3.26.80 \(\int \frac {(5-x) \sqrt {2+5 x+3 x^2}}{(3+2 x)^{3/2}} \, dx\) [2580]

3.26.80.1 Optimal result

Integrand size = 29, antiderivative size = 141 \[ \int \frac {(5-x) \sqrt {2+5 x+3 x^2}}{(3+2 x)^{3/2}} \, dx=-\frac {(21+x) \sqrt {2+5 x+3 x^2}}{3 \sqrt {3+2 x}}+\frac {121 \sqrt {-2-5 x-3 x^2} E\left (\arcsin \left (\sqrt {3} \sqrt {1+x}\right )|-\frac {2}{3}\right )}{6 \sqrt {3} \sqrt {2+5 x+3 x^2}}-\frac {161 \sqrt {-2-5 x-3 x^2} \operatorname {EllipticF}\left (\arcsin \left (\sqrt {3} \sqrt {1+x}\right ),-\frac {2}{3}\right )}{6 \sqrt {3} \sqrt {2+5 x+3 x^2}} \]

output

121/18*EllipticE(3^(1/2)*(1+x)^(1/2),1/3*I*6^(1/2))*(-3*x^2-5*x-2)^(1/2)*3 
^(1/2)/(3*x^2+5*x+2)^(1/2)-161/18*EllipticF(3^(1/2)*(1+x)^(1/2),1/3*I*6^(1 
/2))*(-3*x^2-5*x-2)^(1/2)*3^(1/2)/(3*x^2+5*x+2)^(1/2)-1/3*(21+x)*(3*x^2+5* 
x+2)^(1/2)/(3+2*x)^(1/2)
 

3.26.80.2 Mathematica [A] (verified)

Time = 24.82 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.31 \[ \int \frac {(5-x) \sqrt {2+5 x+3 x^2}}{(3+2 x)^{3/2}} \, dx=\frac {10 \left (116+284 x+159 x^2-9 x^3\right )+605 \sqrt {5} \sqrt {\frac {1+x}{3+2 x}} (3+2 x)^{3/2} \sqrt {\frac {2+3 x}{3+2 x}} E\left (\arcsin \left (\frac {\sqrt {\frac {5}{3}}}{\sqrt {3+2 x}}\right )|\frac {3}{5}\right )-122 \sqrt {5} \sqrt {\frac {1+x}{3+2 x}} (3+2 x)^{3/2} \sqrt {\frac {2+3 x}{3+2 x}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {\frac {5}{3}}}{\sqrt {3+2 x}}\right ),\frac {3}{5}\right )}{90 \sqrt {3+2 x} \sqrt {2+5 x+3 x^2}} \]

input

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

output

(10*(116 + 284*x + 159*x^2 - 9*x^3) + 605*Sqrt[5]*Sqrt[(1 + x)/(3 + 2*x)]* 
(3 + 2*x)^(3/2)*Sqrt[(2 + 3*x)/(3 + 2*x)]*EllipticE[ArcSin[Sqrt[5/3]/Sqrt[ 
3 + 2*x]], 3/5] - 122*Sqrt[5]*Sqrt[(1 + x)/(3 + 2*x)]*(3 + 2*x)^(3/2)*Sqrt 
[(2 + 3*x)/(3 + 2*x)]*EllipticF[ArcSin[Sqrt[5/3]/Sqrt[3 + 2*x]], 3/5])/(90 
*Sqrt[3 + 2*x]*Sqrt[2 + 5*x + 3*x^2])
 

3.26.80.3 Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.01, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {1230, 25, 1269, 1172, 27, 321, 327}

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.

