3.1050 \(\int \frac {(2-5 x) (2+5 x+3 x^2)^{3/2}}{x^{7/2}} \, dx\)

Optimal. Leaf size=185 \[ \frac {2 (89-35 x) \sqrt {3 x^2+5 x+2}}{5 \sqrt {x}}-\frac {1418 \sqrt {x} (3 x+2)}{15 \sqrt {3 x^2+5 x+2}}-\frac {117 \sqrt {2} (x+1) \sqrt {\frac {3 x+2}{x+1}} F\left (\tan ^{-1}\left (\sqrt {x}\right )|-\frac {1}{2}\right )}{\sqrt {3 x^2+5 x+2}}+\frac {1418 \sqrt {2} (x+1) \sqrt {\frac {3 x+2}{x+1}} E\left (\tan ^{-1}\left (\sqrt {x}\right )|-\frac {1}{2}\right )}{15 \sqrt {3 x^2+5 x+2}}-\frac {4 (3-5 x) \left (3 x^2+5 x+2\right )^{3/2}}{15 x^{5/2}} \]

[Out]

-4/15*(3-5*x)*(3*x^2+5*x+2)^(3/2)/x^(5/2)-1418/15*(2+3*x)*x^(1/2)/(3*x^2+5*x+2)^(1/2)+1418/15*(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)-117
*(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)+2/5*(89-35*x)*(3*x^2+5*x+2)^(1/2)/x^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.12, antiderivative size = 185, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {810, 812, 839, 1189, 1100, 1136} \[ -\frac {4 (3-5 x) \left (3 x^2+5 x+2\right )^{3/2}}{15 x^{5/2}}+\frac {2 (89-35 x) \sqrt {3 x^2+5 x+2}}{5 \sqrt {x}}-\frac {1418 \sqrt {x} (3 x+2)}{15 \sqrt {3 x^2+5 x+2}}-\frac {117 \sqrt {2} (x+1) \sqrt {\frac {3 x+2}{x+1}} F\left (\tan ^{-1}\left (\sqrt {x}\right )|-\frac {1}{2}\right )}{\sqrt {3 x^2+5 x+2}}+\frac {1418 \sqrt {2} (x+1) \sqrt {\frac {3 x+2}{x+1}} E\left (\tan ^{-1}\left (\sqrt {x}\right )|-\frac {1}{2}\right )}{15 \sqrt {3 x^2+5 x+2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-1418*Sqrt[x]*(2 + 3*x))/(15*Sqrt[2 + 5*x + 3*x^2]) + (2*(89 - 35*x)*Sqrt[2 + 5*x + 3*x^2])/(5*Sqrt[x]) - (4*
(3 - 5*x)*(2 + 5*x + 3*x^2)^(3/2))/(15*x^(5/2)) + (1418*Sqrt[2]*(1 + x)*Sqrt[(2 + 3*x)/(1 + x)]*EllipticE[ArcT
an[Sqrt[x]], -1/2])/(15*Sqrt[2 + 5*x + 3*x^2]) - (117*Sqrt[2]*(1 + x)*Sqrt[(2 + 3*x)/(1 + x)]*EllipticF[ArcTan
[Sqrt[x]], -1/2])/Sqrt[2 + 5*x + 3*x^2]

Rule 810

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Si
mp[((d + e*x)^(m + 1)*(a + b*x + c*x^2)^p*((d*g - e*f*(m + 2))*(c*d^2 - b*d*e + a*e^2) - d*p*(2*c*d - b*e)*(e*
f - d*g) - e*(g*(m + 1)*(c*d^2 - b*d*e + a*e^2) + p*(2*c*d - b*e)*(e*f - d*g))*x))/(e^2*(m + 1)*(m + 2)*(c*d^2
- b*d*e + a*e^2)), x] - Dist[p/(e^2*(m + 1)*(m + 2)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^(m + 2)*(a + b*x
+ c*x^2)^(p - 1)*Simp[2*a*c*e*(e*f - d*g)*(m + 2) + b^2*e*(d*g*(p + 1) - e*f*(m + p + 2)) + b*(a*e^2*g*(m + 1)
- c*d*(d*g*(2*p + 1) - e*f*(m + 2*p + 2))) - c*(2*c*d*(d*g*(2*p + 1) - e*f*(m + 2*p + 2)) - e*(2*a*e*g*(m + 1
) - b*(d*g*(m - 2*p) + e*f*(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] && GtQ[p, 0] && LtQ[m, -2] && LtQ[m + 2*p, 0] &&  !ILtQ[m + 2*p + 3, 0]

