\(\int \frac {(2+3 x)^3 (3+5 x)^{5/2}}{(1-2 x)^{5/2}} \, dx\) [1208]

Optimal result

Integrand size = 26, antiderivative size = 160 \[ \int \frac {(2+3 x)^3 (3+5 x)^{5/2}}{(1-2 x)^{5/2}} \, dx=-\frac {9444023 \sqrt {1-2 x} \sqrt {3+5 x}}{4096}-\frac {9444023 \sqrt {1-2 x} (3+5 x)^{3/2}}{33792}-\frac {21413 (3+5 x)^{5/2}}{264 \sqrt {1-2 x}}-\frac {603}{128} \sqrt {1-2 x} (3+5 x)^{5/2}+\frac {27}{64} (1-2 x)^{3/2} (3+5 x)^{5/2}+\frac {343 (3+5 x)^{7/2}}{132 (1-2 x)^{3/2}}+\frac {103884253 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )}{4096 \sqrt {10}} \] Output:

-9444023/4096*(1-2*x)^(1/2)*(3+5*x)^(1/2)-9444023/33792*(1-2*x)^(1/2)*(3+5 
*x)^(3/2)-21413/264*(3+5*x)^(5/2)/(1-2*x)^(1/2)-603/128*(1-2*x)^(1/2)*(3+5 
*x)^(5/2)+27/64*(1-2*x)^(3/2)*(3+5*x)^(5/2)+343/132*(3+5*x)^(7/2)/(1-2*x)^ 
(3/2)[phone]/40960*arcsin(1/11*22^(1/2)*(3+5*x)^(1/2))*10^(1/2)
 

Mathematica [A] (verified)

Time = 0.56 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.54 \[ \int \frac {(2+3 x)^3 (3+5 x)^{5/2}}{(1-2 x)^{5/2}} \, dx=\frac {-\frac {5 \sqrt {3+5 x} \left (47216961-129940960 x+40614996 x^2+15301008 x^3+5477760 x^4+1036800 x^5\right )}{(1-2 x)^{3/2}}-311652759 \sqrt {10} \arctan \left (\frac {\sqrt {6+10 x}}{\sqrt {11}-\sqrt {5-10 x}}\right )}{61440} \] Input:

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

Output:

((-5*Sqrt[3 + 5*x]*(47216961 - 129940960*x + 40614996*x^2 + 15301008*x^3 + 
 5477760*x^4 + 1036800*x^5))/(1 - 2*x)^(3/2) - 311652759*Sqrt[10]*ArcTan[S 
qrt[6 + 10*x]/(Sqrt[11] - Sqrt[5 - 10*x])])/61440
 

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.11, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.346, Rules used = {108, 27, 167, 27, 164, 60, 60, 64, 223}

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.

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

\(\Big \downarrow \) 108

\(\displaystyle \frac {(3 x+2)^3 (5 x+3)^{5/2}}{3 (1-2 x)^{3/2}}-\frac {1}{3} \int \frac {(3 x+2)^2 (5 x+3)^{3/2} (165 x+104)}{2 (1-2 x)^{3/2}}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {(3 x+2)^3 (5 x+3)^{5/2}}{3 (1-2 x)^{3/2}}-\frac {1}{6} \int \frac {(3 x+2)^2 (5 x+3)^{3/2} (165 x+104)}{(1-2 x)^{3/2}}dx\)

\(\Big \downarrow \) 167

\(\displaystyle \frac {1}{6} \left (-\frac {1}{11} \int -\frac {(3 x+2) (5 x+3)^{3/2} (50205 x+31978)}{2 \sqrt {1-2 x}}dx-\frac {373 (3 x+2)^2 (5 x+3)^{5/2}}{11 \sqrt {1-2 x}}\right )+\frac {(5 x+3)^{5/2} (3 x+2)^3}{3 (1-2 x)^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{6} \left (\frac {1}{22} \int \frac {(3 x+2) (5 x+3)^{3/2} (50205 x+31978)}{\sqrt {1-2 x}}dx-\frac {373 (3 x+2)^2 (5 x+3)^{5/2}}{11 \sqrt {1-2 x}}\right )+\frac {(5 x+3)^{5/2} (3 x+2)^3}{3 (1-2 x)^{3/2}}\)

