\(\int \frac {1}{(1-2 x)^{3/2} (2+3 x)^3 (3+5 x)^{3/2}} \, dx\) [1173]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [B] (verified)
Fricas [A] (verification not implemented)
Sympy [F]
Maxima [A] (verification not implemented)
Giac [B] (verification not implemented)
Mupad [F(-1)]
Reduce [B] (verification not implemented)

Optimal result

Integrand size = 26, antiderivative size = 137 \[ \int \frac {1}{(1-2 x)^{3/2} (2+3 x)^3 (3+5 x)^{3/2}} \, dx=-\frac {6205}{7546 \sqrt {1-2 x} \sqrt {3+5 x}}-\frac {3125575 \sqrt {1-2 x}}{166012 \sqrt {3+5 x}}+\frac {3}{14 \sqrt {1-2 x} (2+3 x)^2 \sqrt {3+5 x}}+\frac {555}{196 \sqrt {1-2 x} (2+3 x) \sqrt {3+5 x}}+\frac {177255 \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right )}{1372 \sqrt {7}} \] Output:

-6205/7546/(1-2*x)^(1/2)/(3+5*x)^(1/2)-3125575/166012*(1-2*x)^(1/2)/(3+5*x 
)^(1/2)+3/14/(1-2*x)^(1/2)/(2+3*x)^2/(3+5*x)^(1/2)+555/196/(1-2*x)^(1/2)/( 
2+3*x)/(3+5*x)^(1/2)+177255/9604*7^(1/2)*arctan(1/7*(1-2*x)^(1/2)*7^(1/2)/ 
(3+5*x)^(1/2))
 

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.58 \[ \int \frac {1}{(1-2 x)^{3/2} (2+3 x)^3 (3+5 x)^{3/2}} \, dx=\frac {\frac {7 \left (-12072596-12730165 x+45655035 x^2+56260350 x^3\right )}{\sqrt {1-2 x} (2+3 x)^2 \sqrt {3+5 x}}+21447855 \sqrt {7} \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right )}{1162084} \] Input:

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

Output:

((7*(-12072596 - 12730165*x + 45655035*x^2 + 56260350*x^3))/(Sqrt[1 - 2*x] 
*(2 + 3*x)^2*Sqrt[3 + 5*x]) + 21447855*Sqrt[7]*ArcTan[Sqrt[1 - 2*x]/(Sqrt[ 
7]*Sqrt[3 + 5*x])])/1162084
 

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.09, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {114, 27, 168, 27, 169, 27, 169, 27, 104, 217}

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.

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

\(\Big \downarrow \) 114

\(\displaystyle \frac {1}{14} \int \frac {5 (13-36 x)}{2 (1-2 x)^{3/2} (3 x+2)^2 (5 x+3)^{3/2}}dx+\frac {3}{14 \sqrt {1-2 x} (3 x+2)^2 \sqrt {5 x+3}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {5}{28} \int \frac {13-36 x}{(1-2 x)^{3/2} (3 x+2)^2 (5 x+3)^{3/2}}dx+\frac {3}{14 \sqrt {1-2 x} (3 x+2)^2 \sqrt {5 x+3}}\)

\(\Big \downarrow \) 168

\(\displaystyle \frac {5}{28} \left (\frac {1}{7} \int \frac {979-4440 x}{2 (1-2 x)^{3/2} (3 x+2) (5 x+3)^{3/2}}dx+\frac {111}{7 \sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}\right )+\frac {3}{14 \sqrt {1-2 x} (3 x+2)^2 \sqrt {5 x+3}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {5}{28} \left (\frac {1}{14} \int \frac {979-4440 x}{(1-2 x)^{3/2} (3 x+2) (5 x+3)^{3/2}}dx+\frac {111}{7 \sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}\right )+\frac {3}{14 \sqrt {1-2 x} (3 x+2)^2 \sqrt {5 x+3}}\)

