3.2030 \(\int \frac{(3+5 x)^3}{\sqrt{1-2 x} (2+3 x)^2} \, dx\)

Optimal. Leaf size=73 \[ \frac{\sqrt{1-2 x} (5 x+3)^2}{21 (3 x+2)}-\frac{10}{189} \sqrt{1-2 x} (95 x+214)-\frac{208 \tanh ^{-1}\left (\sqrt{\frac{3}{7}} \sqrt{1-2 x}\right )}{189 \sqrt{21}} \]

[Out]

(Sqrt[1 - 2*x]*(3 + 5*x)^2)/(21*(2 + 3*x)) - (10*Sqrt[1 - 2*x]*(214 + 95*x))/189 - (208*ArcTanh[Sqrt[3/7]*Sqrt
[1 - 2*x]])/(189*Sqrt[21])

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Rubi [A]  time = 0.0175322, antiderivative size = 73, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {98, 147, 63, 206} \[ \frac{\sqrt{1-2 x} (5 x+3)^2}{21 (3 x+2)}-\frac{10}{189} \sqrt{1-2 x} (95 x+214)-\frac{208 \tanh ^{-1}\left (\sqrt{\frac{3}{7}} \sqrt{1-2 x}\right )}{189 \sqrt{21}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[1 - 2*x]*(3 + 5*x)^2)/(21*(2 + 3*x)) - (10*Sqrt[1 - 2*x]*(214 + 95*x))/189 - (208*ArcTanh[Sqrt[3/7]*Sqrt
[1 - 2*x]])/(189*Sqrt[21])

Rule 98

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c -
a*d)*(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^(p + 1))/(b*(b*e - a*f)*(m + 1)), x] + Dist[1/(b*(b*e - a*
f)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 2)*(e + f*x)^p*Simp[a*d*(d*e*(n - 1) + c*f*(p + 1)) + b*c*(d
*e*(m - n + 2) - c*f*(m + p + 2)) + d*(a*d*f*(n + p) + b*(d*e*(m + 1) - c*f*(m + n + p + 1)))*x, x], x], x] /;
FreeQ[{a, b, c, d, e, f, p}, x] && LtQ[m, -1] && GtQ[n, 1] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p
] || IntegersQ[p, m + n])

Rule 147

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol]
:> -Simp[((a*d*f*h*(n + 2) + b*c*f*h*(m + 2) - b*d*(f*g + e*h)*(m + n + 3) - b*d*f*h*(m + n + 2)*x)*(a + b*x)^
(m + 1)*(c + d*x)^(n + 1))/(b^2*d^2*(m + n + 2)*(m + n + 3)), x] + Dist[(a^2*d^2*f*h*(n + 1)*(n + 2) + a*b*d*(
n + 1)*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b^2*(c^2*f*h*(m + 1)*(m + 2) - c*d*(f*g + e*h)*(m + 1)*
(m + n + 3) + d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2*d^2*(m + n + 2)*(m + n + 3)), Int[(a + b*x)^m*(c + d*x)^n
, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && NeQ[m + n + 2, 0] && NeQ[m + n + 3, 0]

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)^3}{\sqrt{1-2 x} (2+3 x)^2} \, dx &=\frac{\sqrt{1-2 x} (3+5 x)^2}{21 (2+3 x)}-\frac{1}{21} \int \frac{(-92-190 x) (3+5 x)}{\sqrt{1-2 x} (2+3 x)} \, dx\\ &=\frac{\sqrt{1-2 x} (3+5 x)^2}{21 (2+3 x)}-\frac{10}{189} \sqrt{1-2 x} (214+95 x)+\frac{104}{189} \int \frac{1}{\sqrt{1-2 x} (2+3 x)} \, dx\\ &=\frac{\sqrt{1-2 x} (3+5 x)^2}{21 (2+3 x)}-\frac{10}{189} \sqrt{1-2 x} (214+95 x)-\frac{104}{189} \operatorname{Subst}\left (\int \frac{1}{\frac{7}{2}-\frac{3 x^2}{2}} \, dx,x,\sqrt{1-2 x}\right )\\ &=\frac{\sqrt{1-2 x} (3+5 x)^2}{21 (2+3 x)}-\frac{10}{189} \sqrt{1-2 x} (214+95 x)-\frac{208 \tanh ^{-1}\left (\sqrt{\frac{3}{7}} \sqrt{1-2 x}\right )}{189 \sqrt{21}}\\ \end{align*}

