3.2405 \(\int \frac {(1-2 x)^{5/2} (3+5 x)^{3/2}}{(2+3 x)^3} \, dx\)

Optimal. Leaf size=164 \[ -\frac {(5 x+3)^{3/2} (1-2 x)^{5/2}}{6 (3 x+2)^2}+\frac {115 (5 x+3)^{3/2} (1-2 x)^{3/2}}{36 (3 x+2)}+\frac {41}{18} (5 x+3)^{3/2} \sqrt {1-2 x}-\frac {1649}{108} \sqrt {5 x+3} \sqrt {1-2 x}-\frac {6829 \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )}{162 \sqrt {10}}-\frac {1945}{324} \sqrt {7} \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right ) \]

[Out]

-1/6*(1-2*x)^(5/2)*(3+5*x)^(3/2)/(2+3*x)^2+115/36*(1-2*x)^(3/2)*(3+5*x)^(3/2)/(2+3*x)-1945/324*arctan(1/7*(1-2
*x)^(1/2)*7^(1/2)/(3+5*x)^(1/2))*7^(1/2)-6829/1620*arcsin(1/11*22^(1/2)*(3+5*x)^(1/2))*10^(1/2)+41/18*(3+5*x)^
(3/2)*(1-2*x)^(1/2)-1649/108*(1-2*x)^(1/2)*(3+5*x)^(1/2)

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Rubi [A]  time = 0.07, antiderivative size = 164, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {97, 149, 154, 157, 54, 216, 93, 204} \[ -\frac {(5 x+3)^{3/2} (1-2 x)^{5/2}}{6 (3 x+2)^2}+\frac {115 (5 x+3)^{3/2} (1-2 x)^{3/2}}{36 (3 x+2)}+\frac {41}{18} (5 x+3)^{3/2} \sqrt {1-2 x}-\frac {1649}{108} \sqrt {5 x+3} \sqrt {1-2 x}-\frac {6829 \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )}{162 \sqrt {10}}-\frac {1945}{324} \sqrt {7} \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-1649*Sqrt[1 - 2*x]*Sqrt[3 + 5*x])/108 + (41*Sqrt[1 - 2*x]*(3 + 5*x)^(3/2))/18 - ((1 - 2*x)^(5/2)*(3 + 5*x)^(
3/2))/(6*(2 + 3*x)^2) + (115*(1 - 2*x)^(3/2)*(3 + 5*x)^(3/2))/(36*(2 + 3*x)) - (6829*ArcSin[Sqrt[2/11]*Sqrt[3
+ 5*x]])/(162*Sqrt[10]) - (1945*Sqrt[7]*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/324

Rule 54

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

Rule 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 97

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((a + b
*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p)/(b*(m + 1)), x] - Dist[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 149

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> 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] - Dist[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] /; Free
Q[{a, b, c, d, e, f, g, h, p}, x] && LtQ[m, -1] && GtQ[n, 0] && IntegerQ[m]

Rule 154

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[(h*(a + b*x)^m*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d*f*(m + n + p + 2)), x] + Dist[1/(d*f*(m + n
+ p + 2)), Int[(a + b*x)^(m - 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*g*(m + n + p + 2) - h*(b*c*e*m + a*(d*e*(
n + 1) + c*f*(p + 1))) + (b*d*f*g*(m + n + p + 2) + h*(a*d*f*m - b*(d*e*(m + n + 1) + c*f*(m + p + 1))))*x, x]
, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && GtQ[m, 0] && NeQ[m + n + p + 2, 0] && IntegersQ[2*m, 2
*n, 2*p]

Rule 157

Int[(((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/((a_.) + (b_.)*(x_)), x_Symbol]
:> Dist[h/b, Int[(c + d*x)^n*(e + f*x)^p, x], x] + Dist[(b*g - a*h)/b, Int[((c + d*x)^n*(e + f*x)^p)/(a + b*x
), x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 216

