\(\int \frac {(2+3 x)^3}{(1-2 x)^{5/2} (3+5 x)^3} \, dx\) [838]

Optimal result

Integrand size = 24, antiderivative size = 94 \[ \int \frac {(2+3 x)^3}{(1-2 x)^{5/2} (3+5 x)^3} \, dx=\frac {343}{3993 (1-2 x)^{3/2}}+\frac {294}{14641 \sqrt {1-2 x}}-\frac {\sqrt {1-2 x}}{13310 (3+5 x)^2}-\frac {19 \sqrt {1-2 x}}{13310 (3+5 x)}-\frac {7559 \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{73205 \sqrt {55}} \] Output:

343/3993/(1-2*x)^(3/2)+294/14641/(1-2*x)^(1/2)-1/13310*(1-2*x)^(1/2)/(3+5* 
x)^2-19*(1-2*x)^(1/2)/(39930+66550*x)-7559/4026275*55^(1/2)*arctanh(1/11*5 
5^(1/2)*(1-2*x)^(1/2))
 

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.67 \[ \int \frac {(2+3 x)^3}{(1-2 x)^{5/2} (3+5 x)^3} \, dx=\frac {-\frac {55 \left (-417036-1242261 x-639434 x^2+453540 x^3\right )}{(1-2 x)^{3/2} (3+5 x)^2}-45354 \sqrt {55} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{24157650} \] Input:

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

Output:

((-55*(-417036 - 1242261*x - 639434*x^2 + 453540*x^3))/((1 - 2*x)^(3/2)*(3 
 + 5*x)^2) - 45354*Sqrt[55]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/24157650
 

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.17, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {109, 25, 161, 52, 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.

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

\(\Big \downarrow \) 109

\(\displaystyle \frac {7 (3 x+2)^2}{33 (1-2 x)^{3/2} (5 x+3)^2}-\frac {1}{33} \int -\frac {(3 x+2) (9 x+20)}{(1-2 x)^{3/2} (5 x+3)^3}dx\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{33} \int \frac {(3 x+2) (9 x+20)}{(1-2 x)^{3/2} (5 x+3)^3}dx+\frac {7 (3 x+2)^2}{33 (1-2 x)^{3/2} (5 x+3)^2}\)

\(\Big \downarrow \) 161

\(\displaystyle \frac {1}{33} \left (\frac {22677 \int \frac {1}{\sqrt {1-2 x} (5 x+3)^2}dx}{1210}+\frac {17296 x+10217}{1210 \sqrt {1-2 x} (5 x+3)^2}\right )+\frac {7 (3 x+2)^2}{33 (1-2 x)^{3/2} (5 x+3)^2}\)

\(\Big \downarrow \) 52

\(\displaystyle \frac {1}{33} \left (\frac {22677 \left (\frac {1}{11} \int \frac {1}{\sqrt {1-2 x} (5 x+3)}dx-\frac {\sqrt {1-2 x}}{11 (5 x+3)}\right )}{1210}+\frac {17296 x+10217}{1210 \sqrt {1-2 x} (5 x+3)^2}\right )+\frac {7 (3 x+2)^2}{33 (1-2 x)^{3/2} (5 x+3)^2}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {1}{33} \left (\frac {22677 \left (-\frac {1}{11} \int \frac {1}{\frac {11}{2}-\frac {5}{2} (1-2 x)}d\sqrt {1-2 x}-\frac {\sqrt {1-2 x}}{11 (5 x+3)}\right )}{1210}+\frac {17296 x+10217}{1210 \sqrt {1-2 x} (5 x+3)^2}\right )+\frac {7 (3 x+2)^2}{33 (1-2 x)^{3/2} (5 x+3)^2}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{33} \left (\frac {22677 \left (-\frac {2 \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{11 \sqrt {55}}-\frac {\sqrt {1-2 x}}{11 (5 x+3)}\right )}{1210}+\frac {17296 x+10217}{1210 \sqrt {1-2 x} (5 x+3)^2}\right )+\frac {7 (3 x+2)^2}{33 (1-2 x)^{3/2} (5 x+3)^2}\)

Input:

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

Output:

(7*(2 + 3*x)^2)/(33*(1 - 2*x)^(3/2)*(3 + 5*x)^2) + ((10217 + 17296*x)/(121 
0*Sqrt[1 - 2*x]*(3 + 5*x)^2) + (22677*(-1/11*Sqrt[1 - 2*x]/(3 + 5*x) - (2* 
ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(11*Sqrt[55])))/1210)/33
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( 
m + n + 2)/((b*c - a*d)*(m + 1)))   Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], 
x] /; FreeQ[{a, b, c, d, n}, x] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
 

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 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 161

