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

Optimal. Leaf size=67 \[ -\frac {27}{100} (1-2 x)^{5/2}+\frac {54}{25} (1-2 x)^{3/2}-\frac {3897}{500} \sqrt {1-2 x}-\frac {2 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{125 \sqrt {55}} \]

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Rubi [A]  time = 0.02, 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 = {88, 63, 206} \begin {gather*} -\frac {27}{100} (1-2 x)^{5/2}+\frac {54}{25} (1-2 x)^{3/2}-\frac {3897}{500} \sqrt {1-2 x}-\frac {2 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{125 \sqrt {55}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3897*Sqrt[1 - 2*x])/500 + (54*(1 - 2*x)^(3/2))/25 - (27*(1 - 2*x)^(5/2))/100 - (2*ArcTanh[Sqrt[5/11]*Sqrt[1
- 2*x]])/(125*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 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

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}{\sqrt {1-2 x} (3+5 x)} \, dx &=\int \left (\frac {3897}{500 \sqrt {1-2 x}}-\frac {162}{25} \sqrt {1-2 x}+\frac {27}{20} (1-2 x)^{3/2}+\frac {1}{125 \sqrt {1-2 x} (3+5 x)}\right ) \, dx\\ &=-\frac {3897}{500} \sqrt {1-2 x}+\frac {54}{25} (1-2 x)^{3/2}-\frac {27}{100} (1-2 x)^{5/2}+\frac {1}{125} \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx\\ &=-\frac {3897}{500} \sqrt {1-2 x}+\frac {54}{25} (1-2 x)^{3/2}-\frac {27}{100} (1-2 x)^{5/2}-\frac {1}{125} \operatorname {Subst}\left (\int \frac {1}{\frac {11}{2}-\frac {5 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )\\ &=-\frac {3897}{500} \sqrt {1-2 x}+\frac {54}{25} (1-2 x)^{3/2}-\frac {27}{100} (1-2 x)^{5/2}-\frac {2 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{125 \sqrt {55}}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 51, normalized size = 0.76 \begin {gather*} \frac {-495 \sqrt {1-2 x} \left (15 x^2+45 x+82\right )-2 \sqrt {55} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{6875} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-495*Sqrt[1 - 2*x]*(82 + 45*x + 15*x^2) - 2*Sqrt[55]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/6875

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IntegrateAlgebraic [A]  time = 0.06, size = 59, normalized size = 0.88 \begin {gather*} -\frac {9}{500} \sqrt {1-2 x} \left (15 (1-2 x)^2-120 (1-2 x)+433\right )-\frac {2 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{125 \sqrt {55}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-9*(433 - 120*(1 - 2*x) + 15*(1 - 2*x)^2)*Sqrt[1 - 2*x])/500 - (2*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(125*Sqr
t[55])

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fricas [A]  time = 1.52, size = 50, normalized size = 0.75 \begin {gather*} -\frac {9}{125} \, {\left (15 \, x^{2} + 45 \, x + 82\right )} \sqrt {-2 \, x + 1} + \frac {1}{6875} \, \sqrt {55} \log \left (\frac {5 \, x + \sqrt {55} \sqrt {-2 \, x + 1} - 8}{5 \, x + 3}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-9/125*(15*x^2 + 45*x + 82)*sqrt(-2*x + 1) + 1/6875*sqrt(55)*log((5*x + sqrt(55)*sqrt(-2*x + 1) - 8)/(5*x + 3)
)

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giac [A]  time = 1.22, size = 74, normalized size = 1.10 \begin {gather*} -\frac {27}{100} \, {\left (2 \, x - 1\right )}^{2} \sqrt {-2 \, x + 1} + \frac {54}{25} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {1}{6875} \, \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 {3897}{500} \, \sqrt {-2 \, x + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-27/100*(2*x - 1)^2*sqrt(-2*x + 1) + 54/25*(-2*x + 1)^(3/2) + 1/6875*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqr
t(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 3897/500*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 )}{6875}+\frac {54 \left (-2 x +1\right )^{\frac {3}{2}}}{25}-\frac {27 \left (-2 x +1\right )^{\frac {5}{2}}}{100}-\frac {3897 \sqrt {-2 x +1}}{500} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

54/25*(-2*x+1)^(3/2)-27/100*(-2*x+1)^(5/2)-2/6875*arctanh(1/11*55^(1/2)*(-2*x+1)^(1/2))*55^(1/2)-3897/500*(-2*
x+1)^(1/2)

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maxima [A]  time = 1.11, size = 64, normalized size = 0.96 \begin {gather*} -\frac {27}{100} \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} + \frac {54}{25} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {1}{6875} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) - \frac {3897}{500} \, \sqrt {-2 \, x + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-27/100*(-2*x + 1)^(5/2) + 54/25*(-2*x + 1)^(3/2) + 1/6875*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(5
5) + 5*sqrt(-2*x + 1))) - 3897/500*sqrt(-2*x + 1)

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mupad [B]  time = 0.06, size = 48, normalized size = 0.72 \begin {gather*} \frac {54\,{\left (1-2\,x\right )}^{3/2}}{25}-\frac {3897\,\sqrt {1-2\,x}}{500}-\frac {27\,{\left (1-2\,x\right )}^{5/2}}{100}+\frac {\sqrt {55}\,\mathrm {atan}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}\,1{}\mathrm {i}}{11}\right )\,2{}\mathrm {i}}{6875} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(55^(1/2)*atan((55^(1/2)*(1 - 2*x)^(1/2)*1i)/11)*2i)/6875 - (3897*(1 - 2*x)^(1/2))/500 + (54*(1 - 2*x)^(3/2))/
25 - (27*(1 - 2*x)^(5/2))/100

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sympy [A]  time = 37.26, size = 102, normalized size = 1.52 \begin {gather*} - \frac {27 \left (1 - 2 x\right )^{\frac {5}{2}}}{100} + \frac {54 \left (1 - 2 x\right )^{\frac {3}{2}}}{25} - \frac {3897 \sqrt {1 - 2 x}}{500} + \frac {2 \left (\begin {cases} - \frac {\sqrt {55} \operatorname {acoth}{\left (\frac {\sqrt {55}}{5 \sqrt {1 - 2 x}} \right )}}{55} & \text {for}\: \frac {1}{1 - 2 x} > \frac {5}{11} \\- \frac {\sqrt {55} \operatorname {atanh}{\left (\frac {\sqrt {55}}{5 \sqrt {1 - 2 x}} \right )}}{55} & \text {for}\: \frac {1}{1 - 2 x} < \frac {5}{11} \end {cases}\right )}{125} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-27*(1 - 2*x)**(5/2)/100 + 54*(1 - 2*x)**(3/2)/25 - 3897*sqrt(1 - 2*x)/500 + 2*Piecewise((-sqrt(55)*acoth(sqrt
(55)/(5*sqrt(1 - 2*x)))/55, 1/(1 - 2*x) > 5/11), (-sqrt(55)*atanh(sqrt(55)/(5*sqrt(1 - 2*x)))/55, 1/(1 - 2*x)
< 5/11))/125

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