3.19.51 \(\int \frac {\sqrt {1-2 x} (2+3 x)}{(3+5 x)^3} \, dx\) [1851]

Optimal. Leaf size=68 \[ -\frac {(1-2 x)^{3/2}}{110 (3+5 x)^2}-\frac {67 \sqrt {1-2 x}}{550 (3+5 x)}+\frac {67 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{275 \sqrt {55}} \]

[Out]

-1/110*(1-2*x)^(3/2)/(3+5*x)^2+67/15125*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)-67/550*(1-2*x)^(1/2)/(3+
5*x)

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Rubi [A]
time = 0.01, antiderivative size = 68, 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 = {79, 43, 65, 212} \begin {gather*} -\frac {(1-2 x)^{3/2}}{110 (5 x+3)^2}-\frac {67 \sqrt {1-2 x}}{550 (5 x+3)}+\frac {67 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{275 \sqrt {55}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/110*(1 - 2*x)^(3/2)/(3 + 5*x)^2 - (67*Sqrt[1 - 2*x])/(550*(3 + 5*x)) + (67*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]
])/(275*Sqrt[55])

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + 1))), x] - Dist[d*(n/(b*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n
}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && GtQ[n, 0]

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 79

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] || L
tQ[p, n]))))

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 {\sqrt {1-2 x} (2+3 x)}{(3+5 x)^3} \, dx &=-\frac {(1-2 x)^{3/2}}{110 (3+5 x)^2}+\frac {67}{110} \int \frac {\sqrt {1-2 x}}{(3+5 x)^2} \, dx\\ &=-\frac {(1-2 x)^{3/2}}{110 (3+5 x)^2}-\frac {67 \sqrt {1-2 x}}{550 (3+5 x)}-\frac {67}{550} \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx\\ &=-\frac {(1-2 x)^{3/2}}{110 (3+5 x)^2}-\frac {67 \sqrt {1-2 x}}{550 (3+5 x)}+\frac {67}{550} \text {Subst}\left (\int \frac {1}{\frac {11}{2}-\frac {5 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )\\ &=-\frac {(1-2 x)^{3/2}}{110 (3+5 x)^2}-\frac {67 \sqrt {1-2 x}}{550 (3+5 x)}+\frac {67 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{275 \sqrt {55}}\\ \end {align*}

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Mathematica [A]
time = 0.15, size = 53, normalized size = 0.78 \begin {gather*} -\frac {\sqrt {1-2 x} (206+325 x)}{550 (3+5 x)^2}+\frac {67 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{275 \sqrt {55}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/550*(Sqrt[1 - 2*x]*(206 + 325*x))/(3 + 5*x)^2 + (67*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(275*Sqrt[55])

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Maple [A]
time = 0.11, size = 48, normalized size = 0.71

method	result	size
risch	\(\frac {650 x^{2}+87 x -206}{550 \left (3+5 x \right )^{2} \sqrt {1-2 x}}+\frac {67 \arctanh \left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{15125}\)	\(46\)
derivativedivides	\(-\frac {100 \left (-\frac {13 \left (1-2 x \right )^{\frac {3}{2}}}{1100}+\frac {67 \sqrt {1-2 x}}{2500}\right )}{\left (-6-10 x \right )^{2}}+\frac {67 \arctanh \left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{15125}\)	\(48\)
default	\(-\frac {100 \left (-\frac {13 \left (1-2 x \right )^{\frac {3}{2}}}{1100}+\frac {67 \sqrt {1-2 x}}{2500}\right )}{\left (-6-10 x \right )^{2}}+\frac {67 \arctanh \left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{15125}\)	\(48\)
trager	\(-\frac {\left (325 x +206\right ) \sqrt {1-2 x}}{550 \left (3+5 x \right )^{2}}-\frac {67 \RootOf \left (\textit {\_Z}^{2}-55\right ) \ln \left (\frac {5 \RootOf \left (\textit {\_Z}^{2}-55\right ) x +55 \sqrt {1-2 x}-8 \RootOf \left (\textit {\_Z}^{2}-55\right )}{3+5 x}\right )}{30250}\)	\(67\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-100*(-13/1100*(1-2*x)^(3/2)+67/2500*(1-2*x)^(1/2))/(-6-10*x)^2+67/15125*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*
55^(1/2)

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Maxima [A]
time = 0.50, size = 74, normalized size = 1.09 \begin {gather*} -\frac {67}{30250} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) + \frac {325 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 737 \, \sqrt {-2 \, x + 1}}{275 \, {\left (25 \, {\left (2 \, x - 1\right )}^{2} + 220 \, x + 11\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-67/30250*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 1/275*(325*(-2*x + 1)^(
3/2) - 737*sqrt(-2*x + 1))/(25*(2*x - 1)^2 + 220*x + 11)

