\(\int \frac {(2+3 x)^4}{(1-2 x)^{3/2} \sqrt {3+5 x}} \, dx\) [1155]

Optimal result

Integrand size = 26, antiderivative size = 116 \[ \int \frac {(2+3 x)^4}{(1-2 x)^{3/2} \sqrt {3+5 x}} \, dx=\frac {2401 \sqrt {3+5 x}}{88 \sqrt {1-2 x}}+\frac {13149}{640} \sqrt {1-2 x} \sqrt {3+5 x}-\frac {243}{64} (1-2 x)^{3/2} \sqrt {3+5 x}+\frac {27}{80} (1-2 x)^{5/2} \sqrt {3+5 x}-\frac {184641 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )}{640 \sqrt {10}} \] Output:

2401/88*(3+5*x)^(1/2)/(1-2*x)^(1/2)+13149/640*(1-2*x)^(1/2)*(3+5*x)^(1/2)- 
243/64*(1-2*x)^(3/2)*(3+5*x)^(1/2)+27/80*(1-2*x)^(5/2)*(3+5*x)^(1/2)-18464 
1/6400*arcsin(1/11*22^(1/2)*(3+5*x)^(1/2))*10^(1/2)
 

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.63 \[ \int \frac {(2+3 x)^4}{(1-2 x)^{3/2} \sqrt {3+5 x}} \, dx=\frac {-10 \sqrt {3+5 x} \left (-312365+196614 x+78408 x^2+19008 x^3\right )+2031051 \sqrt {10-20 x} \arctan \left (\frac {\sqrt {\frac {5}{2}-5 x}}{\sqrt {3+5 x}}\right )}{70400 \sqrt {1-2 x}} \] Input:

Integrate[(2 + 3*x)^4/((1 - 2*x)^(3/2)*Sqrt[3 + 5*x]),x]
 

Output:

(-10*Sqrt[3 + 5*x]*(-312365 + 196614*x + 78408*x^2 + 19008*x^3) + 2031051* 
Sqrt[10 - 20*x]*ArcTan[Sqrt[5/2 - 5*x]/Sqrt[3 + 5*x]])/(70400*Sqrt[1 - 2*x 
])
 

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 = {109, 27, 170, 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.

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

\(\Big \downarrow \) 109

\(\displaystyle \frac {7 (3 x+2)^3 \sqrt {5 x+3}}{11 \sqrt {1-2 x}}-\frac {1}{11} \int \frac {3 (3 x+2)^2 (243 x+148)}{2 \sqrt {1-2 x} \sqrt {5 x+3}}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {7 (3 x+2)^3 \sqrt {5 x+3}}{11 \sqrt {1-2 x}}-\frac {3}{22} \int \frac {(3 x+2)^2 (243 x+148)}{\sqrt {1-2 x} \sqrt {5 x+3}}dx\)

\(\Big \downarrow \) 170

\(\displaystyle \frac {7 (3 x+2)^3 \sqrt {5 x+3}}{11 \sqrt {1-2 x}}-\frac {3}{22} \left (-\frac {1}{30} \int -\frac {3 (3 x+2) (14145 x+8674)}{2 \sqrt {1-2 x} \sqrt {5 x+3}}dx-\frac {81}{10} \sqrt {1-2 x} \sqrt {5 x+3} (3 x+2)^2\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {7 (3 x+2)^3 \sqrt {5 x+3}}{11 \sqrt {1-2 x}}-\frac {3}{22} \left (\frac {1}{20} \int \frac {(3 x+2) (14145 x+8674)}{\sqrt {1-2 x} \sqrt {5 x+3}}dx-\frac {81}{10} \sqrt {1-2 x} (3 x+2)^2 \sqrt {5 x+3}\right )\)

\(\Big \downarrow \) 164

\(\displaystyle \frac {7 (3 x+2)^3 \sqrt {5 x+3}}{11 \sqrt {1-2 x}}-\frac {3}{22} \left (\frac {1}{20} \left (\frac {677017}{32} \int \frac {1}{\sqrt {1-2 x} \sqrt {5 x+3}}dx-\frac {3}{16} \sqrt {1-2 x} \sqrt {5 x+3} (11316 x+27269)\right )-\frac {81}{10} \sqrt {1-2 x} (3 x+2)^2 \sqrt {5 x+3}\right )\)

