\(\int \frac {(2-5 x) x^{7/2}}{(2+5 x+3 x^2)^{5/2}} \, dx\) [1075]

Optimal result

Integrand size = 25, antiderivative size = 187 \[ \int \frac {(2-5 x) x^{7/2}}{\left (2+5 x+3 x^2\right )^{5/2}} \, dx=\frac {2 x^{5/2} (74+95 x)}{9 \left (2+5 x+3 x^2\right )^{3/2}}+\frac {8020 \sqrt {x} (2+3 x)}{81 \sqrt {2+5 x+3 x^2}}-\frac {40 \sqrt {x} (167+206 x)}{27 \sqrt {2+5 x+3 x^2}}-\frac {8020 \sqrt {2} (1+x) \sqrt {\frac {2+3 x}{1+x}} E\left (\arctan \left (\sqrt {x}\right )|-\frac {1}{2}\right )}{81 \sqrt {2+5 x+3 x^2}}+\frac {3340 \sqrt {2} (1+x) \sqrt {\frac {2+3 x}{1+x}} \operatorname {EllipticF}\left (\arctan \left (\sqrt {x}\right ),-\frac {1}{2}\right )}{27 \sqrt {2+5 x+3 x^2}} \]

[Out]

2/9*x^(5/2)*(74+95*x)/(3*x^2+5*x+2)^(3/2)+8020/81*(2+3*x)*x^(1/2)/(3*x^2+5*x+2)^(1/2)-40/27*(167+206*x)*x^(1/2
)/(3*x^2+5*x+2)^(1/2)-8020/81*(1+x)^(3/2)*(1/(1+x))^(1/2)*EllipticE(x^(1/2)/(1+x)^(1/2),1/2*I*2^(1/2))*2^(1/2)
*((2+3*x)/(1+x))^(1/2)/(3*x^2+5*x+2)^(1/2)+3340/27*(1+x)^(3/2)*(1/(1+x))^(1/2)*EllipticF(x^(1/2)/(1+x)^(1/2),1
/2*I*2^(1/2))*2^(1/2)*((2+3*x)/(1+x))^(1/2)/(3*x^2+5*x+2)^(1/2)

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {832, 853, 1203, 1114, 1150} \[ \int \frac {(2-5 x) x^{7/2}}{\left (2+5 x+3 x^2\right )^{5/2}} \, dx=\frac {3340 \sqrt {2} (x+1) \sqrt {\frac {3 x+2}{x+1}} \operatorname {EllipticF}\left (\arctan \left (\sqrt {x}\right ),-\frac {1}{2}\right )}{27 \sqrt {3 x^2+5 x+2}}-\frac {8020 \sqrt {2} (x+1) \sqrt {\frac {3 x+2}{x+1}} E\left (\arctan \left (\sqrt {x}\right )|-\frac {1}{2}\right )}{81 \sqrt {3 x^2+5 x+2}}+\frac {8020 (3 x+2) \sqrt {x}}{81 \sqrt {3 x^2+5 x+2}}-\frac {40 (206 x+167) \sqrt {x}}{27 \sqrt {3 x^2+5 x+2}}+\frac {2 (95 x+74) x^{5/2}}{9 \left (3 x^2+5 x+2\right )^{3/2}} \]

[In]

Int[((2 - 5*x)*x^(7/2))/(2 + 5*x + 3*x^2)^(5/2),x]

[Out]

(2*x^(5/2)*(74 + 95*x))/(9*(2 + 5*x + 3*x^2)^(3/2)) + (8020*Sqrt[x]*(2 + 3*x))/(81*Sqrt[2 + 5*x + 3*x^2]) - (4
0*Sqrt[x]*(167 + 206*x))/(27*Sqrt[2 + 5*x + 3*x^2]) - (8020*Sqrt[2]*(1 + x)*Sqrt[(2 + 3*x)/(1 + x)]*EllipticE[
ArcTan[Sqrt[x]], -1/2])/(81*Sqrt[2 + 5*x + 3*x^2]) + (3340*Sqrt[2]*(1 + x)*Sqrt[(2 + 3*x)/(1 + x)]*EllipticF[A
rcTan[Sqrt[x]], -1/2])/(27*Sqrt[2 + 5*x + 3*x^2])

