3.2630 \(\int \frac{(2+3 x)^3}{(1-2 x)^{5/2} (3+5 x)^{5/2}} \, dx\)

Optimal. Leaf size=96 \[ \frac{2 (3 x+2)^3}{33 (1-2 x)^{3/2} (5 x+3)^{3/2}}-\frac{8182 \sqrt{1-2 x}}{219615 \sqrt{5 x+3}}-\frac{3679 \sqrt{1-2 x}}{19965 (5 x+3)^{3/2}}+\frac{49}{121 \sqrt{1-2 x} (5 x+3)^{3/2}} \]

[Out]

49/(121*Sqrt[1 - 2*x]*(3 + 5*x)^(3/2)) - (3679*Sqrt[1 - 2*x])/(19965*(3 + 5*x)^(3/2)) + (2*(2 + 3*x)^3)/(33*(1
- 2*x)^(3/2)*(3 + 5*x)^(3/2)) - (8182*Sqrt[1 - 2*x])/(219615*Sqrt[3 + 5*x])

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Rubi [A]  time = 0.0178367, antiderivative size = 96, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {94, 89, 78, 37} \[ \frac{2 (3 x+2)^3}{33 (1-2 x)^{3/2} (5 x+3)^{3/2}}-\frac{8182 \sqrt{1-2 x}}{219615 \sqrt{5 x+3}}-\frac{3679 \sqrt{1-2 x}}{19965 (5 x+3)^{3/2}}+\frac{49}{121 \sqrt{1-2 x} (5 x+3)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

49/(121*Sqrt[1 - 2*x]*(3 + 5*x)^(3/2)) - (3679*Sqrt[1 - 2*x])/(19965*(3 + 5*x)^(3/2)) + (2*(2 + 3*x)^3)/(33*(1
- 2*x)^(3/2)*(3 + 5*x)^(3/2)) - (8182*Sqrt[1 - 2*x])/(219615*Sqrt[3 + 5*x])

Rule 94

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((a + b
*x)^(m + 1)*(c + d*x)^n*(e + f*x)^(p + 1))/((m + 1)*(b*e - a*f)), x] - Dist[(n*(d*e - c*f))/((m + 1)*(b*e - a*
f)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[
m + n + p + 2, 0] && GtQ[n, 0] &&  !(SumSimplerQ[p, 1] &&  !SumSimplerQ[m, 1])

Rule 89

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c - a*
d)^2*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d^2*(d*e - c*f)*(n + 1)), x] - Dist[1/(d^2*(d*e - c*f)*(n + 1)), In
t[(c + d*x)^(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*(p + 1)) - 2*a*b*d*(d*e*
(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ
[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))

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 37

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] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rubi steps

\begin{align*} \int \frac{(2+3 x)^3}{(1-2 x)^{5/2} (3+5 x)^{5/2}} \, dx &=\frac{2 (2+3 x)^3}{33 (1-2 x)^{3/2} (3+5 x)^{3/2}}+\frac{2}{11} \int \frac{(2+3 x)^2}{(1-2 x)^{3/2} (3+5 x)^{5/2}} \, dx\\ &=\frac{49}{121 \sqrt{1-2 x} (3+5 x)^{3/2}}+\frac{2 (2+3 x)^3}{33 (1-2 x)^{3/2} (3+5 x)^{3/2}}-\frac{1}{121} \int \frac{-\frac{617}{2}+99 x}{\sqrt{1-2 x} (3+5 x)^{5/2}} \, dx\\ &=\frac{49}{121 \sqrt{1-2 x} (3+5 x)^{3/2}}-\frac{3679 \sqrt{1-2 x}}{19965 (3+5 x)^{3/2}}+\frac{2 (2+3 x)^3}{33 (1-2 x)^{3/2} (3+5 x)^{3/2}}+\frac{4091 \int \frac{1}{\sqrt{1-2 x} (3+5 x)^{3/2}} \, dx}{19965}\\ &=\frac{49}{121 \sqrt{1-2 x} (3+5 x)^{3/2}}-\frac{3679 \sqrt{1-2 x}}{19965 (3+5 x)^{3/2}}+\frac{2 (2+3 x)^3}{33 (1-2 x)^{3/2} (3+5 x)^{3/2}}-\frac{8182 \sqrt{1-2 x}}{219615 \sqrt{3+5 x}}\\ \end{align*}

