3.1044 \(\int (2-5 x) x^{5/2} (2+5 x+3 x^2)^{3/2} \, dx\)

Optimal. Leaf size=256 \[ -\frac{61736 \sqrt{2} (x+1) \sqrt{\frac{3 x+2}{x+1}} \text{EllipticF}\left (\tan ^{-1}\left (\sqrt{x}\right ),-\frac{1}{2}\right )}{2189187 \sqrt{3 x^2+5 x+2}}-\frac{2}{9} x^{5/2} \left (3 x^2+5 x+2\right )^{5/2}+\frac{136}{351} x^{3/2} \left (3 x^2+5 x+2\right )^{5/2}-\frac{4660 \sqrt{x} \left (3 x^2+5 x+2\right )^{5/2}}{11583}+\frac{8 \sqrt{x} (32921 x+27010) \left (3 x^2+5 x+2\right )^{3/2}}{243243}-\frac{8 \sqrt{x} (205407 x+190465) \sqrt{3 x^2+5 x+2}}{10945935}-\frac{497824 \sqrt{x} (3 x+2)}{32837805 \sqrt{3 x^2+5 x+2}}+\frac{497824 \sqrt{2} (x+1) \sqrt{\frac{3 x+2}{x+1}} E\left (\tan ^{-1}\left (\sqrt{x}\right )|-\frac{1}{2}\right )}{32837805 \sqrt{3 x^2+5 x+2}} \]

[Out]

(-497824*Sqrt[x]*(2 + 3*x))/(32837805*Sqrt[2 + 5*x + 3*x^2]) - (8*Sqrt[x]*(190465 + 205407*x)*Sqrt[2 + 5*x + 3
*x^2])/10945935 + (8*Sqrt[x]*(27010 + 32921*x)*(2 + 5*x + 3*x^2)^(3/2))/243243 - (4660*Sqrt[x]*(2 + 5*x + 3*x^
2)^(5/2))/11583 + (136*x^(3/2)*(2 + 5*x + 3*x^2)^(5/2))/351 - (2*x^(5/2)*(2 + 5*x + 3*x^2)^(5/2))/9 + (497824*
Sqrt[2]*(1 + x)*Sqrt[(2 + 3*x)/(1 + x)]*EllipticE[ArcTan[Sqrt[x]], -1/2])/(32837805*Sqrt[2 + 5*x + 3*x^2]) - (
61736*Sqrt[2]*(1 + x)*Sqrt[(2 + 3*x)/(1 + x)]*EllipticF[ArcTan[Sqrt[x]], -1/2])/(2189187*Sqrt[2 + 5*x + 3*x^2]
)

