\(\int \frac {(2-5 x) x^{5/2}}{\sqrt {2+5 x+3 x^2}} \, dx\) [202]

Optimal result

Integrand size = 25, antiderivative size = 196 \[ \int \frac {(2-5 x) x^{5/2}}{\sqrt {2+5 x+3 x^2}} \, dx=\frac {13688 \sqrt {x} (2+3 x)}{2835 \sqrt {2+5 x+3 x^2}}-\frac {412}{189} \sqrt {x} \sqrt {2+5 x+3 x^2}+\frac {128}{105} x^{3/2} \sqrt {2+5 x+3 x^2}-\frac {10}{21} x^{5/2} \sqrt {2+5 x+3 x^2}-\frac {13688 \sqrt {2} \sqrt {2+5 x+3 x^2} E\left (\arctan \left (\sqrt {x}\right )|-\frac {1}{2}\right )}{2835 \sqrt {1+x} \sqrt {2+3 x}}+\frac {412 \sqrt {2} \sqrt {1+x} \sqrt {2+3 x} \operatorname {EllipticF}\left (\arctan \left (\sqrt {x}\right ),-\frac {1}{2}\right )}{189 \sqrt {2+5 x+3 x^2}} \] Output:

13688/2835*x^(1/2)*(2+3*x)/(3*x^2+5*x+2)^(1/2)-412/189*x^(1/2)*(3*x^2+5*x+ 
2)^(1/2)+128/105*x^(3/2)*(3*x^2+5*x+2)^(1/2)-10/21*x^(5/2)*(3*x^2+5*x+2)^( 
1/2)-13688/2835*2^(1/2)*(3*x^2+5*x+2)^(1/2)*EllipticE(x^(1/2)/(1+x)^(1/2), 
1/2*I*2^(1/2))/(1+x)^(1/2)/(2+3*x)^(1/2)+412/189*2^(1/2)*(1+x)^(1/2)*(2+3* 
x)^(1/2)*InverseJacobiAM(arctan(x^(1/2)),1/2*I*2^(1/2))/(3*x^2+5*x+2)^(1/2 
)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 21.21 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.82 \[ \int \frac {(2-5 x) x^{5/2}}{\sqrt {2+5 x+3 x^2}} \, dx=\frac {27376+56080 x+17076 x^2-3960 x^3+3618 x^4-4050 x^5+13688 i \sqrt {2} \sqrt {1+\frac {1}{x}} \sqrt {3+\frac {2}{x}} x^{3/2} E\left (i \text {arcsinh}\left (\frac {\sqrt {\frac {2}{3}}}{\sqrt {x}}\right )|\frac {3}{2}\right )-7508 i \sqrt {2} \sqrt {1+\frac {1}{x}} \sqrt {3+\frac {2}{x}} x^{3/2} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\frac {\sqrt {\frac {2}{3}}}{\sqrt {x}}\right ),\frac {3}{2}\right )}{2835 \sqrt {x} \sqrt {2+5 x+3 x^2}} \] Input:

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

Output:

(27376 + 56080*x + 17076*x^2 - 3960*x^3 + 3618*x^4 - 4050*x^5 + (13688*I)* 
Sqrt[2]*Sqrt[1 + x^(-1)]*Sqrt[3 + 2/x]*x^(3/2)*EllipticE[I*ArcSinh[Sqrt[2/ 
3]/Sqrt[x]], 3/2] - (7508*I)*Sqrt[2]*Sqrt[1 + x^(-1)]*Sqrt[3 + 2/x]*x^(3/2 
)*EllipticF[I*ArcSinh[Sqrt[2/3]/Sqrt[x]], 3/2])/(2835*Sqrt[x]*Sqrt[2 + 5*x 
 + 3*x^2])
 

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.10, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {1236, 1236, 27, 1236, 25, 1240, 1503, 1413, 1456}

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.

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

\(\Big \downarrow \) 1236

\(\displaystyle \frac {2}{21} \int \frac {x^{3/2} (96 x+25)}{\sqrt {3 x^2+5 x+2}}dx-\frac {10}{21} x^{5/2} \sqrt {3 x^2+5 x+2}\)

\(\Big \downarrow \) 1236

\(\displaystyle \frac {2}{21} \left (\frac {2}{15} \int -\frac {3 \sqrt {x} (515 x+192)}{2 \sqrt {3 x^2+5 x+2}}dx+\frac {64}{5} \sqrt {3 x^2+5 x+2} x^{3/2}\right )-\frac {10}{21} x^{5/2} \sqrt {3 x^2+5 x+2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2}{21} \left (\frac {64}{5} x^{3/2} \sqrt {3 x^2+5 x+2}-\frac {1}{5} \int \frac {\sqrt {x} (515 x+192)}{\sqrt {3 x^2+5 x+2}}dx\right )-\frac {10}{21} x^{5/2} \sqrt {3 x^2+5 x+2}\)

