3.1036 \(\int (2-5 x) x^{3/2} \sqrt{2+5 x+3 x^2} \, dx\)

Optimal. Leaf size=205 \[ \frac{668 \sqrt{2} (x+1) \sqrt{\frac{3 x+2}{x+1}} \text{EllipticF}\left (\tan ^{-1}\left (\sqrt{x}\right ),-\frac{1}{2}\right )}{1701 \sqrt{3 x^2+5 x+2}}-\frac{10}{27} x^{3/2} \left (3 x^2+5 x+2\right )^{3/2}+\frac{136}{189} \sqrt{x} \left (3 x^2+5 x+2\right )^{3/2}-\frac{4 \sqrt{x} (1035 x+779) \sqrt{3 x^2+5 x+2}}{1701}+\frac{2360 \sqrt{x} (3 x+2)}{5103 \sqrt{3 x^2+5 x+2}}-\frac{2360 \sqrt{2} (x+1) \sqrt{\frac{3 x+2}{x+1}} E\left (\tan ^{-1}\left (\sqrt{x}\right )|-\frac{1}{2}\right )}{5103 \sqrt{3 x^2+5 x+2}} \]

[Out]

(2360*Sqrt[x]*(2 + 3*x))/(5103*Sqrt[2 + 5*x + 3*x^2]) - (4*Sqrt[x]*(779 + 1035*x)*Sqrt[2 + 5*x + 3*x^2])/1701
+ (136*Sqrt[x]*(2 + 5*x + 3*x^2)^(3/2))/189 - (10*x^(3/2)*(2 + 5*x + 3*x^2)^(3/2))/27 - (2360*Sqrt[2]*(1 + x)*
Sqrt[(2 + 3*x)/(1 + x)]*EllipticE[ArcTan[Sqrt[x]], -1/2])/(5103*Sqrt[2 + 5*x + 3*x^2]) + (668*Sqrt[2]*(1 + x)*
Sqrt[(2 + 3*x)/(1 + x)]*EllipticF[ArcTan[Sqrt[x]], -1/2])/(1701*Sqrt[2 + 5*x + 3*x^2])

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Rubi [A]  time = 0.147138, antiderivative size = 205, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {832, 814, 839, 1189, 1100, 1136} \[ -\frac{10}{27} x^{3/2} \left (3 x^2+5 x+2\right )^{3/2}+\frac{136}{189} \sqrt{x} \left (3 x^2+5 x+2\right )^{3/2}-\frac{4 \sqrt{x} (1035 x+779) \sqrt{3 x^2+5 x+2}}{1701}+\frac{2360 \sqrt{x} (3 x+2)}{5103 \sqrt{3 x^2+5 x+2}}+\frac{668 \sqrt{2} (x+1) \sqrt{\frac{3 x+2}{x+1}} F\left (\tan ^{-1}\left (\sqrt{x}\right )|-\frac{1}{2}\right )}{1701 \sqrt{3 x^2+5 x+2}}-\frac{2360 \sqrt{2} (x+1) \sqrt{\frac{3 x+2}{x+1}} E\left (\tan ^{-1}\left (\sqrt{x}\right )|-\frac{1}{2}\right )}{5103 \sqrt{3 x^2+5 x+2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2360*Sqrt[x]*(2 + 3*x))/(5103*Sqrt[2 + 5*x + 3*x^2]) - (4*Sqrt[x]*(779 + 1035*x)*Sqrt[2 + 5*x + 3*x^2])/1701
+ (136*Sqrt[x]*(2 + 5*x + 3*x^2)^(3/2))/189 - (10*x^(3/2)*(2 + 5*x + 3*x^2)^(3/2))/27 - (2360*Sqrt[2]*(1 + x)*
Sqrt[(2 + 3*x)/(1 + x)]*EllipticE[ArcTan[Sqrt[x]], -1/2])/(5103*Sqrt[2 + 5*x + 3*x^2]) + (668*Sqrt[2]*(1 + x)*
Sqrt[(2 + 3*x)/(1 + x)]*EllipticF[ArcTan[Sqrt[x]], -1/2])/(1701*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[(g*(d + e*x)^m*(a + b*x + c*x^2)^(p + 1))/(c*(m + 2*p + 2)), x] + Dist[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] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 -
b*d*e + a*e^2, 0] && GtQ[m, 0] && NeQ[m + 2*p + 2, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])
&&  !(IGtQ[m, 0] && EqQ[f, 0])

