3.1045 \(\int (2-5 x) x^{3/2} (2+5 x+3 x^2)^{3/2} \, dx\)

Optimal. Leaf size=233 \[ \frac{25448 \sqrt{2} (x+1) \sqrt{\frac{3 x+2}{x+1}} \text{EllipticF}\left (\tan ^{-1}\left (\sqrt{x}\right ),-\frac{1}{2}\right )}{243243 \sqrt{3 x^2+5 x+2}}-\frac{10}{39} x^{3/2} \left (3 x^2+5 x+2\right )^{5/2}+\frac{556 \sqrt{x} \left (3 x^2+5 x+2\right )^{5/2}}{1287}-\frac{4 \sqrt{x} (8575 x+6959) \left (3 x^2+5 x+2\right )^{3/2}}{27027}+\frac{8 \sqrt{x} (6381 x+6908) \sqrt{3 x^2+5 x+2}}{243243}+\frac{55112 \sqrt{x} (3 x+2)}{729729 \sqrt{3 x^2+5 x+2}}-\frac{55112 \sqrt{2} (x+1) \sqrt{\frac{3 x+2}{x+1}} E\left (\tan ^{-1}\left (\sqrt{x}\right )|-\frac{1}{2}\right )}{729729 \sqrt{3 x^2+5 x+2}} \]

[Out]

(55112*Sqrt[x]*(2 + 3*x))/(729729*Sqrt[2 + 5*x + 3*x^2]) + (8*Sqrt[x]*(6908 + 6381*x)*Sqrt[2 + 5*x + 3*x^2])/2
43243 - (4*Sqrt[x]*(6959 + 8575*x)*(2 + 5*x + 3*x^2)^(3/2))/27027 + (556*Sqrt[x]*(2 + 5*x + 3*x^2)^(5/2))/1287
- (10*x^(3/2)*(2 + 5*x + 3*x^2)^(5/2))/39 - (55112*Sqrt[2]*(1 + x)*Sqrt[(2 + 3*x)/(1 + x)]*EllipticE[ArcTan[S
qrt[x]], -1/2])/(729729*Sqrt[2 + 5*x + 3*x^2]) + (25448*Sqrt[2]*(1 + x)*Sqrt[(2 + 3*x)/(1 + x)]*EllipticF[ArcT
an[Sqrt[x]], -1/2])/(243243*Sqrt[2 + 5*x + 3*x^2])

