3.1057 \(\int \frac {(2-5 x) x^{5/2}}{\sqrt {2+5 x+3 x^2}} \, dx\)

Optimal. Leaf size=200 \[ -\frac {412}{189} \sqrt {3 x^2+5 x+2} \sqrt {x}+\frac {13688 (3 x+2) \sqrt {x}}{2835 \sqrt {3 x^2+5 x+2}}+\frac {412 \sqrt {2} (x+1) \sqrt {\frac {3 x+2}{x+1}} F\left (\tan ^{-1}\left (\sqrt {x}\right )|-\frac {1}{2}\right )}{189 \sqrt {3 x^2+5 x+2}}-\frac {13688 \sqrt {2} (x+1) \sqrt {\frac {3 x+2}{x+1}} E\left (\tan ^{-1}\left (\sqrt {x}\right )|-\frac {1}{2}\right )}{2835 \sqrt {3 x^2+5 x+2}}-\frac {10}{21} \sqrt {3 x^2+5 x+2} x^{5/2}+\frac {128}{105} \sqrt {3 x^2+5 x+2} x^{3/2} \]

[Out]

13688/2835*(2+3*x)*x^(1/2)/(3*x^2+5*x+2)^(1/2)-13688/2835*(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)+412/189*(1+x)^(3/2)*(1/(1+x))^(1/2)*Ell
ipticF(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)+128/105*x^(3/2)*(3
*x^2+5*x+2)^(1/2)-10/21*x^(5/2)*(3*x^2+5*x+2)^(1/2)-412/189*x^(1/2)*(3*x^2+5*x+2)^(1/2)

