3.1071 \(\int \frac{2-5 x}{x^{7/2} (2+5 x+3 x^2)^{3/2}} \, dx\)

Optimal. Leaf size=224 \[ \frac{157 (x+1) \sqrt{\frac{3 x+2}{x+1}} \text{EllipticF}\left (\tan ^{-1}\left (\sqrt{x}\right ),-\frac{1}{2}\right )}{\sqrt{2} \sqrt{3 x^2+5 x+2}}+\frac{2693 \sqrt{x} (3 x+2)}{30 \sqrt{3 x^2+5 x+2}}-\frac{2693 \sqrt{3 x^2+5 x+2}}{30 \sqrt{x}}+\frac{157 \sqrt{3 x^2+5 x+2}}{3 x^{3/2}}-\frac{191 \sqrt{3 x^2+5 x+2}}{5 x^{5/2}}+\frac{2 (45 x+38)}{x^{5/2} \sqrt{3 x^2+5 x+2}}-\frac{2693 (x+1) \sqrt{\frac{3 x+2}{x+1}} E\left (\tan ^{-1}\left (\sqrt{x}\right )|-\frac{1}{2}\right )}{15 \sqrt{2} \sqrt{3 x^2+5 x+2}} \]

[Out]

(2693*Sqrt[x]*(2 + 3*x))/(30*Sqrt[2 + 5*x + 3*x^2]) + (2*(38 + 45*x))/(x^(5/2)*Sqrt[2 + 5*x + 3*x^2]) - (191*S
qrt[2 + 5*x + 3*x^2])/(5*x^(5/2)) + (157*Sqrt[2 + 5*x + 3*x^2])/(3*x^(3/2)) - (2693*Sqrt[2 + 5*x + 3*x^2])/(30
*Sqrt[x]) - (2693*(1 + x)*Sqrt[(2 + 3*x)/(1 + x)]*EllipticE[ArcTan[Sqrt[x]], -1/2])/(15*Sqrt[2]*Sqrt[2 + 5*x +
3*x^2]) + (157*(1 + x)*Sqrt[(2 + 3*x)/(1 + x)]*EllipticF[ArcTan[Sqrt[x]], -1/2])/(Sqrt[2]*Sqrt[2 + 5*x + 3*x^
2])

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Rubi [A]  time = 0.143889, antiderivative size = 224, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {822, 834, 839, 1189, 1100, 1136} \[ \frac{2693 \sqrt{x} (3 x+2)}{30 \sqrt{3 x^2+5 x+2}}-\frac{2693 \sqrt{3 x^2+5 x+2}}{30 \sqrt{x}}+\frac{157 \sqrt{3 x^2+5 x+2}}{3 x^{3/2}}-\frac{191 \sqrt{3 x^2+5 x+2}}{5 x^{5/2}}+\frac{2 (45 x+38)}{x^{5/2} \sqrt{3 x^2+5 x+2}}+\frac{157 (x+1) \sqrt{\frac{3 x+2}{x+1}} F\left (\tan ^{-1}\left (\sqrt{x}\right )|-\frac{1}{2}\right )}{\sqrt{2} \sqrt{3 x^2+5 x+2}}-\frac{2693 (x+1) \sqrt{\frac{3 x+2}{x+1}} E\left (\tan ^{-1}\left (\sqrt{x}\right )|-\frac{1}{2}\right )}{15 \sqrt{2} \sqrt{3 x^2+5 x+2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2693*Sqrt[x]*(2 + 3*x))/(30*Sqrt[2 + 5*x + 3*x^2]) + (2*(38 + 45*x))/(x^(5/2)*Sqrt[2 + 5*x + 3*x^2]) - (191*S
qrt[2 + 5*x + 3*x^2])/(5*x^(5/2)) + (157*Sqrt[2 + 5*x + 3*x^2])/(3*x^(3/2)) - (2693*Sqrt[2 + 5*x + 3*x^2])/(30
*Sqrt[x]) - (2693*(1 + x)*Sqrt[(2 + 3*x)/(1 + x)]*EllipticE[ArcTan[Sqrt[x]], -1/2])/(15*Sqrt[2]*Sqrt[2 + 5*x +
3*x^2]) + (157*(1 + x)*Sqrt[(2 + 3*x)/(1 + x)]*EllipticF[ArcTan[Sqrt[x]], -1/2])/(Sqrt[2]*Sqrt[2 + 5*x + 3*x^
2])

