Integrand size = 26, antiderivative size = 72 \[ \int \frac {(2+3 x)^2}{(1-2 x)^{3/2} \sqrt {3+5 x}} \, dx=\frac {49 \sqrt {3+5 x}}{22 \sqrt {1-2 x}}+\frac {9}{20} \sqrt {1-2 x} \sqrt {3+5 x}-\frac {321 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )}{20 \sqrt {10}} \] Output:
49/22*(3+5*x)^(1/2)/(1-2*x)^(1/2)+9/20*(1-2*x)^(1/2)*(3+5*x)^(1/2)-321/200 *arcsin(1/11*22^(1/2)*(3+5*x)^(1/2))*10^(1/2)
Time = 0.10 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.88 \[ \int \frac {(2+3 x)^2}{(1-2 x)^{3/2} \sqrt {3+5 x}} \, dx=\frac {10 (589-198 x) \sqrt {3+5 x}+3531 \sqrt {10-20 x} \arctan \left (\frac {\sqrt {\frac {5}{2}-5 x}}{\sqrt {3+5 x}}\right )}{2200 \sqrt {1-2 x}} \] Input:
Integrate[(2 + 3*x)^2/((1 - 2*x)^(3/2)*Sqrt[3 + 5*x]),x]
Output:
(10*(589 - 198*x)*Sqrt[3 + 5*x] + 3531*Sqrt[10 - 20*x]*ArcTan[Sqrt[5/2 - 5 *x]/Sqrt[3 + 5*x]])/(2200*Sqrt[1 - 2*x])
Time = 0.17 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.07, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {100, 27, 90, 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.
\(\displaystyle \int \frac {(3 x+2)^2}{(1-2 x)^{3/2} \sqrt {5 x+3}} \, dx\) |
\(\Big \downarrow \) 100 |
\(\displaystyle \frac {49 \sqrt {5 x+3}}{22 \sqrt {1-2 x}}-\frac {1}{22} \int \frac {33 (6 x+11)}{2 \sqrt {1-2 x} \sqrt {5 x+3}}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {49 \sqrt {5 x+3}}{22 \sqrt {1-2 x}}-\frac {3}{4} \int \frac {6 x+11}{\sqrt {1-2 x} \sqrt {5 x+3}}dx\) |
\(\Big \downarrow \) 90 |
\(\displaystyle \frac {49 \sqrt {5 x+3}}{22 \sqrt {1-2 x}}-\frac {3}{4} \left (\frac {107}{10} \int \frac {1}{\sqrt {1-2 x} \sqrt {5 x+3}}dx-\frac {3}{5} \sqrt {1-2 x} \sqrt {5 x+3}\right )\) |
\(\Big \downarrow \) 64 |
\(\displaystyle \frac {49 \sqrt {5 x+3}}{22 \sqrt {1-2 x}}-\frac {3}{4} \left (\frac {107}{25} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}-\frac {3}{5} \sqrt {1-2 x} \sqrt {5 x+3}\right )\) |
\(\Big \downarrow \) 223 |
\(\displaystyle \frac {49 \sqrt {5 x+3}}{22 \sqrt {1-2 x}}-\frac {3}{4} \left (\frac {107 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )}{5 \sqrt {10}}-\frac {3}{5} \sqrt {1-2 x} \sqrt {5 x+3}\right )\) |
Input:
Int[(2 + 3*x)^2/((1 - 2*x)^(3/2)*Sqrt[3 + 5*x]),x]
Output:
(49*Sqrt[3 + 5*x])/(22*Sqrt[1 - 2*x]) - (3*((-3*Sqrt[1 - 2*x]*Sqrt[3 + 5*x ])/5 + (107*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]])/(5*Sqrt[10])))/4
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, 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[((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[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]
Time = 0.24 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.14
method | result | size |
default | \(\frac {\left (7062 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x -3531 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )-3960 x \sqrt {-10 x^{2}-x +3}+11780 \sqrt {-10 x^{2}-x +3}\right ) \sqrt {3+5 x}}{4400 \sqrt {1-2 x}\, \sqrt {-10 x^{2}-x +3}}\) | \(82\) |
Input:
int((2+3*x)^2/(1-2*x)^(3/2)/(3+5*x)^(1/2),x,method=_RETURNVERBOSE)
Output:
1/4400*(7062*10^(1/2)*arcsin(20/11*x+1/11)*x-3531*10^(1/2)*arcsin(20/11*x+ 1/11)-3960*x*(-10*x^2-x+3)^(1/2)+11780*(-10*x^2-x+3)^(1/2))*(3+5*x)^(1/2)/ (1-2*x)^(1/2)/(-10*x^2-x+3)^(1/2)
Time = 0.08 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.06 \[ \int \frac {(2+3 x)^2}{(1-2 x)^{3/2} \sqrt {3+5 x}} \, dx=\frac {3531 \, \sqrt {10} {\left (2 \, x - 1\right )} \arctan \left (\frac {\sqrt {10} {\left (20 \, x + 1\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{20 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) + 20 \, {\left (198 \, x - 589\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{4400 \, {\left (2 \, x - 1\right )}} \] Input:
integrate((2+3*x)^2/(1-2*x)^(3/2)/(3+5*x)^(1/2),x, algorithm="fricas")
Output:
1/4400*(3531*sqrt(10)*(2*x - 1)*arctan(1/20*sqrt(10)*(20*x + 1)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x - 3)) + 20*(198*x - 589)*sqrt(5*x + 3)*sqrt (-2*x + 1))/(2*x - 1)
\[ \int \frac {(2+3 x)^2}{(1-2 x)^{3/2} \sqrt {3+5 x}} \, dx=\int \frac {\left (3 x + 2\right )^{2}}{\left (1 - 2 x\right )^{\frac {3}{2}} \sqrt {5 x + 3}}\, dx \] Input:
integrate((2+3*x)**2/(1-2*x)**(3/2)/(3+5*x)**(1/2),x)
Output:
Integral((3*x + 2)**2/((1 - 2*x)**(3/2)*sqrt(5*x + 3)), x)
Time = 0.13 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.69 \[ \int \frac {(2+3 x)^2}{(1-2 x)^{3/2} \sqrt {3+5 x}} \, dx=-\frac {321}{400} \, \sqrt {5} \sqrt {2} \arcsin \left (\frac {20}{11} \, x + \frac {1}{11}\right ) + \frac {9}{20} \, \sqrt {-10 \, x^{2} - x + 3} - \frac {49 \, \sqrt {-10 \, x^{2} - x + 3}}{22 \, {\left (2 \, x - 1\right )}} \] Input:
integrate((2+3*x)^2/(1-2*x)^(3/2)/(3+5*x)^(1/2),x, algorithm="maxima")
Output:
-321/400*sqrt(5)*sqrt(2)*arcsin(20/11*x + 1/11) + 9/20*sqrt(-10*x^2 - x + 3) - 49/22*sqrt(-10*x^2 - x + 3)/(2*x - 1)
Time = 0.14 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.81 \[ \int \frac {(2+3 x)^2}{(1-2 x)^{3/2} \sqrt {3+5 x}} \, dx=-\frac {321}{200} \, \sqrt {10} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right ) + \frac {{\left (198 \, \sqrt {5} {\left (5 \, x + 3\right )} - 3539 \, \sqrt {5}\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5}}{5500 \, {\left (2 \, x - 1\right )}} \] Input:
integrate((2+3*x)^2/(1-2*x)^(3/2)/(3+5*x)^(1/2),x, algorithm="giac")
Output:
-321/200*sqrt(10)*arcsin(1/11*sqrt(22)*sqrt(5*x + 3)) + 1/5500*(198*sqrt(5 )*(5*x + 3) - 3539*sqrt(5))*sqrt(5*x + 3)*sqrt(-10*x + 5)/(2*x - 1)
Timed out. \[ \int \frac {(2+3 x)^2}{(1-2 x)^{3/2} \sqrt {3+5 x}} \, dx=\int \frac {{\left (3\,x+2\right )}^2}{{\left (1-2\,x\right )}^{3/2}\,\sqrt {5\,x+3}} \,d x \] Input:
int((3*x + 2)^2/((1 - 2*x)^(3/2)*(5*x + 3)^(1/2)),x)
Output:
int((3*x + 2)^2/((1 - 2*x)^(3/2)*(5*x + 3)^(1/2)), x)
Time = 0.24 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.72 \[ \int \frac {(2+3 x)^2}{(1-2 x)^{3/2} \sqrt {3+5 x}} \, dx=\frac {3531 \sqrt {-2 x +1}\, \sqrt {10}\, \mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )-1980 \sqrt {5 x +3}\, x +5890 \sqrt {5 x +3}}{2200 \sqrt {-2 x +1}} \] Input:
int((2+3*x)^2/(1-2*x)^(3/2)/(3+5*x)^(1/2),x)
Output:
(3531*sqrt( - 2*x + 1)*sqrt(10)*asin((sqrt( - 2*x + 1)*sqrt(5))/sqrt(11)) - 1980*sqrt(5*x + 3)*x + 5890*sqrt(5*x + 3))/(2200*sqrt( - 2*x + 1))