3.25.58 \(\int \frac {(2+3 x)^4 \sqrt {3+5 x}}{\sqrt {1-2 x}} \, dx\) [2458]

3.25.58.1 Optimal result
3.25.58.2 Mathematica [A] (verified)
3.25.58.3 Rubi [A] (verified)
3.25.58.4 Maple [A] (verified)
3.25.58.5 Fricas [A] (verification not implemented)
3.25.58.6 Sympy [F]
3.25.58.7 Maxima [A] (verification not implemented)
3.25.58.8 Giac [A] (verification not implemented)
3.25.58.9 Mupad [B] (verification not implemented)

3.25.58.1 Optimal result

Integrand size = 26, antiderivative size = 135 \[ \int \frac {(2+3 x)^4 \sqrt {3+5 x}}{\sqrt {1-2 x}} \, dx=-\frac {97032047 \sqrt {1-2 x} \sqrt {3+5 x}}{2560000}-\frac {987 \sqrt {1-2 x} (2+3 x)^2 (3+5 x)^{3/2}}{4000}-\frac {3}{50} \sqrt {1-2 x} (2+3 x)^3 (3+5 x)^{3/2}-\frac {21 \sqrt {1-2 x} (3+5 x)^{3/2} (194923+92040 x)}{640000}+\frac {1067352517 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )}{2560000 \sqrt {10}} \]

output
1067352517/25600000*arcsin(1/11*22^(1/2)*(3+5*x)^(1/2))*10^(1/2)-987/4000* 
(2+3*x)^2*(3+5*x)^(3/2)*(1-2*x)^(1/2)-3/50*(2+3*x)^3*(3+5*x)^(3/2)*(1-2*x) 
^(1/2)-21/640000*(3+5*x)^(3/2)*(194923+92040*x)*(1-2*x)^(1/2)-97032047/256 
0000*(1-2*x)^(1/2)*(3+5*x)^(1/2)
 
3.25.58.2 Mathematica [A] (verified)

Time = 0.47 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.64 \[ \int \frac {(2+3 x)^4 \sqrt {3+5 x}}{\sqrt {1-2 x}} \, dx=\frac {-\sqrt {5-10 x} \sqrt {3+5 x} \left (157419203+163168620 x+146144160 x^2+82339200 x^3+20736000 x^4\right )-1067352517 \sqrt {2} \arctan \left (\frac {\sqrt {6+10 x}}{\sqrt {11}-\sqrt {5-10 x}}\right )}{2560000 \sqrt {5}} \]

input
Integrate[((2 + 3*x)^4*Sqrt[3 + 5*x])/Sqrt[1 - 2*x],x]
 
output
(-(Sqrt[5 - 10*x]*Sqrt[3 + 5*x]*(157419203 + 163168620*x + 146144160*x^2 + 
 82339200*x^3 + 20736000*x^4)) - 1067352517*Sqrt[2]*ArcTan[Sqrt[6 + 10*x]/ 
(Sqrt[11] - Sqrt[5 - 10*x])])/(2560000*Sqrt[5])
 
3.25.58.3 Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.11, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {111, 27, 170, 27, 164, 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.

\(\displaystyle \int \frac {(3 x+2)^4 \sqrt {5 x+3}}{\sqrt {1-2 x}} \, dx\)

\(\Big \downarrow \) 111

\(\displaystyle -\frac {1}{50} \int -\frac {7 (3 x+2)^2 \sqrt {5 x+3} (141 x+88)}{2 \sqrt {1-2 x}}dx-\frac {3}{50} \sqrt {1-2 x} (5 x+3)^{3/2} (3 x+2)^3\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {7}{100} \int \frac {(3 x+2)^2 \sqrt {5 x+3} (141 x+88)}{\sqrt {1-2 x}}dx-\frac {3}{50} \sqrt {1-2 x} (3 x+2)^3 (5 x+3)^{3/2}\)

\(\Big \downarrow \) 170

\(\displaystyle \frac {7}{100} \left (-\frac {1}{40} \int -\frac {(3 x+2) \sqrt {5 x+3} (34515 x+21694)}{2 \sqrt {1-2 x}}dx-\frac {141}{40} \sqrt {1-2 x} (5 x+3)^{3/2} (3 x+2)^2\right )-\frac {3}{50} \sqrt {1-2 x} (3 x+2)^3 (5 x+3)^{3/2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {7}{100} \left (\frac {1}{80} \int \frac {(3 x+2) \sqrt {5 x+3} (34515 x+21694)}{\sqrt {1-2 x}}dx-\frac {141}{40} \sqrt {1-2 x} (3 x+2)^2 (5 x+3)^{3/2}\right )-\frac {3}{50} \sqrt {1-2 x} (3 x+2)^3 (5 x+3)^{3/2}\)

