Integrand size = 27, antiderivative size = 94 \[ \int \frac {\sqrt {4+3 x} \left (2+3 x+x^2\right )}{\sqrt {5-2 x}} \, dx=-\frac {2857}{576} \sqrt {5-2 x} \sqrt {4+3 x}-\frac {73}{144} \sqrt {5-2 x} (4+3 x)^{3/2}+\frac {1}{36} (5-2 x)^{3/2} (4+3 x)^{3/2}+\frac {65711 \arcsin \left (\sqrt {\frac {2}{23}} \sqrt {4+3 x}\right )}{576 \sqrt {6}} \] Output:
-2857/576*(5-2*x)^(1/2)*(4+3*x)^(1/2)-73/144*(5-2*x)^(1/2)*(4+3*x)^(3/2)+1 /36*(5-2*x)^(3/2)*(4+3*x)^(3/2)+65711/3456*arcsin(1/23*46^(1/2)*(4+3*x)^(1 /2))*6^(1/2)
Time = 0.49 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.82 \[ \int \frac {\sqrt {4+3 x} \left (2+3 x+x^2\right )}{\sqrt {5-2 x}} \, dx=\frac {-\sqrt {15-6 x} \sqrt {4+3 x} \left (3705+764 x+96 x^2\right )-65711 \sqrt {2} \arctan \left (\frac {\sqrt {8+6 x}}{\sqrt {23}-\sqrt {15-6 x}}\right )}{576 \sqrt {3}} \] Input:
Integrate[(Sqrt[4 + 3*x]*(2 + 3*x + x^2))/Sqrt[5 - 2*x],x]
Output:
(-(Sqrt[15 - 6*x]*Sqrt[4 + 3*x]*(3705 + 764*x + 96*x^2)) - 65711*Sqrt[2]*A rcTan[Sqrt[8 + 6*x]/(Sqrt[23] - Sqrt[15 - 6*x])])/(576*Sqrt[3])
Time = 0.31 (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 = {1194, 27, 90, 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.
\(\displaystyle \int \frac {\sqrt {3 x+4} \left (x^2+3 x+2\right )}{\sqrt {5-2 x}} \, dx\) |
\(\Big \downarrow \) 1194 |
\(\displaystyle \frac {1}{36} \int \frac {3 \sqrt {3 x+4} (146 x+13)}{2 \sqrt {5-2 x}}dx+\frac {1}{36} (5-2 x)^{3/2} (3 x+4)^{3/2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{24} \int \frac {\sqrt {3 x+4} (146 x+13)}{\sqrt {5-2 x}}dx+\frac {1}{36} (5-2 x)^{3/2} (3 x+4)^{3/2}\) |
\(\Big \downarrow \) 90 |
\(\displaystyle \frac {1}{24} \left (\frac {2857}{12} \int \frac {\sqrt {3 x+4}}{\sqrt {5-2 x}}dx-\frac {73}{6} \sqrt {5-2 x} (3 x+4)^{3/2}\right )+\frac {1}{36} (5-2 x)^{3/2} (3 x+4)^{3/2}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {1}{24} \left (\frac {2857}{12} \left (\frac {23}{4} \int \frac {1}{\sqrt {5-2 x} \sqrt {3 x+4}}dx-\frac {1}{2} \sqrt {5-2 x} \sqrt {3 x+4}\right )-\frac {73}{6} \sqrt {5-2 x} (3 x+4)^{3/2}\right )+\frac {1}{36} (5-2 x)^{3/2} (3 x+4)^{3/2}\) |
\(\Big \downarrow \) 64 |
\(\displaystyle \frac {1}{24} \left (\frac {2857}{12} \left (\frac {23}{6} \int \frac {1}{\sqrt {\frac {23}{3}-\frac {2}{3} (3 x+4)}}d\sqrt {3 x+4}-\frac {1}{2} \sqrt {5-2 x} \sqrt {3 x+4}\right )-\frac {73}{6} \sqrt {5-2 x} (3 x+4)^{3/2}\right )+\frac {1}{36} (5-2 x)^{3/2} (3 x+4)^{3/2}\) |
