3.20.41 \(\int \frac {3+5 x}{(1-2 x)^{3/2} (2+3 x)^4} \, dx\)

Optimal. Leaf size=101 \[ \frac {10}{147 \sqrt {1-2 x}}-\frac {5}{63 \sqrt {1-2 x} (3 x+2)}-\frac {1}{9 \sqrt {1-2 x} (3 x+2)^2}+\frac {1}{63 \sqrt {1-2 x} (3 x+2)^3}-\frac {10 \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{49 \sqrt {21}} \]

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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 = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {78, 51, 63, 206} \begin {gather*} -\frac {5 \sqrt {1-2 x}}{49 (3 x+2)}-\frac {5 \sqrt {1-2 x}}{21 (3 x+2)^2}+\frac {4}{9 \sqrt {1-2 x} (3 x+2)^2}+\frac {1}{63 \sqrt {1-2 x} (3 x+2)^3}-\frac {10 \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)^4),x]

[Out]

1/(63*Sqrt[1 - 2*x]*(2 + 3*x)^3) + 4/(9*Sqrt[1 - 2*x]*(2 + 3*x)^2) - (5*Sqrt[1 - 2*x])/(21*(2 + 3*x)^2) - (5*S
qrt[1 - 2*x])/(49*(2 + 3*x)) - (10*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)^4} \, dx &=\frac {1}{63 \sqrt {1-2 x} (2+3 x)^3}+\frac {14}{9} \int \frac {1}{(1-2 x)^{3/2} (2+3 x)^3} \, dx\\ &=\frac {1}{63 \sqrt {1-2 x} (2+3 x)^3}+\frac {4}{9 \sqrt {1-2 x} (2+3 x)^2}+\frac {10}{3} \int \frac {1}{\sqrt {1-2 x} (2+3 x)^3} \, dx\\ &=\frac {1}{63 \sqrt {1-2 x} (2+3 x)^3}+\frac {4}{9 \sqrt {1-2 x} (2+3 x)^2}-\frac {5 \sqrt {1-2 x}}{21 (2+3 x)^2}+\frac {5}{7} \int \frac {1}{\sqrt {1-2 x} (2+3 x)^2} \, dx\\ &=\frac {1}{63 \sqrt {1-2 x} (2+3 x)^3}+\frac {4}{9 \sqrt {1-2 x} (2+3 x)^2}-\frac {5 \sqrt {1-2 x}}{21 (2+3 x)^2}-\frac {5 \sqrt {1-2 x}}{49 (2+3 x)}+\frac {5}{49} \int \frac {1}{\sqrt {1-2 x} (2+3 x)} \, dx\\ &=\frac {1}{63 \sqrt {1-2 x} (2+3 x)^3}+\frac {4}{9 \sqrt {1-2 x} (2+3 x)^2}-\frac {5 \sqrt {1-2 x}}{21 (2+3 x)^2}-\frac {5 \sqrt {1-2 x}}{49 (2+3 x)}-\frac {5}{49} \operatorname {Subst}\left (\int \frac {1}{\frac {7}{2}-\frac {3 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )\\ &=\frac {1}{63 \sqrt {1-2 x} (2+3 x)^3}+\frac {4}{9 \sqrt {1-2 x} (2+3 x)^2}-\frac {5 \sqrt {1-2 x}}{21 (2+3 x)^2}-\frac {5 \sqrt {1-2 x}}{49 (2+3 x)}-\frac {10 \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 = 42, normalized size = 0.42 \begin {gather*} \frac {16 \, _2F_1\left (-\frac {1}{2},3;\frac {1}{2};\frac {3}{7}-\frac {6 x}{7}\right )+\frac {7}{(3 x+2)^3}}{441 \sqrt {1-2 x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(3 + 5*x)/((1 - 2*x)^(3/2)*(2 + 3*x)^4),x]

[Out]

(7/(2 + 3*x)^3 + 16*Hypergeometric2F1[-1/2, 3, 1/2, 3/7 - (6*x)/7])/(441*Sqrt[1 - 2*x])

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IntegrateAlgebraic [A]  time = 0.24, size = 79, normalized size = 0.78 \begin {gather*} \frac {2 \left (45 (1-2 x)^3-280 (1-2 x)^2+539 (1-2 x)-308\right )}{49 (3 (1-2 x)-7)^3 \sqrt {1-2 x}}-\frac {10 \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)^4),x]

[Out]

(2*(-308 + 539*(1 - 2*x) - 280*(1 - 2*x)^2 + 45*(1 - 2*x)^3))/(49*(-7 + 3*(1 - 2*x))^3*Sqrt[1 - 2*x]) - (10*Ar
cTanh[Sqrt[3/7]*Sqrt[1 - 2*x]])/(49*Sqrt[21])

