3.26.41 \(\int \frac {(2+3 x)^2 (3+5 x)^{5/2}}{(1-2 x)^{3/2}} \, dx\) [2541]

3.26.41.1 Optimal result

Integrand size = 26, antiderivative size = 138 \[ \int \frac {(2+3 x)^2 (3+5 x)^{5/2}}{(1-2 x)^{3/2}} \, dx=\frac {838101 \sqrt {1-2 x} \sqrt {3+5 x}}{2048}+\frac {25397}{512} \sqrt {1-2 x} (3+5 x)^{3/2}+\frac {25397 \sqrt {1-2 x} (3+5 x)^{5/2}}{3520}+\frac {49 (3+5 x)^{7/2}}{22 \sqrt {1-2 x}}+\frac {9}{80} \sqrt {1-2 x} (3+5 x)^{7/2}-\frac {9219111 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )}{2048 \sqrt {10}} \]

output

-9219111/20480*arcsin(1/11*22^(1/2)*(3+5*x)^(1/2))*10^(1/2)+49/22*(3+5*x)^ 
(7/2)/(1-2*x)^(1/2)+25397/512*(3+5*x)^(3/2)*(1-2*x)^(1/2)+25397/3520*(3+5* 
x)^(5/2)*(1-2*x)^(1/2)+9/80*(3+5*x)^(7/2)*(1-2*x)^(1/2)+838101/2048*(1-2*x 
)^(1/2)*(3+5*x)^(1/2)
 

3.26.41.2 Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.57 \[ \int \frac {(2+3 x)^2 (3+5 x)^{5/2}}{(1-2 x)^{3/2}} \, dx=\frac {-10 \sqrt {3+5 x} \left (-1405233+966014 x+517096 x^2+243520 x^3+57600 x^4\right )+9219111 \sqrt {10-20 x} \arctan \left (\frac {\sqrt {\frac {5}{2}-5 x}}{\sqrt {3+5 x}}\right )}{20480 \sqrt {1-2 x}} \]

input

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

output

(-10*Sqrt[3 + 5*x]*(-1405233 + 966014*x + 517096*x^2 + 243520*x^3 + 57600* 
x^4) + 9219111*Sqrt[10 - 20*x]*ArcTan[Sqrt[5/2 - 5*x]/Sqrt[3 + 5*x]])/(204 
80*Sqrt[1 - 2*x])
 

3.26.41.3 Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.14, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {100, 27, 90, 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.

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

\(\Big \downarrow \) 100

\(\displaystyle \frac {49 (5 x+3)^{7/2}}{22 \sqrt {1-2 x}}-\frac {1}{22} \int \frac {3 (5 x+3)^{5/2} (66 x+611)}{2 \sqrt {1-2 x}}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {49 (5 x+3)^{7/2}}{22 \sqrt {1-2 x}}-\frac {3}{44} \int \frac {(5 x+3)^{5/2} (66 x+611)}{\sqrt {1-2 x}}dx\)

\(\Big \downarrow \) 90

\(\displaystyle \frac {49 (5 x+3)^{7/2}}{22 \sqrt {1-2 x}}-\frac {3}{44} \left (\frac {25397}{40} \int \frac {(5 x+3)^{5/2}}{\sqrt {1-2 x}}dx-\frac {33}{20} \sqrt {1-2 x} (5 x+3)^{7/2}\right )\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {49 (5 x+3)^{7/2}}{22 \sqrt {1-2 x}}-\frac {3}{44} \left (\frac {25397}{40} \left (\frac {55}{12} \int \frac {(5 x+3)^{3/2}}{\sqrt {1-2 x}}dx-\frac {1}{6} \sqrt {1-2 x} (5 x+3)^{5/2}\right )-\frac {33}{20} \sqrt {1-2 x} (5 x+3)^{7/2}\right )\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {49 (5 x+3)^{7/2}}{22 \sqrt {1-2 x}}-\frac {3}{44} \left (\frac {25397}{40} \left (\frac {55}{12} \left (\frac {33}{8} \int \frac {\sqrt {5 x+3}}{\sqrt {1-2 x}}dx-\frac {1}{4} \sqrt {1-2 x} (5 x+3)^{3/2}\right )-\frac {1}{6} \sqrt {1-2 x} (5 x+3)^{5/2}\right )-\frac {33}{20} \sqrt {1-2 x} (5 x+3)^{7/2}\right )\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {49 (5 x+3)^{7/2}}{22 \sqrt {1-2 x}}-\frac {3}{44} \left (\frac {25397}{40} \left (\frac {55}{12} \left (\frac {33}{8} \left (\frac {11}{4} \int \frac {1}{\sqrt {1-2 x} \sqrt {5 x+3}}dx-\frac {1}{2} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {1}{4} \sqrt {1-2 x} (5 x+3)^{3/2}\right )-\frac {1}{6} \sqrt {1-2 x} (5 x+3)^{5/2}\right )-\frac {33}{20} \sqrt {1-2 x} (5 x+3)^{7/2}\right )\)

