Integrand size = 27, antiderivative size = 94 \[ \int \frac {\sqrt {5-2 x} \left (2+3 x+x^2\right )}{(4+3 x)^{5/2}} \, dx=\frac {4 (5-2 x)^{3/2}}{621 (4+3 x)^{3/2}}-\frac {2 \sqrt {5-2 x}}{27 \sqrt {4+3 x}}+\frac {1}{27} \sqrt {5-2 x} \sqrt {4+3 x}+\frac {19 \arcsin \left (\sqrt {\frac {2}{23}} \sqrt {4+3 x}\right )}{27 \sqrt {6}} \] Output:
4/621*(5-2*x)^(3/2)/(4+3*x)^(3/2)-2/27*(5-2*x)^(1/2)/(4+3*x)^(1/2)+1/27*(5 -2*x)^(1/2)*(4+3*x)^(1/2)+19/162*arcsin(1/23*46^(1/2)*(4+3*x)^(1/2))*6^(1/ 2)
Time = 0.19 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.68 \[ \int \frac {\sqrt {5-2 x} \left (2+3 x+x^2\right )}{(4+3 x)^{5/2}} \, dx=\frac {\sqrt {5-2 x} \left (204+406 x+207 x^2\right )}{621 (4+3 x)^{3/2}}-\frac {19 \arctan \left (\frac {\sqrt {\frac {15}{2}-3 x}}{\sqrt {4+3 x}}\right )}{27 \sqrt {6}} \] Input:
Integrate[(Sqrt[5 - 2*x]*(2 + 3*x + x^2))/(4 + 3*x)^(5/2),x]
Output:
(Sqrt[5 - 2*x]*(204 + 406*x + 207*x^2))/(621*(4 + 3*x)^(3/2)) - (19*ArcTan [Sqrt[15/2 - 3*x]/Sqrt[4 + 3*x]])/(27*Sqrt[6])
Time = 0.32 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.11, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {1193, 27, 87, 60, 64, 223}
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.
\(\displaystyle \int \frac {\sqrt {5-2 x} \left (x^2+3 x+2\right )}{(3 x+4)^{5/2}} \, dx\) |
\(\Big \downarrow \) 1193 |
\(\displaystyle \frac {4 (5-2 x)^{3/2}}{621 (3 x+4)^{3/2}}-\frac {2}{69} \int -\frac {23 \sqrt {5-2 x} (3 x+5)}{6 (3 x+4)^{3/2}}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{9} \int \frac {\sqrt {5-2 x} (3 x+5)}{(3 x+4)^{3/2}}dx+\frac {4 (5-2 x)^{3/2}}{621 (3 x+4)^{3/2}}\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {1}{9} \left (\frac {19}{23} \int \frac {\sqrt {5-2 x}}{\sqrt {3 x+4}}dx-\frac {2 (5-2 x)^{3/2}}{23 \sqrt {3 x+4}}\right )+\frac {4 (5-2 x)^{3/2}}{621 (3 x+4)^{3/2}}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {1}{9} \left (\frac {19}{23} \left (\frac {23}{6} \int \frac {1}{\sqrt {5-2 x} \sqrt {3 x+4}}dx+\frac {1}{3} \sqrt {5-2 x} \sqrt {3 x+4}\right )-\frac {2 (5-2 x)^{3/2}}{23 \sqrt {3 x+4}}\right )+\frac {4 (5-2 x)^{3/2}}{621 (3 x+4)^{3/2}}\) |
\(\Big \downarrow \) 64 |
\(\displaystyle \frac {1}{9} \left (\frac {19}{23} \left (\frac {23}{9} \int \frac {1}{\sqrt {\frac {23}{3}-\frac {2}{3} (3 x+4)}}d\sqrt {3 x+4}+\frac {1}{3} \sqrt {5-2 x} \sqrt {3 x+4}\right )-\frac {2 (5-2 x)^{3/2}}{23 \sqrt {3 x+4}}\right )+\frac {4 (5-2 x)^{3/2}}{621 (3 x+4)^{3/2}}\) |
\(\Big \downarrow \) 223 |
\(\displaystyle \frac {1}{9} \left (\frac {19}{23} \left (\frac {23 \arcsin \left (\sqrt {\frac {2}{23}} \sqrt {3 x+4}\right )}{3 \sqrt {6}}+\frac {1}{3} \sqrt {5-2 x} \sqrt {3 x+4}\right )-\frac {2 (5-2 x)^{3/2}}{23 \sqrt {3 x+4}}\right )+\frac {4 (5-2 x)^{3/2}}{621 (3 x+4)^{3/2}}\) |
Input:
Int[(Sqrt[5 - 2*x]*(2 + 3*x + x^2))/(4 + 3*x)^(5/2),x]
Output:
(4*(5 - 2*x)^(3/2))/(621*(4 + 3*x)^(3/2)) + ((-2*(5 - 2*x)^(3/2))/(23*Sqrt [4 + 3*x]) + (19*((Sqrt[5 - 2*x]*Sqrt[4 + 3*x])/3 + (23*ArcSin[Sqrt[2/23]* Sqrt[4 + 3*x]])/(3*Sqrt[6])))/23)/9
