\(\int \frac {(2+3 x)^3 (3+5 x)^{5/2}}{\sqrt {1-2 x}} \, dx\) [1082]

Optimal result

Integrand size = 26, antiderivative size = 160 \[ \int \frac {(2+3 x)^3 (3+5 x)^{5/2}}{\sqrt {1-2 x}} \, dx=-\frac {243487211 \sqrt {1-2 x} \sqrt {3+5 x}}{819200}-\frac {22135201 \sqrt {1-2 x} (3+5 x)^{3/2}}{614400}-\frac {2012291 \sqrt {1-2 x} (3+5 x)^{5/2}}{384000}-\frac {66519 \sqrt {1-2 x} (3+5 x)^{7/2}}{32000}+\frac {657}{800} (1-2 x)^{3/2} (3+5 x)^{7/2}-\frac {9}{80} (1-2 x)^{5/2} (3+5 x)^{7/2}+\frac {2678359321 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )}{819200 \sqrt {10}} \] Output:

-243487211/819200*(1-2*x)^(1/2)*(3+5*x)^(1/2)-22135201/614400*(1-2*x)^(1/2 
)*(3+5*x)^(3/2)-2012291/384000*(1-2*x)^(1/2)*(3+5*x)^(5/2)-66519/32000*(1- 
2*x)^(1/2)*(3+5*x)^(7/2)+657/800*(1-2*x)^(3/2)*(3+5*x)^(7/2)-9/80*(1-2*x)^ 
(5/2)*(3+5*x)^(7/2)[phone]/8192000*arcsin(1/11*22^(1/2)*(3+5*x)^(1/2)) 
*10^(1/2)
 

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.55 \[ \int \frac {(2+3 x)^3 (3+5 x)^{5/2}}{\sqrt {1-2 x}} \, dx=\frac {-10 \sqrt {1-2 x} \left (3608689671+10102628445 x+11328597700 x^2+11213711200 x^3+7993296000 x^4+3490560000 x^5+691200000 x^6\right )-8035077963 \sqrt {30+50 x} \arctan \left (\frac {\sqrt {\frac {5}{2}-5 x}}{\sqrt {3+5 x}}\right )}{24576000 \sqrt {3+5 x}} \] Input:

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

Output:

(-10*Sqrt[1 - 2*x]*(3608689671 + 10102628445*x + 11328597700*x^2 + 1121371 
1200*x^3 + 7993296000*x^4 + 3490560000*x^5 + 691200000*x^6) - 8035077963*S 
qrt[30 + 50*x]*ArcTan[Sqrt[5/2 - 5*x]/Sqrt[3 + 5*x]])/(24576000*Sqrt[3 + 5 
*x])
 

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.06, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {111, 27, 164, 60, 60, 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.

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

\(\Big \downarrow \) 111

\(\displaystyle -\frac {1}{60} \int -\frac {3 (3 x+2) (5 x+3)^{5/2} (395 x+254)}{2 \sqrt {1-2 x}}dx-\frac {1}{20} \sqrt {1-2 x} (3 x+2)^2 (5 x+3)^{7/2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{40} \int \frac {(3 x+2) (5 x+3)^{5/2} (395 x+254)}{\sqrt {1-2 x}}dx-\frac {1}{20} \sqrt {1-2 x} (3 x+2)^2 (5 x+3)^{7/2}\)

\(\Big \downarrow \) 164

\(\displaystyle \frac {1}{40} \left (\frac {2012291 \int \frac {(5 x+3)^{5/2}}{\sqrt {1-2 x}}dx}{1600}-\frac {1}{800} \sqrt {1-2 x} (5 x+3)^{7/2} (18960 x+37439)\right )-\frac {1}{20} \sqrt {1-2 x} (3 x+2)^2 (5 x+3)^{7/2}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {1}{40} \left (\frac {2012291 \left (\frac {55}{12} \int \frac {(5 x+3)^{3/2}}{\sqrt {1-2 x}}dx-\frac {1}{6} \sqrt {1-2 x} (5 x+3)^{5/2}\right )}{1600}-\frac {1}{800} \sqrt {1-2 x} (5 x+3)^{7/2} (18960 x+37439)\right )-\frac {1}{20} \sqrt {1-2 x} (3 x+2)^2 (5 x+3)^{7/2}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {1}{40} \left (\frac {2012291 \left (\frac {55}{12} \left (\frac {33}{8} \int \frac {\sqrt {5 x+3}}{\sqrt {1-2 x}}dx-\frac {1}{4} \sqrt {1-2 x} (5 x+3)^{3/2}\right )-\frac {1}{6} \sqrt {1-2 x} (5 x+3)^{5/2}\right )}{1600}-\frac {1}{800} \sqrt {1-2 x} (5 x+3)^{7/2} (18960 x+37439)\right )-\frac {1}{20} \sqrt {1-2 x} (3 x+2)^2 (5 x+3)^{7/2}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {1}{40} \left (\frac {2012291 \left (\frac {55}{12} \left (\frac {33}{8} \left (\frac {11}{4} \int \frac {1}{\sqrt {1-2 x} \sqrt {5 x+3}}dx-\frac {1}{2} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {1}{4} \sqrt {1-2 x} (5 x+3)^{3/2}\right )-\frac {1}{6} \sqrt {1-2 x} (5 x+3)^{5/2}\right )}{1600}-\frac {1}{800} \sqrt {1-2 x} (5 x+3)^{7/2} (18960 x+37439)\right )-\frac {1}{20} \sqrt {1-2 x} (3 x+2)^2 (5 x+3)^{7/2}\)

