3.22.8 \(\int \frac {(2+3 x)^2}{(1-2 x)^{3/2} (3+5 x)} \, dx\) [2108]

Optimal. Leaf size=54 \[ \frac {49}{22 \sqrt {1-2 x}}+\frac {9}{10} \sqrt {1-2 x}-\frac {2 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{55 \sqrt {55}} \]

[Out]

-2/3025*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)+49/22/(1-2*x)^(1/2)+9/10*(1-2*x)^(1/2)

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Rubi [A]
time = 0.02, antiderivative size = 54, 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 = {89, 65, 212} \begin {gather*} \frac {9}{10} \sqrt {1-2 x}+\frac {49}{22 \sqrt {1-2 x}}-\frac {2 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{55 \sqrt {55}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

49/(22*Sqrt[1 - 2*x]) + (9*Sqrt[1 - 2*x])/10 - (2*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(55*Sqrt[55])

Rule 65

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 89

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 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin {align*} \int \frac {(2+3 x)^2}{(1-2 x)^{3/2} (3+5 x)} \, dx &=\int \left (\frac {49}{22 (1-2 x)^{3/2}}-\frac {9}{10 \sqrt {1-2 x}}+\frac {1}{55 \sqrt {1-2 x} (3+5 x)}\right ) \, dx\\ &=\frac {49}{22 \sqrt {1-2 x}}+\frac {9}{10} \sqrt {1-2 x}+\frac {1}{55} \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx\\ &=\frac {49}{22 \sqrt {1-2 x}}+\frac {9}{10} \sqrt {1-2 x}-\frac {1}{55} \text {Subst}\left (\int \frac {1}{\frac {11}{2}-\frac {5 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )\\ &=\frac {49}{22 \sqrt {1-2 x}}+\frac {9}{10} \sqrt {1-2 x}-\frac {2 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{55 \sqrt {55}}\\ \end {align*}

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Mathematica [A]
time = 0.06, size = 46, normalized size = 0.85 \begin {gather*} \frac {172-99 x}{55 \sqrt {1-2 x}}-\frac {2 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{55 \sqrt {55}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(172 - 99*x)/(55*Sqrt[1 - 2*x]) - (2*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(55*Sqrt[55])

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Maple [A]
time = 0.12, size = 38, normalized size = 0.70

method	result	size
risch	\(-\frac {99 x -172}{55 \sqrt {1-2 x}}-\frac {2 \arctanh \left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{3025}\)	\(34\)
derivativedivides	\(-\frac {2 \arctanh \left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{3025}+\frac {49}{22 \sqrt {1-2 x}}+\frac {9 \sqrt {1-2 x}}{10}\)	\(38\)
default	\(-\frac {2 \arctanh \left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{3025}+\frac {49}{22 \sqrt {1-2 x}}+\frac {9 \sqrt {1-2 x}}{10}\)	\(38\)
trager	\(\frac {\left (99 x -172\right ) \sqrt {1-2 x}}{-55+110 x}+\frac {\RootOf \left (\textit {\_Z}^{2}-55\right ) \ln \left (\frac {5 \RootOf \left (\textit {\_Z}^{2}-55\right ) x +55 \sqrt {1-2 x}-8 \RootOf \left (\textit {\_Z}^{2}-55\right )}{3+5 x}\right )}{3025}\)	\(67\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3025*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)+49/22/(1-2*x)^(1/2)+9/10*(1-2*x)^(1/2)

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Maxima [A]
time = 0.51, size = 55, normalized size = 1.02 \begin {gather*} \frac {1}{3025} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) + \frac {9}{10} \, \sqrt {-2 \, x + 1} + \frac {49}{22 \, \sqrt {-2 \, x + 1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3025*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 9/10*sqrt(-2*x + 1) + 49/2
2/sqrt(-2*x + 1)

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Fricas [A]
time = 0.62, size = 58, normalized size = 1.07 \begin {gather*} \frac {\sqrt {55} {\left (2 \, x - 1\right )} \log \left (\frac {5 \, x + \sqrt {55} \sqrt {-2 \, x + 1} - 8}{5 \, x + 3}\right ) + 55 \, {\left (99 \, x - 172\right )} \sqrt {-2 \, x + 1}}{3025 \, {\left (2 \, x - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3025*(sqrt(55)*(2*x - 1)*log((5*x + sqrt(55)*sqrt(-2*x + 1) - 8)/(5*x + 3)) + 55*(99*x - 172)*sqrt(-2*x + 1)
)/(2*x - 1)

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Sympy [A]
time = 20.86, size = 83, normalized size = 1.54 \begin {gather*} \frac {9 \sqrt {1 - 2 x}}{10} + \frac {2 \left (\begin {cases} - \frac {\sqrt {55} \operatorname {acoth}{\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} \right )}}{55} & \text {for}\: x < - \frac {3}{5} \\- \frac {\sqrt {55} \operatorname {atanh}{\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} \right )}}{55} & \text {for}\: x > - \frac {3}{5} \end {cases}\right )}{55} + \frac {49}{22 \sqrt {1 - 2 x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

9*sqrt(1 - 2*x)/10 + 2*Piecewise((-sqrt(55)*acoth(sqrt(55)*sqrt(1 - 2*x)/11)/55, x < -3/5), (-sqrt(55)*atanh(s
qrt(55)*sqrt(1 - 2*x)/11)/55, x > -3/5))/55 + 49/(22*sqrt(1 - 2*x))

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Giac [A]
time = 2.35, size = 58, normalized size = 1.07 \begin {gather*} \frac {1}{3025} \, \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 {9}{10} \, \sqrt {-2 \, x + 1} + \frac {49}{22 \, \sqrt {-2 \, x + 1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3025*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 9/10*sqrt(-2*x +
1) + 49/22/sqrt(-2*x + 1)

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Mupad [B]
time = 1.23, size = 37, normalized size = 0.69 \begin {gather*} \frac {49}{22\,\sqrt {1-2\,x}}-\frac {2\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )}{3025}+\frac {9\,\sqrt {1-2\,x}}{10} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

49/(22*(1 - 2*x)^(1/2)) - (2*55^(1/2)*atanh((55^(1/2)*(1 - 2*x)^(1/2))/11))/3025 + (9*(1 - 2*x)^(1/2))/10

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