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Rubi [A]  time = 0.172243, antiderivative size = 233, 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 = {832, 814, 839, 1189, 1100, 1136} \[ -\frac{10}{39} x^{3/2} \left (3 x^2+5 x+2\right )^{5/2}+\frac{556 \sqrt{x} \left (3 x^2+5 x+2\right )^{5/2}}{1287}-\frac{4 \sqrt{x} (8575 x+6959) \left (3 x^2+5 x+2\right )^{3/2}}{27027}+\frac{8 \sqrt{x} (6381 x+6908) \sqrt{3 x^2+5 x+2}}{243243}+\frac{55112 \sqrt{x} (3 x+2)}{729729 \sqrt{3 x^2+5 x+2}}+\frac{25448 \sqrt{2} (x+1) \sqrt{\frac{3 x+2}{x+1}} F\left (\tan ^{-1}\left (\sqrt{x}\right )|-\frac{1}{2}\right )}{243243 \sqrt{3 x^2+5 x+2}}-\frac{55112 \sqrt{2} (x+1) \sqrt{\frac{3 x+2}{x+1}} E\left (\tan ^{-1}\left (\sqrt{x}\right )|-\frac{1}{2}\right )}{729729 \sqrt{3 x^2+5 x+2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(55112*Sqrt[x]*(2 + 3*x))/(729729*Sqrt[2 + 5*x + 3*x^2]) + (8*Sqrt[x]*(6908 + 6381*x)*Sqrt[2 + 5*x + 3*x^2])/2
43243 - (4*Sqrt[x]*(6959 + 8575*x)*(2 + 5*x + 3*x^2)^(3/2))/27027 + (556*Sqrt[x]*(2 + 5*x + 3*x^2)^(5/2))/1287
- (10*x^(3/2)*(2 + 5*x + 3*x^2)^(5/2))/39 - (55112*Sqrt[2]*(1 + x)*Sqrt[(2 + 3*x)/(1 + x)]*EllipticE[ArcTan[S
qrt[x]], -1/2])/(729729*Sqrt[2 + 5*x + 3*x^2]) + (25448*Sqrt[2]*(1 + x)*Sqrt[(2 + 3*x)/(1 + x)]*EllipticF[ArcT
an[Sqrt[x]], -1/2])/(243243*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} \left (2+5 x+3 x^2\right )^{3/2} \, dx &=-\frac{10}{39} x^{3/2} \left (2+5 x+3 x^2\right )^{5/2}+\frac{2}{39} \int \sqrt{x} (15+139 x) \left (2+5 x+3 x^2\right )^{3/2} \, dx\\ &=\frac{556 \sqrt{x} \left (2+5 x+3 x^2\right )^{5/2}}{1287}-\frac{10}{39} x^{3/2} \left (2+5 x+3 x^2\right )^{5/2}+\frac{4 \int \frac{\left (-139-\frac{3675 x}{2}\right ) \left (2+5 x+3 x^2\right )^{3/2}}{\sqrt{x}} \, dx}{1287}\\ &=-\frac{4 \sqrt{x} (6959+8575 x) \left (2+5 x+3 x^2\right )^{3/2}}{27027}+\frac{556 \sqrt{x} \left (2+5 x+3 x^2\right )^{5/2}}{1287}-\frac{10}{39} x^{3/2} \left (2+5 x+3 x^2\right )^{5/2}-\frac{8 \int \frac{\left (-\frac{3363}{2}-\frac{10635 x}{2}\right ) \sqrt{2+5 x+3 x^2}}{\sqrt{x}} \, dx}{81081}\\ &=\frac{8 \sqrt{x} (6908+6381 x) \sqrt{2+5 x+3 x^2}}{243243}-\frac{4 \sqrt{x} (6959+8575 x) \left (2+5 x+3 x^2\right )^{3/2}}{27027}+\frac{556 \sqrt{x} \left (2+5 x+3 x^2\right )^{5/2}}{1287}-\frac{10}{39} x^{3/2} \left (2+5 x+3 x^2\right )^{5/2}+\frac{16 \int \frac{\frac{47715}{2}+\frac{103335 x}{4}}{\sqrt{x} \sqrt{2+5 x+3 x^2}} \, dx}{3648645}\\ &=\frac{8 \sqrt{x} (6908+6381 x) \sqrt{2+5 x+3 x^2}}{243243}-\frac{4 \sqrt{x} (6959+8575 x) \left (2+5 x+3 x^2\right )^{3/2}}{27027}+\frac{556 \sqrt{x} \left (2+5 x+3 x^2\right )^{5/2}}{1287}-\frac{10}{39} x^{3/2} \left (2+5 x+3 x^2\right )^{5/2}+\frac{32 \operatorname{Subst}\left (\int \frac{\frac{47715}{2}+\frac{103335 x^2}{4}}{\sqrt{2+5 x^2+3 x^4}} \, dx,x,\sqrt{x}\right )}{3648645}\\ &=\frac{8 \sqrt{x} (6908+6381 x) \sqrt{2+5 x+3 x^2}}{243243}-\frac{4 \sqrt{x} (6959+8575 x) \left (2+5 x+3 x^2\right )^{3/2}}{27027}+\frac{556 \sqrt{x} \left (2+5 x+3 x^2\right )^{5/2}}{1287}-\frac{10}{39} x^{3/2} \left (2+5 x+3 x^2\right )^{5/2}+\frac{50896 \operatorname{Subst}\left (\int \frac{1}{\sqrt{2+5 x^2+3 x^4}} \, dx,x,\sqrt{x}\right )}{243243}+\frac{55112 \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{2+5 x^2+3 x^4}} \, dx,x,\sqrt{x}\right )}{243243}\\ &=\frac{55112 \sqrt{x} (2+3 x)}{729729 \sqrt{2+5 x+3 x^2}}+\frac{8 \sqrt{x} (6908+6381 x) \sqrt{2+5 x+3 x^2}}{243243}-\frac{4 \sqrt{x} (6959+8575 x) \left (2+5 x+3 x^2\right )^{3/2}}{27027}+\frac{556 \sqrt{x} \left (2+5 x+3 x^2\right )^{5/2}}{1287}-\frac{10}{39} x^{3/2} \left (2+5 x+3 x^2\right )^{5/2}-\frac{55112 \sqrt{2} (1+x) \sqrt{\frac{2+3 x}{1+x}} E\left (\tan ^{-1}\left (\sqrt{x}\right )|-\frac{1}{2}\right )}{729729 \sqrt{2+5 x+3 x^2}}+\frac{25448 \sqrt{2} (1+x) \sqrt{\frac{2+3 x}{1+x}} F\left (\tan ^{-1}\left (\sqrt{x}\right )|-\frac{1}{2}\right )}{243243 \sqrt{2+5 x+3 x^2}}\\ \end{align*}