\(\Big \downarrow \) 64

\(\displaystyle \frac {7 (3 x+2)^3 \sqrt {5 x+3}}{11 \sqrt {1-2 x}}-\frac {3}{22} \left (\frac {1}{20} \left (\frac {677017}{80} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}-\frac {3}{16} \sqrt {1-2 x} \sqrt {5 x+3} (11316 x+27269)\right )-\frac {81}{10} \sqrt {1-2 x} (3 x+2)^2 \sqrt {5 x+3}\right )\)

\(\Big \downarrow \) 223

\(\displaystyle \frac {7 (3 x+2)^3 \sqrt {5 x+3}}{11 \sqrt {1-2 x}}-\frac {3}{22} \left (\frac {1}{20} \left (\frac {677017 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )}{16 \sqrt {10}}-\frac {3}{16} \sqrt {1-2 x} \sqrt {5 x+3} (11316 x+27269)\right )-\frac {81}{10} \sqrt {1-2 x} (3 x+2)^2 \sqrt {5 x+3}\right )\)

Input:

Int[(2 + 3*x)^4/((1 - 2*x)^(3/2)*Sqrt[3 + 5*x]),x]
 

Output:

(7*(2 + 3*x)^3*Sqrt[3 + 5*x])/(11*Sqrt[1 - 2*x]) - (3*((-81*Sqrt[1 - 2*x]* 
(2 + 3*x)^2*Sqrt[3 + 5*x])/10 + ((-3*Sqrt[1 - 2*x]*Sqrt[3 + 5*x]*(27269 + 
11316*x))/16 + (677017*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]])/(16*Sqrt[10]))/20 
))/22
 

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 109

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

Maple [A] (verified)

Time = 0.24 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.00

method	result	size
default	\(\frac {\left (-380160 x^{3} \sqrt {-10 x^{2}-x +3}+4062102 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x -1568160 x^{2} \sqrt {-10 x^{2}-x +3}-2031051 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )-3932280 x \sqrt {-10 x^{2}-x +3}+6247300 \sqrt {-10 x^{2}-x +3}\right ) \sqrt {3+5 x}}{140800 \sqrt {1-2 x}\, \sqrt {-10 x^{2}-x +3}}\)	\(116\)

Input:

int((2+3*x)^4/(1-2*x)^(3/2)/(3+5*x)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

1/140800*(-380160*x^3*(-10*x^2-x+3)^(1/2)+4062102*10^(1/2)*arcsin(20/11*x+ 
1/11)*x-1568160*x^2*(-10*x^2-x+3)^(1/2)-2031051*10^(1/2)*arcsin(20/11*x+1/ 
11)-3932280*x*(-10*x^2-x+3)^(1/2)+6247300*(-10*x^2-x+3)^(1/2))*(3+5*x)^(1/ 
2)/(1-2*x)^(1/2)/(-10*x^2-x+3)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.74 \[ \int \frac {(2+3 x)^4}{(1-2 x)^{3/2} \sqrt {3+5 x}} \, dx=\frac {2031051 \, \sqrt {10} {\left (2 \, x - 1\right )} \arctan \left (\frac {\sqrt {10} {\left (20 \, x + 1\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{20 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) + 20 \, {\left (19008 \, x^{3} + 78408 \, x^{2} + 196614 \, x - 312365\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{140800 \, {\left (2 \, x - 1\right )}} \] Input:

integrate((2+3*x)^4/(1-2*x)^(3/2)/(3+5*x)^(1/2),x, algorithm="fricas")
                                                                                    
                                                                                    
 

Output:

1/140800*(2031051*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*(19008*x^3 + 78408*x^2 + 19 
6614*x - 312365)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(2*x - 1)
 

Sympy [F]

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

integrate((2+3*x)**4/(1-2*x)**(3/2)/(3+5*x)**(1/2),x)
 

Output:

