3.2183 \(\int \frac{(2+3 x)^3}{(1-2 x)^{5/2} (3+5 x)^2} \, dx\)

Optimal. Leaf size=80 \[ \frac{7 (3 x+2)^2}{33 (1-2 x)^{3/2} (5 x+3)}-\frac{2 (17112 x+10309)}{19965 \sqrt{1-2 x} (5 x+3)}-\frac{208 \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )}{6655 \sqrt{55}} \]

[Out]

(7*(2 + 3*x)^2)/(33*(1 - 2*x)^(3/2)*(3 + 5*x)) - (2*(10309 + 17112*x))/(19965*Sqrt[1 - 2*x]*(3 + 5*x)) - (208*
ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(6655*Sqrt[55])

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Rubi [A]  time = 0.018714, antiderivative size = 80, 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, 144, 63, 206} \[ \frac{7 (3 x+2)^2}{33 (1-2 x)^{3/2} (5 x+3)}-\frac{2 (17112 x+10309)}{19965 \sqrt{1-2 x} (5 x+3)}-\frac{208 \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )}{6655 \sqrt{55}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(7*(2 + 3*x)^2)/(33*(1 - 2*x)^(3/2)*(3 + 5*x)) - (2*(10309 + 17112*x))/(19965*Sqrt[1 - 2*x]*(3 + 5*x)) - (208*
ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(6655*Sqrt[55])

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 144

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol] :>
Simp[((b^2*c*d*e*g*(n + 1) + a^2*c*d*f*h*(n + 1) + a*b*(d^2*e*g*(m + 1) + c^2*f*h*(m + 1) - c*d*(f*g + e*h)*(
m + n + 2)) + (a^2*d^2*f*h*(n + 1) - a*b*d^2*(f*g + e*h)*(n + 1) + b^2*(c^2*f*h*(m + 1) - c*d*(f*g + e*h)*(m +
1) + d^2*e*g*(m + n + 2)))*x)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1))/(b*d*(b*c - a*d)^2*(m + 1)*(n + 1)), x] -
Dist[(a^2*d^2*f*h*(2 + 3*n + 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
*(2 + 3*m + m^2) - c*d*(f*g + e*h)*(m + 1)*(m + n + 3) + d^2*e*g*(6 + m^2 + 5*n + n^2 + m*(2*n + 5))))/(b*d*(b
*c - a*d)^2*(m + 1)*(n + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f, g, h
}, x] && LtQ[m, -1] && LtQ[n, -1]

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{(2+3 x)^3}{(1-2 x)^{5/2} (3+5 x)^2} \, dx &=\frac{7 (2+3 x)^2}{33 (1-2 x)^{3/2} (3+5 x)}-\frac{1}{33} \int \frac{(2+3 x) (50+96 x)}{(1-2 x)^{3/2} (3+5 x)^2} \, dx\\ &=\frac{7 (2+3 x)^2}{33 (1-2 x)^{3/2} (3+5 x)}-\frac{2 (10309+17112 x)}{19965 \sqrt{1-2 x} (3+5 x)}+\frac{104 \int \frac{1}{\sqrt{1-2 x} (3+5 x)} \, dx}{6655}\\ &=\frac{7 (2+3 x)^2}{33 (1-2 x)^{3/2} (3+5 x)}-\frac{2 (10309+17112 x)}{19965 \sqrt{1-2 x} (3+5 x)}-\frac{104 \operatorname{Subst}\left (\int \frac{1}{\frac{11}{2}-\frac{5 x^2}{2}} \, dx,x,\sqrt{1-2 x}\right )}{6655}\\ &=\frac{7 (2+3 x)^2}{33 (1-2 x)^{3/2} (3+5 x)}-\frac{2 (10309+17112 x)}{19965 \sqrt{1-2 x} (3+5 x)}-\frac{208 \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )}{6655 \sqrt{55}}\\ \end{align*}