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

\(\Big \downarrow \) 1230

\(\displaystyle -\frac {1}{6} \int -\frac {121 x+101}{\sqrt {2 x+3} \sqrt {3 x^2+5 x+2}}dx-\frac {\sqrt {3 x^2+5 x+2} (x+21)}{3 \sqrt {2 x+3}}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{6} \int \frac {121 x+101}{\sqrt {2 x+3} \sqrt {3 x^2+5 x+2}}dx-\frac {(x+21) \sqrt {3 x^2+5 x+2}}{3 \sqrt {2 x+3}}\)

\(\Big \downarrow \) 1269

\(\displaystyle \frac {1}{6} \left (\frac {121}{2} \int \frac {\sqrt {2 x+3}}{\sqrt {3 x^2+5 x+2}}dx-\frac {161}{2} \int \frac {1}{\sqrt {2 x+3} \sqrt {3 x^2+5 x+2}}dx\right )-\frac {(x+21) \sqrt {3 x^2+5 x+2}}{3 \sqrt {2 x+3}}\)

\(\Big \downarrow \) 1172

\(\displaystyle \frac {1}{6} \left (\frac {121 \sqrt {-3 x^2-5 x-2} \int \frac {\sqrt {6 (x+1)+3}}{\sqrt {3} \sqrt {1-3 (x+1)}}d\left (\sqrt {3} \sqrt {x+1}\right )}{\sqrt {3} \sqrt {3 x^2+5 x+2}}-\frac {161 \sqrt {-3 x^2-5 x-2} \int \frac {\sqrt {3}}{\sqrt {1-3 (x+1)} \sqrt {6 (x+1)+3}}d\left (\sqrt {3} \sqrt {x+1}\right )}{\sqrt {3} \sqrt {3 x^2+5 x+2}}\right )-\frac {(x+21) \sqrt {3 x^2+5 x+2}}{3 \sqrt {2 x+3}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{6} \left (\frac {121 \sqrt {-3 x^2-5 x-2} \int \frac {\sqrt {6 (x+1)+3}}{\sqrt {1-3 (x+1)}}d\left (\sqrt {3} \sqrt {x+1}\right )}{3 \sqrt {3 x^2+5 x+2}}-\frac {161 \sqrt {-3 x^2-5 x-2} \int \frac {1}{\sqrt {1-3 (x+1)} \sqrt {6 (x+1)+3}}d\left (\sqrt {3} \sqrt {x+1}\right )}{\sqrt {3 x^2+5 x+2}}\right )-\frac {(x+21) \sqrt {3 x^2+5 x+2}}{3 \sqrt {2 x+3}}\)

\(\Big \downarrow \) 321

\(\displaystyle \frac {1}{6} \left (\frac {121 \sqrt {-3 x^2-5 x-2} \int \frac {\sqrt {6 (x+1)+3}}{\sqrt {1-3 (x+1)}}d\left (\sqrt {3} \sqrt {x+1}\right )}{3 \sqrt {3 x^2+5 x+2}}-\frac {161 \sqrt {-3 x^2-5 x-2} \operatorname {EllipticF}\left (\arcsin \left (\sqrt {3} \sqrt {x+1}\right ),-\frac {2}{3}\right )}{\sqrt {3} \sqrt {3 x^2+5 x+2}}\right )-\frac {(x+21) \sqrt {3 x^2+5 x+2}}{3 \sqrt {2 x+3}}\)

\(\Big \downarrow \) 327

\(\displaystyle \frac {1}{6} \left (\frac {121 \sqrt {-3 x^2-5 x-2} E\left (\arcsin \left (\sqrt {3} \sqrt {x+1}\right )|-\frac {2}{3}\right )}{\sqrt {3} \sqrt {3 x^2+5 x+2}}-\frac {161 \sqrt {-3 x^2-5 x-2} \operatorname {EllipticF}\left (\arcsin \left (\sqrt {3} \sqrt {x+1}\right ),-\frac {2}{3}\right )}{\sqrt {3} \sqrt {3 x^2+5 x+2}}\right )-\frac {(x+21) \sqrt {3 x^2+5 x+2}}{3 \sqrt {2 x+3}}\)

input

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

output

-1/3*((21 + x)*Sqrt[2 + 5*x + 3*x^2])/Sqrt[3 + 2*x] + ((121*Sqrt[-2 - 5*x 
- 3*x^2]*EllipticE[ArcSin[Sqrt[3]*Sqrt[1 + x]], -2/3])/(Sqrt[3]*Sqrt[2 + 5 
*x + 3*x^2]) - (161*Sqrt[-2 - 5*x - 3*x^2]*EllipticF[ArcSin[Sqrt[3]*Sqrt[1 
 + x]], -2/3])/(Sqrt[3]*Sqrt[2 + 5*x + 3*x^2]))/6
 

