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

Optimal. Leaf size=166 \[ \frac {181 \sqrt {1-2 x} (5 x+3)^{5/2}}{36 (3 x+2)}-\frac {(1-2 x)^{3/2} (5 x+3)^{5/2}}{6 (3 x+2)^2}-\frac {35}{4} \sqrt {1-2 x} (5 x+3)^{3/2}+\frac {185}{27} \sqrt {1-2 x} \sqrt {5 x+3}+\frac {1945}{324} \sqrt {\frac {5}{2}} \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )+\frac {6829 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{324 \sqrt {7}} \]

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Rubi [A]  time = 0.06, antiderivative size = 166, 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} \begin {gather*} \frac {181 \sqrt {1-2 x} (5 x+3)^{5/2}}{36 (3 x+2)}-\frac {(1-2 x)^{3/2} (5 x+3)^{5/2}}{6 (3 x+2)^2}-\frac {35}{4} \sqrt {1-2 x} (5 x+3)^{3/2}+\frac {185}{27} \sqrt {1-2 x} \sqrt {5 x+3}+\frac {1945}{324} \sqrt {\frac {5}{2}} \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )+\frac {6829 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{324 \sqrt {7}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(185*Sqrt[1 - 2*x]*Sqrt[3 + 5*x])/27 - (35*Sqrt[1 - 2*x]*(3 + 5*x)^(3/2))/4 - ((1 - 2*x)^(3/2)*(3 + 5*x)^(5/2)
)/(6*(2 + 3*x)^2) + (181*Sqrt[1 - 2*x]*(3 + 5*x)^(5/2))/(36*(2 + 3*x)) + (1945*Sqrt[5/2]*ArcSin[Sqrt[2/11]*Sqr
t[3 + 5*x]])/324 + (6829*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/(324*Sqrt[7])

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)^{3/2} (3+5 x)^{5/2}}{(2+3 x)^3} \, dx &=-\frac {(1-2 x)^{3/2} (3+5 x)^{5/2}}{6 (2+3 x)^2}+\frac {1}{6} \int \frac {\left (\frac {7}{2}-40 x\right ) \sqrt {1-2 x} (3+5 x)^{3/2}}{(2+3 x)^2} \, dx\\ &=-\frac {(1-2 x)^{3/2} (3+5 x)^{5/2}}{6 (2+3 x)^2}+\frac {181 \sqrt {1-2 x} (3+5 x)^{5/2}}{36 (2+3 x)}-\frac {1}{18} \int \frac {\left (\frac {1789}{4}-1890 x\right ) (3+5 x)^{3/2}}{\sqrt {1-2 x} (2+3 x)} \, dx\\ &=-\frac {35}{4} \sqrt {1-2 x} (3+5 x)^{3/2}-\frac {(1-2 x)^{3/2} (3+5 x)^{5/2}}{6 (2+3 x)^2}+\frac {181 \sqrt {1-2 x} (3+5 x)^{5/2}}{36 (2+3 x)}+\frac {1}{216} \int \frac {(909-8880 x) \sqrt {3+5 x}}{\sqrt {1-2 x} (2+3 x)} \, dx\\ &=\frac {185}{27} \sqrt {1-2 x} \sqrt {3+5 x}-\frac {35}{4} \sqrt {1-2 x} (3+5 x)^{3/2}-\frac {(1-2 x)^{3/2} (3+5 x)^{5/2}}{6 (2+3 x)^2}+\frac {181 \sqrt {1-2 x} (3+5 x)^{5/2}}{36 (2+3 x)}-\frac {\int \frac {-25242-58350 x}{\sqrt {1-2 x} (2+3 x) \sqrt {3+5 x}} \, dx}{1296}\\ &=\frac {185}{27} \sqrt {1-2 x} \sqrt {3+5 x}-\frac {35}{4} \sqrt {1-2 x} (3+5 x)^{3/2}-\frac {(1-2 x)^{3/2} (3+5 x)^{5/2}}{6 (2+3 x)^2}+\frac {181 \sqrt {1-2 x} (3+5 x)^{5/2}}{36 (2+3 x)}-\frac {6829}{648} \int \frac {1}{\sqrt {1-2 x} (2+3 x) \sqrt {3+5 x}} \, dx+\frac {9725}{648} \int \frac {1}{\sqrt {1-2 x} \sqrt {3+5 x}} \, dx\\ &=\frac {185}{27} \sqrt {1-2 x} \sqrt {3+5 x}-\frac {35}{4} \sqrt {1-2 x} (3+5 x)^{3/2}-\frac {(1-2 x)^{3/2} (3+5 x)^{5/2}}{6 (2+3 x)^2}+\frac {181 \sqrt {1-2 x} (3+5 x)^{5/2}}{36 (2+3 x)}-\frac {6829}{324} \operatorname {Subst}\left (\int \frac {1}{-7-x^2} \, dx,x,\frac {\sqrt {1-2 x}}{\sqrt {3+5 x}}\right )+\frac {1}{324} \left (1945 \sqrt {5}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {11-2 x^2}} \, dx,x,\sqrt {3+5 x}\right )\\ &=\frac {185}{27} \sqrt {1-2 x} \sqrt {3+5 x}-\frac {35}{4} \sqrt {1-2 x} (3+5 x)^{3/2}-\frac {(1-2 x)^{3/2} (3+5 x)^{5/2}}{6 (2+3 x)^2}+\frac {181 \sqrt {1-2 x} (3+5 x)^{5/2}}{36 (2+3 x)}+\frac {1945}{324} \sqrt {\frac {5}{2}} \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )+\frac {6829 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right )}{324 \sqrt {7}}\\ \end {align*}