\(\Big \downarrow \) 223 |
\(\displaystyle \frac {1}{24} \left (\frac {2857}{12} \left (\frac {23 \arcsin \left (\sqrt {\frac {2}{23}} \sqrt {3 x+4}\right )}{2 \sqrt {6}}-\frac {1}{2} \sqrt {5-2 x} \sqrt {3 x+4}\right )-\frac {73}{6} \sqrt {5-2 x} (3 x+4)^{3/2}\right )+\frac {1}{36} (5-2 x)^{3/2} (3 x+4)^{3/2}\) |
Input:
Int[(Sqrt[4 + 3*x]*(2 + 3*x + x^2))/Sqrt[5 - 2*x],x]
Output:
((5 - 2*x)^(3/2)*(4 + 3*x)^(3/2))/36 + ((-73*Sqrt[5 - 2*x]*(4 + 3*x)^(3/2) )/6 + (2857*(-1/2*(Sqrt[5 - 2*x]*Sqrt[4 + 3*x]) + (23*ArcSin[Sqrt[2/23]*Sq rt[4 + 3*x]])/(2*Sqrt[6])))/12)/24
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*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(d*f*(n + p + 2)) Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0]
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] :> Simp[c^p*(d + e*x)^(m + 2*p)*((f + g*x )^(n + 1)/(g*e^(2*p)*(m + n + 2*p + 1))), x] + Simp[1/(g*e^(2*p)*(m + n + 2 *p + 1)) Int[(d + e*x)^m*(f + g*x)^n*ExpandToSum[g*(m + n + 2*p + 1)*(e^( 2*p)*(a + b*x + c*x^2)^p - c^p*(d + e*x)^(2*p)) - c^p*(e*f - d*g)*(m + 2*p) *(d + e*x)^(2*p - 1), x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && IGtQ [p, 0] && !IntegerQ[m] && !IntegerQ[n] && NeQ[m + n + 2*p + 1, 0]
Time = 0.98 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.93
method | result | size |
default | \(\frac {\sqrt {3 x +4}\, \sqrt {5-2 x}\, \left (-1152 x^{2} \sqrt {-6 x^{2}+7 x +20}+65711 \sqrt {6}\, \arcsin \left (\frac {12 x}{23}-\frac {7}{23}\right )-9168 x \sqrt {-6 x^{2}+7 x +20}-44460 \sqrt {-6 x^{2}+7 x +20}\right )}{6912 \sqrt {-6 x^{2}+7 x +20}}\) | \(87\) |
risch | \(\frac {\left (96 x^{2}+764 x +3705\right ) \sqrt {3 x +4}\, \left (-5+2 x \right ) \sqrt {\left (5-2 x \right ) \left (3 x +4\right )}}{576 \sqrt {-\left (3 x +4\right ) \left (-5+2 x \right )}\, \sqrt {5-2 x}}+\frac {65711 \sqrt {6}\, \arcsin \left (\frac {12 x}{23}-\frac {7}{23}\right ) \sqrt {\left (5-2 x \right ) \left (3 x +4\right )}}{6912 \sqrt {5-2 x}\, \sqrt {3 x +4}}\) | \(98\) |
Input:
int((3*x+4)^(1/2)*(x^2+3*x+2)/(5-2*x)^(1/2),x,method=_RETURNVERBOSE)
Output:
1/6912*(3*x+4)^(1/2)*(5-2*x)^(1/2)*(-1152*x^2*(-6*x^2+7*x+20)^(1/2)+65711* 6^(1/2)*arcsin(12/23*x-7/23)-9168*x*(-6*x^2+7*x+20)^(1/2)-44460*(-6*x^2+7* x+20)^(1/2))/(-6*x^2+7*x+20)^(1/2)
