3.1047 \(\int \frac {(2-5 x) (2+5 x+3 x^2)^{3/2}}{\sqrt {x}} \, dx\)

Optimal. Leaf size=187 \[ -\frac {2}{9} \sqrt {x} (5 x+1) \left (3 x^2+5 x+2\right )^{3/2}+\frac {4}{81} \sqrt {x} (45 x+82) \sqrt {3 x^2+5 x+2}+\frac {860 \sqrt {x} (3 x+2)}{243 \sqrt {3 x^2+5 x+2}}+\frac {356 \sqrt {2} (x+1) \sqrt {\frac {3 x+2}{x+1}} F\left (\tan ^{-1}\left (\sqrt {x}\right )|-\frac {1}{2}\right )}{81 \sqrt {3 x^2+5 x+2}}-\frac {860 \sqrt {2} (x+1) \sqrt {\frac {3 x+2}{x+1}} E\left (\tan ^{-1}\left (\sqrt {x}\right )|-\frac {1}{2}\right )}{243 \sqrt {3 x^2+5 x+2}} \]

[Out]

-2/9*(1+5*x)*(3*x^2+5*x+2)^(3/2)*x^(1/2)+860/243*(2+3*x)*x^(1/2)/(3*x^2+5*x+2)^(1/2)-860/243*(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)+356/
81*(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)+4/81*(82+45*x)*x^(1/2)*(3*x^2+5*x+2)^(1/2)

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Rubi [A]  time = 0.12, antiderivative size = 187, 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 = {814, 839, 1189, 1100, 1136} \[ -\frac {2}{9} \sqrt {x} (5 x+1) \left (3 x^2+5 x+2\right )^{3/2}+\frac {4}{81} \sqrt {x} (45 x+82) \sqrt {3 x^2+5 x+2}+\frac {860 \sqrt {x} (3 x+2)}{243 \sqrt {3 x^2+5 x+2}}+\frac {356 \sqrt {2} (x+1) \sqrt {\frac {3 x+2}{x+1}} F\left (\tan ^{-1}\left (\sqrt {x}\right )|-\frac {1}{2}\right )}{81 \sqrt {3 x^2+5 x+2}}-\frac {860 \sqrt {2} (x+1) \sqrt {\frac {3 x+2}{x+1}} E\left (\tan ^{-1}\left (\sqrt {x}\right )|-\frac {1}{2}\right )}{243 \sqrt {3 x^2+5 x+2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(860*Sqrt[x]*(2 + 3*x))/(243*Sqrt[2 + 5*x + 3*x^2]) + (4*Sqrt[x]*(82 + 45*x)*Sqrt[2 + 5*x + 3*x^2])/81 - (2*Sq
rt[x]*(1 + 5*x)*(2 + 5*x + 3*x^2)^(3/2))/9 - (860*Sqrt[2]*(1 + x)*Sqrt[(2 + 3*x)/(1 + x)]*EllipticE[ArcTan[Sqr
t[x]], -1/2])/(243*Sqrt[2 + 5*x + 3*x^2]) + (356*Sqrt[2]*(1 + x)*Sqrt[(2 + 3*x)/(1 + x)]*EllipticF[ArcTan[Sqrt
[x]], -1/2])/(81*Sqrt[2 + 5*x + 3*x^2])