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fricas [A]  time = 0.97, size = 99, normalized size = 0.98 \begin {gather*} \frac {5 \, \sqrt {21} {\left (54 \, x^{4} + 81 \, x^{3} + 18 \, x^{2} - 20 \, x - 8\right )} \log \left (\frac {3 \, x + \sqrt {21} \sqrt {-2 \, x + 1} - 5}{3 \, x + 2}\right ) - 21 \, {\left (90 \, x^{3} + 145 \, x^{2} + 57 \, x + 1\right )} \sqrt {-2 \, x + 1}}{1029 \, {\left (54 \, x^{4} + 81 \, x^{3} + 18 \, x^{2} - 20 \, x - 8\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)^4,x, algorithm="fricas")

[Out]

1/1029*(5*sqrt(21)*(54*x^4 + 81*x^3 + 18*x^2 - 20*x - 8)*log((3*x + sqrt(21)*sqrt(-2*x + 1) - 5)/(3*x + 2)) -
21*(90*x^3 + 145*x^2 + 57*x + 1)*sqrt(-2*x + 1))/(54*x^4 + 81*x^3 + 18*x^2 - 20*x - 8)

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giac [A]  time = 1.28, size = 93, normalized size = 0.92 \begin {gather*} \frac {5}{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 {88}{2401 \, \sqrt {-2 \, x + 1}} - \frac {1017 \, {\left (2 \, x - 1\right )}^{2} \sqrt {-2 \, x + 1} - 5404 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + 7007 \, \sqrt {-2 \, x + 1}}{9604 \, {\left (3 \, x + 2\right )}^{3}} \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)^4,x, algorithm="giac")

[Out]

5/1029*sqrt(21)*log(1/2*abs(-2*sqrt(21) + 6*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) + 88/2401/sqrt(-2*x
+ 1) - 1/9604*(1017*(2*x - 1)^2*sqrt(-2*x + 1) - 5404*(-2*x + 1)^(3/2) + 7007*sqrt(-2*x + 1))/(3*x + 2)^3

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maple [A]  time = 0.01, size = 66, normalized size = 0.65 \begin {gather*} -\frac {10 \sqrt {21}\, \arctanh \left (\frac {\sqrt {21}\, \sqrt {-2 x +1}}{7}\right )}{1029}+\frac {88}{2401 \sqrt {-2 x +1}}+\frac {\frac {2034 \left (-2 x +1\right )^{\frac {5}{2}}}{2401}-\frac {1544 \left (-2 x +1\right )^{\frac {3}{2}}}{343}+\frac {286 \sqrt {-2 x +1}}{49}}{\left (-6 x -4\right )^{3}} \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)^4,x)

[Out]

88/2401/(-2*x+1)^(1/2)+216/2401*(113/12*(-2*x+1)^(5/2)-1351/27*(-2*x+1)^(3/2)+7007/108*(-2*x+1)^(1/2))/(-6*x-4
)^3-10/1029*arctanh(1/7*21^(1/2)*(-2*x+1)^(1/2))*21^(1/2)

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maxima [A]  time = 1.47, size = 101, normalized size = 1.00 \begin {gather*} \frac {5}{1029} \, \sqrt {21} \log \left (-\frac {\sqrt {21} - 3 \, \sqrt {-2 \, x + 1}}{\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}}\right ) - \frac {2 \, {\left (45 \, {\left (2 \, x - 1\right )}^{3} + 280 \, {\left (2 \, x - 1\right )}^{2} + 1078 \, x - 231\right )}}{49 \, {\left (27 \, {\left (-2 \, x + 1\right )}^{\frac {7}{2}} - 189 \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} + 441 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 343 \, \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)^4,x, algorithm="maxima")

[Out]

5/1029*sqrt(21)*log(-(sqrt(21) - 3*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) - 2/49*(45*(2*x - 1)^3 + 280
*(2*x - 1)^2 + 1078*x - 231)/(27*(-2*x + 1)^(7/2) - 189*(-2*x + 1)^(5/2) + 441*(-2*x + 1)^(3/2) - 343*sqrt(-2*
x + 1))

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mupad [B]  time = 0.07, size = 82, normalized size = 0.81 \begin {gather*} \frac {\frac {44\,x}{27}+\frac {80\,{\left (2\,x-1\right )}^2}{189}+\frac {10\,{\left (2\,x-1\right )}^3}{147}-\frac {22}{63}}{\frac {343\,\sqrt {1-2\,x}}{27}-\frac {49\,{\left (1-2\,x\right )}^{3/2}}{3}+7\,{\left (1-2\,x\right )}^{5/2}-{\left (1-2\,x\right )}^{7/2}}-\frac {10\,\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)^4),x)

[Out]

((44*x)/27 + (80*(2*x - 1)^2)/189 + (10*(2*x - 1)^3)/147 - 22/63)/((343*(1 - 2*x)^(1/2))/27 - (49*(1 - 2*x)^(3
/2))/3 + 7*(1 - 2*x)^(5/2) - (1 - 2*x)^(7/2)) - (10*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)**4,x)

[Out]

Timed out

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