Rule 822

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[((d + e*x)^(m + 1)*(f*(b*c*d - b^2*e + 2*a*c*e) - a*g*(2*c*d - b*e) + c*(f*(2*c*d - b*e) - g*(b*d - 2*a*e))*x
)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2)), x] + Dist[1/((p + 1)*(b^2 - 4*a*
c)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^m*(a + b*x + c*x^2)^(p + 1)*Simp[f*(b*c*d*e*(2*p - m + 2) + b^2*e^2
*(p + m + 2) - 2*c^2*d^2*(2*p + 3) - 2*a*c*e^2*(m + 2*p + 3)) - g*(a*e*(b*e - 2*c*d*m + b*e*m) - b*d*(3*c*d -
b*e + 2*c*d*p - b*e*p)) + c*e*(g*(b*d - 2*a*e) - f*(2*c*d - b*e))*(m + 2*p + 4)*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] && LtQ[p, -1] && (IntegerQ[m] ||
IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 834

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

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]

Rubi steps

\begin{align*} \int \frac{2-5 x}{x^{7/2} \left (2+5 x+3 x^2\right )^{3/2}} \, dx &=\frac{2 (38+45 x)}{x^{5/2} \sqrt{2+5 x+3 x^2}}-\int \frac{-191-225 x}{x^{7/2} \sqrt{2+5 x+3 x^2}} \, dx\\ &=\frac{2 (38+45 x)}{x^{5/2} \sqrt{2+5 x+3 x^2}}-\frac{191 \sqrt{2+5 x+3 x^2}}{5 x^{5/2}}+\frac{1}{5} \int \frac{-785-\frac{1719 x}{2}}{x^{5/2} \sqrt{2+5 x+3 x^2}} \, dx\\ &=\frac{2 (38+45 x)}{x^{5/2} \sqrt{2+5 x+3 x^2}}-\frac{191 \sqrt{2+5 x+3 x^2}}{5 x^{5/2}}+\frac{157 \sqrt{2+5 x+3 x^2}}{3 x^{3/2}}-\frac{1}{15} \int \frac{-\frac{2693}{2}-\frac{2355 x}{2}}{x^{3/2} \sqrt{2+5 x+3 x^2}} \, dx\\ &=\frac{2 (38+45 x)}{x^{5/2} \sqrt{2+5 x+3 x^2}}-\frac{191 \sqrt{2+5 x+3 x^2}}{5 x^{5/2}}+\frac{157 \sqrt{2+5 x+3 x^2}}{3 x^{3/2}}-\frac{2693 \sqrt{2+5 x+3 x^2}}{30 \sqrt{x}}+\frac{1}{15} \int \frac{\frac{2355}{2}+\frac{8079 x}{4}}{\sqrt{x} \sqrt{2+5 x+3 x^2}} \, dx\\ &=\frac{2 (38+45 x)}{x^{5/2} \sqrt{2+5 x+3 x^2}}-\frac{191 \sqrt{2+5 x+3 x^2}}{5 x^{5/2}}+\frac{157 \sqrt{2+5 x+3 x^2}}{3 x^{3/2}}-\frac{2693 \sqrt{2+5 x+3 x^2}}{30 \sqrt{x}}+\frac{2}{15} \operatorname{Subst}\left (\int \frac{\frac{2355}{2}+\frac{8079 x^2}{4}}{\sqrt{2+5 x^2+3 x^4}} \, dx,x,\sqrt{x}\right )\\ &=\frac{2 (38+45 x)}{x^{5/2} \sqrt{2+5 x+3 x^2}}-\frac{191 \sqrt{2+5 x+3 x^2}}{5 x^{5/2}}+\frac{157 \sqrt{2+5 x+3 x^2}}{3 x^{3/2}}-\frac{2693 \sqrt{2+5 x+3 x^2}}{30 \sqrt{x}}+157 \operatorname{Subst}\left (\int \frac{1}{\sqrt{2+5 x^2+3 x^4}} \, dx,x,\sqrt{x}\right )+\frac{2693}{10} \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{2+5 x^2+3 x^4}} \, dx,x,\sqrt{x}\right )\\ &=\frac{2693 \sqrt{x} (2+3 x)}{30 \sqrt{2+5 x+3 x^2}}+\frac{2 (38+45 x)}{x^{5/2} \sqrt{2+5 x+3 x^2}}-\frac{191 \sqrt{2+5 x+3 x^2}}{5 x^{5/2}}+\frac{157 \sqrt{2+5 x+3 x^2}}{3 x^{3/2}}-\frac{2693 \sqrt{2+5 x+3 x^2}}{30 \sqrt{x}}-\frac{2693 (1+x) \sqrt{\frac{2+3 x}{1+x}} E\left (\tan ^{-1}\left (\sqrt{x}\right )|-\frac{1}{2}\right )}{15 \sqrt{2} \sqrt{2+5 x+3 x^2}}+\frac{157 (1+x) \sqrt{\frac{2+3 x}{1+x}} F\left (\tan ^{-1}\left (\sqrt{x}\right )|-\frac{1}{2}\right )}{\sqrt{2} \sqrt{2+5 x+3 x^2}}\\ \end{align*}