Time = 0.09 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.73 \[ \int \frac {\sqrt {4+3 x} \left (2+3 x+x^2\right )}{\sqrt {5-2 x}} \, dx=-\frac {1}{576} \, {\left (96 \, x^{2} + 764 \, x + 3705\right )} \sqrt {3 \, x + 4} \sqrt {-2 \, x + 5} - \frac {65711}{6912} \, \sqrt {6} \arctan \left (\frac {\sqrt {6} {\left (12 \, x - 7\right )} \sqrt {3 \, x + 4} \sqrt {-2 \, x + 5}}{12 \, {\left (6 \, x^{2} - 7 \, x - 20\right )}}\right ) \] Input:
integrate((4+3*x)^(1/2)*(x^2+3*x+2)/(5-2*x)^(1/2),x, algorithm="fricas")
Output:
-1/576*(96*x^2 + 764*x + 3705)*sqrt(3*x + 4)*sqrt(-2*x + 5) - 65711/6912*s qrt(6)*arctan(1/12*sqrt(6)*(12*x - 7)*sqrt(3*x + 4)*sqrt(-2*x + 5)/(6*x^2 - 7*x - 20))
\[ \int \frac {\sqrt {4+3 x} \left (2+3 x+x^2\right )}{\sqrt {5-2 x}} \, dx=\int \frac {\left (x + 1\right ) \left (x + 2\right ) \sqrt {3 x + 4}}{\sqrt {5 - 2 x}}\, dx \] Input:
integrate((4+3*x)**(1/2)*(x**2+3*x+2)/(5-2*x)**(1/2),x)
Output:
Integral((x + 1)*(x + 2)*sqrt(3*x + 4)/sqrt(5 - 2*x), x)
Time = 0.11 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.62 \[ \int \frac {\sqrt {4+3 x} \left (2+3 x+x^2\right )}{\sqrt {5-2 x}} \, dx=\frac {65711}{6912} \, \sqrt {3} \sqrt {2} \arcsin \left (\frac {12}{23} \, x - \frac {7}{23}\right ) + \frac {1}{36} \, {\left (-6 \, x^{2} + 7 \, x + 20\right )}^{\frac {3}{2}} - \frac {73}{48} \, \sqrt {-6 \, x^{2} + 7 \, x + 20} x - \frac {4025}{576} \, \sqrt {-6 \, x^{2} + 7 \, x + 20} \] Input:
integrate((4+3*x)^(1/2)*(x^2+3*x+2)/(5-2*x)^(1/2),x, algorithm="maxima")
Output:
65711/6912*sqrt(3)*sqrt(2)*arcsin(12/23*x - 7/23) + 1/36*(-6*x^2 + 7*x + 2 0)^(3/2) - 73/48*sqrt(-6*x^2 + 7*x + 20)*x - 4025/576*sqrt(-6*x^2 + 7*x + 20)
Time = 0.51 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.57 \[ \int \frac {\sqrt {4+3 x} \left (2+3 x+x^2\right )}{\sqrt {5-2 x}} \, dx=-\frac {1}{3456} \, \sqrt {3} {\left (2 \, {\left (4 \, {\left (8 \, x + 53\right )} {\left (3 \, x + 4\right )} + 2857\right )} \sqrt {3 \, x + 4} \sqrt {-6 \, x + 15} - 65711 \, \sqrt {2} \arcsin \left (\frac {1}{23} \, \sqrt {46} \sqrt {3 \, x + 4}\right )\right )} \] Input:
integrate((4+3*x)^(1/2)*(x^2+3*x+2)/(5-2*x)^(1/2),x, algorithm="giac")
Output:
-1/3456*sqrt(3)*(2*(4*(8*x + 53)*(3*x + 4) + 2857)*sqrt(3*x + 4)*sqrt(-6*x + 15) - 65711*sqrt(2)*arcsin(1/23*sqrt(46)*sqrt(3*x + 4)))
