3.22.32 \(\int \frac {1}{(1-2 x)^{3/2} (2+3 x) (3+5 x)^3} \, dx\) [2132]

3.22.32.1 Optimal result
3.22.32.2 Mathematica [A] (verified)
3.22.32.3 Rubi [A] (verified)
3.22.32.4 Maple [A] (verified)
3.22.32.5 Fricas [A] (verification not implemented)
3.22.32.6 Sympy [C] (verification not implemented)
3.22.32.7 Maxima [A] (verification not implemented)
3.22.32.8 Giac [A] (verification not implemented)
3.22.32.9 Mupad [B] (verification not implemented)

3.22.32.1 Optimal result

Integrand size = 24, antiderivative size = 112 \[ \int \frac {1}{(1-2 x)^{3/2} (2+3 x) (3+5 x)^3} \, dx=-\frac {2049}{9317 \sqrt {1-2 x}}-\frac {5}{22 \sqrt {1-2 x} (3+5 x)^2}+\frac {305}{242 \sqrt {1-2 x} (3+5 x)}+\frac {54}{7} \sqrt {\frac {3}{7}} \text {arctanh}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )-\frac {9975 \sqrt {\frac {5}{11}} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{1331} \]

output
54/49*arctanh(1/7*21^(1/2)*(1-2*x)^(1/2))*21^(1/2)-9975/14641*arctanh(1/11 
*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)-2049/9317/(1-2*x)^(1/2)-5/22/(3+5*x)^2/( 
1-2*x)^(1/2)+305/242/(3+5*x)/(1-2*x)^(1/2)
 
3.22.32.2 Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.79 \[ \int \frac {1}{(1-2 x)^{3/2} (2+3 x) (3+5 x)^3} \, dx=\frac {54}{7} \sqrt {\frac {3}{7}} \text {arctanh}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )+\frac {-\frac {11 \left (-29338+5515 x+102450 x^2\right )}{\sqrt {1-2 x} (3+5 x)^2}-139650 \sqrt {55} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{204974} \]

input
Integrate[1/((1 - 2*x)^(3/2)*(2 + 3*x)*(3 + 5*x)^3),x]
 
output
(54*Sqrt[3/7]*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]])/7 + ((-11*(-29338 + 5515*x 
 + 102450*x^2))/(Sqrt[1 - 2*x]*(3 + 5*x)^2) - 139650*Sqrt[55]*ArcTanh[Sqrt 
[5/11]*Sqrt[1 - 2*x]])/204974
 
3.22.32.3 Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.10, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {114, 168, 27, 169, 27, 174, 73, 219}

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.

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

\(\Big \downarrow \) 114

\(\displaystyle -\frac {1}{22} \int \frac {16-75 x}{(1-2 x)^{3/2} (3 x+2) (5 x+3)^2}dx-\frac {5}{22 \sqrt {1-2 x} (5 x+3)^2}\)

\(\Big \downarrow \) 168

\(\displaystyle \frac {1}{22} \left (\frac {1}{11} \int \frac {3 (116-915 x)}{(1-2 x)^{3/2} (3 x+2) (5 x+3)}dx+\frac {305}{11 \sqrt {1-2 x} (5 x+3)}\right )-\frac {5}{22 \sqrt {1-2 x} (5 x+3)^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{22} \left (\frac {3}{11} \int \frac {116-915 x}{(1-2 x)^{3/2} (3 x+2) (5 x+3)}dx+\frac {305}{11 \sqrt {1-2 x} (5 x+3)}\right )-\frac {5}{22 \sqrt {1-2 x} (5 x+3)^2}\)

\(\Big \downarrow \) 169

\(\displaystyle \frac {1}{22} \left (\frac {3}{11} \left (-\frac {2}{77} \int -\frac {17128-10245 x}{2 \sqrt {1-2 x} (3 x+2) (5 x+3)}dx-\frac {1366}{77 \sqrt {1-2 x}}\right )+\frac {305}{11 \sqrt {1-2 x} (5 x+3)}\right )-\frac {5}{22 \sqrt {1-2 x} (5 x+3)^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{22} \left (\frac {3}{11} \left (\frac {1}{77} \int \frac {17128-10245 x}{\sqrt {1-2 x} (3 x+2) (5 x+3)}dx-\frac {1366}{77 \sqrt {1-2 x}}\right )+\frac {305}{11 \sqrt {1-2 x} (5 x+3)}\right )-\frac {5}{22 \sqrt {1-2 x} (5 x+3)^2}\)

