Integrand size = 24, antiderivative size = 96 \[ \int \frac {(1-2 x)^{3/2} (2+3 x)}{(3+5 x)^{5/2}} \, dx=-\frac {2 (1-2 x)^{3/2}}{75 (3+5 x)^{3/2}}-\frac {62 \sqrt {1-2 x}}{125 \sqrt {3+5 x}}-\frac {6}{125} \sqrt {1-2 x} \sqrt {3+5 x}-\frac {19}{25} \sqrt {\frac {2}{5}} \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right ) \] Output:
-2/75*(1-2*x)^(3/2)/(3+5*x)^(3/2)-62/125*(1-2*x)^(1/2)/(3+5*x)^(1/2)-6/125 *(1-2*x)^(1/2)*(3+5*x)^(1/2)-19/125*arcsin(1/11*22^(1/2)*(3+5*x)^(1/2))*10 ^(1/2)
Time = 0.13 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.69 \[ \int \frac {(1-2 x)^{3/2} (2+3 x)}{(3+5 x)^{5/2}} \, dx=-\frac {2 \sqrt {1-2 x} \left (73+145 x+45 x^2\right )}{75 (3+5 x)^{3/2}}+\frac {19}{25} \sqrt {\frac {2}{5}} \arctan \left (\frac {\sqrt {\frac {5}{2}-5 x}}{\sqrt {3+5 x}}\right ) \] Input:
Integrate[((1 - 2*x)^(3/2)*(2 + 3*x))/(3 + 5*x)^(5/2),x]
Output:
(-2*Sqrt[1 - 2*x]*(73 + 145*x + 45*x^2))/(75*(3 + 5*x)^(3/2)) + (19*Sqrt[2 /5]*ArcTan[Sqrt[5/2 - 5*x]/Sqrt[3 + 5*x]])/25
Time = 0.18 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.08, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {87, 57, 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.
\(\displaystyle \int \frac {(1-2 x)^{3/2} (3 x+2)}{(5 x+3)^{5/2}} \, dx\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {19}{33} \int \frac {(1-2 x)^{3/2}}{(5 x+3)^{3/2}}dx-\frac {2 (1-2 x)^{5/2}}{165 (5 x+3)^{3/2}}\) |
\(\Big \downarrow \) 57 |
\(\displaystyle \frac {19}{33} \left (-\frac {6}{5} \int \frac {\sqrt {1-2 x}}{\sqrt {5 x+3}}dx-\frac {2 (1-2 x)^{3/2}}{5 \sqrt {5 x+3}}\right )-\frac {2 (1-2 x)^{5/2}}{165 (5 x+3)^{3/2}}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {19}{33} \left (-\frac {6}{5} \left (\frac {11}{10} \int \frac {1}{\sqrt {1-2 x} \sqrt {5 x+3}}dx+\frac {1}{5} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {2 (1-2 x)^{3/2}}{5 \sqrt {5 x+3}}\right )-\frac {2 (1-2 x)^{5/2}}{165 (5 x+3)^{3/2}}\) |
\(\Big \downarrow \) 64 |
\(\displaystyle \frac {19}{33} \left (-\frac {6}{5} \left (\frac {11}{25} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}+\frac {1}{5} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {2 (1-2 x)^{3/2}}{5 \sqrt {5 x+3}}\right )-\frac {2 (1-2 x)^{5/2}}{165 (5 x+3)^{3/2}}\) |
\(\Big \downarrow \) 223 |
\(\displaystyle \frac {19}{33} \left (-\frac {6}{5} \left (\frac {11 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )}{5 \sqrt {10}}+\frac {1}{5} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {2 (1-2 x)^{3/2}}{5 \sqrt {5 x+3}}\right )-\frac {2 (1-2 x)^{5/2}}{165 (5 x+3)^{3/2}}\) |
Input:
Int[((1 - 2*x)^(3/2)*(2 + 3*x))/(3 + 5*x)^(5/2),x]
Output:
(-2*(1 - 2*x)^(5/2))/(165*(3 + 5*x)^(3/2)) + (19*((-2*(1 - 2*x)^(3/2))/(5* Sqrt[3 + 5*x]) - (6*((Sqrt[1 - 2*x]*Sqrt[3 + 5*x])/5 + (11*ArcSin[Sqrt[2/1 1]*Sqrt[3 + 5*x]])/(5*Sqrt[10])))/5))/33
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[((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]
Time = 0.20 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.18
