3.2395 \(\int \frac {(1-2 x)^{5/2} \sqrt {3+5 x}}{(2+3 x)^4} \, dx\)

Optimal. Leaf size=149 \[ -\frac {\sqrt {5 x+3} (1-2 x)^{5/2}}{9 (3 x+2)^3}+\frac {5 \sqrt {5 x+3} (1-2 x)^{3/2}}{12 (3 x+2)^2}+\frac {925 \sqrt {5 x+3} \sqrt {1-2 x}}{216 (3 x+2)}-\frac {8}{81} \sqrt {10} \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )-\frac {32765 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{648 \sqrt {7}} \]

[Out]

-32765/4536*arctan(1/7*(1-2*x)^(1/2)*7^(1/2)/(3+5*x)^(1/2))*7^(1/2)-8/81*arcsin(1/11*22^(1/2)*(3+5*x)^(1/2))*1
0^(1/2)-1/9*(1-2*x)^(5/2)*(3+5*x)^(1/2)/(2+3*x)^3+5/12*(1-2*x)^(3/2)*(3+5*x)^(1/2)/(2+3*x)^2+925/216*(1-2*x)^(
1/2)*(3+5*x)^(1/2)/(2+3*x)

________________________________________________________________________________________

Rubi [A]  time = 0.05, antiderivative size = 149, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {97, 149, 157, 54, 216, 93, 204} \[ -\frac {\sqrt {5 x+3} (1-2 x)^{5/2}}{9 (3 x+2)^3}+\frac {5 \sqrt {5 x+3} (1-2 x)^{3/2}}{12 (3 x+2)^2}+\frac {925 \sqrt {5 x+3} \sqrt {1-2 x}}{216 (3 x+2)}-\frac {8}{81} \sqrt {10} \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )-\frac {32765 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{648 \sqrt {7}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((1 - 2*x)^(5/2)*Sqrt[3 + 5*x])/(9*(2 + 3*x)^3) + (5*(1 - 2*x)^(3/2)*Sqrt[3 + 5*x])/(12*(2 + 3*x)^2) + (925*S
qrt[1 - 2*x]*Sqrt[3 + 5*x])/(216*(2 + 3*x)) - (8*Sqrt[10]*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]])/81 - (32765*ArcTan
[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/(648*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 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} \sqrt {3+5 x}}{(2+3 x)^4} \, dx &=-\frac {(1-2 x)^{5/2} \sqrt {3+5 x}}{9 (2+3 x)^3}+\frac {1}{9} \int \frac {\left (-\frac {25}{2}-30 x\right ) (1-2 x)^{3/2}}{(2+3 x)^3 \sqrt {3+5 x}} \, dx\\ &=-\frac {(1-2 x)^{5/2} \sqrt {3+5 x}}{9 (2+3 x)^3}+\frac {5 (1-2 x)^{3/2} \sqrt {3+5 x}}{12 (2+3 x)^2}-\frac {1}{54} \int \frac {\left (-\frac {1245}{4}-120 x\right ) \sqrt {1-2 x}}{(2+3 x)^2 \sqrt {3+5 x}} \, dx\\ &=-\frac {(1-2 x)^{5/2} \sqrt {3+5 x}}{9 (2+3 x)^3}+\frac {5 (1-2 x)^{3/2} \sqrt {3+5 x}}{12 (2+3 x)^2}+\frac {925 \sqrt {1-2 x} \sqrt {3+5 x}}{216 (2+3 x)}+\frac {1}{162} \int \frac {\frac {31485}{8}-240 x}{\sqrt {1-2 x} (2+3 x) \sqrt {3+5 x}} \, dx\\ &=-\frac {(1-2 x)^{5/2} \sqrt {3+5 x}}{9 (2+3 x)^3}+\frac {5 (1-2 x)^{3/2} \sqrt {3+5 x}}{12 (2+3 x)^2}+\frac {925 \sqrt {1-2 x} \sqrt {3+5 x}}{216 (2+3 x)}-\frac {40}{81} \int \frac {1}{\sqrt {1-2 x} \sqrt {3+5 x}} \, dx+\frac {32765 \int \frac {1}{\sqrt {1-2 x} (2+3 x) \sqrt {3+5 x}} \, dx}{1296}\\ &=-\frac {(1-2 x)^{5/2} \sqrt {3+5 x}}{9 (2+3 x)^3}+\frac {5 (1-2 x)^{3/2} \sqrt {3+5 x}}{12 (2+3 x)^2}+\frac {925 \sqrt {1-2 x} \sqrt {3+5 x}}{216 (2+3 x)}+\frac {32765}{648} \operatorname {Subst}\left (\int \frac {1}{-7-x^2} \, dx,x,\frac {\sqrt {1-2 x}}{\sqrt {3+5 x}}\right )-\frac {1}{81} \left (16 \sqrt {5}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {11-2 x^2}} \, dx,x,\sqrt {3+5 x}\right )\\ &=-\frac {(1-2 x)^{5/2} \sqrt {3+5 x}}{9 (2+3 x)^3}+\frac {5 (1-2 x)^{3/2} \sqrt {3+5 x}}{12 (2+3 x)^2}+\frac {925 \sqrt {1-2 x} \sqrt {3+5 x}}{216 (2+3 x)}-\frac {8}{81} \sqrt {10} \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )-\frac {32765 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right )}{648 \sqrt {7}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.24, size = 134, normalized size = 0.90 \[ \frac {21 \sqrt {-(1-2 x)^2} \sqrt {5 x+3} \left (7689 x^2+11106 x+3856\right )-32765 \sqrt {14 x-7} (3 x+2)^3 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )+448 \sqrt {10-20 x} (3 x+2)^3 \sinh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {2 x-1}\right )}{4536 \sqrt {2 x-1} (3 x+2)^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(21*Sqrt[-(1 - 2*x)^2]*Sqrt[3 + 5*x]*(3856 + 11106*x + 7689*x^2) + 448*Sqrt[10 - 20*x]*(2 + 3*x)^3*ArcSinh[Sqr
t[5/11]*Sqrt[-1 + 2*x]] - 32765*(2 + 3*x)^3*Sqrt[-7 + 14*x]*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/(45
36*Sqrt[-1 + 2*x]*(2 + 3*x)^3)

