3.21.10 \(\int \frac {3+5 x}{\sqrt {1-2 x} (2+3 x)^4} \, dx\) [2010]

Optimal. Leaf size=88 \[ \frac {\sqrt {1-2 x}}{63 (2+3 x)^3}-\frac {50 \sqrt {1-2 x}}{441 (2+3 x)^2}-\frac {50 \sqrt {1-2 x}}{1029 (2+3 x)}-\frac {100 \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{1029 \sqrt {21}} \]

[Out]

-100/21609*arctanh(1/7*21^(1/2)*(1-2*x)^(1/2))*21^(1/2)+1/63*(1-2*x)^(1/2)/(2+3*x)^3-50/441*(1-2*x)^(1/2)/(2+3
*x)^2-50/1029*(1-2*x)^(1/2)/(2+3*x)

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Rubi [A]
time = 0.01, antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 5, 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 {50 \sqrt {1-2 x}}{1029 (3 x+2)}-\frac {50 \sqrt {1-2 x}}{441 (3 x+2)^2}+\frac {\sqrt {1-2 x}}{63 (3 x+2)^3}-\frac {100 \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{1029 \sqrt {21}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(3 + 5*x)/(Sqrt[1 - 2*x]*(2 + 3*x)^4),x]

[Out]

Sqrt[1 - 2*x]/(63*(2 + 3*x)^3) - (50*Sqrt[1 - 2*x])/(441*(2 + 3*x)^2) - (50*Sqrt[1 - 2*x])/(1029*(2 + 3*x)) -
(100*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]])/(1029*Sqrt[21])

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 {3+5 x}{\sqrt {1-2 x} (2+3 x)^4} \, dx &=\frac {\sqrt {1-2 x}}{63 (2+3 x)^3}+\frac {100}{63} \int \frac {1}{\sqrt {1-2 x} (2+3 x)^3} \, dx\\ &=\frac {\sqrt {1-2 x}}{63 (2+3 x)^3}-\frac {50 \sqrt {1-2 x}}{441 (2+3 x)^2}+\frac {50}{147} \int \frac {1}{\sqrt {1-2 x} (2+3 x)^2} \, dx\\ &=\frac {\sqrt {1-2 x}}{63 (2+3 x)^3}-\frac {50 \sqrt {1-2 x}}{441 (2+3 x)^2}-\frac {50 \sqrt {1-2 x}}{1029 (2+3 x)}+\frac {50 \int \frac {1}{\sqrt {1-2 x} (2+3 x)} \, dx}{1029}\\ &=\frac {\sqrt {1-2 x}}{63 (2+3 x)^3}-\frac {50 \sqrt {1-2 x}}{441 (2+3 x)^2}-\frac {50 \sqrt {1-2 x}}{1029 (2+3 x)}-\frac {50 \text {Subst}\left (\int \frac {1}{\frac {7}{2}-\frac {3 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )}{1029}\\ &=\frac {\sqrt {1-2 x}}{63 (2+3 x)^3}-\frac {50 \sqrt {1-2 x}}{441 (2+3 x)^2}-\frac {50 \sqrt {1-2 x}}{1029 (2+3 x)}-\frac {100 \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{1029 \sqrt {21}}\\ \end {align*}

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Mathematica [A]
time = 0.16, size = 60, normalized size = 0.68 \begin {gather*} \frac {4 \left (-\frac {21 \sqrt {1-2 x} \left (417+950 x+450 x^2\right )}{4 (2+3 x)^3}-25 \sqrt {21} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )\right )}{21609} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(3 + 5*x)/(Sqrt[1 - 2*x]*(2 + 3*x)^4),x]

[Out]

(4*((-21*Sqrt[1 - 2*x]*(417 + 950*x + 450*x^2))/(4*(2 + 3*x)^3) - 25*Sqrt[21]*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]]
))/21609

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Maple [A]
time = 0.11, size = 57, normalized size = 0.65

method	result	size
risch	\(\frac {900 x^{3}+1450 x^{2}-116 x -417}{1029 \left (2+3 x \right )^{3} \sqrt {1-2 x}}-\frac {100 \arctanh \left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{21609}\)	\(51\)
derivativedivides	\(\frac {\frac {300 \left (1-2 x \right )^{\frac {5}{2}}}{343}-\frac {800 \left (1-2 x \right )^{\frac {3}{2}}}{147}+\frac {164 \sqrt {1-2 x}}{21}}{\left (-4-6 x \right )^{3}}-\frac {100 \arctanh \left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{21609}\)	\(57\)
default	\(\frac {\frac {300 \left (1-2 x \right )^{\frac {5}{2}}}{343}-\frac {800 \left (1-2 x \right )^{\frac {3}{2}}}{147}+\frac {164 \sqrt {1-2 x}}{21}}{\left (-4-6 x \right )^{3}}-\frac {100 \arctanh \left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{21609}\)	\(57\)
trager	\(-\frac {\left (450 x^{2}+950 x +417\right ) \sqrt {1-2 x}}{1029 \left (2+3 x \right )^{3}}-\frac {50 \RootOf \left (\textit {\_Z}^{2}-21\right ) \ln \left (\frac {-3 \RootOf \left (\textit {\_Z}^{2}-21\right ) x +21 \sqrt {1-2 x}+5 \RootOf \left (\textit {\_Z}^{2}-21\right )}{2+3 x}\right )}{21609}\)	\(72\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3+5*x)/(2+3*x)^4/(1-2*x)^(1/2),x,method=_RETURNVERBOSE)

