3.19.90 \(\int \frac {(1-2 x)^{3/2} (3+5 x)^3}{(2+3 x)^4} \, dx\) [1890]

3.19.90.1 Optimal result
3.19.90.2 Mathematica [A] (verified)
3.19.90.3 Rubi [A] (verified)
3.19.90.4 Maple [A] (verified)
3.19.90.5 Fricas [A] (verification not implemented)
3.19.90.6 Sympy [F(-1)]
3.19.90.7 Maxima [A] (verification not implemented)
3.19.90.8 Giac [A] (verification not implemented)
3.19.90.9 Mupad [B] (verification not implemented)

3.19.90.1 Optimal result

Integrand size = 24, antiderivative size = 125 \[ \int \frac {(1-2 x)^{3/2} (3+5 x)^3}{(2+3 x)^4} \, dx=\frac {251 \sqrt {1-2 x} (3+5 x)^2}{63 (2+3 x)}-\frac {(1-2 x)^{3/2} (3+5 x)^3}{9 (2+3 x)^3}+\frac {2 \sqrt {1-2 x} (3+5 x)^3}{(2+3 x)^2}-\frac {5}{567} \sqrt {1-2 x} (2323+7265 x)-\frac {36038 \text {arctanh}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{567 \sqrt {21}} \]

output
-1/9*(1-2*x)^(3/2)*(3+5*x)^3/(2+3*x)^3-36038/11907*arctanh(1/7*21^(1/2)*(1 
-2*x)^(1/2))*21^(1/2)+251/63*(3+5*x)^2*(1-2*x)^(1/2)/(2+3*x)+2*(3+5*x)^3*( 
1-2*x)^(1/2)/(2+3*x)^2-5/567*(2323+7265*x)*(1-2*x)^(1/2)
 
3.19.90.2 Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.54 \[ \int \frac {(1-2 x)^{3/2} (3+5 x)^3}{(2+3 x)^4} \, dx=\frac {\sqrt {1-2 x} \left (47939+199243 x+259614 x^2+81900 x^3-31500 x^4\right )}{567 (2+3 x)^3}-\frac {36038 \text {arctanh}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{567 \sqrt {21}} \]

input
Integrate[((1 - 2*x)^(3/2)*(3 + 5*x)^3)/(2 + 3*x)^4,x]
 
output
(Sqrt[1 - 2*x]*(47939 + 199243*x + 259614*x^2 + 81900*x^3 - 31500*x^4))/(5 
67*(2 + 3*x)^3) - (36038*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]])/(567*Sqrt[21])
 
3.19.90.3 Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.08, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {108, 27, 166, 27, 166, 164, 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.

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

\(\Big \downarrow \) 108

\(\displaystyle \frac {1}{9} \int \frac {3 (2-15 x) \sqrt {1-2 x} (5 x+3)^2}{(3 x+2)^3}dx-\frac {(1-2 x)^{3/2} (5 x+3)^3}{9 (3 x+2)^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{3} \int \frac {(2-15 x) \sqrt {1-2 x} (5 x+3)^2}{(3 x+2)^3}dx-\frac {(1-2 x)^{3/2} (5 x+3)^3}{9 (3 x+2)^3}\)

\(\Big \downarrow \) 166

\(\displaystyle \frac {1}{3} \left (\frac {6 \sqrt {1-2 x} (5 x+3)^3}{(3 x+2)^2}-\frac {1}{6} \int \frac {6 (17-100 x) (5 x+3)^2}{\sqrt {1-2 x} (3 x+2)^2}dx\right )-\frac {(1-2 x)^{3/2} (5 x+3)^3}{9 (3 x+2)^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{3} \left (\frac {6 \sqrt {1-2 x} (5 x+3)^3}{(3 x+2)^2}-\int \frac {(17-100 x) (5 x+3)^2}{\sqrt {1-2 x} (3 x+2)^2}dx\right )-\frac {(1-2 x)^{3/2} (5 x+3)^3}{9 (3 x+2)^3}\)

