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

Optimal. Leaf size=81 \[ \frac {213}{6655 \sqrt {1-2 x}}-\frac {71}{1210 \sqrt {1-2 x} (5 x+3)}-\frac {1}{110 \sqrt {1-2 x} (5 x+3)^2}-\frac {213 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{1331 \sqrt {55}} \]

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Rubi [A]  time = 0.02, antiderivative size = 88, normalized size of antiderivative = 1.09, number of steps used = 5, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {78, 51, 63, 206} \begin {gather*} -\frac {213 \sqrt {1-2 x}}{2662 (5 x+3)}+\frac {71}{605 \sqrt {1-2 x} (5 x+3)}-\frac {1}{110 \sqrt {1-2 x} (5 x+3)^2}-\frac {213 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{1331 \sqrt {55}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/(110*Sqrt[1 - 2*x]*(3 + 5*x)^2) + 71/(605*Sqrt[1 - 2*x]*(3 + 5*x)) - (213*Sqrt[1 - 2*x])/(2662*(3 + 5*x)) -
(213*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(1331*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 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}{(1-2 x)^{3/2} (3+5 x)^3} \, dx &=-\frac {1}{110 \sqrt {1-2 x} (3+5 x)^2}+\frac {71}{110} \int \frac {1}{(1-2 x)^{3/2} (3+5 x)^2} \, dx\\ &=-\frac {1}{110 \sqrt {1-2 x} (3+5 x)^2}+\frac {71}{605 \sqrt {1-2 x} (3+5 x)}+\frac {213}{242} \int \frac {1}{\sqrt {1-2 x} (3+5 x)^2} \, dx\\ &=-\frac {1}{110 \sqrt {1-2 x} (3+5 x)^2}+\frac {71}{605 \sqrt {1-2 x} (3+5 x)}-\frac {213 \sqrt {1-2 x}}{2662 (3+5 x)}+\frac {213 \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx}{2662}\\ &=-\frac {1}{110 \sqrt {1-2 x} (3+5 x)^2}+\frac {71}{605 \sqrt {1-2 x} (3+5 x)}-\frac {213 \sqrt {1-2 x}}{2662 (3+5 x)}-\frac {213 \operatorname {Subst}\left (\int \frac {1}{\frac {11}{2}-\frac {5 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )}{2662}\\ &=-\frac {1}{110 \sqrt {1-2 x} (3+5 x)^2}+\frac {71}{605 \sqrt {1-2 x} (3+5 x)}-\frac {213 \sqrt {1-2 x}}{2662 (3+5 x)}-\frac {213 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{1331 \sqrt {55}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 48, normalized size = 0.59 \begin {gather*} \frac {284 (5 x+3)^2 \, _2F_1\left (-\frac {1}{2},2;\frac {1}{2};-\frac {5}{11} (2 x-1)\right )-121}{13310 \sqrt {1-2 x} (5 x+3)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-121 + 284*(3 + 5*x)^2*Hypergeometric2F1[-1/2, 2, 1/2, (-5*(-1 + 2*x))/11])/(13310*Sqrt[1 - 2*x]*(3 + 5*x)^2)

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IntegrateAlgebraic [A]  time = 0.19, size = 70, normalized size = 0.86 \begin {gather*} \frac {1065 (1-2 x)^2-3905 (1-2 x)+3388}{1331 (5 (1-2 x)-11)^2 \sqrt {1-2 x}}-\frac {213 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{1331 \sqrt {55}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3388 - 3905*(1 - 2*x) + 1065*(1 - 2*x)^2)/(1331*(-11 + 5*(1 - 2*x))^2*Sqrt[1 - 2*x]) - (213*ArcTanh[Sqrt[5/11
]*Sqrt[1 - 2*x]])/(1331*Sqrt[55])

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fricas [A]  time = 1.07, size = 84, normalized size = 1.04 \begin {gather*} \frac {213 \, \sqrt {55} {\left (50 \, x^{3} + 35 \, x^{2} - 12 \, x - 9\right )} \log \left (\frac {5 \, x + \sqrt {55} \sqrt {-2 \, x + 1} - 8}{5 \, x + 3}\right ) - 55 \, {\left (2130 \, x^{2} + 1775 \, x + 274\right )} \sqrt {-2 \, x + 1}}{146410 \, {\left (50 \, x^{3} + 35 \, x^{2} - 12 \, x - 9\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/146410*(213*sqrt(55)*(50*x^3 + 35*x^2 - 12*x - 9)*log((5*x + sqrt(55)*sqrt(-2*x + 1) - 8)/(5*x + 3)) - 55*(2
130*x^2 + 1775*x + 274)*sqrt(-2*x + 1))/(50*x^3 + 35*x^2 - 12*x - 9)

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giac [A]  time = 1.20, size = 77, normalized size = 0.95 \begin {gather*} \frac {213}{146410} \, \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}{1331 \, \sqrt {-2 \, x + 1}} + \frac {5 \, {\left (73 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 165 \, \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)/(1-2*x)^(3/2)/(3+5*x)^3,x, algorithm="giac")

[Out]

213/146410*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 28/1331/sqrt
(-2*x + 1) + 5/5324*(73*(-2*x + 1)^(3/2) - 165*sqrt(-2*x + 1))/(5*x + 3)^2

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maple [A]  time = 0.01, size = 57, normalized size = 0.70 \begin {gather*} -\frac {213 \sqrt {55}\, \arctanh \left (\frac {\sqrt {55}\, \sqrt {-2 x +1}}{11}\right )}{73205}+\frac {28}{1331 \sqrt {-2 x +1}}+\frac {\frac {365 \left (-2 x +1\right )^{\frac {3}{2}}}{1331}-\frac {75 \sqrt {-2 x +1}}{121}}{\left (-10 x -6\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

28/1331/(-2*x+1)^(1/2)+100/1331*(73/20*(-2*x+1)^(3/2)-33/4*(-2*x+1)^(1/2))/(-10*x-6)^2-213/73205*arctanh(1/11*
55^(1/2)*(-2*x+1)^(1/2))*55^(1/2)

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maxima [A]  time = 1.38, size = 83, normalized size = 1.02 \begin {gather*} \frac {213}{146410} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) + \frac {1065 \, {\left (2 \, x - 1\right )}^{2} + 7810 \, x - 517}{1331 \, {\left (25 \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} - 110 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + 121 \, \sqrt {-2 \, x + 1}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

213/146410*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 1/1331*(1065*(2*x - 1)
^2 + 7810*x - 517)/(25*(-2*x + 1)^(5/2) - 110*(-2*x + 1)^(3/2) + 121*sqrt(-2*x + 1))

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mupad [B]  time = 0.06, size = 62, normalized size = 0.77 \begin {gather*} \frac {\frac {142\,x}{605}+\frac {213\,{\left (2\,x-1\right )}^2}{6655}-\frac {47}{3025}}{\frac {121\,\sqrt {1-2\,x}}{25}-\frac {22\,{\left (1-2\,x\right )}^{3/2}}{5}+{\left (1-2\,x\right )}^{5/2}}-\frac {213\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )}{73205} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((142*x)/605 + (213*(2*x - 1)^2)/6655 - 47/3025)/((121*(1 - 2*x)^(1/2))/25 - (22*(1 - 2*x)^(3/2))/5 + (1 - 2*x
)^(5/2)) - (213*55^(1/2)*atanh((55^(1/2)*(1 - 2*x)^(1/2))/11))/73205

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

[Out]

Timed out

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