Integrand size = 24, antiderivative size = 74 \[ \int \frac {(2+3 x) \sqrt {3+5 x}}{(1-2 x)^{5/2}} \, dx=-\frac {3 \sqrt {3+5 x}}{2 \sqrt {1-2 x}}+\frac {7 (3+5 x)^{3/2}}{33 (1-2 x)^{3/2}}+\frac {3}{2} \sqrt {\frac {5}{2}} \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right ) \] Output:
-3/2*(3+5*x)^(1/2)/(1-2*x)^(1/2)+7/33*(3+5*x)^(3/2)/(1-2*x)^(3/2)+3/4*arcs in(1/11*22^(1/2)*(3+5*x)^(1/2))*10^(1/2)
Time = 0.10 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.92 \[ \int \frac {(2+3 x) \sqrt {3+5 x}}{(1-2 x)^{5/2}} \, dx=\frac {2 \sqrt {3+5 x} (-57+268 x)+99 \sqrt {10-20 x} (-1+2 x) \arctan \left (\frac {\sqrt {\frac {5}{2}-5 x}}{\sqrt {3+5 x}}\right )}{132 (1-2 x)^{3/2}} \] Input:
Integrate[((2 + 3*x)*Sqrt[3 + 5*x])/(1 - 2*x)^(5/2),x]
Output:
(2*Sqrt[3 + 5*x]*(-57 + 268*x) + 99*Sqrt[10 - 20*x]*(-1 + 2*x)*ArcTan[Sqrt [5/2 - 5*x]/Sqrt[3 + 5*x]])/(132*(1 - 2*x)^(3/2))
Time = 0.17 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {87, 57, 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.
\(\displaystyle \int \frac {(3 x+2) \sqrt {5 x+3}}{(1-2 x)^{5/2}} \, dx\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {7 (5 x+3)^{3/2}}{33 (1-2 x)^{3/2}}-\frac {3}{2} \int \frac {\sqrt {5 x+3}}{(1-2 x)^{3/2}}dx\) |
\(\Big \downarrow \) 57 |
\(\displaystyle \frac {7 (5 x+3)^{3/2}}{33 (1-2 x)^{3/2}}-\frac {3}{2} \left (\frac {\sqrt {5 x+3}}{\sqrt {1-2 x}}-\frac {5}{2} \int \frac {1}{\sqrt {1-2 x} \sqrt {5 x+3}}dx\right )\) |
\(\Big \downarrow \) 64 |
\(\displaystyle \frac {7 (5 x+3)^{3/2}}{33 (1-2 x)^{3/2}}-\frac {3}{2} \left (\frac {\sqrt {5 x+3}}{\sqrt {1-2 x}}-\int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}\right )\) |
\(\Big \downarrow \) 223 |
\(\displaystyle \frac {7 (5 x+3)^{3/2}}{33 (1-2 x)^{3/2}}-\frac {3}{2} \left (\frac {\sqrt {5 x+3}}{\sqrt {1-2 x}}-\sqrt {\frac {5}{2}} \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )\right )\) |
Input:
Int[((2 + 3*x)*Sqrt[3 + 5*x])/(1 - 2*x)^(5/2),x]
Output:
(7*(3 + 5*x)^(3/2))/(33*(1 - 2*x)^(3/2)) - (3*(Sqrt[3 + 5*x]/Sqrt[1 - 2*x] - Sqrt[5/2]*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]]))/2
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + 1))), x] - Simp[d*(n/(b*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] & & GtQ[n, 0] && LtQ[m, -1] && !(IntegerQ[n] && !IntegerQ[m]) && !(ILeQ[m + n + 2, 0] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[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]
Time = 0.21 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.30
method | result | size |
default | \(\frac {\left (396 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x^{2}-396 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x +99 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )+1072 x \sqrt {-10 x^{2}-x +3}-228 \sqrt {-10 x^{2}-x +3}\right ) \sqrt {3+5 x}}{264 \left (1-2 x \right )^{\frac {3}{2}} \sqrt {-10 x^{2}-x +3}}\) | \(96\) |
Input:
int((2+3*x)*(3+5*x)^(1/2)/(1-2*x)^(5/2),x,method=_RETURNVERBOSE)
Output:
1/264*(396*10^(1/2)*arcsin(20/11*x+1/11)*x^2-396*10^(1/2)*arcsin(20/11*x+1 /11)*x+99*10^(1/2)*arcsin(20/11*x+1/11)+1072*x*(-10*x^2-x+3)^(1/2)-228*(-1 0*x^2-x+3)^(1/2))*(3+5*x)^(1/2)/(1-2*x)^(3/2)/(-10*x^2-x+3)^(1/2)