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 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}}{\sqrt {x}} \, dx &=-\frac {2}{9} \sqrt {x} (1+5 x) \left (2+5 x+3 x^2\right )^{3/2}-\frac {2}{63} \int \frac {(-133-175 x) \sqrt {2+5 x+3 x^2}}{\sqrt {x}} \, dx\\ &=\frac {4}{81} \sqrt {x} (82+45 x) \sqrt {2+5 x+3 x^2}-\frac {2}{9} \sqrt {x} (1+5 x) \left (2+5 x+3 x^2\right )^{3/2}+\frac {4 \int \frac {3115+\frac {7525 x}{2}}{\sqrt {x} \sqrt {2+5 x+3 x^2}} \, dx}{2835}\\ &=\frac {4}{81} \sqrt {x} (82+45 x) \sqrt {2+5 x+3 x^2}-\frac {2}{9} \sqrt {x} (1+5 x) \left (2+5 x+3 x^2\right )^{3/2}+\frac {8 \operatorname {Subst}\left (\int \frac {3115+\frac {7525 x^2}{2}}{\sqrt {2+5 x^2+3 x^4}} \, dx,x,\sqrt {x}\right )}{2835}\\ &=\frac {4}{81} \sqrt {x} (82+45 x) \sqrt {2+5 x+3 x^2}-\frac {2}{9} \sqrt {x} (1+5 x) \left (2+5 x+3 x^2\right )^{3/2}+\frac {712}{81} \operatorname {Subst}\left (\int \frac {1}{\sqrt {2+5 x^2+3 x^4}} \, dx,x,\sqrt {x}\right )+\frac {860}{81} \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {2+5 x^2+3 x^4}} \, dx,x,\sqrt {x}\right )\\ &=\frac {860 \sqrt {x} (2+3 x)}{243 \sqrt {2+5 x+3 x^2}}+\frac {4}{81} \sqrt {x} (82+45 x) \sqrt {2+5 x+3 x^2}-\frac {2}{9} \sqrt {x} (1+5 x) \left (2+5 x+3 x^2\right )^{3/2}-\frac {860 \sqrt {2} (1+x) \sqrt {\frac {2+3 x}{1+x}} E\left (\tan ^{-1}\left (\sqrt {x}\right )|-\frac {1}{2}\right )}{243 \sqrt {2+5 x+3 x^2}}+\frac {356 \sqrt {2} (1+x) \sqrt {\frac {2+3 x}{1+x}} F\left (\tan ^{-1}\left (\sqrt {x}\right )|-\frac {1}{2}\right )}{81 \sqrt {2+5 x+3 x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.16, size = 165, normalized size = 0.88 \[ \frac {208 i \sqrt {2} \sqrt {\frac {1}{x}+1} \sqrt {\frac {2}{x}+3} x^{3/2} F\left (i \sinh ^{-1}\left (\frac {\sqrt {\frac {2}{3}}}{\sqrt {x}}\right )|\frac {3}{2}\right )+860 i \sqrt {2} \sqrt {\frac {1}{x}+1} \sqrt {\frac {2}{x}+3} x^{3/2} E\left (i \sinh ^{-1}\left (\frac {\sqrt {\frac {2}{3}}}{\sqrt {x}}\right )|\frac {3}{2}\right )-2430 x^6-8586 x^5-9990 x^4-1746 x^3+6420 x^2+6052 x+1720}{243 \sqrt {x} \sqrt {3 x^2+5 x+2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(1720 + 6052*x + 6420*x^2 - 1746*x^3 - 9990*x^4 - 8586*x^5 - 2430*x^6 + (860*I)*Sqrt[2]*Sqrt[1 + x^(-1)]*Sqrt[
3 + 2/x]*x^(3/2)*EllipticE[I*ArcSinh[Sqrt[2/3]/Sqrt[x]], 3/2] + (208*I)*Sqrt[2]*Sqrt[1 + x^(-1)]*Sqrt[3 + 2/x]
*x^(3/2)*EllipticF[I*ArcSinh[Sqrt[2/3]/Sqrt[x]], 3/2])/(243*Sqrt[x]*Sqrt[2 + 5*x + 3*x^2])

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fricas [F]  time = 0.95, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (15 \, x^{3} + 19 \, x^{2} - 4\right )} \sqrt {3 \, x^{2} + 5 \, x + 2}}{\sqrt {x}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.07, size = 127, normalized size = 0.68 \[ -\frac {2 \left (3645 x^{6}+12879 x^{5}+14985 x^{4}+2619 x^{3}-5760 x^{2}-2628 x -215 \sqrt {6 x +4}\, \sqrt {3 x +3}\, \sqrt {6}\, \sqrt {-x}\, \EllipticE \left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )+111 \sqrt {6 x +4}\, \sqrt {3 x +3}\, \sqrt {6}\, \sqrt {-x}\, \EllipticF \left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )\right )}{729 \sqrt {3 x^{2}+5 x +2}\, \sqrt {x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/729/(3*x^2+5*x+2)^(1/2)/x^(1/2)*(3645*x^6+12879*x^5+111*(6*x+4)^(1/2)*(3*x+3)^(1/2)*6^(1/2)*(-x)^(1/2)*Elli
pticF(1/2*(6*x+4)^(1/2),I*2^(1/2))-215*(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))+14985*x^4+2619*x^3-5760*x^2-2628*x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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