Rule 832

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[(-(d + e*x)^(m - 1))*(a + b*x + c*x^2)^(p + 1)*((2*a*c*(e*f + d*g) - b*(c*d*f + a*e*g) - (2*c^2*d*f + b^2*e*
g - c*(b*e*f + b*d*g + 2*a*e*g))*x)/(c*(p + 1)*(b^2 - 4*a*c))), x] - Dist[1/(c*(p + 1)*(b^2 - 4*a*c)), Int[(d
+ e*x)^(m - 2)*(a + b*x + c*x^2)^(p + 1)*Simp[2*c^2*d^2*f*(2*p + 3) + b*e*g*(a*e*(m - 1) + b*d*(p + 2)) - c*(2
*a*e*(e*f*(m - 1) + d*g*m) + b*d*(d*g*(2*p + 3) - e*f*(m - 2*p - 4))) + e*(b^2*e*g*(m + p + 1) + 2*c^2*d*f*(m
+ 2*p + 2) - c*(2*a*e*g*m + b*(e*f + d*g)*(m + 2*p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] &&
NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 1] && ((EqQ[m, 2] && EqQ[p, -3] &
& RationalQ[a, b, c, d, e, f, g]) ||  !ILtQ[m + 2*p + 3, 0])

Rule 853

Int[((f_) + (g_.)*(x_))/(Sqrt[x_]*Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[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] && NeQ[b^2 - 4*a*c, 0]

Rule 1114

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]))*Elli
pticF[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 1150

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 1203

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}
, Dist[d, Int[1/Sqrt[a + b*x^2 + c*x^4], x], x] + Dist[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]

Rubi steps \begin{align*} \text {integral}& = \frac {2 x^{5/2} (74+95 x)}{9 \left (2+5 x+3 x^2\right )^{3/2}}+\frac {2}{9} \int \frac {(-185-55 x) x^{3/2}}{\left (2+5 x+3 x^2\right )^{3/2}} \, dx \\ & = \frac {2 x^{5/2} (74+95 x)}{9 \left (2+5 x+3 x^2\right )^{3/2}}-\frac {40 \sqrt {x} (167+206 x)}{27 \sqrt {2+5 x+3 x^2}}+\frac {4}{27} \int \frac {835+\frac {2005 x}{2}}{\sqrt {x} \sqrt {2+5 x+3 x^2}} \, dx \\ & = \frac {2 x^{5/2} (74+95 x)}{9 \left (2+5 x+3 x^2\right )^{3/2}}-\frac {40 \sqrt {x} (167+206 x)}{27 \sqrt {2+5 x+3 x^2}}+\frac {8}{27} \text {Subst}\left (\int \frac {835+\frac {2005 x^2}{2}}{\sqrt {2+5 x^2+3 x^4}} \, dx,x,\sqrt {x}\right ) \\ & = \frac {2 x^{5/2} (74+95 x)}{9 \left (2+5 x+3 x^2\right )^{3/2}}-\frac {40 \sqrt {x} (167+206 x)}{27 \sqrt {2+5 x+3 x^2}}+\frac {6680}{27} \text {Subst}\left (\int \frac {1}{\sqrt {2+5 x^2+3 x^4}} \, dx,x,\sqrt {x}\right )+\frac {8020}{27} \text {Subst}\left (\int \frac {x^2}{\sqrt {2+5 x^2+3 x^4}} \, dx,x,\sqrt {x}\right ) \\ & = \frac {2 x^{5/2} (74+95 x)}{9 \left (2+5 x+3 x^2\right )^{3/2}}+\frac {8020 \sqrt {x} (2+3 x)}{81 \sqrt {2+5 x+3 x^2}}-\frac {40 \sqrt {x} (167+206 x)}{27 \sqrt {2+5 x+3 x^2}}-\frac {8020 \sqrt {2} (1+x) \sqrt {\frac {2+3 x}{1+x}} E\left (\tan ^{-1}\left (\sqrt {x}\right )|-\frac {1}{2}\right )}{81 \sqrt {2+5 x+3 x^2}}+\frac {3340 \sqrt {2} (1+x) \sqrt {\frac {2+3 x}{1+x}} F\left (\tan ^{-1}\left (\sqrt {x}\right )|-\frac {1}{2}\right )}{27 \sqrt {2+5 x+3 x^2}} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 21.21 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.90 \[ \int \frac {(2-5 x) x^{7/2}}{\left (2+5 x+3 x^2\right )^{5/2}} \, dx=\frac {32080+120320 x+147100 x^2+58212 x^3-270 x^4+8020 i \sqrt {2+\frac {2}{x}} \sqrt {3+\frac {2}{x}} x^{3/2} \left (2+5 x+3 x^2\right ) E\left (i \text {arcsinh}\left (\frac {\sqrt {\frac {2}{3}}}{\sqrt {x}}\right )|\frac {3}{2}\right )+2000 i \sqrt {2+\frac {2}{x}} \sqrt {3+\frac {2}{x}} x^{3/2} \left (2+5 x+3 x^2\right ) \operatorname {EllipticF}\left (i \text {arcsinh}\left (\frac {\sqrt {\frac {2}{3}}}{\sqrt {x}}\right ),\frac {3}{2}\right )}{81 \sqrt {x} \left (2+5 x+3 x^2\right )^{3/2}} \]

