3.307 \(\int \frac {(7+5 x^2)^5}{(2+3 x^2+x^4)^{3/2}} \, dx\)

Optimal. Leaf size=189 \[ \frac {5000}{3} \sqrt {x^4+3 x^2+2} x+\frac {7679 \left (x^2+2\right ) x}{2 \sqrt {x^4+3 x^2+2}}-\frac {\left (179 x^2+115\right ) x}{2 \sqrt {x^4+3 x^2+2}}+\frac {15383 \left (x^2+1\right ) \sqrt {\frac {x^2+2}{x^2+1}} F\left (\tan ^{-1}(x)|\frac {1}{2}\right )}{3 \sqrt {2} \sqrt {x^4+3 x^2+2}}-\frac {7679 \left (x^2+1\right ) \sqrt {\frac {x^2+2}{x^2+1}} E\left (\tan ^{-1}(x)|\frac {1}{2}\right )}{\sqrt {2} \sqrt {x^4+3 x^2+2}}+625 \sqrt {x^4+3 x^2+2} x^3 \]

[Out]

7679/2*x*(x^2+2)/(x^4+3*x^2+2)^(1/2)-1/2*x*(179*x^2+115)/(x^4+3*x^2+2)^(1/2)-7679/2*(x^2+1)^(3/2)*(1/(x^2+1))^
(1/2)*EllipticE(x/(x^2+1)^(1/2),1/2*2^(1/2))*2^(1/2)*((x^2+2)/(x^2+1))^(1/2)/(x^4+3*x^2+2)^(1/2)+15383/6*(x^2+
1)^(3/2)*(1/(x^2+1))^(1/2)*EllipticF(x/(x^2+1)^(1/2),1/2*2^(1/2))*2^(1/2)*((x^2+2)/(x^2+1))^(1/2)/(x^4+3*x^2+2
)^(1/2)+5000/3*x*(x^4+3*x^2+2)^(1/2)+625*x^3*(x^4+3*x^2+2)^(1/2)

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Rubi [A]  time = 0.11, antiderivative size = 189, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {1205, 1679, 1189, 1099, 1135} \[ 625 \sqrt {x^4+3 x^2+2} x^3+\frac {5000}{3} \sqrt {x^4+3 x^2+2} x+\frac {7679 \left (x^2+2\right ) x}{2 \sqrt {x^4+3 x^2+2}}-\frac {\left (179 x^2+115\right ) x}{2 \sqrt {x^4+3 x^2+2}}+\frac {15383 \left (x^2+1\right ) \sqrt {\frac {x^2+2}{x^2+1}} F\left (\tan ^{-1}(x)|\frac {1}{2}\right )}{3 \sqrt {2} \sqrt {x^4+3 x^2+2}}-\frac {7679 \left (x^2+1\right ) \sqrt {\frac {x^2+2}{x^2+1}} E\left (\tan ^{-1}(x)|\frac {1}{2}\right )}{\sqrt {2} \sqrt {x^4+3 x^2+2}} \]

Antiderivative was successfully verified.

[In]

Int[(7 + 5*x^2)^5/(2 + 3*x^2 + x^4)^(3/2),x]

[Out]

(7679*x*(2 + x^2))/(2*Sqrt[2 + 3*x^2 + x^4]) - (x*(115 + 179*x^2))/(2*Sqrt[2 + 3*x^2 + x^4]) + (5000*x*Sqrt[2
+ 3*x^2 + x^4])/3 + 625*x^3*Sqrt[2 + 3*x^2 + x^4] - (7679*(1 + x^2)*Sqrt[(2 + x^2)/(1 + x^2)]*EllipticE[ArcTan
[x], 1/2])/(Sqrt[2]*Sqrt[2 + 3*x^2 + x^4]) + (15383*(1 + x^2)*Sqrt[(2 + x^2)/(1 + x^2)]*EllipticF[ArcTan[x], 1
/2])/(3*Sqrt[2]*Sqrt[2 + 3*x^2 + x^4])

Rule 1099

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] &&  !(PosQ[(b - q)/a] && SimplerSq
rtQ[(b - q)/(2*a), (b + q)/(2*a)])] /; FreeQ[{a, b, c}, x] && GtQ[b^2 - 4*a*c, 0]

Rule 1135

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] &&  !(PosQ[(b - q)/a] && SimplerSqrtQ[(b - q)/(2*a), (b + q)/(2*a)])] /; Fre
eQ[{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]

