\(\int \frac {(2-5 x) \sqrt {2+5 x+3 x^2}}{x^{3/2}} \, dx\) [184]

Optimal result

Integrand size = 25, antiderivative size = 155 \[ \int \frac {(2-5 x) \sqrt {2+5 x+3 x^2}}{x^{3/2}} \, dx=\frac {22 \sqrt {x} (2+3 x)}{9 \sqrt {2+5 x+3 x^2}}-\frac {2 (6+5 x) \sqrt {2+5 x+3 x^2}}{3 \sqrt {x}}-\frac {22 \sqrt {2} \sqrt {2+5 x+3 x^2} E\left (\arctan \left (\sqrt {x}\right )|-\frac {1}{2}\right )}{9 \sqrt {1+x} \sqrt {2+3 x}}+\frac {10 \sqrt {2} \sqrt {1+x} \sqrt {2+3 x} \operatorname {EllipticF}\left (\arctan \left (\sqrt {x}\right ),-\frac {1}{2}\right )}{3 \sqrt {2+5 x+3 x^2}} \] Output:

22/9*x^(1/2)*(2+3*x)/(3*x^2+5*x+2)^(1/2)-2/3*(6+5*x)*(3*x^2+5*x+2)^(1/2)/x 
^(1/2)-22/9*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)+10/3*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 = 153, normalized size of antiderivative = 0.99 \[ \int \frac {(2-5 x) \sqrt {2+5 x+3 x^2}}{x^{3/2}} \, dx=\frac {-2 \left (14+65 x+96 x^2+45 x^3\right )+22 i \sqrt {2} \sqrt {1+\frac {1}{x}} \sqrt {3+\frac {2}{x}} x^{3/2} E\left (i \text {arcsinh}\left (\frac {\sqrt {\frac {2}{3}}}{\sqrt {x}}\right )|\frac {3}{2}\right )+8 i \sqrt {2} \sqrt {1+\frac {1}{x}} \sqrt {3+\frac {2}{x}} x^{3/2} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\frac {\sqrt {\frac {2}{3}}}{\sqrt {x}}\right ),\frac {3}{2}\right )}{9 \sqrt {x} \sqrt {2+5 x+3 x^2}} \] Input:

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

Output:

(-2*(14 + 65*x + 96*x^2 + 45*x^3) + (22*I)*Sqrt[2]*Sqrt[1 + x^(-1)]*Sqrt[3 
 + 2/x]*x^(3/2)*EllipticE[I*ArcSinh[Sqrt[2/3]/Sqrt[x]], 3/2] + (8*I)*Sqrt[ 
2]*Sqrt[1 + x^(-1)]*Sqrt[3 + 2/x]*x^(3/2)*EllipticF[I*ArcSinh[Sqrt[2/3]/Sq 
rt[x]], 3/2])/(9*Sqrt[x]*Sqrt[2 + 5*x + 3*x^2])
 

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.06, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {1230, 27, 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.

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

\(\Big \downarrow \) 1230

\(\displaystyle -\frac {2}{3} \int -\frac {11 x+10}{2 \sqrt {x} \sqrt {3 x^2+5 x+2}}dx-\frac {2 \sqrt {3 x^2+5 x+2} (5 x+6)}{3 \sqrt {x}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{3} \int \frac {11 x+10}{\sqrt {x} \sqrt {3 x^2+5 x+2}}dx-\frac {2 (5 x+6) \sqrt {3 x^2+5 x+2}}{3 \sqrt {x}}\)

\(\Big \downarrow \) 1240

\(\displaystyle \frac {2}{3} \int \frac {11 x+10}{\sqrt {3 x^2+5 x+2}}d\sqrt {x}-\frac {2 (5 x+6) \sqrt {3 x^2+5 x+2}}{3 \sqrt {x}}\)

\(\Big \downarrow \) 1503

\(\displaystyle \frac {2}{3} \left (10 \int \frac {1}{\sqrt {3 x^2+5 x+2}}d\sqrt {x}+11 \int \frac {x}{\sqrt {3 x^2+5 x+2}}d\sqrt {x}\right )-\frac {2 (5 x+6) \sqrt {3 x^2+5 x+2}}{3 \sqrt {x}}\)

\(\Big \downarrow \) 1413

\(\displaystyle \frac {2}{3} \left (11 \int \frac {x}{\sqrt {3 x^2+5 x+2}}d\sqrt {x}+\frac {5 \sqrt {2} (x+1) \sqrt {\frac {3 x+2}{x+1}} \operatorname {EllipticF}\left (\arctan \left (\sqrt {x}\right ),-\frac {1}{2}\right )}{\sqrt {3 x^2+5 x+2}}\right )-\frac {2 (5 x+6) \sqrt {3 x^2+5 x+2}}{3 \sqrt {x}}\)