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/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( b*(m + n + 1))) Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[a, b, c, d, m, n, x]
Int[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_.) + (d_.)*(x_)]), x_Symbol] :> Simp [2/b Subst[Int[1/Sqrt[c - a*(d/b) + d*(x^2/b)], x], x, Sqrt[a + b*x]], x] /; FreeQ[{a, b, c, d}, x] && GtQ[c - a*(d/b), 0] && ( !GtQ[a - c*(b/d), 0] || PosQ[b])
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[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt [a])]/Rt[-b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[b]
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 = 0.95 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.20
method | result | size |
default | \(\frac {\left (3933 \sqrt {6}\, \arcsin \left (\frac {12 x}{23}-\frac {7}{23}\right ) x^{2}+10488 \sqrt {6}\, \arcsin \left (\frac {12 x}{23}-\frac {7}{23}\right ) x +2484 x^{2} \sqrt {-6 x^{2}+7 x +20}+6992 \sqrt {6}\, \arcsin \left (\frac {12 x}{23}-\frac {7}{23}\right )+4872 x \sqrt {-6 x^{2}+7 x +20}+2448 \sqrt {-6 x^{2}+7 x +20}\right ) \sqrt {5-2 x}}{7452 \sqrt {-6 x^{2}+7 x +20}\, \left (3 x +4\right )^{\frac {3}{2}}}\) | \(113\) |
Input:
int((5-2*x)^(1/2)*(x^2+3*x+2)/(3*x+4)^(5/2),x,method=_RETURNVERBOSE)
Output:
1/7452*(3933*6^(1/2)*arcsin(12/23*x-7/23)*x^2+10488*6^(1/2)*arcsin(12/23*x -7/23)*x+2484*x^2*(-6*x^2+7*x+20)^(1/2)+6992*6^(1/2)*arcsin(12/23*x-7/23)+ 4872*x*(-6*x^2+7*x+20)^(1/2)+2448*(-6*x^2+7*x+20)^(1/2))*(5-2*x)^(1/2)/(-6 *x^2+7*x+20)^(1/2)/(3*x+4)^(3/2)
Time = 0.08 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.99 \[ \int \frac {\sqrt {5-2 x} \left (2+3 x+x^2\right )}{(4+3 x)^{5/2}} \, dx=-\frac {437 \, \sqrt {6} {\left (9 \, x^{2} + 24 \, x + 16\right )} \arctan \left (\frac {\sqrt {6} {\left (12 \, x - 7\right )} \sqrt {3 \, x + 4} \sqrt {-2 \, x + 5}}{12 \, {\left (6 \, x^{2} - 7 \, x - 20\right )}}\right ) - 12 \, {\left (207 \, x^{2} + 406 \, x + 204\right )} \sqrt {3 \, x + 4} \sqrt {-2 \, x + 5}}{7452 \, {\left (9 \, x^{2} + 24 \, x + 16\right )}} \] Input:
integrate((5-2*x)^(1/2)*(x^2+3*x+2)/(4+3*x)^(5/2),x, algorithm="fricas")
Output:
-1/7452*(437*sqrt(6)*(9*x^2 + 24*x + 16)*arctan(1/12*sqrt(6)*(12*x - 7)*sq rt(3*x + 4)*sqrt(-2*x + 5)/(6*x^2 - 7*x - 20)) - 12*(207*x^2 + 406*x + 204 )*sqrt(3*x + 4)*sqrt(-2*x + 5))/(9*x^2 + 24*x + 16)
\[ \int \frac {\sqrt {5-2 x} \left (2+3 x+x^2\right )}{(4+3 x)^{5/2}} \, dx=\int \frac {\sqrt {5 - 2 x} \left (x + 1\right ) \left (x + 2\right )}{\left (3 x + 4\right )^{\frac {5}{2}}}\, dx \] Input:
integrate((5-2*x)**(1/2)*(x**2+3*x+2)/(4+3*x)**(5/2),x)
Output:
Integral(sqrt(5 - 2*x)*(x + 1)*(x + 2)/(3*x + 4)**(5/2), x)
Timed out. \[ \int \frac {\sqrt {5-2 x} \left (2+3 x+x^2\right )}{(4+3 x)^{5/2}} \, dx=\text {Timed out} \] Input:
integrate((5-2*x)^(1/2)*(x^2+3*x+2)/(4+3*x)^(5/2),x, algorithm="maxima")
Output:
Timed out
Leaf count of result is larger than twice the leaf count of optimal. 159 vs. \(2 (67) = 134\).