\(\Big \downarrow \) 164

\(\displaystyle \frac {7}{100} \left (\frac {1}{80} \left (\frac {13861721}{160} \int \frac {\sqrt {5 x+3}}{\sqrt {1-2 x}}dx-\frac {3}{80} \sqrt {1-2 x} (5 x+3)^{3/2} (92040 x+194923)\right )-\frac {141}{40} \sqrt {1-2 x} (3 x+2)^2 (5 x+3)^{3/2}\right )-\frac {3}{50} \sqrt {1-2 x} (3 x+2)^3 (5 x+3)^{3/2}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {7}{100} \left (\frac {1}{80} \left (\frac {13861721}{160} \left (\frac {11}{4} \int \frac {1}{\sqrt {1-2 x} \sqrt {5 x+3}}dx-\frac {1}{2} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {3}{80} \sqrt {1-2 x} (5 x+3)^{3/2} (92040 x+194923)\right )-\frac {141}{40} \sqrt {1-2 x} (3 x+2)^2 (5 x+3)^{3/2}\right )-\frac {3}{50} \sqrt {1-2 x} (3 x+2)^3 (5 x+3)^{3/2}\)

\(\Big \downarrow \) 64

\(\displaystyle \frac {7}{100} \left (\frac {1}{80} \left (\frac {13861721}{160} \left (\frac {11}{10} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}-\frac {1}{2} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {3}{80} \sqrt {1-2 x} (5 x+3)^{3/2} (92040 x+194923)\right )-\frac {141}{40} \sqrt {1-2 x} (3 x+2)^2 (5 x+3)^{3/2}\right )-\frac {3}{50} \sqrt {1-2 x} (3 x+2)^3 (5 x+3)^{3/2}\)

\(\Big \downarrow \) 223

\(\displaystyle \frac {7}{100} \left (\frac {1}{80} \left (\frac {13861721}{160} \left (\frac {11 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )}{2 \sqrt {10}}-\frac {1}{2} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {3}{80} \sqrt {1-2 x} (5 x+3)^{3/2} (92040 x+194923)\right )-\frac {141}{40} \sqrt {1-2 x} (3 x+2)^2 (5 x+3)^{3/2}\right )-\frac {3}{50} \sqrt {1-2 x} (3 x+2)^3 (5 x+3)^{3/2}\)

input
Int[((2 + 3*x)^4*Sqrt[3 + 5*x])/Sqrt[1 - 2*x],x]
 
output
(-3*Sqrt[1 - 2*x]*(2 + 3*x)^3*(3 + 5*x)^(3/2))/50 + (7*((-141*Sqrt[1 - 2*x 
]*(2 + 3*x)^2*(3 + 5*x)^(3/2))/40 + ((-3*Sqrt[1 - 2*x]*(3 + 5*x)^(3/2)*(19 
4923 + 92040*x))/80 + (13861721*(-1/2*(Sqrt[1 - 2*x]*Sqrt[3 + 5*x]) + (11* 
ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]])/(2*Sqrt[10])))/160)/80))/100
 

3.25.58.3.1 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 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 170
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_)*((g_.) + (h_.)*(x_)), x_] :> Simp[h*(a + b*x)^m*(c + d*x)^(n + 1)*(( 
e + f*x)^(p + 1)/(d*f*(m + n + p + 2))), x] + Simp[1/(d*f*(m + n + p + 2)) 
  Int[(a + b*x)^(m - 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*g*(m + n + p + 2 
) - h*(b*c*e*m + a*(d*e*(n + 1) + c*f*(p + 1))) + (b*d*f*g*(m + n + p + 2) 
+ h*(a*d*f*m - b*(d*e*(m + n + 1) + c*f*(m + p + 1))))*x, x], x], x] /; Fre 
eQ[{a, b, c, d, e, f, g, h, n, p}, x] && GtQ[m, 0] && NeQ[m + n + p + 2, 0] 
 && IntegerQ[m]
 

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]
 