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Fricas [A]
time = 1.15, size = 70, normalized size = 1.03 \begin {gather*} \frac {67 \, \sqrt {55} {\left (25 \, x^{2} + 30 \, x + 9\right )} \log \left (\frac {5 \, x - \sqrt {55} \sqrt {-2 \, x + 1} - 8}{5 \, x + 3}\right ) - 55 \, {\left (325 \, x + 206\right )} \sqrt {-2 \, x + 1}}{30250 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/30250*(67*sqrt(55)*(25*x^2 + 30*x + 9)*log((5*x - sqrt(55)*sqrt(-2*x + 1) - 8)/(5*x + 3)) - 55*(325*x + 206)
*sqrt(-2*x + 1))/(25*x^2 + 30*x + 9)

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 121 vs. \(2 (56) = 112\).
time = 173.79, size = 348, normalized size = 5.12 \begin {gather*} - \frac {124 \left (\begin {cases} \frac {\sqrt {55} \left (- \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1 \right )}}{4} + \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1 \right )}}{4} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1\right )} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1\right )}\right )}{605} & \text {for}\: \sqrt {1 - 2 x} > - \frac {\sqrt {55}}{5} \wedge \sqrt {1 - 2 x} < \frac {\sqrt {55}}{5} \end {cases}\right )}{25} + \frac {88 \left (\begin {cases} \frac {\sqrt {55} \cdot \left (\frac {3 \log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1 \right )}}{16} - \frac {3 \log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1 \right )}}{16} + \frac {3}{16 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1\right )} + \frac {1}{16 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1\right )^{2}} + \frac {3}{16 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1\right )} - \frac {1}{16 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1\right )^{2}}\right )}{6655} & \text {for}\: \sqrt {1 - 2 x} > - \frac {\sqrt {55}}{5} \wedge \sqrt {1 - 2 x} < \frac {\sqrt {55}}{5} \end {cases}\right )}{25} - \frac {12 \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 )}{25} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-124*Piecewise((sqrt(55)*(-log(sqrt(55)*sqrt(1 - 2*x)/11 - 1)/4 + log(sqrt(55)*sqrt(1 - 2*x)/11 + 1)/4 - 1/(4*
(sqrt(55)*sqrt(1 - 2*x)/11 + 1)) - 1/(4*(sqrt(55)*sqrt(1 - 2*x)/11 - 1)))/605, (sqrt(1 - 2*x) > -sqrt(55)/5) &
(sqrt(1 - 2*x) < sqrt(55)/5)))/25 + 88*Piecewise((sqrt(55)*(3*log(sqrt(55)*sqrt(1 - 2*x)/11 - 1)/16 - 3*log(s
qrt(55)*sqrt(1 - 2*x)/11 + 1)/16 + 3/(16*(sqrt(55)*sqrt(1 - 2*x)/11 + 1)) + 1/(16*(sqrt(55)*sqrt(1 - 2*x)/11 +
1)**2) + 3/(16*(sqrt(55)*sqrt(1 - 2*x)/11 - 1)) - 1/(16*(sqrt(55)*sqrt(1 - 2*x)/11 - 1)**2))/6655, (sqrt(1 -
2*x) > -sqrt(55)/5) & (sqrt(1 - 2*x) < sqrt(55)/5)))/25 - 12*Piecewise((-sqrt(55)*acoth(sqrt(55)*sqrt(1 - 2*x)
/11)/55, x < -3/5), (-sqrt(55)*atanh(sqrt(55)*sqrt(1 - 2*x)/11)/55, x > -3/5))/25

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Giac [A]
time = 1.28, size = 68, normalized size = 1.00 \begin {gather*} -\frac {67}{30250} \, \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 {325 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 737 \, \sqrt {-2 \, x + 1}}{1100 \, {\left (5 \, x + 3\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-67/30250*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 1/1100*(325*(
-2*x + 1)^(3/2) - 737*sqrt(-2*x + 1))/(5*x + 3)^2

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Mupad [B]
time = 1.18, size = 54, normalized size = 0.79 \begin {gather*} \frac {67\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )}{15125}-\frac {\frac {67\,\sqrt {1-2\,x}}{625}-\frac {13\,{\left (1-2\,x\right )}^{3/2}}{275}}{\frac {44\,x}{5}+{\left (2\,x-1\right )}^2+\frac {11}{25}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(67*55^(1/2)*atanh((55^(1/2)*(1 - 2*x)^(1/2))/11))/15125 - ((67*(1 - 2*x)^(1/2))/625 - (13*(1 - 2*x)^(3/2))/27
5)/((44*x)/5 + (2*x - 1)^2 + 11/25)

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