3.26.80.3.1 Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 321

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> S 
imp[(1/(Sqrt[a]*Sqrt[c]*Rt[-d/c, 2]))*EllipticF[ArcSin[Rt[-d/c, 2]*x], b*(c 
/(a*d))], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[a, 
0] &&  !(NegQ[b/a] && SimplerSqrtQ[-b/a, -d/c])
 

rule 327

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

rule 1172

Int[((d_.) + (e_.)*(x_))^(m_)/Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2], x_Sy 
mbol] :> Simp[2*Rt[b^2 - 4*a*c, 2]*(d + e*x)^m*(Sqrt[(-c)*((a + b*x + c*x^2 
)/(b^2 - 4*a*c))]/(c*Sqrt[a + b*x + c*x^2]*(2*c*((d + e*x)/(2*c*d - b*e - e 
*Rt[b^2 - 4*a*c, 2])))^m))   Subst[Int[(1 + 2*e*Rt[b^2 - 4*a*c, 2]*(x^2/(2* 
c*d - b*e - e*Rt[b^2 - 4*a*c, 2])))^m/Sqrt[1 - x^2], x], x, Sqrt[(b + Rt[b^ 
2 - 4*a*c, 2] + 2*c*x)/(2*Rt[b^2 - 4*a*c, 2])]], x] /; FreeQ[{a, b, c, d, e 
}, x] && EqQ[m^2, 1/4]
 

rule 1230

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c 
_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(d + e*x)^(m + 1)*(e*f*(m + 2*p + 2) - 
 d*g*(2*p + 1) + e*g*(m + 1)*x)*((a + b*x + c*x^2)^p/(e^2*(m + 1)*(m + 2*p 
+ 2))), x] + Simp[p/(e^2*(m + 1)*(m + 2*p + 2))   Int[(d + e*x)^(m + 1)*(a 
+ b*x + c*x^2)^(p - 1)*Simp[g*(b*d + 2*a*e + 2*a*e*m + 2*b*d*p) - f*b*e*(m 
+ 2*p + 2) + (g*(2*c*d + b*e + b*e*m + 4*c*d*p) - 2*c*e*f*(m + 2*p + 2))*x, 
 x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && GtQ[p, 0] && (LtQ[m, - 
1] || EqQ[p, 1] || (IntegerQ[p] &&  !RationalQ[m])) && NeQ[m, -1] &&  !ILtQ 
[m + 2*p + 1, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])
 

rule 1269

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c 
_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[g/e   Int[(d + e*x)^(m + 1)*(a + b*x + 
 c*x^2)^p, x], x] + Simp[(e*f - d*g)/e   Int[(d + e*x)^m*(a + b*x + c*x^2)^ 
p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] &&  !IGtQ[m, 0]
 

