\(\int \frac {(1-2 x)^{5/2} (2+3 x)}{\sqrt {3+5 x}} \, dx\) [2427]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [F]
   Maxima [A] (verification not implemented)
   Giac [B] (verification not implemented)
   Mupad [F(-1)]

Optimal result

Integrand size = 24, antiderivative size = 116 \[ \int \frac {(1-2 x)^{5/2} (2+3 x)}{\sqrt {3+5 x}} \, dx=\frac {5929 \sqrt {1-2 x} \sqrt {3+5 x}}{16000}+\frac {539 (1-2 x)^{3/2} \sqrt {3+5 x}}{4800}+\frac {49 (1-2 x)^{5/2} \sqrt {3+5 x}}{1200}-\frac {3}{40} (1-2 x)^{7/2} \sqrt {3+5 x}+\frac {65219 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )}{16000 \sqrt {10}} \]

[Out]

65219/160000*arcsin(1/11*22^(1/2)*(3+5*x)^(1/2))*10^(1/2)+539/4800*(1-2*x)^(3/2)*(3+5*x)^(1/2)+49/1200*(1-2*x)
^(5/2)*(3+5*x)^(1/2)-3/40*(1-2*x)^(7/2)*(3+5*x)^(1/2)+5929/16000*(1-2*x)^(1/2)*(3+5*x)^(1/2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {81, 52, 56, 222} \[ \int \frac {(1-2 x)^{5/2} (2+3 x)}{\sqrt {3+5 x}} \, dx=\frac {65219 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )}{16000 \sqrt {10}}-\frac {3}{40} \sqrt {5 x+3} (1-2 x)^{7/2}+\frac {49 \sqrt {5 x+3} (1-2 x)^{5/2}}{1200}+\frac {539 \sqrt {5 x+3} (1-2 x)^{3/2}}{4800}+\frac {5929 \sqrt {5 x+3} \sqrt {1-2 x}}{16000} \]

[In]

Int[((1 - 2*x)^(5/2)*(2 + 3*x))/Sqrt[3 + 5*x],x]

[Out]

(5929*Sqrt[1 - 2*x]*Sqrt[3 + 5*x])/16000 + (539*(1 - 2*x)^(3/2)*Sqrt[3 + 5*x])/4800 + (49*(1 - 2*x)^(5/2)*Sqrt
[3 + 5*x])/1200 - (3*(1 - 2*x)^(7/2)*Sqrt[3 + 5*x])/40 + (65219*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]])/(16000*Sqrt[
10])

Rule 52

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] + Dist[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] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 56

Int[1/(Sqrt[(a_.) + (b_.)*(x_)]*Sqrt[(c_.) + (d_.)*(x_)]), x_Symbol] :> Dist[2/Sqrt[b], Subst[Int[1/Sqrt[b*c -
 a*d + d*x^2], x], x, Sqrt[a + b*x]], x] /; FreeQ[{a, b, c, d}, x] && GtQ[b*c - a*d, 0] && GtQ[b, 0]

Rule 81

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(c + d*x)^
(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Dist[(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]

Rule 222

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]