Rule 1205

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> With[{f = Coeff[Polynom
ialRemainder[(d + e*x^2)^q, a + b*x^2 + c*x^4, x], x, 0], g = Coeff[PolynomialRemainder[(d + e*x^2)^q, a + b*x
^2 + c*x^4, x], x, 2]}, Simp[(x*(a + b*x^2 + c*x^4)^(p + 1)*(a*b*g - f*(b^2 - 2*a*c) - c*(b*f - 2*a*g)*x^2))/(
2*a*(p + 1)*(b^2 - 4*a*c)), x] + Dist[1/(2*a*(p + 1)*(b^2 - 4*a*c)), Int[(a + b*x^2 + c*x^4)^(p + 1)*ExpandToS
um[2*a*(p + 1)*(b^2 - 4*a*c)*PolynomialQuotient[(d + e*x^2)^q, a + b*x^2 + c*x^4, x] + b^2*f*(2*p + 3) - 2*a*c
*f*(4*p + 5) - a*b*g + c*(4*p + 7)*(b*f - 2*a*g)*x^2, x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*
a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[q, 1] && LtQ[p, -1]

Rule 1679

Int[(Pq_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> With[{q = Expon[Pq, x^2], e = Coeff[Pq, x^2,
Expon[Pq, x^2]]}, Simp[(e*x^(2*q - 3)*(a + b*x^2 + c*x^4)^(p + 1))/(c*(2*q + 4*p + 1)), x] + Dist[1/(c*(2*q +
4*p + 1)), Int[(a + b*x^2 + c*x^4)^p*ExpandToSum[c*(2*q + 4*p + 1)*Pq - a*e*(2*q - 3)*x^(2*q - 4) - b*e*(2*q
+ 2*p - 1)*x^(2*q - 2) - c*e*(2*q + 4*p + 1)*x^(2*q), x], x], x]] /; FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x^2]
&& Expon[Pq, x^2] > 1 && NeQ[b^2 - 4*a*c, 0] &&  !LtQ[p, -1]

Rubi steps

\begin {align*} \int \frac {\left (7+5 x^2\right )^5}{\left (2+3 x^2+x^4\right )^{3/2}} \, dx &=-\frac {x \left (115+179 x^2\right )}{2 \sqrt {2+3 x^2+x^4}}-\frac {1}{2} \int \frac {-16922-35179 x^2-25000 x^4-6250 x^6}{\sqrt {2+3 x^2+x^4}} \, dx\\ &=-\frac {x \left (115+179 x^2\right )}{2 \sqrt {2+3 x^2+x^4}}+625 x^3 \sqrt {2+3 x^2+x^4}-\frac {1}{10} \int \frac {-84610-138395 x^2-50000 x^4}{\sqrt {2+3 x^2+x^4}} \, dx\\ &=-\frac {x \left (115+179 x^2\right )}{2 \sqrt {2+3 x^2+x^4}}+\frac {5000}{3} x \sqrt {2+3 x^2+x^4}+625 x^3 \sqrt {2+3 x^2+x^4}-\frac {1}{30} \int \frac {-153830-115185 x^2}{\sqrt {2+3 x^2+x^4}} \, dx\\ &=-\frac {x \left (115+179 x^2\right )}{2 \sqrt {2+3 x^2+x^4}}+\frac {5000}{3} x \sqrt {2+3 x^2+x^4}+625 x^3 \sqrt {2+3 x^2+x^4}+\frac {7679}{2} \int \frac {x^2}{\sqrt {2+3 x^2+x^4}} \, dx+\frac {15383}{3} \int \frac {1}{\sqrt {2+3 x^2+x^4}} \, dx\\ &=\frac {7679 x \left (2+x^2\right )}{2 \sqrt {2+3 x^2+x^4}}-\frac {x \left (115+179 x^2\right )}{2 \sqrt {2+3 x^2+x^4}}+\frac {5000}{3} x \sqrt {2+3 x^2+x^4}+625 x^3 \sqrt {2+3 x^2+x^4}-\frac {7679 \left (1+x^2\right ) \sqrt {\frac {2+x^2}{1+x^2}} E\left (\tan ^{-1}(x)|\frac {1}{2}\right )}{\sqrt {2} \sqrt {2+3 x^2+x^4}}+\frac {15383 \left (1+x^2\right ) \sqrt {\frac {2+x^2}{1+x^2}} F\left (\tan ^{-1}(x)|\frac {1}{2}\right )}{3 \sqrt {2} \sqrt {2+3 x^2+x^4}}\\ \end {align*}

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Mathematica [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {\$Aborted} \]

Verification is Not applicable to the result.