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Rubi [A]  time = 0.196785, antiderivative size = 256, normalized size of antiderivative = 1., number of steps used = 9, 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} \[ -\frac{2}{9} x^{5/2} \left (3 x^2+5 x+2\right )^{5/2}+\frac{136}{351} x^{3/2} \left (3 x^2+5 x+2\right )^{5/2}-\frac{4660 \sqrt{x} \left (3 x^2+5 x+2\right )^{5/2}}{11583}+\frac{8 \sqrt{x} (32921 x+27010) \left (3 x^2+5 x+2\right )^{3/2}}{243243}-\frac{8 \sqrt{x} (205407 x+190465) \sqrt{3 x^2+5 x+2}}{10945935}-\frac{497824 \sqrt{x} (3 x+2)}{32837805 \sqrt{3 x^2+5 x+2}}-\frac{61736 \sqrt{2} (x+1) \sqrt{\frac{3 x+2}{x+1}} F\left (\tan ^{-1}\left (\sqrt{x}\right )|-\frac{1}{2}\right )}{2189187 \sqrt{3 x^2+5 x+2}}+\frac{497824 \sqrt{2} (x+1) \sqrt{\frac{3 x+2}{x+1}} E\left (\tan ^{-1}\left (\sqrt{x}\right )|-\frac{1}{2}\right )}{32837805 \sqrt{3 x^2+5 x+2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-497824*Sqrt[x]*(2 + 3*x))/(32837805*Sqrt[2 + 5*x + 3*x^2]) - (8*Sqrt[x]*(190465 + 205407*x)*Sqrt[2 + 5*x + 3
*x^2])/10945935 + (8*Sqrt[x]*(27010 + 32921*x)*(2 + 5*x + 3*x^2)^(3/2))/243243 - (4660*Sqrt[x]*(2 + 5*x + 3*x^
2)^(5/2))/11583 + (136*x^(3/2)*(2 + 5*x + 3*x^2)^(5/2))/351 - (2*x^(5/2)*(2 + 5*x + 3*x^2)^(5/2))/9 + (497824*
Sqrt[2]*(1 + x)*Sqrt[(2 + 3*x)/(1 + x)]*EllipticE[ArcTan[Sqrt[x]], -1/2])/(32837805*Sqrt[2 + 5*x + 3*x^2]) - (
61736*Sqrt[2]*(1 + x)*Sqrt[(2 + 3*x)/(1 + x)]*EllipticF[ArcTan[Sqrt[x]], -1/2])/(2189187*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^{5/2} \left (2+5 x+3 x^2\right )^{3/2} \, dx &=-\frac{2}{9} x^{5/2} \left (2+5 x+3 x^2\right )^{5/2}+\frac{2}{45} \int x^{3/2} (25+170 x) \left (2+5 x+3 x^2\right )^{3/2} \, dx\\ &=\frac{136}{351} x^{3/2} \left (2+5 x+3 x^2\right )^{5/2}-\frac{2}{9} x^{5/2} \left (2+5 x+3 x^2\right )^{5/2}+\frac{4 \int \left (-510-\frac{5825 x}{2}\right ) \sqrt{x} \left (2+5 x+3 x^2\right )^{3/2} \, dx}{1755}\\ &=-\frac{4660 \sqrt{x} \left (2+5 x+3 x^2\right )^{5/2}}{11583}+\frac{136}{351} x^{3/2} \left (2+5 x+3 x^2\right )^{5/2}-\frac{2}{9} x^{5/2} \left (2+5 x+3 x^2\right )^{5/2}+\frac{8 \int \frac{\left (\frac{5825}{2}+\frac{70545 x}{2}\right ) \left (2+5 x+3 x^2\right )^{3/2}}{\sqrt{x}} \, dx}{57915}\\ &=\frac{8 \sqrt{x} (27010+32921 x) \left (2+5 x+3 x^2\right )^{3/2}}{243243}-\frac{4660 \sqrt{x} \left (2+5 x+3 x^2\right )^{5/2}}{11583}+\frac{136}{351} x^{3/2} \left (2+5 x+3 x^2\right )^{5/2}-\frac{2}{9} x^{5/2} \left (2+5 x+3 x^2\right )^{5/2}-\frac{16 \int \frac{\left (\frac{38175}{2}+\frac{342345 x}{4}\right ) \sqrt{2+5 x+3 x^2}}{\sqrt{x}} \, dx}{3648645}\\ &=-\frac{8 \sqrt{x} (190465+205407 x) \sqrt{2+5 x+3 x^2}}{10945935}+\frac{8 \sqrt{x} (27010+32921 x) \left (2+5 x+3 x^2\right )^{3/2}}{243243}-\frac{4660 \sqrt{x} \left (2+5 x+3 x^2\right )^{5/2}}{11583}+\frac{136}{351} x^{3/2} \left (2+5 x+3 x^2\right )^{5/2}-\frac{2}{9} x^{5/2} \left (2+5 x+3 x^2\right )^{5/2}+\frac{32 \int \frac{-\frac{578775}{4}-\frac{233355 x}{2}}{\sqrt{x} \sqrt{2+5 x+3 x^2}} \, dx}{164189025}\\ &=-\frac{8 \sqrt{x} (190465+205407 x) \sqrt{2+5 x+3 x^2}}{10945935}+\frac{8 \sqrt{x} (27010+32921 x) \left (2+5 x+3 x^2\right )^{3/2}}{243243}-\frac{4660 \sqrt{x} \left (2+5 x+3 x^2\right )^{5/2}}{11583}+\frac{136}{351} x^{3/2} \left (2+5 x+3 x^2\right )^{5/2}-\frac{2}{9} x^{5/2} \left (2+5 x+3 x^2\right )^{5/2}+\frac{64 \operatorname{Subst}\left (\int \frac{-\frac{578775}{4}-\frac{233355 x^2}{2}}{\sqrt{2+5 x^2+3 x^4}} \, dx,x,\sqrt{x}\right )}{164189025}\\ &=-\frac{8 \sqrt{x} (190465+205407 x) \sqrt{2+5 x+3 x^2}}{10945935}+\frac{8 \sqrt{x} (27010+32921 x) \left (2+5 x+3 x^2\right )^{3/2}}{243243}-\frac{4660 \sqrt{x} \left (2+5 x+3 x^2\right )^{5/2}}{11583}+\frac{136}{351} x^{3/2} \left (2+5 x+3 x^2\right )^{5/2}-\frac{2}{9} x^{5/2} \left (2+5 x+3 x^2\right )^{5/2}-\frac{497824 \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{2+5 x^2+3 x^4}} \, dx,x,\sqrt{x}\right )}{10945935}-\frac{123472 \operatorname{Subst}\left (\int \frac{1}{\sqrt{2+5 x^2+3 x^4}} \, dx,x,\sqrt{x}\right )}{2189187}\\ &=-\frac{497824 \sqrt{x} (2+3 x)}{32837805 \sqrt{2+5 x+3 x^2}}-\frac{8 \sqrt{x} (190465+205407 x) \sqrt{2+5 x+3 x^2}}{10945935}+\frac{8 \sqrt{x} (27010+32921 x) \left (2+5 x+3 x^2\right )^{3/2}}{243243}-\frac{4660 \sqrt{x} \left (2+5 x+3 x^2\right )^{5/2}}{11583}+\frac{136}{351} x^{3/2} \left (2+5 x+3 x^2\right )^{5/2}-\frac{2}{9} x^{5/2} \left (2+5 x+3 x^2\right )^{5/2}+\frac{497824 \sqrt{2} (1+x) \sqrt{\frac{2+3 x}{1+x}} E\left (\tan ^{-1}\left (\sqrt{x}\right )|-\frac{1}{2}\right )}{32837805 \sqrt{2+5 x+3 x^2}}-\frac{61736 \sqrt{2} (1+x) \sqrt{\frac{2+3 x}{1+x}} F\left (\tan ^{-1}\left (\sqrt{x}\right )|-\frac{1}{2}\right )}{2189187 \sqrt{2+5 x+3 x^2}}\\ \end{align*}