Mathematica [A]  time = 0.0303162, size = 58, normalized size = 0.79 \[ \frac{-\frac{21 \sqrt{1-2 x} \left (2625 x^2+8050 x+4199\right )}{3 x+2}-208 \sqrt{21} \tanh ^{-1}\left (\sqrt{\frac{3}{7}} \sqrt{1-2 x}\right )}{3969} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-21*Sqrt[1 - 2*x]*(4199 + 8050*x + 2625*x^2))/(2 + 3*x) - 208*Sqrt[21]*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]])/396
9

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Maple [A]  time = 0.009, size = 54, normalized size = 0.7 \begin{align*}{\frac{125}{54} \left ( 1-2\,x \right ) ^{{\frac{3}{2}}}}-{\frac{725}{54}\sqrt{1-2\,x}}-{\frac{2}{567}\sqrt{1-2\,x} \left ( -2\,x-{\frac{4}{3}} \right ) ^{-1}}-{\frac{208\,\sqrt{21}}{3969}{\it Artanh} \left ({\frac{\sqrt{21}}{7}\sqrt{1-2\,x}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

125/54*(1-2*x)^(3/2)-725/54*(1-2*x)^(1/2)-2/567*(1-2*x)^(1/2)/(-2*x-4/3)-208/3969*arctanh(1/7*21^(1/2)*(1-2*x)
^(1/2))*21^(1/2)

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Maxima [A]  time = 2.70011, size = 96, normalized size = 1.32 \begin{align*} \frac{125}{54} \,{\left (-2 \, x + 1\right )}^{\frac{3}{2}} + \frac{104}{3969} \, \sqrt{21} \log \left (-\frac{\sqrt{21} - 3 \, \sqrt{-2 \, x + 1}}{\sqrt{21} + 3 \, \sqrt{-2 \, x + 1}}\right ) - \frac{725}{54} \, \sqrt{-2 \, x + 1} + \frac{\sqrt{-2 \, x + 1}}{189 \,{\left (3 \, x + 2\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

125/54*(-2*x + 1)^(3/2) + 104/3969*sqrt(21)*log(-(sqrt(21) - 3*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1)))
- 725/54*sqrt(-2*x + 1) + 1/189*sqrt(-2*x + 1)/(3*x + 2)

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Fricas [A]  time = 1.65918, size = 190, normalized size = 2.6 \begin{align*} \frac{104 \, \sqrt{21}{\left (3 \, x + 2\right )} \log \left (\frac{3 \, x + \sqrt{21} \sqrt{-2 \, x + 1} - 5}{3 \, x + 2}\right ) - 21 \,{\left (2625 \, x^{2} + 8050 \, x + 4199\right )} \sqrt{-2 \, x + 1}}{3969 \,{\left (3 \, x + 2\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3969*(104*sqrt(21)*(3*x + 2)*log((3*x + sqrt(21)*sqrt(-2*x + 1) - 5)/(3*x + 2)) - 21*(2625*x^2 + 8050*x + 41
99)*sqrt(-2*x + 1))/(3*x + 2)

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

[Out]

Exception raised: ValueError

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Giac [A]  time = 1.24405, size = 100, normalized size = 1.37 \begin{align*} \frac{125}{54} \,{\left (-2 \, x + 1\right )}^{\frac{3}{2}} + \frac{104}{3969} \, \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{725}{54} \, \sqrt{-2 \, x + 1} + \frac{\sqrt{-2 \, x + 1}}{189 \,{\left (3 \, x + 2\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

125/54*(-2*x + 1)^(3/2) + 104/3969*sqrt(21)*log(1/2*abs(-2*sqrt(21) + 6*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*
x + 1))) - 725/54*sqrt(-2*x + 1) + 1/189*sqrt(-2*x + 1)/(3*x + 2)