3.2128 \(\int \frac{(2+3 x)^3}{(1-2 x)^{3/2} (3+5 x)^3} \, dx\)

Optimal. Leaf size=80 \[ \frac{7 (3 x+2)^2}{11 \sqrt{1-2 x} (5 x+3)^2}-\frac{\sqrt{1-2 x} (24825 x+15676)}{66550 (5 x+3)^2}-\frac{7143 \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )}{33275 \sqrt{55}} \]

[Out]

(7*(2 + 3*x)^2)/(11*Sqrt[1 - 2*x]*(3 + 5*x)^2) - (Sqrt[1 - 2*x]*(15676 + 24825*x))/(66550*(3 + 5*x)^2) - (7143
*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(33275*Sqrt[55])

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Rubi [A]  time = 0.0190146, 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, 145, 63, 206} \[ \frac{7 (3 x+2)^2}{11 \sqrt{1-2 x} (5 x+3)^2}-\frac{\sqrt{1-2 x} (24825 x+15676)}{66550 (5 x+3)^2}-\frac{7143 \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )}{33275 \sqrt{55}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(7*(2 + 3*x)^2)/(11*Sqrt[1 - 2*x]*(3 + 5*x)^2) - (Sqrt[1 - 2*x]*(15676 + 24825*x))/(66550*(3 + 5*x)^2) - (7143
*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(33275*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 145

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol] :
> Simp[((b^3*c*e*g*(m + 2) - a^3*d*f*h*(n + 2) - a^2*b*(c*f*h*m - d*(f*g + e*h)*(m + n + 3)) - a*b^2*(c*(f*g +
e*h) + d*e*g*(2*m + n + 4)) + b*(a^2*d*f*h*(m - n) - a*b*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(n + 1)) + b^2*(c*(
f*g + e*h)*(m + 1) - d*e*g*(m + n + 2)))*x)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1))/(b^2*(b*c - a*d)^2*(m + 1)*(m
+ 2)), x] + Dist[(f*h)/b^2 - (d*(m + n + 3)*(a^2*d*f*h*(m - n) - a*b*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(n + 1)
) + b^2*(c*(f*g + e*h)*(m + 1) - d*e*g*(m + n + 2))))/(b^2*(b*c - a*d)^2*(m + 1)*(m + 2)), Int[(a + b*x)^(m +
2)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && (LtQ[m, -2] || (EqQ[m + n + 3, 0] &&  !L
tQ[n, -2]))

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)^{3/2} (3+5 x)^3} \, dx &=\frac{7 (2+3 x)^2}{11 \sqrt{1-2 x} (3+5 x)^2}-\frac{1}{11} \int \frac{(-16-3 x) (2+3 x)}{\sqrt{1-2 x} (3+5 x)^3} \, dx\\ &=\frac{7 (2+3 x)^2}{11 \sqrt{1-2 x} (3+5 x)^2}-\frac{\sqrt{1-2 x} (15676+24825 x)}{66550 (3+5 x)^2}+\frac{7143 \int \frac{1}{\sqrt{1-2 x} (3+5 x)} \, dx}{66550}\\ &=\frac{7 (2+3 x)^2}{11 \sqrt{1-2 x} (3+5 x)^2}-\frac{\sqrt{1-2 x} (15676+24825 x)}{66550 (3+5 x)^2}-\frac{7143 \operatorname{Subst}\left (\int \frac{1}{\frac{11}{2}-\frac{5 x^2}{2}} \, dx,x,\sqrt{1-2 x}\right )}{66550}\\ &=\frac{7 (2+3 x)^2}{11 \sqrt{1-2 x} (3+5 x)^2}-\frac{\sqrt{1-2 x} (15676+24825 x)}{66550 (3+5 x)^2}-\frac{7143 \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )}{33275 \sqrt{55}}\\ \end{align*}