Time = 0.07 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.16 \[ \int \frac {(2+3 x) \sqrt {3+5 x}}{(1-2 x)^{5/2}} \, dx=-\frac {99 \, \sqrt {\frac {5}{2}} {\left (4 \, x^{2} - 4 \, x + 1\right )} \arctan \left (\frac {\sqrt {\frac {5}{2}} {\left (20 \, x + 1\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{10 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) - 2 \, {\left (268 \, x - 57\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{132 \, {\left (4 \, x^{2} - 4 \, x + 1\right )}} \] Input:
integrate((2+3*x)*(3+5*x)^(1/2)/(1-2*x)^(5/2),x, algorithm="fricas")
Output:
-1/132*(99*sqrt(5/2)*(4*x^2 - 4*x + 1)*arctan(1/10*sqrt(5/2)*(20*x + 1)*sq rt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x - 3)) - 2*(268*x - 57)*sqrt(5*x + 3 )*sqrt(-2*x + 1))/(4*x^2 - 4*x + 1)
\[ \int \frac {(2+3 x) \sqrt {3+5 x}}{(1-2 x)^{5/2}} \, dx=\int \frac {\left (3 x + 2\right ) \sqrt {5 x + 3}}{\left (1 - 2 x\right )^{\frac {5}{2}}}\, dx \] Input:
integrate((2+3*x)*(3+5*x)**(1/2)/(1-2*x)**(5/2),x)
Output:
Integral((3*x + 2)*sqrt(5*x + 3)/(1 - 2*x)**(5/2), x)
Time = 0.14 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.65 \[ \int \frac {(2+3 x) \sqrt {3+5 x}}{(1-2 x)^{5/2}} \, dx=\frac {2 \, \sqrt {-10 \, x^{2} - x + 3}}{3 \, {\left (4 \, x^{2} - 4 \, x + 1\right )}} + \frac {10 \, \sqrt {-10 \, x^{2} - x + 3}}{33 \, {\left (2 \, x - 1\right )}} \] Input:
integrate((2+3*x)*(3+5*x)^(1/2)/(1-2*x)^(5/2),x, algorithm="maxima")
Output:
2/3*sqrt(-10*x^2 - x + 3)/(4*x^2 - 4*x + 1) + 10/33*sqrt(-10*x^2 - x + 3)/ (2*x - 1)
Time = 0.14 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.78 \[ \int \frac {(2+3 x) \sqrt {3+5 x}}{(1-2 x)^{5/2}} \, dx=\frac {3}{4} \, \sqrt {10} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right ) + \frac {{\left (268 \, \sqrt {5} {\left (5 \, x + 3\right )} - 1089 \, \sqrt {5}\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5}}{1650 \, {\left (2 \, x - 1\right )}^{2}} \] Input:
integrate((2+3*x)*(3+5*x)^(1/2)/(1-2*x)^(5/2),x, algorithm="giac")
Output:
3/4*sqrt(10)*arcsin(1/11*sqrt(22)*sqrt(5*x + 3)) + 1/1650*(268*sqrt(5)*(5* x + 3) - 1089*sqrt(5))*sqrt(5*x + 3)*sqrt(-10*x + 5)/(2*x - 1)^2
Timed out. \[ \int \frac {(2+3 x) \sqrt {3+5 x}}{(1-2 x)^{5/2}} \, dx=\int \frac {\left (3\,x+2\right )\,\sqrt {5\,x+3}}{{\left (1-2\,x\right )}^{5/2}} \,d x \] Input:
int(((3*x + 2)*(5*x + 3)^(1/2))/(1 - 2*x)^(5/2),x)
Output:
int(((3*x + 2)*(5*x + 3)^(1/2))/(1 - 2*x)^(5/2), x)
Time = 0.19 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.14 \[ \int \frac {(2+3 x) \sqrt {3+5 x}}{(1-2 x)^{5/2}} \, dx=\frac {-198 \sqrt {-2 x +1}\, \sqrt {10}\, \mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right ) x +99 \sqrt {-2 x +1}\, \sqrt {10}\, \mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )-536 \sqrt {5 x +3}\, x +114 \sqrt {5 x +3}}{132 \sqrt {-2 x +1}\, \left (2 x -1\right )} \] Input:
int((2+3*x)*(3+5*x)^(1/2)/(1-2*x)^(5/2),x)
Output:
( - 198*sqrt( - 2*x + 1)*sqrt(10)*asin((sqrt( - 2*x + 1)*sqrt(5))/sqrt(11) )*x + 99*sqrt( - 2*x + 1)*sqrt(10)*asin((sqrt( - 2*x + 1)*sqrt(5))/sqrt(11 )) - 536*sqrt(5*x + 3)*x + 114*sqrt(5*x + 3))/(132*sqrt( - 2*x + 1)*(2*x - 1))