3.2079 \(\int \frac{3+5 x}{(1-2 x)^{3/2} (2+3 x)^3} \, dx\)

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

[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

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Rubi [A]  time = 0.0215629, antiderivative size = 90, normalized size of antiderivative = 1.08, 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 = {78, 51, 63, 206} \[ -\frac{195 \sqrt{1-2 x}}{686 (3 x+2)}+\frac{65}{147 \sqrt{1-2 x} (3 x+2)}+\frac{1}{42 \sqrt{1-2 x} (3 x+2)^2}-\frac{65}{343} \sqrt{\frac{3}{7}} \tanh ^{-1}\left (\sqrt{\frac{3}{7}} \sqrt{1-2 x}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

1/(42*Sqrt[1 - 2*x]*(2 + 3*x)^2) + 65/(147*Sqrt[1 - 2*x]*(2 + 3*x)) - (195*Sqrt[1 - 2*x])/(686*(2 + 3*x)) - (6
5*Sqrt[3/7]*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]])/343

Rule 78

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] || LtQ
[p, n]))))

Rule 51

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 63

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 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 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} \operatorname{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*}

Mathematica [C]  time = 0.0161998, size = 48, normalized size = 0.58 \[ \frac{260 (3 x+2)^2 \, _2F_1\left (-\frac{1}{2},2;\frac{1}{2};\frac{3}{7}-\frac{6 x}{7}\right )+49}{2058 \sqrt{1-2 x} (3 x+2)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(49 + 260*(2 + 3*x)^2*Hypergeometric2F1[-1/2, 2, 1/2, 3/7 - (6*x)/7])/(2058*Sqrt[1 - 2*x]*(2 + 3*x)^2)

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Maple [A]  time = 0.012, size = 57, normalized size = 0.7 \begin{align*}{\frac{36}{343\, \left ( -6\,x-4 \right ) ^{2}} \left ({\frac{21}{4} \left ( 1-2\,x \right ) ^{{\frac{3}{2}}}}-{\frac{427}{36}\sqrt{1-2\,x}} \right ) }-{\frac{65\,\sqrt{21}}{2401}{\it Artanh} \left ({\frac{\sqrt{21}}{7}\sqrt{1-2\,x}} \right ) }+{\frac{44}{343}{\frac{1}{\sqrt{1-2\,x}}}} \end{align*}

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)

[Out]

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

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Maxima [A]  time = 1.66757, size = 112, normalized size = 1.35 \begin{align*} \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{align*}

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 = 1.61962, size = 254, normalized size = 3.06 \begin{align*} \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{align*}

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(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Giac [A]  time = 2.55653, size = 104, normalized size = 1.25 \begin{align*} \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{align*}

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