Mathematica [A]  time = 0.0163435, size = 37, normalized size = 0.39 \[ \frac{2 \left (19573 x^3+62232 x^2+52044 x+13040\right )}{43923 (1-2 x)^{3/2} (5 x+3)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(13040 + 52044*x + 62232*x^2 + 19573*x^3))/(43923*(1 - 2*x)^(3/2)*(3 + 5*x)^(3/2))

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Maple [A]  time = 0.003, size = 32, normalized size = 0.3 \begin{align*}{\frac{39146\,{x}^{3}+124464\,{x}^{2}+104088\,x+26080}{43923} \left ( 1-2\,x \right ) ^{-{\frac{3}{2}}} \left ( 3+5\,x \right ) ^{-{\frac{3}{2}}}} \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)^(5/2),x)

[Out]

2/43923*(19573*x^3+62232*x^2+52044*x+13040)/(3+5*x)^(3/2)/(1-2*x)^(3/2)

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Maxima [A]  time = 1.13424, size = 103, normalized size = 1.07 \begin{align*} -\frac{19573 \, x}{219615 \, \sqrt{-10 \, x^{2} - x + 3}} + \frac{27 \, x^{2}}{10 \,{\left (-10 \, x^{2} - x + 3\right )}^{\frac{3}{2}}} - \frac{19573}{4392300 \, \sqrt{-10 \, x^{2} - x + 3}} + \frac{95567 \, x}{36300 \,{\left (-10 \, x^{2} - x + 3\right )}^{\frac{3}{2}}} + \frac{22039}{36300 \,{\left (-10 \, x^{2} - x + 3\right )}^{\frac{3}{2}}} \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)^(5/2),x, algorithm="maxima")

[Out]

-19573/219615*x/sqrt(-10*x^2 - x + 3) + 27/10*x^2/(-10*x^2 - x + 3)^(3/2) - 19573/4392300/sqrt(-10*x^2 - x + 3
) + 95567/36300*x/(-10*x^2 - x + 3)^(3/2) + 22039/36300/(-10*x^2 - x + 3)^(3/2)

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Fricas [A]  time = 1.69847, size = 159, normalized size = 1.66 \begin{align*} \frac{2 \,{\left (19573 \, x^{3} + 62232 \, x^{2} + 52044 \, x + 13040\right )} \sqrt{5 \, x + 3} \sqrt{-2 \, x + 1}}{43923 \,{\left (100 \, x^{4} + 20 \, x^{3} - 59 \, x^{2} - 6 \, x + 9\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)^(5/2),x, algorithm="fricas")

[Out]

2/43923*(19573*x^3 + 62232*x^2 + 52044*x + 13040)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(100*x^4 + 20*x^3 - 59*x^2 - 6*
x + 9)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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)**(5/2),x)

[Out]

Timed out

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Giac [B]  time = 2.55302, size = 223, normalized size = 2.32 \begin{align*} -\frac{\sqrt{10}{\left (\sqrt{2} \sqrt{-10 \, x + 5} - \sqrt{22}\right )}^{3}}{17569200 \,{\left (5 \, x + 3\right )}^{\frac{3}{2}}} - \frac{19 \, \sqrt{10}{\left (\sqrt{2} \sqrt{-10 \, x + 5} - \sqrt{22}\right )}}{133100 \, \sqrt{5 \, x + 3}} + \frac{98 \,{\left (17 \, \sqrt{5}{\left (5 \, x + 3\right )} + 99 \, \sqrt{5}\right )} \sqrt{5 \, x + 3} \sqrt{-10 \, x + 5}}{1098075 \,{\left (2 \, x - 1\right )}^{2}} + \frac{{\left (\frac{627 \, \sqrt{10}{\left (\sqrt{2} \sqrt{-10 \, x + 5} - \sqrt{22}\right )}^{2}}{5 \, x + 3} + 4 \, \sqrt{10}\right )}{\left (5 \, x + 3\right )}^{\frac{3}{2}}}{1098075 \,{\left (\sqrt{2} \sqrt{-10 \, x + 5} - \sqrt{22}\right )}^{3}} \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)^(5/2),x, algorithm="giac")

[Out]

-1/17569200*sqrt(10)*(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))^3/(5*x + 3)^(3/2) - 19/133100*sqrt(10)*(sqrt(2)*sqrt
(-10*x + 5) - sqrt(22))/sqrt(5*x + 3) + 98/1098075*(17*sqrt(5)*(5*x + 3) + 99*sqrt(5))*sqrt(5*x + 3)*sqrt(-10*
x + 5)/(2*x - 1)^2 + 1/1098075*(627*sqrt(10)*(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))^2/(5*x + 3) + 4*sqrt(10))*(5
*x + 3)^(3/2)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))^3