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]

Rubi steps

\begin {align*} \int \frac {(1-2 x)^{5/2} (3+5 x)^{3/2}}{(2+3 x)^3} \, dx &=-\frac {(1-2 x)^{5/2} (3+5 x)^{3/2}}{6 (2+3 x)^2}+\frac {1}{6} \int \frac {\left (-\frac {15}{2}-40 x\right ) (1-2 x)^{3/2} \sqrt {3+5 x}}{(2+3 x)^2} \, dx\\ &=-\frac {(1-2 x)^{5/2} (3+5 x)^{3/2}}{6 (2+3 x)^2}+\frac {115 (1-2 x)^{3/2} (3+5 x)^{3/2}}{36 (2+3 x)}-\frac {1}{18} \int \frac {\left (-\frac {1335}{4}-1230 x\right ) \sqrt {1-2 x} \sqrt {3+5 x}}{2+3 x} \, dx\\ &=\frac {41}{18} \sqrt {1-2 x} (3+5 x)^{3/2}-\frac {(1-2 x)^{5/2} (3+5 x)^{3/2}}{6 (2+3 x)^2}+\frac {115 (1-2 x)^{3/2} (3+5 x)^{3/2}}{36 (2+3 x)}-\frac {1}{540} \int \frac {\left (\frac {2115}{2}-49470 x\right ) \sqrt {3+5 x}}{\sqrt {1-2 x} (2+3 x)} \, dx\\ &=-\frac {1649}{108} \sqrt {1-2 x} \sqrt {3+5 x}+\frac {41}{18} \sqrt {1-2 x} (3+5 x)^{3/2}-\frac {(1-2 x)^{5/2} (3+5 x)^{3/2}}{6 (2+3 x)^2}+\frac {115 (1-2 x)^{3/2} (3+5 x)^{3/2}}{36 (2+3 x)}+\frac {\int \frac {-68505-204870 x}{\sqrt {1-2 x} (2+3 x) \sqrt {3+5 x}} \, dx}{3240}\\ &=-\frac {1649}{108} \sqrt {1-2 x} \sqrt {3+5 x}+\frac {41}{18} \sqrt {1-2 x} (3+5 x)^{3/2}-\frac {(1-2 x)^{5/2} (3+5 x)^{3/2}}{6 (2+3 x)^2}+\frac {115 (1-2 x)^{3/2} (3+5 x)^{3/2}}{36 (2+3 x)}+\frac {13615}{648} \int \frac {1}{\sqrt {1-2 x} (2+3 x) \sqrt {3+5 x}} \, dx-\frac {6829}{324} \int \frac {1}{\sqrt {1-2 x} \sqrt {3+5 x}} \, dx\\ &=-\frac {1649}{108} \sqrt {1-2 x} \sqrt {3+5 x}+\frac {41}{18} \sqrt {1-2 x} (3+5 x)^{3/2}-\frac {(1-2 x)^{5/2} (3+5 x)^{3/2}}{6 (2+3 x)^2}+\frac {115 (1-2 x)^{3/2} (3+5 x)^{3/2}}{36 (2+3 x)}+\frac {13615}{324} \operatorname {Subst}\left (\int \frac {1}{-7-x^2} \, dx,x,\frac {\sqrt {1-2 x}}{\sqrt {3+5 x}}\right )-\frac {6829 \operatorname {Subst}\left (\int \frac {1}{\sqrt {11-2 x^2}} \, dx,x,\sqrt {3+5 x}\right )}{162 \sqrt {5}}\\ &=-\frac {1649}{108} \sqrt {1-2 x} \sqrt {3+5 x}+\frac {41}{18} \sqrt {1-2 x} (3+5 x)^{3/2}-\frac {(1-2 x)^{5/2} (3+5 x)^{3/2}}{6 (2+3 x)^2}+\frac {115 (1-2 x)^{3/2} (3+5 x)^{3/2}}{36 (2+3 x)}-\frac {6829 \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )}{162 \sqrt {10}}-\frac {1945}{324} \sqrt {7} \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right )\\ \end {align*}

