Integrand size = 27, antiderivative size = 67 \[ \int \frac {\sqrt {4+3 x} \left (2+3 x+x^2\right )}{(5-2 x)^{9/2}} \, dx=\frac {9 (4+3 x)^{3/2}}{46 (5-2 x)^{7/2}}-\frac {26 (4+3 x)^{3/2}}{529 (5-2 x)^{5/2}}+\frac {217 (4+3 x)^{3/2}}{73002 (5-2 x)^{3/2}} \] Output:
9/46*(4+3*x)^(3/2)/(5-2*x)^(7/2)-26/529*(4+3*x)^(3/2)/(5-2*x)^(5/2)+217/73 002*(4+3*x)^(3/2)/(5-2*x)^(3/2)
Time = 0.13 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.48 \[ \int \frac {\sqrt {4+3 x} \left (2+3 x+x^2\right )}{(5-2 x)^{9/2}} \, dx=\frac {2 (4+3 x)^{3/2} \left (442+709 x+217 x^2\right )}{36501 (5-2 x)^{7/2}} \] Input:
Integrate[(Sqrt[4 + 3*x]*(2 + 3*x + x^2))/(5 - 2*x)^(9/2),x]
Output:
(2*(4 + 3*x)^(3/2)*(442 + 709*x + 217*x^2))/(36501*(5 - 2*x)^(7/2))
Time = 0.28 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.07, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {1193, 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.
\(\displaystyle \int \frac {\sqrt {3 x+4} \left (x^2+3 x+2\right )}{(5-2 x)^{9/2}} \, dx\) |
\(\Big \downarrow \) 1193 |
\(\displaystyle \frac {2}{161} \int -\frac {7 \sqrt {3 x+4} (46 x+145)}{8 (5-2 x)^{7/2}}dx+\frac {9 (3 x+4)^{3/2}}{46 (5-2 x)^{7/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {9 (3 x+4)^{3/2}}{46 (5-2 x)^{7/2}}-\frac {1}{92} \int \frac {\sqrt {3 x+4} (46 x+145)}{(5-2 x)^{7/2}}dx\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {1}{92} \left (\frac {217}{23} \int \frac {\sqrt {3 x+4}}{(5-2 x)^{5/2}}dx-\frac {104 (3 x+4)^{3/2}}{23 (5-2 x)^{5/2}}\right )+\frac {9 (3 x+4)^{3/2}}{46 (5-2 x)^{7/2}}\) |
\(\Big \downarrow \) 48 |
\(\displaystyle \frac {9 (3 x+4)^{3/2}}{46 (5-2 x)^{7/2}}+\frac {1}{92} \left (\frac {434 (3 x+4)^{3/2}}{1587 (5-2 x)^{3/2}}-\frac {104 (3 x+4)^{3/2}}{23 (5-2 x)^{5/2}}\right )\) |
Input:
Int[(Sqrt[4 + 3*x]*(2 + 3*x + x^2))/(5 - 2*x)^(9/2),x]
Output:
(9*(4 + 3*x)^(3/2))/(46*(5 - 2*x)^(7/2)) + ((-104*(4 + 3*x)^(3/2))/(23*(5 - 2*x)^(5/2)) + (434*(4 + 3*x)^(3/2))/(1587*(5 - 2*x)^(3/2)))/92
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[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> With[{Qx = PolynomialQuotient[(a + b*x + c*x^2)^p, d + e*x, x], R = PolynomialRemainder[(a + b*x + c*x^2)^p, d + e*x, x]}, Simp[R*(d + e*x)^(m + 1)*((f + g*x)^(n + 1)/((m + 1)*(e*f - d*g)) ), x] + Simp[1/((m + 1)*(e*f - d*g)) Int[(d + e*x)^(m + 1)*(f + g*x)^n*Ex pandToSum[(m + 1)*(e*f - d*g)*Qx - g*R*(m + n + 2), x], x], x]] /; FreeQ[{a , b, c, d, e, f, g, n}, x] && IGtQ[p, 0] && ILtQ[2*m, -2] && !IntegerQ[n] && !(EqQ[m, -2] && EqQ[p, 1] && EqQ[2*c*d - b*e, 0])
Time = 0.98 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.40
method | result | size |
gosper | \(\frac {2 \left (3 x +4\right )^{\frac {3}{2}} \left (217 x^{2}+709 x +442\right )}{36501 \left (5-2 x \right )^{\frac {7}{2}}}\) | \(27\) |
default | \(\frac {2 \left (3 x +4\right )^{\frac {3}{2}} \left (217 x^{2}+709 x +442\right )}{36501 \left (5-2 x \right )^{\frac {7}{2}}}\) | \(27\) |
orering | \(-\frac {2 \left (217 x^{2}+709 x +442\right ) \left (-5+2 x \right ) \left (3 x +4\right )^{\frac {3}{2}} \left (x^{2}+3 x +2\right )}{36501 \left (2+x \right ) \left (x +1\right ) \left (5-2 x \right )^{\frac {9}{2}}}\) | \(50\) |
Input:
int((3*x+4)^(1/2)*(x^2+3*x+2)/(5-2*x)^(9/2),x,method=_RETURNVERBOSE)
Output:
2/36501/(5-2*x)^(7/2)*(3*x+4)^(3/2)*(217*x^2+709*x+442)
Time = 0.09 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.79 \[ \int \frac {\sqrt {4+3 x} \left (2+3 x+x^2\right )}{(5-2 x)^{9/2}} \, dx=\frac {2 \, {\left (651 \, x^{3} + 2995 \, x^{2} + 4162 \, x + 1768\right )} \sqrt {3 \, x + 4} \sqrt {-2 \, x + 5}}{36501 \, {\left (16 \, x^{4} - 160 \, x^{3} + 600 \, x^{2} - 1000 \, x + 625\right )}} \] Input:
integrate((4+3*x)^(1/2)*(x^2+3*x+2)/(5-2*x)^(9/2),x, algorithm="fricas")
Output:
2/36501*(651*x^3 + 2995*x^2 + 4162*x + 1768)*sqrt(3*x + 4)*sqrt(-2*x + 5)/ (16*x^4 - 160*x^3 + 600*x^2 - 1000*x + 625)
\[ \int \frac {\sqrt {4+3 x} \left (2+3 x+x^2\right )}{(5-2 x)^{9/2}} \, dx=\int \frac {\left (x + 1\right ) \left (x + 2\right ) \sqrt {3 x + 4}}{\left (5 - 2 x\right )^{\frac {9}{2}}}\, dx \] Input:
integrate((4+3*x)**(1/2)*(x**2+3*x+2)/(5-2*x)**(9/2),x)
Output:
Integral((x + 1)*(x + 2)*sqrt(3*x + 4)/(5 - 2*x)**(9/2), x)
Leaf count of result is larger than twice the leaf count of optimal. 115 vs. \(2 (49) = 98\).