________________________________________________________________________________________

fricas [A]  time = 0.97, size = 156, normalized size = 1.05 \[ -\frac {32765 \, \sqrt {7} {\left (27 \, x^{3} + 54 \, x^{2} + 36 \, x + 8\right )} \arctan \left (\frac {\sqrt {7} {\left (37 \, x + 20\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{14 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) - 448 \, \sqrt {10} {\left (27 \, x^{3} + 54 \, x^{2} + 36 \, x + 8\right )} \arctan \left (\frac {\sqrt {10} {\left (20 \, x + 1\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{20 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) - 42 \, {\left (7689 \, x^{2} + 11106 \, x + 3856\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{9072 \, {\left (27 \, x^{3} + 54 \, x^{2} + 36 \, x + 8\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/9072*(32765*sqrt(7)*(27*x^3 + 54*x^2 + 36*x + 8)*arctan(1/14*sqrt(7)*(37*x + 20)*sqrt(5*x + 3)*sqrt(-2*x +
1)/(10*x^2 + x - 3)) - 448*sqrt(10)*(27*x^3 + 54*x^2 + 36*x + 8)*arctan(1/20*sqrt(10)*(20*x + 1)*sqrt(5*x + 3)
*sqrt(-2*x + 1)/(10*x^2 + x - 3)) - 42*(7689*x^2 + 11106*x + 3856)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(27*x^3 + 54*
x^2 + 36*x + 8)

________________________________________________________________________________________

giac [B]  time = 2.82, size = 377, normalized size = 2.53 \[ \frac {6553}{18144} \, \sqrt {70} \sqrt {10} {\left (\pi + 2 \, \arctan \left (-\frac {\sqrt {70} \sqrt {5 \, x + 3} {\left (\frac {{\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} - 4\right )}}{140 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}\right )\right )} - \frac {4}{81} \, \sqrt {10} {\left (\pi + 2 \, \arctan \left (-\frac {\sqrt {5 \, x + 3} {\left (\frac {{\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} - 4\right )}}{4 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}\right )\right )} - \frac {11 \, \sqrt {10} {\left (989 \, {\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}^{5} - 795200 \, {\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}^{3} - \frac {72520000 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}{\sqrt {5 \, x + 3}} + \frac {290080000 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}}{108 \, {\left ({\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}^{2} + 280\right )}^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

