3.21.79 \(\int \frac {3+5 x}{(1-2 x)^{3/2} (2+3 x)^3} \, dx\) [2079]

Optimal. Leaf size=83 \[ \frac {65}{343 \sqrt {1-2 x}}+\frac {1}{42 \sqrt {1-2 x} (2+3 x)^2}-\frac {65}{294 \sqrt {1-2 x} (2+3 x)}-\frac {65}{343} \sqrt {\frac {3}{7}} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right ) \]

[Out]

-65/2401*arctanh(1/7*21^(1/2)*(1-2*x)^(1/2))*21^(1/2)+65/343/(1-2*x)^(1/2)+1/42/(2+3*x)^2/(1-2*x)^(1/2)-65/294
/(2+3*x)/(1-2*x)^(1/2)

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Rubi [A]
time = 0.01, antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {79, 44, 53, 65, 212} \begin {gather*} \frac {65}{343 \sqrt {1-2 x}}-\frac {65}{294 \sqrt {1-2 x} (3 x+2)}+\frac {1}{42 \sqrt {1-2 x} (3 x+2)^2}-\frac {65}{343} \sqrt {\frac {3}{7}} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

65/(343*Sqrt[1 - 2*x]) + 1/(42*Sqrt[1 - 2*x]*(2 + 3*x)^2) - 65/(294*Sqrt[1 - 2*x]*(2 + 3*x)) - (65*Sqrt[3/7]*A
rcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]])/343

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 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)^3} \, dx &=\frac {1}{42 \sqrt {1-2 x} (2+3 x)^2}+\frac {65}{42} \int \frac {1}{(1-2 x)^{3/2} (2+3 x)^2} \, dx\\ &=\frac {1}{42 \sqrt {1-2 x} (2+3 x)^2}+\frac {65}{147 \sqrt {1-2 x} (2+3 x)}+\frac {195}{98} \int \frac {1}{\sqrt {1-2 x} (2+3 x)^2} \, dx\\ &=\frac {1}{42 \sqrt {1-2 x} (2+3 x)^2}+\frac {65}{147 \sqrt {1-2 x} (2+3 x)}-\frac {195 \sqrt {1-2 x}}{686 (2+3 x)}+\frac {195}{686} \int \frac {1}{\sqrt {1-2 x} (2+3 x)} \, dx\\ &=\frac {1}{42 \sqrt {1-2 x} (2+3 x)^2}+\frac {65}{147 \sqrt {1-2 x} (2+3 x)}-\frac {195 \sqrt {1-2 x}}{686 (2+3 x)}-\frac {195}{686} \text {Subst}\left (\int \frac {1}{\frac {7}{2}-\frac {3 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )\\ &=\frac {1}{42 \sqrt {1-2 x} (2+3 x)^2}+\frac {65}{147 \sqrt {1-2 x} (2+3 x)}-\frac {195 \sqrt {1-2 x}}{686 (2+3 x)}-\frac {65}{343} \sqrt {\frac {3}{7}} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )\\ \end {align*}

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Mathematica [A]
time = 0.15, size = 66, normalized size = 0.80 \begin {gather*} \frac {1631+7735 x+8190 x^2-130 \sqrt {21-42 x} (2+3 x)^2 \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{4802 \sqrt {1-2 x} (2+3 x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(1631 + 7735*x + 8190*x^2 - 130*Sqrt[21 - 42*x]*(2 + 3*x)^2*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]])/(4802*Sqrt[1 - 2
*x]*(2 + 3*x)^2)

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Maple [A]
time = 0.13, size = 57, normalized size = 0.69

method	result	size
risch	\(\frac {1170 x^{2}+1105 x +233}{686 \left (2+3 x \right )^{2} \sqrt {1-2 x}}-\frac {65 \arctanh \left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{2401}\)	\(46\)
derivativedivides	\(\frac {44}{343 \sqrt {1-2 x}}+\frac {\frac {27 \left (1-2 x \right )^{\frac {3}{2}}}{49}-\frac {61 \sqrt {1-2 x}}{49}}{\left (-4-6 x \right )^{2}}-\frac {65 \arctanh \left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{2401}\)	\(57\)
default	\(\frac {44}{343 \sqrt {1-2 x}}+\frac {\frac {27 \left (1-2 x \right )^{\frac {3}{2}}}{49}-\frac {61 \sqrt {1-2 x}}{49}}{\left (-4-6 x \right )^{2}}-\frac {65 \arctanh \left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{2401}\)	\(57\)
trager	\(-\frac {\left (1170 x^{2}+1105 x +233\right ) \sqrt {1-2 x}}{686 \left (2+3 x \right )^{2} \left (-1+2 x \right )}-\frac {65 \RootOf \left (\textit {\_Z}^{2}-21\right ) \ln \left (\frac {-3 \RootOf \left (\textit {\_Z}^{2}-21\right ) x +21 \sqrt {1-2 x}+5 \RootOf \left (\textit {\_Z}^{2}-21\right )}{2+3 x}\right )}{4802}\)	\(79\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

