Integrand size = 26, antiderivative size = 89 \[ \int \frac {(2+3 x)^3}{(1-2 x)^{5/2} (3+5 x)^{5/2}} \, dx=\frac {343}{132 (1-2 x)^{3/2} (3+5 x)^{3/2}}-\frac {42883 \sqrt {1-2 x}}{79860 (3+5 x)^{3/2}}-\frac {1421}{2662 \sqrt {1-2 x} \sqrt {3+5 x}}+\frac {19573 \sqrt {1-2 x}}{439230 \sqrt {3+5 x}} \] Output:
343/132/(1-2*x)^(3/2)/(3+5*x)^(3/2)-42883/79860*(1-2*x)^(1/2)/(3+5*x)^(3/2 )-1421/2662/(1-2*x)^(1/2)/(3+5*x)^(1/2)+19573/439230*(1-2*x)^(1/2)/(3+5*x) ^(1/2)
Time = 0.08 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.42 \[ \int \frac {(2+3 x)^3}{(1-2 x)^{5/2} (3+5 x)^{5/2}} \, dx=\frac {2 \left (13040+52044 x+62232 x^2+19573 x^3\right )}{43923 (1-2 x)^{3/2} (3+5 x)^{3/2}} \] Input:
Integrate[(2 + 3*x)^3/((1 - 2*x)^(5/2)*(3 + 5*x)^(5/2)),x]
Output:
(2*(13040 + 52044*x + 62232*x^2 + 19573*x^3))/(43923*(1 - 2*x)^(3/2)*(3 + 5*x)^(3/2))
Time = 0.21 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.19, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {105, 100, 27, 87, 48}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {(3 x+2)^3}{(1-2 x)^{5/2} (5 x+3)^{5/2}} \, dx\) |
\(\Big \downarrow \) 105 |
\(\displaystyle \frac {2}{11} \int \frac {(3 x+2)^2}{(1-2 x)^{3/2} (5 x+3)^{5/2}}dx+\frac {2 (3 x+2)^3}{33 (1-2 x)^{3/2} (5 x+3)^{3/2}}\) |
\(\Big \downarrow \) 100 |
\(\displaystyle \frac {2}{11} \left (\frac {49}{22 \sqrt {1-2 x} (5 x+3)^{3/2}}-\frac {1}{22} \int -\frac {617-198 x}{2 \sqrt {1-2 x} (5 x+3)^{5/2}}dx\right )+\frac {2 (3 x+2)^3}{33 (1-2 x)^{3/2} (5 x+3)^{3/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2}{11} \left (\frac {1}{44} \int \frac {617-198 x}{\sqrt {1-2 x} (5 x+3)^{5/2}}dx+\frac {49}{22 \sqrt {1-2 x} (5 x+3)^{3/2}}\right )+\frac {2 (3 x+2)^3}{33 (1-2 x)^{3/2} (5 x+3)^{3/2}}\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {2}{11} \left (\frac {1}{44} \left (\frac {8182}{165} \int \frac {1}{\sqrt {1-2 x} (5 x+3)^{3/2}}dx-\frac {7358 \sqrt {1-2 x}}{165 (5 x+3)^{3/2}}\right )+\frac {49}{22 \sqrt {1-2 x} (5 x+3)^{3/2}}\right )+\frac {2 (3 x+2)^3}{33 (1-2 x)^{3/2} (5 x+3)^{3/2}}\) |
\(\Big \downarrow \) 48 |
\(\displaystyle \frac {2 (3 x+2)^3}{33 (1-2 x)^{3/2} (5 x+3)^{3/2}}+\frac {2}{11} \left (\frac {1}{44} \left (-\frac {16364 \sqrt {1-2 x}}{1815 \sqrt {5 x+3}}-\frac {7358 \sqrt {1-2 x}}{165 (5 x+3)^{3/2}}\right )+\frac {49}{22 \sqrt {1-2 x} (5 x+3)^{3/2}}\right )\) |
Input:
Int[(2 + 3*x)^3/((1 - 2*x)^(5/2)*(3 + 5*x)^(5/2)),x]
Output:
(2*(2 + 3*x)^3)/(33*(1 - 2*x)^(3/2)*(3 + 5*x)^(3/2)) + (2*(49/(22*Sqrt[1 - 2*x]*(3 + 5*x)^(3/2)) + ((-7358*Sqrt[1 - 2*x])/(165*(3 + 5*x)^(3/2)) - (1 6364*Sqrt[1 - 2*x])/(1815*Sqrt[3 + 5*x]))/44))/11
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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] && EqQ[m + n + 2, 0] && NeQ[m, -1]
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Simp[(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] || Intege rQ[p] || !(IntegerQ[n] || !(EqQ[e, 0] || !(EqQ[c, 0] || LtQ[p, n]))))
Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^( p_), x_] :> 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] - Simp[1/(d^2*(d*e - c*f)*(n + 1)) Int[(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])))
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Simp[(a + b*x)^(m + 1)*(c + d*x)^n*((e + f*x)^(p + 1)/((m + 1)*(b*e - a*f))), x] - Simp[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]
Time = 0.27 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.36
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 | \(\frac {\frac {39146}{43923} x^{3}+\frac {41488}{14641} x^{2}+\frac {34696}{14641} x +\frac {26080}{43923}}{\left (1-2 x \right )^{\frac {3}{2}} \left (3+5 x \right )^{\frac {3}{2}}}\) | \(32\) |