6553/18144*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)))) - 4/81*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)))) - 11/108*sqrt(10)*
(989*((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)
))^5 - 795200*((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 - 72520000*(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))/sqrt(5*x + 3) + 290080000*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)^3

________________________________________________________________________________________

maple [B]  time = 0.01, size = 253, normalized size = 1.70 \[ \frac {\sqrt {-2 x +1}\, \sqrt {5 x +3}\, \left (-12096 \sqrt {10}\, x^{3} \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )+884655 \sqrt {7}\, x^{3} \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )-24192 \sqrt {10}\, x^{2} \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )+1769310 \sqrt {7}\, x^{2} \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )+322938 \sqrt {-10 x^{2}-x +3}\, x^{2}-16128 \sqrt {10}\, x \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )+1179540 \sqrt {7}\, x \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )+466452 \sqrt {-10 x^{2}-x +3}\, x -3584 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )+262120 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )+161952 \sqrt {-10 x^{2}-x +3}\right )}{9072 \sqrt {-10 x^{2}-x +3}\, \left (3 x +2\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/9072*(-2*x+1)^(1/2)*(5*x+3)^(1/2)*(884655*7^(1/2)*x^3*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))-120
96*10^(1/2)*x^3*arcsin(20/11*x+1/11)+1769310*7^(1/2)*x^2*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))-24
192*10^(1/2)*x^2*arcsin(20/11*x+1/11)+1179540*7^(1/2)*x*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))-161
28*10^(1/2)*x*arcsin(20/11*x+1/11)+322938*(-10*x^2-x+3)^(1/2)*x^2+262120*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)
/(-10*x^2-x+3)^(1/2))-3584*10^(1/2)*arcsin(20/11*x+1/11)+466452*(-10*x^2-x+3)^(1/2)*x+161952*(-10*x^2-x+3)^(1/
2))/(-10*x^2-x+3)^(1/2)/(3*x+2)^3

________________________________________________________________________________________

maxima [A]  time = 1.17, size = 132, normalized size = 0.89 \[ -\frac {4}{81} \, \sqrt {10} \arcsin \left (\frac {20}{11} \, x + \frac {1}{11}\right ) + \frac {32765}{9072} \, \sqrt {7} \arcsin \left (\frac {37 \, x}{11 \, {\left | 3 \, x + 2 \right |}} + \frac {20}{11 \, {\left | 3 \, x + 2 \right |}}\right ) + \frac {145}{54} \, \sqrt {-10 \, x^{2} - x + 3} + \frac {7 \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}}}{9 \, {\left (27 \, x^{3} + 54 \, x^{2} + 36 \, x + 8\right )}} + \frac {29 \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}}}{12 \, {\left (9 \, x^{2} + 12 \, x + 4\right )}} - \frac {1105 \, \sqrt {-10 \, x^{2} - x + 3}}{216 \, {\left (3 \, x + 2\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/81*sqrt(10)*arcsin(20/11*x + 1/11) + 32765/9072*sqrt(7)*arcsin(37/11*x/abs(3*x + 2) + 20/11/abs(3*x + 2)) +
145/54*sqrt(-10*x^2 - x + 3) + 7/9*(-10*x^2 - x + 3)^(3/2)/(27*x^3 + 54*x^2 + 36*x + 8) + 29/12*(-10*x^2 - x
+ 3)^(3/2)/(9*x^2 + 12*x + 4) - 1105/216*sqrt(-10*x^2 - x + 3)/(3*x + 2)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (1-2\,x\right )}^{5/2}\,\sqrt {5\,x+3}}{{\left (3\,x+2\right )}^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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)**(1/2)/(2+3*x)**4,x)

[Out]

Timed out

________________________________________________________________________________________