3.1049 \(\int \frac {(2-5 x) (2+5 x+3 x^2)^{3/2}}{x^{5/2}} \, dx\)

Optimal. Leaf size=183 \[ \frac {2 (2-x) \sqrt {3 x^2+5 x+2}}{\sqrt {x}}-\frac {34 \sqrt {x} (3 x+2)}{3 \sqrt {3 x^2+5 x+2}}-\frac {14 \sqrt {2} (x+1) \sqrt {\frac {3 x+2}{x+1}} F\left (\tan ^{-1}\left (\sqrt {x}\right )|-\frac {1}{2}\right )}{\sqrt {3 x^2+5 x+2}}+\frac {34 \sqrt {2} (x+1) \sqrt {\frac {3 x+2}{x+1}} E\left (\tan ^{-1}\left (\sqrt {x}\right )|-\frac {1}{2}\right )}{3 \sqrt {3 x^2+5 x+2}}-\frac {2 (3 x+2) \left (3 x^2+5 x+2\right )^{3/2}}{3 x^{3/2}} \]

[Out]

-2/3*(2+3*x)*(3*x^2+5*x+2)^(3/2)/x^(3/2)-34/3*(2+3*x)*x^(1/2)/(3*x^2+5*x+2)^(1/2)+34/3*(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)-14*(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*(2-x)*(3*x^2+5*x+2)^(1/2)/x^(1/2)

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Rubi [A]  time = 0.12, antiderivative size = 183, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {812, 839, 1189, 1100, 1136} \[ -\frac {2 (3 x+2) \left (3 x^2+5 x+2\right )^{3/2}}{3 x^{3/2}}+\frac {2 (2-x) \sqrt {3 x^2+5 x+2}}{\sqrt {x}}-\frac {34 \sqrt {x} (3 x+2)}{3 \sqrt {3 x^2+5 x+2}}-\frac {14 \sqrt {2} (x+1) \sqrt {\frac {3 x+2}{x+1}} F\left (\tan ^{-1}\left (\sqrt {x}\right )|-\frac {1}{2}\right )}{\sqrt {3 x^2+5 x+2}}+\frac {34 \sqrt {2} (x+1) \sqrt {\frac {3 x+2}{x+1}} E\left (\tan ^{-1}\left (\sqrt {x}\right )|-\frac {1}{2}\right )}{3 \sqrt {3 x^2+5 x+2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-34*Sqrt[x]*(2 + 3*x))/(3*Sqrt[2 + 5*x + 3*x^2]) + (2*(2 - x)*Sqrt[2 + 5*x + 3*x^2])/Sqrt[x] - (2*(2 + 3*x)*(
2 + 5*x + 3*x^2)^(3/2))/(3*x^(3/2)) + (34*Sqrt[2]*(1 + x)*Sqrt[(2 + 3*x)/(1 + x)]*EllipticE[ArcTan[Sqrt[x]], -
1/2])/(3*Sqrt[2 + 5*x + 3*x^2]) - (14*Sqrt[2]*(1 + x)*Sqrt[(2 + 3*x)/(1 + x)]*EllipticF[ArcTan[Sqrt[x]], -1/2]
)/Sqrt[2 + 5*x + 3*x^2]