\(\Big \downarrow \) 166

\(\displaystyle \frac {1}{3} \left (-\frac {1}{21} \int \frac {(1163-7265 x) (5 x+3)}{\sqrt {1-2 x} (3 x+2)}dx+\frac {6 \sqrt {1-2 x} (5 x+3)^3}{(3 x+2)^2}+\frac {251 \sqrt {1-2 x} (5 x+3)^2}{21 (3 x+2)}\right )-\frac {(1-2 x)^{3/2} (5 x+3)^3}{9 (3 x+2)^3}\)

\(\Big \downarrow \) 164

\(\displaystyle \frac {1}{3} \left (\frac {1}{21} \left (\frac {18019}{9} \int \frac {1}{\sqrt {1-2 x} (3 x+2)}dx-\frac {5}{9} \sqrt {1-2 x} (7265 x+2323)\right )+\frac {6 \sqrt {1-2 x} (5 x+3)^3}{(3 x+2)^2}+\frac {251 \sqrt {1-2 x} (5 x+3)^2}{21 (3 x+2)}\right )-\frac {(1-2 x)^{3/2} (5 x+3)^3}{9 (3 x+2)^3}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {1}{3} \left (\frac {1}{21} \left (-\frac {18019}{9} \int \frac {1}{\frac {7}{2}-\frac {3}{2} (1-2 x)}d\sqrt {1-2 x}-\frac {5}{9} \sqrt {1-2 x} (7265 x+2323)\right )+\frac {6 \sqrt {1-2 x} (5 x+3)^3}{(3 x+2)^2}+\frac {251 \sqrt {1-2 x} (5 x+3)^2}{21 (3 x+2)}\right )-\frac {(1-2 x)^{3/2} (5 x+3)^3}{9 (3 x+2)^3}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{3} \left (\frac {1}{21} \left (-\frac {36038 \text {arctanh}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{9 \sqrt {21}}-\frac {5}{9} \sqrt {1-2 x} (7265 x+2323)\right )+\frac {6 \sqrt {1-2 x} (5 x+3)^3}{(3 x+2)^2}+\frac {251 \sqrt {1-2 x} (5 x+3)^2}{21 (3 x+2)}\right )-\frac {(1-2 x)^{3/2} (5 x+3)^3}{9 (3 x+2)^3}\)

input
Int[((1 - 2*x)^(3/2)*(3 + 5*x)^3)/(2 + 3*x)^4,x]
 
output
-1/9*((1 - 2*x)^(3/2)*(3 + 5*x)^3)/(2 + 3*x)^3 + ((251*Sqrt[1 - 2*x]*(3 + 
5*x)^2)/(21*(2 + 3*x)) + (6*Sqrt[1 - 2*x]*(3 + 5*x)^3)/(2 + 3*x)^2 + ((-5* 
Sqrt[1 - 2*x]*(2323 + 7265*x))/9 - (36038*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]] 
)/(9*Sqrt[21]))/21)/3
 

3.19.90.3.1 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 108
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Simp[(a + b*x)^(m + 1)*(c + d*x)^n*((e + f*x)^p/(b*(m + 1))) 
, x] - Simp[1/(b*(m + 1))   Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f* 
x)^(p - 1)*Simp[d*e*n + c*f*p + d*f*(n + p)*x, x], x], x] /; FreeQ[{a, b, c 
, d, e, f}, x] && LtQ[m, -1] && GtQ[n, 0] && GtQ[p, 0] && (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 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])
 
