3.26.81 \(\int \frac {(2+3 x)^3 \sqrt {3+5 x}}{(1-2 x)^{5/2}} \, dx\) [2581]

3.26.81.1 Optimal result
3.26.81.2 Mathematica [A] (verified)
3.26.81.3 Rubi [A] (verified)
3.26.81.4 Maple [A] (verified)
3.26.81.5 Fricas [A] (verification not implemented)
3.26.81.6 Sympy [F]
3.26.81.7 Maxima [F(-1)]
3.26.81.8 Giac [A] (verification not implemented)
3.26.81.9 Mupad [F(-1)]

3.26.81.1 Optimal result

Integrand size = 26, antiderivative size = 113 \[ \int \frac {(2+3 x)^3 \sqrt {3+5 x}}{(1-2 x)^{5/2}} \, dx=-\frac {233 (2+3 x)^2 \sqrt {3+5 x}}{66 \sqrt {1-2 x}}+\frac {(2+3 x)^3 \sqrt {3+5 x}}{3 (1-2 x)^{3/2}}-\frac {\sqrt {1-2 x} \sqrt {3+5 x} (168157+69780 x)}{3520}+\frac {126513 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )}{320 \sqrt {10}} \]

output
126513/3200*arcsin(1/11*22^(1/2)*(3+5*x)^(1/2))*10^(1/2)+1/3*(2+3*x)^3*(3+ 
5*x)^(1/2)/(1-2*x)^(3/2)-233/66*(2+3*x)^2*(3+5*x)^(1/2)/(1-2*x)^(1/2)-1/35 
20*(168157+69780*x)*(1-2*x)^(1/2)*(3+5*x)^(1/2)
 
3.26.81.2 Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.69 \[ \int \frac {(2+3 x)^3 \sqrt {3+5 x}}{(1-2 x)^{5/2}} \, dx=\frac {-10 \sqrt {3+5 x} \left (625431-1786144 x+431244 x^2+71280 x^3\right )+4174929 \sqrt {10-20 x} (-1+2 x) \arctan \left (\frac {\sqrt {\frac {5}{2}-5 x}}{\sqrt {3+5 x}}\right )}{105600 (1-2 x)^{3/2}} \]

input
Integrate[((2 + 3*x)^3*Sqrt[3 + 5*x])/(1 - 2*x)^(5/2),x]
 
output
(-10*Sqrt[3 + 5*x]*(625431 - 1786144*x + 431244*x^2 + 71280*x^3) + 4174929 
*Sqrt[10 - 20*x]*(-1 + 2*x)*ArcTan[Sqrt[5/2 - 5*x]/Sqrt[3 + 5*x]])/(105600 
*(1 - 2*x)^(3/2))
 
3.26.81.3 Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.09, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {108, 27, 167, 27, 164, 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.

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

\(\Big \downarrow \) 108

\(\displaystyle \frac {(3 x+2)^3 \sqrt {5 x+3}}{3 (1-2 x)^{3/2}}-\frac {1}{3} \int \frac {(3 x+2)^2 (105 x+64)}{2 (1-2 x)^{3/2} \sqrt {5 x+3}}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {(3 x+2)^3 \sqrt {5 x+3}}{3 (1-2 x)^{3/2}}-\frac {1}{6} \int \frac {(3 x+2)^2 (105 x+64)}{(1-2 x)^{3/2} \sqrt {5 x+3}}dx\)

\(\Big \downarrow \) 167

\(\displaystyle \frac {1}{6} \left (-\frac {1}{11} \int -\frac {3 (3 x+2) (5815 x+3566)}{2 \sqrt {1-2 x} \sqrt {5 x+3}}dx-\frac {233 \sqrt {5 x+3} (3 x+2)^2}{11 \sqrt {1-2 x}}\right )+\frac {\sqrt {5 x+3} (3 x+2)^3}{3 (1-2 x)^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{6} \left (\frac {3}{22} \int \frac {(3 x+2) (5815 x+3566)}{\sqrt {1-2 x} \sqrt {5 x+3}}dx-\frac {233 (3 x+2)^2 \sqrt {5 x+3}}{11 \sqrt {1-2 x}}\right )+\frac {\sqrt {5 x+3} (3 x+2)^3}{3 (1-2 x)^{3/2}}\)