Time = 0.12 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.72 \[ \int \frac {\sqrt {4+3 x} \left (2+3 x+x^2\right )}{(5-2 x)^{9/2}} \, dx=\frac {9 \, \sqrt {-6 \, x^{2} + 7 \, x + 20}}{4 \, {\left (16 \, x^{4} - 160 \, x^{3} + 600 \, x^{2} - 1000 \, x + 625\right )}} + \frac {79 \, \sqrt {-6 \, x^{2} + 7 \, x + 20}}{92 \, {\left (8 \, x^{3} - 60 \, x^{2} + 150 \, x - 125\right )}} + \frac {685 \, \sqrt {-6 \, x^{2} + 7 \, x + 20}}{6348 \, {\left (4 \, x^{2} - 20 \, x + 25\right )}} + \frac {217 \, \sqrt {-6 \, x^{2} + 7 \, x + 20}}{48668 \, {\left (2 \, x - 5\right )}} \] Input:
integrate((4+3*x)^(1/2)*(x^2+3*x+2)/(5-2*x)^(9/2),x, algorithm="maxima")
Output:
9/4*sqrt(-6*x^2 + 7*x + 20)/(16*x^4 - 160*x^3 + 600*x^2 - 1000*x + 625) + 79/92*sqrt(-6*x^2 + 7*x + 20)/(8*x^3 - 60*x^2 + 150*x - 125) + 685/6348*sq rt(-6*x^2 + 7*x + 20)/(4*x^2 - 20*x + 25) + 217/48668*sqrt(-6*x^2 + 7*x + 20)/(2*x - 5)
Time = 0.31 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.76 \[ \int \frac {\sqrt {4+3 x} \left (2+3 x+x^2\right )}{(5-2 x)^{9/2}} \, dx=\frac {2 \, {\left ({\left (217 \, \sqrt {3} {\left (3 \, x + 4\right )} + 391 \, \sqrt {3}\right )} {\left (3 \, x + 4\right )} - 1058 \, \sqrt {3}\right )} {\left (3 \, x + 4\right )}^{\frac {3}{2}} \sqrt {-6 \, x + 15}}{985527 \, {\left (2 \, x - 5\right )}^{4}} \] Input:
integrate((4+3*x)^(1/2)*(x^2+3*x+2)/(5-2*x)^(9/2),x, algorithm="giac")
Output:
2/985527*((217*sqrt(3)*(3*x + 4) + 391*sqrt(3))*(3*x + 4) - 1058*sqrt(3))* (3*x + 4)^(3/2)*sqrt(-6*x + 15)/(2*x - 5)^4
Time = 0.24 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.07 \[ \int \frac {\sqrt {4+3 x} \left (2+3 x+x^2\right )}{(5-2 x)^{9/2}} \, dx=\frac {\sqrt {5-2\,x}\,\left (\frac {2081\,x\,\sqrt {3\,x+4}}{146004}+\frac {221\,\sqrt {3\,x+4}}{36501}+\frac {2995\,x^2\,\sqrt {3\,x+4}}{292008}+\frac {217\,x^3\,\sqrt {3\,x+4}}{97336}\right )}{x^4-10\,x^3+\frac {75\,x^2}{2}-\frac {125\,x}{2}+\frac {625}{16}} \] Input:
int(((3*x + 4)^(1/2)*(3*x + x^2 + 2))/(5 - 2*x)^(9/2),x)
Output:
((5 - 2*x)^(1/2)*((2081*x*(3*x + 4)^(1/2))/146004 + (221*(3*x + 4)^(1/2))/ 36501 + (2995*x^2*(3*x + 4)^(1/2))/292008 + (217*x^3*(3*x + 4)^(1/2))/9733 6))/((75*x^2)/2 - (125*x)/2 - 10*x^3 + x^4 + 625/16)
Time = 0.17 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.72 \[ \int \frac {\sqrt {4+3 x} \left (2+3 x+x^2\right )}{(5-2 x)^{9/2}} \, dx=\frac {2 \sqrt {3 x +4}\, \left (-651 x^{3}-2995 x^{2}-4162 x -1768\right )}{36501 \sqrt {-2 x +5}\, \left (8 x^{3}-60 x^{2}+150 x -125\right )} \] Input:
int((4+3*x)^(1/2)*(x^2+3*x+2)/(5-2*x)^(9/2),x)
Output:
(2*sqrt(3*x + 4)*( - 651*x**3 - 2995*x**2 - 4162*x - 1768))/(36501*sqrt( - 2*x + 5)*(8*x**3 - 60*x**2 + 150*x - 125))