Rubi steps \begin{align*} \text {integral}& = -\frac {3}{40} (1-2 x)^{7/2} \sqrt {3+5 x}+\frac {49}{80} \int \frac {(1-2 x)^{5/2}}{\sqrt {3+5 x}} \, dx \\ & = \frac {49 (1-2 x)^{5/2} \sqrt {3+5 x}}{1200}-\frac {3}{40} (1-2 x)^{7/2} \sqrt {3+5 x}+\frac {539}{480} \int \frac {(1-2 x)^{3/2}}{\sqrt {3+5 x}} \, dx \\ & = \frac {539 (1-2 x)^{3/2} \sqrt {3+5 x}}{4800}+\frac {49 (1-2 x)^{5/2} \sqrt {3+5 x}}{1200}-\frac {3}{40} (1-2 x)^{7/2} \sqrt {3+5 x}+\frac {5929 \int \frac {\sqrt {1-2 x}}{\sqrt {3+5 x}} \, dx}{3200} \\ & = \frac {5929 \sqrt {1-2 x} \sqrt {3+5 x}}{16000}+\frac {539 (1-2 x)^{3/2} \sqrt {3+5 x}}{4800}+\frac {49 (1-2 x)^{5/2} \sqrt {3+5 x}}{1200}-\frac {3}{40} (1-2 x)^{7/2} \sqrt {3+5 x}+\frac {65219 \int \frac {1}{\sqrt {1-2 x} \sqrt {3+5 x}} \, dx}{32000} \\ & = \frac {5929 \sqrt {1-2 x} \sqrt {3+5 x}}{16000}+\frac {539 (1-2 x)^{3/2} \sqrt {3+5 x}}{4800}+\frac {49 (1-2 x)^{5/2} \sqrt {3+5 x}}{1200}-\frac {3}{40} (1-2 x)^{7/2} \sqrt {3+5 x}+\frac {65219 \text {Subst}\left (\int \frac {1}{\sqrt {11-2 x^2}} \, dx,x,\sqrt {3+5 x}\right )}{16000 \sqrt {5}} \\ & = \frac {5929 \sqrt {1-2 x} \sqrt {3+5 x}}{16000}+\frac {539 (1-2 x)^{3/2} \sqrt {3+5 x}}{4800}+\frac {49 (1-2 x)^{5/2} \sqrt {3+5 x}}{1200}-\frac {3}{40} (1-2 x)^{7/2} \sqrt {3+5 x}+\frac {65219 \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )}{16000 \sqrt {10}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.30 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.67 \[ \int \frac {(1-2 x)^{5/2} (2+3 x)}{\sqrt {3+5 x}} \, dx=\frac {10 \sqrt {1-2 x} \left (64611+116625 x-91180 x^2-90400 x^3+144000 x^4\right )-195657 \sqrt {30+50 x} \arctan \left (\frac {\sqrt {\frac {5}{2}-5 x}}{\sqrt {3+5 x}}\right )}{480000 \sqrt {3+5 x}} \]

[In]

Integrate[((1 - 2*x)^(5/2)*(2 + 3*x))/Sqrt[3 + 5*x],x]

[Out]

(10*Sqrt[1 - 2*x]*(64611 + 116625*x - 91180*x^2 - 90400*x^3 + 144000*x^4) - 195657*Sqrt[30 + 50*x]*ArcTan[Sqrt
[5/2 - 5*x]/Sqrt[3 + 5*x]])/(480000*Sqrt[3 + 5*x])

Maple [A] (verified)

Time = 3.87 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.89

method result size
risch \(-\frac {\left (28800 x^{3}-35360 x^{2}+2980 x +21537\right ) \left (-1+2 x \right ) \sqrt {3+5 x}\, \sqrt {\left (1-2 x \right ) \left (3+5 x \right )}}{48000 \sqrt {-\left (-1+2 x \right ) \left (3+5 x \right )}\, \sqrt {1-2 x}}+\frac {65219 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) \sqrt {\left (1-2 x \right ) \left (3+5 x \right )}}{320000 \sqrt {1-2 x}\, \sqrt {3+5 x}}\) \(103\)
default \(\frac {\sqrt {1-2 x}\, \sqrt {3+5 x}\, \left (576000 x^{3} \sqrt {-10 x^{2}-x +3}-707200 x^{2} \sqrt {-10 x^{2}-x +3}+195657 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )+59600 x \sqrt {-10 x^{2}-x +3}+430740 \sqrt {-10 x^{2}-x +3}\right )}{960000 \sqrt {-10 x^{2}-x +3}}\) \(104\)

[In]

int((1-2*x)^(5/2)*(2+3*x)/(3+5*x)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/48000*(28800*x^3-35360*x^2+2980*x+21537)*(-1+2*x)*(3+5*x)^(1/2)/(-(-1+2*x)*(3+5*x))^(1/2)*((1-2*x)*(3+5*x))
^(1/2)/(1-2*x)^(1/2)+65219/320000*10^(1/2)*arcsin(20/11*x+1/11)*((1-2*x)*(3+5*x))^(1/2)/(1-2*x)^(1/2)/(3+5*x)^
(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.62 \[ \int \frac {(1-2 x)^{5/2} (2+3 x)}{\sqrt {3+5 x}} \, dx=\frac {1}{48000} \, {\left (28800 \, x^{3} - 35360 \, x^{2} + 2980 \, x + 21537\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1} - \frac {65219}{320000} \, \sqrt {10} \arctan \left (\frac {\sqrt {10} {\left (20 \, x + 1\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{20 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) \]

[In]

integrate((1-2*x)^(5/2)*(2+3*x)/(3+5*x)^(1/2),x, algorithm="fricas")

[Out]

1/48000*(28800*x^3 - 35360*x^2 + 2980*x + 21537)*sqrt(5*x + 3)*sqrt(-2*x + 1) - 65219/320000*sqrt(10)*arctan(1
/20*sqrt(10)*(20*x + 1)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x - 3))