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Rubi [A]  time = 0.14, antiderivative size = 200, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {832, 839, 1189, 1100, 1136} \[ -\frac {10}{21} \sqrt {3 x^2+5 x+2} x^{5/2}+\frac {128}{105} \sqrt {3 x^2+5 x+2} x^{3/2}-\frac {412}{189} \sqrt {3 x^2+5 x+2} \sqrt {x}+\frac {13688 (3 x+2) \sqrt {x}}{2835 \sqrt {3 x^2+5 x+2}}+\frac {412 \sqrt {2} (x+1) \sqrt {\frac {3 x+2}{x+1}} F\left (\tan ^{-1}\left (\sqrt {x}\right )|-\frac {1}{2}\right )}{189 \sqrt {3 x^2+5 x+2}}-\frac {13688 \sqrt {2} (x+1) \sqrt {\frac {3 x+2}{x+1}} E\left (\tan ^{-1}\left (\sqrt {x}\right )|-\frac {1}{2}\right )}{2835 \sqrt {3 x^2+5 x+2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(13688*Sqrt[x]*(2 + 3*x))/(2835*Sqrt[2 + 5*x + 3*x^2]) - (412*Sqrt[x]*Sqrt[2 + 5*x + 3*x^2])/189 + (128*x^(3/2
)*Sqrt[2 + 5*x + 3*x^2])/105 - (10*x^(5/2)*Sqrt[2 + 5*x + 3*x^2])/21 - (13688*Sqrt[2]*(1 + x)*Sqrt[(2 + 3*x)/(
1 + x)]*EllipticE[ArcTan[Sqrt[x]], -1/2])/(2835*Sqrt[2 + 5*x + 3*x^2]) + (412*Sqrt[2]*(1 + x)*Sqrt[(2 + 3*x)/(
1 + x)]*EllipticF[ArcTan[Sqrt[x]], -1/2])/(189*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 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) x^{5/2}}{\sqrt {2+5 x+3 x^2}} \, dx &=-\frac {10}{21} x^{5/2} \sqrt {2+5 x+3 x^2}+\frac {2}{21} \int \frac {x^{3/2} (25+96 x)}{\sqrt {2+5 x+3 x^2}} \, dx\\ &=\frac {128}{105} x^{3/2} \sqrt {2+5 x+3 x^2}-\frac {10}{21} x^{5/2} \sqrt {2+5 x+3 x^2}+\frac {4}{315} \int \frac {\left (-288-\frac {1545 x}{2}\right ) \sqrt {x}}{\sqrt {2+5 x+3 x^2}} \, dx\\ &=-\frac {412}{189} \sqrt {x} \sqrt {2+5 x+3 x^2}+\frac {128}{105} x^{3/2} \sqrt {2+5 x+3 x^2}-\frac {10}{21} x^{5/2} \sqrt {2+5 x+3 x^2}+\frac {8 \int \frac {\frac {1545}{2}+\frac {5133 x}{2}}{\sqrt {x} \sqrt {2+5 x+3 x^2}} \, dx}{2835}\\ &=-\frac {412}{189} \sqrt {x} \sqrt {2+5 x+3 x^2}+\frac {128}{105} x^{3/2} \sqrt {2+5 x+3 x^2}-\frac {10}{21} x^{5/2} \sqrt {2+5 x+3 x^2}+\frac {16 \operatorname {Subst}\left (\int \frac {\frac {1545}{2}+\frac {5133 x^2}{2}}{\sqrt {2+5 x^2+3 x^4}} \, dx,x,\sqrt {x}\right )}{2835}\\ &=-\frac {412}{189} \sqrt {x} \sqrt {2+5 x+3 x^2}+\frac {128}{105} x^{3/2} \sqrt {2+5 x+3 x^2}-\frac {10}{21} x^{5/2} \sqrt {2+5 x+3 x^2}+\frac {824}{189} \operatorname {Subst}\left (\int \frac {1}{\sqrt {2+5 x^2+3 x^4}} \, dx,x,\sqrt {x}\right )+\frac {13688}{945} \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {2+5 x^2+3 x^4}} \, dx,x,\sqrt {x}\right )\\ &=\frac {13688 \sqrt {x} (2+3 x)}{2835 \sqrt {2+5 x+3 x^2}}-\frac {412}{189} \sqrt {x} \sqrt {2+5 x+3 x^2}+\frac {128}{105} x^{3/2} \sqrt {2+5 x+3 x^2}-\frac {10}{21} x^{5/2} \sqrt {2+5 x+3 x^2}-\frac {13688 \sqrt {2} (1+x) \sqrt {\frac {2+3 x}{1+x}} E\left (\tan ^{-1}\left (\sqrt {x}\right )|-\frac {1}{2}\right )}{2835 \sqrt {2+5 x+3 x^2}}+\frac {412 \sqrt {2} (1+x) \sqrt {\frac {2+3 x}{1+x}} F\left (\tan ^{-1}\left (\sqrt {x}\right )|-\frac {1}{2}\right )}{189 \sqrt {2+5 x+3 x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.17, size = 160, normalized size = 0.80 \[ \frac {-7508 i \sqrt {2} \sqrt {\frac {1}{x}+1} \sqrt {\frac {2}{x}+3} x^{3/2} F\left (i \sinh ^{-1}\left (\frac {\sqrt {\frac {2}{3}}}{\sqrt {x}}\right )|\frac {3}{2}\right )+13688 i \sqrt {2} \sqrt {\frac {1}{x}+1} \sqrt {\frac {2}{x}+3} x^{3/2} E\left (i \sinh ^{-1}\left (\frac {\sqrt {\frac {2}{3}}}{\sqrt {x}}\right )|\frac {3}{2}\right )-4050 x^5+3618 x^4-3960 x^3+17076 x^2+56080 x+27376}{2835 \sqrt {x} \sqrt {3 x^2+5 x+2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(27376 + 56080*x + 17076*x^2 - 3960*x^3 + 3618*x^4 - 4050*x^5 + (13688*I)*Sqrt[2]*Sqrt[1 + x^(-1)]*Sqrt[3 + 2/
x]*x^(3/2)*EllipticE[I*ArcSinh[Sqrt[2/3]/Sqrt[x]], 3/2] - (7508*I)*Sqrt[2]*Sqrt[1 + x^(-1)]*Sqrt[3 + 2/x]*x^(3
/2)*EllipticF[I*ArcSinh[Sqrt[2/3]/Sqrt[x]], 3/2])/(2835*Sqrt[x]*Sqrt[2 + 5*x + 3*x^2])

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fricas [F]  time = 0.91, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (5 \, x^{3} - 2 \, x^{2}\right )} \sqrt {x}}{\sqrt {3 \, x^{2} + 5 \, x + 2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.15, size = 122, normalized size = 0.61 \[ -\frac {2 \left (6075 x^{5}-5427 x^{4}+5940 x^{3}+35982 x^{2}+18540 x -3422 \sqrt {6 x +4}\, \sqrt {3 x +3}\, \sqrt {6}\, \sqrt {-x}\, \EllipticE \left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )+7176 \sqrt {6 x +4}\, \sqrt {3 x +3}\, \sqrt {6}\, \sqrt {-x}\, \EllipticF \left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )\right )}{8505 \sqrt {3 x^{2}+5 x +2}\, \sqrt {x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/8505/x^(1/2)/(3*x^2+5*x+2)^(1/2)*(6075*x^5+7176*(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))-3422*(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))-5427*x^4+5940*x^3+35982*x^2+18540*x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \int \left (- \frac {2 x^{\frac {5}{2}}}{\sqrt {3 x^{2} + 5 x + 2}}\right )\, dx - \int \frac {5 x^{\frac {7}{2}}}{\sqrt {3 x^{2} + 5 x + 2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral(-2*x**(5/2)/sqrt(3*x**2 + 5*x + 2), x) - Integral(5*x**(7/2)/sqrt(3*x**2 + 5*x + 2), x)

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