\(\Big \downarrow \) 164

\(\displaystyle \frac {1}{6} \left (\frac {3}{22} \left (\frac {1391643}{160} \int \frac {1}{\sqrt {1-2 x} \sqrt {5 x+3}}dx-\frac {1}{80} \sqrt {1-2 x} \sqrt {5 x+3} (69780 x+168157)\right )-\frac {233 (3 x+2)^2 \sqrt {5 x+3}}{11 \sqrt {1-2 x}}\right )+\frac {\sqrt {5 x+3} (3 x+2)^3}{3 (1-2 x)^{3/2}}\)

\(\Big \downarrow \) 64

\(\displaystyle \frac {1}{6} \left (\frac {3}{22} \left (\frac {1391643}{400} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}-\frac {1}{80} \sqrt {1-2 x} \sqrt {5 x+3} (69780 x+168157)\right )-\frac {233 (3 x+2)^2 \sqrt {5 x+3}}{11 \sqrt {1-2 x}}\right )+\frac {\sqrt {5 x+3} (3 x+2)^3}{3 (1-2 x)^{3/2}}\)

\(\Big \downarrow \) 223

\(\displaystyle \frac {1}{6} \left (\frac {3}{22} \left (\frac {1391643 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )}{80 \sqrt {10}}-\frac {1}{80} \sqrt {1-2 x} \sqrt {5 x+3} (69780 x+168157)\right )-\frac {233 (3 x+2)^2 \sqrt {5 x+3}}{11 \sqrt {1-2 x}}\right )+\frac {\sqrt {5 x+3} (3 x+2)^3}{3 (1-2 x)^{3/2}}\)

input
Int[((2 + 3*x)^3*Sqrt[3 + 5*x])/(1 - 2*x)^(5/2),x]
 
output
((2 + 3*x)^3*Sqrt[3 + 5*x])/(3*(1 - 2*x)^(3/2)) + ((-233*(2 + 3*x)^2*Sqrt[ 
3 + 5*x])/(11*Sqrt[1 - 2*x]) + (3*(-1/80*(Sqrt[1 - 2*x]*Sqrt[3 + 5*x]*(168 
157 + 69780*x)) + (1391643*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]])/(80*Sqrt[10]) 
))/22)/6
 

3.26.81.3.1 Defintions of rubi rules used

rule 27
Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 64
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])
 

rule 108
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Simp[(a + b*x)^(m + 1)*(c + d*x)^n*((e + f*x)^p/(b*(m + 1))) 
, x] - Simp[1/(b*(m + 1))   Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f* 
x)^(p - 1)*Simp[d*e*n + c*f*p + d*f*(n + p)*x, x], x], x] /; FreeQ[{a, b, c 
, d, e, f}, x] && LtQ[m, -1] && GtQ[n, 0] && GtQ[p, 0] && (IntegersQ[2*m, 2 
*n, 2*p] || IntegersQ[m, n + p] || IntegersQ[p, m + n])
 

rule 164
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_ 
))*((g_.) + (h_.)*(x_)), x_] :> Simp[(-(a*d*f*h*(n + 2) + b*c*f*h*(m + 2) - 
 b*d*(f*g + e*h)*(m + n + 3) - b*d*f*h*(m + n + 2)*x))*(a + b*x)^(m + 1)*(( 
c + d*x)^(n + 1)/(b^2*d^2*(m + n + 2)*(m + n + 3))), x] + Simp[(a^2*d^2*f*h 
*(n + 1)*(n + 2) + a*b*d*(n + 1)*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 
3)) + b^2*(c^2*f*h*(m + 1)*(m + 2) - c*d*(f*g + e*h)*(m + 1)*(m + n + 3) + 
d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2*d^2*(m + n + 2)*(m + n + 3))   Int[( 
a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] 
&& NeQ[m + n + 2, 0] && NeQ[m + n + 3, 0]
 

rule 167
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_)*((g_.) + (h_.)*(x_)), x_] :> Simp[(b*g - a*h)*(a + b*x)^(m + 1)*(c + 
 d*x)^n*((e + f*x)^(p + 1)/(b*(b*e - a*f)*(m + 1))), x] - Simp[1/(b*(b*e - 
a*f)*(m + 1))   Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p*Simp[b* 
c*(f*g - e*h)*(m + 1) + (b*g - a*h)*(d*e*n + c*f*(p + 1)) + d*(b*(f*g - e*h 
)*(m + 1) + f*(b*g - a*h)*(n + p + 1))*x, x], x], x] /; FreeQ[{a, b, c, d, 
e, f, g, h, p}, x] && LtQ[m, -1] && GtQ[n, 0] && IntegersQ[2*m, 2*n, 2*p]
 

rule 223
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]
 