44/343/(1-2*x)^(1/2)+36/343*(21/4*(1-2*x)^(3/2)-427/36*(1-2*x)^(1/2))/(-4-6*x)^2-65/2401*arctanh(1/7*21^(1/2)*
(1-2*x)^(1/2))*21^(1/2)

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Maxima [A]
time = 0.51, size = 83, normalized size = 1.00 \begin {gather*} \frac {65}{4802} \, \sqrt {21} \log \left (-\frac {\sqrt {21} - 3 \, \sqrt {-2 \, x + 1}}{\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}}\right ) + \frac {585 \, {\left (2 \, x - 1\right )}^{2} + 4550 \, x - 119}{343 \, {\left (9 \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} - 42 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + 49 \, \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)^3,x, algorithm="maxima")

[Out]

65/4802*sqrt(21)*log(-(sqrt(21) - 3*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) + 1/343*(585*(2*x - 1)^2 +
4550*x - 119)/(9*(-2*x + 1)^(5/2) - 42*(-2*x + 1)^(3/2) + 49*sqrt(-2*x + 1))

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Fricas [A]
time = 1.40, size = 90, normalized size = 1.08 \begin {gather*} \frac {65 \, \sqrt {7} \sqrt {3} {\left (18 \, x^{3} + 15 \, x^{2} - 4 \, x - 4\right )} \log \left (\frac {\sqrt {7} \sqrt {3} \sqrt {-2 \, x + 1} + 3 \, x - 5}{3 \, x + 2}\right ) - 7 \, {\left (1170 \, x^{2} + 1105 \, x + 233\right )} \sqrt {-2 \, x + 1}}{4802 \, {\left (18 \, x^{3} + 15 \, x^{2} - 4 \, x - 4\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)^3,x, algorithm="fricas")

[Out]

1/4802*(65*sqrt(7)*sqrt(3)*(18*x^3 + 15*x^2 - 4*x - 4)*log((sqrt(7)*sqrt(3)*sqrt(-2*x + 1) + 3*x - 5)/(3*x + 2
)) - 7*(1170*x^2 + 1105*x + 233)*sqrt(-2*x + 1))/(18*x^3 + 15*x^2 - 4*x - 4)

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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)**3,x)

[Out]

Timed out

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Giac [A]
time = 1.26, size = 77, normalized size = 0.93 \begin {gather*} \frac {65}{4802} \, \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 {44}{343 \, \sqrt {-2 \, x + 1}} + \frac {27 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 61 \, \sqrt {-2 \, x + 1}}{196 \, {\left (3 \, x + 2\right )}^{2}} \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)^3,x, algorithm="giac")

[Out]

65/4802*sqrt(21)*log(1/2*abs(-2*sqrt(21) + 6*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) + 44/343/sqrt(-2*x
+ 1) + 1/196*(27*(-2*x + 1)^(3/2) - 61*sqrt(-2*x + 1))/(3*x + 2)^2

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Mupad [B]
time = 0.08, size = 62, normalized size = 0.75 \begin {gather*} \frac {\frac {650\,x}{441}+\frac {65\,{\left (2\,x-1\right )}^2}{343}-\frac {17}{441}}{\frac {49\,\sqrt {1-2\,x}}{9}-\frac {14\,{\left (1-2\,x\right )}^{3/2}}{3}+{\left (1-2\,x\right )}^{5/2}}-\frac {65\,\sqrt {21}\,\mathrm {atanh}\left (\frac {\sqrt {21}\,\sqrt {1-2\,x}}{7}\right )}{2401} \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)^3),x)

[Out]

((650*x)/441 + (65*(2*x - 1)^2)/343 - 17/441)/((49*(1 - 2*x)^(1/2))/9 - (14*(1 - 2*x)^(3/2))/3 + (1 - 2*x)^(5/
2)) - (65*21^(1/2)*atanh((21^(1/2)*(1 - 2*x)^(1/2))/7))/2401

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