3.2631 \(\int \frac{(2+3 x)^2}{(1-2 x)^{5/2} (3+5 x)^{5/2}} \, dx\)

Optimal. Leaf size=89 \[ -\frac{3298 \sqrt{1-2 x}}{43923 \sqrt{5 x+3}}-\frac{1649 \sqrt{1-2 x}}{7986 (5 x+3)^{3/2}}+\frac{14}{121 (5 x+3)^{3/2} \sqrt{1-2 x}}+\frac{49}{66 (5 x+3)^{3/2} (1-2 x)^{3/2}} \]

[Out]

49/(66*(1 - 2*x)^(3/2)*(3 + 5*x)^(3/2)) + 14/(121*Sqrt[1 - 2*x]*(3 + 5*x)^(3/2)) - (1649*Sqrt[1 - 2*x])/(7986*
(3 + 5*x)^(3/2)) - (3298*Sqrt[1 - 2*x])/(43923*Sqrt[3 + 5*x])

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Rubi [A]  time = 0.0169524, antiderivative size = 89, 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 = {89, 78, 45, 37} \[ -\frac{3298 \sqrt{1-2 x}}{43923 \sqrt{5 x+3}}-\frac{1649 \sqrt{1-2 x}}{7986 (5 x+3)^{3/2}}+\frac{14}{121 (5 x+3)^{3/2} \sqrt{1-2 x}}+\frac{49}{66 (5 x+3)^{3/2} (1-2 x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

49/(66*(1 - 2*x)^(3/2)*(3 + 5*x)^(3/2)) + 14/(121*Sqrt[1 - 2*x]*(3 + 5*x)^(3/2)) - (1649*Sqrt[1 - 2*x])/(7986*
(3 + 5*x)^(3/2)) - (3298*Sqrt[1 - 2*x])/(43923*Sqrt[3 + 5*x])

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 45

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*Simplify[m + n + 2])/((b*c - a*d)*(m + 1)), Int[(a + b*x)^Simplify[m +
1]*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[Simplify[m + n + 2], 0] &&
NeQ[m, -1] &&  !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (
SumSimplerQ[m, 1] ||  !SumSimplerQ[n, 1])

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)^2}{(1-2 x)^{5/2} (3+5 x)^{5/2}} \, dx &=\frac{49}{66 (1-2 x)^{3/2} (3+5 x)^{3/2}}-\frac{1}{66} \int \frac{-\frac{381}{2}+297 x}{(1-2 x)^{3/2} (3+5 x)^{5/2}} \, dx\\ &=\frac{49}{66 (1-2 x)^{3/2} (3+5 x)^{3/2}}+\frac{14}{121 \sqrt{1-2 x} (3+5 x)^{3/2}}+\frac{1649}{484} \int \frac{1}{\sqrt{1-2 x} (3+5 x)^{5/2}} \, dx\\ &=\frac{49}{66 (1-2 x)^{3/2} (3+5 x)^{3/2}}+\frac{14}{121 \sqrt{1-2 x} (3+5 x)^{3/2}}-\frac{1649 \sqrt{1-2 x}}{7986 (3+5 x)^{3/2}}+\frac{1649 \int \frac{1}{\sqrt{1-2 x} (3+5 x)^{3/2}} \, dx}{3993}\\ &=\frac{49}{66 (1-2 x)^{3/2} (3+5 x)^{3/2}}+\frac{14}{121 \sqrt{1-2 x} (3+5 x)^{3/2}}-\frac{1649 \sqrt{1-2 x}}{7986 (3+5 x)^{3/2}}-\frac{3298 \sqrt{1-2 x}}{43923 \sqrt{3+5 x}}\\ \end{align*}

Mathematica [A]  time = 0.0149803, size = 37, normalized size = 0.42 \[ \frac{-65960 x^3-9894 x^2+49200 x+18728}{43923 (1-2 x)^{3/2} (5 x+3)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(18728 + 49200*x - 9894*x^2 - 65960*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 = 0.4 \begin{align*} -{\frac{65960\,{x}^{3}+9894\,{x}^{2}-49200\,x-18728}{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)^2/(1-2*x)^(5/2)/(3+5*x)^(5/2),x)

[Out]

-2/43923*(32980*x^3+4947*x^2-24600*x-9364)/(3+5*x)^(3/2)/(1-2*x)^(3/2)

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Maxima [A]  time = 1.10382, size = 80, normalized size = 0.9 \begin{align*} \frac{6596 \, x}{43923 \, \sqrt{-10 \, x^{2} - x + 3}} + \frac{1649}{219615 \, \sqrt{-10 \, x^{2} - x + 3}} + \frac{1229 \, x}{1815 \,{\left (-10 \, x^{2} - x + 3\right )}^{\frac{3}{2}}} + \frac{733}{1815 \,{\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)^2/(1-2*x)^(5/2)/(3+5*x)^(5/2),x, algorithm="maxima")

[Out]

6596/43923*x/sqrt(-10*x^2 - x + 3) + 1649/219615/sqrt(-10*x^2 - x + 3) + 1229/1815*x/(-10*x^2 - x + 3)^(3/2) +
733/1815/(-10*x^2 - x + 3)^(3/2)

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Fricas [A]  time = 1.70644, size = 158, normalized size = 1.78 \begin{align*} -\frac{2 \,{\left (32980 \, x^{3} + 4947 \, x^{2} - 24600 \, x - 9364\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)^2/(1-2*x)^(5/2)/(3+5*x)^(5/2),x, algorithm="fricas")

[Out]

-2/43923*(32980*x^3 + 4947*x^2 - 24600*x - 9364)*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]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (3 x + 2\right )^{2}}{\left (1 - 2 x\right )^{\frac{5}{2}} \left (5 x + 3\right )^{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 2.67135, size = 223, normalized size = 2.51 \begin{align*} -\frac{\sqrt{10}{\left (\sqrt{2} \sqrt{-10 \, x + 5} - \sqrt{22}\right )}^{3}}{3513840 \,{\left (5 \, x + 3\right )}^{\frac{3}{2}}} - \frac{13 \, \sqrt{10}{\left (\sqrt{2} \sqrt{-10 \, x + 5} - \sqrt{22}\right )}}{26620 \, \sqrt{5 \, x + 3}} - \frac{14 \,{\left (164 \, \sqrt{5}{\left (5 \, x + 3\right )} - 1287 \, \sqrt{5}\right )} \sqrt{5 \, x + 3} \sqrt{-10 \, x + 5}}{1098075 \,{\left (2 \, x - 1\right )}^{2}} + \frac{{\left (\frac{429 \, \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}}}{219615 \,{\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)^2/(1-2*x)^(5/2)/(3+5*x)^(5/2),x, algorithm="giac")

[Out]

-1/3513840*sqrt(10)*(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))^3/(5*x + 3)^(3/2) - 13/26620*sqrt(10)*(sqrt(2)*sqrt(-
10*x + 5) - sqrt(22))/sqrt(5*x + 3) - 14/1098075*(164*sqrt(5)*(5*x + 3) - 1287*sqrt(5))*sqrt(5*x + 3)*sqrt(-10
*x + 5)/(2*x - 1)^2 + 1/219615*(429*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