\(\Big \downarrow \) 1236

\(\displaystyle \frac {2}{21} \left (\frac {1}{5} \left (-\frac {2}{9} \int -\frac {1711 x+515}{\sqrt {x} \sqrt {3 x^2+5 x+2}}dx-\frac {1030}{9} \sqrt {x} \sqrt {3 x^2+5 x+2}\right )+\frac {64}{5} \sqrt {3 x^2+5 x+2} x^{3/2}\right )-\frac {10}{21} x^{5/2} \sqrt {3 x^2+5 x+2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {2}{21} \left (\frac {1}{5} \left (\frac {2}{9} \int \frac {1711 x+515}{\sqrt {x} \sqrt {3 x^2+5 x+2}}dx-\frac {1030}{9} \sqrt {x} \sqrt {3 x^2+5 x+2}\right )+\frac {64}{5} \sqrt {3 x^2+5 x+2} x^{3/2}\right )-\frac {10}{21} x^{5/2} \sqrt {3 x^2+5 x+2}\)

\(\Big \downarrow \) 1240

\(\displaystyle \frac {2}{21} \left (\frac {1}{5} \left (\frac {4}{9} \int \frac {1711 x+515}{\sqrt {3 x^2+5 x+2}}d\sqrt {x}-\frac {1030}{9} \sqrt {x} \sqrt {3 x^2+5 x+2}\right )+\frac {64}{5} \sqrt {3 x^2+5 x+2} x^{3/2}\right )-\frac {10}{21} x^{5/2} \sqrt {3 x^2+5 x+2}\)

\(\Big \downarrow \) 1503

\(\displaystyle \frac {2}{21} \left (\frac {1}{5} \left (\frac {4}{9} \left (515 \int \frac {1}{\sqrt {3 x^2+5 x+2}}d\sqrt {x}+1711 \int \frac {x}{\sqrt {3 x^2+5 x+2}}d\sqrt {x}\right )-\frac {1030}{9} \sqrt {x} \sqrt {3 x^2+5 x+2}\right )+\frac {64}{5} \sqrt {3 x^2+5 x+2} x^{3/2}\right )-\frac {10}{21} x^{5/2} \sqrt {3 x^2+5 x+2}\)

\(\Big \downarrow \) 1413

\(\displaystyle \frac {2}{21} \left (\frac {1}{5} \left (\frac {4}{9} \left (1711 \int \frac {x}{\sqrt {3 x^2+5 x+2}}d\sqrt {x}+\frac {515 (x+1) \sqrt {\frac {3 x+2}{x+1}} \operatorname {EllipticF}\left (\arctan \left (\sqrt {x}\right ),-\frac {1}{2}\right )}{\sqrt {2} \sqrt {3 x^2+5 x+2}}\right )-\frac {1030}{9} \sqrt {x} \sqrt {3 x^2+5 x+2}\right )+\frac {64}{5} \sqrt {3 x^2+5 x+2} x^{3/2}\right )-\frac {10}{21} x^{5/2} \sqrt {3 x^2+5 x+2}\)

\(\Big \downarrow \) 1456

\(\displaystyle \frac {2}{21} \left (\frac {1}{5} \left (\frac {4}{9} \left (\frac {515 (x+1) \sqrt {\frac {3 x+2}{x+1}} \operatorname {EllipticF}\left (\arctan \left (\sqrt {x}\right ),-\frac {1}{2}\right )}{\sqrt {2} \sqrt {3 x^2+5 x+2}}+1711 \left (\frac {\sqrt {x} (3 x+2)}{3 \sqrt {3 x^2+5 x+2}}-\frac {\sqrt {2} (x+1) \sqrt {\frac {3 x+2}{x+1}} E\left (\arctan \left (\sqrt {x}\right )|-\frac {1}{2}\right )}{3 \sqrt {3 x^2+5 x+2}}\right )\right )-\frac {1030}{9} \sqrt {x} \sqrt {3 x^2+5 x+2}\right )+\frac {64}{5} \sqrt {3 x^2+5 x+2} x^{3/2}\right )-\frac {10}{21} x^{5/2} \sqrt {3 x^2+5 x+2}\)

Input:

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

Output:

(-10*x^(5/2)*Sqrt[2 + 5*x + 3*x^2])/21 + (2*((64*x^(3/2)*Sqrt[2 + 5*x + 3* 
x^2])/5 + ((-1030*Sqrt[x]*Sqrt[2 + 5*x + 3*x^2])/9 + (4*(1711*((Sqrt[x]*(2 
 + 3*x))/(3*Sqrt[2 + 5*x + 3*x^2]) - (Sqrt[2]*(1 + x)*Sqrt[(2 + 3*x)/(1 + 
x)]*EllipticE[ArcTan[Sqrt[x]], -1/2])/(3*Sqrt[2 + 5*x + 3*x^2])) + (515*(1 
 + x)*Sqrt[(2 + 3*x)/(1 + x)]*EllipticF[ArcTan[Sqrt[x]], -1/2])/(Sqrt[2]*S 
qrt[2 + 5*x + 3*x^2])))/9)/5))/21
 

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 1236

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

rule 1240

Int[((f_) + (g_.)*(x_))/(Sqrt[x_]*Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2]), 
x_Symbol] :> Simp[2   Subst[Int[(f + g*x^2)/Sqrt[a + b*x^2 + c*x^4], x], x, 
 Sqrt[x]], x] /; FreeQ[{a, b, c, f, g}, x]
 

rule 1413

Int[1/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b 
^2 - 4*a*c, 2]}, Simp[(2*a + (b - q)*x^2)*(Sqrt[(2*a + (b + q)*x^2)/(2*a + 
(b - q)*x^2)]/(2*a*Rt[(b - q)/(2*a), 2]*Sqrt[a + b*x^2 + c*x^4]))*EllipticF 
[ArcTan[Rt[(b - q)/(2*a), 2]*x], -2*(q/(b - q))], x] /; PosQ[(b - q)/a]] /; 
 FreeQ[{a, b, c}, x] && GtQ[b^2 - 4*a*c, 0]
 

rule 1456

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = 
 Rt[b^2 - 4*a*c, 2]}, Simp[x*((b - q + 2*c*x^2)/(2*c*Sqrt[a + b*x^2 + c*x^4 
])), x] - Simp[Rt[(b - q)/(2*a), 2]*(2*a + (b - q)*x^2)*(Sqrt[(2*a + (b + q 
)*x^2)/(2*a + (b - q)*x^2)]/(2*c*Sqrt[a + b*x^2 + c*x^4]))*EllipticE[ArcTan 
[Rt[(b - q)/(2*a), 2]*x], -2*(q/(b - q))], x] /; PosQ[(b - q)/a]] /; FreeQ[ 
{a, b, c}, x] && GtQ[b^2 - 4*a*c, 0]
 

rule 1503

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbo 
l] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[d   Int[1/Sqrt[a + b*x^2 + c*x^4] 
, x], x] + Simp[e   Int[x^2/Sqrt[a + b*x^2 + c*x^4], x], x] /; PosQ[(b + q) 
/a] || PosQ[(b - q)/a]] /; FreeQ[{a, b, c, d, e}, x] && GtQ[b^2 - 4*a*c, 0]
 

Maple [A] (verified)

Time = 1.02 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.62

method	result	size
default	\(-\frac {2 \left (6075 x^{5}+7176 \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {6}\, \sqrt {-x}\, \operatorname {EllipticF}\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )-3422 \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {6}\, \sqrt {-x}\, \operatorname {EllipticE}\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )-5427 x^{4}+5940 x^{3}+35982 x^{2}+18540 x \right )}{8505 \sqrt {x}\, \sqrt {3 x^{2}+5 x +2}}\)	\(122\)
risch	\(-\frac {2 \left (225 x^{2}-576 x +1030\right ) \sqrt {x}\, \sqrt {3 x^{2}+5 x +2}}{945}-\frac {\left (-\frac {412 \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {-6 x}\, \operatorname {EllipticF}\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )}{567 \sqrt {3 x^{3}+5 x^{2}+2 x}}-\frac {6844 \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {-6 x}\, \left (\frac {\operatorname {EllipticE}\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )}{3}-\operatorname {EllipticF}\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )\right )}{2835 \sqrt {3 x^{3}+5 x^{2}+2 x}}\right ) \sqrt {\left (3 x^{2}+5 x +2\right ) x}}{\sqrt {x}\, \sqrt {3 x^{2}+5 x +2}}\)	\(188\)
elliptic	\(\frac {\sqrt {\left (3 x^{2}+5 x +2\right ) x}\, \left (-\frac {10 x^{2} \sqrt {3 x^{3}+5 x^{2}+2 x}}{21}+\frac {128 x \sqrt {3 x^{3}+5 x^{2}+2 x}}{105}-\frac {412 \sqrt {3 x^{3}+5 x^{2}+2 x}}{189}+\frac {412 \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {-6 x}\, \operatorname {EllipticF}\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )}{567 \sqrt {3 x^{3}+5 x^{2}+2 x}}+\frac {6844 \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {-6 x}\, \left (\frac {\operatorname {EllipticE}\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )}{3}-\operatorname {EllipticF}\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )\right )}{2835 \sqrt {3 x^{3}+5 x^{2}+2 x}}\right )}{\sqrt {x}\, \sqrt {3 x^{2}+5 x +2}}\)	\(217\)