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_)) 
*((g_.) + (h_.)*(x_)), x_] :> Simp[((b^2*c*d*e*g*(n + 1) + a^2*c*d*f*h*(n + 
 1) + a*b*(d^2*e*g*(m + 1) + c^2*f*h*(m + 1) - c*d*(f*g + e*h)*(m + n + 2)) 
 + (a^2*d^2*f*h*(n + 1) - a*b*d^2*(f*g + e*h)*(n + 1) + b^2*(c^2*f*h*(m + 1 
) - c*d*(f*g + e*h)*(m + 1) + d^2*e*g*(m + n + 2)))*x)/(b*d*(b*c - a*d)^2*( 
m + 1)*(n + 1)))*(a + b*x)^(m + 1)*(c + d*x)^(n + 1), x] - Simp[(a^2*d^2*f* 
h*(2 + 3*n + 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*(2 + 3*m + m^2) - c*d*(f*g + e*h)*(m + 1)*(m + n + 3) + 
 d^2*e*g*(6 + m^2 + 5*n + n^2 + m*(2*n + 5))))/(b*d*(b*c - a*d)^2*(m + 1)*( 
n + 1))   Int[(a + b*x)^(m + 1)*(c + d*x)^(n + 1), x], x] /; FreeQ[{a, b, c 
, d, e, f, g, h}, x] && LtQ[m, -1] && LtQ[n, -1]
 

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])
 

Maple [A] (verified)

Time = 0.25 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.62

method	result	size
risch	\(\frac {453540 x^{3}-639434 x^{2}-1242261 x -417036}{439230 \sqrt {1-2 x}\, \left (3+5 x \right )^{2} \left (-1+2 x \right )}-\frac {7559 \sqrt {55}\, \operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right )}{4026275}\)	\(58\)
pseudoelliptic	\(-\frac {7559 \left (\frac {\left (1-2 x \right )^{\frac {3}{2}} \sqrt {55}\, \left (3+5 x \right )^{2} \operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right )}{25}+22 x^{3}-\frac {3516887 x^{2}}{113385}-\frac {4554957 x}{75590}-\frac {764566}{37795}\right )}{161051 \left (1-2 x \right )^{\frac {3}{2}} \left (3+5 x \right )^{2}}\)	\(64\)
derivativedivides	\(\frac {\frac {19 \left (1-2 x \right )^{\frac {3}{2}}}{1331}-\frac {211 \sqrt {1-2 x}}{6655}}{\left (-6-10 x \right )^{2}}-\frac {7559 \sqrt {55}\, \operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right )}{4026275}+\frac {343}{3993 \left (1-2 x \right )^{\frac {3}{2}}}+\frac {294}{14641 \sqrt {1-2 x}}\)	\(66\)
default	\(\frac {\frac {19 \left (1-2 x \right )^{\frac {3}{2}}}{1331}-\frac {211 \sqrt {1-2 x}}{6655}}{\left (-6-10 x \right )^{2}}-\frac {7559 \sqrt {55}\, \operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right )}{4026275}+\frac {343}{3993 \left (1-2 x \right )^{\frac {3}{2}}}+\frac {294}{14641 \sqrt {1-2 x}}\)	\(66\)
trager	\(-\frac {\left (453540 x^{3}-639434 x^{2}-1242261 x -417036\right ) \sqrt {1-2 x}}{439230 \left (10 x^{2}+x -3\right )^{2}}-\frac {7559 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) \ln \left (\frac {-5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) x +55 \sqrt {1-2 x}+8 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right )}{3+5 x}\right )}{8052550}\)	\(80\)

Input:

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

Output:

1/439230*(453540*x^3-639434*x^2-1242261*x-417036)/(1-2*x)^(1/2)/(3+5*x)^2/ 
(-1+2*x)-7559/4026275*55^(1/2)*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.05 \[ \int \frac {(2+3 x)^3}{(1-2 x)^{5/2} (3+5 x)^3} \, dx=\frac {22677 \, \sqrt {55} {\left (100 \, x^{4} + 20 \, x^{3} - 59 \, x^{2} - 6 \, x + 9\right )} \log \left (\frac {5 \, x + \sqrt {55} \sqrt {-2 \, x + 1} - 8}{5 \, x + 3}\right ) - 55 \, {\left (453540 \, x^{3} - 639434 \, x^{2} - 1242261 \, x - 417036\right )} \sqrt {-2 \, x + 1}}{24157650 \, {\left (100 \, x^{4} + 20 \, x^{3} - 59 \, x^{2} - 6 \, x + 9\right )}} \] Input:

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

Output:

1/24157650*(22677*sqrt(55)*(100*x^4 + 20*x^3 - 59*x^2 - 6*x + 9)*log((5*x 
+ sqrt(55)*sqrt(-2*x + 1) - 8)/(5*x + 3)) - 55*(453540*x^3 - 639434*x^2 - 
1242261*x - 417036)*sqrt(-2*x + 1))/(100*x^4 + 20*x^3 - 59*x^2 - 6*x + 9)
 