Mathematica [A]  time = 0.0357541, size = 75, normalized size = 0.94 \[ -\frac{-55 \left (106563 x^2+57832 x-3678\right )-624 \sqrt{55} \sqrt{1-2 x} \left (10 x^2+x-3\right ) \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )}{1098075 (1-2 x)^{3/2} (5 x+3)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(-55*(-3678 + 57832*x + 106563*x^2) - 624*Sqrt[55]*Sqrt[1 - 2*x]*(-3 + x + 10*x^2)*ArcTanh[Sqrt[5/11]*Sqrt[1
- 2*x]])/(1098075*(1 - 2*x)^(3/2)*(3 + 5*x))

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Maple [A]  time = 0.012, size = 54, normalized size = 0.7 \begin{align*}{\frac{343}{726} \left ( 1-2\,x \right ) ^{-{\frac{3}{2}}}}-{\frac{1421}{2662}{\frac{1}{\sqrt{1-2\,x}}}}+{\frac{2}{33275}\sqrt{1-2\,x} \left ( -2\,x-{\frac{6}{5}} \right ) ^{-1}}-{\frac{208\,\sqrt{55}}{366025}{\it Artanh} \left ({\frac{\sqrt{55}}{11}\sqrt{1-2\,x}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

343/726/(1-2*x)^(3/2)-1421/2662/(1-2*x)^(1/2)+2/33275*(1-2*x)^(1/2)/(-2*x-6/5)-208/366025*arctanh(1/11*55^(1/2
)*(1-2*x)^(1/2))*55^(1/2)

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Maxima [A]  time = 1.58471, size = 100, normalized size = 1.25 \begin{align*} \frac{104}{366025} \, \sqrt{55} \log \left (-\frac{\sqrt{55} - 5 \, \sqrt{-2 \, x + 1}}{\sqrt{55} + 5 \, \sqrt{-2 \, x + 1}}\right ) - \frac{106563 \,{\left (2 \, x - 1\right )}^{2} + 657580 \, x - 121275}{39930 \,{\left (5 \,{\left (-2 \, x + 1\right )}^{\frac{5}{2}} - 11 \,{\left (-2 \, x + 1\right )}^{\frac{3}{2}}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

104/366025*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 1/39930*(106563*(2*x -
1)^2 + 657580*x - 121275)/(5*(-2*x + 1)^(5/2) - 11*(-2*x + 1)^(3/2))

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Fricas [A]  time = 1.40257, size = 244, normalized size = 3.05 \begin{align*} \frac{312 \, \sqrt{55}{\left (20 \, x^{3} - 8 \, x^{2} - 7 \, x + 3\right )} \log \left (\frac{5 \, x + \sqrt{55} \sqrt{-2 \, x + 1} - 8}{5 \, x + 3}\right ) + 55 \,{\left (106563 \, x^{2} + 57832 \, x - 3678\right )} \sqrt{-2 \, x + 1}}{1098075 \,{\left (20 \, x^{3} - 8 \, x^{2} - 7 \, x + 3\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1098075*(312*sqrt(55)*(20*x^3 - 8*x^2 - 7*x + 3)*log((5*x + sqrt(55)*sqrt(-2*x + 1) - 8)/(5*x + 3)) + 55*(10
6563*x^2 + 57832*x - 3678)*sqrt(-2*x + 1))/(20*x^3 - 8*x^2 - 7*x + 3)

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

[Out]

Exception raised: ValueError

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Giac [A]  time = 1.79907, size = 104, normalized size = 1.3 \begin{align*} \frac{104}{366025} \, \sqrt{55} \log \left (\frac{{\left | -2 \, \sqrt{55} + 10 \, \sqrt{-2 \, x + 1} \right |}}{2 \,{\left (\sqrt{55} + 5 \, \sqrt{-2 \, x + 1}\right )}}\right ) - \frac{49 \,{\left (87 \, x - 5\right )}}{3993 \,{\left (2 \, x - 1\right )} \sqrt{-2 \, x + 1}} - \frac{\sqrt{-2 \, x + 1}}{6655 \,{\left (5 \, x + 3\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

104/366025*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 49/3993*(87*
x - 5)/((2*x - 1)*sqrt(-2*x + 1)) - 1/6655*sqrt(-2*x + 1)/(5*x + 3)