Rule 812

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[((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] + Dist[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] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2
, 0] && RationalQ[p] && 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 839

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 1100

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)]*EllipticF[ArcTan[Rt[(b - q)/(2*a), 2]*x], (-2*q)/(b - q)
])/(2*a*Rt[(b - q)/(2*a), 2]*Sqrt[a + b*x^2 + c*x^4]), x] /; PosQ[(b - q)/a]] /; FreeQ[{a, b, c}, x] && GtQ[b^
2 - 4*a*c, 0]

Rule 1136

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)]*EllipticE[ArcTan[Rt[(b - q)/(2*a), 2]*x], (-2*q)/(b - q)])/(2*c*Sqrt[a + b*x^
2 + c*x^4]), x] /; PosQ[(b - q)/a]] /; FreeQ[{a, b, c}, x] && GtQ[b^2 - 4*a*c, 0]

Rule 1189

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*} \int \frac {(2-5 x) \left (2+5 x+3 x^2\right )^{3/2}}{x^{7/2}} \, dx &=-\frac {4 (3-5 x) \left (2+5 x+3 x^2\right )^{3/2}}{15 x^{5/2}}-\frac {1}{5} \int \frac {(89+105 x) \sqrt {2+5 x+3 x^2}}{x^{3/2}} \, dx\\ &=\frac {2 (89-35 x) \sqrt {2+5 x+3 x^2}}{5 \sqrt {x}}-\frac {4 (3-5 x) \left (2+5 x+3 x^2\right )^{3/2}}{15 x^{5/2}}+\frac {2}{15} \int \frac {-\frac {1755}{2}-\frac {2127 x}{2}}{\sqrt {x} \sqrt {2+5 x+3 x^2}} \, dx\\ &=\frac {2 (89-35 x) \sqrt {2+5 x+3 x^2}}{5 \sqrt {x}}-\frac {4 (3-5 x) \left (2+5 x+3 x^2\right )^{3/2}}{15 x^{5/2}}+\frac {4}{15} \operatorname {Subst}\left (\int \frac {-\frac {1755}{2}-\frac {2127 x^2}{2}}{\sqrt {2+5 x^2+3 x^4}} \, dx,x,\sqrt {x}\right )\\ &=\frac {2 (89-35 x) \sqrt {2+5 x+3 x^2}}{5 \sqrt {x}}-\frac {4 (3-5 x) \left (2+5 x+3 x^2\right )^{3/2}}{15 x^{5/2}}-234 \operatorname {Subst}\left (\int \frac {1}{\sqrt {2+5 x^2+3 x^4}} \, dx,x,\sqrt {x}\right )-\frac {1418}{5} \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {2+5 x^2+3 x^4}} \, dx,x,\sqrt {x}\right )\\ &=-\frac {1418 \sqrt {x} (2+3 x)}{15 \sqrt {2+5 x+3 x^2}}+\frac {2 (89-35 x) \sqrt {2+5 x+3 x^2}}{5 \sqrt {x}}-\frac {4 (3-5 x) \left (2+5 x+3 x^2\right )^{3/2}}{15 x^{5/2}}+\frac {1418 \sqrt {2} (1+x) \sqrt {\frac {2+3 x}{1+x}} E\left (\tan ^{-1}\left (\sqrt {x}\right )|-\frac {1}{2}\right )}{15 \sqrt {2+5 x+3 x^2}}-\frac {117 \sqrt {2} (1+x) \sqrt {\frac {2+3 x}{1+x}} F\left (\tan ^{-1}\left (\sqrt {x}\right )|-\frac {1}{2}\right )}{\sqrt {2+5 x+3 x^2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.19, size = 163, normalized size = 0.88 \[ \frac {-337 i \sqrt {2} \sqrt {\frac {1}{x}+1} \sqrt {\frac {2}{x}+3} x^{7/2} F\left (i \sinh ^{-1}\left (\frac {\sqrt {\frac {2}{3}}}{\sqrt {x}}\right )|\frac {3}{2}\right )-1418 i \sqrt {2} \sqrt {\frac {1}{x}+1} \sqrt {\frac {2}{x}+3} x^{7/2} E\left (i \sinh ^{-1}\left (\frac {\sqrt {\frac {2}{3}}}{\sqrt {x}}\right )|\frac {3}{2}\right )-2 \left (225 x^5+1605 x^4+2230 x^3+906 x^2+80 x+24\right )}{15 x^{5/2} \sqrt {3 x^2+5 x+2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(24 + 80*x + 906*x^2 + 2230*x^3 + 1605*x^4 + 225*x^5) - (1418*I)*Sqrt[2]*Sqrt[1 + x^(-1)]*Sqrt[3 + 2/x]*x^
(7/2)*EllipticE[I*ArcSinh[Sqrt[2/3]/Sqrt[x]], 3/2] - (337*I)*Sqrt[2]*Sqrt[1 + x^(-1)]*Sqrt[3 + 2/x]*x^(7/2)*El
lipticF[I*ArcSinh[Sqrt[2/3]/Sqrt[x]], 3/2])/(15*x^(5/2)*Sqrt[2 + 5*x + 3*x^2])