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^{5/2}} \, dx &=-\frac {2 (2+3 x) \left (2+5 x+3 x^2\right )^{3/2}}{3 x^{3/2}}-\frac {2}{5} \int \frac {\left (5+\frac {15 x}{2}\right ) \sqrt {2+5 x+3 x^2}}{x^{3/2}} \, dx\\ &=\frac {2 (2-x) \sqrt {2+5 x+3 x^2}}{\sqrt {x}}-\frac {2 (2+3 x) \left (2+5 x+3 x^2\right )^{3/2}}{3 x^{3/2}}+\frac {4}{15} \int \frac {-\frac {105}{2}-\frac {255 x}{4}}{\sqrt {x} \sqrt {2+5 x+3 x^2}} \, dx\\ &=\frac {2 (2-x) \sqrt {2+5 x+3 x^2}}{\sqrt {x}}-\frac {2 (2+3 x) \left (2+5 x+3 x^2\right )^{3/2}}{3 x^{3/2}}+\frac {8}{15} \operatorname {Subst}\left (\int \frac {-\frac {105}{2}-\frac {255 x^2}{4}}{\sqrt {2+5 x^2+3 x^4}} \, dx,x,\sqrt {x}\right )\\ &=\frac {2 (2-x) \sqrt {2+5 x+3 x^2}}{\sqrt {x}}-\frac {2 (2+3 x) \left (2+5 x+3 x^2\right )^{3/2}}{3 x^{3/2}}-28 \operatorname {Subst}\left (\int \frac {1}{\sqrt {2+5 x^2+3 x^4}} \, dx,x,\sqrt {x}\right )-34 \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {2+5 x^2+3 x^4}} \, dx,x,\sqrt {x}\right )\\ &=-\frac {34 \sqrt {x} (2+3 x)}{3 \sqrt {2+5 x+3 x^2}}+\frac {2 (2-x) \sqrt {2+5 x+3 x^2}}{\sqrt {x}}-\frac {2 (2+3 x) \left (2+5 x+3 x^2\right )^{3/2}}{3 x^{3/2}}+\frac {34 \sqrt {2} (1+x) \sqrt {\frac {2+3 x}{1+x}} E\left (\tan ^{-1}\left (\sqrt {x}\right )|-\frac {1}{2}\right )}{3 \sqrt {2+5 x+3 x^2}}-\frac {14 \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*}

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Mathematica [C]  time = 0.20, size = 163, normalized size = 0.89 \[ \frac {-34 i \sqrt {2} \sqrt {\frac {1}{x}+1} \sqrt {\frac {2}{x}+3} x^{5/2} E\left (i \sinh ^{-1}\left (\frac {\sqrt {\frac {2}{3}}}{\sqrt {x}}\right )|\frac {3}{2}\right )-2 \left (4 i \sqrt {2} \sqrt {\frac {1}{x}+1} \sqrt {\frac {2}{x}+3} x^{5/2} F\left (i \sinh ^{-1}\left (\frac {\sqrt {\frac {2}{3}}}{\sqrt {x}}\right )|\frac {3}{2}\right )+27 x^5+117 x^4+219 x^3+195 x^2+74 x+8\right )}{3 x^{3/2} \sqrt {3 x^2+5 x+2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-34*I)*Sqrt[2]*Sqrt[1 + x^(-1)]*Sqrt[3 + 2/x]*x^(5/2)*EllipticE[I*ArcSinh[Sqrt[2/3]/Sqrt[x]], 3/2] - 2*(8 +
74*x + 195*x^2 + 219*x^3 + 117*x^4 + 27*x^5 + (4*I)*Sqrt[2]*Sqrt[1 + x^(-1)]*Sqrt[3 + 2/x]*x^(5/2)*EllipticF[I
*ArcSinh[Sqrt[2/3]/Sqrt[x]], 3/2]))/(3*x^(3/2)*Sqrt[2 + 5*x + 3*x^2])

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fricas [F]  time = 0.92, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (15 \, x^{3} + 19 \, x^{2} - 4\right )} \sqrt {3 \, x^{2} + 5 \, x + 2}}{x^{\frac {5}{2}}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.09, size = 125, normalized size = 0.68 \[ \frac {-162 x^{5}-702 x^{4}-1008 x^{3}-660 x^{2}-17 \sqrt {6 x +4}\, \sqrt {3 x +3}\, \sqrt {6}\, \sqrt {-x}\, x \EllipticE \left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )+9 \sqrt {6 x +4}\, \sqrt {3 x +3}\, \sqrt {6}\, \sqrt {-x}\, x \EllipticF \left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )-240 x -48}{9 \sqrt {3 x^{2}+5 x +2}\, x^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/9*(9*(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-17*(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-162*x^5-702*x^4-1008*x^3-660*x^2-2
40*x-48)/(3*x^2+5*x+2)^(1/2)/x^(3/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral(-4*sqrt(3*x**2 + 5*x + 2)/x**(5/2), x) - Integral(19*sqrt(3*x**2 + 5*x + 2)/sqrt(x), x) - Integral(1
5*sqrt(x)*sqrt(3*x**2 + 5*x + 2), x)

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