Mathematica [C]  time = 0.203789, size = 183, normalized size = 0.71 \[ \frac{-2 \left (214108 i \sqrt{2} \sqrt{\frac{1}{x}+1} \sqrt{\frac{2}{x}+3} x^{3/2} \text{EllipticF}\left (i \sinh ^{-1}\left (\frac{\sqrt{\frac{2}{3}}}{\sqrt{x}}\right ),\frac{3}{2}\right )[phone] x^9+320800095 x^8+337486905 x^7+69664455 x^6-83323080 x^5-37601118 x^4+91620 x^3-273876 x^2+318520 x+497824\right )-497824 i \sqrt{2} \sqrt{\frac{1}{x}+1} \sqrt{\frac{2}{x}+3} x^{3/2} E\left (i \sinh ^{-1}\left (\frac{\sqrt{\frac{2}{3}}}{\sqrt{x}}\right )|\frac{3}{2}\right )}{32837805 \sqrt{x} \sqrt{3 x^2+5 x+2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-497824*I)*Sqrt[2]*Sqrt[1 + x^(-1)]*Sqrt[3 + 2/x]*x^(3/2)*EllipticE[I*ArcSinh[Sqrt[2/3]/Sqrt[x]], 3/2] - 2*(
497824 + 318520*x - 273876*x^2 + 91620*x^3 - 37601118*x^4 - 83323080*x^5 + 69664455*x^6 + 337486905*x^7 + 3208
00095*x^8 + 98513415*x^9 + (214108*I)*Sqrt[2]*Sqrt[1 + x^(-1)]*Sqrt[3 + 2/x]*x^(3/2)*EllipticF[I*ArcSinh[Sqrt[
2/3]/Sqrt[x]], 3/2]))/(32837805*Sqrt[x]*Sqrt[2 + 5*x + 3*x^2])

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Maple [A]  time = 0.019, size = 142, normalized size = 0.6 \begin{align*} -{\frac{2}{98513415} \left ( 295540245\,{x}^{9}[phone]\,{x}^{8}[phone]\,{x}^{7}[phone]\,{x}^{6}-249969240\,{x}^{5}+89652\,\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 ) +124456\,\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 ) -112803354\,{x}^{4}+274860\,{x}^{3}-3061836\,{x}^{2}-2778120\,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^(5/2)*(3*x^2+5*x+2)^(3/2),x)

[Out]

-2/98513415/x^(1/2)/(3*x^2+5*x+2)^(1/2)*(295540245*x^9+962400285*x^8+1012460715*x^7+208993365*x^6-249969240*x^
5+89652*(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))+124456*(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))-112803354*x^4+274860*x^3-3061836*
x^2-2778120*x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-{\left (15 \, x^{5} + 19 \, x^{4} - 4 \, x^{2}\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^(5/2)*(3*x^2+5*x+2)^(3/2),x, algorithm="fricas")

[Out]

integral(-(15*x^5 + 19*x^4 - 4*x^2)*sqrt(3*x^2 + 5*x + 2)*sqrt(x), 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**(5/2)*(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 -{\left (3 \, x^{2} + 5 \, x + 2\right )}^{\frac{3}{2}}{\left (5 \, x - 2\right )} x^{\frac{5}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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