Rule 814

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[((d + e*x)^(m + 1)*(c*e*f*(m + 2*p + 2) - g*(c*d + 2*c*d*p - b*e*p) + g*c*e*(m + 2*p + 1)*x)*(a + b*x + c*x^
2)^p)/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2)), x] - Dist[p/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2)), Int[(d + e*x)^m*(a
+ b*x + c*x^2)^(p - 1)*Simp[c*e*f*(b*d - 2*a*e)*(m + 2*p + 2) + g*(a*e*(b*e - 2*c*d*m + b*e*m) + b*d*(b*e*p -
c*d - 2*c*d*p)) + (c*e*f*(2*c*d - b*e)*(m + 2*p + 2) + g*(b^2*e^2*(p + m + 1) - 2*c^2*d^2*(1 + 2*p) - c*e*(b*
d*(m - 2*p) + 2*a*e*(m + 2*p + 1))))*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] && GtQ[p, 0] && (IntegerQ[p] ||  !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 0])
) &&  !ILtQ[m + 2*p, 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 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 (2-5 x) x^{3/2} \sqrt{2+5 x+3 x^2} \, dx &=-\frac{10}{27} x^{3/2} \left (2+5 x+3 x^2\right )^{3/2}+\frac{2}{27} \int \sqrt{x} (15+102 x) \sqrt{2+5 x+3 x^2} \, dx\\ &=\frac{136}{189} \sqrt{x} \left (2+5 x+3 x^2\right )^{3/2}-\frac{10}{27} x^{3/2} \left (2+5 x+3 x^2\right )^{3/2}+\frac{4}{567} \int \frac{\left (-102-\frac{1725 x}{2}\right ) \sqrt{2+5 x+3 x^2}}{\sqrt{x}} \, dx\\ &=-\frac{4 \sqrt{x} (779+1035 x) \sqrt{2+5 x+3 x^2}}{1701}+\frac{136}{189} \sqrt{x} \left (2+5 x+3 x^2\right )^{3/2}-\frac{10}{27} x^{3/2} \left (2+5 x+3 x^2\right )^{3/2}-\frac{8 \int \frac{-\frac{2505}{2}-\frac{4425 x}{2}}{\sqrt{x} \sqrt{2+5 x+3 x^2}} \, dx}{25515}\\ &=-\frac{4 \sqrt{x} (779+1035 x) \sqrt{2+5 x+3 x^2}}{1701}+\frac{136}{189} \sqrt{x} \left (2+5 x+3 x^2\right )^{3/2}-\frac{10}{27} x^{3/2} \left (2+5 x+3 x^2\right )^{3/2}-\frac{16 \operatorname{Subst}\left (\int \frac{-\frac{2505}{2}-\frac{4425 x^2}{2}}{\sqrt{2+5 x^2+3 x^4}} \, dx,x,\sqrt{x}\right )}{25515}\\ &=-\frac{4 \sqrt{x} (779+1035 x) \sqrt{2+5 x+3 x^2}}{1701}+\frac{136}{189} \sqrt{x} \left (2+5 x+3 x^2\right )^{3/2}-\frac{10}{27} x^{3/2} \left (2+5 x+3 x^2\right )^{3/2}+\frac{1336 \operatorname{Subst}\left (\int \frac{1}{\sqrt{2+5 x^2+3 x^4}} \, dx,x,\sqrt{x}\right )}{1701}+\frac{2360 \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{2+5 x^2+3 x^4}} \, dx,x,\sqrt{x}\right )}{1701}\\ &=\frac{2360 \sqrt{x} (2+3 x)}{5103 \sqrt{2+5 x+3 x^2}}-\frac{4 \sqrt{x} (779+1035 x) \sqrt{2+5 x+3 x^2}}{1701}+\frac{136}{189} \sqrt{x} \left (2+5 x+3 x^2\right )^{3/2}-\frac{10}{27} x^{3/2} \left (2+5 x+3 x^2\right )^{3/2}-\frac{2360 \sqrt{2} (1+x) \sqrt{\frac{2+3 x}{1+x}} E\left (\tan ^{-1}\left (\sqrt{x}\right )|-\frac{1}{2}\right )}{5103 \sqrt{2+5 x+3 x^2}}+\frac{668 \sqrt{2} (1+x) \sqrt{\frac{2+3 x}{1+x}} F\left (\tan ^{-1}\left (\sqrt{x}\right )|-\frac{1}{2}\right )}{1701 \sqrt{2+5 x+3 x^2}}\\ \end{align*}

