\(\int \frac {(2+3 x)^4}{(1-2 x)^{5/2} (3+5 x)^3} \, dx\) [837]

Optimal result

Integrand size = 24, antiderivative size = 94 \[ \int \frac {(2+3 x)^4}{(1-2 x)^{5/2} (3+5 x)^3} \, dx=\frac {2401}{7986 (1-2 x)^{3/2}}-\frac {9261}{29282 \sqrt {1-2 x}}-\frac {\sqrt {1-2 x}}{66550 (3+5 x)^2}-\frac {\sqrt {1-2 x}}{2662 (3+5 x)}-\frac {14423 \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{366025 \sqrt {55}} \] Output:

2401/7986/(1-2*x)^(3/2)-9261/29282/(1-2*x)^(1/2)-1/66550*(1-2*x)^(1/2)/(3+ 
5*x)^2-(1-2*x)^(1/2)/(7986+13310*x)-14423/20131375*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 \[ \int \frac {(2+3 x)^4}{(1-2 x)^{5/2} (3+5 x)^3} \, dx=\frac {\frac {55 \left (-311208+11479257 x+40823468 x^2+34712250 x^3\right )}{(1-2 x)^{3/2} (3+5 x)^2}-86538 \sqrt {55} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{120788250} \] Input:

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

Output:

((55*(-311208 + 11479257*x + 40823468*x^2 + 34712250*x^3))/((1 - 2*x)^(3/2 
)*(3 + 5*x)^2) - 86538*Sqrt[55]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/1207882 
50
 

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.24, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {109, 166, 161, 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.

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

\(\Big \downarrow \) 109

\(\displaystyle \frac {7 (3 x+2)^3}{33 (1-2 x)^{3/2} (5 x+3)^2}-\frac {1}{33} \int \frac {(3 x+2)^2 (96 x+43)}{(1-2 x)^{3/2} (5 x+3)^3}dx\)

\(\Big \downarrow \) 166

\(\displaystyle \frac {1}{33} \left (-\frac {1}{110} \int \frac {(3 x+2) (6117 x+3056)}{(1-2 x)^{3/2} (5 x+3)^2}dx-\frac {73 (3 x+2)^2}{110 \sqrt {1-2 x} (5 x+3)^2}\right )+\frac {7 (3 x+2)^3}{33 (1-2 x)^{3/2} (5 x+3)^2}\)

\(\Big \downarrow \) 161

\(\displaystyle \frac {1}{33} \left (\frac {1}{110} \left (\frac {43269}{605} \int \frac {1}{\sqrt {1-2 x} (5 x+3)}dx-\frac {2133933 x+1287116}{605 \sqrt {1-2 x} (5 x+3)}\right )-\frac {73 (3 x+2)^2}{110 \sqrt {1-2 x} (5 x+3)^2}\right )+\frac {7 (3 x+2)^3}{33 (1-2 x)^{3/2} (5 x+3)^2}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {1}{33} \left (\frac {1}{110} \left (-\frac {43269}{605} \int \frac {1}{\frac {11}{2}-\frac {5}{2} (1-2 x)}d\sqrt {1-2 x}-\frac {2133933 x+1287116}{605 \sqrt {1-2 x} (5 x+3)}\right )-\frac {73 (3 x+2)^2}{110 \sqrt {1-2 x} (5 x+3)^2}\right )+\frac {7 (3 x+2)^3}{33 (1-2 x)^{3/2} (5 x+3)^2}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{33} \left (\frac {1}{110} \left (-\frac {86538 \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{605 \sqrt {55}}-\frac {2133933 x+1287116}{605 \sqrt {1-2 x} (5 x+3)}\right )-\frac {73 (3 x+2)^2}{110 \sqrt {1-2 x} (5 x+3)^2}\right )+\frac {7 (3 x+2)^3}{33 (1-2 x)^{3/2} (5 x+3)^2}\)

Input:

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

Output:

(7*(2 + 3*x)^3)/(33*(1 - 2*x)^(3/2)*(3 + 5*x)^2) + ((-73*(2 + 3*x)^2)/(110 
*Sqrt[1 - 2*x]*(3 + 5*x)^2) + (-1/605*(1287116 + 2133933*x)/(Sqrt[1 - 2*x] 
*(3 + 5*x)) - (86538*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(605*Sqrt[55]))/11 
0)/33
 

Defintions of rubi rules used

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 166

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*((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 - 1)*(e + f*x)^p*Simp[b* 
c*(f*g - e*h)*(m + 1) + (b*g - a*h)*(d*e*n + c*f*(p + 1)) + d*(b*(f*g - e*h 
)*(m + 1) + f*(b*g - a*h)*(n + p + 1))*x, x], x], x] /; FreeQ[{a, b, c, d, 
e, f, g, h, p}, x] && ILtQ[m, -1] && GtQ[n, 0]
 