Sympy [F]

\[ \int \frac {(1-2 x)^{5/2} (2+3 x)}{\sqrt {3+5 x}} \, dx=\int \frac {\left (1 - 2 x\right )^{\frac {5}{2}} \cdot \left (3 x + 2\right )}{\sqrt {5 x + 3}}\, dx \]

[In]

integrate((1-2*x)**(5/2)*(2+3*x)/(3+5*x)**(1/2),x)

[Out]

Integral((1 - 2*x)**(5/2)*(3*x + 2)/sqrt(5*x + 3), x)

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.65 \[ \int \frac {(1-2 x)^{5/2} (2+3 x)}{\sqrt {3+5 x}} \, dx=\frac {3}{5} \, \sqrt {-10 \, x^{2} - x + 3} x^{3} - \frac {221}{300} \, \sqrt {-10 \, x^{2} - x + 3} x^{2} + \frac {149}{2400} \, \sqrt {-10 \, x^{2} - x + 3} x - \frac {65219}{320000} \, \sqrt {10} \arcsin \left (-\frac {20}{11} \, x - \frac {1}{11}\right ) + \frac {7179}{16000} \, \sqrt {-10 \, x^{2} - x + 3} \]

[In]

integrate((1-2*x)^(5/2)*(2+3*x)/(3+5*x)^(1/2),x, algorithm="maxima")

[Out]

3/5*sqrt(-10*x^2 - x + 3)*x^3 - 221/300*sqrt(-10*x^2 - x + 3)*x^2 + 149/2400*sqrt(-10*x^2 - x + 3)*x - 65219/3
20000*sqrt(10)*arcsin(-20/11*x - 1/11) + 7179/16000*sqrt(-10*x^2 - x + 3)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 203 vs. \(2 (83) = 166\).

Time = 0.31 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.75 \[ \int \frac {(1-2 x)^{5/2} (2+3 x)}{\sqrt {3+5 x}} \, dx=\frac {1}{800000} \, \sqrt {5} {\left (2 \, {\left (4 \, {\left (8 \, {\left (60 \, x - 119\right )} {\left (5 \, x + 3\right )} + 6163\right )} {\left (5 \, x + 3\right )} - 66189\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5} - 184305 \, \sqrt {2} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right )\right )} - \frac {1}{30000} \, \sqrt {5} {\left (2 \, {\left (4 \, {\left (40 \, x - 59\right )} {\left (5 \, x + 3\right )} + 1293\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5} + 4785 \, \sqrt {2} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right )\right )} - \frac {1}{400} \, \sqrt {5} {\left (2 \, {\left (20 \, x - 23\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5} - 143 \, \sqrt {2} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right )\right )} + \frac {1}{25} \, \sqrt {5} {\left (11 \, \sqrt {2} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right ) + 2 \, \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5}\right )} \]

[In]

integrate((1-2*x)^(5/2)*(2+3*x)/(3+5*x)^(1/2),x, algorithm="giac")

[Out]

1/800000*sqrt(5)*(2*(4*(8*(60*x - 119)*(5*x + 3) + 6163)*(5*x + 3) - 66189)*sqrt(5*x + 3)*sqrt(-10*x + 5) - 18
4305*sqrt(2)*arcsin(1/11*sqrt(22)*sqrt(5*x + 3))) - 1/30000*sqrt(5)*(2*(4*(40*x - 59)*(5*x + 3) + 1293)*sqrt(5
*x + 3)*sqrt(-10*x + 5) + 4785*sqrt(2)*arcsin(1/11*sqrt(22)*sqrt(5*x + 3))) - 1/400*sqrt(5)*(2*(20*x - 23)*sqr
t(5*x + 3)*sqrt(-10*x + 5) - 143*sqrt(2)*arcsin(1/11*sqrt(22)*sqrt(5*x + 3))) + 1/25*sqrt(5)*(11*sqrt(2)*arcsi
n(1/11*sqrt(22)*sqrt(5*x + 3)) + 2*sqrt(5*x + 3)*sqrt(-10*x + 5))

Mupad [F(-1)]

Timed out. \[ \int \frac {(1-2 x)^{5/2} (2+3 x)}{\sqrt {3+5 x}} \, dx=\int \frac {{\left (1-2\,x\right )}^{5/2}\,\left (3\,x+2\right )}{\sqrt {5\,x+3}} \,d x \]

[In]

int(((1 - 2*x)^(5/2)*(3*x + 2))/(5*x + 3)^(1/2),x)

[Out]

int(((1 - 2*x)^(5/2)*(3*x + 2))/(5*x + 3)^(1/2), x)