3.26.80.4 Maple [A] (verified)

Time = 0.34 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.96

method	result	size
default	\(\frac {\sqrt {3 x^{2}+5 x +2}\, \sqrt {3+2 x}\, \left (60 \sqrt {-20-30 x}\, \sqrt {3+3 x}\, \sqrt {15}\, \sqrt {3+2 x}\, F\left (\frac {\sqrt {-20-30 x}}{5}, \frac {\sqrt {10}}{2}\right )-121 \sqrt {-20-30 x}\, \sqrt {3+3 x}\, \sqrt {15}\, \sqrt {3+2 x}\, E\left (\frac {\sqrt {-20-30 x}}{5}, \frac {\sqrt {10}}{2}\right )-270 x^{3}-6120 x^{2}-9630 x -3780\right )}{1620 x^{3}+5130 x^{2}+5130 x +1620}\)	\(136\)
elliptic	\(\frac {\sqrt {\left (3+2 x \right ) \left (3 x^{2}+5 x +2\right )}\, \left (-\frac {13 \left (6 x^{2}+10 x +4\right )}{4 \sqrt {\left (x +\frac {3}{2}\right ) \left (6 x^{2}+10 x +4\right )}}-\frac {\sqrt {6 x^{3}+19 x^{2}+19 x +6}}{6}-\frac {101 \sqrt {-20-30 x}\, \sqrt {3+3 x}\, \sqrt {45+30 x}\, F\left (\frac {\sqrt {-20-30 x}}{5}, \frac {\sqrt {10}}{2}\right )}{90 \sqrt {6 x^{3}+19 x^{2}+19 x +6}}-\frac {121 \sqrt {-20-30 x}\, \sqrt {3+3 x}\, \sqrt {45+30 x}\, \left (\frac {E\left (\frac {\sqrt {-20-30 x}}{5}, \frac {\sqrt {10}}{2}\right )}{3}-F\left (\frac {\sqrt {-20-30 x}}{5}, \frac {\sqrt {10}}{2}\right )\right )}{90 \sqrt {6 x^{3}+19 x^{2}+19 x +6}}\right )}{\sqrt {3+2 x}\, \sqrt {3 x^{2}+5 x +2}}\)	\(217\)

input

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

output

1/270*(3*x^2+5*x+2)^(1/2)*(3+2*x)^(1/2)*(60*(-20-30*x)^(1/2)*(3+3*x)^(1/2) 
*15^(1/2)*(3+2*x)^(1/2)*EllipticF(1/5*(-20-30*x)^(1/2),1/2*10^(1/2))-121*( 
-20-30*x)^(1/2)*(3+3*x)^(1/2)*15^(1/2)*(3+2*x)^(1/2)*EllipticE(1/5*(-20-30 
*x)^(1/2),1/2*10^(1/2))-270*x^3-6120*x^2-9630*x-3780)/(6*x^3+19*x^2+19*x+6 
)
 

3.26.80.5 Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.08 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.49 \[ \int \frac {(5-x) \sqrt {2+5 x+3 x^2}}{(3+2 x)^{3/2}} \, dx=-\frac {481 \, \sqrt {6} {\left (2 \, x + 3\right )} {\rm weierstrassPInverse}\left (\frac {19}{27}, -\frac {28}{729}, x + \frac {19}{18}\right ) + 2178 \, \sqrt {6} {\left (2 \, x + 3\right )} {\rm weierstrassZeta}\left (\frac {19}{27}, -\frac {28}{729}, {\rm weierstrassPInverse}\left (\frac {19}{27}, -\frac {28}{729}, x + \frac {19}{18}\right )\right ) + 108 \, \sqrt {3 \, x^{2} + 5 \, x + 2} \sqrt {2 \, x + 3} {\left (x + 21\right )}}{324 \, {\left (2 \, x + 3\right )}} \]

input

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

output

-1/324*(481*sqrt(6)*(2*x + 3)*weierstrassPInverse(19/27, -28/729, x + 19/1 
8) + 2178*sqrt(6)*(2*x + 3)*weierstrassZeta(19/27, -28/729, weierstrassPIn 
verse(19/27, -28/729, x + 19/18)) + 108*sqrt(3*x^2 + 5*x + 2)*sqrt(2*x + 3 
)*(x + 21))/(2*x + 3)
 

3.26.80.6 Sympy [F]

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

input

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

output

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

3.26.80.7 Maxima [F]

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

input

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

output

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

3.26.80.8 Giac [F]

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

input

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

output

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

3.26.80.9 Mupad [F(-1)]

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

input

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

output

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