Time = 0.35 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.69 \[ \int \frac {\sqrt {5-2 x} \left (2+3 x+x^2\right )}{(4+3 x)^{5/2}} \, dx=\frac {\sqrt {6} {\left (\sqrt {2} \sqrt {-6 \, x + 15} - \sqrt {46}\right )}^{3}}{44712 \, {\left (3 \, x + 4\right )}^{\frac {3}{2}}} + \frac {1}{81} \, \sqrt {3} \sqrt {3 \, x + 4} \sqrt {-6 \, x + 15} + \frac {19}{162} \, \sqrt {6} \arcsin \left (\frac {1}{23} \, \sqrt {46} \sqrt {3 \, x + 4}\right ) - \frac {4 \, \sqrt {6} {\left (\sqrt {2} \sqrt {-6 \, x + 15} - \sqrt {46}\right )}}{621 \, \sqrt {3 \, x + 4}} + \frac {8 \, \sqrt {6} {\left (3 \, x + 4\right )}^{\frac {3}{2}} {\left (\frac {18 \, {\left (\sqrt {2} \sqrt {-6 \, x + 15} - \sqrt {46}\right )}^{2}}{3 \, x + 4} - 1\right )}}{5589 \, {\left (\sqrt {2} \sqrt {-6 \, x + 15} - \sqrt {46}\right )}^{3}} \] Input:
integrate((5-2*x)^(1/2)*(x^2+3*x+2)/(4+3*x)^(5/2),x, algorithm="giac")
Output:
1/44712*sqrt(6)*(sqrt(2)*sqrt(-6*x + 15) - sqrt(46))^3/(3*x + 4)^(3/2) + 1 /81*sqrt(3)*sqrt(3*x + 4)*sqrt(-6*x + 15) + 19/162*sqrt(6)*arcsin(1/23*sqr t(46)*sqrt(3*x + 4)) - 4/621*sqrt(6)*(sqrt(2)*sqrt(-6*x + 15) - sqrt(46))/ sqrt(3*x + 4) + 8/5589*sqrt(6)*(3*x + 4)^(3/2)*(18*(sqrt(2)*sqrt(-6*x + 15 ) - sqrt(46))^2/(3*x + 4) - 1)/(sqrt(2)*sqrt(-6*x + 15) - sqrt(46))^3
Timed out. \[ \int \frac {\sqrt {5-2 x} \left (2+3 x+x^2\right )}{(4+3 x)^{5/2}} \, dx=\int \frac {\sqrt {5-2\,x}\,\left (x^2+3\,x+2\right )}{{\left (3\,x+4\right )}^{5/2}} \,d x \] Input:
int(((5 - 2*x)^(1/2)*(3*x + x^2 + 2))/(3*x + 4)^(5/2),x)
Output:
int(((5 - 2*x)^(1/2)*(3*x + x^2 + 2))/(3*x + 4)^(5/2), x)
Time = 0.16 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.01 \[ \int \frac {\sqrt {5-2 x} \left (2+3 x+x^2\right )}{(4+3 x)^{5/2}} \, dx=\frac {-1311 \sqrt {3 x +4}\, \sqrt {6}\, \mathit {asin} \left (\frac {\sqrt {-2 x +5}\, \sqrt {3}}{\sqrt {23}}\right ) x -1748 \sqrt {3 x +4}\, \sqrt {6}\, \mathit {asin} \left (\frac {\sqrt {-2 x +5}\, \sqrt {3}}{\sqrt {23}}\right )+1242 \sqrt {-2 x +5}\, x^{2}+2436 \sqrt {-2 x +5}\, x +1224 \sqrt {-2 x +5}}{3726 \sqrt {3 x +4}\, \left (3 x +4\right )} \] Input:
int((5-2*x)^(1/2)*(x^2+3*x+2)/(4+3*x)^(5/2),x)
Output:
( - 1311*sqrt(3*x + 4)*sqrt(6)*asin((sqrt( - 2*x + 5)*sqrt(3))/sqrt(23))*x - 1748*sqrt(3*x + 4)*sqrt(6)*asin((sqrt( - 2*x + 5)*sqrt(3))/sqrt(23)) + 1242*sqrt( - 2*x + 5)*x**2 + 2436*sqrt( - 2*x + 5)*x + 1224*sqrt( - 2*x + 5))/(3726*sqrt(3*x + 4)*(3*x + 4))