\(\Big \downarrow \) 169

\(\displaystyle \frac {5}{28} \left (\frac {1}{14} \left (-\frac {2}{77} \int -\frac {80347-74460 x}{2 \sqrt {1-2 x} (3 x+2) (5 x+3)^{3/2}}dx-\frac {4964}{77 \sqrt {1-2 x} \sqrt {5 x+3}}\right )+\frac {111}{7 \sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}\right )+\frac {3}{14 \sqrt {1-2 x} (3 x+2)^2 \sqrt {5 x+3}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {5}{28} \left (\frac {1}{14} \left (\frac {1}{77} \int \frac {80347-74460 x}{\sqrt {1-2 x} (3 x+2) (5 x+3)^{3/2}}dx-\frac {4964}{77 \sqrt {1-2 x} \sqrt {5 x+3}}\right )+\frac {111}{7 \sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}\right )+\frac {3}{14 \sqrt {1-2 x} (3 x+2)^2 \sqrt {5 x+3}}\)

\(\Big \downarrow \) 169

\(\displaystyle \frac {5}{28} \left (\frac {1}{14} \left (\frac {1}{77} \left (-\frac {2}{11} \int \frac {4289571}{2 \sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}dx-\frac {1250230 \sqrt {1-2 x}}{11 \sqrt {5 x+3}}\right )-\frac {4964}{77 \sqrt {1-2 x} \sqrt {5 x+3}}\right )+\frac {111}{7 \sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}\right )+\frac {3}{14 \sqrt {1-2 x} (3 x+2)^2 \sqrt {5 x+3}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {5}{28} \left (\frac {1}{14} \left (\frac {1}{77} \left (-389961 \int \frac {1}{\sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}dx-\frac {1250230 \sqrt {1-2 x}}{11 \sqrt {5 x+3}}\right )-\frac {4964}{77 \sqrt {1-2 x} \sqrt {5 x+3}}\right )+\frac {111}{7 \sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}\right )+\frac {3}{14 \sqrt {1-2 x} (3 x+2)^2 \sqrt {5 x+3}}\)

\(\Big \downarrow \) 104

\(\displaystyle \frac {5}{28} \left (\frac {1}{14} \left (\frac {1}{77} \left (-779922 \int \frac {1}{-\frac {1-2 x}{5 x+3}-7}d\frac {\sqrt {1-2 x}}{\sqrt {5 x+3}}-\frac {1250230 \sqrt {1-2 x}}{11 \sqrt {5 x+3}}\right )-\frac {4964}{77 \sqrt {1-2 x} \sqrt {5 x+3}}\right )+\frac {111}{7 \sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}\right )+\frac {3}{14 \sqrt {1-2 x} (3 x+2)^2 \sqrt {5 x+3}}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {5}{28} \left (\frac {1}{14} \left (\frac {1}{77} \left (\frac {779922 \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{\sqrt {7}}-\frac {1250230 \sqrt {1-2 x}}{11 \sqrt {5 x+3}}\right )-\frac {4964}{77 \sqrt {1-2 x} \sqrt {5 x+3}}\right )+\frac {111}{7 \sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}\right )+\frac {3}{14 \sqrt {1-2 x} (3 x+2)^2 \sqrt {5 x+3}}\)

Input:

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

Output:

3/(14*Sqrt[1 - 2*x]*(2 + 3*x)^2*Sqrt[3 + 5*x]) + (5*(111/(7*Sqrt[1 - 2*x]* 
(2 + 3*x)*Sqrt[3 + 5*x]) + (-4964/(77*Sqrt[1 - 2*x]*Sqrt[3 + 5*x]) + ((-12 
50230*Sqrt[1 - 2*x])/(11*Sqrt[3 + 5*x]) + (779922*ArcTan[Sqrt[1 - 2*x]/(Sq 
rt[7]*Sqrt[3 + 5*x])])/Sqrt[7])/77)/14))/28
 

