\(\int \frac {(2+3 x)^3 \sqrt {3+5 x}}{(1-2 x)^{5/2}} \, dx\) [1190]

Optimal result

Integrand size = 26, antiderivative size = 116 \[ \int \frac {(2+3 x)^3 \sqrt {3+5 x}}{(1-2 x)^{5/2}} \, dx=-\frac {441 \sqrt {3+5 x}}{8 \sqrt {1-2 x}}-\frac {4077}{320} \sqrt {1-2 x} \sqrt {3+5 x}+\frac {27}{32} (1-2 x)^{3/2} \sqrt {3+5 x}+\frac {343 (3+5 x)^{3/2}}{132 (1-2 x)^{3/2}}+\frac {126513 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )}{320 \sqrt {10}} \] Output:

-441/8*(3+5*x)^(1/2)/(1-2*x)^(1/2)-4077/320*(1-2*x)^(1/2)*(3+5*x)^(1/2)+27 
/32*(1-2*x)^(3/2)*(3+5*x)^(1/2)+343/132*(3+5*x)^(3/2)/(1-2*x)^(3/2)+126513 
/3200*arcsin(1/11*22^(1/2)*(3+5*x)^(1/2))*10^(1/2)
 

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.67 \[ \int \frac {(2+3 x)^3 \sqrt {3+5 x}}{(1-2 x)^{5/2}} \, dx=\frac {-10 \sqrt {3+5 x} \left (625431-1786144 x+431244 x^2+71280 x^3\right )+4174929 \sqrt {10-20 x} (-1+2 x) \arctan \left (\frac {\sqrt {\frac {5}{2}-5 x}}{\sqrt {3+5 x}}\right )}{105600 (1-2 x)^{3/2}} \] 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))
 

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.06, 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
 

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]
 

Maple [A] (verified)

Time = 0.26 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.12

method	result	size
default	\(\frac {\left (16699716 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x^{2}-1425600 x^{3} \sqrt {-10 x^{2}-x +3}-16699716 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x -8624880 x^{2} \sqrt {-10 x^{2}-x +3}+4174929 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )[phone] x \sqrt {-10 x^{2}-x +3}-12508620 \sqrt {-10 x^{2}-x +3}\right ) \sqrt {3+5 x}}{211200 \left (1-2 x \right )^{\frac {3}{2}} \sqrt {-10 x^{2}-x +3}}\)	\(130\)

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)[phone]*x*(-10*x^2-x+3)^( 
1/2)-12508620*(-10*x^2-x+3)^(1/2))*(3+5*x)^(1/2)/(1-2*x)^(3/2)/(-10*x^2-x+ 
3)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.83 \[ \int \frac {(2+3 x)^3 \sqrt {3+5 x}}{(1-2 x)^{5/2}} \, dx=-\frac {4174929 \, \sqrt {10} {\left (4 \, x^{2} - 4 \, x + 1\right )} \arctan \left (\frac {\sqrt {10} {\left (20 \, x + 1\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{20 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) + 20 \, {\left (71280 \, x^{3} + 431244 \, x^{2} - 1786144 \, x + 625431\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{211200 \, {\left (4 \, x^{2} - 4 \, x + 1\right )}} \] 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 
)
 

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)
 

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
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.72 \[ \int \frac {(2+3 x)^3 \sqrt {3+5 x}}{(1-2 x)^{5/2}} \, dx=\frac {126513}{3200} \, \sqrt {10} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right ) - \frac {{\left (4 \, {\left (891 \, {\left (4 \, \sqrt {5} {\left (5 \, x + 3\right )} + 85 \, \sqrt {5}\right )} {\left (5 \, x + 3\right )} - 2783318 \, \sqrt {5}\right )} {\left (5 \, x + 3\right )} + 45924219 \, \sqrt {5}\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5}}{1320000 \, {\left (2 \, x - 1\right )}^{2}} \] 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
 

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)
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.91 \[ \int \frac {(2+3 x)^3 \sqrt {3+5 x}}{(1-2 x)^{5/2}} \, dx=\frac {-8349858 \sqrt {-2 x +1}\, \sqrt {10}\, \mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right ) x +4174929 \sqrt {-2 x +1}\, \sqrt {10}\, \mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )+712800 \sqrt {5 x +3}\, x^{3}+4312440 \sqrt {5 x +3}\, x^{2}-17861440 \sqrt {5 x +3}\, x +6254310 \sqrt {5 x +3}}{105600 \sqrt {-2 x +1}\, \left (2 x -1\right )} \] Input:

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

Output:

( - 8349858*sqrt( - 2*x + 1)*sqrt(10)*asin((sqrt( - 2*x + 1)*sqrt(5))/sqrt 
(11))*x + 4174929*sqrt( - 2*x + 1)*sqrt(10)*asin((sqrt( - 2*x + 1)*sqrt(5) 
)/sqrt(11)) + 712800*sqrt(5*x + 3)*x**3 + 4312440*sqrt(5*x + 3)*x**2 - 178 
61440*sqrt(5*x + 3)*x + 6254310*sqrt(5*x + 3))/(105600*sqrt( - 2*x + 1)*(2 
*x - 1))