Sympy [A] (verification not implemented)

Time = 155.95 (sec) , antiderivative size = 354, normalized size of antiderivative = 3.77 \[ \int \frac {(2+3 x)^3}{(1-2 x)^{5/2} (3+5 x)^3} \, dx=\frac {147 \sqrt {55} \left (\log {\left (\sqrt {1 - 2 x} - \frac {\sqrt {55}}{5} \right )} - \log {\left (\sqrt {1 - 2 x} + \frac {\sqrt {55}}{5} \right )}\right )}{161051} - \frac {412 \left (\begin {cases} \frac {\sqrt {55} \left (- \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1 \right )}}{4} + \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1 \right )}}{4} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1\right )} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1\right )}\right )}{605} & \text {for}\: \sqrt {1 - 2 x} > - \frac {\sqrt {55}}{5} \wedge \sqrt {1 - 2 x} < \frac {\sqrt {55}}{5} \end {cases}\right )}{6655} + \frac {8 \left (\begin {cases} \frac {\sqrt {55} \cdot \left (\frac {3 \log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1 \right )}}{16} - \frac {3 \log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1 \right )}}{16} + \frac {3}{16 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1\right )} + \frac {1}{16 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1\right )^{2}} + \frac {3}{16 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1\right )} - \frac {1}{16 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1\right )^{2}}\right )}{6655} & \text {for}\: \sqrt {1 - 2 x} > - \frac {\sqrt {55}}{5} \wedge \sqrt {1 - 2 x} < \frac {\sqrt {55}}{5} \end {cases}\right )}{605} + \frac {294}{14641 \sqrt {1 - 2 x}} + \frac {343}{3993 \left (1 - 2 x\right )^{\frac {3}{2}}} \] Input:

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

Output:

147*sqrt(55)*(log(sqrt(1 - 2*x) - sqrt(55)/5) - log(sqrt(1 - 2*x) + sqrt(5 
5)/5))/161051 - 412*Piecewise((sqrt(55)*(-log(sqrt(55)*sqrt(1 - 2*x)/11 - 
1)/4 + log(sqrt(55)*sqrt(1 - 2*x)/11 + 1)/4 - 1/(4*(sqrt(55)*sqrt(1 - 2*x) 
/11 + 1)) - 1/(4*(sqrt(55)*sqrt(1 - 2*x)/11 - 1)))/605, (sqrt(1 - 2*x) > - 
sqrt(55)/5) & (sqrt(1 - 2*x) < sqrt(55)/5)))/6655 + 8*Piecewise((sqrt(55)* 
(3*log(sqrt(55)*sqrt(1 - 2*x)/11 - 1)/16 - 3*log(sqrt(55)*sqrt(1 - 2*x)/11 
 + 1)/16 + 3/(16*(sqrt(55)*sqrt(1 - 2*x)/11 + 1)) + 1/(16*(sqrt(55)*sqrt(1 
 - 2*x)/11 + 1)**2) + 3/(16*(sqrt(55)*sqrt(1 - 2*x)/11 - 1)) - 1/(16*(sqrt 
(55)*sqrt(1 - 2*x)/11 - 1)**2))/6655, (sqrt(1 - 2*x) > -sqrt(55)/5) & (sqr 
t(1 - 2*x) < sqrt(55)/5)))/605 + 294/(14641*sqrt(1 - 2*x)) + 343/(3993*(1 
- 2*x)**(3/2))
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.98 \[ \int \frac {(2+3 x)^3}{(1-2 x)^{5/2} (3+5 x)^3} \, dx=\frac {7559}{8052550} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) - \frac {113385 \, {\left (2 \, x - 1\right )}^{3} + 20438 \, {\left (2 \, x - 1\right )}^{2} - 3083080 \, x - 741125}{219615 \, {\left (25 \, {\left (-2 \, x + 1\right )}^{\frac {7}{2}} - 110 \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} + 121 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}}\right )}} \] Input:

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

Output:

7559/8052550*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqr 
t(-2*x + 1))) - 1/219615*(113385*(2*x - 1)^3 + 20438*(2*x - 1)^2 - 3083080 
*x - 741125)/(25*(-2*x + 1)^(7/2) - 110*(-2*x + 1)^(5/2) + 121*(-2*x + 1)^ 
(3/2))
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.95 \[ \int \frac {(2+3 x)^3}{(1-2 x)^{5/2} (3+5 x)^3} \, dx=\frac {7559}{8052550} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) + \frac {49 \, {\left (36 \, x - 95\right )}}{43923 \, {\left (2 \, x - 1\right )} \sqrt {-2 \, x + 1}} + \frac {95 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 211 \, \sqrt {-2 \, x + 1}}{26620 \, {\left (5 \, x + 3\right )}^{2}} \] Input:

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

Output:

7559/8052550*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(5 
5) + 5*sqrt(-2*x + 1))) + 49/43923*(36*x - 95)/((2*x - 1)*sqrt(-2*x + 1)) 
+ 1/26620*(95*(-2*x + 1)^(3/2) - 211*sqrt(-2*x + 1))/(5*x + 3)^2
 

Mupad [B] (verification not implemented)

Time = 1.08 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.76 \[ \int \frac {(2+3 x)^3}{(1-2 x)^{5/2} (3+5 x)^3} \, dx=\frac {\frac {5096\,x}{9075}-\frac {1858\,{\left (2\,x-1\right )}^2}{499125}-\frac {7559\,{\left (2\,x-1\right )}^3}{366025}+\frac {49}{363}}{\frac {121\,{\left (1-2\,x\right )}^{3/2}}{25}-\frac {22\,{\left (1-2\,x\right )}^{5/2}}{5}+{\left (1-2\,x\right )}^{7/2}}-\frac {7559\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )}{4026275} \] Input:

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

Output:

((5096*x)/9075 - (1858*(2*x - 1)^2)/499125 - (7559*(2*x - 1)^3)/366025 + 4 
9/363)/((121*(1 - 2*x)^(3/2))/25 - (22*(1 - 2*x)^(5/2))/5 + (1 - 2*x)^(7/2 
)) - (7559*55^(1/2)*atanh((55^(1/2)*(1 - 2*x)^(1/2))/11))/4026275
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 240, normalized size of antiderivative = 2.55 \[ \int \frac {(2+3 x)^3}{(1-2 x)^{5/2} (3+5 x)^3} \, dx=\frac {1133850 \sqrt {-2 x +1}\, \sqrt {55}\, \mathrm {log}\left (5 \sqrt {-2 x +1}-\sqrt {55}\right ) x^{3}+793695 \sqrt {-2 x +1}\, \sqrt {55}\, \mathrm {log}\left (5 \sqrt {-2 x +1}-\sqrt {55}\right ) x^{2}-272124 \sqrt {-2 x +1}\, \sqrt {55}\, \mathrm {log}\left (5 \sqrt {-2 x +1}-\sqrt {55}\right ) x -204093 \sqrt {-2 x +1}\, \sqrt {55}\, \mathrm {log}\left (5 \sqrt {-2 x +1}-\sqrt {55}\right )-1133850 \sqrt {-2 x +1}\, \sqrt {55}\, \mathrm {log}\left (5 \sqrt {-2 x +1}+\sqrt {55}\right ) x^{3}-793695 \sqrt {-2 x +1}\, \sqrt {55}\, \mathrm {log}\left (5 \sqrt {-2 x +1}+\sqrt {55}\right ) x^{2}+272124 \sqrt {-2 x +1}\, \sqrt {55}\, \mathrm {log}\left (5 \sqrt {-2 x +1}+\sqrt {55}\right ) x +204093 \sqrt {-2 x +1}\, \sqrt {55}\, \mathrm {log}\left (5 \sqrt {-2 x +1}+\sqrt {55}\right )[phone] x^{3}-35168870 x^{2}-68324355 x -22936980}{24157650 \sqrt {-2 x +1}\, \left (50 x^{3}+35 x^{2}-12 x -9\right )} \] Input:

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

Output:

(1133850*sqrt( - 2*x + 1)*sqrt(55)*log(5*sqrt( - 2*x + 1) - sqrt(55))*x**3 
 + 793695*sqrt( - 2*x + 1)*sqrt(55)*log(5*sqrt( - 2*x + 1) - sqrt(55))*x** 
2 - 272124*sqrt( - 2*x + 1)*sqrt(55)*log(5*sqrt( - 2*x + 1) - sqrt(55))*x 
- 204093*sqrt( - 2*x + 1)*sqrt(55)*log(5*sqrt( - 2*x + 1) - sqrt(55)) - 11 
33850*sqrt( - 2*x + 1)*sqrt(55)*log(5*sqrt( - 2*x + 1) + sqrt(55))*x**3 - 
793695*sqrt( - 2*x + 1)*sqrt(55)*log(5*sqrt( - 2*x + 1) + sqrt(55))*x**2 + 
 272124*sqrt( - 2*x + 1)*sqrt(55)*log(5*sqrt( - 2*x + 1) + sqrt(55))*x + 2 
04093*sqrt( - 2*x + 1)*sqrt(55)*log(5*sqrt( - 2*x + 1) + sqrt(55)) + 24944 
700*x**3 - 35168870*x**2 - 68324355*x - 22936980)/(24157650*sqrt( - 2*x + 
1)*(50*x**3 + 35*x**2 - 12*x - 9))