\(\Big \downarrow \) 174

\(\displaystyle \frac {1}{22} \left (\frac {3}{11} \left (\frac {1}{77} \left (116375 \int \frac {1}{\sqrt {1-2 x} (5 x+3)}dx-71874 \int \frac {1}{\sqrt {1-2 x} (3 x+2)}dx\right )-\frac {1366}{77 \sqrt {1-2 x}}\right )+\frac {305}{11 \sqrt {1-2 x} (5 x+3)}\right )-\frac {5}{22 \sqrt {1-2 x} (5 x+3)^2}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {1}{22} \left (\frac {3}{11} \left (\frac {1}{77} \left (71874 \int \frac {1}{\frac {7}{2}-\frac {3}{2} (1-2 x)}d\sqrt {1-2 x}-116375 \int \frac {1}{\frac {11}{2}-\frac {5}{2} (1-2 x)}d\sqrt {1-2 x}\right )-\frac {1366}{77 \sqrt {1-2 x}}\right )+\frac {305}{11 \sqrt {1-2 x} (5 x+3)}\right )-\frac {5}{22 \sqrt {1-2 x} (5 x+3)^2}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{22} \left (\frac {3}{11} \left (\frac {1}{77} \left (47916 \sqrt {\frac {3}{7}} \text {arctanh}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )-46550 \sqrt {\frac {5}{11}} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )\right )-\frac {1366}{77 \sqrt {1-2 x}}\right )+\frac {305}{11 \sqrt {1-2 x} (5 x+3)}\right )-\frac {5}{22 \sqrt {1-2 x} (5 x+3)^2}\)

input
Int[1/((1 - 2*x)^(3/2)*(2 + 3*x)*(3 + 5*x)^3),x]
 
output
-5/(22*Sqrt[1 - 2*x]*(3 + 5*x)^2) + (305/(11*Sqrt[1 - 2*x]*(3 + 5*x)) + (3 
*(-1366/(77*Sqrt[1 - 2*x]) + (47916*Sqrt[3/7]*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2 
*x]] - 46550*Sqrt[5/11]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/77))/11)/22
 

3.22.32.3.1 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 73
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, 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 174
Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))* 
((c_.) + (d_.)*(x_))), x_] :> Simp[(b*g - a*h)/(b*c - a*d)   Int[(e + f*x)^ 
p/(a + b*x), x], x] - Simp[(d*g - c*h)/(b*c - a*d)   Int[(e + f*x)^p/(c + d 
*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x]
 

rule 219
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 
3.22.32.4 Maple [A] (verified)

Time = 1.12 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.57

method result size
risch \(-\frac {102450 x^{2}+5515 x -29338}{18634 \left (3+5 x \right )^{2} \sqrt {1-2 x}}+\frac {54 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{49}-\frac {9975 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{14641}\) \(64\)
derivativedivides \(\frac {-\frac {7375 \left (1-2 x \right )^{\frac {3}{2}}}{1331}+\frac {1425 \sqrt {1-2 x}}{121}}{\left (-6-10 x \right )^{2}}-\frac {9975 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{14641}+\frac {54 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{49}+\frac {16}{9317 \sqrt {1-2 x}}\) \(75\)
default \(\frac {-\frac {7375 \left (1-2 x \right )^{\frac {3}{2}}}{1331}+\frac {1425 \sqrt {1-2 x}}{121}}{\left (-6-10 x \right )^{2}}-\frac {9975 \,\operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{14641}+\frac {54 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{49}+\frac {16}{9317 \sqrt {1-2 x}}\) \(75\)
pseudoelliptic \(-\frac {9975 \left (-\frac {161359}{69825}-\frac {263538 \sqrt {1-2 x}\, \operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \left (3+5 x \right )^{2} \sqrt {21}}{162925}+\sqrt {1-2 x}\, \operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \left (3+5 x \right )^{2} \sqrt {55}+\frac {7513 x^{2}}{931}+\frac {12133 x}{27930}\right )}{14641 \sqrt {1-2 x}\, \left (3+5 x \right )^{2}}\) \(90\)
trager \(\frac {\left (102450 x^{2}+5515 x -29338\right ) \sqrt {1-2 x}}{18634 \left (3+5 x \right )^{2} \left (-1+2 x \right )}-\frac {27 \operatorname {RootOf}\left (\textit {\_Z}^{2}-21\right ) \ln \left (\frac {3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-21\right ) x -5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-21\right )+21 \sqrt {1-2 x}}{2+3 x}\right )}{49}+\frac {9975 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) \ln \left (\frac {5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) x +55 \sqrt {1-2 x}-8 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right )}{3+5 x}\right )}{29282}\) \(123\)

input
int(1/(1-2*x)^(3/2)/(2+3*x)/(3+5*x)^3,x,method=_RETURNVERBOSE)
 
output
-1/18634*(102450*x^2+5515*x-29338)/(3+5*x)^2/(1-2*x)^(1/2)+54/49*arctanh(1 
/7*21^(1/2)*(1-2*x)^(1/2))*21^(1/2)-9975/14641*arctanh(1/11*55^(1/2)*(1-2* 
x)^(1/2))*55^(1/2)
 