[In]

Integrate[((2 - 5*x)*x^(7/2))/(2 + 5*x + 3*x^2)^(5/2),x]

[Out]

(32080 + 120320*x + 147100*x^2 + 58212*x^3 - 270*x^4 + (8020*I)*Sqrt[2 + 2/x]*Sqrt[3 + 2/x]*x^(3/2)*(2 + 5*x +
3*x^2)*EllipticE[I*ArcSinh[Sqrt[2/3]/Sqrt[x]], 3/2] + (2000*I)*Sqrt[2 + 2/x]*Sqrt[3 + 2/x]*x^(3/2)*(2 + 5*x +
3*x^2)*EllipticF[I*ArcSinh[Sqrt[2/3]/Sqrt[x]], 3/2])/(81*Sqrt[x]*(2 + 5*x + 3*x^2)^(3/2))

Maple [A] (verified)

Time = 0.21 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.14

method	result	size
elliptic	\(\frac {\sqrt {x \left (3 x^{2}+5 x +2\right )}\, \left (\frac {\left (\frac {1012}{729}+\frac {1390 x}{729}\right ) \sqrt {3 x^{3}+5 x^{2}+2 x}}{\left (x^{2}+\frac {5}{3} x +\frac {2}{3}\right )^{2}}-\frac {2 x \left (\frac {10273}{243}+\frac {4025 x}{81}\right ) \sqrt {3}}{\sqrt {x \left (x^{2}+\frac {5}{3} x +\frac {2}{3}\right )}}+\frac {3340 \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {-6 x}\, F\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )}{81 \sqrt {3 x^{3}+5 x^{2}+2 x}}+\frac {4010 \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {-6 x}\, \left (\frac {E\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )}{3}-F\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )\right )}{81 \sqrt {3 x^{3}+5 x^{2}+2 x}}\right )}{\sqrt {x}\, \sqrt {3 x^{2}+5 x +2}}\)	\(214\)
default	\(-\frac {2 \left (3015 \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {6}\, \sqrt {-x}\, F\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right ) x^{2}-6015 \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {6}\, \sqrt {-x}\, E\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right ) x^{2}+5025 \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {6}\, \sqrt {-x}\, F\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right ) x -10025 \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {6}\, \sqrt {-x}\, E\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right ) x +2010 \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {6}\, \sqrt {-x}\, F\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )-4010 \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {6}\, \sqrt {-x}\, E\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )+108675 x^{4}+273582 x^{3}+224460 x^{2}+60120 x \right ) \sqrt {3 x^{2}+5 x +2}}{243 \sqrt {x}\, \left (2+3 x \right )^{2} \left (1+x \right )^{2}}\)	\(297\)

[In]

int((2-5*x)*x^(7/2)/(3*x^2+5*x+2)^(5/2),x,method=_RETURNVERBOSE)

[Out]