3.19.90.4 Maple [A] (verified)

Time = 1.00 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.49

method result size
risch \(\frac {63000 x^{5}-195300 x^{4}-437328 x^{3}-138872 x^{2}+103365 x +47939}{567 \left (2+3 x \right )^{3} \sqrt {1-2 x}}-\frac {36038 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{11907}\) \(61\)
pseudoelliptic \(\frac {-36038 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \left (2+3 x \right )^{3} \sqrt {21}-21 \sqrt {1-2 x}\, \left (31500 x^{4}-81900 x^{3}-259614 x^{2}-199243 x -47939\right )}{11907 \left (2+3 x \right )^{3}}\) \(65\)
derivativedivides \(\frac {250 \left (1-2 x \right )^{\frac {3}{2}}}{243}+\frac {2050 \sqrt {1-2 x}}{243}+\frac {-\frac {7876 \left (1-2 x \right )^{\frac {5}{2}}}{189}+\frac {46612 \left (1-2 x \right )^{\frac {3}{2}}}{243}-\frac {53648 \sqrt {1-2 x}}{243}}{\left (-4-6 x \right )^{3}}-\frac {36038 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{11907}\) \(75\)
default \(\frac {250 \left (1-2 x \right )^{\frac {3}{2}}}{243}+\frac {2050 \sqrt {1-2 x}}{243}+\frac {-\frac {7876 \left (1-2 x \right )^{\frac {5}{2}}}{189}+\frac {46612 \left (1-2 x \right )^{\frac {3}{2}}}{243}-\frac {53648 \sqrt {1-2 x}}{243}}{\left (-4-6 x \right )^{3}}-\frac {36038 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{11907}\) \(75\)
trager \(-\frac {\left (31500 x^{4}-81900 x^{3}-259614 x^{2}-199243 x -47939\right ) \sqrt {1-2 x}}{567 \left (2+3 x \right )^{3}}+\frac {18019 \operatorname {RootOf}\left (\textit {\_Z}^{2}-21\right ) \ln \left (\frac {3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-21\right ) x -5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-21\right )+21 \sqrt {1-2 x}}{2+3 x}\right )}{11907}\) \(82\)

input
int((1-2*x)^(3/2)*(3+5*x)^3/(2+3*x)^4,x,method=_RETURNVERBOSE)
 
output
1/567*(63000*x^5-195300*x^4-437328*x^3-138872*x^2+103365*x+47939)/(2+3*x)^ 
3/(1-2*x)^(1/2)-36038/11907*arctanh(1/7*21^(1/2)*(1-2*x)^(1/2))*21^(1/2)
 
3.19.90.5 Fricas [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.75 \[ \int \frac {(1-2 x)^{3/2} (3+5 x)^3}{(2+3 x)^4} \, dx=\frac {18019 \, \sqrt {21} {\left (27 \, x^{3} + 54 \, x^{2} + 36 \, x + 8\right )} \log \left (\frac {3 \, x + \sqrt {21} \sqrt {-2 \, x + 1} - 5}{3 \, x + 2}\right ) - 21 \, {\left (31500 \, x^{4} - 81900 \, x^{3} - 259614 \, x^{2} - 199243 \, x - 47939\right )} \sqrt {-2 \, x + 1}}{11907 \, {\left (27 \, x^{3} + 54 \, x^{2} + 36 \, x + 8\right )}} \]

input
integrate((1-2*x)^(3/2)*(3+5*x)^3/(2+3*x)^4,x, algorithm="fricas")
 
output
1/11907*(18019*sqrt(21)*(27*x^3 + 54*x^2 + 36*x + 8)*log((3*x + sqrt(21)*s 
qrt(-2*x + 1) - 5)/(3*x + 2)) - 21*(31500*x^4 - 81900*x^3 - 259614*x^2 - 1 
99243*x - 47939)*sqrt(-2*x + 1))/(27*x^3 + 54*x^2 + 36*x + 8)
 