3.22.32.5 Fricas [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.27 \[ \int \frac {1}{(1-2 x)^{3/2} (2+3 x) (3+5 x)^3} \, dx=\frac {488775 \, \sqrt {11} \sqrt {5} {\left (50 \, x^{3} + 35 \, x^{2} - 12 \, x - 9\right )} \log \left (\frac {\sqrt {11} \sqrt {5} \sqrt {-2 \, x + 1} + 5 \, x - 8}{5 \, x + 3}\right ) + 790614 \, \sqrt {7} \sqrt {3} {\left (50 \, x^{3} + 35 \, x^{2} - 12 \, x - 9\right )} \log \left (-\frac {\sqrt {7} \sqrt {3} \sqrt {-2 \, x + 1} - 3 \, x + 5}{3 \, x + 2}\right ) + 77 \, {\left (102450 \, x^{2} + 5515 \, x - 29338\right )} \sqrt {-2 \, x + 1}}{1434818 \, {\left (50 \, x^{3} + 35 \, x^{2} - 12 \, x - 9\right )}} \]

input
integrate(1/(1-2*x)^(3/2)/(2+3*x)/(3+5*x)^3,x, algorithm="fricas")
 
output
1/1434818*(488775*sqrt(11)*sqrt(5)*(50*x^3 + 35*x^2 - 12*x - 9)*log((sqrt( 
11)*sqrt(5)*sqrt(-2*x + 1) + 5*x - 8)/(5*x + 3)) + 790614*sqrt(7)*sqrt(3)* 
(50*x^3 + 35*x^2 - 12*x - 9)*log(-(sqrt(7)*sqrt(3)*sqrt(-2*x + 1) - 3*x + 
5)/(3*x + 2)) + 77*(102450*x^2 + 5515*x - 29338)*sqrt(-2*x + 1))/(50*x^3 + 
 35*x^2 - 12*x - 9)
 
3.22.32.6 Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 8.18 (sec) , antiderivative size = 1875, normalized size of antiderivative = 16.74 \[ \int \frac {1}{(1-2 x)^{3/2} (2+3 x) (3+5 x)^3} \, dx=\text {Too large to display} \]

input
integrate(1/(1-2*x)**(3/2)/(2+3*x)/(3+5*x)**3,x)
 
output
-9775500000*sqrt(55)*I*(x - 1/2)**(11/2)*atan(sqrt(110)*sqrt(x - 1/2)/11)/ 
(14348180000*(x - 1/2)**(11/2) + 63131992000*(x - 1/2)**(9/2) + 1041677868 
00*(x - 1/2)**(7/2) + 76389710320*(x - 1/2)**(5/2) + 21007170338*(x - 1/2) 
**(3/2)) + 15812280000*sqrt(21)*I*(x - 1/2)**(11/2)*atan(sqrt(42)*sqrt(x - 
 1/2)/7)/(14348180000*(x - 1/2)**(11/2) + 63131992000*(x - 1/2)**(9/2) + 1 
04167786800*(x - 1/2)**(7/2) + 76389710320*(x - 1/2)**(5/2) + 21007170338* 
(x - 1/2)**(3/2)) - 7906140000*sqrt(21)*I*pi*(x - 1/2)**(11/2)/(1434818000 
0*(x - 1/2)**(11/2) + 63131992000*(x - 1/2)**(9/2) + 104167786800*(x - 1/2 
)**(7/2) + 76389710320*(x - 1/2)**(5/2) + 21007170338*(x - 1/2)**(3/2)) + 
4887750000*sqrt(55)*I*pi*(x - 1/2)**(11/2)/(14348180000*(x - 1/2)**(11/2) 
+ 63131992000*(x - 1/2)**(9/2) + 104167786800*(x - 1/2)**(7/2) + 763897103 
20*(x - 1/2)**(5/2) + 21007170338*(x - 1/2)**(3/2)) - 43012200000*sqrt(55) 
*I*(x - 1/2)**(9/2)*atan(sqrt(110)*sqrt(x - 1/2)/11)/(14348180000*(x - 1/2 
)**(11/2) + 63131992000*(x - 1/2)**(9/2) + 104167786800*(x - 1/2)**(7/2) + 
 76389710320*(x - 1/2)**(5/2) + 21007170338*(x - 1/2)**(3/2)) + 6957403200 
0*sqrt(21)*I*(x - 1/2)**(9/2)*atan(sqrt(42)*sqrt(x - 1/2)/7)/(14348180000* 
(x - 1/2)**(11/2) + 63131992000*(x - 1/2)**(9/2) + 104167786800*(x - 1/2)* 
*(7/2) + 76389710320*(x - 1/2)**(5/2) + 21007170338*(x - 1/2)**(3/2)) - 34 
787016000*sqrt(21)*I*pi*(x - 1/2)**(9/2)/(14348180000*(x - 1/2)**(11/2) + 
63131992000*(x - 1/2)**(9/2) + 104167786800*(x - 1/2)**(7/2) + 76389710...
 