method | result | size |
default | \(-\frac {\left (1425 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x^{2}+1710 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x +900 x^{2} \sqrt {-10 x^{2}-x +3}+513 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )+2900 x \sqrt {-10 x^{2}-x +3}+1460 \sqrt {-10 x^{2}-x +3}\right ) \sqrt {1-2 x}}{750 \sqrt {-10 x^{2}-x +3}\, \left (3+5 x \right )^{\frac {3}{2}}}\) | \(113\) |
Input:
int((1-2*x)^(3/2)*(2+3*x)/(3+5*x)^(5/2),x,method=_RETURNVERBOSE)
Output:
-1/750*(1425*10^(1/2)*arcsin(20/11*x+1/11)*x^2+1710*10^(1/2)*arcsin(20/11* x+1/11)*x+900*x^2*(-10*x^2-x+3)^(1/2)+513*10^(1/2)*arcsin(20/11*x+1/11)+29 00*x*(-10*x^2-x+3)^(1/2)+1460*(-10*x^2-x+3)^(1/2))*(1-2*x)^(1/2)/(-10*x^2- x+3)^(1/2)/(3+5*x)^(3/2)
Time = 0.08 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.95 \[ \int \frac {(1-2 x)^{3/2} (2+3 x)}{(3+5 x)^{5/2}} \, dx=\frac {57 \, \sqrt {\frac {2}{5}} {\left (25 \, x^{2} + 30 \, x + 9\right )} \arctan \left (\frac {\sqrt {\frac {2}{5}} {\left (20 \, x + 1\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{4 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) - 4 \, {\left (45 \, x^{2} + 145 \, x + 73\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{150 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} \] Input:
integrate((1-2*x)^(3/2)*(2+3*x)/(3+5*x)^(5/2),x, algorithm="fricas")
Output:
1/150*(57*sqrt(2/5)*(25*x^2 + 30*x + 9)*arctan(1/4*sqrt(2/5)*(20*x + 1)*sq rt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x - 3)) - 4*(45*x^2 + 145*x + 73)*sqr t(5*x + 3)*sqrt(-2*x + 1))/(25*x^2 + 30*x + 9)
\[ \int \frac {(1-2 x)^{3/2} (2+3 x)}{(3+5 x)^{5/2}} \, dx=\int \frac {\left (1 - 2 x\right )^{\frac {3}{2}} \cdot \left (3 x + 2\right )}{\left (5 x + 3\right )^{\frac {5}{2}}}\, dx \] Input:
integrate((1-2*x)**(3/2)*(2+3*x)/(3+5*x)**(5/2),x)
Output:
Integral((1 - 2*x)**(3/2)*(3*x + 2)/(5*x + 3)**(5/2), x)
Time = 0.11 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.24 \[ \int \frac {(1-2 x)^{3/2} (2+3 x)}{(3+5 x)^{5/2}} \, dx=-\frac {19}{250} \, \sqrt {5} \sqrt {2} \arcsin \left (\frac {20}{11} \, x + \frac {1}{11}\right ) - \frac {{\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}}}{75 \, {\left (125 \, x^{3} + 225 \, x^{2} + 135 \, x + 27\right )}} + \frac {3 \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}}}{25 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} - \frac {11 \, \sqrt {-10 \, x^{2} - x + 3}}{375 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} - \frac {283 \, \sqrt {-10 \, x^{2} - x + 3}}{375 \, {\left (5 \, x + 3\right )}} \] Input:
integrate((1-2*x)^(3/2)*(2+3*x)/(3+5*x)^(5/2),x, algorithm="maxima")
Output:
-19/250*sqrt(5)*sqrt(2)*arcsin(20/11*x + 1/11) - 1/75*(-10*x^2 - x + 3)^(3 /2)/(125*x^3 + 225*x^2 + 135*x + 27) + 3/25*(-10*x^2 - x + 3)^(3/2)/(25*x^ 2 + 30*x + 9) - 11/375*sqrt(-10*x^2 - x + 3)/(25*x^2 + 30*x + 9) - 283/375 *sqrt(-10*x^2 - x + 3)/(5*x + 3)
Leaf count of result is larger than twice the leaf count of optimal. 159 vs. \(2 (67) = 134\).