3.19.90.6 Sympy [F(-1)]

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

input
integrate((1-2*x)**(3/2)*(3+5*x)**3/(2+3*x)**4,x)
 
output
Timed out
 
3.19.90.7 Maxima [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.88 \[ \int \frac {(1-2 x)^{3/2} (3+5 x)^3}{(2+3 x)^4} \, dx=\frac {250}{243} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {18019}{11907} \, \sqrt {21} \log \left (-\frac {\sqrt {21} - 3 \, \sqrt {-2 \, x + 1}}{\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}}\right ) + \frac {2050}{243} \, \sqrt {-2 \, x + 1} + \frac {4 \, {\left (17721 \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} - 81571 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + 93884 \, \sqrt {-2 \, x + 1}\right )}}{1701 \, {\left (27 \, {\left (2 \, x - 1\right )}^{3} + 189 \, {\left (2 \, x - 1\right )}^{2} + 882 \, x - 98\right )}} \]

input
integrate((1-2*x)^(3/2)*(3+5*x)^3/(2+3*x)^4,x, algorithm="maxima")
 
output
250/243*(-2*x + 1)^(3/2) + 18019/11907*sqrt(21)*log(-(sqrt(21) - 3*sqrt(-2 
*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) + 2050/243*sqrt(-2*x + 1) + 4/1701 
*(17721*(-2*x + 1)^(5/2) - 81571*(-2*x + 1)^(3/2) + 93884*sqrt(-2*x + 1))/ 
(27*(2*x - 1)^3 + 189*(2*x - 1)^2 + 882*x - 98)
 
3.19.90.8 Giac [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.82 \[ \int \frac {(1-2 x)^{3/2} (3+5 x)^3}{(2+3 x)^4} \, dx=\frac {250}{243} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {18019}{11907} \, \sqrt {21} \log \left (\frac {{\left | -2 \, \sqrt {21} + 6 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}\right )}}\right ) + \frac {2050}{243} \, \sqrt {-2 \, x + 1} + \frac {17721 \, {\left (2 \, x - 1\right )}^{2} \sqrt {-2 \, x + 1} - 81571 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + 93884 \, \sqrt {-2 \, x + 1}}{3402 \, {\left (3 \, x + 2\right )}^{3}} \]

input
integrate((1-2*x)^(3/2)*(3+5*x)^3/(2+3*x)^4,x, algorithm="giac")
 
output
250/243*(-2*x + 1)^(3/2) + 18019/11907*sqrt(21)*log(1/2*abs(-2*sqrt(21) + 
6*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) + 2050/243*sqrt(-2*x + 1) 
 + 1/3402*(17721*(2*x - 1)^2*sqrt(-2*x + 1) - 81571*(-2*x + 1)^(3/2) + 938 
84*sqrt(-2*x + 1))/(3*x + 2)^3
 
3.19.90.9 Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.73 \[ \int \frac {(1-2 x)^{3/2} (3+5 x)^3}{(2+3 x)^4} \, dx=\frac {2050\,\sqrt {1-2\,x}}{243}+\frac {250\,{\left (1-2\,x\right )}^{3/2}}{243}+\frac {\frac {53648\,\sqrt {1-2\,x}}{6561}-\frac {46612\,{\left (1-2\,x\right )}^{3/2}}{6561}+\frac {7876\,{\left (1-2\,x\right )}^{5/2}}{5103}}{\frac {98\,x}{3}+7\,{\left (2\,x-1\right )}^2+{\left (2\,x-1\right )}^3-\frac {98}{27}}+\frac {\sqrt {21}\,\mathrm {atan}\left (\frac {\sqrt {21}\,\sqrt {1-2\,x}\,1{}\mathrm {i}}{7}\right )\,36038{}\mathrm {i}}{11907} \]

input
int(((1 - 2*x)^(3/2)*(5*x + 3)^3)/(3*x + 2)^4,x)
 
output
(21^(1/2)*atan((21^(1/2)*(1 - 2*x)^(1/2)*1i)/7)*36038i)/11907 + (2050*(1 - 
 2*x)^(1/2))/243 + (250*(1 - 2*x)^(3/2))/243 + ((53648*(1 - 2*x)^(1/2))/65 
61 - (46612*(1 - 2*x)^(3/2))/6561 + (7876*(1 - 2*x)^(5/2))/5103)/((98*x)/3 
 + 7*(2*x - 1)^2 + (2*x - 1)^3 - 98/27)