\(\Big \downarrow \) 164

\(\displaystyle \frac {1}{6} \left (\frac {1}{22} \left (\frac {9444023}{64} \int \frac {(5 x+3)^{3/2}}{\sqrt {1-2 x}}dx-\frac {3}{32} \sqrt {1-2 x} (5 x+3)^{5/2} (40164 x+81191)\right )-\frac {373 (3 x+2)^2 (5 x+3)^{5/2}}{11 \sqrt {1-2 x}}\right )+\frac {(5 x+3)^{5/2} (3 x+2)^3}{3 (1-2 x)^{3/2}}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {1}{6} \left (\frac {1}{22} \left (\frac {9444023}{64} \left (\frac {33}{8} \int \frac {\sqrt {5 x+3}}{\sqrt {1-2 x}}dx-\frac {1}{4} \sqrt {1-2 x} (5 x+3)^{3/2}\right )-\frac {3}{32} \sqrt {1-2 x} (5 x+3)^{5/2} (40164 x+81191)\right )-\frac {373 (3 x+2)^2 (5 x+3)^{5/2}}{11 \sqrt {1-2 x}}\right )+\frac {(5 x+3)^{5/2} (3 x+2)^3}{3 (1-2 x)^{3/2}}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {1}{6} \left (\frac {1}{22} \left (\frac {9444023}{64} \left (\frac {33}{8} \left (\frac {11}{4} \int \frac {1}{\sqrt {1-2 x} \sqrt {5 x+3}}dx-\frac {1}{2} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {1}{4} \sqrt {1-2 x} (5 x+3)^{3/2}\right )-\frac {3}{32} \sqrt {1-2 x} (5 x+3)^{5/2} (40164 x+81191)\right )-\frac {373 (3 x+2)^2 (5 x+3)^{5/2}}{11 \sqrt {1-2 x}}\right )+\frac {(5 x+3)^{5/2} (3 x+2)^3}{3 (1-2 x)^{3/2}}\)

\(\Big \downarrow \) 64

\(\displaystyle \frac {1}{6} \left (\frac {1}{22} \left (\frac {9444023}{64} \left (\frac {33}{8} \left (\frac {11}{10} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}-\frac {1}{2} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {1}{4} \sqrt {1-2 x} (5 x+3)^{3/2}\right )-\frac {3}{32} \sqrt {1-2 x} (5 x+3)^{5/2} (40164 x+81191)\right )-\frac {373 (3 x+2)^2 (5 x+3)^{5/2}}{11 \sqrt {1-2 x}}\right )+\frac {(5 x+3)^{5/2} (3 x+2)^3}{3 (1-2 x)^{3/2}}\)

\(\Big \downarrow \) 223

\(\displaystyle \frac {1}{6} \left (\frac {1}{22} \left (\frac {9444023}{64} \left (\frac {33}{8} \left (\frac {11 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )}{2 \sqrt {10}}-\frac {1}{2} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {1}{4} \sqrt {1-2 x} (5 x+3)^{3/2}\right )-\frac {3}{32} \sqrt {1-2 x} (5 x+3)^{5/2} (40164 x+81191)\right )-\frac {373 (3 x+2)^2 (5 x+3)^{5/2}}{11 \sqrt {1-2 x}}\right )+\frac {(5 x+3)^{5/2} (3 x+2)^3}{3 (1-2 x)^{3/2}}\)

Input:

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

Output:

((2 + 3*x)^3*(3 + 5*x)^(5/2))/(3*(1 - 2*x)^(3/2)) + ((-373*(2 + 3*x)^2*(3 
+ 5*x)^(5/2))/(11*Sqrt[1 - 2*x]) + ((-3*Sqrt[1 - 2*x]*(3 + 5*x)^(5/2)*(811 
91 + 40164*x))/32 + (9444023*(-1/4*(Sqrt[1 - 2*x]*(3 + 5*x)^(3/2)) + (33*( 
-1/2*(Sqrt[1 - 2*x]*Sqrt[3 + 5*x]) + (11*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]]) 
/(2*Sqrt[10])))/8))/64)/22)/6
 

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 60

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

rule 64

Int[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_.) + (d_.)*(x_)]), x_Symbol] :> Simp 
[2/b   Subst[Int[1/Sqrt[c - a*(d/b) + d*(x^2/b)], x], x, Sqrt[a + b*x]], x] 
 /; FreeQ[{a, b, c, d}, x] && GtQ[c - a*(d/b), 0] && ( !GtQ[a - c*(b/d), 0] 
 || PosQ[b])
 

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 167

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] && LtQ[m, -1] && GtQ[n, 0] && IntegersQ[2*m, 2*n, 2*p]
 

rule 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt 
[a])]/Rt[-b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[b]
 

Maple [A] (verified)