Time = 24.59 (sec) , antiderivative size = 944, normalized size of antiderivative = 10.04 \[ \int \frac {\sqrt {4+3 x} \left (2+3 x+x^2\right )}{\sqrt {5-2 x}} \, dx=\text {Too large to display} \] Input:
int(((3*x + 4)^(1/2)*(3*x + x^2 + 2))/(5 - 2*x)^(1/2),x)
Output:
(65711*6^(1/2)*atan((6^(1/2)*(5^(1/2) - (5 - 2*x)^(1/2)))/(2*((3*x + 4)^(1 /2) - 2))))/1728 - ((851*(5^(1/2) - (5 - 2*x)^(1/2)))/(81*((3*x + 4)^(1/2) - 2)) - (2597*(5^(1/2) - (5 - 2*x)^(1/2))^3)/(54*((3*x + 4)^(1/2) - 2)^3) + (2597*(5^(1/2) - (5 - 2*x)^(1/2))^5)/(36*((3*x + 4)^(1/2) - 2)^5) - (85 1*(5^(1/2) - (5 - 2*x)^(1/2))^7)/(24*((3*x + 4)^(1/2) - 2)^7) + (256*5^(1/ 2)*(5^(1/2) - (5 - 2*x)^(1/2))^2)/(27*((3*x + 4)^(1/2) - 2)^2) + (704*5^(1 /2)*(5^(1/2) - (5 - 2*x)^(1/2))^4)/(9*((3*x + 4)^(1/2) - 2)^4) + (64*5^(1/ 2)*(5^(1/2) - (5 - 2*x)^(1/2))^6)/(3*((3*x + 4)^(1/2) - 2)^6))/((32*(5^(1/ 2) - (5 - 2*x)^(1/2))^2)/(27*((3*x + 4)^(1/2) - 2)^2) + (8*(5^(1/2) - (5 - 2*x)^(1/2))^4)/(3*((3*x + 4)^(1/2) - 2)^4) + (8*(5^(1/2) - (5 - 2*x)^(1/2 ))^6)/(3*((3*x + 4)^(1/2) - 2)^6) + (5^(1/2) - (5 - 2*x)^(1/2))^8/((3*x + 4)^(1/2) - 2)^8 + 16/81) - ((28*(5^(1/2) - (5 - 2*x)^(1/2)))/(9*((3*x + 4) ^(1/2) - 2)) - (14*(5^(1/2) - (5 - 2*x)^(1/2))^3)/(3*((3*x + 4)^(1/2) - 2) ^3) + (32*5^(1/2)*(5^(1/2) - (5 - 2*x)^(1/2))^2)/(3*((3*x + 4)^(1/2) - 2)^ 2))/((4*(5^(1/2) - (5 - 2*x)^(1/2))^2)/(3*((3*x + 4)^(1/2) - 2)^2) + (5^(1 /2) - (5 - 2*x)^(1/2))^4/((3*x + 4)^(1/2) - 2)^4 + 4/9) - ((21827*(5^(1/2) - (5 - 2*x)^(1/2)))/(6561*((3*x + 4)^(1/2) - 2)) + (125299*(5^(1/2) - (5 - 2*x)^(1/2))^3)/(13122*((3*x + 4)^(1/2) - 2)^3) - (197753*(5^(1/2) - (5 - 2*x)^(1/2))^5)/(1458*((3*x + 4)^(1/2) - 2)^5) + (197753*(5^(1/2) - (5 - 2 *x)^(1/2))^7)/(972*((3*x + 4)^(1/2) - 2)^7) - (125299*(5^(1/2) - (5 - 2...
Time = 0.17 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.69 \[ \int \frac {\sqrt {4+3 x} \left (2+3 x+x^2\right )}{\sqrt {5-2 x}} \, dx=-\frac {65711 \sqrt {6}\, \mathit {asin} \left (\frac {\sqrt {-2 x +5}\, \sqrt {3}}{\sqrt {23}}\right )}{3456}-\frac {\sqrt {3 x +4}\, \sqrt {-2 x +5}\, x^{2}}{6}-\frac {191 \sqrt {3 x +4}\, \sqrt {-2 x +5}\, x}{144}-\frac {1235 \sqrt {3 x +4}\, \sqrt {-2 x +5}}{192} \] Input:
int((4+3*x)^(1/2)*(x^2+3*x+2)/(5-2*x)^(1/2),x)
Output:
( - 65711*sqrt(6)*asin((sqrt( - 2*x + 5)*sqrt(3))/sqrt(23)) - 576*sqrt(3*x + 4)*sqrt( - 2*x + 5)*x**2 - 4584*sqrt(3*x + 4)*sqrt( - 2*x + 5)*x - 2223 0*sqrt(3*x + 4)*sqrt( - 2*x + 5))/3456