\(\Big \downarrow \) 64

\(\displaystyle \frac {1}{40} \left (\frac {2012291 \left (\frac {55}{12} \left (\frac {33}{8} \left (\frac {11}{10} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}-\frac {1}{2} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {1}{4} \sqrt {1-2 x} (5 x+3)^{3/2}\right )-\frac {1}{6} \sqrt {1-2 x} (5 x+3)^{5/2}\right )}{1600}-\frac {1}{800} \sqrt {1-2 x} (5 x+3)^{7/2} (18960 x+37439)\right )-\frac {1}{20} \sqrt {1-2 x} (3 x+2)^2 (5 x+3)^{7/2}\)

\(\Big \downarrow \) 223

\(\displaystyle \frac {1}{40} \left (\frac {2012291 \left (\frac {55}{12} \left (\frac {33}{8} \left (\frac {11 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )}{2 \sqrt {10}}-\frac {1}{2} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {1}{4} \sqrt {1-2 x} (5 x+3)^{3/2}\right )-\frac {1}{6} \sqrt {1-2 x} (5 x+3)^{5/2}\right )}{1600}-\frac {1}{800} \sqrt {1-2 x} (5 x+3)^{7/2} (18960 x+37439)\right )-\frac {1}{20} \sqrt {1-2 x} (3 x+2)^2 (5 x+3)^{7/2}\)

Input:

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

Output:

-1/20*(Sqrt[1 - 2*x]*(2 + 3*x)^2*(3 + 5*x)^(7/2)) + (-1/800*(Sqrt[1 - 2*x] 
*(3 + 5*x)^(7/2)*(37439 + 18960*x)) + (2012291*(-1/6*(Sqrt[1 - 2*x]*(3 + 5 
*x)^(5/2)) + (55*(-1/4*(Sqrt[1 - 2*x]*(3 + 5*x)^(3/2)) + (33*(-1/2*(Sqrt[1 
 - 2*x]*Sqrt[3 + 5*x]) + (11*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]])/(2*Sqrt[10] 
)))/8))/12))/1600)/40
 

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 60

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]
 

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 111

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

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 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 = 0.23 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.71

method	result	size
risch	\(\frac {\left (138240000 x^{5}[phone] x^{4}[phone] x^{3}[phone] x^{2}[phone] x [phone]\right ) \left (-1+2 x \right ) \sqrt {3+5 x}\, \sqrt {\left (1-2 x \right ) \left (3+5 x \right )}}{2457600 \sqrt {-\left (-1+2 x \right ) \left (3+5 x \right )}\, \sqrt {1-2 x}}+\frac {2678359321 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) \sqrt {\left (1-2 x \right ) \left (3+5 x \right )}}{16384000 \sqrt {1-2 x}\, \sqrt {3+5 x}}\)	\(113\)
default	\(\frac {\sqrt {3+5 x}\, \sqrt {1-2 x}\, \left (-2764800000 x^{5} \sqrt {-10 x^{2}-x +3}-12303360000 x^{4} \sqrt {-10 x^{2}-x +3}-24591168000 x^{3} \sqrt {-10 x^{2}-x +3}-30100144000 x^{2} \sqrt {-10 x^{2}-x +3}[phone] \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )-27254304400 x \sqrt {-10 x^{2}-x +3}-24057931140 \sqrt {-10 x^{2}-x +3}\right )}{49152000 \sqrt {-10 x^{2}-x +3}}\)	\(138\)

Input:

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

Output:

1/2457600*(138240000*x^5+615168000*x^4+1229558400*x^3+1505007200*x^2+13627 
15220*x[phone])*(-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)[phone]/16384000*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)

Time = 0.08 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.51 \[ \int \frac {(2+3 x)^3 (3+5 x)^{5/2}}{\sqrt {1-2 x}} \, dx=-\frac {1}{2457600} \, {\left (138240000 \, x^{5} + 615168000 \, x^{4} + 1229558400 \, x^{3} + 1505007200 \, x^{2} + 1362715220 \, x + 1202896557\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1} - \frac {2678359321}{16384000} \, \sqrt {10} \arctan \left (\frac {\sqrt {10} {\left (20 \, x + 1\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{20 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) \] Input:

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

Output:

-1/2457600*(138240000*x^5 + 615168000*x^4 + 1229558400*x^3 + 1505007200*x^ 
2 + 1362715220*x + 1202896557)*sqrt(5*x + 3)*sqrt(-2*x + 1) - 2678359321/1 
6384000*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]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.68 \[ \int \frac {(2+3 x)^3 (3+5 x)^{5/2}}{\sqrt {1-2 x}} \, dx=-\frac {225}{4} \, \sqrt {-10 \, x^{2} - x + 3} x^{5} - \frac {4005}{16} \, \sqrt {-10 \, x^{2} - x + 3} x^{4} - \frac {128079}{256} \, \sqrt {-10 \, x^{2} - x + 3} x^{3} - \frac {1881259}{3072} \, \sqrt {-10 \, x^{2} - x + 3} x^{2} - \frac {68135761}{122880} \, \sqrt {-10 \, x^{2} - x + 3} x - \frac {2678359321}{16384000} \, \sqrt {10} \arcsin \left (-\frac {20}{11} \, x - \frac {1}{11}\right ) - \frac {400965519}{819200} \, \sqrt {-10 \, x^{2} - x + 3} \] Input:

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

Output:

-225/4*sqrt(-10*x^2 - x + 3)*x^5 - 4005/16*sqrt(-10*x^2 - x + 3)*x^4 - 128 
079/256*sqrt(-10*x^2 - x + 3)*x^3 - 1881259/3072*sqrt(-10*x^2 - x + 3)*x^2 
 - 68135761/122880*sqrt(-10*x^2 - x + 3)*x - 2678359321/16384000*sqrt(10)* 
arcsin(-20/11*x - 1/11) - 400965519/819200*sqrt(-10*x^2 - x + 3)
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.51 \[ \int \frac {(2+3 x)^3 (3+5 x)^{5/2}}{\sqrt {1-2 x}} \, dx=-\frac {1}{122880000} \, \sqrt {5} {\left (2 \, {\left (4 \, {\left (8 \, {\left (108 \, {\left (16 \, {\left (20 \, x + 41\right )} {\left (5 \, x + 3\right )} + 2903\right )} {\left (5 \, x + 3\right )} + 2012291\right )} {\left (5 \, x + 3\right )} + 110676005\right )} {\left (5 \, x + 3\right )} + 3652308165\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5} - 40175389815 \, \sqrt {2} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right )\right )} \] Input:

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

Output:

-1/122880000*sqrt(5)*(2*(4*(8*(108*(16*(20*x + 41)*(5*x + 3) + 2903)*(5*x 
+ 3) + 2012291)*(5*x + 3) + 110676005)*(5*x + 3) + 3652308165)*sqrt(5*x + 
3)*sqrt(-10*x + 5) - 40175389815*sqrt(2)*arcsin(1/11*sqrt(22)*sqrt(5*x + 3 
)))
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.72 \[ \int \frac {(2+3 x)^3 (3+5 x)^{5/2}}{\sqrt {1-2 x}} \, dx=-\frac {2678359321 \sqrt {10}\, \mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )}{8192000}-\frac {225 \sqrt {5 x +3}\, \sqrt {-2 x +1}\, x^{5}}{4}-\frac {4005 \sqrt {5 x +3}\, \sqrt {-2 x +1}\, x^{4}}{16}-\frac {128079 \sqrt {5 x +3}\, \sqrt {-2 x +1}\, x^{3}}{256}-\frac {1881259 \sqrt {5 x +3}\, \sqrt {-2 x +1}\, x^{2}}{3072}-\frac {68135761 \sqrt {5 x +3}\, \sqrt {-2 x +1}\, x}{122880}-\frac {400965519 \sqrt {5 x +3}\, \sqrt {-2 x +1}}{819200} \] Input:

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

Output:

( - 8035077963*sqrt(10)*asin((sqrt( - 2*x + 1)*sqrt(5))/sqrt(11)) - 138240 
0000*sqrt(5*x + 3)*sqrt( - 2*x + 1)*x**5 - 6151680000*sqrt(5*x + 3)*sqrt( 
- 2*x + 1)*x**4 - 12295584000*sqrt(5*x + 3)*sqrt( - 2*x + 1)*x**3 - 150500 
72000*sqrt(5*x + 3)*sqrt( - 2*x + 1)*x**2 - 13627152200*sqrt(5*x + 3)*sqrt 
( - 2*x + 1)*x - 12028965570*sqrt(5*x + 3)*sqrt( - 2*x + 1))/24576000