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Mathematica [A]  time = 0.24, size = 139, normalized size = 0.84 \begin {gather*} \frac {-42 \sqrt {-(1-2 x)^2} \sqrt {5 x+3} \left (900 x^3-1095 x^2-2985 x-1232\right )+13658 \sqrt {14 x-7} (3 x+2)^2 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )-13615 \sqrt {10-20 x} (3 x+2)^2 \sinh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {2 x-1}\right )}{4536 \sqrt {2 x-1} (3 x+2)^2} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-42*Sqrt[-(1 - 2*x)^2]*Sqrt[3 + 5*x]*(-1232 - 2985*x - 1095*x^2 + 900*x^3) - 13615*Sqrt[10 - 20*x]*(2 + 3*x)^
2*ArcSinh[Sqrt[5/11]*Sqrt[-1 + 2*x]] + 13658*(2 + 3*x)^2*Sqrt[-7 + 14*x]*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3
+ 5*x])])/(4536*Sqrt[-1 + 2*x]*(2 + 3*x)^2)

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IntegrateAlgebraic [A]  time = 0.31, size = 178, normalized size = 1.07 \begin {gather*} \frac {11 \sqrt {1-2 x} \left (\frac {3700 (1-2 x)^3}{(5 x+3)^3}+\frac {42885 (1-2 x)^2}{(5 x+3)^2}+\frac {98235 (1-2 x)}{5 x+3}+23086\right )}{108 \sqrt {5 x+3} \left (\frac {1-2 x}{5 x+3}+7\right )^2 \left (\frac {5 (1-2 x)}{5 x+3}+2\right )^2}-\frac {1945}{324} \sqrt {\frac {5}{2}} \tan ^{-1}\left (\frac {\sqrt {\frac {5}{2}} \sqrt {1-2 x}}{\sqrt {5 x+3}}\right )+\frac {6829 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{324 \sqrt {7}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(11*Sqrt[1 - 2*x]*(23086 + (3700*(1 - 2*x)^3)/(3 + 5*x)^3 + (42885*(1 - 2*x)^2)/(3 + 5*x)^2 + (98235*(1 - 2*x)
)/(3 + 5*x)))/(108*Sqrt[3 + 5*x]*(7 + (1 - 2*x)/(3 + 5*x))^2*(2 + (5*(1 - 2*x))/(3 + 5*x))^2) - (1945*Sqrt[5/2
]*ArcTan[(Sqrt[5/2]*Sqrt[1 - 2*x])/Sqrt[3 + 5*x]])/324 + (6829*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/
(324*Sqrt[7])

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fricas [A]  time = 0.91, size = 152, normalized size = 0.92 \begin {gather*} -\frac {13615 \, \sqrt {5} \sqrt {2} {\left (9 \, x^{2} + 12 \, x + 4\right )} \arctan \left (\frac {\sqrt {5} \sqrt {2} {\left (20 \, x + 1\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{20 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) - 13658 \, \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 ) + 84 \, {\left (900 \, x^{3} - 1095 \, x^{2} - 2985 \, x - 1232\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{9072 \, {\left (9 \, x^{2} + 12 \, x + 4\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/9072*(13615*sqrt(5)*sqrt(2)*(9*x^2 + 12*x + 4)*arctan(1/20*sqrt(5)*sqrt(2)*(20*x + 1)*sqrt(5*x + 3)*sqrt(-2
*x + 1)/(10*x^2 + x - 3)) - 13658*sqrt(7)*(9*x^2 + 12*x + 4)*arctan(1/14*sqrt(7)*(37*x + 20)*sqrt(5*x + 3)*sqr
t(-2*x + 1)/(10*x^2 + x - 3)) + 84*(900*x^3 - 1095*x^2 - 2985*x - 1232)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(9*x^2 +
12*x + 4)