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 104
Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x 
_)), x_] :> With[{q = Denominator[m]}, Simp[q   Subst[Int[x^(q*(m + 1) - 1) 
/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^(1/q)], x] 
] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && L 
tQ[-1, m, 0] && SimplerQ[a + b*x, c + d*x]
 

rule 114
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 
)/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + Simp[1/((m + 1)*(b*c - a*d)*(b*e 
 - a*f))   Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*(m + 1) 
 - b*(d*e*(m + n + 2) + c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], 
 x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && ILtQ[m, -1] && (IntegerQ[n] || 
 IntegersQ[2*n, 2*p] || ILtQ[m + n + p + 3, 0])
 

rule 168
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 + 1)*((e + f*x)^(p + 1)/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + S 
imp[1/((m + 1)*(b*c - a*d)*(b*e - a*f))   Int[(a + b*x)^(m + 1)*(c + d*x)^n 
*(e + f*x)^p*Simp[(a*d*f*g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a* 
h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p + 3)*x, x], x], 
 x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && ILtQ[m, -1]
 

rule 169
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 + 1)*((e + f*x)^(p + 1)/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + S 
imp[1/((m + 1)*(b*c - a*d)*(b*e - a*f))   Int[(a + b*x)^(m + 1)*(c + d*x)^n 
*(e + f*x)^p*Simp[(a*d*f*g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a* 
h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p + 3)*x, x], x], 
 x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && LtQ[m, -1] && IntegersQ[ 
2*m, 2*n, 2*p]
 

rule 217
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( 
-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & 
& (LtQ[a, 0] || LtQ[b, 0])
 
Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(249\) vs. \(2(104)=208\).

Time = 0.26 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.82

method result size
default \(\frac {1930306950 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right ) x^{4}+2766773295 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right ) x^{3}+536196375 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right ) x^{2}+787644900 x^{3} \sqrt {-10 x^{2}-x +3}-686331360 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right ) x +639170490 x^{2} \sqrt {-10 x^{2}-x +3}-257374260 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )-178222310 x \sqrt {-10 x^{2}-x +3}-169016344 \sqrt {-10 x^{2}-x +3}}{2324168 \left (2+3 x \right )^{2} \sqrt {1-2 x}\, \sqrt {-10 x^{2}-x +3}\, \sqrt {3+5 x}}\) \(250\)

Input:

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

Output:

1/2324168*(1930306950*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^ 
(1/2))*x^4+2766773295*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^ 
(1/2))*x^3+536196375*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^( 
1/2))*x^2+787644900*x^3*(-10*x^2-x+3)^(1/2)-686331360*7^(1/2)*arctan(1/14* 
(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))*x+639170490*x^2*(-10*x^2-x+3)^(1/2) 
-257374260*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))-1782 
22310*x*(-10*x^2-x+3)^(1/2)-169016344*(-10*x^2-x+3)^(1/2))/(2+3*x)^2/(1-2* 
x)^(1/2)/(-10*x^2-x+3)^(1/2)/(3+5*x)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.85 \[ \int \frac {1}{(1-2 x)^{3/2} (2+3 x)^3 (3+5 x)^{3/2}} \, dx=\frac {21447855 \, \sqrt {7} {\left (90 \, x^{4} + 129 \, x^{3} + 25 \, x^{2} - 32 \, x - 12\right )} \arctan \left (\frac {\sqrt {7} {\left (37 \, x + 20\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{14 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) - 14 \, {\left (56260350 \, x^{3} + 45655035 \, x^{2} - 12730165 \, x - 12072596\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{2324168 \, {\left (90 \, x^{4} + 129 \, x^{3} + 25 \, x^{2} - 32 \, x - 12\right )}} \] Input:

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

Output:

1/2324168*(21447855*sqrt(7)*(90*x^4 + 129*x^3 + 25*x^2 - 32*x - 12)*arctan 
(1/14*sqrt(7)*(37*x + 20)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x - 3)) - 
 14*(56260350*x^3 + 45655035*x^2 - 12730165*x - 12072596)*sqrt(5*x + 3)*sq 
rt(-2*x + 1))/(90*x^4 + 129*x^3 + 25*x^2 - 32*x - 12)
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.04 \[ \int \frac {1}{(1-2 x)^{3/2} (2+3 x)^3 (3+5 x)^{3/2}} \, dx=-\frac {177255}{19208} \, \sqrt {7} \arcsin \left (\frac {37 \, x}{11 \, {\left | 3 \, x + 2 \right |}} + \frac {20}{11 \, {\left | 3 \, x + 2 \right |}}\right ) + \frac {3125575 \, x}{83006 \, \sqrt {-10 \, x^{2} - x + 3}} - \frac {3262085}{166012 \, \sqrt {-10 \, x^{2} - x + 3}} + \frac {3}{14 \, {\left (9 \, \sqrt {-10 \, x^{2} - x + 3} x^{2} + 12 \, \sqrt {-10 \, x^{2} - x + 3} x + 4 \, \sqrt {-10 \, x^{2} - x + 3}\right )}} + \frac {555}{196 \, {\left (3 \, \sqrt {-10 \, x^{2} - x + 3} x + 2 \, \sqrt {-10 \, x^{2} - x + 3}\right )}} \] Input:

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

Output:

-177255/19208*sqrt(7)*arcsin(37/11*x/abs(3*x + 2) + 20/11/abs(3*x + 2)) + 
3125575/83006*x/sqrt(-10*x^2 - x + 3) - 3262085/166012/sqrt(-10*x^2 - x + 
3) + 3/14/(9*sqrt(-10*x^2 - x + 3)*x^2 + 12*sqrt(-10*x^2 - x + 3)*x + 4*sq 
rt(-10*x^2 - x + 3)) + 555/196/(3*sqrt(-10*x^2 - x + 3)*x + 2*sqrt(-10*x^2 
 - x + 3))
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 337 vs. \(2 (104) = 208\).

Time = 0.30 (sec) , antiderivative size = 337, normalized size of antiderivative = 2.46 \[ \int \frac {1}{(1-2 x)^{3/2} (2+3 x)^3 (3+5 x)^{3/2}} \, dx=-\frac {35451}{38416} \, \sqrt {70} \sqrt {10} {\left (\pi + 2 \, \arctan \left (-\frac {\sqrt {70} \sqrt {5 \, x + 3} {\left (\frac {{\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} - 4\right )}}{140 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}\right )\right )} - \frac {125}{242} \, \sqrt {10} {\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )} - \frac {32 \, \sqrt {5} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5}}{207515 \, {\left (2 \, x - 1\right )}} - \frac {297 \, \sqrt {10} {\left (47 \, {\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}^{3} + \frac {10520 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}{\sqrt {5 \, x + 3}} - \frac {42080 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}}{98 \, {\left ({\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}^{2} + 280\right )}^{2}} \] Input:

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

Output:

-35451/38416*sqrt(70)*sqrt(10)*(pi + 2*arctan(-1/140*sqrt(70)*sqrt(5*x + 3 
)*((sqrt(2)*sqrt(-10*x + 5) - sqrt(22))^2/(5*x + 3) - 4)/(sqrt(2)*sqrt(-10 
*x + 5) - sqrt(22)))) - 125/242*sqrt(10)*((sqrt(2)*sqrt(-10*x + 5) - sqrt( 
22))/sqrt(5*x + 3) - 4*sqrt(5*x + 3)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))) 
 - 32/207515*sqrt(5)*sqrt(5*x + 3)*sqrt(-10*x + 5)/(2*x - 1) - 297/98*sqrt 
(10)*(47*((sqrt(2)*sqrt(-10*x + 5) - sqrt(22))/sqrt(5*x + 3) - 4*sqrt(5*x 
+ 3)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22)))^3 + 10520*(sqrt(2)*sqrt(-10*x + 
 5) - sqrt(22))/sqrt(5*x + 3) - 42080*sqrt(5*x + 3)/(sqrt(2)*sqrt(-10*x + 
5) - sqrt(22)))/(((sqrt(2)*sqrt(-10*x + 5) - sqrt(22))/sqrt(5*x + 3) - 4*s 
qrt(5*x + 3)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22)))^2 + 280)^2
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.34 (sec) , antiderivative size = 326, normalized size of antiderivative = 2.38 \[ \int \frac {1}{(1-2 x)^{3/2} (2+3 x)^3 (3+5 x)^{3/2}} \, dx=\frac {-193030695 \sqrt {5 x +3}\, \sqrt {-2 x +1}\, \sqrt {7}\, \mathit {atan} \left (\frac {\sqrt {33}-\sqrt {35}\, \tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )}{2}\right )}{\sqrt {2}}\right ) x^{2}-257374260 \sqrt {5 x +3}\, \sqrt {-2 x +1}\, \sqrt {7}\, \mathit {atan} \left (\frac {\sqrt {33}-\sqrt {35}\, \tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )}{2}\right )}{\sqrt {2}}\right ) x -85791420 \sqrt {5 x +3}\, \sqrt {-2 x +1}\, \sqrt {7}\, \mathit {atan} \left (\frac {\sqrt {33}-\sqrt {35}\, \tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )}{2}\right )}{\sqrt {2}}\right )+193030695 \sqrt {5 x +3}\, \sqrt {-2 x +1}\, \sqrt {7}\, \mathit {atan} \left (\frac {\sqrt {33}+\sqrt {35}\, \tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )}{2}\right )}{\sqrt {2}}\right ) x^{2}+257374260 \sqrt {5 x +3}\, \sqrt {-2 x +1}\, \sqrt {7}\, \mathit {atan} \left (\frac {\sqrt {33}+\sqrt {35}\, \tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )}{2}\right )}{\sqrt {2}}\right ) x +85791420 \sqrt {5 x +3}\, \sqrt {-2 x +1}\, \sqrt {7}\, \mathit {atan} \left (\frac {\sqrt {33}+\sqrt {35}\, \tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )}{2}\right )}{\sqrt {2}}\right )+393822450 x^{3}+319585245 x^{2}-89111155 x -84508172}{1162084 \sqrt {5 x +3}\, \sqrt {-2 x +1}\, \left (9 x^{2}+12 x +4\right )} \] Input:

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

Output:

( - 193030695*sqrt(5*x + 3)*sqrt( - 2*x + 1)*sqrt(7)*atan((sqrt(33) - sqrt 
(35)*tan(asin((sqrt( - 2*x + 1)*sqrt(5))/sqrt(11))/2))/sqrt(2))*x**2 - 257 
374260*sqrt(5*x + 3)*sqrt( - 2*x + 1)*sqrt(7)*atan((sqrt(33) - sqrt(35)*ta 
n(asin((sqrt( - 2*x + 1)*sqrt(5))/sqrt(11))/2))/sqrt(2))*x - 85791420*sqrt 
(5*x + 3)*sqrt( - 2*x + 1)*sqrt(7)*atan((sqrt(33) - sqrt(35)*tan(asin((sqr 
t( - 2*x + 1)*sqrt(5))/sqrt(11))/2))/sqrt(2)) + 193030695*sqrt(5*x + 3)*sq 
rt( - 2*x + 1)*sqrt(7)*atan((sqrt(33) + sqrt(35)*tan(asin((sqrt( - 2*x + 1 
)*sqrt(5))/sqrt(11))/2))/sqrt(2))*x**2 + 257374260*sqrt(5*x + 3)*sqrt( - 2 
*x + 1)*sqrt(7)*atan((sqrt(33) + sqrt(35)*tan(asin((sqrt( - 2*x + 1)*sqrt( 
5))/sqrt(11))/2))/sqrt(2))*x + 85791420*sqrt(5*x + 3)*sqrt( - 2*x + 1)*sqr 
t(7)*atan((sqrt(33) + sqrt(35)*tan(asin((sqrt( - 2*x + 1)*sqrt(5))/sqrt(11 
))/2))/sqrt(2)) + 393822450*x**3 + 319585245*x**2 - 89111155*x - 84508172) 
/(1162084*sqrt(5*x + 3)*sqrt( - 2*x + 1)*(9*x**2 + 12*x + 4))