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Mathematica [A]  time = 0.27, size = 139, normalized size = 0.85 \[ \frac {15 \sqrt {-(1-2 x)^2} \sqrt {5 x+3} \left (360 x^3-1230 x^2-3471 x-1628\right )-9725 \sqrt {14 x-7} (3 x+2)^2 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )+6829 \sqrt {10-20 x} (3 x+2)^2 \sinh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {2 x-1}\right )}{1620 \sqrt {2 x-1} (3 x+2)^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(15*Sqrt[-(1 - 2*x)^2]*Sqrt[3 + 5*x]*(-1628 - 3471*x - 1230*x^2 + 360*x^3) + 6829*Sqrt[10 - 20*x]*(2 + 3*x)^2*
ArcSinh[Sqrt[5/11]*Sqrt[-1 + 2*x]] - 9725*(2 + 3*x)^2*Sqrt[-7 + 14*x]*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5
*x])])/(1620*Sqrt[-1 + 2*x]*(2 + 3*x)^2)

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fricas [A]  time = 0.99, size = 146, normalized size = 0.89 \[ -\frac {9725 \, \sqrt {7} {\left (9 \, x^{2} + 12 \, x + 4\right )} \arctan \left (\frac {\sqrt {7} {\left (37 \, x + 20\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{14 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) - 6829 \, \sqrt {10} {\left (9 \, x^{2} + 12 \, x + 4\right )} \arctan \left (\frac {\sqrt {10} {\left (20 \, x + 1\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{20 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) - 30 \, {\left (360 \, x^{3} - 1230 \, x^{2} - 3471 \, x - 1628\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{3240 \, {\left (9 \, x^{2} + 12 \, x + 4\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3240*(9725*sqrt(7)*(9*x^2 + 12*x + 4)*arctan(1/14*sqrt(7)*(37*x + 20)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2
+ x - 3)) - 6829*sqrt(10)*(9*x^2 + 12*x + 4)*arctan(1/20*sqrt(10)*(20*x + 1)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*
x^2 + x - 3)) - 30*(360*x^3 - 1230*x^2 - 3471*x - 1628)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(9*x^2 + 12*x + 4)

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giac [B]  time = 2.83, size = 351, normalized size = 2.14 \[ \frac {389}{1296} \, \sqrt {70} \sqrt {10} {\left (\pi + 2 \, \arctan \left (-\frac {\sqrt {70} \sqrt {5 \, x + 3} {\left (\frac {{\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} - 4\right )}}{140 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}\right )\right )} + \frac {1}{270} \, {\left (4 \, \sqrt {5} {\left (5 \, x + 3\right )} - 107 \, \sqrt {5}\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5} - \frac {6829}{3240} \, \sqrt {10} {\left (\pi + 2 \, \arctan \left (-\frac {\sqrt {5 \, x + 3} {\left (\frac {{\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} - 4\right )}}{4 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}\right )\right )} - \frac {77 \, \sqrt {10} {\left (41 \, {\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}^{3} + \frac {17640 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}{\sqrt {5 \, x + 3}} - \frac {70560 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}}{54 \, {\left ({\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}^{2} + 280\right )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

