-2 x)^{3/2} (5 x+3)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/439230*(-121*(1467 + 2456*x) + 11492*(-1 + 2*x)*(3 + 5*x)^2*Hypergeometric2F1[-1/2, 2, 1/2, (-5*(-1 + 2*x))
/11])/((1 - 2*x)^(3/2)*(3 + 5*x)^2)

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IntegrateAlgebraic [A]  time = 0.19, size = 79, normalized size = 0.78 \begin {gather*} \frac {43095 (1-2 x)^3-158015 (1-2 x)^2+79618 (1-2 x)+130438}{43923 (5 (1-2 x)-11)^2 (1-2 x)^{3/2}}-\frac {2873 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{14641 \sqrt {55}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(130438 + 79618*(1 - 2*x) - 158015*(1 - 2*x)^2 + 43095*(1 - 2*x)^3)/(43923*(-11 + 5*(1 - 2*x))^2*(1 - 2*x)^(3/
2)) - (2873*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(14641*Sqrt[55])

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fricas [A]  time = 1.53, size = 99, normalized size = 0.98 \begin {gather*} \frac {8619 \, \sqrt {55} {\left (100 \, x^{4} + 20 \, x^{3} - 59 \, x^{2} - 6 \, x + 9\right )} \log \left (\frac {5 \, x + \sqrt {55} \sqrt {-2 \, x + 1} - 8}{5 \, x + 3}\right ) - 55 \, {\left (172380 \, x^{3} + 57460 \, x^{2} - 107127 \, x - 47568\right )} \sqrt {-2 \, x + 1}}{4831530 \, {\left (100 \, x^{4} + 20 \, x^{3} - 59 \, x^{2} - 6 \, x + 9\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4831530*(8619*sqrt(55)*(100*x^4 + 20*x^3 - 59*x^2 - 6*x + 9)*log((5*x + sqrt(55)*sqrt(-2*x + 1) - 8)/(5*x +
3)) - 55*(172380*x^3 + 57460*x^2 - 107127*x - 47568)*sqrt(-2*x + 1))/(100*x^4 + 20*x^3 - 59*x^2 - 6*x + 9)

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giac [A]  time = 1.25, size = 89, normalized size = 0.88 \begin {gather*} \frac {2873}{1610510} \, \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 {28 \, {\left (117 \, x - 97\right )}}{43923 \, {\left (2 \, x - 1\right )} \sqrt {-2 \, x + 1}} + \frac {5 \, {\left (13 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 29 \, \sqrt {-2 \, x + 1}\right )}}{5324 \, {\left (5 \, x + 3\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2873/1610510*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 28/43923*(
117*x - 97)/((2*x - 1)*sqrt(-2*x + 1)) + 5/5324*(13*(-2*x + 1)^(3/2) - 29*sqrt(-2*x + 1))/(5*x + 3)^2

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maple [A]  time = 0.02, size = 66, normalized size = 0.65 \begin {gather*} -\frac {2873 \sqrt {55}\, \arctanh \left (\frac {\sqrt {55}\, \sqrt {-2 x +1}}{11}\right )}{805255}+\frac {98}{3993 \left (-2 x +1\right )^{\frac {3}{2}}}+\frac {546}{14641 \sqrt {-2 x +1}}+\frac {\frac {65 \left (-2 x +1\right )^{\frac {3}{2}}}{1331}-\frac {145 \sqrt {-2 x +1}}{1331}}{\left (-10 x -6\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

98/3993/(-2*x+1)^(3/2)+546/14641/(-2*x+1)^(1/2)+50/14641*(143/10*(-2*x+1)^(3/2)-319/10*(-2*x+1)^(1/2))/(-10*x-
6)^2-2873/805255*arctanh(1/11*55^(1/2)*(-2*x+1)^(1/2))*55^(1/2)

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maxima [A]  time = 1.11, size = 92, normalized size = 0.91 \begin {gather*} \frac {2873}{1610510} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) - \frac {43095 \, {\left (2 \, x - 1\right )}^{3} + 158015 \, {\left (2 \, x - 1\right )}^{2} + 159236 \, x - 210056}{43923 \, {\left (25 \, {\left (-2 \, x + 1\right )}^{\frac {7}{2}} - 110 \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} + 121 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2873/1610510*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 1/43923*(43095*(2*x
- 1)^3 + 158015*(2*x - 1)^2 + 159236*x - 210056)/(25*(-2*x + 1)^(7/2) - 110*(-2*x + 1)^(5/2) + 121*(-2*x + 1)^
(3/2))

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mupad [B]  time = 1.22, size = 72, normalized size = 0.71 \begin {gather*} -\frac {2873\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )}{805255}-\frac {\frac {1316\,x}{9075}+\frac {2873\,{\left (2\,x-1\right )}^2}{19965}+\frac {2873\,{\left (2\,x-1\right )}^3}{73205}-\frac {1736}{9075}}{\frac {121\,{\left (1-2\,x\right )}^{3/2}}{25}-\frac {22\,{\left (1-2\,x\right )}^{5/2}}{5}+{\left (1-2\,x\right )}^{7/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- (2873*55^(1/2)*atanh((55^(1/2)*(1 - 2*x)^(1/2))/11))/805255 - ((1316*x)/9075 + (2873*(2*x - 1)^2)/19965 + (2
873*(2*x - 1)^3)/73205 - 1736/9075)/((121*(1 - 2*x)^(3/2))/25 - (22*(1 - 2*x)^(5/2))/5 + (1 - 2*x)^(7/2))

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

[Out]

Timed out

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