\(\Big \downarrow \) 64

\(\displaystyle \frac {49 (5 x+3)^{7/2}}{22 \sqrt {1-2 x}}-\frac {3}{44} \left (\frac {25397}{40} \left (\frac {55}{12} \left (\frac {33}{8} \left (\frac {11}{10} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}-\frac {1}{2} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {1}{4} \sqrt {1-2 x} (5 x+3)^{3/2}\right )-\frac {1}{6} \sqrt {1-2 x} (5 x+3)^{5/2}\right )-\frac {33}{20} \sqrt {1-2 x} (5 x+3)^{7/2}\right )\)

\(\Big \downarrow \) 223

\(\displaystyle \frac {49 (5 x+3)^{7/2}}{22 \sqrt {1-2 x}}-\frac {3}{44} \left (\frac {25397}{40} \left (\frac {55}{12} \left (\frac {33}{8} \left (\frac {11 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )}{2 \sqrt {10}}-\frac {1}{2} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {1}{4} \sqrt {1-2 x} (5 x+3)^{3/2}\right )-\frac {1}{6} \sqrt {1-2 x} (5 x+3)^{5/2}\right )-\frac {33}{20} \sqrt {1-2 x} (5 x+3)^{7/2}\right )\)

input

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

output

(49*(3 + 5*x)^(7/2))/(22*Sqrt[1 - 2*x]) - (3*((-33*Sqrt[1 - 2*x]*(3 + 5*x) 
^(7/2))/20 + (25397*(-1/6*(Sqrt[1 - 2*x]*(3 + 5*x)^(5/2)) + (55*(-1/4*(Sqr 
t[1 - 2*x]*(3 + 5*x)^(3/2)) + (33*(-1/2*(Sqrt[1 - 2*x]*Sqrt[3 + 5*x]) + (1 
1*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]])/(2*Sqrt[10])))/8))/12))/40))/44
 

3.26.41.3.1 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 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 90

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), 
 x] + Simp[(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 100

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^( 
p_), x_] :> Simp[(b*c - a*d)^2*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d^2*(d 
*e - c*f)*(n + 1))), x] - Simp[1/(d^2*(d*e - c*f)*(n + 1))   Int[(c + d*x)^ 
(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*( 
p + 1)) - 2*a*b*d*(d*e*(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x 
, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ[n, -1] || (EqQ[n 
 + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))
 

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]
 

3.26.41.4 Maple [A] (verified)

Time = 1.19 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.01

method	result	size
default	\(-\frac {\left (-1152000 x^{4} \sqrt {-10 x^{2}-x +3}-4870400 x^{3} \sqrt {-10 x^{2}-x +3}[phone] \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x -10341920 x^{2} \sqrt {-10 x^{2}-x +3}-9219111 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )-19320280 x \sqrt {-10 x^{2}-x +3}[phone] \sqrt {-10 x^{2}-x +3}\right ) \sqrt {1-2 x}\, \sqrt {3+5 x}}{40960 \left (-1+2 x \right ) \sqrt {-10 x^{2}-x +3}}\)	\(140\)

input

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

output

-1/40960*(-1152000*x^4*(-10*x^2-x+3)^(1/2)-4870400*x^3*(-10*x^2-x+3)^(1/2) 
[phone]*10^(1/2)*arcsin(20/11*x+1/11)*x-10341920*x^2*(-10*x^2-x+3)^(1/2) 
-9219111*10^(1/2)*arcsin(20/11*x+1/11)-19320280*x*(-10*x^2-x+3)^(1/2)+2810 
4660*(-10*x^2-x+3)^(1/2))*(1-2*x)^(1/2)*(3+5*x)^(1/2)/(-1+2*x)/(-10*x^2-x+ 
3)^(1/2)
 