389/1296*sqrt(70)*sqrt(10)*(pi + 2*arctan(-1/140*sqrt(70)*sqrt(5*x + 3)*((sqrt(2)*sqrt(-10*x + 5) - sqrt(22))^
2/(5*x + 3) - 4)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22)))) + 1/270*(4*sqrt(5)*(5*x + 3) - 107*sqrt(5))*sqrt(5*x +
3)*sqrt(-10*x + 5) - 6829/3240*sqrt(10)*(pi + 2*arctan(-1/4*sqrt(5*x + 3)*((sqrt(2)*sqrt(-10*x + 5) - sqrt(22
))^2/(5*x + 3) - 4)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22)))) - 77/54*sqrt(10)*(41*((sqrt(2)*sqrt(-10*x + 5) - sq
rt(22))/sqrt(5*x + 3) - 4*sqrt(5*x + 3)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22)))^3 + 17640*(sqrt(2)*sqrt(-10*x +
5) - sqrt(22))/sqrt(5*x + 3) - 70560*sqrt(5*x + 3)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22)))/(((sqrt(2)*sqrt(-10*x
+ 5) - sqrt(22))/sqrt(5*x + 3) - 4*sqrt(5*x + 3)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22)))^2 + 280)^2

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maple [A]  time = 0.01, size = 225, normalized size = 1.37 \[ -\frac {\sqrt {-2 x +1}\, \sqrt {5 x +3}\, \left (-10800 \sqrt {-10 x^{2}-x +3}\, x^{3}+61461 \sqrt {10}\, x^{2} \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )-87525 \sqrt {7}\, x^{2} \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )+36900 \sqrt {-10 x^{2}-x +3}\, x^{2}+81948 \sqrt {10}\, x \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )-116700 \sqrt {7}\, x \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )+104130 \sqrt {-10 x^{2}-x +3}\, x +27316 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )-38900 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )+48840 \sqrt {-10 x^{2}-x +3}\right )}{3240 \sqrt {-10 x^{2}-x +3}\, \left (3 x +2\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3240*(-2*x+1)^(1/2)*(5*x+3)^(1/2)*(61461*10^(1/2)*x^2*arcsin(20/11*x+1/11)-87525*7^(1/2)*x^2*arctan(1/14*(3
7*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))-10800*(-10*x^2-x+3)^(1/2)*x^3+81948*10^(1/2)*x*arcsin(20/11*x+1/11)-11670
0*7^(1/2)*x*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))+36900*(-10*x^2-x+3)^(1/2)*x^2+27316*10^(1/2)*ar
csin(20/11*x+1/11)-38900*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))+104130*(-10*x^2-x+3)^(1/2)
*x+48840*(-10*x^2-x+3)^(1/2))/(-10*x^2-x+3)^(1/2)/(3*x+2)^2

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maxima [A]  time = 1.30, size = 130, normalized size = 0.79 \[ \frac {5}{9} \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}} + \frac {{\left (-10 \, x^{2} - x + 3\right )}^{\frac {5}{2}}}{2 \, {\left (9 \, x^{2} + 12 \, x + 4\right )}} + \frac {205}{18} \, \sqrt {-10 \, x^{2} - x + 3} x - \frac {6829}{3240} \, \sqrt {10} \arcsin \left (\frac {20}{11} \, x + \frac {1}{11}\right ) + \frac {1945}{648} \, \sqrt {7} \arcsin \left (\frac {37 \, x}{11 \, {\left | 3 \, x + 2 \right |}} + \frac {20}{11 \, {\left | 3 \, x + 2 \right |}}\right ) - \frac {911}{108} \, \sqrt {-10 \, x^{2} - x + 3} + \frac {5 \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}}}{4 \, {\left (3 \, x + 2\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/9*(-10*x^2 - x + 3)^(3/2) + 1/2*(-10*x^2 - x + 3)^(5/2)/(9*x^2 + 12*x + 4) + 205/18*sqrt(-10*x^2 - x + 3)*x
- 6829/3240*sqrt(10)*arcsin(20/11*x + 1/11) + 1945/648*sqrt(7)*arcsin(37/11*x/abs(3*x + 2) + 20/11/abs(3*x + 2
)) - 911/108*sqrt(-10*x^2 - x + 3) + 5/4*(-10*x^2 - x + 3)^(3/2)/(3*x + 2)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (1-2\,x\right )}^{5/2}\,{\left (5\,x+3\right )}^{3/2}}{{\left (3\,x+2\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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