3.26.81.4 Maple [A] (verified)

Time = 4.05 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.21

method result size
default \(\frac {\left (16699716 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x^{2}-1425600 x^{3} \sqrt {-10 x^{2}-x +3}-16699716 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x -8624880 x^{2} \sqrt {-10 x^{2}-x +3}+4174929 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )+35722880 x \sqrt {-10 x^{2}-x +3}-12508620 \sqrt {-10 x^{2}-x +3}\right ) \sqrt {1-2 x}\, \sqrt {3+5 x}}{211200 \left (-1+2 x \right )^{2} \sqrt {-10 x^{2}-x +3}}\) \(137\)

input
int((2+3*x)^3*(3+5*x)^(1/2)/(1-2*x)^(5/2),x,method=_RETURNVERBOSE)
 
output
1/211200*(16699716*10^(1/2)*arcsin(20/11*x+1/11)*x^2-1425600*x^3*(-10*x^2- 
x+3)^(1/2)-16699716*10^(1/2)*arcsin(20/11*x+1/11)*x-8624880*x^2*(-10*x^2-x 
+3)^(1/2)+4174929*10^(1/2)*arcsin(20/11*x+1/11)+35722880*x*(-10*x^2-x+3)^( 
1/2)-12508620*(-10*x^2-x+3)^(1/2))*(1-2*x)^(1/2)*(3+5*x)^(1/2)/(-1+2*x)^2/ 
(-10*x^2-x+3)^(1/2)
 
3.26.81.5 Fricas [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.85 \[ \int \frac {(2+3 x)^3 \sqrt {3+5 x}}{(1-2 x)^{5/2}} \, dx=-\frac {4174929 \, \sqrt {10} {\left (4 \, x^{2} - 4 \, x + 1\right )} \arctan \left (\frac {\sqrt {10} {\left (20 \, x + 1\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{20 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) + 20 \, {\left (71280 \, x^{3} + 431244 \, x^{2} - 1786144 \, x + 625431\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{211200 \, {\left (4 \, x^{2} - 4 \, x + 1\right )}} \]

input
integrate((2+3*x)^3*(3+5*x)^(1/2)/(1-2*x)^(5/2),x, algorithm="fricas")
 
output
-1/211200*(4174929*sqrt(10)*(4*x^2 - 4*x + 1)*arctan(1/20*sqrt(10)*(20*x + 
 1)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x - 3)) + 20*(71280*x^3 + 43124 
4*x^2 - 1786144*x + 625431)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(4*x^2 - 4*x + 1 
)
 
3.26.81.6 Sympy [F]

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

input
integrate((2+3*x)**3*(3+5*x)**(1/2)/(1-2*x)**(5/2),x)
 
output
Integral((3*x + 2)**3*sqrt(5*x + 3)/(1 - 2*x)**(5/2), x)
 
3.26.81.7 Maxima [F(-1)]

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

input
integrate((2+3*x)^3*(3+5*x)^(1/2)/(1-2*x)^(5/2),x, algorithm="maxima")
 
output
Timed out
 
3.26.81.8 Giac [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.74 \[ \int \frac {(2+3 x)^3 \sqrt {3+5 x}}{(1-2 x)^{5/2}} \, dx=\frac {126513}{3200} \, \sqrt {10} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right ) - \frac {{\left (4 \, {\left (891 \, {\left (4 \, \sqrt {5} {\left (5 \, x + 3\right )} + 85 \, \sqrt {5}\right )} {\left (5 \, x + 3\right )} - 2783318 \, \sqrt {5}\right )} {\left (5 \, x + 3\right )} + 45924219 \, \sqrt {5}\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5}}{1320000 \, {\left (2 \, x - 1\right )}^{2}} \]

input
integrate((2+3*x)^3*(3+5*x)^(1/2)/(1-2*x)^(5/2),x, algorithm="giac")
 
output
126513/3200*sqrt(10)*arcsin(1/11*sqrt(22)*sqrt(5*x + 3)) - 1/1320000*(4*(8 
91*(4*sqrt(5)*(5*x + 3) + 85*sqrt(5))*(5*x + 3) - 2783318*sqrt(5))*(5*x + 
3) + 45924219*sqrt(5))*sqrt(5*x + 3)*sqrt(-10*x + 5)/(2*x - 1)^2
 
3.26.81.9 Mupad [F(-1)]

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

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