(x*(3*x^2+5*x+2))^(1/2)/x^(1/2)/(3*x^2+5*x+2)^(1/2)*((1012/729+1390/729*x)*(3*x^3+5*x^2+2*x)^(1/2)/(x^2+5/3*x+
2/3)^2-2*x*(10273/243+4025/81*x)*3^(1/2)/(x*(x^2+5/3*x+2/3))^(1/2)+3340/81*(6*x+4)^(1/2)*(3+3*x)^(1/2)*(-6*x)^
(1/2)/(3*x^3+5*x^2+2*x)^(1/2)*EllipticF(1/2*(6*x+4)^(1/2),I*2^(1/2))+4010/81*(6*x+4)^(1/2)*(3+3*x)^(1/2)*(-6*x
)^(1/2)/(3*x^3+5*x^2+2*x)^(1/2)*(1/3*EllipticE(1/2*(6*x+4)^(1/2),I*2^(1/2))-EllipticF(1/2*(6*x+4)^(1/2),I*2^(1
/2))))

Fricas [C] (verification not implemented)

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

Time = 0.12 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.65 \[ \int \frac {(2-5 x) x^{7/2}}{\left (2+5 x+3 x^2\right )^{5/2}} \, dx=\frac {2 \, {\left (10010 \, \sqrt {3} {\left (9 \, x^{4} + 30 \, x^{3} + 37 \, x^{2} + 20 \, x + 4\right )} {\rm weierstrassPInverse}\left (\frac {28}{27}, \frac {80}{729}, x + \frac {5}{9}\right ) - 36090 \, \sqrt {3} {\left (9 \, x^{4} + 30 \, x^{3} + 37 \, x^{2} + 20 \, x + 4\right )} {\rm weierstrassZeta}\left (\frac {28}{27}, \frac {80}{729}, {\rm weierstrassPInverse}\left (\frac {28}{27}, \frac {80}{729}, x + \frac {5}{9}\right )\right ) - 27 \, {\left (12075 \, x^{3} + 30398 \, x^{2} + 24940 \, x + 6680\right )} \sqrt {3 \, x^{2} + 5 \, x + 2} \sqrt {x}\right )}}{729 \, {\left (9 \, x^{4} + 30 \, x^{3} + 37 \, x^{2} + 20 \, x + 4\right )}} \]

[In]

integrate((2-5*x)*x^(7/2)/(3*x^2+5*x+2)^(5/2),x, algorithm="fricas")

[Out]

2/729*(10010*sqrt(3)*(9*x^4 + 30*x^3 + 37*x^2 + 20*x + 4)*weierstrassPInverse(28/27, 80/729, x + 5/9) - 36090*
sqrt(3)*(9*x^4 + 30*x^3 + 37*x^2 + 20*x + 4)*weierstrassZeta(28/27, 80/729, weierstrassPInverse(28/27, 80/729,
x + 5/9)) - 27*(12075*x^3 + 30398*x^2 + 24940*x + 6680)*sqrt(3*x^2 + 5*x + 2)*sqrt(x))/(9*x^4 + 30*x^3 + 37*x
^2 + 20*x + 4)

Sympy [F(-1)]

Timed out. \[ \int \frac {(2-5 x) x^{7/2}}{\left (2+5 x+3 x^2\right )^{5/2}} \, dx=\text {Timed out} \]

[In]

integrate((2-5*x)*x**(7/2)/(3*x**2+5*x+2)**(5/2),x)

[Out]

Timed out

Maxima [F]

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

[In]

integrate((2-5*x)*x^(7/2)/(3*x^2+5*x+2)^(5/2),x, algorithm="maxima")

[Out]

-integrate((5*x - 2)*x^(7/2)/(3*x^2 + 5*x + 2)^(5/2), x)

Giac [F]

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

[In]

integrate((2-5*x)*x^(7/2)/(3*x^2+5*x+2)^(5/2),x, algorithm="giac")

[Out]

integrate(-(5*x - 2)*x^(7/2)/(3*x^2 + 5*x + 2)^(5/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {(2-5 x) x^{7/2}}{\left (2+5 x+3 x^2\right )^{5/2}} \, dx=-\int \frac {x^{7/2}\,\left (5\,x-2\right )}{{\left (3\,x^2+5\,x+2\right )}^{5/2}} \,d x \]

[In]

int(-(x^(7/2)*(5*x - 2))/(5*x + 3*x^2 + 2)^(5/2),x)

[Out]

-int((x^(7/2)*(5*x - 2))/(5*x + 3*x^2 + 2)^(5/2), x)