Integral((3*x + 2)**4/((1 - 2*x)**(3/2)*sqrt(5*x + 3)), x)
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.71 \[ \int \frac {(2+3 x)^4}{(1-2 x)^{3/2} \sqrt {3+5 x}} \, dx=\frac {27}{20} \, \sqrt {-10 \, x^{2} - x + 3} x^{2} - \frac {184641}{12800} \, \sqrt {5} \sqrt {2} \arcsin \left (\frac {20}{11} \, x + \frac {1}{11}\right ) + \frac {999}{160} \, \sqrt {-10 \, x^{2} - x + 3} x + \frac {2187}{128} \, \sqrt {-10 \, x^{2} - x + 3} - \frac {2401 \, \sqrt {-10 \, x^{2} - x + 3}}{88 \, {\left (2 \, x - 1\right )}} \] Input:

integrate((2+3*x)^4/(1-2*x)^(3/2)/(3+5*x)^(1/2),x, algorithm="maxima")
 

Output:

27/20*sqrt(-10*x^2 - x + 3)*x^2 - 184641/12800*sqrt(5)*sqrt(2)*arcsin(20/1 
1*x + 1/11) + 999/160*sqrt(-10*x^2 - x + 3)*x + 2187/128*sqrt(-10*x^2 - x 
+ 3) - 2401/88*sqrt(-10*x^2 - x + 3)/(2*x - 1)
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.72 \[ \int \frac {(2+3 x)^4}{(1-2 x)^{3/2} \sqrt {3+5 x}} \, dx=-\frac {184641}{6400} \, \sqrt {10} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right ) + \frac {{\left (594 \, {\left (4 \, {\left (8 \, \sqrt {5} {\left (5 \, x + 3\right )} + 93 \, \sqrt {5}\right )} {\left (5 \, x + 3\right )} + 5179 \, \sqrt {5}\right )} {\left (5 \, x + 3\right )} - 50776531 \, \sqrt {5}\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5}}{4400000 \, {\left (2 \, x - 1\right )}} \] Input:

integrate((2+3*x)^4/(1-2*x)^(3/2)/(3+5*x)^(1/2),x, algorithm="giac")
 

Output:

-184641/6400*sqrt(10)*arcsin(1/11*sqrt(22)*sqrt(5*x + 3)) + 1/4400000*(594 
*(4*(8*sqrt(5)*(5*x + 3) + 93*sqrt(5))*(5*x + 3) + 5179*sqrt(5))*(5*x + 3) 
 - 50776531*sqrt(5))*sqrt(5*x + 3)*sqrt(-10*x + 5)/(2*x - 1)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(2+3 x)^4}{(1-2 x)^{3/2} \sqrt {3+5 x}} \, dx=\int \frac {{\left (3\,x+2\right )}^4}{{\left (1-2\,x\right )}^{3/2}\,\sqrt {5\,x+3}} \,d x \] Input:

int((3*x + 2)^4/((1 - 2*x)^(3/2)*(5*x + 3)^(1/2)),x)
 

Output:

int((3*x + 2)^4/((1 - 2*x)^(3/2)*(5*x + 3)^(1/2)), x)
 

Reduce [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.64 \[ \int \frac {(2+3 x)^4}{(1-2 x)^{3/2} \sqrt {3+5 x}} \, dx=\frac {2031051 \sqrt {-2 x +1}\, \sqrt {10}\, \mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )-190080 \sqrt {5 x +3}\, x^{3}-784080 \sqrt {5 x +3}\, x^{2}-1966140 \sqrt {5 x +3}\, x +3123650 \sqrt {5 x +3}}{70400 \sqrt {-2 x +1}} \] Input:

int((2+3*x)^4/(1-2*x)^(3/2)/(3+5*x)^(1/2),x)
 

Output:

(2031051*sqrt( - 2*x + 1)*sqrt(10)*asin((sqrt( - 2*x + 1)*sqrt(5))/sqrt(11 
)) - 190080*sqrt(5*x + 3)*x**3 - 784080*sqrt(5*x + 3)*x**2 - 1966140*sqrt( 
5*x + 3)*x + 3123650*sqrt(5*x + 3))/(70400*sqrt( - 2*x + 1))