\(\int \frac {(2+3 x)^3}{(1-2 x)^{5/2} (3+5 x)^{5/2}} \, dx\) [2630]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [F]
   Maxima [A] (verification not implemented)
   Giac [B] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 26, antiderivative size = 96 \[ \int \frac {(2+3 x)^3}{(1-2 x)^{5/2} (3+5 x)^{5/2}} \, dx=\frac {49}{121 \sqrt {1-2 x} (3+5 x)^{3/2}}-\frac {3679 \sqrt {1-2 x}}{19965 (3+5 x)^{3/2}}+\frac {2 (2+3 x)^3}{33 (1-2 x)^{3/2} (3+5 x)^{3/2}}-\frac {8182 \sqrt {1-2 x}}{219615 \sqrt {3+5 x}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {96, 91, 79, 37} \[ \int \frac {(2+3 x)^3}{(1-2 x)^{5/2} (3+5 x)^{5/2}} \, dx=\frac {2 (3 x+2)^3}{33 (1-2 x)^{3/2} (5 x+3)^{3/2}}-\frac {8182 \sqrt {1-2 x}}{219615 \sqrt {5 x+3}}-\frac {3679 \sqrt {1-2 x}}{19965 (5 x+3)^{3/2}}+\frac {49}{121 \sqrt {1-2 x} (5 x+3)^{3/2}} \]

[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 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]

Rule 79

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] || L
tQ[p, n]))))

Rule 91

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 96

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[m, 1] ||  !SumSimplerQ[p, 1]) && NeQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = \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] (verified)

Time = 0.09 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.39 \[ \int \frac {(2+3 x)^3}{(1-2 x)^{5/2} (3+5 x)^{5/2}} \, dx=\frac {2 \left (13040+52044 x+62232 x^2+19573 x^3\right )}{43923 (1-2 x)^{3/2} (3+5 x)^{3/2}} \]

[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))

Maple [A] (verified)

Time = 4.15 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.33

method result size
gosper \(\frac {\frac {39146}{43923} x^{3}+\frac {41488}{14641} x^{2}+\frac {34696}{14641} x +\frac {26080}{43923}}{\left (1-2 x \right )^{\frac {3}{2}} \left (3+5 x \right )^{\frac {3}{2}}}\) \(32\)
default \(\frac {2 \sqrt {1-2 x}\, \left (19573 x^{3}+62232 x^{2}+52044 x +13040\right )}{43923 \left (3+5 x \right )^{\frac {3}{2}} \left (-1+2 x \right )^{2}}\) \(39\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.55 \[ \int \frac {(2+3 x)^3}{(1-2 x)^{5/2} (3+5 x)^{5/2}} \, dx=\frac {2 \, {\left (19573 \, x^{3} + 62232 \, x^{2} + 52044 \, x + 13040\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{43923 \, {\left (100 \, x^{4} + 20 \, x^{3} - 59 \, x^{2} - 6 \, x + 9\right )}} \]

[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)

Sympy [F]

\[ \int \frac {(2+3 x)^3}{(1-2 x)^{5/2} (3+5 x)^{5/2}} \, dx=\int \frac {\left (3 x + 2\right )^{3}}{\left (1 - 2 x\right )^{\frac {5}{2}} \left (5 x + 3\right )^{\frac {5}{2}}}\, dx \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.79 \[ \int \frac {(2+3 x)^3}{(1-2 x)^{5/2} (3+5 x)^{5/2}} \, dx=-\frac {19573 \, x}{219615 \, \sqrt {-10 \, x^{2} - x + 3}} + \frac {27 \, x^{2}}{10 \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}}} - \frac {19573}{4392300 \, \sqrt {-10 \, x^{2} - x + 3}} + \frac {95567 \, x}{36300 \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}}} + \frac {22039}{36300 \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}}} \]

[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)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 161 vs. \(2 (72) = 144\).

Time = 0.33 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.68 \[ \int \frac {(2+3 x)^3}{(1-2 x)^{5/2} (3+5 x)^{5/2}} \, dx=-\frac {\sqrt {10} {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{3}}{17569200 \, {\left (5 \, x + 3\right )}^{\frac {3}{2}}} - \frac {19 \, \sqrt {10} {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}{133100 \, \sqrt {5 \, x + 3}} + \frac {98 \, {\left (17 \, \sqrt {5} {\left (5 \, x + 3\right )} + 99 \, \sqrt {5}\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5}}{1098075 \, {\left (2 \, x - 1\right )}^{2}} + \frac {\sqrt {10} {\left (5 \, x + 3\right )}^{\frac {3}{2}} {\left (\frac {627 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} + 4\right )}}{1098075 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{3}} \]

[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*sqrt(10)*(5*x + 3)^(3/2)*(627*(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))^2/(5*x + 3)
+ 4)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))^3

Mupad [B] (verification not implemented)

Time = 0.35 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.72 \[ \int \frac {(2+3 x)^3}{(1-2 x)^{5/2} (3+5 x)^{5/2}} \, dx=\frac {\sqrt {5\,x+3}\,\left (\frac {19573\,x^3}{1098075}+\frac {20744\,x^2}{366025}+\frac {17348\,x}{366025}+\frac {2608}{219615}\right )}{\frac {6\,x\,\sqrt {1-2\,x}}{25}+\frac {9\,\sqrt {1-2\,x}}{50}-\frac {7\,x^2\,\sqrt {1-2\,x}}{10}-x^3\,\sqrt {1-2\,x}} \]

[In]

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

[Out]

((5*x + 3)^(1/2)*((17348*x)/366025 + (20744*x^2)/366025 + (19573*x^3)/1098075 + 2608/219615))/((6*x*(1 - 2*x)^
(1/2))/25 + (9*(1 - 2*x)^(1/2))/50 - (7*x^2*(1 - 2*x)^(1/2))/10 - x^3*(1 - 2*x)^(1/2))