[In]

Integrate[(7 + 5*x^2)^5/(2 + 3*x^2 + x^4)^(3/2),x]

[Out]

$Aborted

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fricas [F]  time = 0.42, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (3125 \, x^{10} + 21875 \, x^{8} + 61250 \, x^{6} + 85750 \, x^{4} + 60025 \, x^{2} + 16807\right )} \sqrt {x^{4} + 3 \, x^{2} + 2}}{x^{8} + 6 \, x^{6} + 13 \, x^{4} + 12 \, x^{2} + 4}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x^2+7)^5/(x^4+3*x^2+2)^(3/2),x, algorithm="fricas")

[Out]

integral((3125*x^10 + 21875*x^8 + 61250*x^6 + 85750*x^4 + 60025*x^2 + 16807)*sqrt(x^4 + 3*x^2 + 2)/(x^8 + 6*x^
6 + 13*x^4 + 12*x^2 + 4), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (5 \, x^{2} + 7\right )}^{5}}{{\left (x^{4} + 3 \, x^{2} + 2\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x^2+7)^5/(x^4+3*x^2+2)^(3/2),x, algorithm="giac")

[Out]

integrate((5*x^2 + 7)^5/(x^4 + 3*x^2 + 2)^(3/2), x)

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maple [C]  time = 0.04, size = 274, normalized size = 1.45 \[ 625 \sqrt {x^{4}+3 x^{2}+2}\, x^{3}+\frac {5000 \sqrt {x^{4}+3 x^{2}+2}\, x}{3}-\frac {15383 i \sqrt {2}\, \sqrt {2 x^{2}+4}\, \sqrt {x^{2}+1}\, \EllipticF \left (\frac {i \sqrt {2}\, x}{2}, \sqrt {2}\right )}{6 \sqrt {x^{4}+3 x^{2}+2}}-\frac {6250 \left (\frac {17}{2} x^{3}+9 x \right )}{\sqrt {x^{4}+3 x^{2}+2}}+\frac {7679 i \sqrt {2}\, \sqrt {2 x^{2}+4}\, \sqrt {x^{2}+1}\, \left (-\EllipticE \left (\frac {i \sqrt {2}\, x}{2}, \sqrt {2}\right )+\EllipticF \left (\frac {i \sqrt {2}\, x}{2}, \sqrt {2}\right )\right )}{4 \sqrt {x^{4}+3 x^{2}+2}}-\frac {43750 \left (-\frac {9}{2} x^{3}-5 x \right )}{\sqrt {x^{4}+3 x^{2}+2}}-\frac {122500 \left (\frac {5}{2} x^{3}+3 x \right )}{\sqrt {x^{4}+3 x^{2}+2}}-\frac {171500 \left (-\frac {3}{2} x^{3}-2 x \right )}{\sqrt {x^{4}+3 x^{2}+2}}-\frac {120050 \left (x^{3}+\frac {3}{2} x \right )}{\sqrt {x^{4}+3 x^{2}+2}}-\frac {33614 \left (-\frac {3}{4} x^{3}-\frac {5}{4} x \right )}{\sqrt {x^{4}+3 x^{2}+2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*x^2+7)^5/(x^4+3*x^2+2)^(3/2),x)

[Out]

-6250*(17/2*x^3+9*x)/(x^4+3*x^2+2)^(1/2)+625*(x^4+3*x^2+2)^(1/2)*x^3+5000/3*(x^4+3*x^2+2)^(1/2)*x+7679/4*I*2^(
1/2)*(2*x^2+4)^(1/2)*(x^2+1)^(1/2)/(x^4+3*x^2+2)^(1/2)*(EllipticF(1/2*I*2^(1/2)*x,2^(1/2))-EllipticE(1/2*I*2^(
1/2)*x,2^(1/2)))-15383/6*I*2^(1/2)*(2*x^2+4)^(1/2)*(x^2+1)^(1/2)/(x^4+3*x^2+2)^(1/2)*EllipticF(1/2*I*2^(1/2)*x
,2^(1/2))-43750*(-9/2*x^3-5*x)/(x^4+3*x^2+2)^(1/2)-122500*(5/2*x^3+3*x)/(x^4+3*x^2+2)^(1/2)-171500*(-3/2*x^3-2
*x)/(x^4+3*x^2+2)^(1/2)-120050*(x^3+3/2*x)/(x^4+3*x^2+2)^(1/2)-33614*(-3/4*x^3-5/4*x)/(x^4+3*x^2+2)^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (5 \, x^{2} + 7\right )}^{5}}{{\left (x^{4} + 3 \, x^{2} + 2\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x^2+7)^5/(x^4+3*x^2+2)^(3/2),x, algorithm="maxima")

[Out]

integrate((5*x^2 + 7)^5/(x^4 + 3*x^2 + 2)^(3/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (5\,x^2+7\right )}^5}{{\left (x^4+3\,x^2+2\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*x^2 + 7)^5/(3*x^2 + x^4 + 2)^(3/2),x)

[Out]

int((5*x^2 + 7)^5/(3*x^2 + x^4 + 2)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (5 x^{2} + 7\right )^{5}}{\left (\left (x^{2} + 1\right ) \left (x^{2} + 2\right )\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x**2+7)**5/(x**4+3*x**2+2)**(3/2),x)

[Out]

Integral((5*x**2 + 7)**5/((x**2 + 1)*(x**2 + 2))**(3/2), x)

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