3.25.58.4 Maple [A] (verified)

Time = 1.18 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.80

method result size
risch \(\frac {\left (20736000 x^{4}+82339200 x^{3}+146144160 x^{2}+163168620 x +157419203\right ) \left (-1+2 x \right ) \sqrt {3+5 x}\, \sqrt {\left (1-2 x \right ) \left (3+5 x \right )}}{2560000 \sqrt {-\left (-1+2 x \right ) \left (3+5 x \right )}\, \sqrt {1-2 x}}+\frac {1067352517 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) \sqrt {\left (1-2 x \right ) \left (3+5 x \right )}}{51200000 \sqrt {1-2 x}\, \sqrt {3+5 x}}\) \(108\)
default \(\frac {\sqrt {3+5 x}\, \sqrt {1-2 x}\, \left (-414720000 x^{4} \sqrt {-10 x^{2}-x +3}-1646784000 x^{3} \sqrt {-10 x^{2}-x +3}-2922883200 x^{2} \sqrt {-10 x^{2}-x +3}+1067352517 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )-3263372400 x \sqrt {-10 x^{2}-x +3}-3148384060 \sqrt {-10 x^{2}-x +3}\right )}{51200000 \sqrt {-10 x^{2}-x +3}}\) \(121\)

input
int((2+3*x)^4*(3+5*x)^(1/2)/(1-2*x)^(1/2),x,method=_RETURNVERBOSE)
 
output
1/2560000*(20736000*x^4+82339200*x^3+146144160*x^2+163168620*x+157419203)* 
(-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)+1067352517/51200000*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)
 
3.25.58.5 Fricas [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.57 \[ \int \frac {(2+3 x)^4 \sqrt {3+5 x}}{\sqrt {1-2 x}} \, dx=-\frac {1}{2560000} \, {\left (20736000 \, x^{4} + 82339200 \, x^{3} + 146144160 \, x^{2} + 163168620 \, x + 157419203\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1} - \frac {1067352517}{51200000} \, \sqrt {10} \arctan \left (\frac {\sqrt {10} {\left (20 \, x + 1\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{20 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) \]

input
integrate((2+3*x)^4*(3+5*x)^(1/2)/(1-2*x)^(1/2),x, algorithm="fricas")
 
output
-1/2560000*(20736000*x^4 + 82339200*x^3 + 146144160*x^2 + 163168620*x + 15 
7419203)*sqrt(5*x + 3)*sqrt(-2*x + 1) - 1067352517/51200000*sqrt(10)*arcta 
n(1/20*sqrt(10)*(20*x + 1)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x - 3))
 
3.25.58.6 Sympy [F]

\[ \int \frac {(2+3 x)^4 \sqrt {3+5 x}}{\sqrt {1-2 x}} \, dx=\int \frac {\left (3 x + 2\right )^{4} \sqrt {5 x + 3}}{\sqrt {1 - 2 x}}\, dx \]

input
integrate((2+3*x)**4*(3+5*x)**(1/2)/(1-2*x)**(1/2),x)
 
output
Integral((3*x + 2)**4*sqrt(5*x + 3)/sqrt(1 - 2*x), x)
 
3.25.58.7 Maxima [A] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.67 \[ \int \frac {(2+3 x)^4 \sqrt {3+5 x}}{\sqrt {1-2 x}} \, dx=\frac {81}{100} \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}} x^{2} + \frac {25083}{8000} \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}} x + \frac {1067352517}{51200000} \, \sqrt {5} \sqrt {2} \arcsin \left (\frac {20}{11} \, x + \frac {1}{11}\right ) + \frac {180423}{32000} \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}} - \frac {8640723}{128000} \, \sqrt {-10 \, x^{2} - x + 3} x - \frac {200720723}{2560000} \, \sqrt {-10 \, x^{2} - x + 3} \]

input
integrate((2+3*x)^4*(3+5*x)^(1/2)/(1-2*x)^(1/2),x, algorithm="maxima")
 
output
81/100*(-10*x^2 - x + 3)^(3/2)*x^2 + 25083/8000*(-10*x^2 - x + 3)^(3/2)*x 
+ 1067352517/51200000*sqrt(5)*sqrt(2)*arcsin(20/11*x + 1/11) + 180423/3200 
0*(-10*x^2 - x + 3)^(3/2) - 8640723/128000*sqrt(-10*x^2 - x + 3)*x - 20072 
0723/2560000*sqrt(-10*x^2 - x + 3)
 