Mathematica [C]  time = 0.176469, size = 150, normalized size = 0.67 \[ \frac{-338 i \sqrt{2} \sqrt{\frac{1}{x}+1} \sqrt{\frac{2}{x}+3} x^{7/2} \text{EllipticF}\left (i \sinh ^{-1}\left (\frac{\sqrt{\frac{2}{3}}}{\sqrt{x}}\right ),\frac{3}{2}\right )+4710 x^3+4412 x^2+2693 i \sqrt{2} \sqrt{\frac{1}{x}+1} \sqrt{\frac{2}{x}+3} x^{7/2} E\left (i \sinh ^{-1}\left (\frac{\sqrt{\frac{2}{3}}}{\sqrt{x}}\right )|\frac{3}{2}\right )+110 x-12}{30 x^{5/2} \sqrt{3 x^2+5 x+2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-12 + 110*x + 4412*x^2 + 4710*x^3 + (2693*I)*Sqrt[2]*Sqrt[1 + x^(-1)]*Sqrt[3 + 2/x]*x^(7/2)*EllipticE[I*ArcSi
nh[Sqrt[2/3]/Sqrt[x]], 3/2] - (338*I)*Sqrt[2]*Sqrt[1 + x^(-1)]*Sqrt[3 + 2/x]*x^(7/2)*EllipticF[I*ArcSinh[Sqrt[
2/3]/Sqrt[x]], 3/2])/(30*x^(5/2)*Sqrt[2 + 5*x + 3*x^2])

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Maple [A]  time = 0.023, size = 124, normalized size = 0.6 \begin{align*} -{\frac{1}{180} \left ( 3369\,\sqrt{6\,x+4}\sqrt{3+3\,x}\sqrt{6}\sqrt{-x}{\it EllipticF} \left ( 1/2\,\sqrt{6\,x+4},i\sqrt{2} \right ){x}^{2}-2693\,\sqrt{6\,x+4}\sqrt{3+3\,x}\sqrt{6}\sqrt{-x}{\it EllipticE} \left ( 1/2\,\sqrt{6\,x+4},i\sqrt{2} \right ){x}^{2}+48474\,{x}^{4}+52530\,{x}^{3}+5844\,{x}^{2}-660\,x+72 \right ){x}^{-{\frac{5}{2}}}{\frac{1}{\sqrt{3\,{x}^{2}+5\,x+2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/180*(3369*(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))*x^2-2693*(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))*x^2+48474*x^4+52530*x^3+58
44*x^2-660*x+72)/x^(5/2)/(3*x^2+5*x+2)^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\int \frac{5 \, x - 2}{{\left (3 \, x^{2} + 5 \, x + 2\right )}^{\frac{3}{2}} x^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{\sqrt{3 \, x^{2} + 5 \, x + 2}{\left (5 \, x - 2\right )} \sqrt{x}}{9 \, x^{8} + 30 \, x^{7} + 37 \, x^{6} + 20 \, x^{5} + 4 \, x^{4}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-sqrt(3*x^2 + 5*x + 2)*(5*x - 2)*sqrt(x)/(9*x^8 + 30*x^7 + 37*x^6 + 20*x^5 + 4*x^4), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int -\frac{5 \, x - 2}{{\left (3 \, x^{2} + 5 \, x + 2\right )}^{\frac{3}{2}} x^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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