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	\(-\frac {34712250 x^{3}[phone] x^{2}[phone] x -311208}{2196150 \sqrt {1-2 x}\, \left (3+5 x \right )^{2} \left (-1+2 x \right )}-\frac {14423 \sqrt {55}\, \operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right )}{20131375}\)	\(58\)
pseudoelliptic	\(\frac {-\frac {51868}{366025}-\frac {14423 \left (1-2 x \right )^{\frac {3}{2}} \sqrt {55}\, \left (3+5 x \right )^{2} \operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right )}{20131375}+\frac {231415 x^{3}}{14641}+\frac {20411734 x^{2}}{1098075}+\frac {3826419 x}{732050}}{\left (1-2 x \right )^{\frac {3}{2}} \left (3+5 x \right )^{2}}\)	\(64\)
derivativedivides	\(\frac {2401}{7986 \left (1-2 x \right )^{\frac {3}{2}}}-\frac {9261}{29282 \sqrt {1-2 x}}+\frac {\frac {5 \left (1-2 x \right )^{\frac {3}{2}}}{1331}-\frac {277 \sqrt {1-2 x}}{33275}}{\left (-6-10 x \right )^{2}}-\frac {14423 \sqrt {55}\, \operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right )}{20131375}\)	\(66\)
default	\(\frac {2401}{7986 \left (1-2 x \right )^{\frac {3}{2}}}-\frac {9261}{29282 \sqrt {1-2 x}}+\frac {\frac {5 \left (1-2 x \right )^{\frac {3}{2}}}{1331}-\frac {277 \sqrt {1-2 x}}{33275}}{\left (-6-10 x \right )^{2}}-\frac {14423 \sqrt {55}\, \operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right )}{20131375}\)	\(66\)
trager	\(\frac {\left (34712250 x^{3}[phone] x^{2}[phone] x -311208\right ) \sqrt {1-2 x}}{2196150 \left (10 x^{2}+x -3\right )^{2}}+\frac {14423 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) \ln \left (\frac {5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) x +55 \sqrt {1-2 x}-8 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right )}{3+5 x}\right )}{40262750}\)	\(80\)

Input:

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

Output:

-1/2196150*(34712250*x^3+40823468*x^2+11479257*x-311208)/(1-2*x)^(1/2)/(3+ 
5*x)^2/(-1+2*x)-14423/20131375*55^(1/2)*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2 
))
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.05 \[ \int \frac {(2+3 x)^4}{(1-2 x)^{5/2} (3+5 x)^3} \, dx=\frac {43269 \, \sqrt {55} {\left (100 \, x^{4} + 20 \, x^{3} - 59 \, x^{2} - 6 \, x + 9\right )} \log \left (\frac {5 \, x + \sqrt {55} \sqrt {-2 \, x + 1} - 8}{5 \, x + 3}\right ) + 55 \, {\left (34712250 \, x^{3} + 40823468 \, x^{2} + 11479257 \, x - 311208\right )} \sqrt {-2 \, x + 1}}{120788250 \, {\left (100 \, x^{4} + 20 \, x^{3} - 59 \, x^{2} - 6 \, x + 9\right )}} \] Input:

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

Output:

1/120788250*(43269*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*(34712250*x^3 + 40823468*x 
^2 + 11479257*x - 311208)*sqrt(-2*x + 1))/(100*x^4 + 20*x^3 - 59*x^2 - 6*x 
 + 9)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.98 \[ \int \frac {(2+3 x)^4}{(1-2 x)^{5/2} (3+5 x)^3} \, dx=\frac {14423}{40262750} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) + \frac {17356125 \, {\left (2 \, x - 1\right )}^{3} + 92891843 \, {\left (2 \, x - 1\right )}^{2} + 313347650 \, x - 76780550}{2196150 \, {\left (25 \, {\left (-2 \, x + 1\right )}^{\frac {7}{2}} - 110 \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} + 121 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}}\right )}} \] Input:

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

Output:

14423/40262750*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*s 
qrt(-2*x + 1))) + 1/2196150*(17356125*(2*x - 1)^3 + 92891843*(2*x - 1)^2 + 
 313347650*x - 76780550)/(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 \[ \int \frac {(2+3 x)^4}{(1-2 x)^{5/2} (3+5 x)^3} \, dx=\frac {14423}{40262750} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) - \frac {343 \, {\left (81 \, x - 2\right )}}{43923 \, {\left (2 \, x - 1\right )} \sqrt {-2 \, x + 1}} + \frac {125 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 277 \, \sqrt {-2 \, x + 1}}{133100 \, {\left (5 \, x + 3\right )}^{2}} \] Input:

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

Output:

14423/40262750*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt 
(55) + 5*sqrt(-2*x + 1))) - 343/43923*(81*x - 2)/((2*x - 1)*sqrt(-2*x + 1) 
) + 1/133100*(125*(-2*x + 1)^(3/2) - 277*sqrt(-2*x + 1))/(5*x + 3)^2
 

Mupad [B] (verification not implemented)

Time = 1.06 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.76 \[ \int \frac {(2+3 x)^4}{(1-2 x)^{5/2} (3+5 x)^3} \, dx=\frac {\frac {51793\,x}{9075}+\frac {8444713\,{\left (2\,x-1\right )}^2}{4991250}+\frac {46283\,{\left (2\,x-1\right )}^3}{146410}-\frac {12691}{9075}}{\frac {121\,{\left (1-2\,x\right )}^{3/2}}{25}-\frac {22\,{\left (1-2\,x\right )}^{5/2}}{5}+{\left (1-2\,x\right )}^{7/2}}-\frac {14423\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )}{20131375} \] Input:

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

Output:

((51793*x)/9075 + (8444713*(2*x - 1)^2)/4991250 + (46283*(2*x - 1)^3)/1464 
10 - 12691/9075)/((121*(1 - 2*x)^(3/2))/25 - (22*(1 - 2*x)^(5/2))/5 + (1 - 
 2*x)^(7/2)) - (14423*55^(1/2)*atanh((55^(1/2)*(1 - 2*x)^(1/2))/11))/20131 
375
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 240, normalized size of antiderivative = 2.55 \[ \int \frac {(2+3 x)^4}{(1-2 x)^{5/2} (3+5 x)^3} \, dx=\frac {2163450 \sqrt {-2 x +1}\, \sqrt {55}\, \mathrm {log}\left (5 \sqrt {-2 x +1}-\sqrt {55}\right ) x^{3}+1514415 \sqrt {-2 x +1}\, \sqrt {55}\, \mathrm {log}\left (5 \sqrt {-2 x +1}-\sqrt {55}\right ) x^{2}-519228 \sqrt {-2 x +1}\, \sqrt {55}\, \mathrm {log}\left (5 \sqrt {-2 x +1}-\sqrt {55}\right ) x -389421 \sqrt {-2 x +1}\, \sqrt {55}\, \mathrm {log}\left (5 \sqrt {-2 x +1}-\sqrt {55}\right )-2163450 \sqrt {-2 x +1}\, \sqrt {55}\, \mathrm {log}\left (5 \sqrt {-2 x +1}+\sqrt {55}\right ) x^{3}-1514415 \sqrt {-2 x +1}\, \sqrt {55}\, \mathrm {log}\left (5 \sqrt {-2 x +1}+\sqrt {55}\right ) x^{2}+519228 \sqrt {-2 x +1}\, \sqrt {55}\, \mathrm {log}\left (5 \sqrt {-2 x +1}+\sqrt {55}\right ) x +389421 \sqrt {-2 x +1}\, \sqrt {55}\, \mathrm {log}\left (5 \sqrt {-2 x +1}+\sqrt {55}\right )-1909173750 x^{3}-2245290740 x^{2}-631359135 x [phone]}{120788250 \sqrt {-2 x +1}\, \left (50 x^{3}+35 x^{2}-12 x -9\right )} \] Input:

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

Output:

(2163450*sqrt( - 2*x + 1)*sqrt(55)*log(5*sqrt( - 2*x + 1) - sqrt(55))*x**3 
 + 1514415*sqrt( - 2*x + 1)*sqrt(55)*log(5*sqrt( - 2*x + 1) - sqrt(55))*x* 
*2 - 519228*sqrt( - 2*x + 1)*sqrt(55)*log(5*sqrt( - 2*x + 1) - sqrt(55))*x 
 - 389421*sqrt( - 2*x + 1)*sqrt(55)*log(5*sqrt( - 2*x + 1) - sqrt(55)) - 2 
163450*sqrt( - 2*x + 1)*sqrt(55)*log(5*sqrt( - 2*x + 1) + sqrt(55))*x**3 - 
 1514415*sqrt( - 2*x + 1)*sqrt(55)*log(5*sqrt( - 2*x + 1) + sqrt(55))*x**2 
 + 519228*sqrt( - 2*x + 1)*sqrt(55)*log(5*sqrt( - 2*x + 1) + sqrt(55))*x + 
 389421*sqrt( - 2*x + 1)*sqrt(55)*log(5*sqrt( - 2*x + 1) + sqrt(55)) - 190 
9173750*x**3 - 2245290740*x**2 - 631359135*x + 17116440)/(120788250*sqrt( 
- 2*x + 1)*(50*x**3 + 35*x**2 - 12*x - 9))