3.25.58.8 Giac [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.53 \[ \int \frac {(2+3 x)^4 \sqrt {3+5 x}}{\sqrt {1-2 x}} \, dx=-\frac {1}{128000000} \, \sqrt {5} {\left (2 \, {\left (12 \, {\left (24 \, {\left (12 \, {\left (240 \, x + 521\right )} {\left (5 \, x + 3\right )} + 29669\right )} {\left (5 \, x + 3\right )} + 4900505\right )} {\left (5 \, x + 3\right )} + 485160235\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5} - 5336762585 \, \sqrt {2} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right )\right )} \]

input
integrate((2+3*x)^4*(3+5*x)^(1/2)/(1-2*x)^(1/2),x, algorithm="giac")
 
output
-1/128000000*sqrt(5)*(2*(12*(24*(12*(240*x + 521)*(5*x + 3) + 29669)*(5*x 
+ 3) + 4900505)*(5*x + 3) + 485160235)*sqrt(5*x + 3)*sqrt(-10*x + 5) - 533 
6762585*sqrt(2)*arcsin(1/11*sqrt(22)*sqrt(5*x + 3)))
 
3.25.58.9 Mupad [B] (verification not implemented)

Time = 15.39 (sec) , antiderivative size = 882, normalized size of antiderivative = 6.53 \[ \int \frac {(2+3 x)^4 \sqrt {3+5 x}}{\sqrt {1-2 x}} \, dx=\text {Too large to display} \]

input
int(((3*x + 2)^4*(5*x + 3)^(1/2))/(1 - 2*x)^(1/2),x)
 
output
(1067352517*10^(1/2)*atan((10^(1/2)*((1 - 2*x)^(1/2) - 1))/(2*(3^(1/2) - ( 
5*x + 3)^(1/2)))))/12800000 - ((821592517*((1 - 2*x)^(1/2) - 1))/(24414062 
500*(3^(1/2) - (5*x + 3)^(1/2))) - (1047293669*((1 - 2*x)^(1/2) - 1)^3)/(9 
765625000*(3^(1/2) - (5*x + 3)^(1/2))^3) - (47930155877*((1 - 2*x)^(1/2) - 
 1)^5)/(4882812500*(3^(1/2) - (5*x + 3)^(1/2))^5) - (3319241183*((1 - 2*x) 
^(1/2) - 1)^7)/(390625000*(3^(1/2) - (5*x + 3)^(1/2))^7) - (7192921169*((1 
 - 2*x)^(1/2) - 1)^9)/(312500000*(3^(1/2) - (5*x + 3)^(1/2))^9) + (7192921 
169*((1 - 2*x)^(1/2) - 1)^11)/(125000000*(3^(1/2) - (5*x + 3)^(1/2))^11) + 
 (3319241183*((1 - 2*x)^(1/2) - 1)^13)/(25000000*(3^(1/2) - (5*x + 3)^(1/2 
))^13) + (47930155877*((1 - 2*x)^(1/2) - 1)^15)/(50000000*(3^(1/2) - (5*x 
+ 3)^(1/2))^15) + (1047293669*((1 - 2*x)^(1/2) - 1)^17)/(16000000*(3^(1/2) 
 - (5*x + 3)^(1/2))^17) - (821592517*((1 - 2*x)^(1/2) - 1)^19)/(6400000*(3 
^(1/2) - (5*x + 3)^(1/2))^19) + (753664*3^(1/2)*((1 - 2*x)^(1/2) - 1)^2)/( 
9765625*(3^(1/2) - (5*x + 3)^(1/2))^2) + (1359872*3^(1/2)*((1 - 2*x)^(1/2) 
 - 1)^4)/(390625*(3^(1/2) - (5*x + 3)^(1/2))^4) + (54837248*3^(1/2)*((1 - 
2*x)^(1/2) - 1)^6)/(1953125*(3^(1/2) - (5*x + 3)^(1/2))^6) + (768075776*3^ 
(1/2)*((1 - 2*x)^(1/2) - 1)^8)/(9765625*(3^(1/2) - (5*x + 3)^(1/2))^8) + ( 
721424384*3^(1/2)*((1 - 2*x)^(1/2) - 1)^10)/(1953125*(3^(1/2) - (5*x + 3)^ 
(1/2))^10) + (192018944*3^(1/2)*((1 - 2*x)^(1/2) - 1)^12)/(390625*(3^(1/2) 
 - (5*x + 3)^(1/2))^12) + (3427328*3^(1/2)*((1 - 2*x)^(1/2) - 1)^14)/(3...