3.21.78 \(\int \frac {3+5 x}{(1-2 x)^{3/2} (2+3 x)^2} \, dx\) [2078]

Optimal. Leaf size=61 \[ \frac {64}{147 \sqrt {1-2 x}}+\frac {1}{21 \sqrt {1-2 x} (2+3 x)}-\frac {64 \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{49 \sqrt {21}} \]

[Out]

-64/1029*arctanh(1/7*21^(1/2)*(1-2*x)^(1/2))*21^(1/2)+64/147/(1-2*x)^(1/2)+1/21/(2+3*x)/(1-2*x)^(1/2)

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Rubi [A]
time = 0.01, antiderivative size = 61, 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, 53, 65, 212} \begin {gather*} \frac {64}{147 \sqrt {1-2 x}}+\frac {1}{21 \sqrt {1-2 x} (3 x+2)}-\frac {64 \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{49 \sqrt {21}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

64/(147*Sqrt[1 - 2*x]) + 1/(21*Sqrt[1 - 2*x]*(2 + 3*x)) - (64*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]])/(49*Sqrt[21])

Rule 53

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] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

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}{(1-2 x)^{3/2} (2+3 x)^2} \, dx &=\frac {1}{21 \sqrt {1-2 x} (2+3 x)}+\frac {32}{21} \int \frac {1}{(1-2 x)^{3/2} (2+3 x)} \, dx\\ &=\frac {64}{147 \sqrt {1-2 x}}+\frac {1}{21 \sqrt {1-2 x} (2+3 x)}+\frac {32}{49} \int \frac {1}{\sqrt {1-2 x} (2+3 x)} \, dx\\ &=\frac {64}{147 \sqrt {1-2 x}}+\frac {1}{21 \sqrt {1-2 x} (2+3 x)}-\frac {32}{49} \text {Subst}\left (\int \frac {1}{\frac {7}{2}-\frac {3 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )\\ &=\frac {64}{147 \sqrt {1-2 x}}+\frac {1}{21 \sqrt {1-2 x} (2+3 x)}-\frac {64 \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{49 \sqrt {21}}\\ \end {align*}

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Mathematica [A]
time = 0.13, size = 53, normalized size = 0.87 \begin {gather*} \frac {45+64 x}{49 \sqrt {1-2 x} (2+3 x)}-\frac {64 \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{49 \sqrt {21}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(45 + 64*x)/(49*Sqrt[1 - 2*x]*(2 + 3*x)) - (64*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]])/(49*Sqrt[21])

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Maple [A]
time = 0.10, size = 45, normalized size = 0.74

method	result	size
risch	\(\frac {64 x +45}{49 \left (2+3 x \right ) \sqrt {1-2 x}}-\frac {64 \arctanh \left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{1029}\)	\(41\)
derivativedivides	\(\frac {22}{49 \sqrt {1-2 x}}-\frac {2 \sqrt {1-2 x}}{147 \left (-\frac {4}{3}-2 x \right )}-\frac {64 \arctanh \left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{1029}\)	\(45\)
default	\(\frac {22}{49 \sqrt {1-2 x}}-\frac {2 \sqrt {1-2 x}}{147 \left (-\frac {4}{3}-2 x \right )}-\frac {64 \arctanh \left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{1029}\)	\(45\)
trager	\(-\frac {\left (64 x +45\right ) \sqrt {1-2 x}}{49 \left (6 x^{2}+x -2\right )}-\frac {32 \RootOf \left (\textit {\_Z}^{2}-21\right ) \ln \left (\frac {-3 \RootOf \left (\textit {\_Z}^{2}-21\right ) x +21 \sqrt {1-2 x}+5 \RootOf \left (\textit {\_Z}^{2}-21\right )}{2+3 x}\right )}{1029}\)	\(70\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

22/49/(1-2*x)^(1/2)-2/147*(1-2*x)^(1/2)/(-4/3-2*x)-64/1029*arctanh(1/7*21^(1/2)*(1-2*x)^(1/2))*21^(1/2)

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Maxima [A]
time = 0.52, size = 65, normalized size = 1.07 \begin {gather*} \frac {32}{1029} \, \sqrt {21} \log \left (-\frac {\sqrt {21} - 3 \, \sqrt {-2 \, x + 1}}{\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}}\right ) - \frac {2 \, {\left (64 \, x + 45\right )}}{49 \, {\left (3 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 7 \, \sqrt {-2 \, x + 1}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

32/1029*sqrt(21)*log(-(sqrt(21) - 3*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) - 2/49*(64*x + 45)/(3*(-2*x
+ 1)^(3/2) - 7*sqrt(-2*x + 1))

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Fricas [A]
time = 0.88, size = 65, normalized size = 1.07 \begin {gather*} \frac {32 \, \sqrt {21} {\left (6 \, x^{2} + x - 2\right )} \log \left (\frac {3 \, x + \sqrt {21} \sqrt {-2 \, x + 1} - 5}{3 \, x + 2}\right ) - 21 \, {\left (64 \, x + 45\right )} \sqrt {-2 \, x + 1}}{1029 \, {\left (6 \, x^{2} + x - 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1029*(32*sqrt(21)*(6*x^2 + x - 2)*log((3*x + sqrt(21)*sqrt(-2*x + 1) - 5)/(3*x + 2)) - 21*(64*x + 45)*sqrt(-
2*x + 1))/(6*x^2 + x - 2)

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

[Out]

Timed out

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Giac [A]
time = 1.29, size = 68, normalized size = 1.11 \begin {gather*} \frac {32}{1029} \, \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 {2 \, {\left (64 \, x + 45\right )}}{49 \, {\left (3 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 7 \, \sqrt {-2 \, x + 1}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

32/1029*sqrt(21)*log(1/2*abs(-2*sqrt(21) + 6*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) - 2/49*(64*x + 45)
/(3*(-2*x + 1)^(3/2) - 7*sqrt(-2*x + 1))

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Mupad [B]
time = 1.24, size = 46, normalized size = 0.75 \begin {gather*} \frac {\frac {128\,x}{147}+\frac {30}{49}}{\frac {7\,\sqrt {1-2\,x}}{3}-{\left (1-2\,x\right )}^{3/2}}-\frac {64\,\sqrt {21}\,\mathrm {atanh}\left (\frac {\sqrt {21}\,\sqrt {1-2\,x}}{7}\right )}{1029} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((128*x)/147 + 30/49)/((7*(1 - 2*x)^(1/2))/3 - (1 - 2*x)^(3/2)) - (64*21^(1/2)*atanh((21^(1/2)*(1 - 2*x)^(1/2)
)/7))/1029

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