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

Optimal. Leaf size=61 \[ \frac {64}{147 \sqrt {1-2 x}}+\frac {1}{21 \sqrt {1-2 x} (3 x+2)}-\frac {64 \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{49 \sqrt {21}} \]

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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 {64}{147 \sqrt {1-2 x}}+\frac {1}{21 \sqrt {1-2 x} (3 x+2)}-\frac {64 \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{49 \sqrt {21}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

64/(147*Sqrt[1 - 2*x]) + 1/(21*Sqrt[1 - 2*x]*(2 + 3*x)) - (64*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]])/(49*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}{(1-2 x)^{3/2} (2+3 x)^2} \, dx &=\frac {1}{21 \sqrt {1-2 x} (2+3 x)}+\frac {32}{21} \int \frac {1}{(1-2 x)^{3/2} (2+3 x)} \, dx\\ &=\frac {64}{147 \sqrt {1-2 x}}+\frac {1}{21 \sqrt {1-2 x} (2+3 x)}+\frac {32}{49} \int \frac {1}{\sqrt {1-2 x} (2+3 x)} \, dx\\ &=\frac {64}{147 \sqrt {1-2 x}}+\frac {1}{21 \sqrt {1-2 x} (2+3 x)}-\frac {32}{49} \operatorname {Subst}\left (\int \frac {1}{\frac {7}{2}-\frac {3 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )\\ &=\frac {64}{147 \sqrt {1-2 x}}+\frac {1}{21 \sqrt {1-2 x} (2+3 x)}-\frac {64 \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{49 \sqrt {21}}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 46, normalized size = 0.75 \begin {gather*} \frac {64 (3 x+2) \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\frac {3}{7}-\frac {6 x}{7}\right )+7}{147 \sqrt {1-2 x} (3 x+2)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(7 + 64*(2 + 3*x)*Hypergeometric2F1[-1/2, 1, 1/2, 3/7 - (6*x)/7])/(147*Sqrt[1 - 2*x]*(2 + 3*x))

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IntegrateAlgebraic [A]  time = 0.13, size = 61, normalized size = 1.00 \begin {gather*} \frac {2 (32 (1-2 x)-77)}{49 (3 (1-2 x)-7) \sqrt {1-2 x}}-\frac {64 \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{49 \sqrt {21}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(-77 + 32*(1 - 2*x)))/(49*(-7 + 3*(1 - 2*x))*Sqrt[1 - 2*x]) - (64*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]])/(49*Sqr
t[21])

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fricas [A]  time = 1.71, size = 65, normalized size = 1.07 \begin {gather*} \frac {32 \, \sqrt {21} {\left (6 \, x^{2} + x - 2\right )} \log \left (\frac {3 \, x + \sqrt {21} \sqrt {-2 \, x + 1} - 5}{3 \, x + 2}\right ) - 21 \, {\left (64 \, x + 45\right )} \sqrt {-2 \, x + 1}}{1029 \, {\left (6 \, x^{2} + x - 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1029*(32*sqrt(21)*(6*x^2 + x - 2)*log((3*x + sqrt(21)*sqrt(-2*x + 1) - 5)/(3*x + 2)) - 21*(64*x + 45)*sqrt(-
2*x + 1))/(6*x^2 + x - 2)

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giac [A]  time = 1.21, size = 68, normalized size = 1.11 \begin {gather*} \frac {32}{1029} \, \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 {2 \, {\left (64 \, x + 45\right )}}{49 \, {\left (3 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 7 \, \sqrt {-2 \, x + 1}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

32/1029*sqrt(21)*log(1/2*abs(-2*sqrt(21) + 6*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) - 2/49*(64*x + 45)
/(3*(-2*x + 1)^(3/2) - 7*sqrt(-2*x + 1))

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maple [A]  time = 0.01, size = 45, normalized size = 0.74 \begin {gather*} -\frac {64 \sqrt {21}\, \arctanh \left (\frac {\sqrt {21}\, \sqrt {-2 x +1}}{7}\right )}{1029}+\frac {22}{49 \sqrt {-2 x +1}}-\frac {2 \sqrt {-2 x +1}}{147 \left (-2 x -\frac {4}{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

22/49/(-2*x+1)^(1/2)-2/147*(-2*x+1)^(1/2)/(-2*x-4/3)-64/1029*arctanh(1/7*21^(1/2)*(-2*x+1)^(1/2))*21^(1/2)

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maxima [A]  time = 1.14, size = 65, normalized size = 1.07 \begin {gather*} \frac {32}{1029} \, \sqrt {21} \log \left (-\frac {\sqrt {21} - 3 \, \sqrt {-2 \, x + 1}}{\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}}\right ) - \frac {2 \, {\left (64 \, x + 45\right )}}{49 \, {\left (3 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 7 \, \sqrt {-2 \, x + 1}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

32/1029*sqrt(21)*log(-(sqrt(21) - 3*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) - 2/49*(64*x + 45)/(3*(-2*x
+ 1)^(3/2) - 7*sqrt(-2*x + 1))

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mupad [B]  time = 1.24, size = 46, normalized size = 0.75 \begin {gather*} \frac {\frac {128\,x}{147}+\frac {30}{49}}{\frac {7\,\sqrt {1-2\,x}}{3}-{\left (1-2\,x\right )}^{3/2}}-\frac {64\,\sqrt {21}\,\mathrm {atanh}\left (\frac {\sqrt {21}\,\sqrt {1-2\,x}}{7}\right )}{1029} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((128*x)/147 + 30/49)/((7*(1 - 2*x)^(1/2))/3 - (1 - 2*x)^(3/2)) - (64*21^(1/2)*atanh((21^(1/2)*(1 - 2*x)^(1/2)
)/7))/1029

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

[Out]

Timed out

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