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

Optimal. Leaf size=61 \[ \frac {72}{605 \sqrt {1-2 x}}-\frac {1}{55 \sqrt {1-2 x} (5 x+3)}-\frac {72 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{121 \sqrt {55}} \]

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Rubi [A]  time = 0.01, antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 4, 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 {72}{605 \sqrt {1-2 x}}-\frac {1}{55 \sqrt {1-2 x} (5 x+3)}-\frac {72 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{121 \sqrt {55}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

72/(605*Sqrt[1 - 2*x]) - 1/(55*Sqrt[1 - 2*x]*(3 + 5*x)) - (72*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(121*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)^2} \, dx &=-\frac {1}{55 \sqrt {1-2 x} (3+5 x)}+\frac {36}{55} \int \frac {1}{(1-2 x)^{3/2} (3+5 x)} \, dx\\ &=\frac {72}{605 \sqrt {1-2 x}}-\frac {1}{55 \sqrt {1-2 x} (3+5 x)}+\frac {36}{121} \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx\\ &=\frac {72}{605 \sqrt {1-2 x}}-\frac {1}{55 \sqrt {1-2 x} (3+5 x)}-\frac {36}{121} \operatorname {Subst}\left (\int \frac {1}{\frac {11}{2}-\frac {5 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )\\ &=\frac {72}{605 \sqrt {1-2 x}}-\frac {1}{55 \sqrt {1-2 x} (3+5 x)}-\frac {72 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{121 \sqrt {55}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 46, normalized size = 0.75 \begin {gather*} \frac {72 (5 x+3) \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};-\frac {5}{11} (2 x-1)\right )-11}{605 \sqrt {1-2 x} (5 x+3)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-11 + 72*(3 + 5*x)*Hypergeometric2F1[-1/2, 1, 1/2, (-5*(-1 + 2*x))/11])/(605*Sqrt[1 - 2*x]*(3 + 5*x))

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IntegrateAlgebraic [A]  time = 0.12, size = 61, normalized size = 1.00 \begin {gather*} \frac {2 (36 (1-2 x)-77)}{121 (5 (1-2 x)-11) \sqrt {1-2 x}}-\frac {72 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{121 \sqrt {55}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(-77 + 36*(1 - 2*x)))/(121*(-11 + 5*(1 - 2*x))*Sqrt[1 - 2*x]) - (72*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(121
*Sqrt[55])

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fricas [A]  time = 1.31, size = 65, normalized size = 1.07 \begin {gather*} \frac {36 \, \sqrt {55} {\left (10 \, x^{2} + x - 3\right )} \log \left (\frac {5 \, x + \sqrt {55} \sqrt {-2 \, x + 1} - 8}{5 \, x + 3}\right ) - 55 \, {\left (72 \, x + 41\right )} \sqrt {-2 \, x + 1}}{6655 \, {\left (10 \, x^{2} + x - 3\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)^2,x, algorithm="fricas")

[Out]

1/6655*(36*sqrt(55)*(10*x^2 + x - 3)*log((5*x + sqrt(55)*sqrt(-2*x + 1) - 8)/(5*x + 3)) - 55*(72*x + 41)*sqrt(
-2*x + 1))/(10*x^2 + x - 3)

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giac [A]  time = 1.21, size = 68, normalized size = 1.11 \begin {gather*} \frac {36}{6655} \, \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 {2 \, {\left (72 \, x + 41\right )}}{121 \, {\left (5 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 11 \, \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)^2,x, algorithm="giac")

[Out]

36/6655*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 2/121*(72*x + 4
1)/(5*(-2*x + 1)^(3/2) - 11*sqrt(-2*x + 1))

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maple [A]  time = 0.01, size = 45, normalized size = 0.74 \begin {gather*} -\frac {72 \sqrt {55}\, \arctanh \left (\frac {\sqrt {55}\, \sqrt {-2 x +1}}{11}\right )}{6655}+\frac {14}{121 \sqrt {-2 x +1}}+\frac {2 \sqrt {-2 x +1}}{605 \left (-2 x -\frac {6}{5}\right )} \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)^2,x)

[Out]

14/121/(-2*x+1)^(1/2)+2/605*(-2*x+1)^(1/2)/(-2*x-6/5)-72/6655*arctanh(1/11*55^(1/2)*(-2*x+1)^(1/2))*55^(1/2)

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maxima [A]  time = 1.20, size = 65, normalized size = 1.07 \begin {gather*} \frac {36}{6655} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) - \frac {2 \, {\left (72 \, x + 41\right )}}{121 \, {\left (5 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 11 \, \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)^2,x, algorithm="maxima")

[Out]

36/6655*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 2/121*(72*x + 41)/(5*(-2*
x + 1)^(3/2) - 11*sqrt(-2*x + 1))

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mupad [B]  time = 0.06, size = 46, normalized size = 0.75 \begin {gather*} \frac {\frac {144\,x}{605}+\frac {82}{605}}{\frac {11\,\sqrt {1-2\,x}}{5}-{\left (1-2\,x\right )}^{3/2}}-\frac {72\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )}{6655} \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)^2),x)

[Out]

((144*x)/605 + 82/605)/((11*(1 - 2*x)^(1/2))/5 - (1 - 2*x)^(3/2)) - (72*55^(1/2)*atanh((55^(1/2)*(1 - 2*x)^(1/
2))/11))/6655

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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)**2,x)

[Out]

Timed out

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