\(\int \frac {(2+3 x)^4}{(1-2 x)^{3/2} (3+5 x)^3} \, dx\) [781]

Optimal result

Integrand size = 24, antiderivative size = 94 \[ \int \frac {(2+3 x)^4}{(1-2 x)^{3/2} (3+5 x)^3} \, dx=\frac {2401}{2662 \sqrt {1-2 x}}+\frac {81}{250} \sqrt {1-2 x}-\frac {\sqrt {1-2 x}}{30250 (3+5 x)^2}-\frac {271 \sqrt {1-2 x}}{332750 (3+5 x)}-\frac {111 \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{1331 \sqrt {55}} \] Output:

2401/2662/(1-2*x)^(1/2)+81/250*(1-2*x)^(1/2)-1/30250*(1-2*x)^(1/2)/(3+5*x) 
^2-271*(1-2*x)^(1/2)/(998250+1663750*x)-111/73205*55^(1/2)*arctanh(1/11*55 
^(1/2)*(1-2*x)^(1/2))
 

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.67 \[ \int \frac {(2+3 x)^4}{(1-2 x)^{3/2} (3+5 x)^3} \, dx=\frac {\frac {11 \left (146824+411911 x+149298 x^2-215622 x^3\right )}{\sqrt {1-2 x} (3+5 x)^2}-222 \sqrt {55} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{146410} \] Input:

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

Output:

((11*(146824 + 411911*x + 149298*x^2 - 215622*x^3))/(Sqrt[1 - 2*x]*(3 + 5* 
x)^2) - 222*Sqrt[55]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/146410
 

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.27, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {109, 166, 27, 163, 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.

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

\(\Big \downarrow \) 109

\(\displaystyle \frac {7 (3 x+2)^3}{11 \sqrt {1-2 x} (5 x+3)^2}-\frac {1}{11} \int \frac {(3 x+2)^2 (102 x+47)}{\sqrt {1-2 x} (5 x+3)^3}dx\)

\(\Big \downarrow \) 166

\(\displaystyle \frac {1}{11} \left (-\frac {1}{110} \int \frac {3 (3 x+2) (2315 x+1212)}{\sqrt {1-2 x} (5 x+3)^2}dx-\frac {71 \sqrt {1-2 x} (3 x+2)^2}{110 (5 x+3)^2}\right )+\frac {7 (3 x+2)^3}{11 \sqrt {1-2 x} (5 x+3)^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{11} \left (-\frac {3}{110} \int \frac {(3 x+2) (2315 x+1212)}{\sqrt {1-2 x} (5 x+3)^2}dx-\frac {71 \sqrt {1-2 x} (3 x+2)^2}{110 (5 x+3)^2}\right )+\frac {7 (3 x+2)^3}{11 \sqrt {1-2 x} (5 x+3)^2}\)

\(\Big \downarrow \) 163

\(\displaystyle \frac {1}{11} \left (-\frac {3}{110} \left (-\frac {185}{11} \int \frac {1}{\sqrt {1-2 x} (5 x+3)}dx-\frac {3 \sqrt {1-2 x} (5093 x+3044)}{11 (5 x+3)}\right )-\frac {71 \sqrt {1-2 x} (3 x+2)^2}{110 (5 x+3)^2}\right )+\frac {7 (3 x+2)^3}{11 \sqrt {1-2 x} (5 x+3)^2}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {1}{11} \left (-\frac {3}{110} \left (\frac {185}{11} \int \frac {1}{\frac {11}{2}-\frac {5}{2} (1-2 x)}d\sqrt {1-2 x}-\frac {3 \sqrt {1-2 x} (5093 x+3044)}{11 (5 x+3)}\right )-\frac {71 \sqrt {1-2 x} (3 x+2)^2}{110 (5 x+3)^2}\right )+\frac {7 (3 x+2)^3}{11 \sqrt {1-2 x} (5 x+3)^2}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{11} \left (-\frac {3}{110} \left (\frac {74}{11} \sqrt {\frac {5}{11}} \text {arctanh}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )-\frac {3 \sqrt {1-2 x} (5093 x+3044)}{11 (5 x+3)}\right )-\frac {71 \sqrt {1-2 x} (3 x+2)^2}{110 (5 x+3)^2}\right )+\frac {7 (3 x+2)^3}{11 \sqrt {1-2 x} (5 x+3)^2}\)

Input:

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

Output:

(7*(2 + 3*x)^3)/(11*Sqrt[1 - 2*x]*(3 + 5*x)^2) + ((-71*Sqrt[1 - 2*x]*(2 + 
3*x)^2)/(110*(3 + 5*x)^2) - (3*((-3*Sqrt[1 - 2*x]*(3044 + 5093*x))/(11*(3 
+ 5*x)) + (74*Sqrt[5/11]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/11))/110)/11
 

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 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 163

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

Maple [A] (verified)

Time = 0.25 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.54

method	result	size
risch	\(-\frac {215622 x^{3}-149298 x^{2}-411911 x -146824}{13310 \left (3+5 x \right )^{2} \sqrt {1-2 x}}-\frac {111 \sqrt {55}\, \operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right )}{73205}\)	\(51\)
pseudoelliptic	\(-\frac {111 \left (-\frac {807532}{111}+\sqrt {1-2 x}\, \operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \left (3+5 x \right )^{2} \sqrt {55}+\frac {395307 x^{3}}{37}-\frac {273713 x^{2}}{37}-\frac {4531021 x}{222}\right )}{73205 \sqrt {1-2 x}\, \left (3+5 x \right )^{2}}\)	\(63\)
derivativedivides	\(\frac {81 \sqrt {1-2 x}}{250}+\frac {2401}{2662 \sqrt {1-2 x}}+\frac {\frac {271 \left (1-2 x \right )^{\frac {3}{2}}}{33275}-\frac {273 \sqrt {1-2 x}}{15125}}{\left (-6-10 x \right )^{2}}-\frac {111 \sqrt {55}\, \operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right )}{73205}\)	\(66\)
default	\(\frac {81 \sqrt {1-2 x}}{250}+\frac {2401}{2662 \sqrt {1-2 x}}+\frac {\frac {271 \left (1-2 x \right )^{\frac {3}{2}}}{33275}-\frac {273 \sqrt {1-2 x}}{15125}}{\left (-6-10 x \right )^{2}}-\frac {111 \sqrt {55}\, \operatorname {arctanh}\left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right )}{73205}\)	\(66\)
trager	\(\frac {\left (215622 x^{3}-149298 x^{2}-411911 x -146824\right ) \sqrt {1-2 x}}{13310 \left (3+5 x \right )^{2} \left (-1+2 x \right )}-\frac {111 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) \ln \left (\frac {-5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right ) x +55 \sqrt {1-2 x}+8 \operatorname {RootOf}\left (\textit {\_Z}^{2}-55\right )}{3+5 x}\right )}{146410}\)	\(84\)

Input:

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

Output:

-1/13310*(215622*x^3-149298*x^2-411911*x-146824)/(3+5*x)^2/(1-2*x)^(1/2)-1 
11/73205*55^(1/2)*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.95 \[ \int \frac {(2+3 x)^4}{(1-2 x)^{3/2} (3+5 x)^3} \, dx=\frac {111 \, \sqrt {55} {\left (50 \, x^{3} + 35 \, x^{2} - 12 \, x - 9\right )} \log \left (\frac {5 \, x + \sqrt {55} \sqrt {-2 \, x + 1} - 8}{5 \, x + 3}\right ) + 11 \, {\left (215622 \, x^{3} - 149298 \, x^{2} - 411911 \, x - 146824\right )} \sqrt {-2 \, x + 1}}{146410 \, {\left (50 \, x^{3} + 35 \, x^{2} - 12 \, x - 9\right )}} \] Input:

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

Output:

1/146410*(111*sqrt(55)*(50*x^3 + 35*x^2 - 12*x - 9)*log((5*x + sqrt(55)*sq 
rt(-2*x + 1) - 8)/(5*x + 3)) + 11*(215622*x^3 - 149298*x^2 - 411911*x - 14 
6824)*sqrt(-2*x + 1))/(50*x^3 + 35*x^2 - 12*x - 9)
 

Sympy [F(-1)]

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

integrate((2+3*x)**4/(1-2*x)**(3/2)/(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 \[ \int \frac {(2+3 x)^4}{(1-2 x)^{3/2} (3+5 x)^3} \, dx=\frac {111}{146410} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) + \frac {81}{250} \, \sqrt {-2 \, x + 1} + \frac {7505835 \, {\left (2 \, x - 1\right )}^{2} + 66039512 \, x + 3295369}{332750 \, {\left (25 \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} - 110 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + 121 \, \sqrt {-2 \, x + 1}\right )}} \] Input:

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

Output:

111/146410*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt( 
-2*x + 1))) + 81/250*sqrt(-2*x + 1) + 1/332750*(7505835*(2*x - 1)^2 + 6603 
9512*x + 3295369)/(25*(-2*x + 1)^(5/2) - 110*(-2*x + 1)^(3/2) + 121*sqrt(- 
2*x + 1))
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.91 \[ \int \frac {(2+3 x)^4}{(1-2 x)^{3/2} (3+5 x)^3} \, dx=\frac {111}{146410} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) + \frac {81}{250} \, \sqrt {-2 \, x + 1} + \frac {2401}{2662 \, \sqrt {-2 \, x + 1}} + \frac {1355 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 3003 \, \sqrt {-2 \, x + 1}}{665500 \, {\left (5 \, x + 3\right )}^{2}} \] Input:

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

Output:

111/146410*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) 
 + 5*sqrt(-2*x + 1))) + 81/250*sqrt(-2*x + 1) + 2401/2662/sqrt(-2*x + 1) + 
 1/665500*(1355*(-2*x + 1)^(3/2) - 3003*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 \[ \int \frac {(2+3 x)^4}{(1-2 x)^{3/2} (3+5 x)^3} \, dx=\frac {81\,\sqrt {1-2\,x}}{250}-\frac {111\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )}{73205}+\frac {\frac {3001796\,x}{378125}+\frac {1501167\,{\left (2\,x-1\right )}^2}{1663750}+\frac {299579}{756250}}{\frac {121\,\sqrt {1-2\,x}}{25}-\frac {22\,{\left (1-2\,x\right )}^{3/2}}{5}+{\left (1-2\,x\right )}^{5/2}} \] Input:

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

Output:

(81*(1 - 2*x)^(1/2))/250 - (111*55^(1/2)*atanh((55^(1/2)*(1 - 2*x)^(1/2))/ 
11))/73205 + ((3001796*x)/378125 + (1501167*(2*x - 1)^2)/1663750 + 299579/ 
756250)/((121*(1 - 2*x)^(1/2))/25 - (22*(1 - 2*x)^(3/2))/5 + (1 - 2*x)^(5/ 
2))
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.95 \[ \int \frac {(2+3 x)^4}{(1-2 x)^{3/2} (3+5 x)^3} \, dx=\frac {2775 \sqrt {-2 x +1}\, \sqrt {55}\, \mathrm {log}\left (5 \sqrt {-2 x +1}-\sqrt {55}\right ) x^{2}+3330 \sqrt {-2 x +1}\, \sqrt {55}\, \mathrm {log}\left (5 \sqrt {-2 x +1}-\sqrt {55}\right ) x +999 \sqrt {-2 x +1}\, \sqrt {55}\, \mathrm {log}\left (5 \sqrt {-2 x +1}-\sqrt {55}\right )-2775 \sqrt {-2 x +1}\, \sqrt {55}\, \mathrm {log}\left (5 \sqrt {-2 x +1}+\sqrt {55}\right ) x^{2}-3330 \sqrt {-2 x +1}\, \sqrt {55}\, \mathrm {log}\left (5 \sqrt {-2 x +1}+\sqrt {55}\right ) x -999 \sqrt {-2 x +1}\, \sqrt {55}\, \mathrm {log}\left (5 \sqrt {-2 x +1}+\sqrt {55}\right )-2371842 x^{3}+1642278 x^{2}+4531021 x +1615064}{146410 \sqrt {-2 x +1}\, \left (25 x^{2}+30 x +9\right )} \] Input:

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

Output:

(2775*sqrt( - 2*x + 1)*sqrt(55)*log(5*sqrt( - 2*x + 1) - sqrt(55))*x**2 + 
3330*sqrt( - 2*x + 1)*sqrt(55)*log(5*sqrt( - 2*x + 1) - sqrt(55))*x + 999* 
sqrt( - 2*x + 1)*sqrt(55)*log(5*sqrt( - 2*x + 1) - sqrt(55)) - 2775*sqrt( 
- 2*x + 1)*sqrt(55)*log(5*sqrt( - 2*x + 1) + sqrt(55))*x**2 - 3330*sqrt( - 
 2*x + 1)*sqrt(55)*log(5*sqrt( - 2*x + 1) + sqrt(55))*x - 999*sqrt( - 2*x 
+ 1)*sqrt(55)*log(5*sqrt( - 2*x + 1) + sqrt(55)) - 2371842*x**3 + 1642278* 
x**2 + 4531021*x + 1615064)/(146410*sqrt( - 2*x + 1)*(25*x**2 + 30*x + 9))