3.22.32.7 Maxima [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.06 \[ \int \frac {1}{(1-2 x)^{3/2} (2+3 x) (3+5 x)^3} \, dx=\frac {9975}{29282} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) - \frac {27}{49} \, \sqrt {21} \log \left (-\frac {\sqrt {21} - 3 \, \sqrt {-2 \, x + 1}}{\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}}\right ) - \frac {51225 \, {\left (2 \, x - 1\right )}^{2} + 215930 \, x - 109901}{9317 \, {\left (25 \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} - 110 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + 121 \, \sqrt {-2 \, x + 1}\right )}} \]

input
integrate(1/(1-2*x)^(3/2)/(2+3*x)/(3+5*x)^3,x, algorithm="maxima")
 
output
9975/29282*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt( 
-2*x + 1))) - 27/49*sqrt(21)*log(-(sqrt(21) - 3*sqrt(-2*x + 1))/(sqrt(21) 
+ 3*sqrt(-2*x + 1))) - 1/9317*(51225*(2*x - 1)^2 + 215930*x - 109901)/(25* 
(-2*x + 1)^(5/2) - 110*(-2*x + 1)^(3/2) + 121*sqrt(-2*x + 1))
 
3.22.32.8 Giac [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.04 \[ \int \frac {1}{(1-2 x)^{3/2} (2+3 x) (3+5 x)^3} \, dx=\frac {9975}{29282} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) - \frac {27}{49} \, \sqrt {21} \log \left (\frac {{\left | -2 \, \sqrt {21} + 6 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}\right )}}\right ) + \frac {16}{9317 \, \sqrt {-2 \, x + 1}} - \frac {25 \, {\left (295 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 627 \, \sqrt {-2 \, x + 1}\right )}}{5324 \, {\left (5 \, x + 3\right )}^{2}} \]

input
integrate(1/(1-2*x)^(3/2)/(2+3*x)/(3+5*x)^3,x, algorithm="giac")
 
output
9975/29282*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) 
 + 5*sqrt(-2*x + 1))) - 27/49*sqrt(21)*log(1/2*abs(-2*sqrt(21) + 6*sqrt(-2 
*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) + 16/9317/sqrt(-2*x + 1) - 25/5324 
*(295*(-2*x + 1)^(3/2) - 627*sqrt(-2*x + 1))/(5*x + 3)^2
 
3.22.32.9 Mupad [B] (verification not implemented)

Time = 1.51 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.72 \[ \int \frac {1}{(1-2 x)^{3/2} (2+3 x) (3+5 x)^3} \, dx=\frac {54\,\sqrt {21}\,\mathrm {atanh}\left (\frac {\sqrt {21}\,\sqrt {1-2\,x}}{7}\right )}{49}-\frac {9975\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )}{14641}-\frac {\frac {3926\,x}{4235}+\frac {2049\,{\left (2\,x-1\right )}^2}{9317}-\frac {9991}{21175}}{\frac {121\,\sqrt {1-2\,x}}{25}-\frac {22\,{\left (1-2\,x\right )}^{3/2}}{5}+{\left (1-2\,x\right )}^{5/2}} \]

input
int(1/((1 - 2*x)^(3/2)*(3*x + 2)*(5*x + 3)^3),x)
 
output
(54*21^(1/2)*atanh((21^(1/2)*(1 - 2*x)^(1/2))/7))/49 - (9975*55^(1/2)*atan 
h((55^(1/2)*(1 - 2*x)^(1/2))/11))/14641 - ((3926*x)/4235 + (2049*(2*x - 1) 
^2)/9317 - 9991/21175)/((121*(1 - 2*x)^(1/2))/25 - (22*(1 - 2*x)^(3/2))/5 
+ (1 - 2*x)^(5/2))