Mathematica [C]  time = 0.166106, size = 165, normalized size = 0.8 \[ \frac{-356 i \sqrt{2} \sqrt{\frac{1}{x}+1} \sqrt{\frac{2}{x}+3} x^{3/2} \text{EllipticF}\left (i \sinh ^{-1}\left (\frac{\sqrt{\frac{2}{3}}}{\sqrt{x}}\right ),\frac{3}{2}\right )-17010 x^6-23652 x^5+2970 x^4+7920 x^3+1380 x^2+2360 i \sqrt{2} \sqrt{\frac{1}{x}+1} \sqrt{\frac{2}{x}+3} x^{3/2} E\left (i \sinh ^{-1}\left (\frac{\sqrt{\frac{2}{3}}}{\sqrt{x}}\right )|\frac{3}{2}\right )+7792 x+4720}{5103 \sqrt{x} \sqrt{3 x^2+5 x+2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4720 + 7792*x + 1380*x^2 + 7920*x^3 + 2970*x^4 - 23652*x^5 - 17010*x^6 + (2360*I)*Sqrt[2]*Sqrt[1 + x^(-1)]*Sq
rt[3 + 2/x]*x^(3/2)*EllipticE[I*ArcSinh[Sqrt[2/3]/Sqrt[x]], 3/2] - (356*I)*Sqrt[2]*Sqrt[1 + x^(-1)]*Sqrt[3 + 2
/x]*x^(3/2)*EllipticF[I*ArcSinh[Sqrt[2/3]/Sqrt[x]], 3/2])/(5103*Sqrt[x]*Sqrt[2 + 5*x + 3*x^2])

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Maple [A]  time = 0.017, size = 127, normalized size = 0.6 \begin{align*} -{\frac{2}{15309} \left ( 25515\,{x}^{6}+35478\,{x}^{5}+768\,\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 ) -590\,\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 ) -4455\,{x}^{4}-11880\,{x}^{3}+8550\,{x}^{2}+6012\,x \right ){\frac{1}{\sqrt{x}}}{\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^(3/2)*(3*x^2+5*x+2)^(1/2),x)

[Out]

-2/15309/x^(1/2)/(3*x^2+5*x+2)^(1/2)*(25515*x^6+35478*x^5+768*(6*x+4)^(1/2)*(3+3*x)^(1/2)*6^(1/2)*(-x)^(1/2)*E
llipticF(1/2*(6*x+4)^(1/2),I*2^(1/2))-590*(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))-4455*x^4-11880*x^3+8550*x^2+6012*x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-{\left (5 \, x^{2} - 2 \, x\right )} \sqrt{3 \, x^{2} + 5 \, x + 2} \sqrt{x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \int - 2 x^{\frac{3}{2}} \sqrt{3 x^{2} + 5 x + 2}\, dx - \int 5 x^{\frac{5}{2}} \sqrt{3 x^{2} + 5 x + 2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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