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giac [B]  time = 2.81, size = 351, normalized size = 2.11 \begin {gather*} -\frac {6829}{45360} \, \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}{108} \, {\left (4 \, \sqrt {5} {\left (5 \, x + 3\right )} - 63 \, \sqrt {5}\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5} + \frac {1945}{1296} \, \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 {55 \, \sqrt {10} {\left (17 \, {\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 {5992 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}{\sqrt {5 \, x + 3}} - \frac {23968 \, \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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-6829/45360*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)))) - 1/108*(4*sqrt(5)*(5*x + 3) - 63*sqrt(5))*sqrt(5*x
+ 3)*sqrt(-10*x + 5) + 1945/1296*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)))) + 55/54*sqrt(10)*(17*((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)))^3 + 5992*(sqrt(2)*sqrt(-10*x +
5) - sqrt(22))/sqrt(5*x + 3) - 23968*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 = 1.36 \begin {gather*} \frac {\sqrt {-2 x +1}\, \sqrt {5 x +3}\, \left (-75600 \sqrt {-10 x^{2}-x +3}\, x^{3}+122535 \sqrt {10}\, x^{2} \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )-122922 \sqrt {7}\, x^{2} \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )+91980 \sqrt {-10 x^{2}-x +3}\, x^{2}+163380 \sqrt {10}\, x \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )-163896 \sqrt {7}\, x \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )+250740 \sqrt {-10 x^{2}-x +3}\, x +54460 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )-54632 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )+103488 \sqrt {-10 x^{2}-x +3}\right )}{9072 \sqrt {-10 x^{2}-x +3}\, \left (3 x +2\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/9072*(-2*x+1)^(1/2)*(5*x+3)^(1/2)*(122535*10^(1/2)*x^2*arcsin(20/11*x+1/11)-122922*7^(1/2)*x^2*arctan(1/14*(
37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))-75600*(-10*x^2-x+3)^(1/2)*x^3+163380*10^(1/2)*x*arcsin(20/11*x+1/11)-163
896*7^(1/2)*x*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))+91980*(-10*x^2-x+3)^(1/2)*x^2+54460*10^(1/2)*
arcsin(20/11*x+1/11)-54632*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))+250740*(-10*x^2-x+3)^(1/
2)*x+103488*(-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 = 1.35, size = 130, normalized size = 0.78 \begin {gather*} -\frac {5}{63} \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}} - \frac {{\left (-10 \, x^{2} - x + 3\right )}^{\frac {5}{2}}}{14 \, {\left (9 \, x^{2} + 12 \, x + 4\right )}} - \frac {535}{126} \, \sqrt {-10 \, x^{2} - x + 3} x + \frac {1945}{1296} \, \sqrt {10} \arcsin \left (\frac {20}{11} \, x + \frac {1}{11}\right ) - \frac {6829}{4536} \, \sqrt {7} \arcsin \left (\frac {37 \, x}{11 \, {\left | 3 \, x + 2 \right |}} + \frac {20}{11 \, {\left | 3 \, x + 2 \right |}}\right ) + \frac {1627}{378} \, \sqrt {-10 \, x^{2} - x + 3} - \frac {59 \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}}}{84 \, {\left (3 \, x + 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5/63*(-10*x^2 - x + 3)^(3/2) - 1/14*(-10*x^2 - x + 3)^(5/2)/(9*x^2 + 12*x + 4) - 535/126*sqrt(-10*x^2 - x + 3
)*x + 1945/1296*sqrt(10)*arcsin(20/11*x + 1/11) - 6829/4536*sqrt(7)*arcsin(37/11*x/abs(3*x + 2) + 20/11/abs(3*
x + 2)) + 1627/378*sqrt(-10*x^2 - x + 3) - 59/84*(-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 \begin {gather*} \int \frac {{\left (1-2\,x\right )}^{3/2}\,{\left (5\,x+3\right )}^{5/2}}{{\left (3\,x+2\right )}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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