orering | \(-\frac {2 \left (-1+2 x \right ) \left (19573 x^{3}+62232 x^{2}+52044 x +13040\right )}{43923 \left (3+5 x \right )^{\frac {3}{2}} \left (1-2 x \right )^{\frac {5}{2}}}\) | \(37\) |
Input:
int((2+3*x)^3/(1-2*x)^(5/2)/(3+5*x)^(5/2),x,method=_RETURNVERBOSE)
Output:
2/43923/(1-2*x)^(3/2)/(3+5*x)^(3/2)*(19573*x^3+62232*x^2+52044*x+13040)
Time = 0.07 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.60 \[ \int \frac {(2+3 x)^3}{(1-2 x)^{5/2} (3+5 x)^{5/2}} \, dx=\frac {2 \, {\left (19573 \, x^{3} + 62232 \, x^{2} + 52044 \, x + 13040\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{43923 \, {\left (100 \, x^{4} + 20 \, x^{3} - 59 \, x^{2} - 6 \, x + 9\right )}} \] Input:
integrate((2+3*x)^3/(1-2*x)^(5/2)/(3+5*x)^(5/2),x, algorithm="fricas")
Output:
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)
\[ \int \frac {(2+3 x)^3}{(1-2 x)^{5/2} (3+5 x)^{5/2}} \, dx=\int \frac {\left (3 x + 2\right )^{3}}{\left (1 - 2 x\right )^{\frac {5}{2}} \left (5 x + 3\right )^{\frac {5}{2}}}\, dx \] Input:
integrate((2+3*x)**3/(1-2*x)**(5/2)/(3+5*x)**(5/2),x)
Output:
Integral((3*x + 2)**3/((1 - 2*x)**(5/2)*(5*x + 3)**(5/2)), x)
Time = 0.04 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.85 \[ \int \frac {(2+3 x)^3}{(1-2 x)^{5/2} (3+5 x)^{5/2}} \, dx=-\frac {19573 \, x}{219615 \, \sqrt {-10 \, x^{2} - x + 3}} + \frac {27 \, x^{2}}{10 \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}}} - \frac {19573}{4392300 \, \sqrt {-10 \, x^{2} - x + 3}} + \frac {95567 \, x}{36300 \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}}} + \frac {22039}{36300 \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}}} \] Input:
integrate((2+3*x)^3/(1-2*x)^(5/2)/(3+5*x)^(5/2),x, algorithm="maxima")
Output:
-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)
Leaf count of result is larger than twice the leaf count of optimal. 161 vs. \(2 (65) = 130\).
Time = 0.18 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.81 \[ \int \frac {(2+3 x)^3}{(1-2 x)^{5/2} (3+5 x)^{5/2}} \, dx=-\frac {\sqrt {10} {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{3}}{17569200 \, {\left (5 \, x + 3\right )}^{\frac {3}{2}}} - \frac {19 \, \sqrt {10} {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}{133100 \, \sqrt {5 \, x + 3}} + \frac {98 \, {\left (17 \, \sqrt {5} {\left (5 \, x + 3\right )} + 99 \, \sqrt {5}\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5}}{1098075 \, {\left (2 \, x - 1\right )}^{2}} + \frac {\sqrt {10} {\left (5 \, x + 3\right )}^{\frac {3}{2}} {\left (\frac {627 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} + 4\right )}}{1098075 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{3}} \] Input:
integrate((2+3*x)^3/(1-2*x)^(5/2)/(3+5*x)^(5/2),x, algorithm="giac")
Output:
-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
Time = 0.20 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.78 \[ \int \frac {(2+3 x)^3}{(1-2 x)^{5/2} (3+5 x)^{5/2}} \, dx=\frac {\sqrt {5\,x+3}\,\left (\frac {19573\,x^3}{1098075}+\frac {20744\,x^2}{366025}+\frac {17348\,x}{366025}+\frac {2608}{219615}\right )}{\frac {6\,x\,\sqrt {1-2\,x}}{25}+\frac {9\,\sqrt {1-2\,x}}{50}-\frac {7\,x^2\,\sqrt {1-2\,x}}{10}-x^3\,\sqrt {1-2\,x}} \] Input:
int((3*x + 2)^3/((1 - 2*x)^(5/2)*(5*x + 3)^(5/2)),x)
Output:
((5*x + 3)^(1/2)*((17348*x)/366025 + (20744*x^2)/366025 + (19573*x^3)/1098 075 + 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))
Time = 0.18 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.48 \[ \int \frac {(2+3 x)^3}{(1-2 x)^{5/2} (3+5 x)^{5/2}} \, dx=\frac {-\frac {39146}{43923} x^{3}-\frac {41488}{14641} x^{2}-\frac {34696}{14641} x -\frac {26080}{43923}}{\sqrt {5 x +3}\, \sqrt {-2 x +1}\, \left (10 x^{2}+x -3\right )} \] Input:
int((2+3*x)^3/(1-2*x)^(5/2)/(3+5*x)^(5/2),x)
Output:
(2*( - 19573*x**3 - 62232*x**2 - 52044*x - 13040))/(43923*sqrt(5*x + 3)*sq rt( - 2*x + 1)*(10*x**2 + x - 3))