Time = 0.25 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.02

method	result	size
default	\(\frac {\left (-20736000 x^{5} \sqrt {-10 x^{2}-x +3}-109555200 x^{4} \sqrt {-10 x^{2}-x +3}[phone] \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x^{2}-306020160 x^{3} \sqrt {-10 x^{2}-x +3}-1246611036 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x -812299920 x^{2} \sqrt {-10 x^{2}-x +3}[phone] \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )[phone] x \sqrt {-10 x^{2}-x +3}-944339220 \sqrt {-10 x^{2}-x +3}\right ) \sqrt {3+5 x}}{245760 \left (1-2 x \right )^{\frac {3}{2}} \sqrt {-10 x^{2}-x +3}}\)	\(164\)

Input:

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

Output:

1/245760*(-20736000*x^5*(-10*x^2-x+3)^(1/2)-109555200*x^4*(-10*x^2-x+3)^(1 
/2)[phone]*10^(1/2)*arcsin(20/11*x+1/11)*x^2-306020160*x^3*(-10*x^2-x+ 
3)^(1/2)-1246611036*10^(1/2)*arcsin(20/11*x+1/11)*x-812299920*x^2*(-10*x^2 
-x+3)^(1/2)[phone]*10^(1/2)*arcsin(20/11*x+1/11)[phone]*x*(-10*x^2- 
x+3)^(1/2)-944339220*(-10*x^2-x+3)^(1/2))*(3+5*x)^(1/2)/(1-2*x)^(3/2)/(-10 
*x^2-x+3)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.66 \[ \int \frac {(2+3 x)^3 (3+5 x)^{5/2}}{(1-2 x)^{5/2}} \, dx=-\frac {311652759 \, \sqrt {10} {\left (4 \, x^{2} - 4 \, x + 1\right )} \arctan \left (\frac {\sqrt {10} {\left (20 \, x + 1\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{20 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) + 20 \, {\left (1036800 \, x^{5} + 5477760 \, x^{4} + 15301008 \, x^{3} + 40614996 \, x^{2} - 129940960 \, x + 47216961\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{245760 \, {\left (4 \, x^{2} - 4 \, x + 1\right )}} \] Input:

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

Output:

-1/245760*(311652759*sqrt(10)*(4*x^2 - 4*x + 1)*arctan(1/20*sqrt(10)*(20*x 
 + 1)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x - 3)) + 20*(1036800*x^5 + 5 
477760*x^4 + 15301008*x^3 + 40614996*x^2 - 129940960*x + 47216961)*sqrt(5* 
x + 3)*sqrt(-2*x + 1))/(4*x^2 - 4*x + 1)
 

Sympy [F]

\[ \int \frac {(2+3 x)^3 (3+5 x)^{5/2}}{(1-2 x)^{5/2}} \, dx=\int \frac {\left (3 x + 2\right )^{3} \left (5 x + 3\right )^{\frac {5}{2}}}{\left (1 - 2 x\right )^{\frac {5}{2}}}\, dx \] Input:

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

Output:

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

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.13 (sec) , antiderivative size = 325, normalized size of antiderivative = 2.03 \[ \int \frac {(2+3 x)^3 (3+5 x)^{5/2}}{(1-2 x)^{5/2}} \, dx=\frac {2606989}{2048} \, \sqrt {5} \sqrt {2} \arcsin \left (\frac {20}{11} \, x + \frac {1}{11}\right ) + \frac {395307}{81920} i \, \sqrt {5} \sqrt {2} \arcsin \left (\frac {20}{11} \, x - \frac {21}{11}\right ) + \frac {495}{256} \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}} - \frac {343 \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {5}{2}}}{16 \, {\left (16 \, x^{4} - 32 \, x^{3} + 24 \, x^{2} - 8 \, x + 1\right )}} - \frac {441 \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {5}{2}}}{32 \, {\left (8 \, x^{3} - 12 \, x^{2} + 6 \, x - 1\right )}} - \frac {63 \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {5}{2}}}{16 \, {\left (4 \, x^{2} - 4 \, x + 1\right )}} - \frac {27 \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {5}{2}}}{64 \, {\left (2 \, x - 1\right )}} - \frac {16335}{1024} \, \sqrt {10 \, x^{2} - 21 \, x + 8} x + \frac {68607}{4096} \, \sqrt {10 \, x^{2} - 21 \, x + 8} - \frac {114345}{512} \, \sqrt {-10 \, x^{2} - x + 3} - \frac {18865 \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}}}{192 \, {\left (8 \, x^{3} - 12 \, x^{2} + 6 \, x - 1\right )}} + \frac {24255 \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}}}{128 \, {\left (4 \, x^{2} - 4 \, x + 1\right )}} + \frac {3465 \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}}}{128 \, {\left (2 \, x - 1\right )}} + \frac {207515 \, \sqrt {-10 \, x^{2} - x + 3}}{384 \, {\left (4 \, x^{2} - 4 \, x + 1\right )}} + \frac {3721795 \, \sqrt {-10 \, x^{2} - x + 3}}{768 \, {\left (2 \, x - 1\right )}} \] Input:

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

Output:

2606989/2048*sqrt(5)*sqrt(2)*arcsin(20/11*x + 1/11) + 395307/81920*I*sqrt( 
5)*sqrt(2)*arcsin(20/11*x - 21/11) + 495/256*(-10*x^2 - x + 3)^(3/2) - 343 
/16*(-10*x^2 - x + 3)^(5/2)/(16*x^4 - 32*x^3 + 24*x^2 - 8*x + 1) - 441/32* 
(-10*x^2 - x + 3)^(5/2)/(8*x^3 - 12*x^2 + 6*x - 1) - 63/16*(-10*x^2 - x + 
3)^(5/2)/(4*x^2 - 4*x + 1) - 27/64*(-10*x^2 - x + 3)^(5/2)/(2*x - 1) - 163 
35/1024*sqrt(10*x^2 - 21*x + 8)*x + 68607/4096*sqrt(10*x^2 - 21*x + 8) - 1 
14345/512*sqrt(-10*x^2 - x + 3) - 18865/192*(-10*x^2 - x + 3)^(3/2)/(8*x^3 
 - 12*x^2 + 6*x - 1) + 24255/128*(-10*x^2 - x + 3)^(3/2)/(4*x^2 - 4*x + 1) 
 + 3465/128*(-10*x^2 - x + 3)^(3/2)/(2*x - 1) + 207515/384*sqrt(-10*x^2 - 
x + 3)/(4*x^2 - 4*x + 1) + 3721795/768*sqrt(-10*x^2 - x + 3)/(2*x - 1)
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.69 \[ \int \frac {(2+3 x)^3 (3+5 x)^{5/2}}{(1-2 x)^{5/2}} \, dx=\frac {103884253}{40960} \, \sqrt {10} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right ) - \frac {{\left (4 \, {\left (3 \, {\left (36 \, {\left (8 \, {\left (12 \, \sqrt {5} {\left (5 \, x + 3\right )} + 137 \, \sqrt {5}\right )} {\left (5 \, x + 3\right )} + 13627 \, \sqrt {5}\right )} {\left (5 \, x + 3\right )} + 9444023 \, \sqrt {5}\right )} {\left (5 \, x + 3\right )} - 1038842530 \, \sqrt {5}\right )} {\left (5 \, x + 3\right )} + 17140901745 \, \sqrt {5}\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5}}{7680000 \, {\left (2 \, x - 1\right )}^{2}} \] Input:

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

Output:

103884253/40960*sqrt(10)*arcsin(1/11*sqrt(22)*sqrt(5*x + 3)) - 1/7680000*( 
4*(3*(36*(8*(12*sqrt(5)*(5*x + 3) + 137*sqrt(5))*(5*x + 3) + 13627*sqrt(5) 
)*(5*x + 3) + 9444023*sqrt(5))*(5*x + 3) - 1038842530*sqrt(5))*(5*x + 3) + 
 17140901745*sqrt(5))*sqrt(5*x + 3)*sqrt(-10*x + 5)/(2*x - 1)^2
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.80 \[ \int \frac {(2+3 x)^3 (3+5 x)^{5/2}}{(1-2 x)^{5/2}} \, dx=\frac {-623305518 \sqrt {-2 x +1}\, \sqrt {10}\, \mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right ) x [phone] \sqrt {-2 x +1}\, \sqrt {10}\, \mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )[phone] \sqrt {5 x +3}\, x^{5}[phone] \sqrt {5 x +3}\, x^{4}[phone] \sqrt {5 x +3}\, x^{3}[phone] \sqrt {5 x +3}\, x^{2}-1299409600 \sqrt {5 x +3}\, x [phone] \sqrt {5 x +3}}{122880 \sqrt {-2 x +1}\, \left (2 x -1\right )} \] Input:

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

Output:

( - 623305518*sqrt( - 2*x + 1)*sqrt(10)*asin((sqrt( - 2*x + 1)*sqrt(5))/sq 
rt(11))*x + 311652759*sqrt( - 2*x + 1)*sqrt(10)*asin((sqrt( - 2*x + 1)*sqr 
t(5))/sqrt(11)) + 10368000*sqrt(5*x + 3)*x**5 + 54777600*sqrt(5*x + 3)*x** 
4 + 153010080*sqrt(5*x + 3)*x**3 + 406149960*sqrt(5*x + 3)*x**2 - 12994096 
00*sqrt(5*x + 3)*x + 472169610*sqrt(5*x + 3))/(122880*sqrt( - 2*x + 1)*(2* 
x - 1))