3.26.41.5 Fricas [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.66 \[ \int \frac {(2+3 x)^2 (3+5 x)^{5/2}}{(1-2 x)^{3/2}} \, dx=\frac {9219111 \, \sqrt {10} {\left (2 \, x - 1\right )} \arctan \left (\frac {\sqrt {10} {\left (20 \, x + 1\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{20 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) + 20 \, {\left (57600 \, x^{4} + 243520 \, x^{3} + 517096 \, x^{2} + 966014 \, x - 1405233\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{40960 \, {\left (2 \, x - 1\right )}} \]

input

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

output

1/40960*(9219111*sqrt(10)*(2*x - 1)*arctan(1/20*sqrt(10)*(20*x + 1)*sqrt(5 
*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x - 3)) + 20*(57600*x^4 + 243520*x^3 + 51 
7096*x^2 + 966014*x - 1405233)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(2*x - 1)
 

3.26.41.6 Sympy [F]

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

input

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

output

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

3.26.41.7 Maxima [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.79 \[ \int \frac {(2+3 x)^2 (3+5 x)^{5/2}}{(1-2 x)^{3/2}} \, dx=-\frac {1125 \, x^{5}}{8 \, \sqrt {-10 \, x^{2} - x + 3}} - \frac {21725 \, x^{4}}{32 \, \sqrt {-10 \, x^{2} - x + 3}} - \frac {414505 \, x^{3}}{256 \, \sqrt {-10 \, x^{2} - x + 3}} - \frac {3190679 \, x^{2}}{1024 \, \sqrt {-10 \, x^{2} - x + 3}} + \frac {9219111}{40960} \, \sqrt {10} \arcsin \left (-\frac {20}{11} \, x - \frac {1}{11}\right ) + \frac {4128123 \, x}{2048 \, \sqrt {-10 \, x^{2} - x + 3}} + \frac {4215699}{2048 \, \sqrt {-10 \, x^{2} - x + 3}} \]

input

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

output

-1125/8*x^5/sqrt(-10*x^2 - x + 3) - 21725/32*x^4/sqrt(-10*x^2 - x + 3) - 4 
14505/256*x^3/sqrt(-10*x^2 - x + 3) - 3190679/1024*x^2/sqrt(-10*x^2 - x + 
3) + 9219111/40960*sqrt(10)*arcsin(-20/11*x - 1/11) + 4128123/2048*x/sqrt( 
-10*x^2 - x + 3) + 4215699/2048/sqrt(-10*x^2 - x + 3)
 

3.26.41.8 Giac [A] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.70 \[ \int \frac {(2+3 x)^2 (3+5 x)^{5/2}}{(1-2 x)^{3/2}} \, dx=-\frac {9219111}{20480} \, \sqrt {10} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right ) + \frac {{\left (2 \, {\left (4 \, {\left (8 \, {\left (36 \, \sqrt {5} {\left (5 \, x + 3\right )} + 329 \, \sqrt {5}\right )} {\left (5 \, x + 3\right )} + 25397 \, \sqrt {5}\right )} {\left (5 \, x + 3\right )} + 1396835 \, \sqrt {5}\right )} {\left (5 \, x + 3\right )} - 46095555 \, \sqrt {5}\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5}}{256000 \, {\left (2 \, x - 1\right )}} \]

input

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

output

-9219111/20480*sqrt(10)*arcsin(1/11*sqrt(22)*sqrt(5*x + 3)) + 1/256000*(2* 
(4*(8*(36*sqrt(5)*(5*x + 3) + 329*sqrt(5))*(5*x + 3) + 25397*sqrt(5))*(5*x 
 + 3) + 1396835*sqrt(5))*(5*x + 3) - 46095555*sqrt(5))*sqrt(5*x + 3)*sqrt( 
-10*x + 5)/(2*x - 1)
 

3.26.41.9 Mupad [F(-1)]

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

input

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

output

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