[Out]

216*(25/6174*(1-2*x)^(5/2)-100/3969*(1-2*x)^(3/2)+41/1134*(1-2*x)^(1/2))/(-4-6*x)^3-100/21609*arctanh(1/7*21^(
1/2)*(1-2*x)^(1/2))*21^(1/2)

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Maxima [A]
time = 0.48, size = 92, normalized size = 1.05 \begin {gather*} \frac {50}{21609} \, \sqrt {21} \log \left (-\frac {\sqrt {21} - 3 \, \sqrt {-2 \, x + 1}}{\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}}\right ) - \frac {4 \, {\left (225 \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} - 1400 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + 2009 \, \sqrt {-2 \, x + 1}\right )}}{1029 \, {\left (27 \, {\left (2 \, x - 1\right )}^{3} + 189 \, {\left (2 \, x - 1\right )}^{2} + 882 \, x - 98\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3+5*x)/(2+3*x)^4/(1-2*x)^(1/2),x, algorithm="maxima")

[Out]

50/21609*sqrt(21)*log(-(sqrt(21) - 3*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) - 4/1029*(225*(-2*x + 1)^(
5/2) - 1400*(-2*x + 1)^(3/2) + 2009*sqrt(-2*x + 1))/(27*(2*x - 1)^3 + 189*(2*x - 1)^2 + 882*x - 98)

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Fricas [A]
time = 1.20, size = 84, normalized size = 0.95 \begin {gather*} \frac {50 \, \sqrt {21} {\left (27 \, x^{3} + 54 \, x^{2} + 36 \, x + 8\right )} \log \left (\frac {3 \, x + \sqrt {21} \sqrt {-2 \, x + 1} - 5}{3 \, x + 2}\right ) - 21 \, {\left (450 \, x^{2} + 950 \, x + 417\right )} \sqrt {-2 \, x + 1}}{21609 \, {\left (27 \, x^{3} + 54 \, x^{2} + 36 \, x + 8\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3+5*x)/(2+3*x)^4/(1-2*x)^(1/2),x, algorithm="fricas")

[Out]

1/21609*(50*sqrt(21)*(27*x^3 + 54*x^2 + 36*x + 8)*log((3*x + sqrt(21)*sqrt(-2*x + 1) - 5)/(3*x + 2)) - 21*(450
*x^2 + 950*x + 417)*sqrt(-2*x + 1))/(27*x^3 + 54*x^2 + 36*x + 8)

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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((3+5*x)/(2+3*x)**4/(1-2*x)**(1/2),x)

[Out]

Timed out

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Giac [A]
time = 0.65, size = 84, normalized size = 0.95 \begin {gather*} \frac {50}{21609} \, \sqrt {21} \log \left (\frac {{\left | -2 \, \sqrt {21} + 6 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}\right )}}\right ) - \frac {225 \, {\left (2 \, x - 1\right )}^{2} \sqrt {-2 \, x + 1} - 1400 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + 2009 \, \sqrt {-2 \, x + 1}}{2058 \, {\left (3 \, x + 2\right )}^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3+5*x)/(2+3*x)^4/(1-2*x)^(1/2),x, algorithm="giac")

[Out]

50/21609*sqrt(21)*log(1/2*abs(-2*sqrt(21) + 6*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) - 1/2058*(225*(2*
x - 1)^2*sqrt(-2*x + 1) - 1400*(-2*x + 1)^(3/2) + 2009*sqrt(-2*x + 1))/(3*x + 2)^3

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Mupad [B]
time = 1.19, size = 72, normalized size = 0.82 \begin {gather*} -\frac {100\,\sqrt {21}\,\mathrm {atanh}\left (\frac {\sqrt {21}\,\sqrt {1-2\,x}}{7}\right )}{21609}-\frac {\frac {164\,\sqrt {1-2\,x}}{567}-\frac {800\,{\left (1-2\,x\right )}^{3/2}}{3969}+\frac {100\,{\left (1-2\,x\right )}^{5/2}}{3087}}{\frac {98\,x}{3}+7\,{\left (2\,x-1\right )}^2+{\left (2\,x-1\right )}^3-\frac {98}{27}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*x + 3)/((1 - 2*x)^(1/2)*(3*x + 2)^4),x)

[Out]

- (100*21^(1/2)*atanh((21^(1/2)*(1 - 2*x)^(1/2))/7))/21609 - ((164*(1 - 2*x)^(1/2))/567 - (800*(1 - 2*x)^(3/2)
)/3969 + (100*(1 - 2*x)^(5/2))/3087)/((98*x)/3 + 7*(2*x - 1)^2 + (2*x - 1)^3 - 98/27)

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