Time = 0.18 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.66 \[ \int \frac {(1-2 x)^{3/2} (2+3 x)}{(3+5 x)^{5/2}} \, dx=-\frac {\sqrt {10} {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{3}}{30000 \, {\left (5 \, x + 3\right )}^{\frac {3}{2}}} - \frac {6}{625} \, \sqrt {5} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5} - \frac {19}{125} \, \sqrt {10} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right ) - \frac {61 \, \sqrt {10} {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}{2500 \, \sqrt {5 \, x + 3}} + \frac {\sqrt {10} {\left (5 \, x + 3\right )}^{\frac {3}{2}} {\left (\frac {183 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} + 4\right )}}{1875 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{3}} \] Input:
integrate((1-2*x)^(3/2)*(2+3*x)/(3+5*x)^(5/2),x, algorithm="giac")
Output:
-1/30000*sqrt(10)*(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))^3/(5*x + 3)^(3/2) - 6/625*sqrt(5)*sqrt(5*x + 3)*sqrt(-10*x + 5) - 19/125*sqrt(10)*arcsin(1/11 *sqrt(22)*sqrt(5*x + 3)) - 61/2500*sqrt(10)*(sqrt(2)*sqrt(-10*x + 5) - sqr t(22))/sqrt(5*x + 3) + 1/1875*sqrt(10)*(5*x + 3)^(3/2)*(183*(sqrt(2)*sqrt( -10*x + 5) - sqrt(22))^2/(5*x + 3) + 4)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22 ))^3
Timed out. \[ \int \frac {(1-2 x)^{3/2} (2+3 x)}{(3+5 x)^{5/2}} \, dx=\int \frac {{\left (1-2\,x\right )}^{3/2}\,\left (3\,x+2\right )}{{\left (5\,x+3\right )}^{5/2}} \,d x \] Input:
int(((1 - 2*x)^(3/2)*(3*x + 2))/(5*x + 3)^(5/2),x)
Output:
int(((1 - 2*x)^(3/2)*(3*x + 2))/(5*x + 3)^(5/2), x)
Time = 0.26 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.99 \[ \int \frac {(1-2 x)^{3/2} (2+3 x)}{(3+5 x)^{5/2}} \, dx=\frac {285 \sqrt {5 x +3}\, \sqrt {10}\, \mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right ) x +171 \sqrt {5 x +3}\, \sqrt {10}\, \mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )-450 \sqrt {-2 x +1}\, x^{2}-1450 \sqrt {-2 x +1}\, x -730 \sqrt {-2 x +1}}{375 \sqrt {5 x +3}\, \left (5 x +3\right )} \] Input:
int((1-2*x)^(3/2)*(2+3*x)/(3+5*x)^(5/2),x)
Output:
(285*sqrt(5*x + 3)*sqrt(10)*asin((sqrt( - 2*x + 1)*sqrt(5))/sqrt(11))*x + 171*sqrt(5*x + 3)*sqrt(10)*asin((sqrt( - 2*x + 1)*sqrt(5))/sqrt(11)) - 450 *sqrt( - 2*x + 1)*x**2 - 1450*sqrt( - 2*x + 1)*x - 730*sqrt( - 2*x + 1))/( 375*sqrt(5*x + 3)*(5*x + 3))