________________________________________________________________________________________

fricas [F]  time = 0.96, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (15 \, x^{3} + 19 \, x^{2} - 4\right )} \sqrt {3 \, x^{2} + 5 \, x + 2}}{x^{\frac {7}{2}}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-(15*x^3 + 19*x^2 - 4)*sqrt(3*x^2 + 5*x + 2)/x^(7/2), x)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int -\frac {{\left (3 \, x^{2} + 5 \, x + 2\right )}^{\frac {3}{2}} {\left (5 \, x - 2\right )}}{x^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.11, size = 129, normalized size = 0.70 \[ \frac {-1350 x^{5}+3132 x^{4}+7890 x^{3}-709 \sqrt {6 x +4}\, \sqrt {3 x +3}\, \sqrt {6}\, \sqrt {-x}\, x^{2} \EllipticE \left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )+372 \sqrt {6 x +4}\, \sqrt {3 x +3}\, \sqrt {6}\, \sqrt {-x}\, x^{2} \EllipticF \left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )+3072 x^{2}-480 x -144}{45 \sqrt {3 x^{2}+5 x +2}\, x^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/45*(372*(6*x+4)^(1/2)*(3*x+3)^(1/2)*6^(1/2)*(-x)^(1/2)*EllipticF(1/2*(6*x+4)^(1/2),I*2^(1/2))*x^2-709*(6*x+4
)^(1/2)*(3*x+3)^(1/2)*6^(1/2)*(-x)^(1/2)*EllipticE(1/2*(6*x+4)^(1/2),I*2^(1/2))*x^2-1350*x^5+3132*x^4+7890*x^3
+3072*x^2-480*x-144)/(3*x^2+5*x+2)^(1/2)/x^(5/2)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\int \frac {{\left (3 \, x^{2} + 5 \, x + 2\right )}^{\frac {3}{2}} {\left (5 \, x - 2\right )}}{x^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int -\frac {\left (5\,x-2\right )\,{\left (3\,x^2+5\,x+2\right )}^{3/2}}{x^{7/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \int \left (- \frac {4 \sqrt {3 x^{2} + 5 x + 2}}{x^{\frac {7}{2}}}\right )\, dx - \int \frac {19 \sqrt {3 x^{2} + 5 x + 2}}{x^{\frac {3}{2}}}\, dx - \int \frac {15 \sqrt {3 x^{2} + 5 x + 2}}{\sqrt {x}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral(-4*sqrt(3*x**2 + 5*x + 2)/x**(7/2), x) - Integral(19*sqrt(3*x**2 + 5*x + 2)/x**(3/2), x) - Integral(
15*sqrt(3*x**2 + 5*x + 2)/sqrt(x), x)

________________________________________________________________________________________