Mathematica [C]  time = 0.195838, size = 178, normalized size = 0.76 \[ \frac{21232 i \sqrt{2} \sqrt{\frac{1}{x}+1} \sqrt{\frac{2}{x}+3} x^{3/2} \text{EllipticF}\left (i \sinh ^{-1}\left (\frac{\sqrt{\frac{2}{3}}}{\sqrt{x}}\right ),\frac{3}{2}\right )-2 \left (2525985 x^8+8374023 x^7+8989785 x^6+1830195 x^5-2497986 x^4-1171602 x^3+8508 x^2-61436 x-55112\right )+55112 i \sqrt{2} \sqrt{\frac{1}{x}+1} \sqrt{\frac{2}{x}+3} x^{3/2} E\left (i \sinh ^{-1}\left (\frac{\sqrt{\frac{2}{3}}}{\sqrt{x}}\right )|\frac{3}{2}\right )}{729729 \sqrt{x} \sqrt{3 x^2+5 x+2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(-55112 - 61436*x + 8508*x^2 - 1171602*x^3 - 2497986*x^4 + 1830195*x^5 + 8989785*x^6 + 8374023*x^7 + 25259
85*x^8) + (55112*I)*Sqrt[2]*Sqrt[1 + x^(-1)]*Sqrt[3 + 2/x]*x^(3/2)*EllipticE[I*ArcSinh[Sqrt[2/3]/Sqrt[x]], 3/2
] + (21232*I)*Sqrt[2]*Sqrt[1 + x^(-1)]*Sqrt[3 + 2/x]*x^(3/2)*EllipticF[I*ArcSinh[Sqrt[2/3]/Sqrt[x]], 3/2])/(72
9729*Sqrt[x]*Sqrt[2 + 5*x + 3*x^2])

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Maple [A]  time = 0.01, size = 137, normalized size = 0.6 \begin{align*} -{\frac{2}{2189187} \left ( 7577955\,{x}^{8}[phone]\,{x}^{7}[phone]\,{x}^{6}+5490585\,{x}^{5}+3162\,\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 ) -13778\,\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 ) -7493958\,{x}^{4}-3514806\,{x}^{3}+273528\,{x}^{2}+229032\,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)^(3/2),x)

[Out]

-2/2189187/x^(1/2)/(3*x^2+5*x+2)^(1/2)*(7577955*x^8+25122069*x^7+26969355*x^6+5490585*x^5+3162*(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))-13778*(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))-7493958*x^4-3514806*x^3+273528*x^2+229032*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{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)^(3/2),x, algorithm="maxima")

[Out]

-integrate((3*x^2 + 5*x + 2)^(3/2)*(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 (15 \, x^{4} + 19 \, x^{3} - 4 \, 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)^(3/2),x, algorithm="fricas")

[Out]

integral(-(15*x^4 + 19*x^3 - 4*x)*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**(3/2)*(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{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)^(3/2),x, algorithm="giac")

[Out]

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