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

Optimal. Leaf size=67 \[ -\frac {27}{20} \sqrt {1-2 x}-\frac {784}{121 \sqrt {1-2 x}}+\frac {343}{132 (1-2 x)^{3/2}}-\frac {2 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{605 \sqrt {55}} \]

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Rubi [A]  time = 0.03, antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {87, 63, 206} \begin {gather*} -\frac {27}{20} \sqrt {1-2 x}-\frac {784}{121 \sqrt {1-2 x}}+\frac {343}{132 (1-2 x)^{3/2}}-\frac {2 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{605 \sqrt {55}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

343/(132*(1 - 2*x)^(3/2)) - 784/(121*Sqrt[1 - 2*x]) - (27*Sqrt[1 - 2*x])/20 - (2*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2
*x]])/(605*Sqrt[55])

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 87

Int[(((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_))/((a_.) + (b_.)*(x_)), x_Symbol] :> Int[ExpandIntegr
and[(e + f*x)^FractionalPart[p], ((c + d*x)^n*(e + f*x)^IntegerPart[p])/(a + b*x), x], x] /; FreeQ[{a, b, c, d
, e, f}, x] && IGtQ[n, 0] && LtQ[p, -1] && FractionQ[p]

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)^3}{(1-2 x)^{5/2} (3+5 x)} \, dx &=\int \left (\frac {343}{44 (1-2 x)^{5/2}}-\frac {784}{121 (1-2 x)^{3/2}}+\frac {27}{20 \sqrt {1-2 x}}+\frac {1}{605 \sqrt {1-2 x} (3+5 x)}\right ) \, dx\\ &=\frac {343}{132 (1-2 x)^{3/2}}-\frac {784}{121 \sqrt {1-2 x}}-\frac {27}{20} \sqrt {1-2 x}+\frac {1}{605} \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx\\ &=\frac {343}{132 (1-2 x)^{3/2}}-\frac {784}{121 \sqrt {1-2 x}}-\frac {27}{20} \sqrt {1-2 x}-\frac {1}{605} \operatorname {Subst}\left (\int \frac {1}{\frac {11}{2}-\frac {5 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )\\ &=\frac {343}{132 (1-2 x)^{3/2}}-\frac {784}{121 \sqrt {1-2 x}}-\frac {27}{20} \sqrt {1-2 x}-\frac {2 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{605 \sqrt {55}}\\ \end {align*}

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Mathematica [C]  time = 0.04, size = 45, normalized size = 0.67 \begin {gather*} \frac {2 \, _2F_1\left (-\frac {3}{2},1;-\frac {1}{2};\frac {5}{11} (1-2 x)\right )-99 \left (225 x^2-765 x+218\right )}{4125 (1-2 x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-99*(218 - 765*x + 225*x^2) + 2*Hypergeometric2F1[-3/2, 1, -1/2, (5*(1 - 2*x))/11])/(4125*(1 - 2*x)^(3/2))

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IntegrateAlgebraic [A]  time = 0.08, size = 59, normalized size = 0.88 \begin {gather*} \frac {-9801 (1-2 x)^2-47040 (1-2 x)+18865}{7260 (1-2 x)^{3/2}}-\frac {2 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{605 \sqrt {55}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(18865 - 47040*(1 - 2*x) - 9801*(1 - 2*x)^2)/(7260*(1 - 2*x)^(3/2)) - (2*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(6
05*Sqrt[55])

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fricas [A]  time = 1.33, size = 74, normalized size = 1.10 \begin {gather*} \frac {3 \, \sqrt {55} {\left (4 \, x^{2} - 4 \, x + 1\right )} \log \left (\frac {5 \, x + \sqrt {55} \sqrt {-2 \, x + 1} - 8}{5 \, x + 3}\right ) - 55 \, {\left (9801 \, x^{2} - 33321 \, x + 9494\right )} \sqrt {-2 \, x + 1}}{99825 \, {\left (4 \, x^{2} - 4 \, x + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/99825*(3*sqrt(55)*(4*x^2 - 4*x + 1)*log((5*x + sqrt(55)*sqrt(-2*x + 1) - 8)/(5*x + 3)) - 55*(9801*x^2 - 3332
1*x + 9494)*sqrt(-2*x + 1))/(4*x^2 - 4*x + 1)

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giac [A]  time = 1.54, size = 70, normalized size = 1.04 \begin {gather*} \frac {1}{33275} \, \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 {27}{20} \, \sqrt {-2 \, x + 1} - \frac {49 \, {\left (384 \, x - 115\right )}}{1452 \, {\left (2 \, x - 1\right )} \sqrt {-2 \, x + 1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/33275*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 27/20*sqrt(-2*x
+ 1) - 49/1452*(384*x - 115)/((2*x - 1)*sqrt(-2*x + 1))

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maple [A]  time = 0.01, size = 47, normalized size = 0.70 \begin {gather*} -\frac {2 \sqrt {55}\, \arctanh \left (\frac {\sqrt {55}\, \sqrt {-2 x +1}}{11}\right )}{33275}+\frac {343}{132 \left (-2 x +1\right )^{\frac {3}{2}}}-\frac {784}{121 \sqrt {-2 x +1}}-\frac {27 \sqrt {-2 x +1}}{20} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

343/132/(-2*x+1)^(3/2)-2/33275*arctanh(1/11*55^(1/2)*(-2*x+1)^(1/2))*55^(1/2)-784/121/(-2*x+1)^(1/2)-27/20*(-2
*x+1)^(1/2)

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maxima [A]  time = 1.33, size = 60, normalized size = 0.90 \begin {gather*} \frac {1}{33275} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) - \frac {27}{20} \, \sqrt {-2 \, x + 1} + \frac {49 \, {\left (384 \, x - 115\right )}}{1452 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/33275*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 27/20*sqrt(-2*x + 1) + 49
/1452*(384*x - 115)/(-2*x + 1)^(3/2)

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mupad [B]  time = 1.20, size = 41, normalized size = 0.61 \begin {gather*} \frac {\frac {1568\,x}{121}-\frac {5635}{1452}}{{\left (1-2\,x\right )}^{3/2}}-\frac {2\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )}{33275}-\frac {27\,\sqrt {1-2\,x}}{20} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((1568*x)/121 - 5635/1452)/(1 - 2*x)^(3/2) - (2*55^(1/2)*atanh((55^(1/2)*(1 - 2*x)^(1/2))/11))/33275 - (27*(1
- 2*x)^(1/2))/20

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sympy [A]  time = 60.07, size = 102, normalized size = 1.52 \begin {gather*} - \frac {27 \sqrt {1 - 2 x}}{20} + \frac {2 \left (\begin {cases} - \frac {\sqrt {55} \operatorname {acoth}{\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} \right )}}{55} & \text {for}\: 2 x - 1 < - \frac {11}{5} \\- \frac {\sqrt {55} \operatorname {atanh}{\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} \right )}}{55} & \text {for}\: 2 x - 1 > - \frac {11}{5} \end {cases}\right )}{605} - \frac {784}{121 \sqrt {1 - 2 x}} + \frac {343}{132 \left (1 - 2 x\right )^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-27*sqrt(1 - 2*x)/20 + 2*Piecewise((-sqrt(55)*acoth(sqrt(55)*sqrt(1 - 2*x)/11)/55, 2*x - 1 < -11/5), (-sqrt(55
)*atanh(sqrt(55)*sqrt(1 - 2*x)/11)/55, 2*x - 1 > -11/5))/605 - 784/(121*sqrt(1 - 2*x)) + 343/(132*(1 - 2*x)**(
3/2))

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