\(\Big \downarrow \) 1456

\(\displaystyle \frac {2}{3} \left (\frac {5 \sqrt {2} (x+1) \sqrt {\frac {3 x+2}{x+1}} \operatorname {EllipticF}\left (\arctan \left (\sqrt {x}\right ),-\frac {1}{2}\right )}{\sqrt {3 x^2+5 x+2}}+11 \left (\frac {\sqrt {x} (3 x+2)}{3 \sqrt {3 x^2+5 x+2}}-\frac {\sqrt {2} (x+1) \sqrt {\frac {3 x+2}{x+1}} E\left (\arctan \left (\sqrt {x}\right )|-\frac {1}{2}\right )}{3 \sqrt {3 x^2+5 x+2}}\right )\right )-\frac {2 (5 x+6) \sqrt {3 x^2+5 x+2}}{3 \sqrt {x}}\)

Input:

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

Output:

(-2*(6 + 5*x)*Sqrt[2 + 5*x + 3*x^2])/(3*Sqrt[x]) + (2*(11*((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])) + (5*Sqrt[2]* 
(1 + x)*Sqrt[(2 + 3*x)/(1 + x)]*EllipticF[ArcTan[Sqrt[x]], -1/2])/Sqrt[2 + 
 5*x + 3*x^2]))/3
 

Defintions of rubi rules used

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 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 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 = 1.03 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.73

method	result	size
default	\(-\frac {3 \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {6}\, \sqrt {-x}\, \operatorname {EllipticF}\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )-11 \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {6}\, \sqrt {-x}\, \operatorname {EllipticE}\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )+270 x^{3}+774 x^{2}+720 x +216}{27 \sqrt {3 x^{2}+5 x +2}\, \sqrt {x}}\)	\(113\)
elliptic	\(\frac {\sqrt {\left (3 x^{2}+5 x +2\right ) x}\, \left (-\frac {4 \left (3 x^{2}+5 x +2\right )}{\sqrt {\left (3 x^{2}+5 x +2\right ) x}}-\frac {10 \sqrt {3 x^{3}+5 x^{2}+2 x}}{3}+\frac {10 \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {-6 x}\, \operatorname {EllipticF}\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )}{9 \sqrt {3 x^{3}+5 x^{2}+2 x}}+\frac {11 \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {-6 x}\, \left (\frac {\operatorname {EllipticE}\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )}{3}-\operatorname {EllipticF}\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )\right )}{9 \sqrt {3 x^{3}+5 x^{2}+2 x}}\right )}{\sqrt {x}\, \sqrt {3 x^{2}+5 x +2}}\)	\(203\)

Input:

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

Output:

-1/27*(3*(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))-11*(6*x+4)^(1/2)*(3+3*x)^(1/2)*6^(1/2)*(-x)^(1/2)*Ell 
ipticE(1/2*(6*x+4)^(1/2),I*2^(1/2))+270*x^3+774*x^2+720*x+216)/(3*x^2+5*x+ 
2)^(1/2)/x^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.35 \[ \int \frac {(2-5 x) \sqrt {2+5 x+3 x^2}}{x^{3/2}} \, dx=\frac {2 \, {\left (35 \, \sqrt {3} x {\rm weierstrassPInverse}\left (\frac {28}{27}, \frac {80}{729}, x + \frac {5}{9}\right ) - 99 \, \sqrt {3} x {\rm weierstrassZeta}\left (\frac {28}{27}, \frac {80}{729}, {\rm weierstrassPInverse}\left (\frac {28}{27}, \frac {80}{729}, x + \frac {5}{9}\right )\right ) - 27 \, \sqrt {3 \, x^{2} + 5 \, x + 2} {\left (5 \, x + 6\right )} \sqrt {x}\right )}}{81 \, x} \] Input:

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

Output:

2/81*(35*sqrt(3)*x*weierstrassPInverse(28/27, 80/729, x + 5/9) - 99*sqrt(3 
)*x*weierstrassZeta(28/27, 80/729, weierstrassPInverse(28/27, 80/729, x + 
5/9)) - 27*sqrt(3*x^2 + 5*x + 2)*(5*x + 6)*sqrt(x))/x
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {(2-5 x) \sqrt {2+5 x+3 x^2}}{x^{3/2}} \, dx=\frac {-10 \sqrt {3 x^{2}+5 x +2}\, x -12 \sqrt {3 x^{2}+5 x +2}+10 \sqrt {x}\, \left (\int \frac {\sqrt {x}\, \sqrt {3 x^{2}+5 x +2}}{3 x^{3}+5 x^{2}+2 x}d x \right )+11 \sqrt {x}\, \left (\int \frac {\sqrt {x}\, \sqrt {3 x^{2}+5 x +2}}{3 x^{2}+5 x +2}d x \right )}{3 \sqrt {x}} \] Input:

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

Output:

( - 10*sqrt(3*x**2 + 5*x + 2)*x - 12*sqrt(3*x**2 + 5*x + 2) + 10*sqrt(x)*i 
nt((sqrt(x)*sqrt(3*x**2 + 5*x + 2))/(3*x**3 + 5*x**2 + 2*x),x) + 11*sqrt(x 
)*int((sqrt(x)*sqrt(3*x**2 + 5*x + 2))/(3*x**2 + 5*x + 2),x))/(3*sqrt(x))