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

Optimal. Leaf size=68 \[ -\frac {67 \sqrt {1-2 x}}{294 (3 x+2)}+\frac {\sqrt {1-2 x}}{42 (3 x+2)^2}-\frac {67 \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{147 \sqrt {21}} \]

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Rubi [A]  time = 0.01, antiderivative size = 68, 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 {67 \sqrt {1-2 x}}{294 (3 x+2)}+\frac {\sqrt {1-2 x}}{42 (3 x+2)^2}-\frac {67 \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{147 \sqrt {21}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

Sqrt[1 - 2*x]/(42*(2 + 3*x)^2) - (67*Sqrt[1 - 2*x])/(294*(2 + 3*x)) - (67*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]])/(1
47*Sqrt[21])

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 {3+5 x}{\sqrt {1-2 x} (2+3 x)^3} \, dx &=\frac {\sqrt {1-2 x}}{42 (2+3 x)^2}+\frac {67}{42} \int \frac {1}{\sqrt {1-2 x} (2+3 x)^2} \, dx\\ &=\frac {\sqrt {1-2 x}}{42 (2+3 x)^2}-\frac {67 \sqrt {1-2 x}}{294 (2+3 x)}+\frac {67}{294} \int \frac {1}{\sqrt {1-2 x} (2+3 x)} \, dx\\ &=\frac {\sqrt {1-2 x}}{42 (2+3 x)^2}-\frac {67 \sqrt {1-2 x}}{294 (2+3 x)}-\frac {67}{294} \operatorname {Subst}\left (\int \frac {1}{\frac {7}{2}-\frac {3 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )\\ &=\frac {\sqrt {1-2 x}}{42 (2+3 x)^2}-\frac {67 \sqrt {1-2 x}}{294 (2+3 x)}-\frac {67 \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{147 \sqrt {21}}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 62, normalized size = 0.91 \begin {gather*} \frac {\sqrt {1-2 x} \left (-\frac {21 (201 x+127)}{(3 x+2)^2}-\frac {134 \sqrt {21} \tan ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {2 x-1}\right )}{\sqrt {2 x-1}}\right )}{6174} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[1 - 2*x]*((-21*(127 + 201*x))/(2 + 3*x)^2 - (134*Sqrt[21]*ArcTan[Sqrt[3/7]*Sqrt[-1 + 2*x]])/Sqrt[-1 + 2*
x]))/6174

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IntegrateAlgebraic [A]  time = 0.16, size = 61, normalized size = 0.90 \begin {gather*} \frac {(201 (1-2 x)-455) \sqrt {1-2 x}}{147 (3 (1-2 x)-7)^2}-\frac {67 \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{147 \sqrt {21}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-455 + 201*(1 - 2*x))*Sqrt[1 - 2*x])/(147*(-7 + 3*(1 - 2*x))^2) - (67*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]])/(147
*Sqrt[21])

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fricas [A]  time = 1.53, size = 69, normalized size = 1.01 \begin {gather*} \frac {67 \, \sqrt {21} {\left (9 \, x^{2} + 12 \, x + 4\right )} \log \left (\frac {3 \, x + \sqrt {21} \sqrt {-2 \, x + 1} - 5}{3 \, x + 2}\right ) - 21 \, {\left (201 \, x + 127\right )} \sqrt {-2 \, x + 1}}{6174 \, {\left (9 \, x^{2} + 12 \, x + 4\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6174*(67*sqrt(21)*(9*x^2 + 12*x + 4)*log((3*x + sqrt(21)*sqrt(-2*x + 1) - 5)/(3*x + 2)) - 21*(201*x + 127)*s
qrt(-2*x + 1))/(9*x^2 + 12*x + 4)

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giac [A]  time = 0.91, size = 68, normalized size = 1.00 \begin {gather*} \frac {67}{6174} \, \sqrt {21} \log \left (\frac {{\left | -2 \, \sqrt {21} + 6 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}\right )}}\right ) + \frac {201 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 455 \, \sqrt {-2 \, x + 1}}{588 \, {\left (3 \, x + 2\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

67/6174*sqrt(21)*log(1/2*abs(-2*sqrt(21) + 6*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) + 1/588*(201*(-2*x
 + 1)^(3/2) - 455*sqrt(-2*x + 1))/(3*x + 2)^2

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maple [A]  time = 0.01, size = 48, normalized size = 0.71 \begin {gather*} -\frac {67 \sqrt {21}\, \arctanh \left (\frac {\sqrt {21}\, \sqrt {-2 x +1}}{7}\right )}{3087}-\frac {36 \left (-\frac {67 \left (-2 x +1\right )^{\frac {3}{2}}}{1764}+\frac {65 \sqrt {-2 x +1}}{756}\right )}{\left (-6 x -4\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-36*(-67/1764*(-2*x+1)^(3/2)+65/756*(-2*x+1)^(1/2))/(-6*x-4)^2-67/3087*arctanh(1/7*21^(1/2)*(-2*x+1)^(1/2))*21
^(1/2)

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maxima [A]  time = 1.26, size = 74, normalized size = 1.09 \begin {gather*} \frac {67}{6174} \, \sqrt {21} \log \left (-\frac {\sqrt {21} - 3 \, \sqrt {-2 \, x + 1}}{\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}}\right ) + \frac {201 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 455 \, \sqrt {-2 \, x + 1}}{147 \, {\left (9 \, {\left (2 \, x - 1\right )}^{2} + 84 \, x + 7\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

67/6174*sqrt(21)*log(-(sqrt(21) - 3*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) + 1/147*(201*(-2*x + 1)^(3/
2) - 455*sqrt(-2*x + 1))/(9*(2*x - 1)^2 + 84*x + 7)

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mupad [B]  time = 0.06, size = 54, normalized size = 0.79 \begin {gather*} -\frac {67\,\sqrt {21}\,\mathrm {atanh}\left (\frac {\sqrt {21}\,\sqrt {1-2\,x}}{7}\right )}{3087}-\frac {\frac {65\,\sqrt {1-2\,x}}{189}-\frac {67\,{\left (1-2\,x\right )}^{3/2}}{441}}{\frac {28\,x}{3}+{\left (2\,x-1\right )}^2+\frac {7}{9}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- (67*21^(1/2)*atanh((21^(1/2)*(1 - 2*x)^(1/2))/7))/3087 - ((65*(1 - 2*x)^(1/2))/189 - (67*(1 - 2*x)^(3/2))/44
1)/((28*x)/3 + (2*x - 1)^2 + 7/9)

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

[Out]

Timed out

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