Mathematica [C]  time = 0.0141634, size = 59, normalized size = 0.74 \[ \frac{14286 (5 x+3)^2 \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};-\frac{5}{11} (2 x-1)\right )+11 \left (163350 x^2+195005 x+58186\right )}{332750 \sqrt{1-2 x} (5 x+3)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(11*(58186 + 195005*x + 163350*x^2) + 14286*(3 + 5*x)^2*Hypergeometric2F1[-1/2, 1, 1/2, (-5*(-1 + 2*x))/11])/(
332750*Sqrt[1 - 2*x]*(3 + 5*x)^2)

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Maple [A]  time = 0.011, size = 57, normalized size = 0.7 \begin{align*}{\frac{343}{1331}{\frac{1}{\sqrt{1-2\,x}}}}+{\frac{50}{1331\, \left ( -10\,x-6 \right ) ^{2}} \left ({\frac{41}{50} \left ( 1-2\,x \right ) ^{{\frac{3}{2}}}}-{\frac{2277}{1250}\sqrt{1-2\,x}} \right ) }-{\frac{7143\,\sqrt{55}}{1830125}{\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)^(3/2)/(3+5*x)^3,x)

[Out]

343/1331/(1-2*x)^(1/2)+50/1331*(41/50*(1-2*x)^(3/2)-2277/1250*(1-2*x)^(1/2))/(-10*x-6)^2-7143/1830125*arctanh(
1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)

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Maxima [A]  time = 2.20935, size = 112, normalized size = 1.4 \begin{align*} \frac{7143}{3660250} \, \sqrt{55} \log \left (-\frac{\sqrt{55} - 5 \, \sqrt{-2 \, x + 1}}{\sqrt{55} + 5 \, \sqrt{-2 \, x + 1}}\right ) + \frac{2 \,{\left (107700 \,{\left (2 \, x - 1\right )}^{2} + 945527 \, x + 46024\right )}}{33275 \,{\left (25 \,{\left (-2 \, x + 1\right )}^{\frac{5}{2}} - 110 \,{\left (-2 \, x + 1\right )}^{\frac{3}{2}} + 121 \, \sqrt{-2 \, x + 1}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

7143/3660250*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 2/33275*(107700*(2*x
- 1)^2 + 945527*x + 46024)/(25*(-2*x + 1)^(5/2) - 110*(-2*x + 1)^(3/2) + 121*sqrt(-2*x + 1))

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Fricas [A]  time = 1.5551, size = 255, normalized size = 3.19 \begin{align*} \frac{7143 \, \sqrt{55}{\left (50 \, x^{3} + 35 \, x^{2} - 12 \, x - 9\right )} \log \left (\frac{5 \, x + \sqrt{55} \sqrt{-2 \, x + 1} - 8}{5 \, x + 3}\right ) - 55 \,{\left (430800 \, x^{2} + 514727 \, x + 153724\right )} \sqrt{-2 \, x + 1}}{3660250 \,{\left (50 \, x^{3} + 35 \, x^{2} - 12 \, x - 9\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3660250*(7143*sqrt(55)*(50*x^3 + 35*x^2 - 12*x - 9)*log((5*x + sqrt(55)*sqrt(-2*x + 1) - 8)/(5*x + 3)) - 55*
(430800*x^2 + 514727*x + 153724)*sqrt(-2*x + 1))/(50*x^3 + 35*x^2 - 12*x - 9)

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

[Out]

Exception raised: ValueError

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Giac [A]  time = 2.00578, size = 104, normalized size = 1.3 \begin{align*} \frac{7143}{3660250} \, \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{343}{1331 \, \sqrt{-2 \, x + 1}} + \frac{1025 \,{\left (-2 \, x + 1\right )}^{\frac{3}{2}} - 2277 \, \sqrt{-2 \, x + 1}}{133100 \,{\left (5 \, x + 3\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

7143/3660250*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 343/1331/s
qrt(-2*x + 1) + 1/133100*(1025*(-2*x + 1)^(3/2) - 2277*sqrt(-2*x + 1))/(5*x + 3)^2