Input:

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

Output:

-2/8505/x^(1/2)/(3*x^2+5*x+2)^(1/2)*(6075*x^5+7176*(6*x+4)^(1/2)*(3+3*x)^( 
1/2)*6^(1/2)*(-x)^(1/2)*EllipticF(1/2*(6*x+4)^(1/2),I*2^(1/2))-3422*(6*x+4 
)^(1/2)*(3+3*x)^(1/2)*6^(1/2)*(-x)^(1/2)*EllipticE(1/2*(6*x+4)^(1/2),I*2^( 
1/2))-5427*x^4+5940*x^3+35982*x^2+18540*x)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.27 \[ \int \frac {(2-5 x) x^{5/2}}{\sqrt {2+5 x+3 x^2}} \, dx=-\frac {2}{945} \, {\left (225 \, x^{2} - 576 \, x + 1030\right )} \sqrt {3 \, x^{2} + 5 \, x + 2} \sqrt {x} - \frac {896}{729} \, \sqrt {3} {\rm weierstrassPInverse}\left (\frac {28}{27}, \frac {80}{729}, x + \frac {5}{9}\right ) - \frac {13688}{2835} \, \sqrt {3} {\rm weierstrassZeta}\left (\frac {28}{27}, \frac {80}{729}, {\rm weierstrassPInverse}\left (\frac {28}{27}, \frac {80}{729}, x + \frac {5}{9}\right )\right ) \] Input:

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

Output:

-2/945*(225*x^2 - 576*x + 1030)*sqrt(3*x^2 + 5*x + 2)*sqrt(x) - 896/729*sq 
rt(3)*weierstrassPInverse(28/27, 80/729, x + 5/9) - 13688/2835*sqrt(3)*wei 
erstrassZeta(28/27, 80/729, weierstrassPInverse(28/27, 80/729, x + 5/9))
 

Sympy [F]

\[ \int \frac {(2-5 x) x^{5/2}}{\sqrt {2+5 x+3 x^2}} \, dx=- \int \left (- \frac {2 x^{\frac {5}{2}}}{\sqrt {3 x^{2} + 5 x + 2}}\right )\, dx - \int \frac {5 x^{\frac {7}{2}}}{\sqrt {3 x^{2} + 5 x + 2}}\, dx \] Input:

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

Output:

-Integral(-2*x**(5/2)/sqrt(3*x**2 + 5*x + 2), x) - Integral(5*x**(7/2)/sqr 
t(3*x**2 + 5*x + 2), x)
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {(2-5 x) x^{5/2}}{\sqrt {2+5 x+3 x^2}} \, dx=-\frac {10 \sqrt {x}\, \sqrt {3 x^{2}+5 x +2}\, x^{2}}{21}+\frac {128 \sqrt {x}\, \sqrt {3 x^{2}+5 x +2}\, x}{105}-\frac {128 \sqrt {x}\, \sqrt {3 x^{2}+5 x +2}}{175}-\frac {3422 \left (\int \frac {\sqrt {x}\, \sqrt {3 x^{2}+5 x +2}\, x}{3 x^{2}+5 x +2}d x \right )}{525}+\frac {128 \left (\int \frac {\sqrt {x}\, \sqrt {3 x^{2}+5 x +2}}{3 x^{3}+5 x^{2}+2 x}d x \right )}{175} \] Input:

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

Output:

(2*( - 125*sqrt(x)*sqrt(3*x**2 + 5*x + 2)*x**2 + 320*sqrt(x)*sqrt(3*x**2 + 
 5*x + 2)*x - 192*sqrt(x)*sqrt(3*x**2 + 5*x + 2) - 1711*int((sqrt(x)*sqrt( 
3*x**2 + 5*x + 2)*x)/(3*x**2 + 5*x + 2),x) + 192*int((sqrt(x)*sqrt(3*x**2 
+ 5*x + 2))/(3*x**3 + 5*x**2 + 2*x),x)))/525