3.21.63 \(\int \frac {2+3 x}{\sqrt {1-2 x} (3+5 x)^3} \, dx\) [2063]

Optimal. Leaf size=68 \[ -\frac {\sqrt {1-2 x}}{110 (3+5 x)^2}-\frac {69 \sqrt {1-2 x}}{1210 (3+5 x)}-\frac {69 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{605 \sqrt {55}} \]

[Out]

-69/33275*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)-1/110*(1-2*x)^(1/2)/(3+5*x)^2-69/1210*(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, 44, 65, 212} \begin {gather*} -\frac {69 \sqrt {1-2 x}}{1210 (5 x+3)}-\frac {\sqrt {1-2 x}}{110 (5 x+3)^2}-\frac {69 \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)/(Sqrt[1 - 2*x]*(3 + 5*x)^3),x]

[Out]

-1/110*Sqrt[1 - 2*x]/(3 + 5*x)^2 - (69*Sqrt[1 - 2*x])/(1210*(3 + 5*x)) - (69*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]]
)/(605*Sqrt[55])

Rule 44

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] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[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 {2+3 x}{\sqrt {1-2 x} (3+5 x)^3} \, dx &=-\frac {\sqrt {1-2 x}}{110 (3+5 x)^2}+\frac {69}{110} \int \frac {1}{\sqrt {1-2 x} (3+5 x)^2} \, dx\\ &=-\frac {\sqrt {1-2 x}}{110 (3+5 x)^2}-\frac {69 \sqrt {1-2 x}}{1210 (3+5 x)}+\frac {69 \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx}{1210}\\ &=-\frac {\sqrt {1-2 x}}{110 (3+5 x)^2}-\frac {69 \sqrt {1-2 x}}{1210 (3+5 x)}-\frac {69 \text {Subst}\left (\int \frac {1}{\frac {11}{2}-\frac {5 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )}{1210}\\ &=-\frac {\sqrt {1-2 x}}{110 (3+5 x)^2}-\frac {69 \sqrt {1-2 x}}{1210 (3+5 x)}-\frac {69 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{605 \sqrt {55}}\\ \end {align*}

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Mathematica [A]
time = 0.14, size = 53, normalized size = 0.78 \begin {gather*} -\frac {\sqrt {1-2 x} (218+345 x)}{1210 (3+5 x)^2}-\frac {69 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{605 \sqrt {55}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/1210*(Sqrt[1 - 2*x]*(218 + 345*x))/(3 + 5*x)^2 - (69*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(605*Sqrt[55])

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

method	result	size
risch	\(\frac {690 x^{2}+91 x -218}{1210 \left (3+5 x \right )^{2} \sqrt {1-2 x}}-\frac {69 \arctanh \left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{33275}\)	\(46\)
derivativedivides	\(-\frac {100 \left (-\frac {69 \left (1-2 x \right )^{\frac {3}{2}}}{12100}+\frac {71 \sqrt {1-2 x}}{5500}\right )}{\left (-6-10 x \right )^{2}}-\frac {69 \arctanh \left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{33275}\)	\(48\)
default	\(-\frac {100 \left (-\frac {69 \left (1-2 x \right )^{\frac {3}{2}}}{12100}+\frac {71 \sqrt {1-2 x}}{5500}\right )}{\left (-6-10 x \right )^{2}}-\frac {69 \arctanh \left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{33275}\)	\(48\)
trager	\(-\frac {\left (345 x +218\right ) \sqrt {1-2 x}}{1210 \left (3+5 x \right )^{2}}+\frac {69 \RootOf \left (\textit {\_Z}^{2}-55\right ) \ln \left (\frac {5 \RootOf \left (\textit {\_Z}^{2}-55\right ) x +55 \sqrt {1-2 x}-8 \RootOf \left (\textit {\_Z}^{2}-55\right )}{3+5 x}\right )}{66550}\)	\(67\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-100*(-69/12100*(1-2*x)^(3/2)+71/5500*(1-2*x)^(1/2))/(-6-10*x)^2-69/33275*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))
*55^(1/2)

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Maxima [A]
time = 0.51, size = 74, normalized size = 1.09 \begin {gather*} \frac {69}{66550} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) + \frac {345 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 781 \, \sqrt {-2 \, x + 1}}{605 \, {\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)/(3+5*x)^3/(1-2*x)^(1/2),x, algorithm="maxima")

[Out]

69/66550*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 1/605*(345*(-2*x + 1)^(3
/2) - 781*sqrt(-2*x + 1))/(25*(2*x - 1)^2 + 220*x + 11)

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Fricas [A]
time = 0.74, size = 69, normalized size = 1.01 \begin {gather*} \frac {69 \, \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 (345 \, x + 218\right )} \sqrt {-2 \, x + 1}}{66550 \, {\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)/(3+5*x)^3/(1-2*x)^(1/2),x, algorithm="fricas")

[Out]

1/66550*(69*sqrt(55)*(25*x^2 + 30*x + 9)*log((5*x + sqrt(55)*sqrt(-2*x + 1) - 8)/(5*x + 3)) - 55*(345*x + 218)
*sqrt(-2*x + 1))/(25*x^2 + 30*x + 9)

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Sympy [F(-1)] Timed out
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((2+3*x)/(3+5*x)**3/(1-2*x)**(1/2),x)

[Out]

Timed out

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Giac [A]
time = 1.28, size = 68, normalized size = 1.00 \begin {gather*} \frac {69}{66550} \, \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 {345 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 781 \, \sqrt {-2 \, x + 1}}{2420 \, {\left (5 \, x + 3\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

69/66550*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 1/2420*(345*(-
2*x + 1)^(3/2) - 781*sqrt(-2*x + 1))/(5*x + 3)^2

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Mupad [B]
time = 0.06, size = 54, normalized size = 0.79 \begin {gather*} -\frac {69\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )}{33275}-\frac {\frac {71\,\sqrt {1-2\,x}}{1375}-\frac {69\,{\left (1-2\,x\right )}^{3/2}}{3025}}{\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((3*x + 2)/((1 - 2*x)^(1/2)*(5*x + 3)^3),x)

[Out]

- (69*55^(1/2)*atanh((55^(1/2)*(1 - 2*x)^(1/2))/11))/33275 - ((71*(1 - 2*x)^(1/2))/1375 - (69*(1 - 2*x)^(3/2))
/3025)/((44*x)/5 + (2*x - 1)^2 + 11/25)

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